Advanced Experimental Organic Chemistry 9781032974491

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Advanced Experimental Organic Chemistry

Advanced Experimental Organic Chemistry

V.K. Ahluwalia

Prof. (Retd.) Department of Chemistry University of Delhi, Delhi-110007 Visiting Professor Dr. B.R. Ambedkar Center for Biomedical Research University of Delhi, Delhi-110007

& Sunita Dhingra

Associate Prof. (Retd.) Miranda House University of Delhi, Delhi-110007

First published 2025 by CRC Press 4 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN and by CRC Press 2385 NW Executive Center Drive, Suite 320, Boca Raton FL 33431 © 2025 Manakin Press Pvt. Ltd. CRC Press is an imprint of Informa UK Limited The right of V.K. Ahluwalia and Sunita Dhingra to be identified as authors of this work has been asserted in accordance with sections 77 and 78 of the Copyright, Designs and Patents Act 1988. All rights reserved. No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Print edition not for sale in South Asia (India, Sri Lanka, Nepal, Bangladesh, Pakistan or Bhutan). ISBN13: 9781032974491 (hbk) ISBN13: 9781032974507 (pbk) ISBN13: 9781003593812 (ebk) DOI: 10.1201/9781003593812 Typeset in Times New Roman by Manakin Press, Delhi

Preface The book entitled “Advanced Experimental Organic Chemistry” has been written for under and post graduate students keeping in mind the curriculum of organic chemistry laboratory in all the universities. We have made an effort to present this book in a simple way and hope it will find a place as a reference book in the laboratories, libraries and in the book shelf of teachers and learners. The topics covered in the book are: Qualitative analysis of organic compounds: This chapter includes systematic procedure adopted for identification of a single organic compound, separation of a mixture of two compounds and their identification. We have made an earnest effort to explain the chemistry of each and every laboratory test, procedure employed for preparation of derivatives. A comprehensive list of m.ps./b.ps. of compounds and their derivatives is provided for comparison with the observed laboratories values (chapter 7). A salient feature of this chapter is the inclusion of SPOT TESTS for identification of extra elements and functional group/s in the unknown organic compound. Microwave assisted synthesis of a few derivatives has also been included in this chapter. Spectrometric methods of qualitative analysis including IR, UV, NMR and Mass are discussed in detail in the second chapter. A large number of examples with spectra will benefit the students in understanding the application of these techniques in elucidating the structures of unknown compounds. Preparation of organic compounds, isolation of compounds from their natural source, quantitative organic estimations and the chromatographic techniques used for purification of compounds forms the subject matter of the third, fourth, fifth and sixth chapters. Laboratory training in this type of work will benefit the students in understanding chemistry and train them to take up the challenges of research and development in chemistry. An elaborate list of questions is given at the end of each chapter for self assessment and better understanding of the subject. We would be grateful for all constructive suggestions for the improvement of the book. V.K. Ahluwalia Sunita Dhingra

Detailed Contents Preface

v

Safe Practices in the Laboratory

1—6

Safety Equipment

2

Safety Symbols

3

1.

Qualitative Organic Analysis

7—146

1.1

Introduction

7

1.2

Experiments Performed to Analyse a Compound

8

1.3

Preliminary Examination

8

1.3.2

Colour of the Compound

8

1.3.3

Odour of the Compound

8

1.3.4

Flame Test

9

1.3.5

Action of Alkali

9

1.3.6

Test with Ferric Chloride

9

1.3.7

Solubility Test

10

1.3.8

Detection of Extra Elements

12

1.3.9

Test for Nitrogen

13

1.3.10

Test for Sulphur

17

1.3.11

Test for Nitrogen and Sulphur Present Together

18

1.3.12

Test for Halogens

19

1.3.13

Determination of Melting Point

22

1.3.14

Determination of Boiling Point

23

1.3.15

Detection of Unsaturation

25

1.3.16

Test with Bromine Solution

25

1.3.17

Baeyer’s Test: Test with Potassium Permanganate

27

1.4

Identification of Functional Group

28

1.4.1

Introduction

28

1.4.2

Carbohydrates

29

1.4.3

Carboxylic Acids

35

1.4.4

Phenols

38

1.4.5

Quinones

43

1.4.6

Aldehydes and Ketones

44

xviii

DetailedContents Contents Detailed

1.5

1.4.7

Alcohols

50

1.4.8

Esters

53

1.4.9

Lactones

56

1.4.10

Acid anhydrides

56

1.4.11

Ethers

57

1.4.12

Hydrocarbons

58

1.4.13

Alkyl, Vinyl and Aryl Halides

61

1.4.14

Acid chlorides

64

1.4.15

Amines

65

1.4.16

Nitro Compounds

73

1.4.17

Amides, Imides and Nitriles

76

1.4.18

Biuret test for urea

79

1.4.19

Anilides and Substituted Amides

80

1.4.20

Amino Acids

81

1.4.21

Sulphonic acids

84

1.4.22

Thiols

85

1.4.23

Sulphonamides

86

Preparation of Derivatives

87

1.5.1

Introduction

87

1.5.2

Crystallisation

88

1.5.3

Derivatives of Carbohydrates

90

1.5.4

Derivatives of Carboxylic Acids

95

1.5.5

Derivatives of Phenols

99

1.5.6

Derivatives of Aldehydes and Ketones

103

1.5.7

Derivatives of Alcohols

107

1.5.8

Derivatives of Esters

110

1.5.9

Derivatives of Anhydrides

113

1.5.10

Derivatives of Hydrocarbons

113

1.5.11

Derivatives of Ethers

114

1.5.12

Derivatives of Amines

115

1.5.13

Primary Amines

115

1.5.14

Secondary Amines

119

1.5.15

Tertiary Amines

119

1.5.16

Derivatives of Nitro Compounds

120

1.5.17

Derivatives of the Amides

123

1.5.18

Derivatives of Urea

123

Detailed Contents Advanced Experimental Organic Chemistry

1.6

1.5.19

Derivatives of Imides

124

1.5.20

Derivatives of Nitriles

124

1.5.21

Derivative of Anilides

124

1.5.22

Derivatives of Amino Acids

125

1.5.23

Derivatives of Sulphonic acids

127

1.5.24

Thiols

127

1.5.25

Sulphonamides

128

1.5.26

Derivatives of Alkyl and Aryl Halides

128

1.5.27

Derivatives of Acid Chloride

129

Identification of a Mixture of Two Compounds

129

1.6.1

Preliminary Examination

130

1.6.2

Separation of the Components of Mixture

130

1.6.3

Solubility

130

1.6.4

Separation of the components of a solid mixture

131

1.6.5

Separation of a Liquid Mixture on the basis of Solubility

131

1.6.6

Separation of Mixture with Sodium Bisulphite

132

1.6.7

Separation by using Hinsberg reagent

132

1.6.8

Separation of Mixture by Sublimation

133

1.6.9

Separation of Mixture by Fractional Distillation

133

1.6.10

Chromatography

133

1.6.11

Purification of the Components

133

1.6.12

Identification of each Component

133

Questions

2.

Spectroscopy 2.1 2.2

xi ix

Introduction IR Spectroscopy 2.2.1 Introduction 2.2.2 Types of Stretching Vibrations 2.2.3 Types of Bending Vibrations 2.2.4 Instrument 2.2.5 Sample Preparation 2.2.6 Uses of IR Spectroscopy 2.2.7 IR Absorptions of Some Common types of Compounds 2.2.8 Hydrocarbons 2.2.9 Saturated Hydrocarbons 2.2.10 Alkenes

133—145

147—244 147 148 148 150 150 151 152 153 153 153 153 154

xii x

DetailedContents Contents Detailed

2.2.11 2.2.12 2.2.13 2.2.14 2.2.15 2.2.16 2.2.17 2.2.18 2.2.19 2.2.20 2.2.21 2.2.22 2.2.23 2.2.24 2.2.25 2.2.26 2.2.27 2.2.28 2.2.29 2.2.30 2.2.31 2.3

Alkynes Aromatic hydrocarbons Alcohols Phenols Ethers Epoxides and Cyclic Ethers IR Spectra of Compounds Containing a C=O bond Aldehydes Ketones Carboxylic Acids Anhydrides Esters Acid Chlorides Acid Amides and Substituted Amides Amines Nitro Compounds Nitriles Amino acids Thioalcohols Sulphonic Acids Sulphonamides Questions

156 157 159 161 162 162 162 164 165 168 169 170 171 171 172 174 175 175 176 176 176 177—178

UV Spectroscopy

182

2.3.1

Introduction

182

2.3.2

Theory

182

2.3.3

Types of Electronic Transitions

183

2.3.4

Instrument

184

2.3.5

Sample Handling

184

2.3.6

Solvents used and their Effect on the Position of Absorption

184

2.3.7

UV Absorption of Common Functional Groups

185

2.3.8

Alkanes

185

2.3.9

Alkenes

185

2.3.10

Woodwards Rules for Calculating the λmax of Polyunsaturated Alkenes

186

Alkynes

187

2.3.12

Aromatic Compounds

187

2.3.13

Calculation of p to p* Transitions of some Benzene Derivatives

189

2.3.14

UV of Compounds Containing a C=O Group

190

2.3.11

Detailed Contents Advanced Experimental Organic Chemistry

2.3.15

Calculation of p–p* Transition of a, b–Unsaturated Carbonyl Compounds (enones) and Dienones Questions

2.4

Nuclear Magnetic Resonance Spectroscopy

xiii xi

190 192—193 194

2.4.1

Introduction

194

2.4.2

Instrument

195

2.4.3

Solvents

195

2.4.4

Uses of 1HNMR (PMR)

196

2.4.5

Principle

196

2.4.6

Chemical Shift

197

2.4.7

Factors Influencing the Chemical Shift

198

2.4.8

Electronegativity

198

2.4.9

Anisotropic Effect

198

2.4.10

Intensity of Peaks

200

2.4.11

Equivalent and Nonequivalent Protons

200

2.4.12

Spin Spin Coupling

202

2.4.13

Coupling Constant

204

2.4.14

NMR of Some Common Functional Groups

204

2.4.15

Alkanes

204

2.4.16

Alkenes

206

2.4.18

Aromatic Compounds

207

2.4.19

Alkyl Halides

210

2.4.20

Alcohols

210

2.4.21

Ethers

212

2.4.22

Aldehydes

212

2.4.23

Ketones

213

2.4.24

Esters

214

2.4.25

Carboxylic Acids

214

2.4.26

Amides

215

2.4.27

Acid Chlorides

216

2.4.28

Nitriles

216

2.4.29

Amino Acids

217

2.4.30

Nitroalkanes

217

2.4.31

Amines

217

2.4.32

Sulphonic Acids

219

2.4.33

Sulphonamides Questions

220 224—232

xiv xii

DetailedContents Contents Detailed

2.5

3.

Mass Spectrometry

232

2.5.1

Introduction

232

2.5.2

Instrument

233

2.5.3

Molecular Formula from Molecular Mass

233

2.5.4

Fragmentation Pattern of some Types of Compounds

234

2.5.5

Alkanes and Cyclic Alkanes

235

2.5.6

Alkenes

235

2.5.7

Alkyl Halides

235

2.5.8

Alcohols

235

2.5.9

Ethers

235

2.5.10

Aldehydes

236

2.5.11

Ketones

236

2.5.12

Carboxylic Acids

237

2.5.13

Esters

237

2.5.14

Amides

237

2.5.15

Nitro Compounds

237

Questions

238—243

Preparation of Organic Compounds

245—355

3.1 3.2

3.3

3.4

Introduction Acetylation 3.2.1 Acetanilide 3.2.2 Aspirin (Acetylsalicylic acid) 3.2.3 Paracetamol (4-Acetamidophenol) 3.2.4 Hydroquinone Diacetate 3.2.5 Hydroxyhydroquinone Triacetate 3.2.6 b-Naphthyl Acetate 3.2.7 a-D-GIucose Pentaacetate 3.2.8 b-D-Glucose Pentaacetate Anhydrides 3.3.1 Phthalic Anhydride 3.3.2 Succinic Anhydride 3.3.3 Maleic Anhydride Benzoylation 3.4.1 Benzanilide 3.4.2 N-Benzoylglycine (Hippuric Acid) 3.4.3 Phenyl benzoate

245 246 249 251 252 252 253 253 254 255 255 255 256 256 257 258 259 260

Detailed Contents Advanced Experimental Organic Chemistry

3.5

3.6

3.7 3.8 3.9 3.10

3.11

3.12

xv xiii

Beckmann Rearrangement

260

3.5.1

Benzanilide from Benzophenone Oxime

261

3.5.2

Acetanilide from Acetophenone Oxime

262

Benzoin Condensation

263

3.6.1

Benzoin

264

3.6.2

Benzil

265

3.6.3

Benzilic Acid

266

Benzidine Rearrangement

266

3.7.1

267

Benzidine

Cannizzaro Reaction

268

3.8.1

269

Cannizzaro’s reaction with Benzaldehyde

Ethyl Acetoacetate/Acetoacetic Ester (Claisen Condensation)

270

3.9.1

270

Ethyl Acetoacetate

Derivatives of Aldehydes and Ketones

271

3.10.1

2,4–Dinitrophenylhydrazone

271

3.10.2

Semicarbazone

273

3.10.3

Oxime

274

3.10.4

Phenylhydrazone

274

Diazonium Salts and their Reactions

275

3.11.1

Diazotisation

275

3.11.2

Substitution of Diazo Group by Hydrogen

278

3.11.3

m-Nitrophenol

280

3.11.4

Iodobenzene

280

3.11.5

p-Chlorotoluene and p-Bromotoluence

281

3.11.6

Phenylhydrazine

284

3.11.7

Coupling Reactions of Diazonium Salts

285

3.11.8

Phenylazo-b-naphthol (Sudan 1)

285

3.11.9

4-(4′-Nitrobenzeneazo)-l-naphthol (Magneson II)

286

3.11.10 Methyl Orange

287

3.11.11 Orange II (b-Naphthol orange)

288

3.11.12 Methyl Red

289

Synthesis of Other Dyes

290

3.12.1

Fluorescein

290

3.12.2

Eosin (Tetrabromo fluorescein)

291

3.12.3

Malachite Green

292

3.12.4

Crystal Violet (Hexamethylpararosaniline)

293

xvi xiv

DetailedContents Contents Detailed

3.13

3.14

3.15 3.16

3.17

3.18

3.19

3.20

Esterification

294

3.13.1

Ethyl Benzoate

295

3.13.2

Methyl Salicylate

296

3.13.3

Ethyl p-aminobenzoate

296

3.13.4

Diethyl Adipate

297

Friedel Crafts Reaction

297

3.14.1

Tert. Butylbenzene

300

3.14.2

Triphenylmethane

300

3.14.3

Acylation

301

Grignard Reaction

301

3.15.1

302

Triphenylcarbinol (Triphenylmethanol)

Halogenation

303

3.16.1

Bromination

304

3.16.2

p-Bromoacetanilide

305

3.16.3

2,4,6-Tribromoaniline

306

3.16.4

2,4,6-Tribromophenol

3.16.5

l-Bromo-2-Naphthol

308

Houben Hoesch and Nencki Reaction

308

3.17.1

2,4-Dihydroxy Acetophenone (β-Resacetophenone)

309

3.17.2

2,4,6- Trihydroxy Acetophenone (Phloracetophenone)

311

Hydrolysis

312

3.18.1

Hydrolysis of Ethyl Benzoate (Preparation of Benzoic Acid)

312

3.18.2

Hydrolysis of Benzamide (Preparation of Benzoic Acid)

313

3.18.3

Hydrolysis of p-nitroacetanilide (Preparation of p-nitroaniline)

314

3.18.4

Hydrolysis of p-bromoacetanilide

315

Methylation

315

3.19.1

b-Methyl naphthyl ether (Nerolin)

316

3.19.2

1,3-Dimethoxybenzene (Resorcinol dimethylether)

317

3.19.3

1,2,4-Trimethoxybenzene (hydroxy hydroquinone trimethylether)

318

Nitration

318

3.20.1

Nitrobenzene

321

3.20.2

1-Nitronaphthalene (a-Nitronaphthalene)

321

3.20.3

p-Bromonitrobenzene

322

3.20.4

o- and p-Nitrophenol

323

3.20.5

p-Nitroacetanilide

324

3.20.6

m-Dinitrobenzene

325

Detailed Contents Advanced Experimental Organic Chemistry

3.20.7 3.21

Osazone 3.21.1

3.22

3.23

3.24

Picric Acid (2,4,6-Trinitrophenol)

xvii xv

325 326

Osazone of Glucose

327

Oxidation Reactions

329

3.22.1

Oxidation Reactions with Potassium permanganate

330

3.22.2

Benzoic Acid

330

3.22.3

Veratric Acid (3,4-dimethoxybenzoic acid)

331

3.22.4

Oxidation with Chromium Trioxide

332

3.22.5

9,10-Anthraquinone

332

3.22.6

p-Nitrobenzaldehyde

333

3.22.7

Oxidation with Potassium Dichromate/Sodium Dichromate

333

3.22.8

Adipic Acid

334

3.22.9

p-Nitrobenzoic Acid

334

3.22.10 9,10-Phenanthraquinone

335

3.22.11 Terephthalic Acid

335

3.22.12 Oxidation with Hydrogen Peroxide

336

3.22.13 Diphenic Acid (Biphenyl-2,2′-dicarboxylic acid)

336

3.22.14 Oxidation with Potassium Bromate

336

3.22.15 p-Benzoquinone

336

3.22.16 Oxidation with Sodium Hypochlorite

337

3.22.17 2-Naphthoic Acid

337

3.22.18 Elbs Persulphate Oxidation

337

3.22.19 2,5- Dihydroxybenzaldehyde (gentisic aldehyde)

338

3.22.20 Oxidation with Peracids (Baeyer Villiger oxidation)

339

3.22.21 3,4-Dimethoxyphenol

339

Photochemical Reactions

340

3.23.1

Introduction

340

3.23.2

Benzopinacol

341

3.23.3

Isomerisation of trans-azobenzene to cis-azobenzene

342

3.23.4

1,4-Naphthaquinone Photodimer

342

Polymerisation

343

3.24.1

Introduction

343

3.24.2

Nylon 66

343

3.24.3

Phenol Formaldehyde Resin (Bakelite)

344

3.24.4

Thiokol Rubber

346

3.24.5

Polystyrene

346

xviii xvi

DetailedContents Contents Detailed

3.25

4.

Isolation of Natural Products 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13

5.

Reductions 3.25.1 Reduction of Nitrobenzene to Aniline 3.25.2 Reduction of Nitrobenzene to Azobenzene 3.25.3 Reduction of m-Dinitrobenzene to m-Nitroaniline Questions Introduction Isolation of Caffeine from Tea Leaves Isolation of Caffeine using Supercritical Carbon Dioxide Isolation of Casein from Milk Isolation of Piperine from Black Pepper Isolation of Lycopene from Tomatoes/Watermelon Isolation of Cholesterol from Gallstones Isolation of Eugenol from Cloves Isolation of Limonene from Orange Peel Isolation of Nicotine from Tobacco Leaves Isolation of DNA (Deoxyribonucleic Acid) from Onion Isolation of Citric Acid from Lemon Juice Isolation of b-Carotene from Carrots Questions

Estimations 5.1

5.2 5.3 5.4 5.5 5.6 5.7 5.8

5.9 5.10 5.11 5.12

Estimation of Amines 5.1.1 Acetylation Method 5.1.2 Bromination Method Estimation of Amides Estimation of Phenols Estimation of Reducing Sugars Saponification Value of a Fat or an Oil Determination of Iodine Number of a Fat or an Oil Estimation of Amino Acids Equivalent Weight of Carboxylic Acids 5.8.1 Volumetric Estimation 5.8.2 Gravimetric Estimation Estimation of Ester Group Estimation of Nitro Group Determination of Dissolved Oxygen (DO) in a Given Sample of Water Determination of Chemical Oxygen Demand (COD) in a Given Sample of Water

348 348 348 349 351—355

357—366 357 357 358 358 358 359 359 360 361 362 363 364 364 366

367—390 367 367 369 372 373 374 375 377 378 379 379 380 381 382 384 385

Detailed Contents Advanced Experimental Organic Chemistry

5.13

Plot the Titration Curve of Glycine

386

5.14

Effect of Temperature on the Activity of Amylase

388

Questions

6.

Chromatography 6.1 6.2 6.3

6.4

6.5 6.6 6.7 6.8 6.9

7.

xix xvii

Introduction Types of Chromatography Paper Chromatography 6.3.1 Ascending Paper Chromatography 6.3.2 Radial Chromatography 6.3.3 Descending Paper Chromatography Thin Layer Chromatography (TLC) 6.4.1 Identification of Number of Components in a mixture of Compounds 6.4.2 Preparative TLC 6.4.3 Monitoring the Progress of a Reaction 6.4.4 Monitoring Fractions in Column Chromatography 6.4.5 Checking the Purity of a Compound Column Chromatography Gas Chromatography High Performance Liquid Chromatography (HPLC) Ion Exchange Chromatography Applications 6.9.1 Chromatography of Amino Acids 6.9.2 Chromatography of Sugars 6.9.3 Chromatography of Phenols 6.9.4 Chromatography of Chlorophyl (Ascending Paper Chromatography) 6.9.5 Chromatography of Dyes 6.9.6 Separation a mixture of o-Nitroaniline and p-Nitroanline by Column Chromatography 6.9.7 Separation of Syn and Anti-azobenzene by Column Chromatography Questions

Tables of Organic Compounds and their Derivatives

389—390

391—406 391 392 392 392 394 395 395 396 397 398 398 398 398 400 401 402 402 402 404 404 405 405 405 406 406

407—522

Table 7.1

Carbohydrates

407

Table 7.2

Carboxylic Acids

410

Table 7.3

Carbonyl Compounds (Aldehydes and Ketones)

421

Table 7.4

Phenols

430

Table 7.5

Quinones

439

Table 7.6 Alcohols

441

xx xviii

DetailedContents Contents Detailed

Table 7.7

Esters

446

Table 7.8

Lactones

457

Table 7.9 Acid Anhydrides

457

Table 7.10

Ethers

458

Table 7.11

Hydrocarbons

464

Table 7.12

Halohydrocarbons (Alkyl and Aryl Halides)

468

Table 7.13

Carboxylic Acid Halides

473

Table 7.14

Nitro Compounds

475

Table 7.15

Primary amines

482

Table 7.16

Secondary amines

492

Table 7.17 Tertiary amines

495

Table 7.18 Amides, Imides, Anilides, Guanidines

498

Table 7.19

Nitriles

511

Table 7.20 Amino Acids, Esters

513

Table 7.21

516

Sulphonic Acids

Table 7.22 Thiols, Mercaptans (Thioalcohols and Thiophenols)

519

Table 7.23

520

Sulphonamides

Appendix

Index

523—530

I.

Preparation of Reagents

525

II.

Normality of Conc. Acids

527

III.

Boiling points of some Common Organic Solvents

529 531—550

SAFE PRACTICES IN THE LABORATORY Like all other science laboratories, chemistry laboratory is a place of learning. Those students who are interested in pursuing chemistry get an opportunity to perform experiments which help them to understand this subject better. In most practical classes in chemistry laboratory, there is one teacher for every 12-15 students. A teacher normally gives write ups about the experiment in advance, discusses the experiment to be performed and gives instructions etc. It is important to remember that inspite of all this because of the nature of the work involved, accidents do happen. Safety of students, the teaching and the non teaching staff present in the laboratory is of utmost importance. For safety of all in the laboratory, it is very important that every one follows good work practices and reduce accidents in the laboratory, here are a few important precautions for the students that if observed will be of great help.

• Listen to the instructions carefully and follow these in letter and spirit.



• Work with concentration, without hurry and keep your work place clean.



• If you are handling a chemical for the first time, find out about its safe use, disposal etc. from the supervisor before starting the experiment.



• Obtain necessary information about all chemicals by referring to official web sites on the internet (i.e. CAS Registry numbers and MSDS data).



• Never work alone in the laboratory.



• Wear a laboratory coat made of cotton, avoid wearing synthetic fabrics in the chemistry practical laboratory.



• Protect your eyes and hands by wearing safety glasses and gloves while working. Do not wear contact lenses in the laboratory as fumes may accumulate under these and cause harm to the eyes.



• Do not wear high heels and short dresses in a chemistry laboratory.



• Tie long hair and never leave them open.



• Never eat or drink in the laboratory.



• Always wash your hands and face on leaving the laboratory.



• Handle all chemicals in the laboratory carefully. Never use your hands for transferring any. chemical. Never smell or taste any chemical.



• Try not to take excess chemicals than required, do not return excess unused chemical to the stock bottles.



• Stopper the reagent bottles after use and never leave these open.



• If you have to dilute a concentrated acid, Always add conc. acid to water, slowly with swirlling



• Never point a tube or a flask in which you are heating a chemical or a reaction mixture towards your or your neighbour’s face in the laboratory.

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SafePractices practicesininthe theLaboratory laboratory Safe



• All organic solvents are inflammable, handle these carefully. Never heat any organic solvent directly on a burner. • Some solvents like diethyl ether or petroleum ether have very low boiling point and catch fire. Never work with solvent ether in the laboratory where other students are using burners. • If you are using a gas burner for heating, check that the tube is securely fitted to the gas outlet and there is no leakage, in case of any leakage put off the burner and immediately report to the laboratory staff. • Never leave burners on while not using them. • Use glass apparatus, which is made from corning or pyrex glass only and never use any apparatus that is chipped or cracked. • Use fume chamber when recommended for any work. • Report to the supervisor in case of any accident. For finding out more about laboratory chemicals and your own safety, consult the following: (i) Material Safety Data Sheets (MSDS): also known as Safety Data Sheets (SDS) or Product Safety Data Sheets (PSDS) which provide complete information about all the compounds • Physical properties like boiling and melting points • Potential health hazard • Flammable properties • Accident release measures and • First aid measures There are about 100 free sites on the internet for reference like: (i) Chemspider ( Royal Society of Chemistry) (ii) Pubchem (US National Institute of Health) and a number of other agencies, chemical manufacturers and suppliers. (ii) CAS Registry Numbers (CASRNs) or CAS ( Chemical Abstract Service, a division of American Chemical Society) provides a list of all known (~150 million in number) organic and inorganic compounds, proteins, DNA sequences, minerals, alloys etc. Every known chemical substance has been assigned a unique identity number. This identity number has been separated by hyphens into three parts, the first number consisting of 2-10 digits, the second of 2 and the third only one digit. CAS number gives a wealth of information about each compound. You can search the compound even if limited information is available to you about the compound like its trivial or common or systematic or commercial name, its molecular formula or even structure diagram. You can find out CAS number of a compound from (i) SciFinder (ii) CAS Publications and Chemical Catalogues (iii) Sigma Aldrich etc.

SAFETY EQUIPMENT Following safety equipment must be easily accessible in a chemistry laboratory 1. Fire extinguisher

2. Eye wash fountain 3. Shower 4. First aid kit

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SAFETY SYMBOLS These symbols warn of possible dangers in the laboratory and remind you to work carefully.

Safety goggles: Wear safety goggles to protect your eyes in any activity involving chemicals, flames or heating etc.

Lab apron: Wear a laboratory apron to protect your skin and clothing from damage.

Breakage: Handle breakable materials, such as glassware, with care. Do not touch broken glassware.

Heat-resistant gloves: Use an oven mitt or other hand protection when handling hot materials such as hot plates or hot glassware.

Plastic gloves: Wear disposable plastic gloves when working with harmful chemicals. Keep your hands away from your face and dispose of the gloves according to your teacher’s instructions.

Heating: Use a clamp or tongs to pick up hot glassware. Do not touch hot objects with your bare hands.

Flames: Before you work with flames, tie hair. Follow instructions from your teacher about lighting and extinguishing flames.

No Flames: When using flammable materials, make sure that there are no flames, sparks, or other exposed heat sources present.

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SafePractices practicesininthe theLaboratory laboratory Safe

Corrosive chemical: Avoid getting acid or other corrosive chemicals on your skin or clothing or in your eyes. Do not inhale the vapours. Wash your hands after the activity.

Poison: Do not let any poisonous chemical come into contact with your skin and do not inhale its vapours. Wash your hands when you are finished with the activity.

Fumes: Work in a ventilated area when harmful vapours may be involved. Avoid inhaling vapours directly. Only test an odour when directed to do so by your teacher, use a wafting motion to direct the vapor toward your nose.

Sharp object: Scissors, scalpels, knives, needles, pins can cut your skin. Always direct a sharp edge or point away from yourself and others.

Electric shock: To avoid electric shock, never use electrical equipment around water, or when the equipment is wet or your hands are wet. Unplug equipment which are not in use.

Disposal: Dispose of chemicals and other laboratory materials safely. Follow the instructions from your teacher.

Hand washing: Wash your hands thoroughly when finished with the practical. Use antibacterial soap and warm water. Rinse well.

General safety awareness: When this symbol appears, follow the instructions provided. When you are asked to develop your own procedure in a lab, have your teacher approve your plan before you go further.

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Corrosive: Avoid contact with the skin, these can rust chemical cupboards.

Irritant or Harmful: This symbol covers a wide range of hazards-avoid contact with the skin, do not inhale, etc. Refer to relevant data sheet for details.

Harmful material that should be kept away from edible material.

Dangerous when wet: This generally means that it will react fairly violently with water.

1 Qualitative Organic Analysis 1.1 INTRODUCTION Qualitative organic analysis of an unknown organic compound is an integral component of chemistry practical curriculum of all universities in India and a large number of universities all over the world. The benefits of this type of exercise in the laboratory training are immense. A student learns to (i) measure physical constants like melting and boiling points, optical rotation (ii) perform chemical tests for detection of functional groups (iii) use techniques for purification of compounds like crystallisation, distillation, sublimation and extraction etc. (iv) carry out small scale preparation of organic compounds while preparing derivatives of given organic compounds (v) interpret the spectral data of the compound (vi) refer to the literature and (vii) finally logically analyse all the observations and arrive at a particular conclusion. Analysis of an unknown compound can be compared to solving a Zigsaw puzzle, it is interesting but challenging and prepares a student to face the challenges of a carrier in R&D in chemistry. Though there are a number of advantages of including organic analysis in the syllabus, there are a few disadvantages associated with it and all other types of experiments in general. A few of these can be listed here: (i) huge amounts of chemicals are used and thus a corresponding amount of chemical waste is produced (ii) the chemicals that were used conventionally were not evaluated for their hazardous and toxic nature because of paucity of awareness. In many countries there are practically no laws for handling the above mentioned concerns and hence the academic institutes cause a lot of harm to the environment and life on the planet. The problems should not be solved by NOT giving good training to the students but should be addressed in a positive way. The question obviously arises HOW? The answer is not simple but the following approaches are being adopted worldwide to tackle the concern • New experiments are being designed keeping in mind the safety of students and others in the laboratory. • The conventionally used hazardous chemicals are being replaced by safer chemicals. • The use of chemicals is being drastically reduced by performing experiments with micro amounts of chemicals. • Safe disposal of chemical waste etc.

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1.2 EXPERIMENTS PERFORMED TO ANALYSE A COMPOUND Following experiments are performed in the laboratory to analyse the given unknown organic compound: (i) Preliminary tests (ii) Detection of functional group/s (iii) Specific tests for a compound (iv) Preparation of suitable crystalline derivative of the given compound. (v) Comparison of results obtained experimentally (specially the m.p./b.p of compound and derivatives) with those reported in literature.

1.3 PRELIMINARY EXAMINATION The preliminary examination of the given unknown compound provides useful information about it as well as helps to decide many a times, the method to be adopted to separate a mixture of two compounds. Following observations and tests are performed.

1.3.1 Physical State Note the physical state of the given compound: whether the compound is a solid or a liquid. If solid, is it crystalline or amorphous? The physical state of the compound is not of much significance. However, it is helpful in locating the compound in the reference books as these are arranged according to their physical state, melting/ boiling point and functional group. 1.3.2 Colour of the Compound Note the colour of the compound. The colour of the compound gives information which may be helpful for its identification. Observation

Inference Colourless Carbohydrates, aldehydes, ketones, carboxylic acids, esters, alcohols, ethers and some simple hydrocarbons. Colourless when freshly distilled or recrystallised Phenols, amines, amino phenols etc. but slowly turning reddish or pinkish (due to slow oxidation by air) Yellow Nitro compounds, some quinones and iodoform etc.

1.3.3 Odour of the Compound Do not smell the compound directly. Some organic compounds have a strong and a characteristic odour which can be sensed easily without smelling the compound directly. Observation Cinnamom-like Vinegar-like Pleasant-fruity Bitter almond odour

Inference Cinnamaldehyde Acetic acid Estres Nitrobenzene, benzaldehyde, benzonitrile

1.3 Preliminary Examination Advanced Experimental Organic Chemistry

Sweet odour Garlic-like Phenolic odour Irritating pungent odour Pyridine-like Fish-like

9

Chloroform, alcohols Thiophenols, thioalcohols Phenols Lower acids, aldehydes and acid chlorides Heterocyclic bases Aliphatic amines

Note: Do not inhale the vapours of the compound directly.

1.3.4 Flame Test This test is normally performed to distinguish between aliphatic and aromatic compounds though there are a number of exceptions. This test also gives a good indication about the presence/absence of a carbohydrate. Procedure

Heat the organic compound (20 mg) on a nickel spatula over a low flame, raise the temperature gradually till the compound starts burning. Observation Non-sooty flame Sooty flame Ammonical odour Charring with smell of burnt sugar

Inference Aliphatic compounds Aromatic compounds Urea, thiourea etc. Carbohydrates, tartrates, citrates etc.

Notes: Some aromatic compounds may produce a non sooty flame and some aliphatic compounds like chloroform, carbon tetrachloride, acetylene etc. give a sooty flame on burning.

1.3.5

Action of Alkali

Procedure Heat the organic compound (~50 mg) with freshly prepared aqueous sodium hydroxide solution (4-5mL, 20%) for about a minute. Observation Smell of ammonia Smell of chloroform Smell of amine on prolonged boiling

1.3.6

Inference Primary amides Chloral Anilides

Test with Ferric Chloride

Procedure To a solution of the organic compound (50 mg) in water or alcohol (2 mL) add a drop of neutral ferric chloride solution through the sides of the test tube. Note the colour immediately and also if there are any subsequent colour changes. Shake the test tube and add a few more drops of neutral ferric chloride solution and note again.

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Observation Intense purple, green, blue or red colour

Qualitative Organic Analysis

Inference Phenols

Initially a red colour and then formation of a dark p-hydroquinone green precipitate Buff coloured precipitate

Some aromatic acids and enols

Red colour (decolourised on addition of dilute Amino acids, lower fatty acids hydrochloride acid) Intense reddish brown colour or a precipitate (no o-and m-phenylene diamines change on addition of hydrochloric acid)

Notes: • The neutral ferric chloride solution is prepared by adding a dilute ammonium hydroxide solution (dropwise) to the ferric chloride solution until a faint precipitate is formed, which does not dissolve on shaking. The solution is filtered and the filtrate is used for the test. • A freshly prepared ferric chloride can also be used directly.

1.3.7 Solubility Test The solubility of the given compound is checked in the following solvents in the order given below. This test provides useful information about the given compound and can be very useful in the separation of compounds present in a mixture. • Water • Sodium bicarbonate solution • Dilute sodium hydroxide solution • Dilute hydrochloric acid • Concentrated sulphuric acid (caution, it is highly corrosive in nature) • Solvent ether (caution, ether is highly inflamable). On the basis of their solubility, the compounds are divided into three categories. Category-A: The given compound is soluble in water as well as in ether. Category-B: The compound is soluble in water but insoluble in ether. Category-C: The compound is insoluble in water. If the compound is soluble in water (compounds in category A and B), check whether the aqueous solution is neutral or acidic or basic to the litmus. (a) Category-A: Compound is soluble in water as well as in either.

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Observation

Inference

The aqueous solution is acidic to litmus

Phenols, aliphatic acids, aromatic acids (more soluble in hot water.)

The aqueous solution is alkaline to litmus

Amines.

The aqueous solution is neutral to litmus

Lower aliphatic alcohols, aldehydes, ketones, nitriles etc.

(b) Category–B: Compound is soluble in water but insoluble in ether. Observation

Inference

The aqueous solution is acidic to litmus

Hydroxy acids, sulphonic acids, salts of amines, amino acids

The aqueous solution is alkaline to litmus

Amino alcohols.

The aqueous solution is neutral to litmus

Carbohydrates, other polyhydric alcohols.

Note: Ether used for checking the solubility must be dry, otherwise in presence of moisture even the carbohydrates may also dissolve. (c) Category-C: If the given compound is insoluble in water, it may be an aromatic acid, phenol, ester, aldehyde, ketone, ether, amine, nitro compound or a hydrocarbon. Check the solubility in dilute sodium bicarbonate, dilute sodium hydroxide and dilute hydrochloric acid. Observation Soluble in sodium bicarbonate solution

Inference Carboxylic acids, polynitro Phenols

Soluble in dilute sodium hydroxide (5%) but insoluble Phenols in sodium bicarbonate Soluble in dilute hydrochloric acid

Amines

If the compound does not respond to any of the above solubility tests, check its solubility in concentrated sulphuric acid. Observation

Inference

Soluble

Neutral compounds like ethers, alcohols, esters, aldehydes and ketones etc.

Insoluble

Inert compounds like benzene, halo substituted benzene, toluene, xylenes etc.

The solubility tests are summarised in Chart 1.1.

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Chart 1.1: Solubility tests

1.3.8 Detection of Extra Elements Organic compounds are defined as compounds of carbon and hydrogen (hydrocarbons) or their derivatives. The elements carbon and hydrogen are always present in all organic compounds. The elements other than these, for example nitrogen, sulphur, oxygen and the halogens (chlorine, bromine and iodine) are known as extra elements and the presence of these in an organic compounds can be established by the following tests. Lassaigne’s test: This is the most widely used test for the detection of nitrogen, sulphur and halogens in an organic compound. In this test the organic compound is fused with sodium to give the corresponding easily identifiable sodium salts. Thus nitrogen, sulphur and halogens when present in an organic compound give sodium cyanide, sodium sulphide and sodium halide respectively on fusion with sodium. Na + C + N → NaCN

2Na + S

→ Na2S



Na + X

→ NaX (X = CI, Br, I)

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If an organic compound contains both nitrogen and sulphur, on fusion with sodium, it may give a mixture of sodium cyanide and sodium sulphide or sodium thiocyanate as shown below:

Na + C + N + S → NaSCN

Procedure Sodium is a very reactive metal and reacts with oxygen, moisture etc. present in air therefore, it is preserved in a bottle in a non reactive solvent like kerosene. For use, take out a small piece of sodium from under the kerosene and dry on a filter paper. It is hard and yellowish brown in colour. Hold the piece of sodium with a pair of forceps and peel off the top yellow brown layer with a knife. A silvery, soft and shinning piece of sodium is obtained. Take a small piece of this sodium (approx. 50–60 mg) in a clean, dry ignition tube. Heat the tube slowly till the sodium melts to form a shiny ball. Remove it from over the flame and add the compound (20–25 mg or a very small drop of the liquid compound). (A paste of the liquid compound in sodium carbonate may be used or solid sodium carbonate may be added after the addition of liquid compound to reduce its loss by evaporation.) First heat the ignition tube on a low flame and then on a stronger flame until its lower end becomes red hot. Keep it in the red-hot condition for 1–2 minutes (until the compound has completely reacted) and then plunge into 10 mL of distilled water kept in a porcelain dish. Carry out two more fusions in separate ignition tubes and plunge them in the same porcelain dish. Boil the contents of the porcelain dish for 2 minutes and filter. Use the filtrate for testing the extra elements. Precautions • Do not touch sodium with your fingers; handle it with forceps. • There may be a slight explosion on heating sodium with compounds like nitroalkanes, azides, diazonium salts, chloroform and carbon tetrachloride. Wear safety goggles to avoid accidents. • Unused pieces of sodium (that might be left because the ignition tube broke before the addition of the compound) should be carefully decomposed by reacting it with a small amount of ethyl alcohol in a test tube and never discarded in the sink or dustbin. • Take freshly cut sodium in one ignition tube at a time and not in three ignition tubes together.

1.3.9 Test for Nitrogen (a) Reaction with Ferrous sulphate As already mentioned compounds containing carbon and nitrogen produce sodium cyanide on fusion with sodium. The addition of freshly prepared ferrous sulphate solution to the Lassaigne’s extract gives a bluish green precipitate due to the formation of ferrous hydroxide. Warming_the solution converts some ferrous ions into ferric ions and acidification gives ferriferro cyanide complex, which is Prussian blue in colour

6NaCN + FeSO4 → Na4[Fe(CN)6] + Na2SO4

3Na4[Fe(CN)6] + 2 Fe2(SO4)3 → Fe4[Fe(CN)6]3 + 6Na2SO4

Procedure To the lassaigne’s or sodium extract (1–2 mL) in a test tube, add solid ferrous sulphate (50–75 mg). A dark greenish grey precipitate of ferrous hydroxide is obtained; in case sulphur is also present, a black precipitate of ferrous sulphide is also obtained. Boil the mixture gently for ~ 30 seconds and acidify with

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dilute sulphuric acid (3–4 mL). A blue precipitate (Prussian blue) or a blue colour indicates the presence of nitrogen in the compound. Notes: • A Prussian blue coloured solution is obtained only when both Fe2+ and Fe3+ ions are present in the reaction mixture. Acidification of the solution without boiling gives a –ve test and excess boiling of it also results in a –ve test in spite of the presence of sodium cyanide in the solution. The reason is that Fe2+ is not oxidised to Fe3+ in the first case and complete oxidation occurs converting all Fe2+ ions to Fe3+ leaving no Fe2+ ions in the solution in the second case. • It is not advisable to add ferric chloride (as a source of Fe3+) to the solution, as it tends to produce green colour in the solution. • Only sulphuric acid should be used for acidification. Nitric acid cannot be used as it is a strong oxidising agent and will therefore oxidise all ferrous to ferric ions. Spot test for Nitrogen The above test for nitrogen can be performed as a spot test instead of conventional test in a test tube. The chemicals used and the chemistry of the test remains the same. General Instructions for Spot Tests • Use concentrated Lassaigne’s extract for detection of extra elements (N and S) and a concentrated solution of the compound obtained by dissolving 3–4 drops of liquid or ~200–300 mg of the solid in a test tube in one of the following solvents (1–2 ml) in the order given below for detection of unsaturation and functional groups Water Alcohol Acetone Ethylacetate Notes: • If the given compound is water-soluble, use its concentrated aqueous solution for all the spot tests except for the hydroxamic acid test for esters (See under the tests for esters). • If the given compound is insoluble in water but soluble in alcohol, use the alcoholic solution for all the spot tests except Baeyer’s test for unsaturation and ceric ammonium nitrate test for alcohol.

Fig. 1.1: Spot Test kit

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• If the given compound is insoluble in water as well as in alcohol, prepare its solution in acetone or ethylacetate. • Do not use: (a) acetone solution of the compound for 2,4-DNP test, fehling’s and tollens’ test etc. (b) ethyl acetate solution of the compound for hydroxamic acid test for esters. • Liquid organic compound may be used directly for spotting, wherever mentioned in the procedure. • The spot tests can be performed on one of the following surfaces. • A thin strip (1″ × 6″) of Whatman filter paper except in those cases in which concentrated sulphuric acid is a reactant because the paper is charred and gets burnt by it. • A silica gel coated TLC plate or a strip (1 × 3″) cut out from a Merck silica gel (60F, 754) coated plate is used in such cases: • Use a fine capillary tube for spotting by starting spotting on end of the strip so that all the tests can be performed on one strip. • Heat the strip of paper or Merck plate or TLC plate (wherever heating is recommended) in an electric oven maintained at ~120°C. • Add the solution of the compound and the reagents using a dropper with a fine nozzle or a capillary tube while using a grooved tile for testing. • A reagent kit (Fig. 1.1) will be available to you. Use only capillary tubes to take out the reagents, do not mix up the reagents and always keep them covered in order to prevent evaporation of the solvent. If a reagent gets discoloured, it should be replaced by the freshly prepared reagent. A wooden box with a lid (Fig. 1.1) is used for holding the reagent bottles. The reagent bottles may be either small transparent injection bottles or homeopathic medicine bottles with a lid. A small hole in the lid is provided for inserting the capillary tube used for spotting the reagents. The kit box should contain the following reagents: • Potassium permanganate solution. • α-Naphthol in alcohol (10%) • Fehling solution A • Fehling solution B • Barfoed reagent (Freshly prepared) • Concentrated sulphuric acid • Saturated sodium bicarbonate solution • Solution of resorcinol in water (saturated) • Dilute sulphuric acid • Neutral ferric chloride solution • Concentrated solution of succinic acid • Alcoholic solution of 2,4-dinitrophenyl hydrazine • Schiff’s base solution

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• Ceric ammonium nitrate solution (saturated) • Saturated solution of hydroxylamine hydrochloride in water • Saturated solution of sodium nitrite in water • Conc. solution of β-naphthol in dilute sodium hydroxide (freshly prepared) • Dilute hydrochloric acid • Conc. hydrochloric acid • Zinc dust • Phenol • Bromine in carbon tetrachloride. Spot test for nitrogen Spot the following solutions one over the other in the order given below on a Whatmann paper strip • Freshly prepared concentrated solution of ferrous sulphate (twice) • A spot of concentrated Laassaigne’s extract (twice) • Dilute sulphuric acid. Appearance of a blue coloured spot shows the presence of nitrogen. (b) Benzidine-Copper Sulphate Test This test is based on the fact that benzidine is oxidised to benzidine blue by cupric ions in presence of cyanide ions. When cupric sulphate solution is added to benzidine solution, a small amount of benzidine is oxidised to benzidine blue but the reaction equilibrium lies mainly towards left and the blue colour of benzidine blue is not seen.     Benzidine blue + Cu22+ Benzidine + Cu2+ 

But in presence of cyanide ions, the reaction equilibrium is shifted to the right and benzidine blue is obtained. This is due to the fact that trace amount of cuprous ions formed in the above reaction, react irreversibly with the cyanide ions, disturbing the equilibrium thus forcing the reaction in the forward direction.

4 CN– + Cu2+ 2 → 2 Cu (CN)2

This reaction does not occur in absence of cyanide ions. Procedure Take the Lassaigne’s extract (1–2 mL) in a test tube and acidify with acetic acid (3–4 drops). Add freshly prepared benzidine solution (2–3 drops of 1% solution in 50% acetic acid) to it and shake the mixture. Treat the resultant solution with copper sulphate solution (1–2 drops, 1%). Appearance of a blue colour or a precipatitate indicates the presence of cyanide and hence nitrogen in the given compound. Notes: If iodine is also present along with nitrogen, a greenish precipitate is obtained instead of a blue colour.

1.3 Preliminary Examination Advanced Experimental Organic Chemistry

17

(c) Reaction with PNB and ODNB Another test that can be used to detect nitrogen (as sodium cyanide) in the Lassaigne’s extract is by its reaction with p-nitrobenzaldehyde (PNB) and o-dinitrobenzene (ODNB) in the presence of sodium hydroxide. Development of a blue-violet colour indicates a positive test.

Acidification of the blue-violet solution gives a yellow colour due to loss of quinonoid structure.

Procedure To 1 ml of the alkaline Lassaigen’s extract (add a few drops of sodium hydroxide solution, if it is not alkaline), add p-nitrobenzaldehyde solution (4-5 drops) and o-dinitrobenzene solution (4-5 drops). Development of a blue-violet colour indicates the presence of nitrogen. Notes: • Use 0.1 M solution of PNB and ODNB in methyl cellosol (2- methoxyethanol). • Perform a blank test simultaneously. • The test is not reliable if the given compound is a carbohydrate. 1.3.10 Test for Sulphur (a) Lead acetate test Sulphur of the organic compound, present as sodium sulphide in the Lassaigne’s extract can detected by acidification of the extract followed by addition of lead acetate. Formation of black precipitate of lead sulphide indicates the presence of sulphur. Na2S + (CH3COO)2Pb → PbS + 2 CH3COONa Black

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Procedure In a test tube, acidify the sodium extract (~ 1 mL) with acetic acid (2 mL) and add a few drops (4–5) of lead acetate solution. Appearance of a black precipitate of lead sulphide indicates the presence of sulphur. Notes: Hydrochloric, sulphuric or nitric acids should not be used for acidification of the extract. Spot test Spot the following solutions one over the other on a Whatmann paper strip Lassaigne’s extract (2–3 times at the same point) Acetic acid (2 times) Lead acetate solution Development of a black spot shows the presence of sulphur (b) Sodium Nitroprusside Test Sodium extract containing sodium sulphide gives a reddish violet/purple colour when treated with a solution of sodium nitroprusside

Na2S + Na2 [Fe(CN)5NO] → Na4 [Fe(CN)5NOS] Sodium nitroprusside

Sodium sulphonitroprusside (Reddish violet)

Procedure To the sodium extract (l–2 mL) in a test tube, add a freshly prepared aqueous solution of sodium nitroprusside (1–2 drops). A deep reddish violet colouration indicates the presence of sulphur. Spot test Procedure Spot the following one over the other in the order given on a Whatmann filter paper strip Concentrated Lassaigne’s extract (twice) Sodium nitroprusside solution (twice) Development of a purple spot shows the presence of sulphur in the compound

1.3.11 Test for Nitrogen and Sulphur Present Together As already mentioned, a mixture of NaCN and Na2S or NaSCN is formed when a compound containing both nitrogen and sulphur is fused with sodium for the preparation of Lassaigne’s extract. The test for NaCN and Na2S have already been discussed above. NaSCN that might be formed during fusion is tested by addition of ferric chloride, formation of blood red colour confirms the presence of nitrogen and sulphur in the given compound. (a) Test with FeCl3

6 NaSCN + FeCl3 → Na3 [Fe(SCN)6] + 3NaCl Blood red colour

1.3 Preliminary Examination Advanced Experimental Organic Chemistry

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Procedure Acidify the sodium extract (2 mL) with dilute hydrochloric acid (5–6 drops) and treat with ferric chloride solution (2–3 drops). Appearance of a blood red colour indicates the presence of nitrogen and sulphur. Notes: If the organic compound containing both nitrogen and sulphur is fused with a slight excess of sodium for the preparation of Lassaigne’s extract, sodium cyanide and sodium sulphide are formed instead of sodium thiocyanate, and the test for NaCN and Na2S can be performed. NaSCN + 2 Na → NaCN + Na2S Spot test for nitrogen and sulphur present together The spot test can be performed on a Whatmann paper strip by spotting the following solutions one over the other • Concentrated Lassaigne’s extract (twice) • dil. hydrochloric acid • Ferric chloride solution Appearance of a blood red coloured spot shows the presence of nitrogen and sulphur in the given compound. (b) Cobalt nitrate test A blue coloured complex of cobaltithiocyanic acid is obtained when the Lassaigne’s extract containing sodium thiocyanate is treated with cobalt nitrate in an acidic medium NaSCN + HCl → HSCN + NaCl 4 HSCN + Co (NO3)2 → H2[Co(SCN)4] + 2 HNO3 Cobaltithiocyanic acid (blue)

Procedure Acidify the sodium extract (2 mL) with dilute hydrochloric acid (1 mL). Add alcohol (2 mL) and then treat with cobalt nitrate solution (4–5 drops). Appearance of a blue colour indicates the presence of both nitrogen and sulphur in the given organic compound.

1.3.12 Test for Halogens (a) Beilstein test The test is based on the fact that halides of copper burn to give a green coloured flame. Procedure Heat a copper wire having a small loop at its end in a flame until the flame is no longer coloured. Cool, dip the loop in the organic compound and heat on the edge of a clear flame. A green coloured flame indicates the presence of a halogen. Notes: • Fluorine containing compounds do not give this test. • This test is extremely sensitive since minute traces of halogen containing impurities give this test. • This test may also fail for very volatile liquids, since these evaporate before the copper wire is heated sufficiently to cause decomposition of the given compound. • Certain compounds which do not contain halogen like derivatives of quinoline, pyridine, organic acids and urea also impart a green colour to the flame.

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(b) Silver nitrate test This is a more reliable test for detection of halogen (Cl, Br and I) in an organic compound and is based on the fact that when a solution (Lassaigne’s extract) containing soluble halides is treated with silver nitrate, an insoluble precipitate of silver halide is formed NaX + AgNO3 → AgX + NaNO3 where X = Cl, Br or I Procedure Acidify the sodium extract (~ 1 mL) with dilute nitric acid (l–2 mL). Boil the solution gently for 1–2 minutes to expel any hydrogen cyanide or hydrogen sulphide that may be presents due to the presence of nitrogen or sulphur in the organic compound. Add silver nitrate solution (0.5 mL) to the above solution. A white precipitate readily soluble in ammonium hydroxide indicates the presence of chlorine. A pale yellow precipitate soluble in excess of ammonium hydroxide solution indicates the presence of bromine. In case a yellow precipitate insoluble in ammonium hydroxide is formed, the presence of iodine is indicated. Notes: • The extract is not boiled with dilute nitric acid if the given compound does not contain N and/or S as extra elements. • Boiling with nitric acid is essential as otherwise both nitrogen (as NaCN) and sulphur (as Na2S) if present in the extract, interfere in the silver nitrate test because NaCN gives a white precipitate of AgCN and Na2S gives a black precipitate of AgS. • The colour of the silver halides and their solubility in ammonium hydroxide are not dependable observation for inferring the presence of a specific halogen (Cl, Br, I) in an organic compound because the amount of sodium halide formed by fusion of the organic compound with sodium is very small. Even silver iodide which is present in a very small amount will be soluble in excess of ammonium hydroxide solution. Thus a positive silver nitrate test only indicates the presence of a halogen. To find out which halogen is present, the layer test given below should be performed. (c) Layer test In this test, the sodium halide present in the Lassaigne’s extract is oxidized to the corresponding halogen by reaction with an oxidising agent like nitric acid or potassium permanganate or chlorine water. The halogens (Br2 and I2) thus formed are coloured and can be dissolved in an organic solvent like carbon tetrachloride and easily detected from the colour obtained in the organic layer. Bromine gives an orange and iodine a purple colour on dissolution in an organic solvent whereas chlorine gives almost a colourless solution. 2 NaI + Cl2 → 2 NaCl + I2







Purple colour in CCl4

NaBr + Cl2 → 2 NaCl + Br2

Reddish brown/orange colour in CCl4

Procedure To the sodium extract (~2 mL), add concentrated nitric acid (1 mL) in a test tube. Heat the mixture for a few seconds. Add carbon tetrachloride or chloroform (1 mL) and shake well. If the organic layer attains a violet colour, the presence of iodine is indicated; on the other hand if the organic layer attains an orange brown colour the presence of bromine is indicated; in case both iodine and bromine are absent and a white precipitate is obtained in the silver nitrate test, the presence of chlorine is indicated.

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Notes: • Do not heat the extract with nitric acid for a long time, as the halogen formed by oxidation of sodium halide may escape from the solution. • This test is helpful in case only one halogen (chlorine, bromine or iodine) is present in the organic compound. (d) Schiffs reagent test/Test with Fluorescein This test is based on oxidation of sodium halide to the corresponding halogen with lead dioxide. The liberated halogen then reacts with fluorescein. Sodium bromide, for example is oxidised to bromine, which then reacts as shown below.

Procedure Acidify the sodium extract (2 mL) with glacial acetic acid (1 mL) and heat to boiling with lead dioxide (0.1 – 0.2 g). Place a piece of filter paper moistened with Schiff’s reagent on the mouth of the test tube. If the paper turns violet–the presence of bromine is indicated, a blue colour indicates the presence of iodine. Alternatively, a piece of filter paper dipped in a 1% alcoholic solution of fluorescein is kept over the mouth of the test tube in place of Schiff’s reagent. Appearance of a rose–pink colour indicates the presence of bromine and the appearance of brown colour indicates the presence of iodine. (e) Test for halogens (when present together) When the given compound contains more than one halogen, the relative ease of oxidation of sodium halide becomes the basis of testing. The ease of oxidation is NaI > NaBr > NaCl i.e. sodium iodide is oxidised most easily. Procedure Acidify the sodium extract (~2 mL) with dilute sulphuric acid. Boil for a minute, cool and add carbon tetrachloride (2 mL). Now add a solution of sodium nitrite (2 mL, 20%) while shaking the test tube cautiously. A purple colour in the organic layer indicates the presence of iodine. If iodine is indicated, add more of sodium nitrite and boil until all the iodine (violet vapours) is removed. Boil for some more time to expel the nitrous fumes. Cool the remaining solution, add an oxidising agent (KMnO4, ~ 1 mL) and carbon tetrachloride (1 mL) and shake well. An orange-brown colour in the organic layer indicates bromine. Boil this solution with glacial acetic acid (1 mL) and lead dioxide (0.5 g) to remove all the bromine. After

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the removal of bromine, filter, acidify the remaining solution with dilute nitric acid and then add silver nitrate solution (1 mL). Appearance of a white precipitate indicates the presence of chlorine. (f) Test for fluorine Acidify the sodium extract (~1mL) with acetic acid, boil and cool. Place a drop of this solution on a piece of zirconium-alizarin test paper. A yellow colour on the red paper indicates the presence of fluorine. (The test paper is prepared by dipping a piece of filter paper into a solution obtained by mixing alcoholic alizarin solution (3 mL, 1%) and zirconium chloride (or nitrate) solution (2 mL, 0.4%). The red filter paper is dried before use and moistened with a drop of acetic acid (50%) while performing the test.)

1.3.13 Determination of Melting Point Melting point of a solid is the temperature at which it becomes a liquid at normal atmospheric pressure. It is also defined as a temperature at which the solid and liquid phases of a substance are in equilibrium at normal atmospheric pressure unless stated otherwise. The melting point of a substance is considerably influenced by the external pressure and the impurities. Pure compound shows sharp melting point whereas an impure compound melts over a range. Therefore, it is important to use a pure compound for determination of melting point. The melting point of the given organic compound is very crucial for its identification as a given compound can only be traced in the literature if its melting point is correctly determined and known. Apart from other precautions which are listed below (notes), it is important to check the calibration of the thermometer to be used for determination of melting point of an unknown compound. The calibration is checked by determining the melting or boiling points of a few selected organic compounds such as p-dichlorobenzene (53°C), benzoic acid (122°C), salicylic acid (159°C) and anisic acid (184°C) as reference compounds. Acetone (B.P. 56°C) and water (B.P. 100°C) are also useful reference compounds. The apparatus used for determination of melting point are shown below:

Fig. 1.2: Electric melting point apparatus

Fig. 1.3: Kjeldahl’s method of melting point determination

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Procedure Finely powder the dry organic compound (100 mg) with help of a spatula on a clean, dry watch glass. Seal one end of a capillary tube (~3′′ long) and fill it (3-5 mm) with the finely powdered sample to form a compact column. Attach the capillary to the thermometer and place it in the Kjeldahl flask (50 mL capacity) containing the bath liquid (~ 20 mL) (concentrated sulphuric acid or paraffin wax) such that the thermometer bulb and the portion of capillary containing the compound are immersed in the bath liquid as shown in the diagram (Fig. 1.3). Heat the Kjeldahl flask gently at a uniform rate (1–2° rise in temperature per minute). Note the temperature at which the solid melts. Notes: • A sand bath should always be kept under the flask while determining the melting point by the Kjeldahl method so that in case the Kjeldahl flask leaks or breaks, the hot bath liquid does not spill over but is absorbed by the sand. • Sulphuric acid is highly corrosive and causes severe burns. It should be handled with utmost care. Injurious to eyes, skin. Causes serious injuries if ingested or inhaled. Immediately get expert help without wasting any time. • It is advisable to determine the approximate melting point first and then the precise melting point. In the second determination, the rate of heating should be carefully controlled (especially near the melting point of the compound.) • The organic compound should be finely powdered. • The compound should from a compact column at the sealed end of the capillary tube. This can be done by either dropping the capillary tube through a glass tube (18–20′′ long) on the working table or tapping the lower end of the capillary. • The lower tip of the capillary tube should be at the same level as the tip of the thermometer bulb in the Kjeldahl flask. • A Thiele’s tube may be used in place of a Kjeldahl flask. • Many organic compounds undergo a change in crystalline structure just before melting, usually because of release of solvent of crystalisation. This may lead to shrinkage of sample in the capillary tube. This should not be taken as the beginning of the melting process. • Some compounds undergo decomposition near their melting point. Darkening near the melting point should not be confused with the melting. • An electric or an electronic melting point apparatus can also be used instead of Kjeldahl’s or thiele’s method. In both the type of apparatus, the finely powdered sample is filled in a capillary tube and is placed in the space provided for it in the apparatus. Melting point is then measured as described in the monograph.

1.3.14 Determination of Boiling Point Like the melting point of a solid, the boiling point of a liquid compound is crucial for its identification. The boiling point of a liquid is defined as the temperature at which its vapour pressure is equal to the normal atmospheric pressure. As a liquid is heated, its vapour pressure increases and at a particular temperature its vapour pressure equals the atmospheric pressure, the liquid starts boiling at this temperature. Though a pure liquid has a sharp boiling point—the converse is not always true. A sharp boiling point may sometimes be caused by a constant boiling mixture (Azeotrope) of two or more liquids. The boiling point rises with increased external pressure and is lowered by low external pressure.

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Fig. 1.4: Determination of Boiling point above 100°C Fig. 1.5: Determination of Boiling point below 100°C

Procedure Take the liquid compound (~20–25 mL) in a clean and a dry round bottomed flask (50 mL). Add a few pieces of pumice stones and set up the distillation apparatus as shown is Fig. 1.4. Use a water condenser for liquids boiling below 150°C and an air condenser for liquids having a higher boiling point.

Fig. 1.6: (a) Ignition tube with the organic compound attached to therometer, (b) Kjeldahl apparatus, (c) Thieles’ tube

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25

Heat the liquid gently first in a boiling water bath (Fig. 1.5). If the liquid does not start boiling in the water bath, it means that the boiling point of the compound is >100°C, in such a case heat the liquid uniformly over a low flame using a Bunsen burner (Fig. 1.4). Collect the distillate, discard first fraction of distillate (3-4 mL). Change the receiver and collect the distillate which comes at constant temperature. Note the temperature i.e. the boiling point of the compound. In case only a small amount of liquid is available, determine the boiling point by Siwoloboff’s method. Take the organic liquid (0.5–1 mL) in a clean and dry ignition tube (4–5 mm in diameter and 80-100 mm long). Introduce a capillary tube sealed at the upper end into the liquid (as shown Fig. 1.6) Attach the ignition tube with the help of a rubber band to the thermometer and fit this into a melting point bath. Heat the bath gently. The temperature at which a rapid and continuous stream of bubbles first emerges from the capillary tube is the boiling point of the liquid. At this temperature the bubbling ceases and the liquid begins to rise in the capillary. Notes: • All the apparatus should be dried before setting up the apparatus for distillation. • If a turbid distillate is obtained or a cracking sound is produced during boiling point determination it generally means that there is moisture in the liquid or the apparatus. • Use a calibrated thermometer for boiling point determination. • Fix all the joints of the boiling point apparatus securely to ensure that there is no leakage of vapours from any part of the apparatus. • In case of any leakage, immediately stop heating.

1.3.15 Detection of Unsaturation Aliphatic organic compounds containing one or more carbon-carbon double or triple bond/s are unsaturated in nature. Aromatic compounds with a side chain containing double or triple bonds also give tests like aliphatic unsaturated compounds. Inspite of high degree of unsaturation present in all aromatic hydrocarbons like benzene, toluene, xylenes etc., these do not react like unsaturated aliphatic compounds. The presence of unsaturation in the given compound can be established by the following reactions. 1.3.16 Test with Bromine Solution Aliphatic compounds which have olefinic or acetylinic bonds, aromatic compounds with a side chain containing multiple bonds and aromatic compounds which have strong activating groups in them (phenols and amines) decolourise the bromine solution. Compounds containing a carbon-carbon multiple bond add bromine and in the process decolourise bromine solution. The reaction is given below:

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However, the presence of electron withdrawing groups on the carbon atoms of an olefinic or acetylinic bond slows down the rate of addition of bromine and in extreme cases i.e. in compounds containing more electron withdrawing groups, the addition may not occur at all under the conditions used for the reaction. Steric hindrance on the multiple bond may also prevent the reaction. For example C6H5CH=CHCOOH reacts slowly with bromine and (C6H5)2C=C(C6H5)2 does not react with bromine in carbon tetrachloride. A positive test for unsaturation is one in which bromine colour is discharged without the evolution of hydrogen bromide. Discharge of the bromine colour accompanied by evolution of hydrogen bromide indicates substitution of hydrogen of the compound by bromine and not addition. Many aromatic compounds which have strong electron donating groups present on the ring undergo this type of reaction. Some examples are phenols and amines.

Certain heterocyclic compounds containing tertiary nitrogen and aliphatic amines add bromine. For example

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As already stated HBr is evolved in a substitution reaction and no HBr is obtained if addition of bromine takes place. Evolution of HBr is observed only if carbon tetrachloride has been used as a solvent. If the reaction is carried out with Br2 water instead, HBr cannot be easily detected as it is soluble in water. The rate of bromination reaction is greatly enhanced in polar solvents like a water, acetic acid etc.

Procedure Dissolve the organic compound (0.1g or 0.2 mL) in carbon tetrachloride (2 mL) and add dropwise with shaking a solution of bromine in carbon tetrachloride (5%). Discharge of bromine colour without evolution of HBr shows the presence of unsaturation in the molecule. Notes: • Bromine is highly injurious, Causes eye burn, skin burns and ulceration, may be fatal if inhaled. In case of contact or burning, wash eyes and skin with plenty of fresh water and seek medical help immediately. Seek urgent medical advice if inhaled. • Bromine water or a solution of bromine in acetic acid can also be used in this test. 1.3.17 Baeyer’s Test: Test with Potassium Permanganate This test gives important information about the presence/absence of unsaturation and the presence of easily oxidisable group in the given organic compound. Olefinic, acetylinic compounds give a positive test but aromatic hydrocarbons (without any unsaturated side chain) and those substituted with a deactivating group do not give this test. Easily oxidisable compounds like some aldehydes, some carbohydrates, formic acid and its esters, aromatic organic compounds which have strong activating groups like phenolic and amino directly attached to the aromatic nucleus also decolourise potassium permanganate solution.

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Procedure To a solution of the organic compound (100 mg) in water or acetone in a test tube, add dropwise a dilute potassium permanganate solution (2–3 drops, 1–2%). Discharge of the purple colour of potassium permanganate indicates the presence of unsaturation or the presence of an easily oxidisable group/s in the organic compound. Spot test Procedure: Use a concentrated solution of the compound in either water or acetone or the liquid compound and spot the following one over the other on a Whatman filter paper strip. (i) Concentrated solution of the compound or the liquid compound (ii) Baeyer’s reagent (from the kit) Disappearance of the purple colour of Baeyer’s reagent shows the presence of unsaturation in the given compound. Notes: • Observe the colour change immediately after spotting of the solutions. • Perform a blank test simultaneously. • Certain compounds like phenols, amines and aldehydes undergo oxidation by KMnO4 and hence decolourise the reagent.

1.4 IDENTIFICATION OF FUNCTIONAL GROUP 1.4.1 Introduction Identification of functional group/s in an unknown organic compound is one of the most significant step of analysis because • The known compounds are classified in literature based on the functional groups present in them so an unknown compound can only be correctly identified if the functional group analysis is performed correctly. • The derivative of the compound, which must be prepared for complete analysis can only be prepared if the nature of functional group/s is correctly established. The functional group/s in a given organic compound can be identified by • Performing chemical tests in the laboratory. • Recording and then analysing the spectra. Analysis of IR spectrum gives information about the functional group present, UV is useful for detection of unsaturation and conjugation, NMR spectrum provides useful information about the functional group as well as the structure of compound and the mass spectrum also gives good information about the structure of the compound and the molecular mass of the compound. Following points should always be kept in mind while performing the functional group tests in laboratory: • Tests for all the functional groups must be systematically performed in the order given below as the given compound may contain one or more than one functional groups. Short cuts should be avoided to save time, confusion and chemicals.

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29

• Some compounds do not give a positive test for the functional group present in them, for example, unlike other phenols both α- and β-naphthols do not give a +ve test with ferric chloride. Other supporting evidence (preliminary and specific tests mentioned in the text etc.) help in the identification of the compound. • Similarly some compounds respond to functional group test not present in them, for example, some amines give a test with ferric chloride in spite of the absence of a phenolic group in them. In such cases, the given compound also responds to the functional group test present in it. Here also logical analysis of all the observation will be helpful in analysis. Hence it is advised that the results of all preliminary and functional group tests should be logically analysed to arrive at the identity of the given organic compound before preparing a derivative.

1.4.2 Carbohydrates Carbohydrates are a class of naturally occuring organic compounds. This group of compounds were so named because most of them have a general formula Cn(H2O)n and were erroneously considered to be hydrates of carbon. Carbohydrates are better known as sugars or saccharides and are now defined as polyhydroxy aldehydes or ketones or compounds which yield these on hydrolysis. Most of the carbohydrates are known by their common names, a few familiar names are Glucose, Sucrose, Starch, Cellulose, Lactose, Fructose etc. Chemically these compounds are called Saccharides and classified as: Monosaccharides: The simplest sugars which cannot be hydrolysed to smaller sugar molecules, Glucose, Fructose, Ribose and Galactose are a few examples of this group. Disaccharides: These are sugars which give two monosachharides on hydrolysis. Some examples are: Sucrose, Maltose, Lactose etc. Trisaccharides yield three molecules of monosacchharides, Tetra saccharides produce four mono sacchharide molecules on complete hydrolysis. Polysaccharides (e.g. Starch, Glycogen, Cellulose etc.) are polymers or macromolecules and give a large number of monosacchharides on complete hydrolysis. In the IUPAC system of nomenclature the name of saccharides ends with the suffix ose. The number of carbon atoms, the functional group present and the number of atoms present in the cyclic ring structure are all indicated in the name. An aldohexose (I), means that the sugar contains six carbon atoms and an aldehyde group, a ketohexose (II) is a ketonic compound containing six carbon atoms. The term pyranose (III) represents a six membered and furanose (IV), a five membered cyclic ring structure of the sugar.

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All carbohydrates are polyhydroxy compounds i.e. they contain —OH groups, hence most simple sugars are water soluble and give a smell of burning sugar on dry heating. They give a positive test (blood red colouration) with ceric ammonium nitrate, the red colour fading slowly. Table 1.1 gives the time taken by different sugars for decolourisation of red colour and it may be helpful in the identification of sugars. Table 1.1: Time taken for decolourisation of ceric ammonium nitrate solution by Carbohydrates S.No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Compound D-(+)-Xylose D-(–)-Lyxose D-Ribose L-(+)-Arabinose D-(–)-Fructose D-(+)-Glucose L-(–)-Sorbose D-Mannitol D-(+)-Galactose D-(+)-Mannose D-(+)-Maltose D-(+)-Mellibiose D-(+)-Cellobiose D-(+)-Lactose Sucrose

25°C 00:27 00:16 00:12 00:12 00:32 01:01 00:51 00:52 00:38 01:05 03:38 03:47 05:56 06:30 19:29

30°C 00:17 00:12 00:11 00:12 00:17 01:00 00:23 00:32 00:22 00:51 03:05 03:01 03:30 03:40 11:55

As mentioned earlier, carbohydrates contain a –CHO or a >CO group, these groups are not present free as in aldehydes and ketones, but are involved in forming the cyclic structure and hence the carbohydrates give a negative test with 2, 4-DNP. Carbohydrates can be either reducing (i.e. they can reduce Fehling or Tollens’ solution, some examples of this type are: glucose, fructose, maltose etc.) or non-reducing in nature (i.e. they cannot reduce Fehling or Tollens’ solutuion, common example is sucrose).

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The presence of a carbohydrate can be established by performing the following tests in the laboratory Molisch’s test: a general test for carbohydrates Barfoed’s test: a test given by monosaccharides Test with Fehling or Benedict solution: test for reducing sugars Seliwanoff’s test: test for ketoses Bial’s test: test for pentoses Molisch’s test This is a general laboratory test for carbohydrates. A violet or sometimes a red violet coloured ring is obtained when an aqueous solution of a carbohydrate is treated with a dilute alcoholic solution of α-naphthol followed by reaction with conc. H2SO4. A carbohydrate undergoes dehydration and cyclisation in presence of concentrated sulphuric acid to produce furfural or substituted furfural which condenses with α-naphthol to produced a violet coloured complex (VII). For example glucose gives 5-hydroxymethylfurfural on reaction with sulphuric acid, which gives a violet colour with α-naphthol.

Procedure In a test tube, add an alcoholic of α-naphthol (10%, 1 mL) to a dilute aqueous solution of carbohydrate (1–2 mL). Then pour cone. H2SO4 (1 mL) carefully along the sides in the test tube. A violet ring at the junction of two layers indicates the presence of a carbohydrate. Notes: • A red violet coloured ring is sometimes obtained which slowly changes to a violet coloured ring. • A dilute solution of α-naphthol should be used. • Though this is a general test for carbohydrates, compounds like furfural and substituted furfural also give this test. This test can be performed as a spot test Spot test Procedure Spot the following solutions one over the other in the order given below on a TLC plate or a strip of Merck plate.

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• Saturated aqueous solution of the compound • α-naphthol solution (provided in the kit) • Concentrated sulphuric acid A purple or a reddish-purple color indicates the presence of a carbohydrate. Note: This test can also be performed in a grooved tile. Add a drop each of sugar solution, α-naphthol in a grooved tile and mix well using a glass rod and then add a drop of concentrated sulphuric acid carefully. Appearance of a purple colour or red colour turning purple indicates the presence of a carbohydrate. Barfoed’s test This test helps to distinguish a monosaccharide form higher saccharides like di-, tri- and tetrasaccharides etc. Monosaccharides react at a faster rate with the Barfoed reagent (Aqueous solution of cupric acetate containing acetic acid) as compared to higher saccharides. A red precipitate of Cu2O is formed within one minute when a monosaccharide is heated with the Barfoed reagent

Procedure Heat an aqueous solution of the compound (1 mL) with the Barfoed reagent (1 mL, freshly prepared) in a test tube in a boiling water bath for bath for 1-2 minutes. Formation of a brick red precipitate of Cu2O indicates the presence of a monosaccharide. Notes: • Barfoed reagent should be preferably freshly prepared for use. • The test tube containing the test solution should be heated in a pre heated boiling water bath to note the time. Spot test Spot the following solutions on a thin strip of Whatman filter paper: • Saturated aqueous solution of the compound • Freshly prepared Barfoed’s reagent Heat the paper strip in the oven maintained at ~120°C for 1–2 min. Formation of a brick-red spot confirms the presence of a monosaccharide. Notes: Use a saturated solution of copper acetate to prepare Barfoed’s reagent. Test with Fehling or Benedict solution This reaction is used to distinguish carbohydrates which are reducing in nature from the non-reducing ones. A brick red precipitate of cuprous oxide is formed when a reducing carbohydrate is heated with any of these solutions.

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Fehling solution for the test is made by mixing equal volume of Fehling solution A (an aqueous solution of cupric sulphate containing dilute sulphuric acid) and Fehling solution B (a solution of sodium potassium tartrate in strong aqueous sodium hydroxide solution) just before test. If the two solutions are mixed and allowed to stand, a precipitate of cupric hydroxide is slowly formed. Though cupric hydroxide can oxidize some carbohydrates, the reaction rate is very slow because of very low solubility of cupric hydroxide in aqueous alkaline solution. Its reactivity is enhanced by the formation of soluble complexes of Cu2+ with either tartrate ions (Fehling solution) or citrate ions (Benedict solution).

Carbohydrates (reducing aldoses and ketoses) both can reduce Fehling and Benedict solution because of the presence of an aldehydic or a ketonic group in them.Whereas simple aldehydes can but ketones cannot reduce Fehling and Benedict solution. The reason is that in carbohydrates, the ketoses undergo base catalysed rearrangement (Lobry de Bruyn-Van Ekenstein rearrangement), in which a ketose is partially converted to an aldose.

Or the alkali may degrade the aldoses and ketoses to simple carbonyl compounds which have reducing properties.

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Procedure Heat an aqueous solution of the compound (1-2 mL) with an equal amount of Benedict or Fehling solution (mix equal amount of Fehling A and B) in a test tube directly on a low flame. Formation of a red precipitate indicates the presence of a reducing sugar. Spot test Spot the following one over the other on a thin strip of Whatmann filter paper • Saturated aqueous solution of compound • Fehling solution A three times (from kit) • Fehling solution B three times (from kit) • Heat in an oven maintained at ~ 120°C for 1-2 minutes • Appearance of a brick red coloured spot shows the presence of reducing sugar. Note: Perform a blank test simultaneously on one side of the strip by spotting Fehling solution A and B and heating as above. Seliwanoff’s test This is a test for ketoses and helps to distinguish them from aldoses and is based on the fact that ketoses are more readily dehydrated to furfural or its derivatives as compared to aldoses. Ketoses on condensation with resorcinol in presence of concentrated hydrochloric acid give a red-blue colouration within two minutes.

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Procedure Add Seliwanoff’s reagent (2 mL) to an aqueous solution of the compound (0.1 g in 2 mL water) in a test tube. Development of cherry red or blue colour indicates the presence of a ketose. Bial’s test Pentoses can be distinguished from hexoses by this test. Pentoses give blue green colour on reaction with Bial’s reagent and hexoses give a muddy brown to grey solution. Bial’s solution contains orcinol which condences with furfural or substituted furfural obtained from the carbohydrate to produce a characteristic colour.

Procedure Heat an aqueous dilute solution of the compound (4-5 drops) with Bial’s reagent (2 mL) for 2 minutes in a boiling water bath. Development of a blue green colour indicates the presence of a pentose. Hexoses give a brown or a grey colour.

1.4.3 Carboxylic Acids Compounds containing a —COOH group are known as carboxylic acids. These compounds are acidic in nature and their strength depends on the ease with which they can liberate a proton and also on the stability of carboxylate ion (conjugate base), which results after the loss of a proton. The acid strength is increased by the presence of electron withdrawing groups which act either due to their–I effect or stabilise the conjugate base by resonance. A few examples of acids given below indicate their relative strength, the lower the pKa value, stronger is the acid. Ethanoic acid (acetic acid) CH3COOH 4.74 Choro ethanoic acid

ClCH2COOH

3.10

Bromo ethanoic acid

BrCH2COOH

2.90

Butanoic acid

CH3CH2CH2COOH

4.82

r-chlorobutanoic acid

ClCH2CH2CH2COOH

4.53

β-Chlorobutanoic acid

CH3CH(Cl)CH2COOH

4.05

α-Chlorobutanoic acid

CH3CH2CH(Cl) COOH

2.89

Benzoic acid

C6H5COOH

4.20

p-Nitrobenzoic acid

3.45

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Carboxylic acids being acidic in nature, dissolve in basic solvents like sodium hydroxide and give carbon dioxide when treated with sodium bicarbonate or sodium carbonate solution. Carboxylic acids do not give a positive test with 2,4-DNP (a test given by compounds containing a group) because of the following resonance

The presence of a -COOH group can be established by the following tests Sodium bicarbonate test a test common to all carboxylic acids Fluorescein test a test given by some dicarboxylic acids Esterification test common to all carboxylic acids Sodium bicarbonate test Carboxylic acids react with an aqueous solution of sodium bicarbonate and give effervescence due to evolution of carbon dioxide. RCOOH + NaHCO3 → RCOONa + H2O + CO2↑ Since the reaction is normally performed in an aqueous solution, water soluble acids give a brisk effervescence and water insoluble acids may give a mild effervescence in this test. Weak acids also produce only a mild reaction. Procedure To a saturated solution of sodium bicarbonate in water (1 mL) in a test tube, add the given compound (pinch of solid or 1-2 drops of liquid). Effervescence indicates the presence of a carboxylic acid. Notes: • If a mild effervescence occurs, take the alcoholic solution of the compound and add aqueous solution of bicarbonate to it. • Use acid free alcohol for the test. • If alcohol is used as a solvent, perform a blank test also. Spot Test Procedure • Take a capillary and dip one end into a concentrated aqueous solution of the given compound. Some solution rises in the capillary. • Now, dip the capillary into a saturated solution of sodium bicarbonate (provided in the kit). • Effervescence in the capillary indicates the presence of an acid. Fluorescein test This test should be performed only if the above test with sodium bicarbonate is positive. It is given by only those dicarboxylic acids which can form a five or a six membered cyclic anhydride on heating (e.g. phthalic, succinic acid etc.). When a dicarboxylic acid or its anhydride is heated with

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resorcinol in presence of concentrated sulphuric acid, a fluorescent dye called fluorescein is obtained which gives green fluorescence in an alkaline medium.

Procedure Heat gently, for ~1 minute, a mixture of the given acid (50 mg), resorcinol (50 mg) and concentrated sulphuric acid (4-5 drops) in a dry test tube on a small bunsen burner flame. Pour the mixture into a beaker containing dilute sodium hydroxide solution (~50 mL water and ~ 20 mL dilute sodium hydroxide). A yellow orange solution with green fluorescence shows the presence of a dicarboxylic acid. Notes: • Over heating should be avoided as the reaction mixture chars and does not give a reliable result. • The solution should be alkaline at the end of test (check with a litmus paper). Spot test Procedure Spot each of the following solutions twice one over the other in the order given below on a TLC plate or a strip of Merck plate. • Concentrated solution of the compound or the given liquid compound. • Concentrated aqueous solution of resorcinol (provided in the kit). • Concentrated suphuric acid. • Heat the strip in the oven maintained at ~ 120°C for ~ 40–50 seconds till the spoted portion is dry. • Scratch the spotted portion of silica gel from the plate and transfer it into a test tube containing dilute sodium hydroxide solution (5-7 drops and 1-2 mL of water). • A fluorescent green colour in the test tube indicates the presence of certain dicarboxylic acids. Esterification test The test is based on the fact that carboxylic acids react with alcohols in presence of concentrated sulphuric acid to produce esters, which can be recognized from their sweet smell.

∆

 → RCOOR′ + H O RCOOH + R′OH  Conc. H 2SO 4 2

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Procedure Heat a mixture of the given acid (50 mg), ethanol (4-5 drops) and concentrated sulphuric acid (2-3 drops) in a dry test tube in a boiling water bath for 2-3 minutes. Pour the reaction mixture carefully into a beaker containing water (~25 mL). Neutralize the sulphuric acid with dilute sodium hydroxide solution. Formation of a sweet smelling substance indicates the presence of an acid. Spectroscopic identification of a –COOH group IR spectrum The IR spectrum of a compound containing a –COOH group shows two strong and broad adsorption bands between 1725-1690 and 3300-2500 cm–1 due to and O—H stretching vibrations respectively, The bands are broad because the acids exist as dimers as a result of hydrogen bonding between and –OH. A number of other absorption bands are seen in the IR spectrum of all organic compounds, these are discussed in detail in chapter 2. NMR spectrum In the PMR spectrum, the hydrogen of –COOH group appears between 10-13δ. Its position is variable and changes with concentration of the compound used. It can be easily recognized by deuterium exchange i.e. the spectrum is first recorded by dissolving the compound in say CDC13, D2O is then added to the solution and spectrum rerun. The absorption signal which was due to H of –OH in the first case slowly diminishes and finally disappears because of acidic proton of –COOH group is exchanged with deuterium of D2O. The hydrogens α- to the –COOH group absorb between 2-3δ.

1.4.4 Phenols Phenols are a class of organic compounds containing –OH group/s directly attached to an aromatic ring for example C6H5OH is a phenol but C6H5CH2OH is an alcohol. Besides being important laboratory chemicals, a large number of them find commercial applications: are widely used as antioxidants, anticeptics, disinfectants, in the manufacture of dyes, drugs and polymers etc. Phenols are acidic in nature because on loss of a proton they produce a phenoxide ion, which is/are more resonance stabilised than the phenol.

Phenols are generally weaker acids than carboxylic acids and that is why most carboxylic acids dissolve in sodium bicarbonate solution (a weaker base) whereas phenols are generally soluble in sodium hydroxide (a stronger base). The acid strength of phenols is enhanced by the presence of electron withdrawing groups, the effect depending on the position of the substituent. Electron donating substituents decrease the strength. Table 1.2 gives the comparative acid strength of a few phenols.

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Table 1.2: pKa values of some Phenols Phenol Nitrophenol Cresol Amino phenol Chloro phenol

pka o7.23 10.28 9.71 8.48

pka m8.40 10.08 9.87 9.02

pka p7.15 10.14 10.30 9.38

Phenol has a pka: 9.98 2.4.6-tri nitrophenol (Picric acid), pka: 0.71 Phenols decolourise bromine and potassium permanganate solutions, because of substitution of aromatic ring by bromine and oxidation by potassium permanganate respectively. Phenols form coloured complexes with ferric chloride and give different coloured dyes on condensation with phthalic anhydride/ acid. The reactions that are used for the detection of phenolic group in the laboratory are: • Solubility in sodium hydroxide solution • Test with neutral ferric chloride • Libermann nitroso test • Phthalein test • Test with bromine solution.

Solubility in Sodium hydroxide Most phenols are weakly acidic in nature and are therefore soluble in strong bases like sodium hydroxide solution but insoluble in sodium bicarbonate solution. This test is helpful only if the given compound is insoluble in water. Test with neutral Ferric chloride Most phenols give a dark coloured solution (blue, green, purple etc) on reaction with a neutral solution of ferric chloride, e.g. simple phenol reacts with FeCl3 as follows. 6C6H5OH + FeCl3 → H3 [Fe(OC6H5)6] + 3HCI Purple

Some phenols do not give this test and hence a negative test must not be taken as an indication of the absence of phenolic group. Results of preliminary examination like solubility in dilute sodium hydroxide and test with bromine and potassium permanganate solution must be considered. Other confirmatory tests like Liebermann nitroso reaction and phthalein test must also be performed. Some phenols react in a different way with ferric chloride, e.g. Hydroquinone is first oxidized to p-benzoquinone which then forms a dark green coloured solid of quinhydron with hydroquinone.

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Both α-and β-Naphthol undergo oxidative coupling on reaction with ferric chloride instead of forming a coloured complex, though there is some observable change on reaction with ferric chloride in these cases which should not be ignored.

Procedure Dissolve the given compound (~50 mg ) in water (4-5 ml) in a test tube, add a solution of neutral ferric chloride slowly (several drops, dropwise) and observe the change in colour. A red, blue, green or purple colouration indicates the presence of a phenol. Enols produce a red-violet or tan colour. Notes: • Use dilute alcohol if the compound is insoluble in water. • Phenols can be distinguished from enols by the following test. Take the given compound (~50 mg) in a test tube and treat it with a solution of mercurous nitrate in nitric acid. Formation of an immediate grey precipitate shows the presence of an enol. • Some amino compounds also give this test. Spot test Spot the following solutions one over the other on a strip of Whatmann filter paper • A concentrated solution of the compound • Concentrated neutral ferric chloride solution (from the kit) • A dark coloured spot (purple, blue, green or red) indicates the presence of a phenol. Liebermann Nitroso test Phenols like simple phenol, resorcinol, o-cresol and o-quinol etc. which have an unsubstituted paraposition with respect to the –OH group undergo this reaction. In this test the given phenol is treated with sodium nitrite in presence of sulphuric acid, p-nitroso phenol first obtained reacts with another molecule of phenol to produce a blue coloured compound.

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Procedure Take the compound (100 mg) in a dry test tube, add sodium nitrite (few crystals) and concentrated sulphuric acid (1 mL), mix well and heat gently. A blue colour is obtained. The colour changes to red on dilution with water and blue on basification with sodium hydroxide solution. Spot test Spot the following solutions one over the other on a TLC plate or a thin strip cut out from a Merck plate • A concentrated solution of the compound. • A concentrated solution of sodium nitrite. • Concentrated suphuric acid. • Heat the strip or plate in a preheated oven maintained at ~120°C for about a minute (the spotted portion should become dry). • Take out the strip from the oven. • Spot the phenol once more. • Heat again for a minute. • Scratch the spotted portion and transfer the scratched silica gel into a tube containing dilute sodium hydroxide solution (~2-3mL). • Development of blue colour shows the presence of a phenol. Phthalein test This test is based on the fact that phenols undergo condensation with phthalic acid in presence of concentrated sulphuric acid to produce compounds called phthalein/xanthene dyes which give characteristic colour in alkaline solution. For example phenol gives phenolphthalein (a phthalein dye) and resorcinol produces fluorescein (a xanthene dye) in this test.

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Procedure In a dry test tube, gently heat the given compound (50 mg) with an equal amount of phthalic acid/ anhydride and concentrated sulphuric acid (2-3 drop), for ~ a minute. Cool and pour the reaction mixture into a beaker containing dilute sodium hydroxide solution (~10 mL dil. sodium hydroxide + ~ 50 mL water ). Appearance of a pink, green, red etc. colour indicates the presence of a phenol. Test with Bromine solution Phenols decolourise bromine solution because they can undergo electrophillic substitution reaction with bromine, decolourisation is accompanied with evolution of HBr gas.

The rate of bromination is greater in water or acetic acid than in carbon tetrachloride solution. Hydrogen bromide is liberated due to substitution reaction but this is not observed when water is used as a solvent.

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Procedure Dissolve the given compound (100 mg) in glacial acetic acid or water in a test tube. Add to it, a saturated solution of bromine in water or acetic acid (dropwise). Disappearance of the orange colour of bromine indicates the presence of a phenol. Notes: • Compound may be dissolved in carbon tetrachloride and treated with the solution of bromine dissolved in the same solvent. • Aliphatic unsaturated compounds, amines and other activated aromatic compound also decolourise bromine solution. Therefore this test is not specific to phenols. Spectroscopic identification of –OH group IR Spectrum Three characteristic stretching bands are observed in the IR spectrum of a phenol • O–H • C–O • Ar–H • The position, intensity and shape of O–H stretching band depends mainly on concentration of the phenol used i.e. whether the O–H is free or hydrogen bonded to other –O–H groups Band O–H Free O–H Bonded

Position

Shape

3650–3600 cm–1

Sharp peak

cm–1

Broad peak

3400–3300

• Absorption due to C–O stretching appears between 1260–1000 cm–1 • Ar–H above 3000 cm–1 i.e. between 3100-3000 cm–1. NMR Spectrum The position of –OH absorption is concentration dependent, it may be observed between 4-12δ and can be easily recognized by exchange with D2O (Section 1.4.3). Aromatic protons absorb near 7δ. The position and splitting pattern of the aromatic protons depends on the nature and number of substituents on the aromatic ring.

1.4.5 Quinones α,β-Unsaturated diketones are known as quinones. These compounds can either be o or p-diketones. Structures of a few quinones are given below:

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Test with KI Quinones are reduced easily to the corresponding hydroxy compounds and this forms the basis of their laboratory test. Quinones on reaction with potassium iodide in presence of dil. H2SO4, liberate I2, which can be tested with starch paper. This test is not specific to quinones. Spectral data should be used to confirm the presence of a quinone.

Procedure Add KI containing dilute sulphuric acid (1 mL of KI solution containing a few drops of dil. H2SO4) to the organic compound (0.1g) in water (2mL). A pale yellow colour is developed due to the formation of iodine. Add a drop of this solution onto a piece of starch paper. Appearance of a blue colour indicates the presence of a quinone. Spectroscopic identification of quinones IR spectrum stretching appear between 1675-1660 cm–1 due to conjugation of C=O with C=C. For the The same reason the C=C is shifted to ~1600 cm–1. NMR spectrum The protons at α-and β-positions absorb between 6.5-7δ.

1.4.6 Aldehydes and Ketones Compounds containing a group are known as carbonyl compounds. Aldehydes (RCHO) and Ketones (RCOR) belong to this class. Many other compounds like carboxylic acids, esters, anhydrides and amides etc. also contain a but do not show reactions typical to aldehydes and ketones and are therefore not called as carbonyl compounds. Aldehydes and ketones show some common reactions. These compounds react with ammonia and substituted ammonia (NH3 and ZNH2) which act as nucleophiles and react with the electrophillic carbon of group to give condensation products, a molecule of water is then eliminated to form imines.

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Thus the reactivity of the carbonyl compounds depends on the effective positive charge on the carbon. Electron donating groups attached to the carbon reduce the reactivity of carbonyl group that is why the reactivity order of some of these compounds is: HCHO > CH3CHO>CH3COCH3>C6H5CHO>C6H5COC6H5 etc. The reaction of aldehydes and ketones with 2,4-Dinitrophenylhydrazine is used for their functional group identification in the laboratory. Aldehydes are much more easily oxidised to the carboxylic acids than ketones and this forms the basis of distinguishing them in the laboratory. Following tests are used to identify and distinguish aldehydes and ketones. 2, 4-Dinitrophenylhydrazine test Tollens’ test Schiffs’ test Fehling test Test with chromic acid lodoform test 2,4 dinitrophenylhydrazine test. Aldehydes and ketones give yellow/orange precipitate when treated with 2,4-dinitrophenylhydrazine. Yellow precipitate is generally obtained from aliphatic carbonyl compounds and orange coloured precipitate from aromatic compounds. The reason is that in aromatic compounds there is an extension of conjugation in 2,4 dinitrophenylhydrazones whereas that is not generally the case in aliphatic compounds.

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Procedure Take the alcoholic solution of 2,4- dinitrophenylhydrazine (2mL) in a test tube. Dissolve the carbonyl compound (100 mg or 1-2 drops) in ethanol (2-3 mL) in a separate test tube. Now add the solution of the carbonyl compound slowly to the solution of 2,4-DNP, shake the mixture well. A yellow/orange coloured precipitate indicates that the given compound is an aldehyde or a ketone. Notes: • The test can also be carried out by dissolving the compound in water and adding it to an aqueous solution of 2, 4-Dinitrophenylhydrazine. • Brady’s reagent (an aqueous alcoholic solution of 2,4-dinitrophenylhydrazine) can be used for both water soluble and water insoluble carbonyl compounds. The Brady’s reagent is prepared as follows: Dissolve 2, 4-Dinitrophenyl hydrazine (0.1 g) in concentrated sulphuric acid (0.5 mL). Add to this, slowly with shaking, mixture of water (1 mL) and ethanol (2.5 mL). Filter to remove any suspended impurities. • The test should be performed by adding the solution of the carbonyl compound to the 2,4-DNP solution and not vice-versa. • It is advisable to carry out a blank test. A blank test is performed by adding all reagents except the given compound. • Some amines like aniline etc. may give a precipitate of the salt of the amine with acid used in the preparation of the 2,4- DNP reagent. However, this precipitated salt is soluble in water, whereas the 2.4-DNP of aldehydes and ketones are insoluble. • At times, the 2, 4-dinitrophenylhydrazine itself may get precipitated out as an orange solid, which dissolves in dilute sulphuric acid whereas the 2, 4-dinitrophenylhydrazones are insoluble. Spot test This test can be performed as a spot test also. Best results are obtained for aromatic aldehydes and ketones. Procedure Spot the following solutions one over the other on a Whatmann filter paper strip • Liquid compound or its concentrated alcoholic solution • Alcoholic solution of 2,4-DNP • Appearance of an orange spot shows the presence of a carbonyl compound. Notes: Perform a blank test as follows: Put a spot of only the DNP reagent simultaneously on the paper strip for comparision of the colour with the above spot. Aldehydes can be distinguished from ketones on the basis of the following tests: Tollens’ test Tollens’ reagent consists of silver ammonia complex. It can be easily reduced by aldehydes (which are oxidized to the corresponding carboxylic acids) to metallic silver which appears as a grey precipitate or a silver mirror. Ketones do not undergo oxidation easily and hence do not reduce Tollens’ reagent. RCHO + 2 [Ag(NH3)2]OH → RCOONH4 + 3NH3 + H2O + 2 Ag↓ Tollens’ reagent

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Procedure To a solution of the given carbonyl compound (1-2 drops or 50 mg) dissolved in aldehyde free alcohol (2-3 mL) in a test tube, add the freshly prepared Tollens’ reagent (1 mL) and warm the solution in a hot water bath for 1-2 minutes. Formation of a grey black precipitate or silver mirror indicates the presence of an aldehyde. Notes: • Tollens’ reagent should be freshly prepared for use. • Reducing carbohydrates, some formates, tartrates, lactates, quinones, β-diketones and phenols also give a positive test. • The test tube containing the test solution or the Tollens’ reagent should not be kept but immediately washed with dilute nitric acid 2-3 times after performing the test. (Silver fulminate, an explosive substance, that might be produced is thus decomposed). Schiff’s test Schiff’s base is a triphenylmethane dye and it dissolves to produce a magenta coloured solution. The Schiff’s reagent is made by saturating the magenta coloured solution with sulphur dioxide when a yellow coloured solution is obtained. Aldehydes restore the magenta colour of the Schiff’s base. Ketones do not react.

Procedure To a solution of the given carbonyl compound (1-2 drops or 50 mg in 2-3 mL of aldehyde free alcohol) in a test tube, add the Schiff’s reagent (2-3 drops). Instant appearance of a magenta colour indicates the presence of an aldehyde. Notes: • Do not heat the solution while performing the test. • Some ketones regenerate a light pink colour of the reagent, which is not considered a positive test.

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Fehling’s test Fehling solution contains a soluble complex of Cu2+ with tartarate (see page 33) in strong alkaline solution. When treated with an aldehyde, Cu2+ is reduces to Cu1+ (as a red precipitate of Cu2O) and the aldehyde is oxidised to the acid. Aromatic aldehydes do not reduce Fehling solution. Procedure Heat the mixture of the given carbonyl compound (50 mg dissolved in 2-3 mL water/aldehyde free alcohol) and Fehling solution (~1 mL, obtained by mixing equal amount of Fehling solution A+ Fehling solution B) in a test tube. Formation of a brick red precipitate indicates the presence of an aliphatic aldehyde. Spot test Procedure Spot the following one over the other on a thin strip of Whatmann filter paper • Liquid compound or its concentrated alcoholic solution. • Concentrated Fehling solution A (from kit). • Concentrated Fehling solution B (from kit). • Heat in an oven for l-2 minutes maintained at ~ 120°C. • Development of a brick red spot confirms the presence of an aliphatic aldehyde. Test with Chromic acid Aldehydes are oxidized by chromic acid to the corresponding carboxylic acids and Cr6+ in chromic acid is reduced to Cr3+ in the redox reaction. Cr3+ compounds are green in colour therefore, a green or a blue precipitate is formed when an aldehyde is treated with chromic acid. Ketones do not react. Procedure Add the carbonyl compound (1 drop or 20 mg of solid) to the chromic acid (1 mL). Appearance of a green-blue precipitate indicates the presence of an aldehyde. Notes: • This test is very sensitive, small amount of impurities (which can reduce chromic acid) produce a colour change. • Primary and secondary alcohols also give a positive test with chromic acid but can be distinguished from aldehydes and ketones on the basis of 2, 4- DNP test. Enols may give a positive test. Phenols give a dark coloured solution entirely different from green-blue colour of a positive test.

Iodoform test When a carbonyl compound containing a CH3CO-group is treated with a solution of iodine in basic medium, iodoform (CHI3), a yellow solid is formed. The hydrogens which are α- to the electron withdrawing group are substituted by iodine in the basic medium producing CI3CO–, which then gives iodoform as shown below. RCOCH3 + 3I2 + 3NaOH → RCOCI3 + 3 NaI + 3 H2O Methyl ketone



RCOCI3 + NaOH → RCOONa + CHI3↓

Iodoform yellow solid

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Procedure Dissolve the given carbonyl compound (2-3 drops or 100 mg) in water in a test tube and add sodium hydroxide (1 mL, 10%). To this add slowly with shaking, a freshly prepared saturated solution of iodine in potassium iodide until the dark colour of iodine persists. Heat the solution in a boiling water bath for 1-2 minutes. Remove the colour of excess iodine by adding a few drops of sodium hydroxide solution. A yellow precipitate of iodoform indicates the presence of the CH3CO– group in the compound. Notes: • Acetaldehyde, all methyl ketones and alcohols containing a CH3CHOH– group give a positive test. • Use dioxane or methanol to dissolve the compound if it is water insoluble. • If the iodoform is dark reddish yellow in colour, dissolve it in 3-4 mL of dioxane and treat it with 1 mL of 10% sodium hydroxide solution, shake until a light yellow solid remains. Dilute with water and filter the solid. Spectroscopic identification of aldehydes and ketones IR spectrum Aldehydes: C=O stretch The absorption due to stretching of carbonyl compounds is one of the strongest absorption in the infrared spectrum, appearing between 1750-1450 cm–1. The position of absorption varies with the nature of substitutes attached to the carbonyl carbon. For example the position of absorption due to stretch varies as shown below: Type of aldehyde

Position of >C=O in cm–1

RCHO

1740–1725

C=C–CHO

1700–1680

ArCHO

1600–1450

C–H stretch of –CHO Two weak bands, one between 2860-2800 and the other between 2760–2700 cm–1 are due to C–H stretch of –CHO group. These bands are absent in the spectrum of a ketone. Ketones A strong absorption due to C=O stretch of ketones is generally observed between 1720–1600 cm–1. Table (1.3) shows stretching absorption values of some ketones Table 1.3 Type of ketone RCOR C=C–COR ArCOR ArCOAr

>C=O stretch in cm–1 1720-1708 1700-1675 1700-1660

1670-1600

NMR spectrum Hydrogens which are alpha to a carbonyl group show absorption between 2-2.5δ. The H of –CHO absorbs between 9-10δ. Coupling of the absorption signal at 9-10 δ is observed if the aldehyde contains α-hydrogen atoms. For example, a singlet is observed for benzaldehyde and a quartet for acetaldehyde between 9-10δ.

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1.4.7 Alcohols Alcohols are a class of compounds represented by the general formula ROH, they may contain one –OH group (monohydric alcohols e.g. CH3OH), two –OH groups (dihydric alcohols HOCH2CH2OH), three –OH groups (trihydric alcohols HOCH2CHOHCH2OH) etc. Alcohols are further classified as primary, secondary and tertiary alcohols represented by general formula RCH2OH, R2CHOH and R3COH respectively. Alcohols are very weakly acidic in nature, the acid strength mainly depending on the stability of alkoxide ion (RO–), resulting after the loss of a proton. The order of acid strength is RCH2OH > R2CHOH > R3COH because the stability order of alkoxide ions is RCH2O– > R2CHO– > R3CO– Some important reactions of alcohols that form the basis of their laboratory identification are: Ceric ammonium nitrate test Xanthate test Test with sodium metal Test with Lucas reagent Oxidation with chromic acid Iodoform test Ceric ammonium nitrate test Alcohols replace the nitrate ion of ceric ammonium nitrate, changing the colour of the solution from yellow to red. (NH4)2 [Ce(NO3)6] + ROH → (NH4)2 [Ce (OR) (NO3)5] + HNO3 Yellow

red

The red colour slowly fades to give a yellowish solution. This is because the Ce4+ ( which is present in the red coloured complex) is reduced to Ce3+. ( NH 4 ) 2 [ Ce(OCH 2 R ) ( NO3 )5 ] → ( NH 4 ) 2 [ Ce( NO3 )5 ] + HNO3 + RCH 2 O − ( Red )

(Yellow)

Or

( NH 4 ) 2 [ Ce( NO3 )6 ] + RCH 2 O → ( NH 4 )2 [ Ce( NO3 )5 ] + RCHO + HNO3 −

( Red )

( Yellow)

The time taken for complete change of colour depends on the nature of alcohol. Procedure Dissolve the given organic compound (50 mg) in a test tube in water (1-2 mL) or dioxane and add ceric ammonium nitrate solution (few drops). Observe the colour change immediately. Appearance of a red colour shows the presence of an alcoholic group. Notes: • A blank test must be performed if dioxane has been used as a solvent. • Some alcohols give only a transient red colouration, so look for red colour immediately on mixing. • Some amines and phenols also give various colours in this test.

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Spot test Procedure Spot the following solutions one over the other on a Whatmann filter paper strip. • Liquid compound or concentrated aqueous/acetone solution of the compound. • Ceric ammonium nitrate solution (2-3 times). • Development of a blood red spot indicates the presence of alcohol. Notes: The blood red colour fades, observe the colour immediately after spotting the reagents. Xanthate test Alcohols produce a yellow precipitate of xanthate on treatment with CS2 in presence of an alkali.

Procedure Warm the compound (0.l g) in a test tube with solid potassium hydroxide (one pellet) until it dissolves. Cool, wash with a little ether, decant the ether and add carbon disulphide (2-3 drops) to the residue in the test tube. Shake well. Formation of a yellow precipitate indicates the presence of an alcohol. Notes: • Ether is highly inflamable, handle it very carefully.

Test with Sodium metal Reactive metals like sodium react with alcohols and liberate hydrogen gas with effervescence. 2 ROH + 2 Na → 2 RONa + H2↑ Procedure In a dry test tube take the dry compound (1 mL of liquid or a few crystals of solid dissolved in dry ether). To this add a very small piece of clean, dry sodium (~ 20 mg). Dissolution of sodium with brisk effervescence indicates the presence of an alcoholic group. Notes: • The reaction must be performed under anhydrous conditions i.e. dry test tube, solvent and compound should be used. • Sodium should be handled carefully with forceps. • Excess of unreacted sodium should not be discarded in the sink or dust bin. It should be decomposed carefully by adding a small amount of alcohol. • A blank test should be performed if the given compound has been dissolved in a solvent. • Many other compounds like acetone, acids etc. also give this test.

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Test with Lucas reagent This test is used for distinction between primary, secondary and tertiary alcohols. Lucas reagent consists of anhydrous zinc chloride dissolved in concentrated hydrochloric acid. The order of reactivity of alcohols with Lucas reagent is: tert. Alcohols > sec. alcohols, primary alcohols do not react Tertiary and secondary alcohols are converted to the corresponding alkyl chlorides by replacement of –OH group by –Cl, the rate of reaction being faster for a tertiary alcohol. Alcohols are soluble in Lucas reagent whereas the alkyl halides are insoluble and separate out as white precipitate. ZnCl



2 → R3CCl + H2O R3COH + HCl  ZnCl2 → R2 CHCl + H2O R2CHOH + HCl  ZnCl2 → No reaction RCH2OH + HCl 

Procedure Add Lucas reagent (5 mL) to the given organic compound (~0.5 mL) in a dry test tube. Shake well and allow to stand. Note the time required for the separation of two distinct layers. Tertiary alcohol – Two layers are formed immediately Secondary alcohol – Layer separation takes 5-10 minutes Primary alcohol – A clear homogeneous solution is obtained. Notes: The role of zinc chloride is to enhance the reactivity of HCl δ+

δ−

H⋅ ⋅ ⋅ Zn Cl3 • Lucas test is applicable to only those alcohols which are soluble in the reagent, as it is based on the appearance of alkyl halides as a second liquid phase. • Primary alcohols which can form a stable carbocation (example benzyl alcohol) after the loss of –OH group also give a positive test. Oxidation (Jones Oxidation) A primary alcohol gives an aldehyde which is oxidised to an acid, a secondary alcohol gives a ketone on oxidation with chromic acids. Tertiary alcohols do not oxidize under these conditions. 3RCH2 OH + 4CrO3 + 6H2 SO4 → 3RCOOH + 2Cr2 (SO4)3 + 9H2O 6R2CHOH + 4CrO3 + 3H2SO4 → 6R2CO + 2 Cr2(SO4)3 + 6H2O 3RCHO + 2CrO3 + 3H2SO4 → 3RCOOH + Cr2(SO4)3 + 3H2O Procedure Mix the given compound (1 mL), water (5 mL), chromic acid (1 mL) and concentrated sulphuric acid (1 mL) in a small test tube and immediately note the visual changes. An opaque suspension with a green/ blue colour indicates the presence of a primary or a secondary alcohol. Spectroscopic identification of alcohols IR spectrum There are two stretching modes of vibration characteristic to alcohols i.e. O–H and C–O. The position and shape of band due to O–H stretching is variable, depending mainly on the concentration of the compound and the polarity of solvent used for recording the spectrum.

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The hydrogen bonded O–H stretching vibration appears between 3400-3300 cm–1 as a broad and a strong band if • Neat compound or its concentrated solution is used for recording the spectrum • A polar solvent is used for dissolving the compound • There is an interamolecular hydrogen bonding in the compound. The free or non hydrogen bonded O–H absorption band is sharp and appears between 3650-3550 cm–1 if • A dilute solution of the compound in a non polar solvent is used for recording the spectrum • There is no intramolecular hydrogen bonding in the compound. Generally both the bands are seen in the IR spectrum of an alcohol. The C–O stretching of alcohols appears between 1200-1000 cm–1. The position of O–H and C–O stretching can be used for distingtion between primary, secondary and tertiary alcohols.

NMR spectrum The H attached to the oxygen of O–H group is acidic in nature and does not stay permanently attached to oxygen hence it does not show the coupling (splitting pattern) expected in the NMR spectrum. For the same reason the hydrogens on the carbon bearing the –OH group do not couple with the –OH. The coupling can be observed if a solvent that is used for spectrum recording forms a strong hydrogen bond with the H of –OH group, thereby preventing the hydrogen exchange. DMSO is ideal for this purpose. Hence primary, secondary and tertiary alcohols can be distinguished from each other by recording the spectrum of the alcohol in DMSO. The –OH absorbs between 0.5–5.0 δ (the position depending on the concentration of solution) and –CH–OH appears between 3.2–4.0 δ. 1.4.8 Esters Esters are o-alkyl/ary1 derivatives of carboxylic acids and are represented by the general formula RCOOR′, where R and R′ can have a variety of structures. A few examples of esters given below for example, are derivatives of benzoic acid. C6H5COOCH3 C6H5COOC6H5 C6H5COOCH2C6H5

Methyl benzoate

Phenyl benzoate

Benzyl benzoate

Esters on hydrolysis produce an acid and an alcohol/a phenol. The hydrolysis is best achieved under alkaline conditions and is known as Saponification C6H5COOCH3 + NaOH → C6H5COONa + CH3OH C6H5COOC6H5 + NaOH → C6H5COONa + C6H5OH The acid catalysed hydrolysis of esters is a reversible reaction H+

  → C H COOH + CH OH C6H5COOCH3 + H2O ←  6 5 3 Trans esterification is a term used for conversion of an ester to another ester by its reaction with an alcohol in presence of a mineral acid like conc. sulphuric acid.

+

H → RCOOR′′ + R′OH RCOOR′ + R′′OH 

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Esters are extensively used in perfumery industry. Tri esters of glycerol of higher fatty acids (known as triglycerides) are important naturally occuring compounds. These compounds are represented by the general formula given below are present in oils and fats.

A triglyceride where R, R′ and R″ may be same or different. The presence of an ester is established by the following laboratory tests: Hydroxamic acid test (Feigl test) Alkaline hydrolysis Hydroxamic acid test (Feigl test) An ester reacts with hydroxyl amine in presence of sodium hydroxide to give hydroxamic acid, which is then characterised by its reaction with ferric chloride. A magenta coloured complex is formed when hydroxamic acid is treated with ferric chloride in an acidic medium.

Procedure Perform the following preliminary test before carrying out the hydroxamic acid test with the given compound. Dissolve the given compound (50 mg) in alcohol (1 mL) in a test tube and add hydrochloric acid (1 M, 1 mL). Shake and then add ferric chlorides solution (few drops). If a red, blue, green or a violet colour is produced, the hydroxamic acid test cannot be applied on the given compound. Otherwise proceed as follows: Mix the compound (a drop or 40-50 mg) with hydroxylamine hydrochloride (1 mL, 1 M solution), sodium hydroxide (0.2 mL, 6 M) and alcohol (l-2 mL) in a test tube and boil the mixture for 1-2 minutes. Cool and add hydrochloric acid (2 mL, 1 M). If the solution is cloudy add ethanol (1-2 mL) to make it clear and then add ferric chloride solution (l-2 drops). Add a little more of ferric chloride if necessary. Formation of a magenta colour confirms the presence of an ester. Notes: Derivatives of acids like anhydrides, chlorides and amides give a positive test. Most nitriles, primary and secondary nitro compounds also give a positive test.

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55

Hydroxamic acid test in a groove tile Procedure Use liquid compound for this test. Add 1-2 drops of each of the following solutions in a grooved tile in the order given below: • Liquid compound. • Saturated solution of sodium hydroxide (from the kit). • Concentrated solution of hydroxylamine hydrochloride (from the kit). Keep the above solution in a grooved tile for 2-3 minutes and then add a drop each of the following solutions: • Concentrated hydrochloric acid (to acidify the solution) • Ferric chloride solution. Development of a magenta or a burgundy colour confirms the presence of an ester. Alkaline Hydrolysis: An ester on hydrolysis with alkali gives an acid (as its sodium salt) and an alcohol or a phenol RCOOR + NaOH → RCOONa + R′OH RCOONa + HCl → RCOOH + HCl Procedure Add the given compound (0.1 mL) to a dilute solution of sodium hydroxide (1%, 2 mL) containing a drop of phenolphthalein in a test tube, the resultant solution should be pink in colour. If the pink colour of the solution is discharged in cold, add more of dilute sodium hydroxide solution (a few drops) to obtain a permanent pink solution. Now boil the solution for a 2-3 minutes, discharge of pink colour indicates the presence of an ester group. Notes: • The carboxylic acid obtained by hydrolysis of ester, neutralises the sodium hydroxide thereby discharging the pink colour of the solution. • Acid anhydrides, lactones and other compounds which give an acid on hydrolysis also give a positive test. Spectroscopy identification of a –COOR group IR spectrum In addition to the other absorption signals, an ester shows a strong absorption peak due to C=O stretching vibrations at 1750–1715 cm–1 and one or two strong bands at 1170–1270 cm–1 due to C–O stretch. The C=O stretching absorption varies with the structure of the ester as shown in the Table (1.4) below: Table 1.4 Structure of ester RCOOR C=C–COOR ArCOR

C=O stretching (cm–1) 1750–1735 1740–1720 1740–1715

NMR spectrum In addition to other signals, the following signals are observed in the NMR spectrum of an ester. • The hydrogens on the carbon attached to an alkyl oxygen (–COOCH ) absorb between 3.7-4.2δ • The hydrogens α- to the carbonyl group absorb between 2-2.6 δ (–CH–CO–OR)

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1.4.9 Lactones Cyclic esters are known as lactones. They produce sodium salt of hydroxy acids on alkaline hydrolysis which undergo cyclisation on acidification regenerating the original compound.

Lactones give a positive hydroxamic acid test (Section 1.4.8) like esters but can be distinguished from them on the basis of products obtained on alkaline hydrolysis. Spectroscopic identification of lactones IR spectrum The absorption due to C=O stretching appears at variable position, depending mainly on the ring size of the cyclic ester. For 5-6 membered ring size, a strong absorption signal is seen at 1750-1735cm–1. For highly strained lactone ring, the absorption may appear at ~1800 cm–1. NMR spectrum The protons attached to the carbon α- to carbonyl carbon appear between 2-3.5δ and protons attached to the carbon attached to the oxygen absorb between 4-4.5 δ.

1.4.10 Acid anhydrides Acid anhydrides as the name suggests are formed by the loss of a water molecule from two –COOH groups of a (i) dicaboxylic acid (cyclic anhydrides, e.g. Succinic anhydride) or (ii) two molecules of the same acid (simple anhydrides e.g. Acetic anhydride) or (iii) one molecule each of two different acids (mixed anhydride e.g. Acetic propionic anhydride)

Anhydrides are normally contaminated with the carboxylic acids which might be formed by slow hydrolysis on exposure to moisture present in air. The pure anhydrides do not produce effervescence but the contaminated sample gives effervescence on reaction with sodium bicarbonate solution. The presence of anhydrides can be detected by the following laboratory tests. Hydrolysis to acids (page 55) Feigl test (page 54) Fluorescein test a test given by cyclic anhydrides (page 36)

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Spectroscopic identification of anhydrides IR spectrum The IR spectrum of anhydrides exhibits two absorption bands in the region 1870-1800 and 1790-1740cm–1 due to asymmetric and symmetric stretching of the two C=O groups. NMR spectrum In the NMR spectrum, apart from other signals, the hydrogens attached to the α-carbon atom absorb between 2-3δ.

1.4.11 Ethers Dialkyl, diaryl or alkyl aryl derivatives of oxygen are known as ethers, represented by the general formula ROR, ArOAr and ROAr respectively. Ethers can form peroxides on standing which are highly explosive in nature. A sample of ether containing solid particles may be containing peroxides. It should not be handled in the lab. Ethers are • Neutral in nature • Generally insoluble in water • Simple ethers show absence of extra elements (N, S, halogens) • Non reactive and do not respond to any of the functional group test given above • Generally soluble in cold concentrated sulphuric acid • Give test with iodine • Identified by analysis of the spectral data Test with cold concentrated sulphuric acid. Ethers normally dissolve in concentrated sulphuric acid at room temperature because of the protonation of the electronegative oxygen and are regenerated on dilution. +

−   R 2 O + H 2SO 4   R 2 O H HSO 4

Some aromatic ethers undergo sulphonation under these conditions and do not separate out on dilution because the formed sulphonic acids are water soluble.

Procedure Carefully shake the given compound (0.l g) with cone. sulphuric acid (~2 mL) in a dry test tube. Dissolution of the compound indicates the presence of an ether. Notes: • This test is not specific to ethers as all basic compounds and all those compounds which can undergo sulphonation under these conditions will dissolve in sulphuric acid. • All the results of the preliminary and functional group tests must be considered before arriving at any conclusion.

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Test with Iodine Ethers form charge transfer complex with iodine to give a tan colour R2O + I2 → R2O: → I2 Procedure Add a dilute solution of iodine in carbon tetrachloride (~ lmL) to the solution of the given compound (2-3 drops in carbon tetrachloride) in a test tube and shake. Formation of a tan colour shows the presence of an ether. Notes: • This test is not specific to ethers. • All the results of the preliminary and functional group tests must be considered before arriving at any conclusion. Spectroscopic identification of ethers IR spectrum The IR spectrum of ethers shows absorption due to C–O stretching at 1300-1000 cm–1 Ether C–O stretch in alkyl ethers C–O stretch in alkyl-aryl ethers

Absorption position in cm–1 ~1120 Two bands at ~ 1250 and ~ 1040

NMR spectrum The protons attached to the carbon linked to oxygen absorb between 3.2-4δ. Diaryl ethers do not show this absorption.

1.4.12 Hydrocarbons Compounds of carbon and hydrogen are referred to as hydrocarbons. These are classified as (i) aliphatic (ii) aromatic hydrocarbons. The aliphatic hydrocarbons can be either saturated or unsaturated. Here is the list of different types of aliphatic hydrocarbons (i) Saturated alicyclic: These are open chain compounds, containing only sp3 hybradized carbons and only single bonds. (ii) Saturated cyclic: Cyclic compounds containing only sp3 carbons in the cyclic structure and thus contain only single bonds between the carbon atoms of ring structure. (iii) Unsaturated : These may be (a) olefins: in addition to sp3 carbons these compounds contain sp2 hybridized carbons and so one or more carbon-carbon double bonds (b) acetylenes, contain sp hybridised carbons and thus one or more carbon-carbon triple bonds. The unsaturated hydrocarbons may be open chain or cyclic in nature.

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Aromatic hydrocarbons on the other hand have following features: (i) All of these have cyclic carbon rings (ii) All are poly unsaturated containing conjugated double bonds (iii) All have delocalised pi electrons (iv) All are stabilised systems, have resonance stabilisation energy and therefore show different properties than unsaturated aliphatic hydrocarbons. For example the aliphatic unsaturated hydrocarbons decolourise potassium permanganate and bromine solution, whereas aromatic hydrocarbons do not except those (i) which contain an aliphatic unsaturated side chain (ii) some polycyclic hydrocarbons like anthracene. Hydrocarbons containing a high carbon: hydrogen ratio like most aromatic compounds normally burn (flame test) giving a sooty flame whereas compounds with low C:H ratio like most aliphatic compounds give a non sooty flame on burning. The presence of a hydrocarbon is generally indicated by the • The absence of any extra element • Absence of the above mentioned functional groups (–ve functional group tests) • Solubility in fuming sulphuric acid in case of aromatic hydrocarbons • Test with chloroform (aromatic hydrocarbons) • Identified by analysis of spectral data. Solubility in Fuming Sulphuric Acid Saturated aliphatic compounds (both open chain and cyclic) do not dissolve in fuming sulphuric acid. Aromatic hydrocarbons dissolve in fuming sulphuric acid and are not precipitated on dilution because they undergo sulphonation and the formed sulphonic acids are water soluble compounds

Procedure Take the given compound (100 mg) in a test tube and add fuming sulphuric acid (1 mL, 20%). Shake the mixture gently and carefully, aromatic hydrocarbons dissolve.

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Notes: • Perform the test in a fume chamber. • Handle fuming sulphuric acid very carefully. • Some aromatic hydrocarbons undergo sulphonation on heating. Test with Chloroform Some aromatic hydrocarbons undergo series of condensation reactions (Friedel-craft alkylation) with chloroform in presence of anhydrous aluminium chloride. The resultant compounds (Ar3C+AlCl4–) are coloured and hence can be used for identification of these hydrocarbons 3 C6 H 6 + CHCl3 (C6 H5 )3 CH + AlCl3 + 2C l

AlCl

3  → (C6 H5 )3 CH + 3HCl

→

(C6 H5 )3 C+ AlCl4− + HCl

The test is positive for many other aromatic compounds. However, the aromatic compounds containing strong deactivating groups do not answer this test. Procedure Heat anhydrous aluminium chloride (100 mg) gently in a pyrex glass test tube by keeping the tube almost horizontal on the flame. Cool the tube and add the organic compound (1 drop or 20 mg) and chloroform (2-3 drops) along the sides into the tube. Appearance of a bright colour indicates the presence of an aromatic compound. Following colours are obtained with some of the common aromatic hydrocarbons: Compound Benzene and its homologs Naphthalene Biphenyl Phenanthrene Anthracene ArX (X = Cl, Br, I)

Colour Orange to red Blue Purple Purple Green Orange to red

Notes: All aromatic compounds do not give this test. Alkanes are non reactive in nature and do not show any reaction described in the functional group tests. These can be best characterised by spectral analysis. Alkenes and Alkynes: Decolourise bromine solution (page 25) without evolution of hydrobromic acid and also react with Baeyer’s reagent (page 27) Spectroscopic identification of hydrocarbons IR Spectrum Some significant absorptions that may be observed in the IR spectrum of hydrocarbons are tabulated below 3

C-C

weak band

C sp − H > 3000 cm −1

C=C

~1650 cm–1

C sp − H < 3000 cm −1

C≡C

2260-2100 cm–1

C sp − H < 3000 cm −1

2

1.4 Identification of Functional Group Advanced Experimental Organic Chemistry

C=C (aromatic) 1600-1500 cm–1

61

C–H (aromatic) 3100-3000 cm–1

CH3– 1465–1460 cm–1 and 1380 cm–1

(CH3)2CH-symmetrical doublet at 1380 cm–1

(CH3)3C-unsymmetrical doublet at 1380 cm–1 NMR spectrum Saturated hydrocarbons Following absorption signals may be seen in the NMR spectrum of a saturated hydrocarbon. Coupling patterns obtained are discussed in detail in chapter 2. Nature of protons CH3 (Primary) RCH2R (Secondary) R3CH (Tertiary)

Position of absorption ~0.9δ ~1.3δ ~1.5δ

Alkenes In addition to the proton absorptions of the saturated hydrocarbon component present in alkenes, the following absorptions are seen in the NMR spectrum Type of proton C=C-H (vinylic protons) C=C-CH (allylic)

Position of absorption 4.6-5.9δ 1.7-2.5δ

Type of proton

Position of absorption 2-3δ 1.8-2.5δ

Alkynes C≡CH C≡C-CH

Aromatic Type of protons Ar-H Ar-CH

Position 6-8.5δ 2.2-3δ

Halogen containing Compounds 1.4.13 Alkyl, Vinyl and Aryl Halides These compounds are represented by the general formula RX, RCH=CHX and ArX respectively. X may be Cl, Br or I. Flourine containing compounds are rarely given for identification. Halogen present in alkyl halides is more easily substituted by nucleophiles, which forms the basis of their detection and distinction from less reactive halogen present in aryl and vinyl halides. The ease of substitution of –X by nucleophilic reagents mainly depends on the stability of the carbocation formed on loss of X– and in some cases on the stearic effects exerted by the groups attached to the carbon bearing the halogen.

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Generally the ease of substitution in alkyl halides is in the order Tertiary > secondary > primary halides (because the order of stability of carbocations is tert> secondary> primary). In aryl and vinyl halides, the carbon-halogen bond is stronger because of the resonance (shown below), the substitution of halogen by nucleophiles is difficult.

The ease of substitution also depends on the nature of halogen, the order being I > Br > Cl. However, there are a few exception to these general trends: • Benzyl chloride (a primary alkyl halide) is more reactive as compared to some tertiary alkyl halides. • Some aryl halides (substituted by electron withdrawing groups at o- and p-positions ) are very reactive. The presence of an alkyl, vinyl and an aryl halide can be detected by the following laboratory tests Test with alcoholic silver nitrate Test with sodium iodide a test for alkyl bromides and chlorides Spectroscopic methods Test with alcoholic silver nitrate. Many halogen containing compounds react with alcoholic solution of silver nitrate to produce water insoluble silver halides. R 3CX + AgNO3 → R 3CNO3 + AgX ↓ t −alkylhalide

ArX + AgNO3 → No reaction Aryl halide

Procedure To an alcoholic solution of silver nitrate (2~3 mL, 0. IM) in a test tube, add the solution of the given compound (a drop in –1 mL alcohol). Keep for 4-5 minutes, if a precipitate is not obtained, warm the solution in a water bath for 1-2 minutes. Separation of a white solid indicates the presence of RX A clear solution indicates the presence of ArX or RCH=CHX

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Notes: • This test should be performed only if a halogen is indicated during the detection of extra elements. • Some substituted Aryl halides (containing strong electron withdrawing groups at o- and p- positions) may also give the test. • Carboxylic acids also react with silver nitrate to form a white precipiate, this however is soluble in dilute nitric acid, whereas the precipitate of AgCl is insoluble, AgBr is sparingly soluble and AgI is yellow in colour and soluble in dilute nitric acid. RCOOH + AgNO3 —→ RCOOAg↓ + HNO3 Test with Sodium iodide This test is based on the following facts • The halogen present in alkyl chlorides and bromides can be replaced by iodine on reaction with sodium iodide. • The product formed i.e. sodium chloride or bromide by the above reaction is insoluble in the medium (acetone). • Hence the formation of a precipitate indicates the presence of RCl or RBr • The rate of reaction is fastest with tertiary halides. • Aromatic halides do not react Acetone RCl + NaI  → RI + NaCl ↓

RBr + NaI  → RI + NaBr ↓

Procedure Take sodium iodide (0.5g) in a test tube and dissolve it in acetone (1 mL). Add to it a solution of the given compound in acetone (a drop of the compound in ~1 mL of acetone). Mix and allow to stand for ~5 minutes, warm in a hot water bath for 3-4 minutes. Formation of a precipitate indicates the presence of RCl or RBr. Spectroscopic identification IR spectrum There are two main absorption signals in the IR spectrum C-X stretching and bending Type of compound

C-X stretch in cm–1

C-X bending in cm–1

RCl

785-540

1300-1200

ArCl

1100-1035

RBr

650-510

ArBr

1075-1030

RI

600-485

1250-1190 1200-1150

The position of absorption also depends on the number of halogens in the molecule. For example, the position of C-Cl in substituted methane is

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CH2C12

739 cm–1

CHC13

759 cm–1

CCl4

785 cm–1

Qualitative Organic Analysis

NMR spectrum Type of compound CH-F CH-C1 CH-Br CH-I

Position of absorption 4.2-4.8δ 3.1-4.1δ 2.7-4.1δ 2.0-4.0δ

The position of absorption is also influenced by number of halogens in the alkyl halide. CH3C1

3δ

CH2Cl2

5.3δ

CHCl3

7.3δ

1.4.14 Acid chlorides Another class of compounds containing halogen are acid chlorides, these could be either aliphatic (RCOCl) or aromatic (ArCOCl) in nature. The acid chlorides can be detected in the laboratory by the following tests Test with silver nitrate Alkaline hydrolysis Test with silver nitrate

RCOCl + AgNO3 → RCOONO2 + AgCl↓

Procedure Take the given compound (l-2 drops) in water (2-3 mL) in a test tube and shake well. Add to it an aqueous solution of silver nitrate (1 mL). Formation of a white precipitate indicates the presence of an acid chloride. Alkaline hydrolysis The acid chlorides get hydrolysed to the corresponding sodium salt of acid on heating with an alkali like sodium hydroxide. The acid can be obtained by acidification of the alkaline solution

RCOCl + NaOH → RCOONa + HCl



2 RCOONa + H2SO4 → 2 RCOOH + Na2SO4

Procedure Take the given compound (2-3 drops) and reflux with sodium hydroxide in a round bottomed flask for 5-10 minutes. Cool the solution and acidify with dilute sulphuric acid. Separation of a solid or a liquid compound indicates the presence of an acid chloride. Notes: If the acid is water soluble, it does not separate on acidification. In such cases, acid may be isolated by extraction.

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Spectroscopic identification IR Spectrum C=O stretch appears at 1810-1775 cm–l –C–C1 stretch at 730-550 cm–1. NMR CHCOC1 absorbs between 2-3δ.

Compounds containing N as extra Elements 1.4.15 Amines Alkyl or Aryl substituted ammonia derivatives are known as amines. Mono substituted ammonias are primary amines represented by a general formula RNH2 or ArNH2.Di substituted ammonia derivatives are secondary amines ( R2NH orAr2NH), tri substituted ammonias are tertiary amines (R3N or Ar3 N etc.) and tetra substituted ammonias are known as the quaternary ammonium salts (R4N+ X or Ar4N+ X–). Amines can further be classified as simple amines i.e. all alkyl or aryl groups present in them are identical or mixed amines in which all the groups are not identical. For example RNHR′ and ArNHR are mixed and R2NH and Ar2NH are simple secondary amines. Amines are basic in nature because of the presence of a lone pair of electrons on nitrogen, which they can share with the acids. The strength of their basic nature depends on the availability of these electrons, which varies with the nature of alkyl or aryl groups bonded to the nitrogen of the amines. If these groups are electron donating (either due to inductive (+I) or resonance effect (+R), the amines are stronger than those having electron withdrawing groups. Steric factors also affect the availability of electron pair of nitrogen and hence the basic strength of amines. Table 1.5 gives the basic strength of a few amines. The higher the pKa value, the stronger is the base. Table 1.5: Basic strength of amines Amine Ammonia Methyl amine Dimethyl amine Trimethyl amine Aniline N-Methyl aniline N, N-Dimethyl aniline p-Nitro aniline p-Anisidine

pKa 9.27 10.62 10.77 9.80 4.58 4.85 5.06 1.02 5.29

The primary, secondary and tertiary amines can be tested and distinguished by the following tests in the laboratory • Solubility in mineral acids • Test with nitrous acid (a preliminary test for primary, secondary and tertiary amines followed by confirmatory tests) • Confirmatory test for primary amines Azo dye test

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Carbylamine test Test with 5-nitrosalicylaldehyde • Conformatory test for secondary amines Liebermann nitroso test • Confirmatory test for tertiary amines ♦ Hinsberg test: Test with benzene sulphonyl chloride/p-toluene sulphonyl chloride (a test for primary, secondary and tertiary amines) ♦ Test with sodium nitroprusside Solubility in Mineral Acids Amines react with mineral acids forming the corresponding salts which are soluble in the medium. The ease of solubility depends on the basic strength of the amine. Some amines are insoluble in acids because of low basic strength. This test should be done only with water insoluble amines Test with Nitrous Acid Primary, secondary and tertiary amines react with nitrous acid giving different products. This reaction is used in the laboratory as a preliminary indicator for the presence of primary, secondary or tertiary amines. Both aliphatic and aromatic primary amines react with nitrous acid to form the corresponding diazonium salts. The diazonium salts obtained from aliphatic primary amines are generally highly unstable and decompose immediately and react with water to form the corresponding alcohol. RNH2 + NaNO2 + 2HCl → [RN=N+ Cl–] + NaCl + 2H2O [RN=N+ Cl–] + H2O → ROH + N2 + HCl Whereas the diazonium salts obtained from aromatic primary amines are comparatively more stable, these remain dissolved in the aqueous medium and their presence is confirmed by coupling with β-naphthol, when an orange coloured azo dye is formed (Azo dye test).

Both aliphatic and aromatic secondary amines react with nitrous acid to form N-Nitroso compounds, which are liquids and insoluble in the reaction medium. Hence these separate as oily droplets or give a turbid solution. For confirmation, the N-Nitroso compounds are extracted with solvent ether, ether removed and residue is subjected to Liebermann nitroso test.

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The aliphatic tertiary amines form the salt of amine in this test and are regenerated on basification of the solution. R3N + HNO2 → R3N ⋅ HNO2 R3N + HCl → R3N ⋅ HCl Aromatic tertiary amines react differently. Those tertiary amines which have an unsubstituted para position, form a p-nitroso derivative, which separates as p-nitroso tertiary amine hydrochloride (an orange coloured solid). Basification of the solution gives p-nitroso derivative of the tertiary amine (a green coloured solid).

Table (1.6) gives the nature of the product formed and the visual change accompanying it. As already stated, depending on the preliminary observation, confirmatory tests are performed to arrive at a specific conclusion regarding the nature of the unknown amine. Procedure Prepare the following solutions and cool separately in an ice bath to 0-5°C. • Given compound (20-30 mg or a small drop) in a mixture of cone. hydrochloric acid (~l mL) and water (~2 mL) or dilute hydrochloric acid (2-3 mL) in a boiling tube. • Sodium nitrite (50 mg in ~2mL water) in a test rube. Add sodium nitrite solution slowly to the solution of the given compound in the boiling tube, mix well and keep in ice bath for 1-2 minutes. Table 1.6: Reaction of amines with nitrous acid Observation A clear solution is obtained with the brisk evolution of a gas A clear solution is obtained without brisk evolution of a gas Oily droplets separate out An organge solid is formed

Inference May be an aliphatic primary amine, Confirm by isocyanide test May be an aromatic primary amine, Confirm by azo dye test Secondary amine Confirm by Liebermann nitroso reaction Aromatic tertiary amine Confirm by basification of reaction

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Observation Inference A white solid is formed which dissolves in water and May be an aliphatic tertiary amine basification of the reaction mixture regenerates the original compound.

Notes: • Formation of a clear solution in the above reaction does not confirm the presence of a primary amine. It can only be confirmed by subsequent confirmatory tests. • The temperature during the reaction must be kept low because the diazonium salt formed from a primary aromatic amine may decompose at a higher temperature. • The reaction mixture should be acidic during the reaction with nitrous acid. • Some free nitrous acid should be present in the reaction mixture at the end of the reaction. This can be tested by adding a drop of the test solution on a starch-iodide paper strip, appearance of a blue colour indicates the presence of free nitrous acid.

2HNO2 + 2H+ + 2I–1 → 2 NO + 2H2O + I2

Liberated iodine gives a blue colour with starch. • If a yellow precipitate is formed in this reaction, it indicates an incomplete diazotisation of a primary aromatic amine. In the acidic medium the excess amine couples with the diazotised amine to give a precipitate of the azo dye, which is yellow in colour. • Some amides also react with nitrous acid with brisk effervescence but do not give +ve confirmatory test for any type of amine.

Confirmatory tests of amines Azodye test (Confirmatory test for aromatic primary amines) If a clear solution is obtained in the above test with nitrous and there was no rapid evolution of a gas, proceed as follows: Procedure To the cooled clear solution (obtained by reaction of acidic solution of the compound with sodium nitrite) in the boiling tube, add slowly with shaking, a cold (0-5°C) solution of β-naphthol (50 mg) in dilute sodium hydroxide (2–3 mL). Formation of an orange/red coloured solid confirms the presence of an aromatic primary amine. Notes: • The solution during diazotisation should be mildly acidic. In a strongly acidic solution the phenoxide ion, which is required for coupling will get converted to phenol and if the solution is too basic C6H5N2+ Cl– will get converted to C6H5N = NOH which does not couple. So both in strongly acidic and basic medium, the coupling reaction does not take place. • If the amino compound contains acidic groups like COOH, –OH (phenolic) etc, the azo dye remains dissolved in the basic medium. In such, cases red/orange solution and not a precipitation is formed. Spot test Procedure Take a whatmann paper strip and apply the following reagents one over the other • Dilute solution of compound in water/ alcohol • Dilute hydrochloric acid • Aqueous solution of sodium nitrite (from the kit) • Alkaline solution of β-naphthol (from the kit) Development of a red-orange spot of azo dye confirms the presence of an aromatic primary amine.

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Carbylamine test/isocyanide test (Confirmatory test for both aliphatic and aromatic primary amines) All primary amines react with chloroform in presence of alkali to produce isocyanides, which have a very strong unpleasant smell, are highly poisonous and hence the test should be performed very cautiously, in a fume chamber. RNH 2

Primary amine

+ CHCl3 + 3KOH →

RNC

Carbylamine

+ 3KCl + 3H 2 O

Procedure Warm carefully, in a fume chamber, the given compound (10 mg) with chloroform (1-2 drops) and alcoholic potassium hydroxide (~ 2mL) in a water bath. A foul smell of isocyanide confirms the presence of a primary amine. Notes: • Never smell directly from the tube. • Perform the reaction with a very small amount of the compound in a fume chamber. • Add concentrated hydrochloric acid immediately after performing the test to the tube so as to decompose the isocyanide.

• Isocyanides effect eyes, skin and respiratory tract. Seek medical help immediately. Test with 5-nitrosalicylaldehyde reagent (Confirmatory test for primary amines) Primary amines react with 5-nitrosalicylaldehyde to produce condensation products which give a precipitate of nickel complex on reaction with nickel chloride present in the reaction (see below). Primary aliphatic amines react at a faster rate than aromatic primary amines.

Procedure Dissolve the given compound (a drop) in a test tube in dilute hydrochloric acid (3-4 mL) and add 5-nitro salicylaldehyde (~2 mL) reagent. Immediate formation of copious precipitate shows the presence of an aliphatic primary amine. Aromatic amines give a precipitate after 2-3 minutes. Notes: • The reagent is made by separately dissolving

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(i) 5-nitro salicylaldehyde (0.5g) in a mixture of triethanol amine (12 mL) and water (20 mL) (ii) Nickel chloride (0.4g) in water (4 mL) Mixing the two solutions and diluting with water (90 mL). Liebermann Nitroso test (Confirmatory test for secondary amines) As already mentioned secondary amines give N-Nitroso derivatives on reaction with nitrous acid, which separate out as water insoluble oily droplets. The presence of a secondary amine is then confirmed by performing Liebermann nitroso test. In this test, the N-nitrosamine is heated with phenol in presence of concentrated sulphuric acid, the nitroso group migrates from amine to phenol to produce p-nitroso phenol which then condenses with another molecule of phenol (see section 1.4.4 for complete reaction) and finally gives a blue colour in the alkaline medium.

Procedure Extract the N-nitrosamine obtained (page 67) with ether (~10mL) carefully. Separate the ether layer, first wash with ~5-7 mL water then with urea solution to remove excess of unreacted free nitrous acid, transfer ether to a boiling tube, dry over anhydrous calcium chloride. Filter ether into another dry boiling tube and evaporate ether (caution). The residue is the N-nitrosamine of the secondary amine. Add phenol (20 mg), concentrated sulphuric acid (1 mL), heat gently over a low flame and follow the procedure in section 1.4.4. Notes: Free nitrous acid should be removed by treating with urea. Otherwise this will react with phenol to form p-nitrosophenol and interfere in the test for secondary amine.

HNO2 + NH2CONH2→ N2 + 2H2O + HCNO

Spot test for secondary amines Reaction with nitrous acid followed by Liebermann’s nitroso test Procedure Use the liquid compound directly or a concentrated solution of solid compound in water or alcohol. Spot the following solutions one over the other in the order given below on a thin strip of Merck plate. • Liquid compound or concentrated solution of the compound • Aqueous solution of sodium nitrite (from the kit) • Concentrated sulfuric acid • Heat the Merck strip in an oven at ~120°C for 1 min and apply the spots of the following solutions on the above spot: • Saturated aqueous solution of urea. • Concentrated alcoholic solution of phenol • Again heat the Merck strip in oven at ~120°C for 30 sec and scratch the silica of the sported portion and transfer it in the test tube containing dilute sodium hydroxide solution and 1-2 mL of water. A blue coloured solution confirms the presence of a secondary amine.

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Confirmatory test for aromatic tertiary amines The tertiary aromatic amines form an orange coloured solid on reaction with nitrous acid (page 67). Basification of the solution gives a green precipitate of p-nitroso tertiary amine

Procedure Take the tube containing the orange coloured solid (page 67) and add enough dilute sodium hydroxide to make it alkaline and shake well. Appearance of a green solid indicates the presence of a tertiary amine. Spot test for tertiary aromatic amines Procedure Use either an aqueous or an alcoholic solution of the given compound for this test. Spot the following solutions one over the other in the order given below on a thin strip of Whatman filter paper: • Concentrated solution of the compound • Concentrated hydrochloric acid • Aqueous solution of sodium nitrite (from the kit) Appearance of an orange spot indicates the presence of aromatic tertiary amines. • Apply dilute sodium hydroxide solution 2-3 times (from the kit) The orange spot turns green confirming the presence of aromatic tertiary amine. Test with Hinsberg reagent Primary amines react with benzene sulphonyl chloride/toluene sulphonyl chloride (Hinsberg reagent) in an alkaline medium to form substituted sulphonamides, which are soluble in the medium. Secondary amines react with the reagent to form sulphonamides which are insoluble in the medium. Tertiary amines do not react. Hence this reaction is also used for identification of a mixture of primary, secondary and tertiary amines.

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Procedure Shake vigorously a mixture of the given compound (0.2 g), benzene sulphonyl chloride (0.7 mL) and potassium hydroxide (10 mL, 2M) in a conical flask for 5-6 minutes till the colour of sulphonamide disappears. Check the alkalinity of the solution, if required, add some potassium hydroxide to make the solution alkaline. (i) If a clear solution is obtained, neutralise with dilute hydrochloric acid. Separation of a solid indicates the presence of a primary amine. (ii) If a solid or an oily product is formed, separate it and treat with dilute hydrochloric acid, a tertiary amine is indicated if the product dissolves and a secondary amine is present if the solid does not dissolve. Notes: Follow the procedure (quantities of reagents and time) as closely as possible to avoid misleading results. For example, in presence of excess amine, a tertiary amine may also react.

Test with Sodium nitroprusside In this test, the amine is treated with sodium nitroprusside in presence of a carbonyl compound. The condensation product obtained between the carbonyl compound and the amine, complexes with sodium nitroprusside to give different colours. Two tests have been developed on this basis namely the Rimini test in which acetone is used and Simons test in which acetaldehyde is used as the carbonyl compound. Rimini test (a test given by aliphatic primary and secondary amines). Procedure Add the given compound (25-30 mg), water (1 mL) acetone (2 mL) to sodium nitroprusside solution (1 mL) in a test tube and mix well. Formation of a deep red colour indicates the presence of a primary or a secondary aliphatic amine. Modified Rimini test (a test given by all three types of aromatic amines). Procedure To the ~1 mL of the modified Rimini reagent (see Appendix 1) in a test tube, add the following solutions in the order given below • Saturated aqueous solution of zinc chloride (1 mL) • Acetone (0.2mL) • Given compound (~30 mg) • Mix well. Primary and secondary aromatic amines give orange-red colour within 5 minutes. Tertiary aromatic amines give an orange-red colour changing to green on standing. Simon test a test given by aliphatic primary and secondary amines.

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Procedure To a mixture of acetaldehyde (aqueous solution, 0.2 mL, 2.5 M), water (1 mL) and sodium nitroprusside solution (1 mL), add the given compound (~30 mg) in a test tube. Mix well. Aliphatic primary amines produce a yellow or brown colour and aliphatic secondary amines give a deep blue colour within ~5 minutes. Modified Simon test Procedure Take the modified Simons reagent (1 mL) (see Appendix 1) add the following solutions in the order given below: • Saturated solution of zinc chloride (1 mL) • Aqueous acetaldehyde solution (0.2 mL, 2.5M) • Given compound (~30 mg) • Mix well and keep for 5 minutes — Primary aromatic amines give orange/red or brown colour — Secondary and tertiary aromatic amines give an orange/red colour changing to green. Spectroscopic Identification of Amines IR Spectrum Amines exhibit the following absorptions in their IR spectrum N-H stretching in the range 3500-3200 cm–1 (position varying with concentration) • Two bands of ~NH2 for a primary amine • One band of –NH for a secondary amine • No absorption for a tertiary amine N-H bending • 1650-1560 cm–1 for primary amines • 1580-1490 cm–1 for secondary amines C-N stretching at 1350-1100 cm–1 NMR Spectrum In addition to the other absorption signals, the following absorptions are due to amino group • N-H: it appears as a broad signal. Its position is variable and depends mainly on the nature of the solvent used and the concentration of the solution. — 0.5–3.0 δ for aliphatic amines — 3.0–5.0 δ for aromatic amines • CH-N : appears between 2.0-3.0 δ

1.4.16 Nitro Compounds Compounds containing –NO2 group are classified as nitro compounds, these may be aliphatic (RNO2) or aromatic (ArNO2) in nature. Nitro compounds are generally yellow in colour. The nitro group in these

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compounds can be reduced to a variety of products by different reducing agents, which forms the basis of their laboratory identification. Given below are some of the important reduction products and the corresponding reducing agents (i) Acidic reducing agent (Zn + HCl or Sn + HCl)



(ii) Neutral reducing agent (Zn + NH4Cl) RNO2 + 4H → RNHOH + H2O



(iii) Alkaline reducing agent ArNO2 + 2H → Ar NO + H2O













Hydroxyl amine

Nitroso compound



ArNO2 + 4H → ArNHOH + H2O



ArNO + ArNHOH → Ar − N = N − Ar + H 2 O ↓ ( Azoxybenzene) O Ar − N = N − Ar + 2H → Ar − N = N − Ar + H 2O ↓ ↓ 2H (Azobenzene) O ArNH − NHAr (Hydrazobenzene)

The presence of a nitro group in an unknown compound is identified by the following tests: Mulliken Barker Test Test with ferrous hydroxide Reduction to amino group followed by Azo dye test Spectroscopic identification Mulliken Barker test The test is based on the fact that the substituted hydroxyl amine obtained by reduction of nitro group with a neutral reducing agent (zinc + ammonium chloride) can reduce the Tollens’ reagent to give a silver mirror or a greyish precipitate, which slowly becomes black. Nitro compounds themselves do not reduce the Tollens reagent. C6 H5 NO 2 + 4H

→

C6 H5 NHOH + H 2O

C6 H5 NHOH + 2Ag( NH3 ) 2 OH

→

C6 H5 NO + 2Ag + 2H 2O + 2 NH3

Tollens reagent

Procedure Dissolve the given compound (a drop or 50 mg) in alcohol (~ 1 mL) in a test tube. Add zinc dust (100 mg) and concentrated ammonium chloride solution (0.5 g in 3-4 mL water) to it. Mix well and boil the mixture with shaking for 1-2 minutes. Filter the hot solution into a tube containing freshly prepared Tollens’ reagent (l-2 mL). Formation of a silver mirror along the inner walls of the tube or grey precipitate turning black indicates the presence of a nitro group.

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Notes: • Freshly prepare Tollens’ reagent should be used. • Dilute nitric acid (l-2 mL) should be added to the reaction mixture before discarding it. This is done to decompose any silver fulminate that might form if the Tollens reagent itself or the solution in – + – + which Mulliken Barker test has been performed is discarded. Silver fulminate (Ag … C ≡ N – O) is an explosive substance. • Tollens’ test should not be performed to detect the presence of nitro group if the given compound contains other groups like an aldehyde or is a carbohydrate etc. Test with Ferrous hydroxide Nitro compounds can oxidise ferrous hydroxide (formed by reaction of ferrous ammonium sulphate with potassium hydroxide) to ferric hydroxide (a brown precipitate). RNO 2 + 6 Fe(OH)2 + 4H 2 O → RNH 2 + Ferrous hydroxide

6Fe (OH)3

Ferric hydroxide ( Brown ppt )

Procedure Take the given compound (a drop or ~20 mg) in a semi micro test tube, add to it a freshly prepared solution of ferrous ammonium sulphate (2 mL, 5%) and a drop of dilute sulphuric acid (3M). Mix well and then add a solution of potassium hydroxide in methanol (~1 mL, 2M). Stopper the tube and shake well. Formation of a brown precipitate indicates the presence of a nitro group. Notes: • To avoid aerial oxidation of ferrous to ferric, the test should be performed in a small test tube which should be filled with the reaction mixture so as to displace air from the tube. • Any compound other than the nitro compound which can oxidise ferrous to ferric will also give a positive test. Reduction to Amino compound followed by Azo dye test Aromatic nitro compounds can be reduced to the aromatic primary amines by a variety of acidic reducing agents like tin or zinc with hydrochloric acid etc. The presence of aromatic amine can then be confirmed in the laboratory by Azo dye test. Procedure Reflux a mixture of given nitro compound (1-2 drops or 100 mg) dissolved in alcohol (2-3 mL), tin metal (2-3 pieces) and concentrated hydrochloric acid (~ 5 mL) in a round bottomed flask, fitted with a water condenser for 15-20 minutes. Cool the reaction mixture. Transfer to a beaker, add a few pieces of ice and neutralise with excess dilute sodium hydroxide. The amino compound separates out. Extract it with ~10 mL ether cautiously, separate the ether layer into a boiling tube, remove ether (caution) and perform azo dye test with the residue (see page 68 for azo dye test). Notes: • This test should only be performed if the given compound does not respond to Azo dye test before reduction • Handle ether carefully.

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Spot test Procedure spot the following solutions in the order given below on a strip of Whatmann filter paper • Alcoholic solution of the given compound • Dilute hydrochloric acid • Scrubn gently ~ 5 mg of zinc dust on the spot using an ear bud or cotton • Wait for ~ 30 seconds • Turn the paper strip upside down On the sported portion, on the reverse side spot the following: • Dilute hydrochloric acid • Sodium nitrite solution from the kit • Dilute alkaline solution of β-naphthol • Development of an orange coloured spot indicates the presence of an aromatic nitro compound. Spectroscopic identification of Nitro group IR spectrum The compounds containing a nitro group show two absorptions due to stretching of the two N=O bonds of the nitro group between 1600-1300 cm–1. Aliphatic nitro compounds

1600-1530 cm–1 (asymmetric stretch)

1390-1300 cm–1 (symmetric stretch)

Aromatic nitro compounds

1550-1490 cm–1 (asymmetric stretch)

1355-1315 cm–1 (symmetric stretch)

NMR spectrum In aliphatic nitro compounds, the protons at the α-carbon show absorption between 4.2-4.6δ. In the aromatic nitro compounds the o-protons are most desheilded appearing near 8.1δ, the m-protons near 7.5 δ and p-at 7.65 δ.

1.4.17 Amides, Imides and Nitriles Amides are derivatives of carboxylic acids formed by replacement of their –OH group by –NH2 group. Such amides are known as primary amides and are represented by the general formula RCONH2 or ArCONH2. Cyclic amides of some dicarboxylic acids containing a –CONHCO- group are known as imides. Compounds containing a C≡N are known as nitriles. Though all these compounds have different structures, all show some common reaction.

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Following reactions are used to identify these compounds in the laboratory Alkaline hydrolysis Hydroxamic acid test Test of imides with potassium hydroxide Biuret test for urea Spectroscopic identification Alkaline hydrolysis All the above mentioned compounds liberate ammonia on alkaline hydrolysis, which is identified in the laboratory on the basis of its smell (do not smell directly) or formation of dense white fumes of ammonium chloride on reaction with hydrochloric acid or formation of a brown precipitate on reaction with Nesseler’s reagent. Formation of carboxylic acids on acidification of the alkaline solution is another indication of the presence of amides, nitriles and imides.

Procedure Take the given organic compound (100 mg) in a test tube and heat it with dilute sodium hydroxide solution (~ 5 mL) for 2-3 minutes. Bring a glass rod which has some hydrochloric acid adhering to it (obtained by dipping an end of a clean glass rod in hydrochloric acid) near the mouth of the test tube while heating the above reaction mixture. Formation of dense white fumes indicates the presence of an amide, imide or a nitrile. Acidification of the remaining solution after cooling and filtration gives the corresponding carboxylic acid, which may separate out as a water insoluble solid. Notes: Ammonia can also be tested by passing the gas into a U tube containing Nessler’s reagent. Formation of a brown precipitate is a positive test. Hydroxamic acid test Amides and nitriles give Hydroxamic acid test. Esters also give this test but do not liberate ammonia on alkaline hydrolysis.

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Procedure Take the solution of the compound (~50 mg in minimum amount of propylene glycol), solution of hydroxyl amine hydrochloride (2 mL in propylene glycol, 1 M) and an aqueous solution of potassium hydroxide (2 mL, 1 M) in a test tube. Heat the reaction mixture for ~2 minutes, cool and acidify with dilute hydrochloric acid. Add a drop or two of ferric chloride, a red or violet colour indicates the presence of amide or nitrile. Spot test for Aromatic amides This lest is based on the conversion of an aromatic amide to primary aromatic amine by Hoffmann bromamide reaction followed by azo dye test. Procedure Use either an aqueous or an alcoholic solution of the given compound for this test. Hoffmann bromamide reaction: Spot the following solutions one over the other in the order given below on a Whatmann filter paper strip or a Merck plate. • Solution of the compound • Dilute sodium hydroxide solution (provided in the kit) • Solution of bromine in carbon tetrachloride (from the shelf) • Heat the paper strip in the oven at ~120°C for 20-30 sec. • The amide is converted to aromatic primary amine. Azo dye test: Take out the paper strip from the oven and apply the following reagents on the above spot: • Aqueous solution of sodium nitrite (from the kit) • Concentrated hydrochloric acid • Alkaline solution of β-naphthol (from the kit). Development of a red-orange spot of azo dye confirms the presence of an aromatic amide group in the compound. Notes: • Perform simultaneously a blank test (using all the reagents except the compound). • An orange coloured spot may appear before spotting alkaline β-naphthol solution. This is due to the liberation of free bromine from sodium hypobromite. The colour disappears on keeping. • The test can be performed only with those compounds which do not contain an aromatic primary amino group. Test of imides with potassium hydroxide Imides form a white precipitate of their potassium salt on reaction with potassium hydroxide.

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Procedure Add a saturated solution of potassium hydroxide in alcohol (~l mL) to a saturated alcoholic solution of the compound (~l mL) in a test tube. Formation of a white precipitate shows the presence of an imide.

1.4.18 Biuret Test for Urea Urea (NH2CONH2) is the amide of carbamic acid (NH2COOH). Biuret test is specific to urea and is based on the fact that two molecules of urea undergo condensation reaction on dry heating, loosing a molecule of ammonia and producing a compound called Biuret. Biuret on reaction with copper sulphate in presence of a base like sodium hydroxide produces a violet coloured complex. ∆

2 NH 2 CONH 2 → NH 2 CONHCONH 2 + NH3 Biuret

Procedure Take the compound (100 mg) in a dry test tube and heat it on a low flame. The compound will melt first and solidify on continued heating. Heat it for ~30 seconds more to eliminate ammonia from the tube. Allow the tube to cool and dissolve the solid in a few drops of dilute sodium hydroxide solution and add 1-2 drops of dilute copper sulphate solution. Formation of a purple coloured solution confirms the presence of urea. Notes: • If the given compound does not melt and solidify on mild heating, do not heat it for very long as the compound will char, stop heating otherwise the tube might crack or break. • Do not add excess of sodium hydroxide, as all the copper sulphate will get precipitate as cupric hydroxide and will not react with biuret. • Do not add excess of copper sulphate as its blue colour will mask the violet colour of the complex formed between biuret and cupric ions. Spectroscopic identification of amides, imides and nitriles IR spectrum Amides The main bands in the IR spectrum of amides are C=O stretch at ~ 1680-1630 cm–1 N–H stretch-two bands at ~3350 and 3180 cm–1 N–H bending at 1640-1550 cm Imides IR – The C=O stretch in succinimide appear as a doublet between 1800-1700 cm–1 and NH stretch at ~ 3200 cm–1. Nitriles The main absorption is a medium intensity band at ~2250 cm–1 due to C≡N stretch of aliphatic nitriles. In conjugated or aromatic nitriles, this band is shifted to lower frequency.

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NMR spectrum Amides There are two important absorption signals due to the amide functional group –CONH2: the position of this absorption varies with concentration, the nature of the solvent used and the temperature. It appears between 5-9δ. –CH–CONH2 show absorption between 2.1-2.5 δ. NMR– The NH proton is highly acidic and is rapidly exchanged. The two –CH2– groups for example in succinimide appear at 2.8 δ as a singlet. Nitriles –CHC ≡ N the hydrogens on α-carbon appear between 2.1-3 δ.

1.4.19 Anilides and Substituted Amides RCONHR, RCONR2 and RCONHC6H5 are the general formulae of secondary, tertiary amides and anilides respectively. The secondary amides give an acid and an aliphatic primary amine on hydrolysis. The tertiary amides give an acid and an aliphatic secondary amine. The anilide is a common name for N- aryl substituted amide represented by the general formula RCONHAr. These compounds produce an acid and a primary aromatic amine on hydrolysis. These compounds can be identified in the laboratory by Acid catalysed hydrolysis Base catalysed hydrolysis Acid catalysed hydrolysis On heating with an acid like sulphuric acid, these compounds produce a carboxylic acid (which separates either as a solid or a liquid or may remain dissolved because of its water soluble property) and an amine (which remains dissolved in the acidic medium). The acid obtained is first separated and the amine is then liberated from the solution by basification with sodium hydroxide. +

C6 H5 NHCOC6 H5 + H 2 O + HX → C6 H5 COOH ↓ + C6 H5 N H3 X − Benzanilide

Benzoic acid

Anilinium salt

Procedure Take the given compound (1 g) in a round bottomed flask (50 mL), add sulphuric acid (50%, 10 mL). Reflux the mixture for ~30 minutes. A solid acid may separate while refluxing. Cool the solution and filter the acid, collect the filtrate (A). Wash the solid with the water till a drop of fresh filtrate gives no effervescence with sodium bicarbonate solution. Now test solid with sodium bicarbonate (see page 36). Basify the filtrate (A) with dilute sodium hydroxide and test in the compound so formed for the presence of an amino group (see page 65) Alkaline Hydrolysis On alkaline hydrolysis, an anilide or a substituted amide produces an amine (which separates out from the solution as a solid or a liquid) and the sodium salt of the carboxylic acid, which remains dissolved in the solution.

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Procedure Take the compound (1 g) and reflux it with alkali (15 mL, 15%) in a round bottomed flask for ~ 30 minutes. Separate the amine (by filtration or extraction) and acidify the filtrate to obtain the acid. Characterise the amine and the acid by the tests given on page 65 & 36.

1.4.20 Amino Acids These are nitrogen containing compounds and contain amino group/s and carboxylic acid group/s, they show reactions due to amino and carboxyl group and some reactions due to the combination of these groups. Amino acids are important natural compound and are the building units of proteins. There are twenty two naturally occurring amino acids, all are optically active (except glycine), belong to L series of compounds. Two out of these twenty two are imino acids (proline and hydroxyproline). A number of synthetic amino acids are also known. As already mentioned amino acids show certain reactions in which both the functional groups participate, one of such property is based on the pH of the solution. In strongly acidic solution, they are present as (i), in alkaline medium as (ii) and at a particular pH (specific to each amino acid) these are present as (iii). The pH at which an amino acid is present as a doubly charged ions is known as its isoelectric point, and the ion as Zwitter ion.

Table 1.7 gives the isoelectric points of a few amino acids. Table 1.7 Name of amino acid Isoelectric point (pI) Alanine 6.11 Arginine 10.75 Aspartic acid 2.98 Glycine 6.06 Leucine 6.04 Proline 6.30 Tyrosine 5.63 Amino acids respond to laboratory tests of both the functional groups present in them. Following tests may be performed in the laboratory Test with Ninhydrin Test with Isatin

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Test with copper sulphate Test with Nitrous acid Carbylamine reaction Sodium bicarbonate test Spectroscopic identification Test with ninhydrin This test is used for the detection of α- and β-amino acids, proteins and peptides. The amino acids react with ninhydrin to produce blue-violet colour. In spite of widely variable structure, all α- and β-amino acids give the same colour because only the nitrogen of amino group is incorporated into the final product. Proline and Hydroxy proline being imino acids do not react. The chemistry of the reaction is given below

Procedure Take the amino acid (20 mg in 1mL water), ninhydrin solution (1mL) in a test tube and add ~ 10mL water to it. Heat the mixture for ~ 30 seconds. Development of a blue-violet colour indicated the presence of an α- or β-amino acid. Proteins and peptides also produce the same colour. Spot test Procedure Spot the following solutions one over the other on a strip cut out of a Whatmann filter paper. • Ninhydrin solution (twice) allow the spot to dry • Concentrated solution of the given compound Development of blue-violet coloured spot within 1-2 min. is a positive test. Notes: Hydroxyl proline produce a pink colour proline gives a yellow colour.

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Test with Isatin Isatin reacts with amino and imino acids to produce different colours, for example alanine gives a pinkred colour, valine red and proline a intense blue colour. The reaction of proline with isatin is given below.

Procedure To a solution of isatin (1 mL, 1%) in a test tube, add the given compound (~20 mg). Appearance of an intense colour shows the presence of an amino acid. Test with Copper sulphate Amino acids form a blue coloured complex on reaction with copper sulphate solution. The reaction between copper sulphate and alanine is given below:

Procedure Dissolve the given compound (20-30 mg) in water (1 mL) in a test tube and add copper sulphate solution (4-5 drops). A blue coloured solution is formed in cold or heating in a hot water bath for 4-5 minutes. Test with Nitrous acid The amino acids which contain an aliphatic primary amino group react with nitrous acid (NaNO2 + HCI) to produce a diazonium salt, which decomposes to liberate nitrogen and hydroxy acid.

Procedure See page 66. Carbylamine reaction All amino acid give this test. Procedure Follow the procedure and all the precautions discussed on page 69.

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Sodium bicarbonate test All amino acids give this test Procedure Follow the procedure described on page 36. Spectroscopic identification IR spectrum Two absorptions due to C=O stretch are seen in the IR Spectrum of the amino acid near 1600 and 1400 cm–1 due to –COO– group. + ) stretch appears between 3300-2600 cm–1 The (–NH 3 And absorption between 1660-1400 cm–1 is due to N-H bending. NMR spectrum A broad band near 7.5δ is due to hydrogens attached to ~ NH3+ group. The hydrogens on α-carbon absorb near 4.2-4.5δ.

Compounds containing sulphur 1.4.21 Sulphonic acids The general formula of sulphonic acid is RSO3H, where R may be aliphatic or aromatic group. These are commonly prepared by sulphonation of a compound with concentrated or fuming sulphuric acid. They are acidic in nature and generally soluble in water. Sulphonation reaction is extensively used in industry, for example for synthesis of (i) dyes (specially to make water soluble dyes) (ii) detergents (iii) drugs like sulphonamides etc. Since sulphonic group can be easily removed (desulphonation), this group is also used as a protecting group in certain preparations in the laboratory and industry.

The presence of sulphonic acid group can be known by performing following tests Reaction with Potassium hydroxide Hydroxamic acid test Reaction with Potassium hydroxide On fusion with potassium hydroxide, sulphonic acids form potassium sulphite, which is oxidised to sulphate. Potassium sulphate is then treated with barium chloride to produce a white precipitate of barium sulphate which is insoluble even in concentrated hydrochloric acid. C6H5SO3H + 2KOH → C6H5OK + K2SO3 + H2O

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K2SO3 + [O] → K2SO4 K2SO4 + BaCl2 → 2 KCl + BaSO4↓ Procedure Fuse the organic compound (100 mg) with potassium hydroxide (200 mg) in a nickel crucible. Extract the fused mass with water (l-2 mL). Acidify the extract with dilute hydrochloric acid and treat with a few drops of barium chloride solution. Formation of a white precipitate, insoluble in concentrated hydrochloric acid shows the presence of a sulphonic acid. Hydroxamic acid test Sulphonic acids produce sulphonyl chloride on reaction with thionyl chloride, which, on reaction with hydroxylamine followed by acetaldehyde yields hydroxamic acid. A magenta coloured complex is formed when hydroxamic acid is treated with ferric chloride in acidic medium.

Procedure Heat the given compound (100 mg) and thionyl chloride (5 drops) in a test tube in hot water bath for a minute. Make a saturated solution of hydroxylamine hydrochloride in methanol (0.5 mL) and add it to the sulphonyl chloride. Now add a drop of acetaldehyde and make the resultant mixture alkaline with alcoholic potassium hydroxide. Boil for a few seconds. Allow the solution to cool, acidify with dilute hydrochloric acid and add ferric chloride solution (l-2 drops). Appearance of a magenta colour shows the presence of a sulphonic acid. Notes: Perform the test in a fume chamber. Spectroscopic identification IR Spectrum S=O shows stretch at 1350 and 1450 cm–1 respectively.

1.4.22 Thiols Thiols also known as mercaptans or thio alcohols are compounds represented by a general formula RSH and contain an -SH functional group. These are sulphur analogues of alcohols. Sulphur present in these compounds is much less electronegative than oxygen and hence the hydrogen bonding in thiols is much weaker as compared to alcohols. Below are listed the laboratory tests of thiols Test with lead acetate

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Sodium nitroprusside test Spectroscopic identification Test with lead acetate Thiols form a yellow precipitate of lead mercaptide on reaction with lead acetate 2RSH + Pb(OCOCH3 )2 → 2CH3COOH +

Pb(SR )2

Yellow precipitate

Procedure Add an alcoholic solution of the given compound (0.5 mL) to the lead acetate solution (0.5 mL) in a test tube. Formation of a yellow precipitate indicates the presence of a thiol. Sodium nitroprusside test Thiols give a purple colouration on reaction with sodium nitroprusside solution. Spectroscopic identification IR spectrum S-H stretch appears in the region 2600-2500 cm–1. NMR spectrum Unlike other acidic hydrogens (attached to electronegative elements) the H of -SH group does not exchange and hence couples with the protons on the α-carbon atoms. ArSH shows absorption between 2.8-3.6 δ RSH shows absorption between 1.2-1.6 δ The position of the protons on the α-carbon is as follows: CH3SH

2.1δ

RCH2SH

2.6δ

1.4.23 Sulphonamides The amides of sulphonic acids are known as sulphonamides and are represented by the general formula RSO2NH2. These can be detected in the laboratory by: Alkaline hydrolysis On alkaline hydrolysis they produce sulphonic acid and liberate ammonia.

RSO2NH2 + KOH → RSO3K + NH3 RSO3K + HCl → ROH + KCl + SO2

Procedure Fuse the given compound (200 mg) with KOH (500 mg) on a nickel spatula. Ammonia is liberated. Extract the residue with water. Heat the residue with dilute hydrochloric acid, evolution of sulphur dioxide shows the presence of a sulphonamide.

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Spectroscopic identification IR spectrum The IR Spectrum of sulphonamide shows the following absorption S=O stretch at 1370-1330 cm–1 and 1170-1155 cm–1 N–H stretch appears as follows: Two bands for –NH2 between 3390-3330 cm–1 One band for –NH at 3265 cm–1.

1.5 PREPARATION OF DERIVATIVES 1.5.1 Introduction After performing the preliminary and functional group tests, prepare a suitable crystalline derivative of the given compound for its identification. Refer to the literature i.e. the list of compounds given on page 407-521, where the common compounds are arranged according to the functional group present in them and in an ascending order of their boiling/ melting points. Let us understand with an example as to what is to be done. Suppose the given compound is a carbonyl compound with a boiling point 155°C. Refer to the Table 7.3 of carbonyl compounds, you will find that at least five compounds have a boiling point close to your compound. These are Compound

Boiling point (°C)

1.

n-Amyl methyl ketone

151

2.

Ethyl pyruvate

155

3.

Cyclohexanone

155

4.

n-Heptaldehyde

156

5.

Furfuraldehyde

161

If the given compound is a ketone (gives-ve test with Tollens’ and Schiff’s reagent), it is neither heptaldehyde nor furfuraldehyde. Out of compounds 1, 2 and 3, two (1 and 2) are methyl ketones and compound 3 is not a methyl ketone. Suppose your compound gave a negative iodoform test, it could be 3 (cyclohexanone). Prepare a crystalline derivative out the derivatives mentioned for cyclohexanone keeping in mind the following points for preparing the derivative/s of any compound. (i) A water insoluble solid derivative should be preferably prepared for a given compound. However, if such a possibility cannot be excercised, a liquid derivative may be first made, which in turn may be converted to a solid derivative. (ii) The melting point of the derivative should neither be too high nor low, it should preferably be between 60-200°C. (iii) The melting point of the derivative should be reasonably lower or higher than that of the compound. (iv) The chosen derivative should be such that it can be easily prepared in the laboratory.

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(v) If there are two or three possible compounds with similar melting/ boiling points, which cannot be distinguished on the basis of chemical tests as mentioned in the above example, then a derivative which has different melting point for each of the possible compound should be prepared. (vi) Finally prepare a suitable derivative, determine its melting point and look up the list of compounds. The compound will be confirmed if its melting/ boiling point and melting point of its derivative tally with those given in the literature. (vii) The procedures described for preparation of derivatives can be used for most compounds. However, the reaction time may have to be prolonged and reaction temperature increased in some cases. (viii) Specific test for the compound should be performed, if available. (ix) The derivative that is prepared for a given compound is generally impure and hence is purified by crystallisation before determination of its melting point.

1.5.2 Crystallisation Below are given the steps performed for crystallisation: (a) Selection of a suitable solvent/s (b) Preparation of a hot saturated solution of the compound (c) Decolourisation (d) Formation of crystals (e) Filtration (f) Drying. (a) Selection of a suitable solvent for crystallisation The points that are important in selecting a suitable solvent are (i) A large variety of solvents can be used for recrystallisation but for under and post graduate students, one out of the following solvents or a mixture of two miscible solvents can be used for crystallisation

Water



Ethanol Methanol



Ethyl acetate. (ii) The compound to be recrystallised should be almost insoluble in the solvent at room temperature and soluble at high temperature. (iii) Impurities should either be insoluble even in the hot solvent or soluble in it at room temperature. (iv) The boiling point of the chosen solvent should be lower than the melting point of the derivative. (v) Organic solvents are inflammable and hence should never be heated directly on a burner, these should be heated in a hot water bath.

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Procedure Take ~50 mg of the compound in a test tube and add 1-2 mL of the solvent (try the solvents in the order given above) to it, if the compound dissolves in it, the solvent is not suitable. However, if the compound is insoluble, heat till the solvent boils. If the compound dissolves in hot and separates out on cooling, the solvent can be used for crystallisation. If a single solvent is unsuitable, a mixture of two miscible solvents can be used. One in which the compound is freely soluble and the other in which the compound is insoluble, for example a mixture of alcohol and water. (b) Preparation of hot saturated solution of the compound Procedure Take ~100 mg of the compound in a boiling tube and add to it ~4-5 mL of the selected solvent. Heat to boiling and stir continuously using a glass rod while heating. (i) The compound dissolves completely, add a small amount of the compound to the tube and heat again to obtain a clear, hot saturated solution of the compound. (ii) The compound does not dissolve completely, add ~l-2 mL of the solvent, heat to dissolve. (iii) If a mixture of two solvents has to be used, take ~100 mg of the compound in a boiling tube and add ~l-2 mL of the solvent in which the compound is soluble. Now add dropwise the solvent in which it is insoluble with shaking till a permanent turbidity appears. Heat to boiling to get a clear solution. If a clear solution is not obtained or a trace amount of the compound remains undissolved, filter the hot solution quickly into another tube. (c) Decolourisation If the hot solution obtained above is dark coloured (blackish), the colour must be removed. This colour is removed by using activated charcoal, which has the capacity to adsorb the colour. Procedure Take the hot solution obtained above in a boiling tube, add to it l-2 mL of the solvent and then add ~100 mg of activated charcoal. Heat the contents of the boiling tube for 2-3 minutes, use a glass rod to stir it. Now filter the hot solution to remove charcoal into another tube. The filtrate should be almost colourless. (d) Crystal formation Since the compound was insoluble in the solvent at room temperature, cooling of the hot solution deposits the solid back. The solution should be allowed to cool slowly otherwise the impurities get co-precipitated and lead to deposition of impure solid. Procedure Allow the hot saturated solution to cool undisturbed. Crystals are formed slowly. If the crystals are not formed, either scratch the inner side of the tube containing the solution of the compound using a glass rod or add a few crystals of the compound to the solution to initiate crystallisation. (e) Filtration Filter the solid using a filter funnel fitted on a filter tube. Allow the solvent to drain completely.

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(f) Drying Transfer the solid from the funnel on to a filter paper, cover it with an inverted funnel or keep in a vacuume dessicator or keep it in an oven maintained at ~ 20-30°C lower than the expected melting point of the compound for drying.

1.5.3 Derivatives of Carbohydrates Following derivatives can be prepared for carbohydrates (i) Acetate (ii) Osazone (i) Acetate: Since carbohydrates are poly hydroxyl compounds, all can be acetylated. Acetylation should be carried out to completion i.e. all hydroxyl (–OH) groups in the given compound should be converted to –OCOCH3 groups. If the acetylation is incomplete, a mixture of partially acetylated compounds are obtained as a result of which the melting point of the derivative is not sharp and does not tally with that reported in literature leading to inconclusive results. Carbohydrates are acetylated with acetic anhydride in presence of either anhydrous sodium acetate or fused zinc chloride to give β-acetate or α-acetate respectively. The –OCOCH3 at carbon no 1 is equatorial in β-acetate and is axial in α-acetate. Given below are the isomeric acetates of glucose and fructose. The β-acetate is kinetically favoured product and the α-acetate is thermodynamically more stable. It is supported by the fact that the β-acetate gets converted to the α-acetate on heating with acetic anhydride and fused zinc chloride.

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Acetic anhydride undergoes very slow hydrolysis on storage but can undergo faster hydrolysis in presence of water to produce acetic acid. The hydrolysis reaction is catalysed by mineral acids. +

H → 2CH3COOH (CH3CO) 2 O + H 2 O 

The acetic anhydride to be used for acetylation should be free from acetic acid. To check this, take a drop of acetic anhydride in a test tube and add to it ~ 1 mL of sodium bicarbonate solution, immediate effervescence indicates the presence of acetic acid in it. If acetic acid is present, treat ~20 mL of acetic anhydride with an aqueous solution of sodium bicarbonate till there is no more effervescence. Remove the acetic anhydride layer using a separating funnel in a dry conical flask, dry over anhydrous calcium chloride for ~15 minutes. Filter and distil acetic anhydride (B.P. 140°C). α-Acetate Procedure Preparation of fused zinc chloride Take ~ 2 g zinc chloride in a clean, dry porcelain dish and heat it on a wire gauze using low flame. Initially it loses the absorbed water (zinc chloride is a hydroscopic substance and hence contains water which it has slowly absorbed from air) and on continuing heating, it melts and starts giving white fumes. Remove from over the flame and continuously stir the liquid mass till it solidifies. Immediately transfer the solid into a dry round bottomed flask (50 mL capacity) and stopper it. Procedure for preparation of α-acetate Add acetic anhydride (10 mL) to the flask containing fused zinc chloride and mix the two well. To this mixture, now add the given carbohydrate (1 g) in small portions and shake to mix. Add a few pieces of pumice stones to the flask. Fix a water condenser on the flask, circulate water through it and heat the reaction mixture in the flask in a hot water bath for about an hour. Allow the reaction to complete (take out a drop or two of the reaction mixture and add it into a 100 mL beaker containing ~ 10 g of crushed ice, stir the mixture with a glass rod, formation of a solid indicates the completion of reaction) otherwise continue heating for 15-20 minutes more and check again. Cool the reaction mixture and pour the contents

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of the flask into a 250 mL beaker containing ~ 100 g of crushed ice. Stir to decompose excess acetic anhydride and allow ice to melt. Filter the solid acetate, dry in air and recrystallise ~ 100 mg of solid using alcohol as a solvent. Filter the crystals, dry and determine the melting point.

β-Acetate Procedure Preparation of anhydrous sodium acetate Sodium acetate is available as trihydrate or monohydrate and as anhydrous form in the laboratory. The anhydrous form slowly absorbs moisture from air and hence all the three forms must be made anhydrous just before starting the acetylation reaction. The trihydrate loses its water of hydration in two stages i.e. ∆ ∆ CH3COONa.3H 2 O  → CH3COONa.H 2 O  → CH3COONa −2 H O −H O

Sodium acetate trihydrate

2

2

Anhydrous sodium acetate from trihydrate Take sodium acetate trihydrate (2 g) in a clean, dry china dish. Heat it on a wire gauze, continuously stirring while heating using a glass rod. The triacetate melts. Continue heating, the melt solidifies and on further heating melts again. Stop heating, allow to cool, stir till a solid (anhydrous sodium acetate) is obtained. Transfer the solid immediately into a dry round bottomed flask and stopper it.

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Anhydrous sodium acetate from monohydrate and anhydrous sodium acetate Take any one of these (2g) in a dry and clean china dish, heat on a wire gauze till the solid melts, allow to cool, stirring continuously while heating and cooling. Transfer to a dry and clean round bottomed flask (50 mL) and stopper. Notes: • Anhydrous sodium acetate is shinning white or very light greyish in colour. • Do not over heat sodium acetate otherwise it chars to a black solid which is not suitable for acetylation reaction. Procedure for preparation of β-acetate Add acetic anhydride (10 mL) to the round bottomed flask containing anhydrous sodium acetate and mix well. Now add the carbohydrate (1g) in small lots and shake to mix all the reactants. Add few pumice stones, Fix up a water condenser on the flask and heat in water bath for 30-40 minutes and follow the procedure as mentioned for α-acetate above. Notes: The heating time for the formation of β-acetate varies from 30 minutes to two hrs. So check for the completion of reaction before working up. Osazone The cyclic structure of some carbohydrates can open up to generate a free –CHO or a C=O group, such carbohydrates are reducing in nature and reduce Fehling and Tollens’ reagents. Carbohydrates containing CHO–CHOH–CHOH–R or CH2OH–CO–CHOH–R i.e. those carbohydrates which have an –OH group present on the α-carbon with respect to –CHO or C=O group react with phenyl hydrazine to form the corresponding osazones. Ribose (V) forms osazone but 2-Deoxy ribose (VI) does not form the osazone as it lacks an –OH group on α-carbon. C6 H5 NH NH 2 ⋅HCl + CH3COONa → C6 H5 NHNH 2 + NaCl + CH3COOH Phenylhydrazine hydrochloride

Phenylhydrazine

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Procedure Take the given reducing carbohydrate (1g) in a clean boiling tube, add water (~5mL) to dissolve the carbohydrate. Separately dissolve phenylhydrazine hydrochloride (2g) and sodium acetate (3g) in water (~10 mL) in a boiling tube. Mix the two solutions in a conical flask (100 mL), cork it loosely and immerse this flask partially in a water bath containing boiling water. Heat the reaction mixture, with occasional swirling, for half an hour.( Note down the time when yellow precipitate first appears in the flask). Allow the reaction mixture to cool to room temperature, filter the solid and allow it to dry in air. Recrystallise ~100 mg of solid using alcohol as a solvent. Dry and determine its melting point. Notes: • Above procedure can be used for the preparation of osazone of reducing mono, di- and tri saccharides. • Weigh all the reactants correctly on a chemical balance otherwise a mixture of products will be formed leading to an incorrect melting point. • Osazones of some carbohydrates like lactose do not separate in hot but are formed only on cooling the reaction mixture after heating it for half an hour. • Time of first separation of yellow solid is important as it may help in identification of some carbohydrates. • Most osazones decompose just before melting i.e. these darken near their melting points. So darkening of the sample should not be confused with melting. Osazone of sucrose: Sucrose is a non reducing sugar and hence does not form osazone but, on hydrolysis it gives an equimolar mixture of glucose and fructose, both of which form the same osazone on reaction with phenylhydrazine. Sucrose can therefore be converted to osazone after hydrolysis. This method is not applicable to those di and trisaccharides which yield a mixture of monosaccharides, which give different osazones having different melting points.

Procedure Take sucrose (1 g) in a round bottomed flask (50 mL), dissolve it in minimum volume of water, add dilute hydrochloric (5 mL) and a few pieces of pumice stones to it. Fix up a water condenser on the flask and heat in a boiling water bath for half an hour. Sucrose is hydrolysed to a mixture of glucose and fructose. Transfer this solution into a conical flask (100 mL), add a solution of phenylhydrazine hydrochloride (2 g) and sodium acetate (3 g), loosely cork the flask and proceed as given above.

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Alternate procedure Alternatively the hydrolysis of sucrose and its conversion to osazone can be done in a microwave oven, where hydrolysis is complete in five minutes at 70°C and subsequent reaction with phenylhydrazine hydrochloride and sodium acetate is completed in just 2-3 minutes of heating. The quantities of the reagents etc. are same as mentioned above. Procedure take sucrose solution as above, add HCl and heat in a microwave at 70°C for five minutes. Take out the hydrolysed solution of sucrose. Add phenylhydrazine hydrochloride and sodium acetate as above. Heat again in microwave for 2 minutes. Work up as given above. Notes: This method is energy efficient as compared to conventional method.

Fig. 1.9: Microwave (Milestone start Microwave Lab Station for Academia)

1.5.4 Derivatives of Carboxylic Acids Carboxylic acids can be converted to the following derivatives: (i) S-Benzylisothiuronium salt (ii) Amide (iii) Anilide (iv) p-Toludide (v) Anhydride (only for a few carboxylic acids) S-Benzylisothiuronium salt Carboxylate ion of the given carboxylic acid can replace the chloride ion of S-Benzylisothiuronium chloride in a neutral medium to produce well defined crystalline salts, which possess sharp melting points when pure and are prepared for characterisation of carboxylic acids

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Procedure Take the unknown acid (500 mg) in conical flask (100 mL), dissolve in minimum volume of water or suspend the acid in water (~ 5-6 mL) and add alcohol (5-6 mL), if it is water insoluble. Add a drop of phenolphthalein indicator. Now add dilute sodium hydroxide solution (10%) drop wise with constant shaking till a permanent light pink coloured solution is obtained. Dissolve S-benzylisothiuronium chloride (2 g) in minimum amount of water (~10 mL) in a separate flask and slowly add it to the light pink coloured solution with constant swirling. A white solid separates. Filter the solid, dry and recrystallise ~200 mg from alcohol. Filter the crystals, air dry and determine melting point. Notes: • The solution of the acid in sodium hydroxide should be almost neutral (indicated by a very light pink colour) at the time of adding S- benzylisothiuronium chloride. • In case it is dark pink in colour, add very dilute hydrochloric acid (1%) drop wise with shaking to make it light pink. • S-benzylisothiuronium chloride undergoes decomposition in strongly acidic or alkaline medium and hence the need of a neutral solution for the reaction. • A strong unpleasant smell during the reaction indicates decomposition of the reagent. Amide Carboxylic acids can be converted to acid amides in two steps (a) converting the acid to acid chloride (b) reacting the acid chloride with ammonia to get the amide because the direct reaction between an acid and ammonia is reversible and does not give good yield of the amide RCOOH + SOCl2 → RCOCl + SO2 + HCl or

RCOOH + PCl5

→ RCOCl + POCl3 + HCl



RCOCl + NH3

→ RCONH2 + HCl

Procedure (a) Preparation of acid chloride: Given acid may be converted to its chloride by two methods (i ) By using thionyl chloride (SOC12): Take the given acid (1 g) in a dry, clean round bottomed flask (50 mL), add (using a dry measuring cylinder) to it, thionyl chloride (5-6 mL) and pumice stones (5-6 pieces). Fix a water condenser on the flask and reflux the reaction mixture in a boiling water bath in a fume chamber for half an hour. Stop heating, allow the reaction mixture to cool (remove the hot water bath from under the flask but do not remove the condenser for faster cooling). Distil excess of unreacted thionyl chloride (boiling point of thionyl chloride is 76°C) in a water bath. The residue in the flask, which does not distil along with thionyl chloride is the acid chloride, which is used for the next step. However, if pure acid chloride is required, it may also be distilled.

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Notes: • All the apparatus should be dried before use. • Acid chloride absorbs moisture from air so it should be stoppered immediately after the above procedure is completed. • Those acid chlorides which have a boiling point close to that of thionyl chloride (76° C) cannot be separated from unreacted thionyl chloride by this method. • Thionyl chloride causes skin and eye irritation. In case of skin contact, wash the affected portion with fresh running water for at least 10 minutes. Do not rub your eyes if there is irritation because of exposure to thionyl chloride, but wash eyes with fresh water and seek medical help immediately. • Sulphur dioxide and hydrochloric acid fumes that are produced as byproducts of the reaction also affect skin, eyes and cause problem in breathing. Eyes and skin should be washed thoroughly with water for at least 10 minutes. In case the irritation continues, seek medical help. For any problem in breathing, immediately take medical help. (ii) Using phosphorus pentachloride (PC15): Take the given acid (1 g) in a clean dry boiling tube, weigh phosphorus pentachloride (4 g) and immediately add this to the tube containing the acid. Mix the two solids well with a glass rod and heat the mixture in a water bath in a fume chamber till the solid mass liquifies or is converted to a paste. Use the crude acid chloride thus formed immediately. Notes: • If required, the acid chloride can be purified by distillation • Use dry apparatus for the reaction • Phosphorus pentachloride is highly hygroscopic so take minimum time for weighing it • Phosphorus pentachloride causes skin irritation, in case of any contact, wash the affected portion with fresh running water for 10-15 minutes • It is harmful for the eyes. In case of irritation in the eyes, do not rub your eyes but wash thoroughly with water for 5 minutes and get medical help immediately • Do not inhale the fumes. In case of any such problem, move out of laboratory and breath in fresh air. Take medical help if necessary. (b) Preparation of amide Take the whole of acid chloride (~l g) obtained by any of the above method, add ammonia (~ 20 mL, specific gravity 0.88) drop by drop to the acid chloride in a fume chamber. Hold the tube with a test tube holder while adding ammonia to the acid chloride as a lot of hydrochloric acid fumes are produced during the reaction. Do not inhale the acid fumes. When whole of ammonia has been added, warm the reaction mixture for ~5 minutes in a hot water bath. Allow the reaction mixture to cool to room temperature and slowly add it to a beaker (100 mL) containing crushed ice (~ 20 g). The amide separates out as a white solid. Filter, dry the solid in air, recrystallise and determine its melting point. Notes: • Avoid using too much of water or ice during work up because some amides are sparingly soluble in water. • Amides of lower fatty acids are difficult to recrystallise, it is better to prepare S-benzylisothiuronium salts as derivatives.

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Anilide Like acid amides, the acid anilides are also made in two steps SOCl2 / PCl5

RCOOH → RCOCl RCOCl + C6H5NH2 —————→ RCONH C6 H5 + HCl Procedure

Anilide

Take the acid chloride (~ 1 g, prepared as given above) in a dry and clean round bottomed flask. Add to it, aniline (0.5 mL) and reflux on a water bath for -10 minutes in a fume chamber. Cool the reaction mixture and pour it over crushed ice (~ 50 g) in a beaker, (i) A solid is formed, filter it. Take the solid in a beaker, add a saturated sodium bicarbonate solution in small lots till the effervescence ceases. Filter the solid again, dry, recrystallise with alcohol and determine its melting point. (ii) If a solid does not separate out, decant the water. First add cold dilute hydrochloric acid to the beaker to remove any unreacted aniline, remove the acid layer and then treat the residue with sodium bicarbonate solution till the effervescence ceases. The treatment with bicarbonate neutralises the hydrochloric acid and also dissolves the carboxylic acid that may be present. Now an insoluble solid that remains in the beaker is the anilide of the given acid. Filter it, wash with water, dry, recrystallise from alcohol and determine its melting point. Notes: Alternatively, dissolve the remaining product in the beaker in ether (caution). Take the ether solution in a separating funnel, wash the ether layer first with dilute hydrochloric acid and then with sodium bicarbonate solution very cautiously. Dry the ether layer over anhydrous sodium sulphate for ~10 minutes. Filter and distil or evaporate ether (do not work with ether in laboratory where other students are using burners for their work). The solid residue is crude anilide, recrystallise and determine its melting point. p-Toludide Like anilides, the p-toludides can also be made for identification of carboxylic acids

Procedure Same as for anilides. Anhydride Only those di-carboxylic acids which can lose a water molecule on simple dry heating to form a solid anhydride can be converted to the anhydride derivative. It is also important to remember that only those anhydrides which have a five or a six membered cyclic ring structure are stable, other cyclic anhydrides are either not formed or decompose on heating. For example, succinic acid forms a stable anhydride, phthalic acid also forms the anhydride but oxalic anhydride and malic anhydride are not obtained. In general those dicarboxylic acids which give a positive fluorescein test can be converted to their anhydrides

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Fig. 1.10: Sublimation

Procedure Take the given acid (~500 mg) in a clean, dry porcelain dish. Take a circular filter paper marginally bigger than the circumference of the porcelain dish, punch small holes in it. Cover the dish containing the acid with it and keep an inverted funnel on it. Loosely plug the open end of the funnel with cotton. Now heat the porcelain dish on a sand bath, kept on a tripod stand using a small flame. The anhydride that is formed collects on the filter paper under the funnel. Stop heating, allow the apparatus to cool, collect the anhydride and determine its melting point.

1.5.5 Derivatives of Phenols Following derivatives can be prepared from a phenol (i) Benzoate (ii) 3,5-dinitrobenzoate (iii) p-Nitrobenzoate (iv) Bromo derivative (v) Acetate Benzoate Almost all phenols can be converted to water insoluble benzoates. These benzoates are made by Schotten Bauman method, in which the phenol to be benzoylated is treated with benzoyl chloride in an alkaline medium. Following reaction takes place between (i) phenol and (ii) resorcinol with benzoyl chloride

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Sodium hydroxide is used during the reaction to perform the following functions: (a) Excesses of benzoyl chloride is required for complete reaction. Even if a small amount of unreacted benzol chloride is left in the reaction mixture, it does not allow separation of the solid product. Sodium hydroxide slowly hydrolysis the unreacted benzol chloride, converting it into sodium benzoate, which remains dissolved in the reaction mixture, thus allowing the separation of solid benzoyl derivative. (b) Sodium hydroxide also activates the phenol by converting it into the phenoxide, which is more reactive than the phenol as a nucleophile and reacts faster with benzoyl chloride.

(c) Sodium hydroxide also keeps the unreacted phenol in solution thereby preventing the benzol derivative from contamination with phenol. Procedure Dissolve the given phenol (500 mg) in a clean conical flask (100 mL) in freshly prepared sodium hydroxide solution (15 mL, 15%). Add benzoyl chloride (1 mL) to the phenol solution in small lots and shake the reaction mixture vigorously after each addition (use a bark cork to stopper the flask for shaking). The completion of reaction is indicated by the separation of a solid product, absence of any smell of benzoyl chloride (do not smell the flask directly) and alkaline pH of reaction mixture (use litmus paper to check). Filter the solid, after the whole of the mother liquor has been filtered, add some water on the solid to wash the adhering alkali and filter. Recrystallise ~ 100 mg of the solid from alcohol and determine its melting point. Notes: • The above quantities of benzoyl chloride are for a monohydric phenol, the quantities are doubled for a dihydric phenol. • A dark colour of the alkaline solution at the completion of the reaction indicates the presence of some unreacted phenol, add some more benzoyl chloride ( a few drops) in such a case and shake well to complete the reaction. • Add some more sodium hydroxide, if the solution is not alkaline. • Break slowly, if lumps of a solid are formed during the reaction to release any benzoyl chloride that gets trapped in these lumps.

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3,5- Dinitrobenzoate Phenols form crystalline 3,5-dinitrobenzoates on reaction with 3,5-dinitrobenzoyl chloride.

Procedure Prepare 3,5-dinitrobenzoyl chloride by any of the method given on page 96. Take the 3,5-dinitrobenzoyl chloride (0.5 g) in a clean and dry boiling tube, add the given phenol (1 g) to it. Warm the contents of the boiling tube for 10-15 minutes in a hot water bath in a fume cupboard, continuously stir the mixture with a glass rod while heating it. When the reaction is complete (the completion of the reaction can be known by taking out a drop of the reaction mixture, after heating it for 15 minutes and adding it into a beaker (100 mL) containing a few pieces of crushed ice and stirring. Formation of a solid indicates the completion of the reaction), transfer the contents of the boiling tube into a beaker containing crushed ice (100 g) and stir. When a solid is formed, allow ice to melt and filter the solid. Take the solid in a beaker and add a saturated solution of sodium bicarbonate (10 mL). Stir till effervescence ceases, add more bicarbonate solution till there is no more effervescence. Filter the solid and recrystallise ~100 mg of it from alcohol and determine its melting point. Notes: • HC1 fumes are released during the reaction and phosphoruspentachloride or thionyl chloride are used for preparation of 3,5-dinitrobenzoyl chloride, follow all first aid precautions mentioned on page 96 and 97. • The 3,5-dinitrobenzoic acid is removed by treating the 3,5-dinitrobenzoate with sodium bicarbonate solution. p-Nitrobenzoate Follow the procedure described above for 3,5-dinitrobenzoates. Bromo derivatives Phenols with an unsubstituted o- and/or p-position with respect to –OH group can undergo electrophillic substitution reaction with bromine at these positions. The bromo substituted compounds should only be prepared as derivatives if no other choice is available because (i) bromine is very corrosive for the skin and causes severe burns in case of an accidental spill (ii) bromine vapours are very harmful if inhaled (iii) many a times the bromo derivatives are difficult to recrystallise. The reaction between bromine and phenol and resorcinol are given below:

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Procedure Take the given phenol (250 mg) in a clean and dry 100 mL conical flask and dissolve in minimum amount of glacial acetic acid (~ 5-7 mL). Add (carefully) to it bromine solution in acetic acid dropwise from a burette with constant swirling. Initially the colour of the bromine fades very fast but as the reaction approaches completion, the speed of the fading of the colour of bromine drops. The completion of reaction is indicated by a permanent yellow colour of the reaction mixture on standing for 15 minutes. In some cases the bromo compounds start separating out as a solid during the reaction while in others, a solid product does not separate. Pour the reaction mixture in both the cases in a beaker containing crushed ice (~ 100 g) and stir. A solid product separates with the supernatant liquid having a light yellow or a darker colour due to the presence of unreacted excess bromine. For removing the colour, make a saturated solution of sodium bisulphite in water and add it to the yellow coloured reaction mixture slowly with stirring till the solution becomes colourless. Filter the bromo derivative, dry and recrystallise from ethanol or ethanol-water. Notes: • Bromine is highly corrosive, handle it very carefully. • In case of irritation on skin or eyes due to exposure to bromine vapours, wash the affected portion with flowing water for 10-15 minutes, do not rub eyes and seek medical help if irritation continues. Immediately move out of the laboratory in case of any problem in breathing and seek medical help. • Add only a slight excess of bromine to ensure complete bromination of phenol. • Any excess of unreacted bromine is removed by adding sodium bisulphite NaHSO3 + Br2 + H2O → NaHSO4 + 2HBr or SO2 + 2H2O + Br2 → H2SO4 + 2HBr or H2SO3 + Br2 + H2O → H2SO4 + 2HBr • Bromine is also a mild oxidising agent, hence bromo derivatives of some phenols which are easily oxidised (like hydroquinone, phloroglucinol and pyrogallol) are difficult to prepare. Acetate Acetates of phenols are best prepared by their reaction with acetic anhydride in presence of sodium hydroxide

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Procedure Dissolve the given phenol (1 g) in freshly prepared sodium hydroxide solution (15 mL 10%) in a 100 mL conical flask. Add a few pieces of crushed ice and then acetic anhydride (0.4 mL) to the phenol solution. Shake till a solid is formed. Filter the solid, dry and recrystallise ~ 100 mg of it from water or dilute alcohol and determine its melting point.

1.5.6 Derivatives of Aldehydes and Ketones Following derivatives can be prepared for both aldehydes and ketones (i) 2,4-dinitrophenylhydrazones (ii) Semicarbazone (iii) Hydrazones (iv) Phenylhydrazone (v) p-Nitrophenylhydrazone (vi) Iodoform (for those aldehydes and ketones which contain a –COCH3 group). (vii) Oxidation with alkaline potassium permanganate. 2,4-Dinitrophenylhydrazone All aldehydes and ketones react with 2,4-dinitrophenylhydrazine to form 2,4-dinitrophenylhydrazones. The reaction occurs in the acidic medium. The acid helps to activate the carbonyl carbon for attack by the weak nucleophile i.e. 2,4-dinitrophenylhydrazine. The acid is also required to help dissolve 2,4-dinitrophenylhydrazine in alcohol in which it is otherwise insoluble and it helps in removal of water molecule in the formation of final product. The reaction takes place as follows:

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Procedure Suspend 2,4-dinitrophenylhydrazine (0.2 g) in methanol/ethanol (10 mL) in a 100 ml conical flask. Add cautiously concentrated hydrochloric acid (1 mL) to the suspension dropwise with constant shaking. The 2,4-dinitrophenylhydrazine dissolves. In case some solid particles do not dissolve, filter to get a clear solution. (if a lot of yellow precipitate is formed due to addition of excess of acid, the reagent is not suitable for further reaction). Add dropwise the solution of carbonyl compound (0.2 g dissolved in minimum amount of methanol/ethanol) to the clear solution of 2,4 dinitrophenylhydrazine. A yellow or orange precipitate is obtained. Allow the reaction mixture to cool down to room temperature, filter the solid and recrystallise ~ 100 mg from alcohol or glacial acetic acid and determine its melting point. Notes: • If the given carbonyl compound is water soluble, its aqueous solution can be treated with an aqueous solution of 2,4-dinitrophenylhydrazine, which has been prepared following the procedure described for the preparation of alcoholic solution of 2,4-dinitrophenylhydrazine. • Never use alcoholic solution of 2,4-dinitrophenylhydrazine for aqueous solution of the compound in the above preparation because the reagent and not the derivative separates out. • Concentrated sulphuric acid can be used in place of hydrochloric acid in the above procedure. • Aliphatic compounds normally give a yellow whereas aromatic compounds give an orange coloured precipitate of 2,4-dinitrophenylhydrazone. Semicarbazone Semicarbazones of aldehydes and ketones are obtained by their reaction with semicarbazide hydrochloride in presence of sodium acetate. The reaction occurs according to the following equations NH2CONHNH2.HCl + CH3COONa → NH2CONHNH2 + NaCl + CH3COOH Semicarbazide hydrochloride

Semicarbazide has two nucleophillic ends marked 1 and 2 in the structure below, which can attack the electrophillic carbon of C=O group. The attack take place only through 1 as it is a stronger nucleophillic position in semicarbazide. The lone pair on –NH2 marked 2 is in resonance with the carbonyl carbon of semicarbazide and hence not available for attack on the C=O group of aldehydes or ketones.

The next question is why is semicarbazide hydrochloride used and not semicarbazide itself? The reason is: free semicabazide is unstable and slowly undergoes oxidation by air and to prevent it from undergoing oxidation, it is stored as its salt like hydrochloride. Whenever required it is liberated from its salt by addition of a base like sodium acetate NH2CONHNH2.HCl + CH3COONa → NH2CONHNH2 + CH3COOH + NaCl The acetic acid produced in the above reaction helps in the formation of semicarbazone.

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Procedure Dissolve the given carbonyl compound (1 g) in minimum amount of water or in minimum amount of methanol (if it is water insoluble) in a conical flask. Separately dissolve semicarbazide hydrochloride (1 g) and sodium acetate (2 g) in minimum amount of water, mix the two solutions immediately and shake. A white coloured semicarbazone separates. Filter the solid and recrystallise ~ 200 mg of it from water or dilute alcohol. Notes: • Do not keep the solution of semicarbazide hydrochloride and sodium acetate for long as the free semicarbazide obtained will be slowly oxidised. • Many semicarbazones are soluble in water or alcohol so use minimum amounts of these in the experiment. • If the semicarbazone does not separate out on shaking as mentioned in the procedure above, heat the reaction mixture in a water bath for ~5 minutes and cool. Prolonged heating should be avoided as some semicarbazide may undergo self condensation and form hydrazodicarbonamide (NH2CONHNHCONH2), m.p. 246°C. Oxime Aldehydes and ketones form oximes on reaction with hydroxylamine hydrochloride and sodium acetate. Like semicarbazide the hydroxylamine is also sensitive to aerial oxidation and preserved as its hydrochloride

Procedure Dissolve the carbonyl compound (1 g) in minimum amount of water or ethanol in a clean round bottomed flask and immediately add a freshly prepared solution of hydroxylamine hydrochloride (1g) and sodium acetate (2g) made in minimum amount of water. Mix and reflux in a water bath for 10-15 minutes. Cool the reaction mixture, a solid oxime separates out. If a solid is not formed, induce solid formation by scratching the inner sides of the flask with a glass rod. Filter, recrystallise from water or dilute alcohol and determine its melting point. Phenylhydrazone Aldehydes and ketones react with phenylhydrazine hydrochloride in presence of sodium acetate to give crystalline phenyl hydrazones.

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Procedure Take the given carbonyl compound (0.5g) dissolve in water or ethanol (~ 5 mL) in a boiling tube and add to it a freshly prepared solution of phenylhydrazine hydrochloride (1 g) and sodium acetate (2 g) in minimum amount of water. Heat in a boiling water bath for 5-10 minutes. Cool, filter, recrystallise and determine the melting point of the phenylhydrazone. Notes: Alternate procedure Take phenyl hydrazine (1 g) and dissolve it in water (~4-5 mL) in a boiling tube. Add to it a solution of the given carbonyl compound (0.5 g) in alcohol (10 mL). Boil the mixture for a minute. Now add glacial acetic acid (5-6 drops) and again heat for ~3-4 minutes. Cool, filter, recrystallise from alcohol and determine its melting point. p-Nitrophenylhydrazone

Procedure Same as the alternative procedure described above for phenyl hydrazone. Iodoform Aldehydes and ketones which contain a –COCH3 group react with iodine in presence of sodium hydroxide to form iodoform, a yellow coloured water insoluble product. Though iodoform is not a specific derivative for any specific compound but may still help in final confirmation of the compound. The reaction is given below:

Procedure Saturated solution of iodine in aqueous potassium iodide: take potassium iodide (5 g) and dissolve it in ~10 mL water in a conical flask. Add solid iodine in small lots to the potassium iodide solution with shaking till no more iodine dissolves. Now take the given compound (250 mg) in a 100 mL conical flask and dissolve in water (~10 mL), if the compound is insoluble in water, dissolve in methanol (5-7 mL). Add sodium hydroxide (3-4 mL, 10%) to the solution of the compound and then add the saturated iodine-potassium iodide slowly with shaking, initially the colour of iodine solution disappears fast and then slowly and finally the colour persists. At this stage stop adding iodine solution and shake for 2-3 minutes. Heat the reaction mixture is a hot water bath for 1-2 minutes, if the colour of iodine still persists, add to it a few drops of sodium hydroxide solution to remove the colour. A colourless solution with yellow precipitate is obtained, cool, filter the precipitate, recrystallise from alcohol and determine melting point. (Lit. m.p. 119°C)

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Oxidation to carboxylic acids Aromatic aldehydes like benzaldehyde can be oxidised to the corresponding acids by oxidation with alkaline potassium permanganate.

Procedure Take the given compound (1 mL) in a small round bottomed flask, add to it a solution of sodium carbonate (1g) and a solution of potassium permanganate (1.25g) slowly with shaking. Initially the colour of potassium permangate fades fast but slows down after some time. When the colour of permanganate persists, fix a condenser and set up the reaction for refluxing on a wire gauze. Reflux for 10-15 minutes and add potassium permanganate slowly till its colour persists. Stop heating, allow the solution to cool down to room temperature. Filter the brown precipitate of manganese dioxide (some fine particles of MnO2 may also pass into the filtrate). Acidify the filtrate containing sodium salt of acid, some dissolved unreacted potassium permanganate and manganese dioxide with dilute hydrochloric and then add a saturated solution of sodium sulphite slowly with stirring till the supernatant solution becomes colourless and a white precipitate of acid separates out. Filter the solid, wash with water and dry. Recrystallise ~100 mg solid from water. Determine its melting point Notes: • Acidified sodium sulphite is added to remove MnO2 and KMnO4. 3KMnO4 + 2HCl + 3NaHSO3 —→ 2MnO2 + 3NaHSO4 + 2KCl + H2O HCl + NaHSO3 —→ H2SO3 + NaCl MnO2 + H2SO3 —→ MnSO4 + H2O • SO2 gas may be passed into the filtrate to remove KMnO4 and MnO2. • Since SO2 is produced or used during the work up, the procedure must be performed in a fume chamber. SO2 is injurious for respiratory system. Do not inhale it. • Potassium permanganate causes irritation in the eyes and skin (also stains the skin brown). Wash the effected area with plenty of fresh water and seek medical help if necessary. • Aromatic aldehydes containing electron withdrawing groups like nitro substituted aldehydes can be oxidised to the corresponding carboxylic acids by this method. • Aromatic aldehydes containing strong electron donating groups like –NH2 or –OH etc. cannot be oxidised by this method. In these compounds the electron rich aromatic ring undergoes oxidative degradation on reaction with alkaline potassium permanganate.

1.5.7 Derivatives of Alcohols Following derivatives are prepared for an unknown alcohol: (i) 3,5-dinitrobenzoates (ii) p-nitrobenzoate (iii) Iodoform 3,5-Dinitrobenzoates These are conventionally prepared in two steps (i) conversion of 3,5-dinitrobenzoic acid to 3,5-dinitro benzoyl chloride (ii) reaction of 3,5-dinitrobenzoyl chloride with the given alcohol.

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The reaction takes place under anhydrous conditions because in presence of moisture both phosphorus pentachloride and thionyl chloride get hydrolysed and if more water is present they react violently producing harmful gases ( see reaction below) and thus do not react with the 3,5-dinitrobenzoic acid. 3,5-dinitrobenzoyl chloride is also slowly hydrolysed by water thereby interfering in its reaction with alcohol. PCl5 + H2O → POCl3 + 2HCl SOCl2 + H2O → SO2 + 2HCl Procedure Preparation of 3,5-dinitrobenzoyl chloride Take 3,5-dinitrobenzoic acid (1 g) and convert it to chloride by any of the methods given on page 96-97. Procedure for preparation of 3,5-dinitrobenzoate Take 3,5-dinitrobenzoyl chloride (prepared using 1.25g of 3,5-dinitrobenzoic acid) in a dry boiling tube and add the given alcohol (1 mL) dropwise with continuous stirring using a glass rod in a fume chamber. Hydrochloric acid fumes are produced continuously, do not inhale these. After the addition of alcohol, warm the reaction mixture in a hot water bath for 5-10 minutes for completion of reaction, (for checking completion, take out a drop of the reaction mixture from the boiling tube and add it over ice (~10 g) in a boiling tube, formation of a solid on stirring indicates the completion of the reaction). If the reaction is complete, pour the remaining reaction mixture over ice (100 g) in a beaker and stir but if the reaction is incomplete, heat the reaction mixture for some more time and then work up. Filter the solid, take it in a 250 mL beaker and treat with saturated sodium bicarbonate solution till the effervescence ceases. Filter again, dry and recrystallise ~100 mg from alcohol. Determine its melting point. Notes: • Use dry apparatus for the reaction. • Excess of unreacted 3,5-dinitrobenzoic acid is removed by reaction with sodium bicarbonate solution. • If the given alcohol is water soluble it causes no difficulty in separation of the 3,5-dinitrobenzoate but if the given alcohol is water insoluble and even if a small amount of it remains unreacted at the end of reaction, it prevents the formation of a solid derivative. Hence for a water insoluble alcohol, a slight excess of 3,5-dinitrobenzoic acid (0.25g) should be used which is finally removed during treatment with sodium bicarbonate solution.

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Alternative method of preparing 3,5-dinitrobenzoate The 3,5-dinitrobenzoates of alcohols can be prepared by direct reaction between 3,5-dinitrobenzoic acid and alcohol in one step

Take 3,5-dinitro benzoic acid (1.5 g) in a dry round bottomed flask (50 mL). Add to it the given alcohol (1 mL) and concentrated sulphuric acid (5-6 drops). Shake to mix well and slide carefully a small magnetic bar in the flask. Now keep the round bottomed flask in the microwave oven and attach a condenser as shown in Fig. 1.9. Circulate water if the given alcohol has a boiling point < 120°C. Reflux the reaction mixture with stirring in the microwave for 2 + 5 minutes at 70°C (2 minutes for reaching 70°C then maintain the reaction mixture at 70°C for 5 minutes). Take out the reaction mixture from inside the microwave, cool and pour the contents of the flask over crushed ice (~ 100 g) in a beaker. Stir using a glass rod. A white solid of crude 3,5-dinitro benzoate separates. Filter it and transfer the solid to a clean beaker. Add a saturated aqueous solution of sodium bicarbonate to the beaker till the reaction ceases (no effervescence is observed). Filter the remaining solid, recrystallise from alcohol and determine its melting point. Notes: • This method eliminates the use of hazardous chemicals like phosphorus penta chloride and thionyl chloride. • No harmful byproducts like hydrochloric acid and sulphur dioxide are produced. • It is a time and energy efficient method. • Concentrated sulphuric acid is the only hazardous chemical that is used but in very small amount. • The major drawback of this modified method is that it works only for primary alcohols ( except benzyl and furfuryl alcohol). • As in conventional method, for water insoluble alcohol, the 3,5-dinitrobenzoic acid should be in slight excess. p-Nitrobenzoate Few alcohols form crystalline p-nitrobenzoates. The p-nitro benzoate, like the 3,5-dinitro benzoate is conventionally prepared in two steps (i) conversion of p-nitrobenzoic acid to p-nitrobenzoyl chloride (ii) reaction of p-nitrobenzoyl chloride with given alcohol

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Procedure The procedure used for preparation of p-nitrobenzoates is exactly the same as described above for 3,5-dinitrobenzoate. Alternative method The procedure adopted for 3,5-dinitrobenzoate can be used for p-nitrobenzoate. Notes: This method works only for primary alcohols. Oxidation of side chain of aromatic alcohols The carbon side chains present in aromatic alcohols like benzyl alcohol (C6H5CH2OH) can be oxidised to the carboxylic acid by alkaline potassium permanganate solution. The resultant acid thus produced is characterised by its melting point. KMnO / NaOH

4 C6 H5CH 2OH  → C6 H5COO − [O ] A side chain must have at least one bezylic hydrogen to be oxidised by alkaline potassium permanganate. For example side chains like –CH 2OH and –CHOH– can be oxidised but not C6H5C(OH)(R2).

Procedure See page 107 lodoform As already mentioned, iodoform is not a specific derivative for any compound, it is formed by all carbonyl compounds containing a –COCH3 group. Similarly, all those alcohols which have a CH3CHOH– group form iodoform. A few examples are ethyl, isopropyl and s-butyl alcohol etc. The CH3CHOH– group of these alcohols is oxidised by sodium hypoiodide (which is formed by reaction between iodine and sodium hydroxide) to CH3CO– group, which then reacts with iodine in presence of sodium hydroxide to form iodoform.

Procedure Follow the procedure given under carbonyl compounds.

1.5.8 Derivatives of Esters The esters are O-alkyl or aryl substituted acids, for their complete identification, both the acid as well as the alkyl or aryl group has to be identified. The acid is obtained and identified by alkaline hydrolysis of the given ester followed by acidification. RCOOR′ + NaOH → RCOONa + R′OH RCOONa + HCl → RCOOH + NaCl The given ester is subjected to trans-esteriflcation in presence of 3,5-dinitrobenzoic acid for identification of –O alkyl or –O aryl part of the ester

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The alkyl or aryl group in the above reaction produces a crystalline 3,5-dinitrobenzoate. Identification of acid part of the ester An ester on alkaline hydrolysis (also known as saponification) gives the sodium salt of the carboxylic acid from which, the acid can be obtained on acidification of the alkaline solution with a mineral acid like dilute sulphuric acid. If the acid thus obtained is a water insoluble solid, it can be filtered, recrystallised and determination of its melting point will established its identity. For example, esters of benzoic acid give benzoic acid—a water insoluble acid which can be identified from its melting point.

However, if the ester gives a water insoluble liquid or a water soluble carboxylic acid, the acid has to be converted to a crystalline derivative for its identification. For example, ethyl acetate on hydrolysis gives a water soluble carboxylic acid i.e. acetic acid, which has to be converted to S-benzyl isothiuronium salt for its characterisation.

In addition to identification of acid, the alkaline hydrolysis reaction can also be used to preliminarly distinguish between two esters like for example ethyl and methyl benzoate. Ethyl benzoate on hydrolysis gives ethyl alcohol which gives a positive iodoform test where as methyl alcohol, which is obtained from methyl benzoate gives a negative iodoform test.

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Procedure Take the given ester (1 mL) in a clean round bottomed flask, add to it freshly prepared sodium hydroxide solution (15 mL, 10%) and few pieces of pumice stones. Clamp the flask and fix a water condenser on it. Circulate water through the water condenser and reflux the reaction mixture on a wire gauze for half an hour or till the reaction is complete. The completion of reaction is indicated by disappearance of the insoluble ester layer and its smell (do not smell directly). In case the ester produces a water insoluble alcohol or phenol, the progress and the completion of the reaction can be monitored by TLC ( see chapter 6, page 398). Allow the reaction mixture to cool, distil the alkaline solution to collect ~ 3-4 mL of the distillate and allow the residue to cool. Transfer the residue from the flask into a beaker (250 mL) containing crushed ice (~100 g) and acidify with dilute sulphuric acid (check with blue litmus paper). Following may be observed on acidification: (i) A water insoluble acid is obtained. Filter, recrystallise with a suitable solvent and determine its melting point. (ii) A water soluble acid is obtained on acidification. Take the solution, add a drop of phenolphthalein indicator and neutralize with a very dilute sodium hydroxide solution to obtain a permanent light pink coloured solution. Add to it the solution of S-benzylisothiuronium chloride as given on page 95 to get the derivative of the acid. Perform iodoform test with the distillate (see procedure on page 106) and note whether the iodoforrn is formed or not. (See on page 110) to know which type of alcohols give a +ve iodoforrn test). The alkaline hydrolysis of the ester can be carried out in a microwave oven to save energy and time. The hydrolysis is complete in most cases in 5-10 minutes of refluxing at ~ 90-100° C. The work up procedure is same as mentioned above. Notes: • Alkaline hydrolysis of the esters may be done with alcoholic alkali (i.e. by making 15% solution of sodium hydroxide in alcohol). The method works better for some water insoluble and sterically hindered esters. However, in this method of hydrolysis, the iodoforrn test cannot be performed. • Instead of acidification of the entire alkaline solution for obtaining the acid. Acidity a test portion of it. If a water insoluble acid is obtained, acidify the entire solution to get the acid. If however, no solid is formed on acidification of the test solution, do not acidify the remaining alkaline solution. Add a drop of phenolphthalein to the alkaline solution, a dark pink colour is obtained. Now neutralise the dark pink solution with a very dilute solution of hydrochloric acid to obtain a light pink coloured solution and treat it with S-Benzylthiuronium chloride as described on page 95. Identification of the -O alkyl or -O aryl group of the ester Procedure Take the given ester (1 mL) in a clean, dry boiling tube, add 3,5-dinitrobenzoic acid (1.25 g) and carefully add concentrated sulphuric acid (5-6 drops) to the boiling tube. Stir the mixture using a dry glass rod. Heat the reaction mixture on a low flame in a fume chamber on a sand bath for ~10-15 minutes. Stir the mixture with a glass rod while heating, it normally becomes slightly dark but should not char on heating. Stop heating and check for the completion of the reaction by taking out a drop of reaction mixture and adding it over a small amount of ice or ice cold water in a test tube. Separation of a solid in the tube indicates that the reaction is complete. Pour the remaining reaction mixture over ice cold water (~70 mL) in a beaker and stir. Filter the solid and treat it with a saturated solution of sodium bicarbonate in a beaker till the effervescence ceases as described on page 108, filter the remaining solid, recrystallise from alcohol and determine its melting point.

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Alternate method Reflux with stirring (use a magnetic stir bar) in a microwave oven a mixture of 3,5-dinitrobenzoic acid (1.25g), ester (1 mL) and conc. sulphuric acid (5-6 drops) in a round bottomed flask for 2+5 minutes at 70°C work up the reaction as mentioned above.

1.5.9

Derivatives of Anhydrides

Hydrolysis Acid anhydrides can be hydrolysed to corresponding sodium salt/s of carboxylic acid. + H2O (RCO)2O + 2NaOH → 2RCOONa + ↓H

RCOOH

Procedure Reflux the anhydride (1 g) with freshly prepared sodium hydroxide (10 mL, 10%) for 15-20 minutes. The completion of the reaction is indicated by complete dissolution of the starting compound. Allow the reaction mixture to cool, transfer to a beaker, add ice and acidify with dil hydrochloric acid. Filter, dry, crystallise and determine melting point if a solid acid is formed. If a water soluble or liquid acid is formed, convert it into S-benzylisothiuronium salt as described on page 95. Notes: Alkaline hydrolysis can be used only for identification of symmetrical or cyclic anhydride.

1.5.10 Derivatives of Hydrocarbons Aromatic hydrocarbons can be converted to the following derivatives (i) Nitro derivatives (ii) Picrate (iii) Oxidation to quinones (iv) Oxidation of the side chain. (See page 107) Nitro derivatives: Most aromatic hydrocarbons can be nitrated with a mixture of nitric and sulphuric acid to produce solid nitro substituted compounds but those like anthracene and phenanthrene tend to undergo oxidation with the nitric acid and hence cannot be nitrated using the method described below. Nitration is carried out with a mixture of conc./fuming nitric acid and conc. sulphuric acid and is very slow in absence of sulphuric acid. The nature of the product obtained depends on the conditions employed for nitration

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Procedure Take the given aromatic hydrocarbon (0.5 mL/0.5 g) in a clean, dry round bottomed flask. In a dry boiling tube, mix carefully concentrated nitric acid (3 mL) and concentrated sulphuric acid (3 mL). Add carefully, dropwise, with shaking the nitration mixture to the compound in the round bottomed flask, fix a water condenser and reflux the reaction mixture on a wire gauze/water bath for ~30 minutes. Check for the completion of the reaction by taking out a drop of the reaction mixture and adding it to crushed ice (10 g) in a boiling tube, formation of a solid on stirring indicates that the reaction is complete. Now work up the rest of the reaction mixture by pouring onto crushed ice (~ 50 g) in a beaker. Filter the solid nitro compound, wash with water (5 mL) twice, filter, dry, recrystallise from alcohol and determine its melting point. Notes: • The time required and the temperature varies for different compounds. • Both conc. nitric and sulphuric acids are corrosive chemicals and should be handled carefully. Picrate Polynuclear hydrocarbons can easily be converted to picrates, most of these derivatives are stable and can be easily crystallized.

Procedure Dissolve separately picric acid (0.5 g) and the given hydrocarbon (1 mL) in minimum amount of ethanol in two boiling tubes and mix the two solutions. A yellow solid (sometimes the picrate may have a different colour) separates out. If a solid is not obtained on mixing, warm the solution in a hot water bath for ~5 minutes and keep aside for cooling. A solid picrate will separate out. Filter the solid, recrystallise from alcohol and determine its melting point. Notes: • Picric acid is stored in dark coloured bottles under a layer of water. So for weighing it, take out the solid from the bottle and dry in air for 3-4 minutes by spreading it on a filter paper. • Piric acid stains the skin, do not allow any skin contact. • Acetone, glacial acetic acid or diethyl ether (ether is highly flammable, do not use it in the laboratory where other students are using burners) can be used to dissolve the hydrocarbon and picric acid instead of alcohol.

1.5.11 Derivatives of Ethers Aliphatic ether Symmetrical dialkyl ethers can be converted to alkyl-3, 5-dinitrobenzoates

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Procedure Reflux the given ether (1 mL) anhydrous zinc chloride (0.15 g) and 3,5-dinitrobenzoyl chloride (0.5 g, see procedure for its preparation on page 96-97) for one hour. Allow the reaction mixture to cool and add sodium carbonate solution (10 mL) warm the mixture in a boiling water bath for ~ 2-3 minutes, cool and filter the solid. Recrystallise from alcohol and determine its melting point. Aromatic ethers These can be converted to crystalline picrates

Procedure Follow the procedure described on page 114 for aromatic hydrocarbons.

1.5.12 Derivatives of Amines Primary and secondary amines, both contain hydrogen/s attached to nitrogen which is/are acidic in nature and can be replaced by electrophiles like –COCH3 and –COC6H5. Hence both primary and secondary amines can be converted to acetyl, benzoyl and p-toluene sulphonyl derivatives. Tertiary amines do not have a replaceable hydrogen and hence cannot be acetylated or benzoylated. All the amines can be converted to quaternary salts, methiodides or ethiodides. Some aromatic tertiary amines with a free p-position can be converted to p-nitroso derivatives. Picrates and bromo substituted compounds can also be prepared. 1.5.13 Primary Amines Primary amines can be converted to the following derivatives: (i) Acetyl derivatives (ii) Benzoyl (iii) p-Toluene sulphonyl (iv) Picrates (v) Bromo derivatives. Acetyl derivatives Almost all aliphatic and aromatic primary amines can be acetylated. Acetic anhydride or acetic acid is used as the acetylating agents and out of the two, acetic anhydride is generally used in the laboratory. Since most amines are insoluble in water (the medium used for acetylation reaction), the given primary amine is first dissolved in dilute acid solution and then treated with acetic anhydride and sodium acetate. Sodium acetate liberates the amine from its salt which then reacts with acetic anhydride to give the acetyl derivative

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C6H5NH2.HCl + CH3COONa → C6H5NH2 + NaCl + CH3COOH

Aniline hydrochloride

C6H5NH2 + (CH3CO)2O → C6H5NHCOCH3 + CH3COOH Acetanilide

Procedure Dissolve the given amine (0.5 mL) in minimum amount of dilute hydrochloric acid (3-4 mL) in a 100 mL conical flask and add (10 mL) of ice cold water to it. Separately prepare a solution of sodium acetate by dissolving 2 g in 4 mL water. First add acetic anhydride (4 mL) to the flask containing amine solution, mix and immediately add the sodium acetate solution also. Shake, a white coloured acetyl derivative separates out. Filter the solid, dry, recrystallise from water or aqueous alcohol and determine its melting point. Notes: • Sometimes a diacetyl derivative i.e. RN(COCH3)2 of the primary amine may also be produced in the reaction. It does remain as such but gets converted to RNHCOCH3 during recrystallisation from water or aqueous alcohol and hence the crystalline product is pure. • If the given amine is not freely soluble in the acid, either suspend the amine in the acid or add some more hydrochloric acid to dissolve the amine and add more sodium acetate so as to neutralize the excess acid used. • If a solid acetyl derivative does not separate on shaking, cool the reaction mixture in ice cold water. • Acetyl derivatives of some amines are sparingly soluble in water and hence avoid use of excess water. • If the amine is water soluble: dissolve it (0.5 g) in ice cold water (10 mL) and add acetic anhydride. Shake to get the acetyl derivative. Benzoyl derivative Primary amines (aliphatic and aromatic) can be easily converted to their crystalline benzoyl derivatives by Schotten Baumann Reaction. In this method, the amine is treated with benzoyl chloride in presence of sodium hydroxide.

Procedure Suspend the amine (0.5 g) in freshly prepared sodium hydroxide solution (10 mL, 10%) in a 100 mL conical flask. Cork the flask with a bark cork. Take benzoyl chloride (1 mL) in a test tube and add it in small lots to the suspension of amine, shaking the flask after each addition. After the addition of benzoyl chloride, break the lumps of solid if these have been formed. This is necessary to release any benzoyl chloride that might have got trapped in these solid lumps. Shake well again, the solution must be alkaline,

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otherwise add some more sodium hydroxide to make it alkaline. There should be no smell of benzoyl chloride at the completion of the reaction. Filter the solid benzoyl derivative, wash with water to remove any adhering alkali, recrystallise from alcohol and determine its melting point. Notes: • The benzoyl derivative may be contaminated with the unreacted solid amine. In this situation, take the crude benzoyl derivative after filtration in a 250 mL beaker and treat it with dilute hydrochloric acid, stir to dissolve the unreacted amine. Filter the purified benzoyl derivative, wash with water, recrystallise and determine melting point. • Sodium hydroxide is added to help hydrolyse the unreacted benzoyl chloride and dissolve the sodium benzoate so obtained C6H5COCl + NaOH → C6 H5COONa + HCl • 2-3 mL of acetone can be added for water insoluble amines before adding benzoyl chloride to enhance the rate of reaction. p-Toluenesulphonyl derivative (p-Tosyl derivative) Both aliphatic and aromatic primary amines react with p-toluene sulphonyl chloride to form alkali-soluble p-toluene sulphonyl derivatives.

This reaction is also used for separation of a mixture of primary, secondary and tertiary amines. Procedure Take the given amine (0.5g), p-toluenesulphonyl chloride (1.5 g) in a clean conical flask (100 mL) and dissolve in minimum amount of acetone. Add dil. sodium hydroxide solution (25 mL, 2 N) and shake the reaction mixture for –10 minutes, making sure that the reaction mixture is alkaline, otherwise add ~5 mL sodium hydroxide to make it alkaline. Acidify the reaction mixture, the tosyl derivative separates, filter, wash with water (~5 mL) and recrystallise from alcohol and determine its melting point. Notes: For substituted amines like nitro amines the p-tosyl derivative is best prepared by refluxing a mixture of amine, tosyl chloride and pyridine (~5 mL) for 20-25 minutes, pouring the cooled reaction mixture into ice cold water.

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Picrates Picrates are charge transfer compounds formed between the electron deficient aromatic ring of picric acid with the electron rich aromatic ring of the amine.

Procedure Dissolve separately picric acid (0.2 g) and the given amine (0.5 mL) in minimum amount of ethanol in two boiling tubes and mix the two solutions. A yellow solid separates out. If a solid is not obtained on mixing, warm the solution in a hot water bath for –5 minutes and keep aside for cooling. A yellow coloured picrate will separate out. Filter the solid, recystallise from alcohol and determine its melting point. Notes: • Picric acid is stored in dark coloured bottles under a layer of water. So for weighing it, take out the solid from the bottle and dry in air for 3-4 minutes by spreading it on a filter paper. (Caution: dry picric acid is explosive in nature) • Piric acid stains the skin so do not allow any skin contact. • Acetone, glacial acetic acid or diethyl ether (ether is highly flammable, do not use it in the laboratory where other students are using burners) can also be used to dissolve amine and picric acid. Bromo derivative All those aromatic primary amines which have an unsubstituted o- and/or p-position can undergo electrophillic substitution reaction at these positions on treatment with bromine to produce bromo substituted derivatives.

Bromo derivatives should only be prepared if the other two derivatives i.e. acetyl or benzoyl derivatives cannot be prepared from the given amine because bromine is harmful and must be handled with a lot of caution. Procedure Take the given amine (0.5 g) in a clean, dry conical flask and dissolve in glacial acetic acid (5-7 mL). Add to it a solution of bromine in glacial acetic acid dropwise carefully from a burette till a permanent light orange colour is obtained, which does not fade on keeping the reaction mixture for ~ 10 minutes. Continuously mix the solutions during addition of bromine. Once the addition is complete, pour the reaction mixture with stirring, into a beaker (250 mL) containing crushed ice (~100 g) or ice cold water (~100 mL). Allow the ice to melt and filter the solid. Dry, recrystallise and determine its melting point.

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Notes See page 102.

1.5.14 Secondary Amines Secondary amines, like primary amines can be converted to the following derivatives: (i) Acetyl (ii) Benzoyl (iii) p-toluene sulphonyl (iv) Picrate (v) Bromo compounds. The procedure followed for acetyl, benzoyl, picrate and bromo derivatives is same as described for primary amines. However, in the preparation of p-tosyl derivative, the alkaline solution is not acidified as the derivative of secondary amines are alkali insoluble and separate out without acidification. 1.5.15 Tertiary Amines Following derivatives can be prepared from tertiary amines: (i) Picrates (ii) p-Nitroso derivatives (iii) Methiodides (iv) Ethiodides. Picrates Follow the procedure given under primary amines (page 114) p-Nitroso derivative These derivatives can be prepared from aromatic tertiary amines in which the para position is free or available for substitution by the nitroso group. Nitrosation is carried out by reaction of the aromatic tertiary amine with nitrous acid (it is generated in situ by reaction between sodium nitrite and dilute hydrochloric acid at low temperature). The p-nitroso derivative first separates as its hydrochloride, as an orange coloured solid, which on basification gives the p-nitroso derivative – a green solid.

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Procedure Take the given tertiary amine (1 g) in a clean conical flask (100 mL) and dissolve it in dilute hydrochloric acid (~ 5 mL). In a boiling tube, dissolve sodium nitrite (0.7 g) in water (4-5 mL). Cool both solutions in ice bath to ~0-5°C. Add the sodium nitrite solution to the amine solution slowly with stirring maintaining the reaction mixture at low temperature. The p-Nitrosamine hydrochloride separates as an orange coloured solid. Filter this solid, take it in a boiling tube and neutralize with sodium hydroxide till green in colour. Extract the green product with ether in a separating funnel, remove the ether layer, wash with water, dry by keeping it over anhydrous sodium sulphate in a dry conical flask for 10-15 minutes. Filter the sodium sulphate. Distil ether and recrystallise the residue from light petroleum ether. (Carry out this work very cautiously away from any burner or flame). Recrystallise the residue from petroleum ether, filter and determine its melting point. Note: The temperature of the reaction mixture should be kept between 0-5°C. Methiodides Almost all aliphatic, aromatic, primary secondary and tertiary amines form quaternary salts on reaction with methyl iodide. These are well defined crystalline salts and are useful for characterization of these amines.

Procedure Take the given tertiary amine (0.5g) in a dry boiling tube and add to it dropwise methyl iodide (2 mL) with shaking. Allow the reaction mixture to stand at room temperature for ~ 5 minutes and then heat it for ~ 5 minutes in a hot water bath. Now cool the reaction mixture in an ice bath to get the solid methiodide. Filter, recrystallise from absolute alcohol or dry acetone or ethyl acetate and determine its melting point. Ethiodides All aliphatic and aromatic primary, secondary and tertiary amines form ethiodides

Procedure Same as for methiodides.

1.5.16 Derivatives of Nitro Compounds Nitro compounds can be characterized by conversion to any of the following derivatives (i) Poly substituted nitro compounds (ii) Reduction to amine (iii) Oxidation of the side chain to carboxylic acids.

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Poly substituted nitro derivatives Most simple mono nitroaromatic compounds can be nitrated to polynitro compounds. For example nitrobenzene can be nitrated to m-dinitro benzene and p-nitrochloro benzene to 2,4-dinitrochlorobenzene.

But nitro compounds containing activating groups like –OH and –NH2 cannot be nitrated directly because these compounds undergo oxidation and react violently with the concentrated nitric acid present in nitrating agent. Nitration is carried out with a mixture of concentrated nitric acid and concentrated sulphuric acid. The reaction is normally slow if nitration is done without sulphuric acid. The mixture of acids produces nitronium cation which nitrates the compounds

Procedure Take the given nitro compound (0.5 mL) in a clean, dry round bottomed flask. In a boiling tube, mix carefully concentrated nitric acid (3 mL) and concentrated sulphuric acid (3 mL). Add carefully, dropwise, with shaking the nitration mixture to the compound in a round bottomed flask, fix a water condensor and reflux the reaction mixture on a wire gauze/ water bath for ~30 minutes. Check for the completion of the reaction by taking out a drop of the reaction mixture and adding it to crushed ice (10 g) in a boiling tube, formation of a solid on stirring indicates that the reaction is complete. Now work up the rest of the reaction mixture by pouring onto crushed ice in a beaker. Filter the solid nitro compound, dry, recrystallise from alcohol and determine its melting point. Notes: • It is important to remember that certain nitro compounds (e.g. o- and p-nitro toluene, m- di nitrobenzene etc.) should not be subjected to further nitration as some of the resultant polynitro compounds may be explosive in nature. • These polynitro compounds should either be partially or completely reduced to the corresponding amines for preparation of derivatives. • Both conc. nitric and sulphuric acids are corrosive chemicals and should be handled carefully. Reduction Nitro compounds can be reduced to a variety of products using different types of reducing agents, which are discussed in detail on page 74.

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However for the purpose of derivatisation, the nitro group is generally reduced to the amino group and the resultant amino compound then characterized by its melting point or conversion to its benzoyl derivative and determination of the melting point. Zn + HCl

→ C H NH + 2H O C6H5NO2 + 6H  6 5 2 2

Nitrobenzene



C6H5NH2 + C6H5COCl + NaOH →



Aniline

C6H5NH COC6H5 + NaCl + H2O Benzanilide

The reduction of the nitro to the amino group is carried out in the laboratory with one of the following reagents • Tin metal and concentrated hydrochloric acid • Zinc dust and concentrated hydrochloric acid • Stannous chloride and concentrated hydrochloric acid. The reduced compound is isolated and crystallized or converted to its benzoyl derivative. Procedure Take the given nitro compound (0.5g) in a clean round bottomed flask (50 mL), add small pieces of tin metal (2 g) and a few pieces of pumice stones to it. Fix a water condenser and set up the apparatus for refluxing on a wire gauze. Add concentrated hydrochloric acid (15 mL) into the flask from the open top end of the condenser in small lots during refluxing (15-20 minutes). Stop heating, cool the reaction mixture and filter any undissolved pieces of tin. To the filtrate, add a few pieces of ice and neutralize with dilute sodium hydroxide. Filter if a solid amine is obtained, recrystallise and determine its melting point, otherwise extract the amine with ether, separate the ether layer, wash the ether layer with water. Distil ether/evaporate (do not work with ether in the laboratory where other students are using burners). Take the residue, add freshly prepared sodium hydroxide (15 mL, 10%), benzoyl chloride (0.5 mL), shake till a solid separates and the reaction mixture is alkaline and has no smell of benzoyl chloride, (see page 116, for more details). Notes: If the melting point of the amine is lower than the reported melting point, it may be contaminated with the starting nitro compound. In such a case, dissolve the solid in dilute hydrochloric acid, filter the insoluble nitro compound and basify the clear acidic solution to get the amino compound. Recrystallise and determine its melting point. Alternative method Procedure Take the given nitro compound (0.5g) and alcohol (4-5 mL) in a clean conical flask, add to it zinc dust (2.5 g) and then concentrated hydrochloric acid (15 mL) in small lots with shaking. As the nitro compound is reduced, it dissolves in the acidic medium and the completion of reaction is indicated by complete dissolution of the initially insoluble nitro compound. Add ice to cool and neutralize and proceed as given above. Notes: • The main advantage of the method is that the energy used (LPG burner or electricity) for refluxing (15-20 minutes) per student is saved. So the modified method is energy efficient. • This method works well for mono substituted nitro compounds.

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Oxidation of the side chain Nitro compounds containing a side chain (–CHO, –CH2OH, –CHOH– and –CH2R etc.) can be oxidized by alkaline potassium permanganate to the corresponding nitro substituted carboxylic acids. Determination of their melting point can help in the characterization of the given nitro compounds.

Procedure Follow the procedure and other details given on page 107.

1.5.17 Derivatives of the Amides Amides can be hydrolysed to the sodium salt of corresponding carboxylic acids with sodium hydroxide solution and the resultant acid obtained by acidification can be either characterized from its melting or converted a suitable crystalline derivative. +

RCONH 2 + NaOH → RCOO − Na + H 2 O 2 RCOONa + H 2SO 4 → 2RCOOH + Na 2SO 4 The amide of carbamic acid i.e. urea (NH2 CONH2) gives only gaseous products (NH3 and CO2) on hydrolysis and hence cannot be characterized by the above method

NH2CONH2 + NaOH → [NH2COONa] + NH3↑



[NH2COOH] → NH3↑ + CO2↑

Urea is identified by converting it to urea oxalate, urea nitrate is another derivative of urea, since it is explosive in nature, it should not be prepared for identification purpose. Procedure for hydrolysis of amides The procedure adopted for alkaline hydrolysis of amides is as described under esters (page 112) • Ammonia is evolved during hydrolysis and the completion of reaction is indicated when the evolution of ammonia ceases. • As in the case of esters, if a solid acid is obtained, filter it, crystallise and determine its melting point. In case a solid acid is not formed on acidification, Convert to S-benzyl isothiuronium salt (see page 96 for details).

1.5.18 Derivatives of Urea Urea oxalate and nitrate are two derivatives of urea. Urea nitrate is an explosive and hence should not be prepared in the laboratory.

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Urea oxalate

Procedure Dissolve urea (0.5 g) in water (~5 mL) in a boiling tube. Dissolve separately oxalic acid (0.6 g) in water (~ 7-8 mL). Mix the two solutions and shake well. The crystals of urea oxalate separate out. Filter, dry and determine melting point. Urea nitrate

Procedure Dissolve urea (0.5 mL) in water (5 mL) in a boiling tube and add to it concentrated nitric acid (1 mL) drop wise with shaking. Urea nitrate separates as crystalline solid. Filter, dry and determine its melting point. Note: It is explosive when left in dry state.

1.5.19 Derivatives of Imides Imides are compounds containing a –CONHCO– group and are cyclic in nature. Some common examples are succinimide, phthalimide etc. These are hydrolysed to the corresponding acids by the procedure described above for amides. 1.5.20 Derivatives of Nitriles The nitriles are hydrolysed to the corresponding carboxylic acids either with a base or an acid. +

−

H /OH → RCONH 2 → RCOOH RC ≡ N 

Alkaline hydrolysis Take the nitrile (1 g) in a round bottomed flask (100 mL), add freshly prepared sodium hydroxide (15 mL, 25%). Add a few pieces of pumice stones and reflux the reaction on a wire gauze for 20-30 minutes. Work up the reaction mixture as mentioned on page 111-112. Acid hydrolysis Take the nitrile (1 g) in a round bottomed flask (50 mL), add conc. sulphuric acid (5 mL) and a few pieces of pumice stones. Reflux the reaction mixture for 30-40 minutes. Filter the solid, recrystallise and determine its melting point.

1.5.21 Derivative of Anilides Following derivatives can be prepared for their identification. (i) Acid hydrolysis (ii) Bromo derivatives.

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Acid hydrolysis Anilides produce a mixture of a carboxylic acid and an amine on hydrolysis

The acid may separate out as a solid on hydrolysis. It is filtered and the filtrate on basification gives the amine. Procedure Take the anilide (1g), conc. sulphuric acid (10 mL, 70%) in a round bottomed flask and reflux the mixture for ~ 30–40 minutes. Cool the reaction mixture (a) a solid acid separates out. Filter it, recrystallise and determine its melting point (b) the acid is soluble and hence does not separate. Extract it with ether, evaporate ether (Ether is highly inflamable) and convert the residue to S-Benzylthiuronium salt (page 96). The acidic filtrate contains the amine, neutralise with dil sodium hydroxide, filter if a solid amine is formed and determine its melting point after recrystallisation or extract the amine and prepare a suitable derivative from it section 1.5.12. Bromo derivative: Some anilides like acetanilide can be converted to bromo derivatives.

Procedure: See section 1.5.12.

1.5.22 Derivatives of Amino Acids Following derivatives can be prepared: (i) Benzoyl derivative (ii) 3,5-Dinitro benzoate (iii) 2,4-Dinitro phenyl derivative Benzoyl derivative Amino acids react with benzoyl chloride in an alkaline medium (Schotten Bauman Reaction) to form the benzoyl derivatives, which remain dissolved in the medium because of the presence of carboxylic acid group. H 2 N − CH − COOH + C6 H5 COCl + 2 NaOH → C6 H5 CONH − C H − COONa + H 2 O + NaCl | R

| R

The benzoyl derivative is isolated by acidification of the solution. 2C6 H5 CONH − CH − COONa + H 2SO 4 → 2C6 H5 CONH − CH − COOH + Na 2SO 4 | | R R Benzoyl derivative

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Procedure Take the amino acid (500 mg) in a conical flask (100 mL), add sodium hydroxide solution (10 mL, 15%) and shake to dissolve. Now add benzoyl chloride (1 mL), 2-3 drops at a time with shaking. Cork the flask and shake the reaction mixture vigorously (release pressure while shaking by removing the cork occasionally). The reaction mixture should be alkaline without any smell of unreacted benzoyl chloride. Acidify the reaction mixture with dilute sulphuric acid. Filter the benzoyl derivative, dry, wash with hot carbon tetrachloride (5 mL, twice) and recrystallise from dilute alcohol. Determine the melting point of dry sample. Notes: Any unreated benzoyl chloride is hydrolysed to benzoic acid, which being insoluble in water gets precipitated along with the benzoyl derivative of amino acid. The benzoic acid is removed by washing the benzoyl derivative with carbon tetrachloride. 3,5-Dinitrobenzoate Amino acids can form 3,5-dinitrobenzoates on reaction with 3,5-dinitrobenzoyl chloride.

Procedure Prepare 3,5-dinitro benzoyl chloride as described on page 96-97. Prepare 3,5-dinitro benzoate following the procedure described above for benzoyl derivative. 2,4-Dinitrophenyl derivative Amino acids form 2,4-dinitro phenyl derivative on reaction with 2,4-dinitro chlorobenzene.

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Procedure To a solution of amino acid (250 mg) in saturated sodium bicarbonate (5g in 10-15 mL water) in a conical flask, add a solution of 2,4-dinitrochlorobenzene (0.5 g in 5 mL alcohol). Shake the reaction mixture vigorously for 5 minutes and allow it to stand with occasional swirling for one hour. Add solid sodium chloride (1 g) to the above reaction mixture, swirl to dissolve and extract with ether. Acidify the aqueous extract with dilute hydrochloric acid, filter the solid, recrystallise from alcohol and determine its melting point.

1.5.23 Derivatives of Sulphonic acids Following derivatives can be prepared: (i) S-Benzylisothiuronium sulphonate (ii) Sulphonamides. S-benzylisothiuronium sulphonate Like carboxylic acids, sulphonic acids also react with S-benzylisothiuronium chloride to form the corresponding salts. RSO3H + NaOH → RSO3Na + H2O RSO3Na + [C6H5CH2SC(NH2)2]+Cl– → [C6H5CH2SC(NH2)2]+RSO3– + NaCl Procedure Take the sulphonic acid (500 mg) in a conical flask and add a drop of phenolphthalein indicator. Now add dilute sodium hydroxide solution (10%) dropwise with thorough shaking to dissolve the acid and obtain a pink coloured solution such that the colour does not fade even on standing. Neutralise excess alkali with a very dilute solution of hydrochloric acid to obtain a light pink coloured solution. Add a solution of S-benzylisothiuronium chloride (prepare by dissolving 1g of S-benzylisothiuronium chloride in minimum amount of water). Mix the two solutions and filter the S-benzylisothiuronium sulphonate that separates as a white solid. Recrystallise from 50% alcohol and determine its melting point. Sulphonamides Sulphonic acids can be converted to sulphonamides by the following steps RSO3H + PC15 → RSO2Cl + POCl3 + HCl 2RSO2C1 + (NH4)2CO3 → 2RSO2NH2 + 2HC1 + H2O+CO2 Procedure Take the given sulphonic acid (500 mg) in a dry round bottomed flask. Add phosphorus pentachloride (1.25g) to it. Heat (150° C) the reaction mixture in a fume chamber in an oil bath for ~30 minutes. Cool, add ammonium carbonate (2g) slowly with shaking and heat (100 °C) again in an oil bath for 30 minutes. Cool, pour the reaction mixture into a beaker containing ~ 30-40 mL ice cold water, stir well, filter the sulphonamide and wash the solid with cold water (10 mL). Dissolve the sulphonamide in dilute sodium hydroxide, filter to remove any insoluble solid and acidify the cooled filtrate with dilute hydrochloric acid. Filter the sulphonamide, recrystallise from dilute alcohol and determine its melting point.

1.5.24 Thiols 2,4-Dinitrophenyl thioethers These are prepared by reaction of a thiol with 2,4-dinitrochlorobenzene.

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Procedure Dissolve the given thiol (500 mg) in alcohol (15 mL) in a round bottomed flask. Add sodium hydroxide solution (2 g in 4-5 mL water) and a solution of 2,4-dinitrochlorobenzene (1 g in 10 mL alcohol). Reflux the mixture for 10 minutes in a water bath. Now filter the hot solution and allow the filtrate to cool to obtain crystals of 2,4 dinitrophenyl ether. Filter, dry and determine melting point.

1.5.25 Sulphonamides Sulphonamides can be converted to the following derivatives Acetyl derivative C6H5 SO2 NH2 + CH3 COCl → C6H5 SO2 NHCOCH3 + HCl Acetyl derivative

Procedure Reflux a mixture of sulphonamide (1 g), acetyl chloride (2.5 mL) in a dry round bottomed flask in a fume chamber for 30 minutes in a water bath. Cool the reaction mixture and pour into ice cold water. Filter, wash with water and recrystallise from alcohol and determine its melting point. Derivatives of Compounds Containing Halogen

1.5.26 Derivatives of Alkyl and Aryl Halides Aryl halides Halogen directly attached to the nucleus. Nitration–Aryl halides can be converted to the nitro compounds on reaction with conc. HNO3 and conc. H2SO4

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Procedure Follow the procedure given on page 113-114. Halogen in the side chain Such compounds can be oxidised to the corresponding carboxylic acid with alkaline potassium permanganate.

Alkyl halides An alkyl halide on reaction with β-naphthol in an alkaline medium gives the corresponding alkyl-βnaphthyl ether.

Procedure Dissolve β-naphthol (0.5 g) in alcoholic sodium hydroxide solution (0.15 g NaOH in 5-6 mL alcohol) in a round bottomed flask. Add the alkyl halide (0.5 mL) and reflux the reaction mixture for 20-30 minutes in a water bath. Cool the reaction mixture, filter the solid, recrystallise from alcohol and determine its melting point.

1.5.27 Derivatives of Acid Chloride Alkaline hydrolysis: Follow the procedure given in section 1.5.8. Amide, Anilide, p-toluidide: Follow the procedures given in section 1.5.4. 1.6 IDENTIFICATION OF A MIXTURE OF TWO COMPOUNDS A mixture of two compounds can be analysed by performing the following procedures in the laboratory • Preliminary examination • Separation of the components of the mixture • Purification of each component • Identification of each component

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1.6.1 Preliminary Examination The given mixture should be subjected to following tests before proceeding for separation of the mixture Solubility Dry heating Test with 2, 4-Dinitrophenyl hydrazine Test with ferric chloride Test with bromine solution Test with potassium permanganate The details of the procedure and the conclusion drawn from each test are given in the previous sections on page 8-12. 1.6.2 Separation of the Components of Mixture No definite scheme of separation can be formulated since a mixture may have endless possibilities of a combinations of two compounds. The scheme to be adopted will depend on the results of preliminary examination. General guidelines are given in the following section which will be helpful in separating the two compounds. One of the following procedures may be adopted • Solubility in one of the following solvents: water, alcohol, acetone or ether. • Solubility in aqueous acidic or basic solvents: dilute sodium bicarbonate, dilute sodium hydroxide and dilute hydrochloric acid. • If one of the component is a carbonyl compound, separation can be achieved by treatment with sodium bisulphite solution. • A mixture of two given amines (primary and secondary etc.) can be separated by using Hinsberg’s reagent. • Sublimation. • Fractional distillation. • Chromatographic separation. 1.6.3 Solubility Check the solubility of the mixture (0.1 g) in the following solvents (2-3 mL) in the order given below Water Solvent ether Alcohol Acetone. The method cannot be used for separation of the components of the mixture, if both the components are soluble or completely insoluble in the solvent. If however, only one of the components of the mixture is soluble, this method can be employed for separation. First try on a small scale (~ 200 mg) of the mixture and then proceed as given below: Procedure Take the mixture (~10 g) in a conical flask (100 mL) and add one of the above solvent (20 mL) which has been selected for separation. Swirl to dissolve. Filter, treat the insoluble solid residue with more of the solvent (2-3 mL) two times and filter into the previous filtrate. The solid is component (1) and the filtrate contains the second component (2).

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Take the filtrate in a china dish, if water has been used and evaporate it on a steam bath, the solid obtained after complete evaporation is the second component of the mixture. Take the filtrate in a round bottomed flask (100 mL) if any organic solvent has been used and distil it in an electric water bath or remove the solvent in a rotary evaporator. Never distil an organic solvent on a burner. Special care must be taken if solvent ether was used as ether is highly inflammable. Notes: • Never work with ether in the laboratory where other students are using burners. • Rotary evaporator should be preferably used for removal of the solvent. Solubility in acidic or basic solvent This is the easiest method that is used for separation of a mixture in cases where • The mixture is insoluble in water • One of the component of the mixture is acidic or basic in nature • A mixture of two solids or liquids can be separated.

1.6.4 Separation of the components of a solid mixture Try with ~100 mg of the mixture and 2-3 mL of the solvent say sodium bicarbonate solution, if only one of the components is soluble, use the method for separation. Procedure • Take the mixture (~10 g) in a conical flask and add sodium bicarbonate solution (10 mL) slowly till the effervescence ceases. Add more of bicarbonate solution till there is no more effervescence. Filter the insoluble component (1). Add a few pieces of ice to the filtrate and acidify slowly with dilute sulphuric acid. Filter the solid (2). • Follow the above mentioned procedure if the components of the mixture are to be separated using dilute sodium hydroxide solution. • If one of the components is basic in nature, shake the mixture (10 g) with dilute hydrochloric acid (~10 mL) in a conical flask, filter the solution and treat the solid residue with some more dilute hydrochloric acid. The insoluble solid is component (1). Add some crushed ice to the filtrate, neutralise with sodium hydroxide solution to obtain the component (2) and filter it.

1.6.5



Separation of a Liquid Mixture on the basis of Solubility • Dissolve the liquid mixture (~15 mL) in ether (30-35 mL) in a conical flask. Add the solution of sodium bicarbonate (2-3mL) to it. Allow the reaction to cease and again add more of sodium bicarbonate till the freshly added sodium bicarbonate gives no effervescence. Now transfer the entire amount from the conicl flask to a separating funnel carefully, stopper the funnel and shake the mixture slowly, releasing pressure every time by opening the nozzle of the separating funnel. Keep the funnel in the tripod/separating funnel stand to allow the two layers to separate. Drain out each layer in a separate conical flask. Remove ether (see above) to get component (1) and neutralise the sodium bicarbonate solution (as described above) to get component (2) Extract component (2) with ether, remove ether to get (2): — Follow same procedure with dilute sodium hydroxide. — When using dilute hydrochloric acid, the same procedure is followed, the only difference is that neutralisation of the aqueous acidic layer is done with dilute sodium hydroxide solution.

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1.6.6 Separation of Mixture with Sodium Bisulphite If one of the component is a carbonyl compound, this can be separated by reaction with sodium bisulphite, the other component does not react. The bisulphite adduct of the carbonyl compound is then converted to the original compound by heating with sodium bicarbonate/carbonate.

• A solid mixture Take the solid mixture (10 g) in a conical flask and dissolve it in ether. Shake it well with a saturated solution of sodium bisulphite (add in small lots of 5 mL each time). — A soluble bisulphite adduct is formed. Separate the ether layer and remove ether to get component (1). Heat the aqueous bisulphite solution with a saturated sodium carbonate solution to get the solid component (2). — An insoluble bisulphite is obtained. Filter and decompose as above to get the carbonyl component (2). Remove ether to get component (1) • A liquid mixture Dissolve 10 mL of mixture in ether. Treat with saturated sodium bisulphite solution as above. Component (1) remains in ether, remove ether to get it. Treat the bisulphite layer as mentioned to get component (2). Notes: Ether is highly inflamable, do not use solvent ether for any type of work where other students are using burners.

1.6.7 Separation by using Hinsberg reagent If both the components of the mixture are amino compounds, these can be separated by using either bezensulphonyl chloride or p-toluenesulphonyl chloride (tosyl chloride). However, a mixture of two primary/secondary or tertiary amines cannot be separated by this method. The separation of a mixture of amines is based on the fact that a primary amine has two hydrogens attached to nitrogen, a secondary amine has only one and a tertiary amine none. When a primary amine reacts with tosyl chloride in an alkaline medium, one hydrogen on the nitrogen is substituted by an electron withdrawing –SO2C6H4-CH3 group whereas the second hydrogen is still present on nitrogen which now becomes acidic and hence the reaction product (sulphonamide) remains dissolved in the alkaline medium. A secondary amine has no hydrogen on nitrogen after the reaction with tosyl chloride and hence the sulphonamide of secondary amine is insoluble in the alkaline medium and tertiary amine does not react. Procedure Take the mixture (10 g) and p-toluene sulphonyl chloride (20g) in a conical flask. Now add sodium hydroxide (150 mL, 2N) and shake the mixture vigorously for 10-15 minutes. Cool the mixture, add ether (~70 mL) and shake again. If a solid is present, which is insoluble in ether as well as in alkali, it is the sulphonamide of secondary amine. Filter it. Now take the filtrate in a separating funnel, separate the ether (containing the tertiary amine) and aqueous alkaline solution (contains the sulphonamide of primary amine). Remove ether (caution) to get the tertiary amine and neutralise the alkaline solution to get the sulphonamide of primary amine. Filter it. Decompose the sulphonamides separately by refluxing with

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hydrochloric acid (50 mL, 25%) or sulphuric acid (50 mL, 70%) for 2-3 hours to get the primary and/or secondary amine in solution. Neutralise the acidic solution to get the amine. Filter if a solid is formed, extract with ether if a liquid separates out. Remove ether (caution) and get the amine.

1.6.8 Separation of Mixture by Sublimation This method can be used if one of the component sublimes on heating. Set up the apparatus as shown in Fig. 1.10. Take ~ 7-8 g of mixture and heat over a low flame. The component which sublimes collects over the paper under the funnel. The other component remains in the china dish. 1.6.9 Separation of Mixture by Fractional Distillation Fractional distillation can be tried if the two components have boiling points differing at least by 30-40°C. Set up the apparatus as shown in Fig. 1.4. Take ~30mL of the mixture in the round bottomed flask and distil it. The lower boiling component will distil first, collect it in a dry boiling tube when the temperature remains constant. Continue heating to distil the higher boiling component. 1.6.10 Chromatography This method can be used for separation of a mixture of two or more compounds. • Column chromatography (see page 398-400) • HPLC (see page 401). 1.6.11 Purification of the Components The solid compound is purified by crystallisation (see page 88) The liquid compounds are purified by distillation during which the boiling point of the liquid compound is also noted (page 23-24). 1.6.12 Identification of each Component Follow the procedure given in this chapter for identification of pure compounds.

Questions Q.l. Q.2. Q.3.

Q.4. Q.5. Q.6. Q.7.

What are organic compounds? List a few important physical properties of organic compounds. What important information do you get about the unknown organic compound from the following preliminary examination (a) Flame test (b) Solubility test (c) Test with potassium permanganate? How is the flame test performed in the laboratory? What is the observable difference between a sooty and a non-sooty flame? Why do most aromatic organic compounds produce a sooty flame and aliphatic compounds a non- sooty flame? Are there any exceptions? If yes, give the names of a few such compounds. Why most organic compounds are water insoluble? Name a few classes of organic compounds which dissolve in water and why? Does solubility decrease with increasing molecular weight even in these classes of compounds?

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Name a types of compounds that are most likely to dissolve in the following: (i) Aq. Sodium bicarbonate solution (ii) Sodium hydroxide solution (iii) Dil. Hydrochloric acid (iv) Conc. Sulphuric acid.

Q.9.

What precautions must be observed while checking the solubility of a compound in solvent ether?

Q.10. What are extra elements? Name a few extra elements that are most commonly present in organic compounds and give the structures of a few such natural compounds which contain them. Q.11.

What is Lassaigne’s extract and how is it prepared in the laboratory? What special precautions MUST be observed while handling sodium in the laboratory?

Q.12. Why is sodium preserved under a non reactive organic solvent and protected from air and water? Q.13. A stored piece of sodium gets covered with a hard yellow brown layer. What is this due to? Q.14. What is the colour of sodium? Q.15. Why should unused pieces of sodium be NEVER discarded in the dust bin or sink? Q.16. Why should anhydrous conditions be maintained while fusing an organic compound with sodium metal for preparation of sodium extract? Q.17. Why is the sodium extract normally alkaline? Q.18. Name the acid that is used for acidification of extract in the following tests: (i) Test for nitrogen with ferrous sulphate. Can HCl or HNO3 be used? Why? (ii) Lead acetate test for sulphur (iii) Silver nitrate test for halogens (iv) Layer test for halogens. Q.19. A nitrogen containing organic compound is fused with sodium for the preparation of Lassaigne’s extract. What does the nitrogen get converted to? Write the detailed procedure and the chemical reactions involved in the test for nitrogen in the extract. Q.20. Which of the following compounds will give a positive test for nitrogen in the Lassaigne’s extract and why? (i) Urea (ii) Nitrobenzene (iii) Hydrazine (iv) Hydroxyl amine (v) Phenyl hydrazine. Q.21. A student prepares the sodium extract of aniline and tests for nitrogen. A negative test is obtained. Tick mark the possible mistakes that the student might have committed (i) Compound was not dry (ii) A freshly cut piece of sodium was not used (iii) Ferrous solution was prepared just before use (iv) The extract was boiled after addition of ferrous sulphate to convert all Fe2+ to Fe3+ (v) Enough dilute sulphuric acid was added to acidify the test solution. Q.22. Name the sodium salts that will be formed when the following compounds are fused with sodium (i) Thiophene

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(ii) Thiourea (iii) p-Nitrochlorobenzene (iv) Bromobenzene (v) Iodoform. Q.23. Why do nitrogen and sulphur interfere in the silver nitrate test for halogens in the Lassaigne’s extract? How is the silver nitrate test for chlorine performed in the extract of o-nitrochlorobenzene? Q.24. Do nitrogen and sulphur interfere in the layer test for bromine and iodine? Why? Q.25. Name a few oxidising agents that are used in the layer test? Q.26. Define the term melting point. Why do most inorganic compounds have a higher melting point than the organic compounds? Q.27. What is the effect of the following on the melting point of a compound? (i) Impurities (ii) External pressure. Q.28. Give the names of the following used in the Kjeldahl’s method for determination of melting point (i) Flask (ii) Conventional bath liquid, its advantages and disadvantages (iii) Modern bath liquid, its advantages and disadvantages (iv) Tube used for holding the compound in the bath liquid. Q.29. Why are following precautions observed during the melting point determination? Explain. (i) The crystalline sample of the compound is powdered (ii) The compound is filled in the capillary to make its compact column (iii) The thermometer is calibrated (iv) The bulb of the thermometer and the lower tip of the capillary are dipped and kept at the same level in the bath liquid in the flask (v) The flask is heated slowly and uniformally and not at one point. Q.30. Why does conc. sulphuric acid turns dark and ultimately blackish on repeated use in the Kjeldahl flask for melting point determination. How can this colour be removed? Q.31. Besides Kjeldahl method, have you used any other apparatus for melting point determination? If yes, explain the method and its advantages and disadvantages over the Kjeldahl method. Q.32. Can the identity or non-identity of two compounds be established by determination of the melting point? How? Q.33. Define boiling point and explain the effect of (i) impurities and (ii) external pressure on it. Q.34. Tick mark the correct statement (i) The boiling point of a liquid is independent of external air pressure (ii) Distillation can be used to purify a liquid from another volatile liquid having a boiling point close to the liquid to be purified (iii) Distillation helps to purify an organic liquid from a non volatile impurity (iv) Distillation is the only method that can be used for purification of an organic liquid compound (v) Fractional, reduced pressure and steam distillation are some other common techniques that can be used for purification of liquid compounds.

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Q.35. Why is mercury metal considered a health hazard? Where have you used it in the laboratory? Can it be replaced by safer chemicals? Q.36. Why are the following precautions observed during determination of the boiling point of an organic liquid? (i) Apparatus is dried before use (ii) A liquid with a boiling point < 100°C is heated in a water bath for distillation (iii) Pumice stones are added before starting heating and NEVER during heating. Q.37. Which of the following compounds will decolourise bromine solution and why? (i) Benzene (ii) Naphthalene (iii) Anthracene (iv) Styrene (v) Butene-2. Q.38. What is Baeyer’s reagent? Explain why the following compounds will decolourise this reagent. (i) Styrene (ii) Butadiene (iii) Butyne-2. Q.39. Explain in brief, why most aromatic hydrocarbons do not react with (i) Baeyer’s reagent (ii) bromine solution? Q.40. What observation would be expected from the compounds listed in column A on performing the preliminary tests in column B. Compound (Coulmn A)

Column B

Benzene

Flame test

Chlorobenzene

Solubility in water

Aniline

Ferrous sulphate test with Lassaigne’s extract

Acetic acid

Test with bromine solution

Glucose

Solubility in dil. HC1

Resorcinol

Silver nitrate test with Lassaigne’s extract

Q.41. Why most carbohydrates are water soluble, insoluble in solvent ether, char on heating and do not have a sharp melting point? Q.42. Name the functional groups that are present in carbohydrates. Q.43. Name an (i) Aldotriose (ii) Aldopentose (iii) Ketohexose (iv) Reducing disaccharide (v) Non reducing disaccharide and a (vi) polysaccharide. Q.44. Which of the following compounds will give a positive Molisch test? (i) Benzaldehyde (ii) furfural (iii) tartaric acid (iv) glucose. Q.45. Which of the following compounds can reduce the Barfoeds reagent on heating within one minute? (i) Galactose (ii) Ribose (iii) sucrose (iv) lactose. Q.46. Why do carbohydrates give +ve test with ceric ammonium nitrate and –ve test with 2,4- DNP reagent? Q.47. How are the following pairs of compounds distinguished from each other (i) Glucose and Fructose (ii) Glucose and Sucrose (iii) Maltose and Sucrose (iv) Ribose and Glucose.

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Q.48. Give common names of the following compounds (i) Invert sugar (ii) Grape sugar (iii) Milk sugar. Q.49. From where do plants and animals get the carbohydrates? Q.50. Why are carboxylic acids acidic in nature? Q.51. Why carboxylic acids and their derivatives give a negative test with 2,4-dinitrophenylhydrazine inspite of the presence of a group in their structure? Q.52. Explain why p-Nitrobenzoic acid is a stronger acid than benzoic acid and α-chlorobutanoic acid stronger than β-chlorobutanoic acid? Q.53. Why the litmus test for an acidic compound not performed using an alcoholic solution of an acid? Q.54. Name two carboxylic acids of each type (i) Which give +ve fluorescein test (ii) Do not give fluorescein test (iii) Form anhydride on sublimation. Q.55. Can a compound other than a dicarboxylic acid give +ve fluorescein test? If yes, give an example. Q.56. A carboxylic acid RC14OOH gives carbon dioxide on reaction with NaHCO3, the gas is CO2 or C14 O2? Q.57. How can you separate a mixture of benzoic acid and naphthalene?

Q.58. Why is there a broad absorption signal between 2500-3000 cm–1 in the IR spectrum of a carboxylic acid? Q.59. Mention the salient differences in the PMR spectrum of acetic acid and benzoic acid. Q.60. Predict the PMR spectra of following: (i) Acetamide (ii) Propionic acid. Q.61. Why most phenols are acidic in nature but generally weaker acids than carboxylic acids? Q.62. Write the structure of two substituted phenols which are stronger acids than phenol and explain your answer. Q.63. Explain why phenols decolourise bromine solution and why the reaction is faster with bromine in acetic acid than with bromine in carbon tetrachloride? Q.64. What is the reaction of phenol with ferric chloride? What will be observed if (i) sodium hydroxide (ii) dilute hydrochloric acid is added to the above test solution? Q.65. Name two phenols which give –ve test with ferric chloride, explain why? Q.66. What product is obtained when the following phenols are subjected to phthaleine test and which colour is obtained when the reaction mixture is basified with dilute sodium hydroxide solution? (i) Phenol (ii) Resorcinol. Q.67. What is picric acid? Why it is called so? Q.68. Give common and distinguishing laboratory tests for aldehydes and ketones. Q.69. Why are the following more reactive? (i) Aliphatic aldehydes than aliphatic ketones (ii) Aliphatic aldehydes than aromatic aldehydes (iii) Formaldehyde than all other carbonyl compounds.

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Q.70. Which compounds will give a positive iodoform test? (i) Formaldehyde (ii) Acetaldehyde (iii) Acetone (iv) Acetophenone (v) Benzophenone (vi) Propanaldehyde (vii) Cyclohexanone. Q.71. Why is sodium hydroxide solution added to the reaction mixture during iodoform test? Q.72. Write the structure of the chemical present in following which reacts with the aldehydes/ketones (i) Fehling solution A+B (ii) Tollens reagent (iii) Schiffs base. (iv) 2,4- DNP in acidic medium. Q.73. Write chemical equations for the following reactions. (i) Acetone with bromine and sodium hydroxide (ii) Benzaldehyde with Tollens’ reagent (iii) Iodoacetaldehyde with iodine and sodium hydroxide (iv) Formaldehyde with Fehling solution A + B. Q.74. Why is sodium acetate used in the reaction between acetone and hydrazine hydrochloride in the preparation of acetone hydrazone? Q.75. What is Brady’s reagent? Q.76. Amines, at times give a precipitate in the DNP test. What is this precipitate due to? How can this precipitate be distinguished from the 2,4-dinitrohydrazone of carbonyl compounds? Q.77. Why the 2,4-DNP obtained from aliphatic compounds like formaldehyde, acetaldehyde and acetone etc. is light coloured than that obtained from aromatic compounds like benzaldehyde, acetophenone etc.? Q.78. A carbonyl compound A (C4H8O) gives 2,4-DNP and iodoform test. What is the structure of A? Justify your answer. Predict the PMR spectrum of A and the position of IR absorption due to the functional group present in it. Q.79. Can you differenciate CH3CH2CHO and CH3COCH3 on the basis of their PMR spectra? Q.80. Can IR spectroscopy be used to distinguish: (i) Cyclohexanone from n-Hexanone (ii) p-Hydroxy acetophenone from o-hydroxy acetophenone? Q.81. What are alcohols? How many isomeric alcohols can have a molecular formula C4H10O? Write their structures and classify them as 1°, 2° and 3° alcohol. Q.82. Write the structure of each of two dihydric and two tri hydric alcohols and write their common and IUPAC names. Q.83. Why are alcohols weaker acids than carboxylic acids and phenols? Q.84. Arrange the following in an increasing order of acid strength n, sec., tert. Butyl alcohol.

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Q.85. A water insoluble alcohol is likely to dissolve in (i) aq. sodium bicarbonate (ii) aq. sodium hydroxide (iii) dil. Hydrochloric acid (iv) an organic solvent like ether. Why? Q.86. An alcohol reacts vigorously with sodium. Write this reaction between butanol-2 and sodium. What special precautions must be observed while using this reaction as a laboratory test for alcohols? Q.87. What is the formula of ceric ammonium nitrate? What is the structure of coloured compound that is formed when ceric ammoniumnitrate is added to (i) ethanol (ii) 1-butanol? Why does the colour of solution fade slowly on keeping? Q.88. Can chromic acid be used to distinguished between 1°, 2° and 3° alcohols? Explain your answer. Q.89. What is Lucas’ reagent? What happens when it is added to (i) Ethanol (ii) propanol-1 (iii) propanol-2 and (iv) tert. butanol? Q.90. Where does the –OH stretch of an alcohol show an absorption signal in the IR spectrum? Is the absorption signal sharp or broad? Why? Q.91. Is the position of absorption of a proton of –OH group in the NMR spectrum independent of the concentration of alcohol solution? If no, how is it identified? Q.92. Define the following terms: Esterification, alcoholysis and saponification. Q.93. What are Triglycerides? Q.94. Why are esters used extensively in perfume industry? Q.95. What is the chemistry of hydroxamic acid test? Q.96. Why is alkaline hydrolysis of an ester preferred to acid catalysed hydrolysis? Q.97. Write the chemical equations for the following: (i) Alkaline hydrolysis of ethyl propionate (ii) Acid catalysed hydrolysis of phenyl benzoate (iii) Base catalysed hydrolysis of benzyl benzoate (iv) Alkaline hydrolysis of methyl benzoate (v) Alkaline hydrolysis of phenyl acetate. Q.98. How can benzyl alcohol be separated from a mixture of acetic acid, benzoic acid and benzyl alcohol? Q.99. What is a possible reason of a turbid distillate resulting during distillation of an ester? What must be done to obtain a clear transparent distillate? Q.100. What are lactones? How can these be distinguished from esters? Q.101. What do you understand from the following? Give examples. (i) simple anhydride (ii) mixed anhydride (iii) cyclic anhydride. Q.102. Suggest at least two laboratory tests to distinguish cyclohexane from cyclohexene. Q.103. Which of the following will give a coloured product with anhydrous aluminium chloride? Write the chemical reactions for positive test. (i) benzene (ii) toluene (iii) naphthalene. Q.104. What alcohol/phenol would result on reaction of following ethers with hydrobromic acid? (i) dimethylether (ii) ethylmethyl ether (iii) methylphenyl ether. Q.105. What is alcoholic silver nitrate? How is it used in the laboratory to distinguish between an alkyl and an aryl halide?

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Q.106. Which of the following compounds will give a white precipitate on reaction with aqueous silver nitrate solution? Why the other two compounds given below do not react? (i) C6H5Cl

(ii) CH3Cl

(iii) C6H5COC1.

Q.107. Write the structures of all isomeric amines having a molecular formula C3H9N and classify them as primary, secondary and tertiary amines. Q.108. Why most amines are soluble in acids? List a few possible reasons of some amines being insoluble. Q.109. Which is a stronger base and why? (i) CH3NH2 or CH3NHCH3 (ii) C6H5NH2 or CH3NH2 (iii) p-NO2- C6H4NH2 or C6H5NH2. Q.110. What is isocyanide (Carbylamine) test? Which of the following will give this test? (i) Aniline (ii) o-Anisidine (iii) Methyl amine (iv) N-Methylaniline (v) Trimethyl amine. Q.111. What special precautions must be observed while performing the isocyanide test in the laboratory? Q.112. How is isocyanide decomposed after performing the test? Q.113. What is Azo dye test? Which of the following will give a positive test? (i) Aniline (ii) N-Methyl aniline (iii) N, N-Dimethyl aniline (iv) Ethyl amine (v) Isopropyl amine (vi) n-Propyl amine. Q.114. Why is nitrous acid not available like nitric acid in the laboratory? How do you prepare it for use in the laboratory? Q.115. Why must the temperature be kept low while performing the azo dye test in a test tube and why that is not required while performing the same test as spot test? Q.116. Write the structure of product resulting in the reaction of the following compounds with nitrous acid at a low temperature (>~5°C) (i) p-Toluidine (ii) p-Phenylene diamine (iii) N-Methyl aniline (iv) N, N-Dimethyl aniline (v) o-Anisidine. Q.117. Indicate the product obtained and the pH required during the coupling reaction between (i) C6H5N=NCl and C6H5NH2 (ii) C6H5N=NCl and C6H5OH.

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Q.118. A student added a cooled aq. solution of sodium nitrite to a cold solution of aniline in dilute HCL, a yellow precipitate was formed instead of a clear solution. Identify the mistake and suggest precautions to correct it. Q.119. An amino compound gave oily droplets on reaction with nitrous acid. What could be the class of amine and which confirmatory test should be performed with the oily droplets and how? Q.120. Give a laboratory test and explain the chemical reactions of the following amines with nitrous acid (i) Aliphatic tertiary amine (ii) Aromatic tertiary amine (iii) Aliphatic primary amine (iv) Aromatic primary amine (v) Aromatic secondary amine. Q.121. Why do most aromatic amines decolourise bromine and potassium permanganate solution? Write the reaction of following amines with these reagents (i) Aniline (ii) N,N-Dimethyl aniline (iii) o-Nitro aniline. Q.122. Can methyl amine, dimethylamine and trimethylamine amine be distinguished from each other using IR spectroscopy? Explain your answer. Q.123. Can PMR spectrum data be used to distinguish two isomeric amines having a molecular formula C2H7N? Q.124. Does the position of IR absorption signal due to —NH2 shift with change in concentration of amine solution? Explain. Q.125. Why do some amines give a positive test with a solution of ferric chloride? Q.126. What are? (i) Amides (ii) Anilides (iii) Toludides (iv) Lactams (v) Imides. Q.127. What will be the products of acid catalysed hydrolysis of the following compounds? Write a balanced equation for each example (i) Acetamide (ii) Benzamide (iii) Benzanilide (iv) N-Ethyl acetamide. Q.128. Suggest a laboratory test to distinguish the two compounds in each pair given below. (i) Acetamide from benzamide (ii) Acetamide from N-methyl acetamide (iii) Benzanilide from N-benzyl acetamide (iv) A lactam from an amide.

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Q.129. Do amides give Hydroxamic acid and 2,4-DNP test? Explain your answer. Q.130. How will you know the presence of a –CONH2 group in the IR spectrum of a compound?

Q.131. Can you distinguish an amide from an imide by using their IR spectra? Q.132. Which of the following group will require least energy for the red radiations? Why?

stretching when irradiated with infra

(i) –COOH (ii) –COOR (iii) CONH2.

Q.133. Can you distinguish acetamide, propanamide and benzamide from each other on the basis of their PMR spectra? Explain. Q.134. What reagents are used for reduction of (a) Nitrobenzene to (i) Aniline (ii) Phenylhydroxylamine (iii) Azobenzene (iv) Hydrazobenzene. (b) m-Dinitrobenzene to (i) m-Phenylene diamine (ii) m-Nitraniline? Q.135. What is Mulliken Baker test? Discuss in detail the chemistry of the test. Q.136. What is Tollens’ reagent? How is it prepared in the laboratory and why it is prepared fresh just before use? What is the risk in storing it in the laboratory? Q.137. How is nitro benzene tested in the laboratory by first reducing it to aniline? Q.138. How can nitro methane be distinguished from nitro benzene in the laboratory? Q.139. Where does the –NO2 group show an absorption signal in the IR spectrum of nitrobenzene?

Q.140. What will be the difference in the IR spectrum of aniline and nitrobenzene?

Q.141. What changes would you expect in the PMR spectrum of nitrobenzene on reduction to the following? (i) Aniline (ii) Hydrazobenzene (iii) Azobenzene? Q.142. Can PMR spectroscopy be used for distingtion between the following compounds? Explain your answer. (i) Nitrobenzene from p-Dinitrobenzene (ii) Nitrobenzene from Nitrosobenzene (iii) Benzidine from Nitrobenzene. Q.143. Where do the following bonds show absorption due to stretching in the IR spectrum?

R-Cl, Ar-Cl, R-Br, R-NH2 Ar-NH2

Q.144. What special points must be considered before starting to prepare a derivative of a given compound? Q.145. Why should the melting point of the derivative be different from that of the given compound? Q.146. Why should the derivative be recrystallised before determining its melting point?

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Q.147. Fill in the blanks • The derivative should be almost----- in the solvent selected for recrystallisation at room temperature. • A ------ solution of the derivative should be prepared in the ---- solvent for recrystallisation. • The impurities tend to get ----- with the derivative, if the hot solution prepared for crystallisation is quenched. • A derivative (x) can be purified from its impurity (y) by crystallisation if (y) is ----- or freely ----- in the selected solvent. Q.148. Name a few common solvents that are used for purification of organic compounds. Q.149. The crystallised sample of Glucosazone prepared from pure sample of glucose by a student shows a lower melting point compared to the literature melting point. Identify the impurities that could have caused lowering of melting point. Q.150. Which of the following cannot form an osazone and why? (i) Ribose (ii) 2-Deoxyribose (iii) Fructose. Q.151. Why Glucose and Fructose form identical osazone but Glucose and Galactose form non-identical osazone? Q.152. What is the difference in structure between α- and β-Glucopentaacetate? Q.153. Which out of the following forms an octaacetate on complete acetylation? (i) Sucrose (ii) Fructose (iii) Starch. Q.154. What is benzoylation reaction? Why is sodium hydroxide added during the reaction? How many moles of benzoyl chloride will be required for complete benzoylation of a mole of each of the following? (i) Resorcinol (ii) Orcinol (iii) Phloroglucinol (iv) p-Cresol. Q.155. Why is acetic anhydride preferred to acetic acid for acetylation reactions? Q.156. Why most of phenols cannot be nitrated directly with nitrating mixture? Q.157. Which of the following phenol can give a bromo derivative on reaction with bromine? Resorcinol or 2,4,6- trinitro phenol. Q.158. Why is alkaline hydrolysis of an ester preferred over acid hydrolysis? Q.159. Which of the following ester will give +ve iodoform test with the distillate and a water insoluble acid on hydrolysis?

Phenyl benzoate, methyl benzoate, ethyl acetate or ethyl benzoate.

Q.160. Two esters A and B are separately subjected to alkaline hydrolysis. In the flask containing ester A, a clear solution is obtained whereas an oily layer is present in the flask containing ester B, even after complete hydrolysis. Explain the observation. Q.161. Why is the 3,5-dinitrobenzoate prepared from an ester treated with a bicarbonate solution before recrystallisation?

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Q.162. Name any three derivatives that can be prepared for acetone. Q.163. Why is sodium acetate used in the reaction between cyclohexanone and hydrazine hydrochloride in the preparation of cyclohexanone hydrazone? Q.164. Acetophenone can form two isomeric oximes. How can these be distinguished? Q.165. Why is conc., sulphuric acid used while preparing 2,4-dinitrophenyl hydrazone of an aldehyde/ketone? Q.166. What derivative should be prepared to differentiate the two alcohols given below? Explain. n-Propyl alcohol

b.p. 97°

3,5-Dinitro benzoate m.p. 74°C

p-Nitro benzoate m.p. 35°C

Sec. Butyl alcohol

b.p. 99°C

3,5-Dinitro benzoate m.p. 76°C

p-Nitro benzoate m.p. 26°C

Q.167. Why should a water insoluble alcohol be never used even in slight excess for the preparation of its 3,5-dinitro benzoate? Q.168. What product results on mono nitration of the following hydrocarbons? Naphthalene Toluene p-Xylene Q.169. Compounds given below are subjected to oxidation with alkaline potassium permanganate. Write the structure of the oxidised compound in each case. Toluene o-Xylene naphthalene Q.170. What is picric acid? Why it is stored under water in a bottle? Write the structure of picrate of naphthalene Q.171. Name any two derivatives that can be prepared for a given ether. Q.172. Name one derivative that can be prepared for both acetic anhydride and aniline. Q.173. Fill in the blanks A lactone gives a -----, a mixed anhydride a ----- each of two ----- and an ----- gives an alcohol and an acid on hydrolysis. Q.174. Why is sulphuric acid used with nitric acid for nitration of compounds? Q.175. Why does a nitro group already present on an aromatic ring directs the incoming nitro group to the meta position? Q.176. Name a few reagents that can be used to convert nitrobenzene to aniline. Q.177. Why 2,4-dinitrotoluene should either be reduced to the corresponding amine or oxidized to the acid and not nitrated for its characterisation? Q.178. Match the compounds in the left column with their derivatives in the right column. Nitrobenzene

p-Nitro chlorobenzene

p-Nitrotoluene

p-Nitrochlorobenzene

m-Dinitrobenzene

p-Nitro benzoic acid

Chloro benzene

m-Nitroaniline

Q.179. Name at least two derivatives each that can be made from primary and secondary amines. Can similar derivatives be prepared from tertiary amines? If your answer is no, suggest an alternate derivative for tertiary amines. Q.180. Which of the following cannot form a derivative on reaction with sodium nitrite and hydrochloric acid and why? Write the structure of derivative that will be formed from the other amines given below.

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Q.181. Q.182. Q.183. Q.184.

Name and write the structure of methiodide and ethiodide of N,N-Dimethyl and N,N-Diethyl aniline. Why is urea converted to urea oxalate and not hydrolysed like other primary amides? What are the products of hydrolysis of (i) benzamide (ii) benzanilide (iii) acetanilide? Both aniline and alanine are converted to their respective benzoyl derivatives. Why is the work up procedure different in two cases? Q.185. What will be the major difference in the IR spectrum of the compounds in column A and their respective derivatives in column B? Column A

Column B

Hydroquinone

Hydroquinone diacetate

Acetic acid

Acetamide

Aniline

2,4,6-Tribromo aniline

Nitrobenzene

Aniline

Benzaldehyde

Benzoic acid

Q.186. How is the progress of a reaction monitored by performing a TLC? Q.187. Which type of spectroscopy (IR or NMR) can assist in differentiating the N-acetyl from N, N-diacetyl derivative of aniline? Q.188. The observed melting point of the following derivatives was found to be lower than the expected melting point. What could be the impurity present in each one of these derivative. Suggest a chemical method for purification. 3,5-dinitrobenzoate of methanol. Anilide of benzoic acid Partial and complete reduction product of m-dinitrobenzene Anhydride of phthalic acid. Q.189. What precautions MUST be observed in the laboratory while using ether as a solvent. Q.190. How can you separate out a water soluble compound from its aqueous solution? Q.191. How will you proceed to separate a mixture of two unknown compounds? Q.192. Is it essential to carry out a few preliminary tests with the mixture of compounds before starting to separate its components? Explain your answer with a suitable example. Q.193. Match the mixture of compounds (that can be separated) in the left hand column with the reagent on the right hand column A water soluble compound from a water insoluble Dilute sodium hydroxide solution, only one component compound will dissolve A water insoluble hydrocarbon from a carboxylic acid

Either using sodium bisulphite or dilute hydrochloric acid solution

A water insoluble amino compound from a water insoluble Either dilute sodium hydroxide or hydrochloric acid ketone A water insoluble acid from a water insoluble amine

Water to dissolve one component

A carbohydrate from a hydrocarbon

Ether, only one compound will dissolve

Q.194. Write the name and structure of one derivative that is prepared for each of the following: N-Methylaniline, Benzoyl chloride, glycine and acetamide Q.195. How will you convert (i) Sulphonic acid to sulphonamide (ii) Thiol to 2,4-dinitrophenyl ether (iii) Sulphonamide to N-acetylsulphonamide.

2 Spectroscopy 2.1 INTRODUCTION Spectroscopic identification of an unknown organic compound is important for training students of chemistry as these techniques are extensively used in R&D and in industry for development of products useful to mankind. In addition, these techniques have the following advantages: • Small quantity (~50 mg or even less) of the compound is sufficient for structure elucidation, that too can be recovered by evaporation of the solvent used for recording the spectrum (except the compound used for recording the mass spectrum). • The techniques are eco friendly because very little chemical waste is produced. • The identity or non identity of samples can be established easily. The chemical methods used for qualitative analysis have the following advantages: • A student gets a good training of performing the chemical tests, determining the physical constants, learns the techniques like chromatography, solvent extraction, sublimation etc. A student gets an opportunity to do synthesis of compounds while preparing the derivatives of the given compound etc. thus a student gets a laboratory training for isolation of compounds from natural sources, their purification, identification and synthesis in the laboratory. • A student also gets trained to handle chemicals and their proper disposal etc. The main disadvantage of this type of work is the consumption of huge volumes of chemicals and their safe disposal. Four types of simple spectroscopic techniques are used at under and post graduate level for structure elucidation of an unknown organic compound. IR Spectroscopy UV Spectroscopy NMR Spectroscopy Mass Spectrometry When a compound is exposed to a beam of electromagnetic radiations, it undergoes changes at the molecular level resulting due to absorption of energy. All wavelengths are not absorbed by all the molecules or a single molecule. Only those which have just the right amount of energy to bring about a specific

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change in the molecule are absorbed.The nature of these changes depends on the energy associated with the electromagnetic radiations. For example, on exposure to UV radiations, generally the π or non bonded (n) electrons are excited from a lower to a higher energy level. When irradiated with IR radiations, the molecule shows changes in the vibration levels of the bonds present in it and when bombarded with high energy electrons (basis of mass spectroscopy) the molecules undergo fragmentation etc. Any electromagnetic radiation is expressed by its: • Wavelength: λ (in meters: symbol m) The other SI units are 1 Km = 1000 m (Km is kilometer) 1 µm = 10–6 m (µm is micrometer) 1 nm = 10–9 m (nm is nanometer) • Wave number (v ) : 1/λ (in reciprocal m–1 or cm–1) • Frequency: v (in seconds, s–1 or Hertz (Hz) 1s–1 = 1 Hz • Energy (in joules J) Table 2.1 below gives a brief account of the radiations and their effect on the molecules. Table 2.1 Type of radiation Ultra violet (λ = 190 – 400 nm) Visible (λ = 400–800 nm) Infra red λ = 2.5 – 25 µm v = 400–4000 cm–1 Microwave (v = 9.5 × l09 Hz) Radio frequency (v = 60 – 600 MHz) Electron beam (energy = 6000 KJ mol–1)

Effect on the molecule Changes in the electronic level within the molecule Changes in the rotational and vibrational levels in the molecule

Induce changes in the magnetic properties of the unpaired electrons Induce changes in the magnetic properties of certain nuclei lonisation and fragmentation of the molecule

Thus when a molecule is irradiated with a beam of electromagnetic radiations, some are absorbed and some pass unchanged. The graph showing the absorption vs the wave length/frequency etc. is known as a spectrum and the instrument used for measuring absorption and recording the spectrum as spectrophotometer.

2.2 IR SPECTROSCOPY 2.2.1 Introduction When a molecule is irradiated with monochromatic beam of IR radiations, some of these are absorbed and result in an absorption signal which is recorded by the spectrophotometer. In a molecule, a covalent bond is compared to a spring and the atoms attached to balls. The atoms and bonds in a molecule are in constant motion (rotation and vibration). The IR radiations which have a frequency less than 100 cm–1 cause a change in the rotational levels of the molecule whereas the IR radiations with a frequency between 4000–600 cm–1 result in vibrational changes in a molecule. It is these changes which give useful

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information about the structure of a molecule specially the functional groups present in it. Only those vibration changes which result in a net change in the dipole of the bond are IR active (these appear as absorption signals), the others which do not lead to a change in the dipole of the bond are IR inactive (these vibrational changes do not result in the absorption of IR radiations). The more the difference in the dipole moment of the excited state and the ground state, the more is the intensity of absorption band. For example, the stretching of O–H, C=O, –N=O bonds produces a strong band, whereas the C–C, C=C, C≡C produce a weak absorption band. There are two basic types of transitions that occur when a molecule absorbs IR radiations, these are: • The covalent bonds in the molecule stretch i.e. the bond length between the atoms changes. • The bonds bend i.e. the bond angle or the plane of the bond changes. More energy is required to stretch a bond than to bend it The energy required for stretching a bond depends on a number of factors but the significant ones are: 3 The bond strength C≡C is stronger than C=C is stronger than a C–C C≡C absorbs at ~ 2150 cm–1 C=C at ~ 1650 cm–1 C–C at ~ 1200 cm–1 3 The bond order C=C in ethene is stronger than C=C in benzene, the carbon–carbon bond in benzene has a partial double bond character because of delocalisation of π electrons C=C of ethene absorbs at 1650 cm–1 C=C of benzene at 1600–1450 cm–1 3 The mass of the atoms attached to the bond C–Cl C–Br C–I –1 –1 600 cm 500 cm–1 750 cm The vibrational frequency of a bond can be calculated with reasonable accuracy by application of Hooke’s law, which correlates the frequency with the bond strength and the atomic masses of the atoms vα

bond strength mass

 k 1  v =  2πC  m1m2 / (m1 + m2 )  v =

 m + m2  1 K, 1  (m1m2 )  2πC 

1

2

where v is frequency of vibration. K–force constant —a constant related to the strength of the bond (dynes/cm) m1 and m2 are masses of two atoms The factor m1m2 (m1 + m2) is called the reduced mass (µ) C is velocity of light

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2.2.2 Types of Stretching Vibrations In a diatomic molecule, only one type of stretching mode is possible as there is only one bond in the molecule, for example in a diatomic molecule A–B, the bond can stretch, whereas in a tri atomic molecule, two types of stretchings can occur: Symmetric and Asymmetric stretching. In a tri atomic molecule A–B–C, if both bonds A–B and B–C stretch to increase the bond length, the stretching is known as symmetric stretching. If one bond length increases, it is known as asymmetric stretching.

Fig. 2.1: Stretching modes in a triatomic molecule A–B–C

The asymmetric stretch requires more energy than symmetric stretch.

2.2.3 Types of Bending Vibrations There can be two types of bending vibrations. These are: In plain bending Out of plain bending In plain bending In this type of bending, as the term suggests, the original plain of the atoms attached through a bond remains the same but the bond angle changes.

Fig. 2.2: In plane bending in a triatomic molecule ABC

Scissoring (movement like two blades of a scissors) and rocking are example of bending in the same plain. Out of plain bending Twisting and Wagging are two examples of out of plain bending. In twisting, one bond moves in front of the original plain and the other towards the back. In wagging both the bonds move either in front or back of the plain.

Fig. 2.3: Out of plane bending in a triatomic molecule ABC

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The IR radiations are either expressed in wavelength (λ, 2.5–15 µm) or wave number (4000–600 Since the energy is inversely proportional to λ, the longer the wavelength, less is the energy and the larger the wave number, larger is the energy associated with it. The IR spectrum or the graph is plotted between absorbance or transmittance and wavelength or wave number. The relationship between absorbance and transmittance is given by the following formula cm–1).

A = log10 (1/T) where A is absorbance and T is transmittance.

The IR spectrum is generally quite complex because most organic molecules are polyatomic and can show a large number of vibrational changes on exposure to IR radiations. The intensity of bands is normally expressed qualitatively as a weak, medium intensity or strong band while describing the IR spectrum of a molecule.

Most organic compounds absorb in the above region. The region between 2.5–7 µm (4000–1500 cm–1) is known as the functional group region as most functional groups present in organic compounds absorb in this region of wavelength/wave number. The region between 7–15 µm (1500–600 cm–1) is known as the finger print region because like the finger print of human beings, it is specific to each compound. The region between 900–675 cm–1 is referred to as aromatic region as the absorptions in this region indicate the number and relative positions of the substituents present in a aromatic ring.

2.2.4 Instrument Two types of spectrophotometers are in use Dispersive spectrophotometer Fourier transform spectrophotometer The spectrophotometer consist of the following: An infrared energy source: any of the below mentioned filament is electrically heated to 1800°C to get the infra red radiations. Nerst filament–made from a mixture of sintered oxides of Cerium, Thorium, Zirconium, Etrium etc. Globar filament–made of silicon carbide Reference and Sample cell: one of this holds the sample under study and the other only the medium for example if CC14 has been used for dissolving the sample, the reference cell contains only the solvent. Monochromator: it resolves the IR radiations produced by the source into beams of monochromatic radiations of increasing frequency/wavelength. Detector: detects the frequency/wave number of IR radiations that are absorbed by the sample. Amplifier: it amplifies the absorption signal. Recorder: automatically records the spectrum.

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Fig. 2.4: Schematic diagram of dispersive infrared spectrometer

2.2.5 Sample Preparation For recording the spectrum, the sample is not taken in a glass or a plastic container as both absorb in the entire IR range. Most organic solvents have a complex spectrum themselves and hence are not suitable for dissolving the compound. Solids The solid compounds are taken in one of the following ways for recording the spectrum Nujol mull KBr pellet Solution in carbon tetrachloride Nujol mull The solid compound is grinded with nujol (mineral oil, it is a mixture of hydrocarbons), the thick suspension is called the mull. The mull is placed on a sodium chloride plate/disk, covered with another sodium chloride plate and placed in the holder in the spectrophotometer. Nujol shows absorption bands at 2924, 1462 and 1377 cm–1 in the IR spectrum and hence the absorptions due to the compound are masked by those of nujol in these regions. KBr pellet The solid sample is thoroughly grinded with potassium bromide, the mixture is pressed under pressure to form a pellet and then dried to expel moisture. The pellet is placed in the holder for recording the spectrum. KBr is transparent in the entire IR region and hence does not interfere, therefore this is the most commonly used method for recording the spectrum. Solution in carbon tetrachloride The organic compound is dissolved in carbon tetrachloride, a drop of this solution is placed between two sodium chloride plates and the spectrum recorded. Carbon tetrachloride absorbs strongly around 785 cm–1. Liquids A liquid compound may be used in one of the following ways:

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Neat liquid A drop of the liquid compound is placed between two sodium chloride plates and the spectrum recorded. KBr pellet As described above for solid compounds.

2.2.6 Uses of IR Spectroscopy IR spectrum provides following information about the compound • The presence or absence of a functional group • Identity or non–identity of two compounds • The substitution pattern of aromatic compounds and olefins etc. • Presence or absence of hydrogen bonding and inter and intramolecular hydrogen bonding. 2.2.7 IR Absorptions of Some Common types of Compounds Following section gives the position and intensity of bands of some important functional groups. It must be emphasised here that the concentration of the sample, the solvent/medium used for recording the spectrum as well as small structural changes in the molecule lead to a shift in the position of absorption signal. It is also important to mention that every organic molecule, besides the functional group, also has carbon skeleton. For example, acetone and benzophenone both show an absorption signal due to C=O stretch but in the IR spectrum of acetone Csp3–H and methyl bending are seen whereas in benzophenone, absorptions due to stretching of Csp2–H and C=C (aromatic) are present. 2.2.8 Hydrocarbons Compounds containing carbon and hydrogen can be saturated, unsaturated (alkenes and alkynes) or aromatic in nature. 2.2.9 Saturated Hydrocarbons Their spectrum is normally simple. These molecules can absorb IR radiations resulting from the following transitions. C–H stretching and bending C–C stretching and bending –CH3, –CH2, –CH(CH3)2 and (CH3)3C–stretching and bending The Csp3–H; stretch occurs around 3000–2840 cm–1 C–C: stretching absorptions are weak and insignificant. –CH2: bending is seen at ~1465 cm–1 and rocking at ~720 cm–1. –CH3 : shows a characteristic symmetrical bending absorption at 1375 cm–1 and asymmetric bending near 1450 cm–1. –CH(CH3)2 shows a symmetrical doublet one at 1385–1380 cm–1 and the other at 1370–1365 cm–1 . –C(CH3)3 shows an unsymmetrical doublet, the longer wavelength band is more intense. Below (Fig. 2.5) is given the IR spectrum of a saturated hydrocarbon (molecular formula C6H12). Suggest the structure of the compound and explain how you arrived at it.

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Fig. 2.5: IR spectrum of a saturated hydrocarbon

Hint: Absence of an absorption signal at 1375 cm–1.

2.2.10 Alkenes The IR spectrum of an alkene is normally a complex spectrum and the exact position of absorption depends on a number of factors like. • The substitution pattern around the double bond • The geometry around the double bond • Conjugation • The size of the ring containing the double bond in cyclic alkenes. =C–H stretch observed above 3000 cm–1 (generally between 3095–3010 cm–1) =C–H bending occurs between 1000–650 cm–1 C=C stretch: In an unconjugated linear alkene, the C=C stretch appears at 1675–1640 cm–1. The C=C stretching is shifted with introduction of alkyl substituent on the sp2 hybridised carbon atom CH3CH=CH2 (CH3)2C=CH2 (CH3)2C=CHCH3 1640 cm–1 1650 cm–1 1670 cm–1 The C=C Stretching is also affected by the geometry around the double bond. In a symmetrically substituted alkene, the C=C stretch in the trans isomer is a very weak absorption. The cis isomer produces a stronger absorption. In a diene, if the double bonds are isolated, the position of absorption is not influenced much by the second double bond, if however the second bond is conjugated, it shifts the C=C stretching to a lower frequency due to resonance, which reduces the double bond character of both the bonds. CH2=CH–CH=CH–CH3 CH2=CH–CH=CH2 CH2=CH–CH2–CH2–CH=CH2 –1 –1 ~1650 cm 1650 and 1600 cm one band at 1600 cm–1

2.2 IR Spectroscopy Advanced Experimental Organic Chemistry

Fig. 2.6: IR spectrum of cyclohexne

Fig. 2.7: IR spectrum of isoprene

The extension of conjugation shifts the absorption further to a lower frequency. The size of the ring in cyclic alkenes greatly influences the position of C=C stretching.

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Introduction of an alkyl group on the double bond carbon of a cyclic alkene has pronounced effect on the position of absorption of C=C streching In Alkenes containing an exocyclic double bond, the strain effects the absorption as shown below.

The C–H bending is also affected by the structure and substitution pattern around the double bond. The C–H scissoring for a terminal alkene appears at 1415 cm–1. The C–H out of plain bending is important as it provides a valuable information about the compound. These absorption bands appear from 1000–650 cm–1. Mono substituted alkene: Two absorption bands, one near 990 cm–1 and the other near 910 cm–1 are seen in the IR spectrum of mono substituted alkenes. An overtone at 1820 cm–1 confirming the presence of a mono substituted alkene. Disubstituted alkene: 1,1–Disubstituted alkene These types of alkenes produce a strong absorption band near 890 cm–1. 1,2–Disubstituted alkenes A cis-disubstituted alkene gives a strong absorption band near 700 cm–1. A trans-disubstituted alkene gives a band near 970 cm–1. Tri substituted alkene A medium intensity band is seen near 815 cm–1. Tetra substituted alkene No absorption band is seen due to the =C–H bond in these alkenes.

2.2.11 Alkynes –C≡C stretching occurs at 2260–2100 cm–1. No absorption signal is observed for a symmetrically substituted alkyne. –C≡C–H stretching occurs in a terminal alkyne at 3330–3260 cm–1. –C≡C–H bending is seen at 700–610 cm–1. As in the case of alkenes, the position of absorption is affected by conjugation, introduction of alkyl groups etc. for example for a monosubstituted alkyne the C≡C appers at 2140–2100 cm–1. The disubstituted alkynes in which the two substituents are different, this stretching appears at 2260–2190 cm–1.

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Fig. 2.8: IR spectrum of 1-Hexyne

2.2.12 Aromatic hydrocarbons Significant absorption signals in the IR spectrum of aromatic organic compounds are =C–H stretching occurs between 3100–3000 cm–1 (>3000 cm–1). =C–H bending (out of plain) is observed between 900–675 cm–1. These bands and the corresponding overtones are of great significance in obtaining information about the substitution pattern in aromatic compounds. =C–H bending (in plain) occurs between 1300–1000 cm–1. These bands are mostly overlapped by other stronger bands and therefore are not of much help in structure elucidation. C=C stretching causes absorptions to occur at 1600–1585 and 1500–1400 cm–1. =C–H out of plain bending absorptions are of a great diagonostic value as these can be used to derive at the substitution pattern of some aromatic compounds (like alkyl, alkoxy, amino, carbonyl and halo substituted compounds). Reliable absorption patterns are not obtained in some aromatic compounds (like nitro, carboxylic, sulphonic acids and their derivatives). • In a mono substituted aromatic compound, two absorption bands are seen, one at 750 and the other at 690 cm–1.

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Fig. 2.9: IR spectrum of Tuluene (neat Liquid)

In disubstituted aromatic compounds, the position of absorption varies for the o-, m- and p-disubstituted aromatic compounds. • For o-disubstituted aromatic ring a strong absorption band is seen neat 750 cm–1

Fig. 2.10: IR spectrum of o-xylene

• Two strong and one medium intensity absorption band at 780, 690 and 880 cm–1 respectively are observed for a m-disubstituted compound.

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Fig. 2.11: IR spectrum of m-xylene

• One single band appears for a p-disubstituted aromatic compound in the region 850–800 cm–1

Fig. 2.12: IR Spectrum of p-xylene

2.2.13 Alcohols O–H stretching is the most significant absorption signal in the IR spectrum of alcohols though its position is variable and mainly dependent on the concentration of the sample. A strong signal due to O–H stretching appears between 3650–3200 cm–1. C–O stretching occurs between 1260–1000 cm–1 and is useful in distinguishing between a primary, secondary and a tertiary alcohol. C–O–H bending occurs between 1440–1220 cm–1.

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• O–H stretching: As already mentioned, the position and pattern of absorption is concentration dependent i.e. presence and absence of hydrogen bonding.The hydrogen bonding reduces the bond strength of O–H bond, the stronger the hydrogen bonding, weaker is the O–H bond, less the energy required to stretch it. A broad and a strong absorption signal is obtained between 3500–3200 cm–1 of an alcohol (bonded O–H). • If the concentration of the compound used for recording the IR spectrum is high. This happens because of intermolecular hydrogen bonding. • If the compound has intramolecular hydrogen bonding. The IR spectrum of t-butanol (neat liquid) Fig. 2.13 shows a broad band at 3380 cm–1 due to O–H stretch

Fig. 2.13: IR spectrum of t-butyl alcohol (neat)

A sharp and a strong band appears at 3650–3580 cm–1 if the concentration of the sample is low so that there is no hydrogen bonding (free O–H). The IR spectrum of t–butanol (CCl4 solution) showing a sharp absorption band of O–H (free) stretching at 3600 cm–1.

Fig. 2.14: IR spectrum of t-butyl alcohol in CCl4 (a free –OH group)

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Generally for an alcohol (except those which contain intramolecular hydrogen bonds) both the absorption signals are obtained. C–O stretching of alcohols is generally observed between 1260–1000 cm–1. The factors influencing this absorption are: • The nature of the alcohol i.e. whether primary, secondary or tertiary. • Hybridisation of carbon atom β– to the –OH group For a primary alcohol it occurs at 1050 cm–1 (free O–H at 3640 cm–1), for a secondary alcohol it at 1100 cm–1 (free O–H at 3630 cm–1) and for a tertiary alcohol at 1150 cm–1 (free O–H at 3620 cm–1). C–O–H bending appears between 1440–1220 but is generally masked by strong absorption signals in this region generally due to –CH3 and –CH2 bending absorptions.

2.2.14 Phenols Like alcohols, phenols also show three significant bands O–H stretching is concentration dependent and the free O–H appears at 3610 cm–1, it at a value lower than most alcohols. Intramolecular hydrogen bonding shifts it to 3200 cm–1. C–O stretching requires more energy to stretch. Because of resonance in phenols, the C–O bond acquires some double character and appears at 1220 cm–1. C–O–H bending occurs between 1450–1220 cm–1. Below is given the IR spectrum of phenol (KBr pellet)

Fig. 2.15: IR Spectrum of phenol (KBr pallet)

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2.2.15 Ethers The main diagonostic peak in ethers is that of C–O stretch. C–O stretch appears between 1300–1000 cm–1. Its position depends on the nature of ether. In a symmetrical ether, the symmetric C–O stretch is IR inactive but the asymmetric stretch is IR active, some of the important absorption positions are ROR: The asymmetric C–O stretch occurs at l120 cm–1 as a strong absorption signal, the symmetric stretch is very weak and appears at 850 cm–1. ArOR: The O–Ar bond is stronger than the O–R bond due to resonance, hence the asymmetric stretch absorption appears at 1250 cm–1. Symmetric stretch is also IR active and is seen near 1040 cm–1. C=COR: The asymmetric and symmetric C–O stretch occur at 1220 and 850 cm–1 respectively.

Fig. 2.16: IR spectrum of diisopropyl ether (neat liquid)

2.2.16 Epoxides and Cyclic Ethers Epoxides are small ring compounds containing an ether linkage and show absorption bands at 1280–1230 cm–1 (weak signal due to ring stretching), 950 and 815 cm–1 (strong signal due to ring deformation). 2.2.17 IR Spectra of Compounds Containing a C=O bond Aldehydes, ketones, carboxylic acids, esters, amides, acid anhydrides and acid halides all contain a C=O group. The absorption signal due to C=O stretch is one of the strongest signal in the IR spectrum. The exact position depending on the nature of substituents attached to the carbon of the double bond which influence the strength of the carbon oxygen double bond, the major factors being the • Resonance effect • Inductive effect • Steric strain • Angle strain

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Table 2.2 gives a brief summary of factors that influence the position of absorption of C=O group in unsubstituted simple compounds containing a C=O group. Table 2.2 Type of compound

Nature of influence

Position of absorption signal in cm–1

Amide RCONH2

~ 1690

Acid RCOOH

~ 1710 (broad signal)

Ketone RCOR

~ 1715

Aldehyde RCHO

~ 1725

Ester RCOOR

~ 1735

Anhydride (RCO)2O

~ 1760, 1810

Acid chloride RCOCl

~ 1800

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2.2.18 Aldehydes Aldehydes show a strong absorption band due to C=O stretching at 1740–1725 cm–1 which is influenced by the nature of R group attached to the carbonyl carbon. In HCHO the C=O stretch appers at 1750 cm–1 and CH3CHO at 1731 cm–1. An electron withdrawing group (–I effect) attached to a–carbon increase the strength of C=O group and hence shifts the absorption, the more the number of such groups, the cumulative is the effect on the position of absorption 1730 cm–1 1768 cm–1 CH3CHO Cl3CCHO

Fig. 2.17

Fig. 2.18

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An electron donating group (+ R effect) decreases the bond strength of C=O group of the aldehyde and thus the energy required to stretch it, shifting the position of absorption to a longer wavelength.

The C–H stretching of –CHO group is specific to aldehydes and no other carbonyl compound shows this absorption. One weak band appears at 2860-2800 cm –1 and the other at 2760-2700 cm –1. The band at ~2800 cm–1 is generally masked by other stronger absorptions in this region.

2.2.19 Ketones Like other carbonyl compounds, the strength of the C=O bond in ketones varies with the nature of the groups attached to the carbon of the C=O group. Presence of electron donating groups and conjugation reduces the C=O bond strength, the absorption correspondingly shifting to lower frequency and the electron withdrawing groups exert an opposite effect. Aliphatic dialkyl ketones : The C=O stretch of RCHO appears at ~1725 cm–1 but for ketones the position of the C=O stretch is shifted to 1715 cm–1 because of the presence of two electron donating groups. 1730 cm–1 1715 cm–1 1715 cm–1 CH3CHO CH3COCH3 cyclohexanone –1 IR spectrum of acetone showing a C=O stretch at 1715 cm .

Fig. 2.19

Alkyl aryl ketones Conjugation with a phenyl group decreases the bond strength and the absorption signal shifts to 1685 cm–1.

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IR of acetophenone showing C=O stretch at 1685 cm–1.

Fig. 2.20

Di aryl ketones There are two aryl groups in conjugation with the C=O group which further shift the absorption to 1665 cm–1. a, β-unsaturated ketones Because of the conjugation, the C=O stretch shifts to 1680 cm–1. Cyclic ketones The position of absorption depends on the ring size of cyclic ketones. The bond order of C=O in cyclic ketones is the key factor influencing the position of absorption due to C=O stretch. In cyclohexanone, the bond C–C(O)–C is not strained and hence the bond order of C=O in cyclohexanone is same as that of acetone and these ketones absorbs at almost the same position. In other cyclic ketones the bond angle (C–C(O)–C) is decreased, for example it is 90° in cyclobutanone and 60° in cyclopropanone compared to cyclohexanone in which it is 120°. Because of this the C–C(O)–C bond is strained and the C of C=O acquires more of s character compared to sp2 character of the C of acyclic ketones, the C=O bond strength is increased and it requires correspondingly more energy to stretch. Ketene > cyclopropanone > cyclobutanone > cyclopentanone > cyclohexanone, hence the absorption position is 2140, 1815, 1780, 1745 and 1715 cm–1 respectively. Diketones 1, 2–diketones (a–diketones) The C=O stretching of the aliphatic and aromatic a-diketones are given below: CH3–CO–CO–CH3 C6H5–CO–CO–C6H5 1716 cm–1

1660 cm–1

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Fig. 2.21

1,3–diketones (β–diketones) These normally exist in equilibrium mixture of keto and enol form. There are two C=O groups in the keto form which result in the asymmetric and symmetric stretching absorptions at 1725 and 1705 cm–1 respectively. The C=O stretch of the enolic form (which is hydrogen bonded with the H of O–H group) occurs at 1620 cm–1 and O–H stretch at ~ 3000 cm–1 CH3–CO–CH2–CO–CH3

Fig. 2.22

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Fig. 2.23

2.2.20 Carboxylic Acids The important absorption signals in the IR spectrum of a carboxylic acid are due to C=O stretch O–H stretch O–H out of plain bending C–O stretch

Fig. 2.24: IR of acetic acid (neat liquid)

The carboxylic acids generally exist as dimmers, in which there is strong hydrogen bond between O of C=O and H of O–H group as a result of which both C=O and O–H bonds are weakend, resulting in shifting of absorption bands to lower frequency.

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In a simple aliphatic carboxylic acid (dimer) C=O stretch (bonded) appears in the range 1730-1700 cm–1 O–H stretch (bonded) extending from 3400–2400 cm–1 (as a broad absorption signal) If the IR is recorded in the vapour phase or by using an extremely low concentration of the acid such that the acid molecules exist as monomers C=O stretch (free) occurs at 1760-1730 and O–H stretch at 3500–3600 cm–1 Conjugation of C=O in aromatic acids and a, β-unsaturated carboxylic acids leads to shifting of this band to a lower frequency. For example in C6H5COOH: C=O stretch is seen at 1700 cm–1. O–H out of plain bending appears at 930 cm–1. The C–O stretch band appears at 1260 cm–1. Dicarboxylic acids There are two absorption signals in the IR spectrum of a dicarboxylic acid for example in the spectrum of phthalic acid, the two C=O stretch appear between 1750–1650 cm–1.

2.2.21 Anhydrides Anhydrides of aliphatic, aromatic and cyclic anhydrides of dicarboxylic acids are some of the common types of anhydrides. In RCOOCOR, there are two absorptions due to the stretching of two C=O groups in the IR spectrum, the asymmetric stretch appears at 1820 and symmetric stretch at 1740 cm–1. In ArCOOCOAr these are seen at 1800 and 1730 cm–1. In cyclic anhydrides, like for example succinic anhydride, the asymmetric and symmetric stretch appear at 1900 and 1800 cm–1 respectively.

Fig. 2.25

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2.2.22 Esters The significant IR absorptions seen in the spectrum of an ester are due to: C=O stretch C–O stretch C=O and C–O stretch, the position of which varies within a range depending on the structure of the ester. Following types of esters are common R–CO–O–R: RC=O stretch at 1750–1735 (e.g. 1738 in ethyl butyrate) O–R stretch at 1250 and 1055 cm–1. Ar–CO–O–R ArC=O stretch at 1740–1715 (1720 in ethyl benzoate) O–R at 1280 and 1100 cm–1. R–CO–O–Ar RC=O at 1770–1750 (1765 in phenyl acetate) O–Ar at 1200 and 1000 cm–1. Ar–CO—O–Ar ArC=O at 1740–1715 (e.g. 1715 in pheny benzoate) O–Ar at 1100 and 1000 cm–1.

Fig. 2.26: IR spectrum of methyl propionate (neat liquid)

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2.2.23 Acid Chlorides In acid chlorides, the electron withdrawing inductive effect of chlorine overpowers its electron donating resonance effect, as a result of which, the C=O is strengthen and it requires more energy to stretch.

Strong electron with drawing effect of Cl in an acid chloride. The C=O stretching occurs at 1780 cm–1 in ArCOCl.

Fig. 2.27

2.2.24 Acid Amides and Substituted Amides In acid amides, the electron withdrawing inductive effect of nitrogen is weaker than its electron donating resonance effect hence the overall effect of nitrogen is electron donation due to which, the C=O bond in amides is weaker as compared to acids, esters and acid chlorides. The OC–N bond also acquires considerable double bond character due to the mesomeric effect. The IR of amides is further complicated by hydrogen bonding.

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Fig. 2.28: IR spectrum of benzamide

In RCONH2, the C=O stretch is broad due to hydrogen bonding and occurs at 1680–1630 cm–1. In ArCONH2 at 1660 (amide 1) and 1620 (amide 11 band) cm–1. N–H stretch Primary amides show two absorptions due to asymmetric and symmetric stretch of N–H bonds at 3400 and 3200 cm–1 in a dilute solution of benzamide. In N–alkyl substituted amides, the C=O bond is further weakened and C–N bond strengthened due to electron donating inductive effect of alkyl group/s. The region of N–H absorption is also different as compared to primary amides (RCONH2). In RCONHR, for example, there is one absorption band due to N–H stretch at 3400-3200 cm–1 and in RCONR2, no absorption band in this region.

2.2.25 Amines Amines give absorption signal due to the vibrational changes in the following bonds N–H stretching N–H bending C–N stretching N–H stretching The position of N–H stretching, as in alcohols varies with the concentration but because of the lower electronegativity of nitrogen as compared to oxygen, the hydrogen bonds in amines are weaker and the

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difference in absorption frequency of free and bonded N–H is less as compared to the O–H group. The position of absorption is also different for aliphatic and aromatic amines. Primary aliphatic amines in dilute solution show two weak absorptions due to N–H stretch at ~ 3400–3330 and 3330–3250 cm–1. These bands are shifted to longer wave length (shorter frequency) if the amine concentration is high. Aromatic primary amines show two bands at a higher frequency as compared to aliphatic amines. Secondary amines show a single weak band between 3350–3310 cm–1. Tertiary amines do not have an N–H bond. N–H bending Absorption due to two types of bending modes may be seen in the IR spectrum of amines. Scissoring: Primary amines show medium to strong intensity band between 1650-1589 cm–1. Aromatic secondary amines absorb near 1515 cm–1. This band is normally not present in aliphatic secondary amines. Wagging: Absorption due to N–H wagging is seen between 900–660 cm–1. C–N stretching The C–N bond is stronger in aromatic amines (due to resonance) as compared to aliphatic amines. In aromatic primary amines the absorption occurs at 1340–1250 cm–1. In aromatic secondary amines the absorption occurs at 1350–1280 cm–1. In aromatic tertiary amines the absorption occurs at 1360–1310 cm–1.

Fig. 2.29: IR spectrum of aniline (neat liquid)

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Fig. 2.30: IR spectrum of N-methyl aniline

Fig. 2.31: IR spectrum of N, N-Dimethyl aniline

2.2.26 Nitro Compounds Two main absorption bands in nitro compounds are due to N=O stretch C–N stretch N=O stretch There are two absorption bands in the IR spectrum of nitro compounds due to asymmetric and symmetric stretch of N=O bonds at 1570-1550 and 1380-1370 cm–1.

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C–N stretch The C–N bond stretch in an aromatic nitro compound results in an absorption at 870 cm–1. In aliphatic nitro compounds it is at a lower frequency.

Fig. 2.32: IR of Nitrobenzene (Liquid film)

2.2.27 Nitriles C≡N stretch in aliphatic nitriles occurs at 2260–2240 cm–1. This occurs at 2240–2220 cm–1 in aromatic nitriles.

Fig. 2.33: IR of phenyl nitrile (Liquid film)

2.2.28 Amino acids Depending on the pH of the solution, the amino acids occur in an equilibrium mixture of the following forms +

+

−  −  H3 N − C H − COOH    H3 N − C H − COO   H 2 N − C H − COO | | | R R R Hence the IR spectrum of amino acids show absorption signals due to following:

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COO–: A strong absorption at 1600 (due to asymmetric stretch) and a weaker signal at 1400 cm–1 (symmetric stretch) is seen in the IR spectrum of amino acids. +

N H 3 stretching appears as a strong absorption band at 3100-2600 cm–1. +

N H 3 bending: A weak bending band occurs at 1660-1610 and a strong band near 1550–1480 cm–1. O–H stretch: A strong and broad band occurs at 3330-2380 cm–1, which is due to NH3+ and O–H stretching.

Fig. 2.34

2.2.29 Thioalcohols The main absorption band in the IR spectrum of a concentrated solution of a thioalcohol is due to S–H stretch which appears at 2600-2550 cm–1 for aliphatic compound as a weak absorption and is sometimes not detectable. In aromatic compounds it appears at ~3000 cm–1. The hydrogen bonding is also insignificant. 2.2.30 Sulphonic Acids The S=O stretch in the IR spectrum of sulphonic acids appears at 1350-1340 and 1165-1150 cm–1 due to asymmetric and symmetric stretch respectively. The absorption due to S=O is much weaker than that due to C=O. 2.2.31 Sulphonamides S=O stretch N–H stretch C–S stretch

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177

Fig. 2.35

Two absorption bands appear in the region 1370-1335 and 1170-1155 cm–1. The N–H stretching band appears at 3390-3330 and 3300-3245 cm–1 for primary sulphonamides. In secondary sulphonamides, the N–H stretch appears at ~ 3265 cm–1. The C–S stretching occurs between 700-600 cm–1.

Questions (iR) Q.1.

What do you understand by the following terms? Explain with a suitable example. Spectrum, Dipole moment, IR active and inactive vibration modes. Symmetric and asymmetric stretching, out of plain bending, nujol, overtone.

Q.2.

What is the range of IR radiations normally used in the IR spectroscopy? What do you understand by the terms:

Q.3.

Why glass containers are not used for holding the sample of a compound while recording the IR spectrum?

Q.4.

Which out of the following will result in the absorption of IR radiation due to bond stretching and why?

Functional group and finger print region and super imposable IR spectra?

O=C=O, CCl4, CH4, CH3Cl Q.5.

Why are IR radiations harmful to the human eye?

Q.6.

Why are some absorption signals strong and others weak? Explain your answer with suitable examples.

Q.7.

Indicate the approximate position of absorption of the following functional groups. C=O stretch in aliphatic aldehydes, C=O stretch in aliphatic amides, C=O stretch in aliphatic acid chlorides. Explain your answer.

Q.8.

What is the most striking difference in the functional group region between the following pairs of compounds? (i) Acetic acid and acetic anhydride (ii) Benzaldehyde and benzophenone (iii) Aniline and N,N-dimethylaniline (iv) Dilute and concentrated samples of ethyl alcohol (v) n-Hexane and cyclohexane.

178

Q.9.

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Which absorption signal will disappear and what new signals will appear in the IR spectrum when the following derivatives are prepared from the compounds given below? (i) Anhydride of phthalic acid (ii) Pentaacetate of glucose (iii) Aniline from nitrobenzene (iv) 3,5-Dinitrobenzoate of methyl alcohol (v) 2,4,6-Tribromo aniline from aniline (vi) Iodoform from acetaldehyde.

Q.10. Give the position of diagonostic absorption signal that will be seen in the functional group region of IR spectrum of each of the following compound. Acetic acid Benzoic acid Cyclohexene Cyclohexane Toluene o-xylene benzamide aniline N-methyl aniline isopropyl alcohol acetyl chloride benzanilide diethyl ketone formaldehyde.

2.2 IR Spectroscopy Advanced Experimental Organic Chemistry

Characteristic absorption frequencies of common functional groups

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181

182

2.3

Chapter 2

Spectroscopy

UV SPECTROSCOPY

2.3.1 Introduction The UV portion of the electromagnetic spectrum is expressed as the wavelength and its range extends from 200-400 nm. Vacuum UV range is from 10-200 and visible light rays have wavelength between 400-800 nm. Some molecules can absorb selectively certain wavelengths in the UV or the visible range. The absorption of these radiations results in the excitation of electrons from a lower energy level to a higher energy level that is why UV is referred to as electronic Spectroscopy. A plot of wavelength of absorption Vs the absorption intensity or absorbance is known as the UV spectrum. As compared to the IR spectrum, a UV spectrum is simpler and has fewer absorption signals. 2.3.2 Theory A molecule basically has two types of bonds i.e. sigma (σ) and pi (π), three types of electrons sigma, pi and non-bonded electrons.The energy associated with the UV radiations is just enough to excite some of these electrons from a lower energy level to the next higher energy level. In some compounds, electronic excitation requires much less energy, in such cases, even visible light can bring about these electronic excitations. The energy required to excite the electrons depends on the strength of the bonds in the molecule. For example, the sigma bonds are stronger than the pi bonds and hence the energy required to excite an electron of the sigma bond is higher than that required for a pi bond. Similarly, the energy required to excite an electron of a delocalised double bond is less than what is required for an isolated double bond. How strongly the non-bonded electrons are held to the nucleus and the energy required to excite these depends on the electronegativity, the size and the number of protons in the nucleus of the atom. For example, chlorine atom is smaller than iodine atom, the non-bonded electrons are more strongly held to its nucleus as compared to iodine, the energy required to excite n electrons of iodine is less than that required for chlorine. The electronic excitations are best explained by using the concepts of molecular orbital theory. In simplified statement, when two atomic orbitals overlap to form a bond, two molecular orbitals result, one having the highest electron density is called the bonding molecular orbital (lower energy) and the other with the least electron density as antibonding molecular orbital (higher energy). Thus in a molecule having sigma, pi and non bonded electrons, there are sigma (bonding and antibonding molecular orbitals represented as σ and σ*), pi (π and π∗) and non bonded electrons (n). The molecular orbital with the highest energy of all the occupied molecular orbitals is called the highest occupied molecular orbital (abbreviated as HOMO) and the antibonding orbital with the least energy is called the lowest unoccupied molecular orbital (LUMO). Before discussing the concepts involved in electronic spectroscopy, it is important to introduce a few other important terms. Chromophore: An unsaturated functional group containing multiple bonds. C=O, C=C, C≡C, –NO2, –SO3, C≡N etc. are all examples of chromophores. Auxochrome: A saturated functional group having nonbonded electrons, which when present in a molecule with a chromophore influences the position of absorption. Examples of auxochromes are –NH2, –OH, –SH, –Cl, –Br etc. λmax: Wavelength at which the absorption of highest intensity occurs.

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Bathochromic shift (Red shift): A shift of absorption of a chromophore to a longer wavelength. Hypsochromic shift (Blue shift): Shift of absorption to a shorter wavelength. Hyperchromic Effect: An increase in the absorption intensity. Hypsochromic Effect: A decrease in the absorption intensity. R band: Absorption resulting due to n– π* transition. K band: Absorptions occurring due to π–π*. B band: Bands present in aromatic compounds. E band: Bands present in aromatic compounds.

2.3.3 Types of Electronic Transitions As already stated, the energy required to excite an electron, depends on • The type of electron i.e. σ, π or n • The strength of a bond • The energy gap between HOMO to LUMO. Fig. 2.35 shows the relative energies of the various types of bonding and anti bonding molecular orbitals that may be present in a molecule and the relative energies required to excite an electron from a bonding orbital to an anti bonding orbital.

Fig. 2.35: Electronic energy levels of σ, π (bonding and antibonding) and non bonded (n) electrons

It is also important to state that • All possible electronic transitions are not allowed. Whether a transition is allowed or forbidden depends on a number of factors and mathematical calculations have to be done but qualitatively three factors play an important role to decide whether an electronic transition is permitted or not. These are; (i) the geometry of the MO in the ground state (ii) the geometry of the MO in the excited state (iii) orientation of the electric dipole of the incident light. • The absorption signals obtained in a UV spectrum are not sharp but broad and this can be explained as follows: A number of vibrational and rotational energy levels are present in each electronic level thus an electron jumping from a lower energy level E0 to a higher energy level E1 can jump to a number of rotational and vibrational energy level having close energy Fig. 2.36.

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Fig. 2.36: Rotational and Vibrational energy levels in each electronic energy level

2.3.4 Instrument The UV spectrophotometers are double beam instruments i.e. one beam of UV radiation is directed to the sample cell and the other to the reference cell. A spectrophotometer consists of the following parts. Source of UV light: A deuterium lamp is used, it emits radiations in the UV range. A tungsten lamp is used for including the visible range. Monochromator: It resolves the radiation into monochromatic waves, these strike a mirror and a system of slits which focuses the desired monochromatic beam on the sample cell containing the sample under examination and the reference cell. The length of the cell is normally 1 cm. Detector: It is a series of photodiods which detect the intensity of transmitted light. Recorder: It automatically records the spectrum or all absorbance values. 2.3.5 Sample Handling Since both glass and plastic absorb UV radiations, these cannot be used for holding the sample. Quartz containers are used as sample containers for recording the UV spectrum of a compound. Glass and plastic containers can be used for recording the absorption in the visible range. 2.3.6 Solvents used and their Effect on the Position of Absorption The sample is dissolved in a suitable solvent for recording its UV spectrum. Water, ethanol and hexane are the most commonly used solvents. Other solvents which can be used are: acetonitrile, chloroform, dioxane, methanol and cyclohexane. The choice of the solvent is mainly based on the following criteria: • The sample should be freely soluble in the solvent • The solvent should be transparent in the region in which the compound absorbs • A non polar solvent does not interact with the sample and helps to produce a fine spectrum • Conjugated alkenes and aromatic compounds experience a very little shifts of absorption signal due to polarity of the solvent • A polar solvent however shifts the position of absorption due to n–π* transition to shorter wave length and π–π* transition to longer wavelength.

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Table 2.3 gives the absorption position of a few solvents. Table 2.3 Solvent Water Ethanol n-hexane Methanol Acetonitrile Dioxane Chloroform

λmax (nm) 190 205 201 205 190 215 240

2.3.7 UV Absorption of Common Functional Groups This section gives a brief account of the organic compounds containing different functional groups, the wavelength of absorption and the intensity of absorption. 2.3.8 Alkanes These compounds contain only σ bonds and hence can absorb electromagnetic radiations which can cause σ– σ* transitions. These transitions require more energy than what the visible and UV radiations can provide and hence alkanes are transparent to UV and visible radiations. 2.3.9 Alkenes Simple alkenes contain π and σ bonds. The π– π* transition require less energy than σ–σ*. Alkenes containing a single double bond or polyunsaturated alkenes in which the double bonds are isolated absorb at ~175 nm. Introduction of an alkyl group shifts the absorption position to a longer wavelength and there is a cumulative effect of introduction of more alkyl groups. Conjugation in polyunsaturated alkenes results in a bathochromic shift of absorption signal due to π–π* transition as conjugation reduces the energy gap between the HOMO to LUMO.

Fig. 2.37: Relative energy difference between HOMO and LUMO of an alkene (ethene) and a conjugated diene (Butadiene)

The position of absorption of some alkenes is

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The position of absorption is also influenced by the geometry of the double bond, Lack of coplanarity due to steric hinderance in the cis isomer results in the absorption to occur at a lower wavelength as is clear from the examples given below:

2.3.10 Woodwards Rules for Calculating the λmax of Polyunsaturated Alkenes Empirical guidlines are available, known as Woodward’s rules for calculating the λmax for various types of simple and substituted conjugated dienes and polyenes Dienes Calculation of π–π* transitions Base values for π–π∗ transitions Acyclic and heteroannular diene = 215 nm Homoannular diene = 253 nm Acyclic triene = 245 nm Increments in the base value of substituted dienes (Substituent attached to sp2 hybridised carbon) are calculated by adding the following values to the base values. Alkyl group or ring residue = +5 –OR group = +6 –Cl or –Br = +5 Extension of double bond = +30 Each exocyclic double bond = +5 Notes: The nature of the solvent does not influence the experimental value of position of absorption and hence solvent corrections are not applied in these cases.

Fig. 2.38: Some common types of alkenes

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Some solved examples: Table 2.4: Calculation of λmax for some dienes Compound CH2=CH–CH=CH2 Alicyclic diene CH2=CH–CH=C(CH3)2 Acyclic diene

Calculation of λmax Base value = 215 No substituents = 0 Total = 215 nm Base value = 215 Alkyl substituents = 2 × 5 = 10 Total = 225 nm Base value = 215 Ring residue = 3 × 5 = 15 Exocyclic double bond = 1 × 5 = 5 Total = 235 Base value Ring residue Exocyclic double bond Total

= = = =

253 3 × 5 = 15 1×5=5 273 nm

Base value (Homoannular) Extension of conjugation Ring residues Exocyclic double bonds Total

= = = = =

253 1 × 30 = 30 4 × 5 = 20 2 × 5 = 10 313 nm

2.3.11 Alkynes A simple alkyne absorbs at 175 nm, introduction of alkyl groups or conjugation results in a bathochromic shift as in alkenes. H3C–C≡C–C≡C–CH3 H3C–C≡C–C≡C–C≡C–CH3 227 nm 268 nm 2, 4-hexadiyne 2,4,6-octatriyne 2.3.12 Aromatic Compounds Benzene shows a fine UV spectrum displaying three absorption bands. An intense band at 180 nm (ε, 60,000) due to allowed π–π* transition. Two weaker absorptions at 204 (ε, 7900) and 256 (ε, 200) respectively. Introduction of alkyl substituent shifts the absorption to a longer wavelength, o-isomer absorbing at shorter and p-isomer at the longest wavelength of the three dialkyl substituted benzene derivatives

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Introduction of polar groups like –OH, –NH2, –NO2 etc. and their relative position has more pronounced effect on the position of absorption. Table 2.5 Compound–Mono substituted Phenol C6H5CHO Acetophenone Aniline Nitrobenzene Disubstituted compound o-nitrophenol m-nitrophenol p-nitrophenol

λmax (k band) (ε) 211 244 (15000) 246 (13000) 230 252 (10000) 279 (6,600) 274 (6,000) 318 (10,000)

Disubstituted benzenes The λmax positions are influenced by the nature and relative position of two substituents. (a) If both groups are electron donating or electron withdrawing, the effect of group causing a larger shift is considered. (b) If one group is electron donating and other electron withdrawing, the effect on the position of λmax is additive.

An increase in number of condensed rings shifts the λmax to a longer wavelength.

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2.3.13 Calculation of π to π* Transitions of Some Benzene Derivatives Base value for C6H5COR 246 nm Base value for C6H5CHO 250 nm Base value for C6H5COOH 230 nm Base value for C6H5COOR 230 nm Increments due to the substituents on the benzene ring are tabulated below and are added to the base value to calculate λmax. Table 2.6 Position o–

m–

p-

Alkyl/ring residue

+3

+3

+10

–OH, –OR

+7

+7

+25

–NH2

+13

+13

+58

+20

+20

+45

–N(CH3)2

+20

+20

+85

–Cl

0

0

+10

–Br

+2

+2

+15

Group

–NHCOCH3

Notes: The calculated values sometimes do not tally with the experimentally observed values due to steric hindrance in the molecule.

Solved examples Table 2.7 Compound

Calculation of λmax (nm) Base value = 230 –Cl at o-position = 0 230 nm

Base value = 246 R at o-position = +3 Br at p-position = +15 264 nm

Base value = 230 p-NH2 group = +58 m-Cl = 2 × 0 = 0 288 nm (Contd...)

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(a) (a)

(b)

(b)

Base value = 250 p-OH group = + 25 275 nm

Base value = 250 o–OH group = + 7 257 nm

2.3.14 UV of Compounds Containing a C=O Group These compounds contain σ, π and n electrons hence can show n – σ*, n – π*, σ – σ*, and π – π*. Out of these, n – σ* and σ – σ* cannot occur in the UV and visible region. The remaining two (n– π* and π – π*) can occur in the UV region. Following are the positions of weak absorption due to n – π* transition of some carbonyl compounds 293 (11.8) Aldehyde (CH3CHO in hexane) 279 (13) Ketone (CH3COCH3 in isooctane) 204 ( 41) Carboxylic acids(CH3COOH in ethanol) 207 ( 69) Esters (CH3COOC2H5 in pet. ether) 220 Acid amide (CH3CONH2 in water) 225 ( 47) Anhydride (CH3COOCOCH3 in isooctane) n – π* is forbidden and occurs as a weak absorption and π – π* is permitted and result in a stronger absorption. 2.3.15 Calculation of π–π* Transition of a, β–Unsaturated Carbonyl Compounds (enones) and Dienones Structure of some chromophores

a, β-unsaturated ketones (acyclic) a, β-unsaturated ketones (cyclic, six membered) a, β-unsaturated ketones (cyclic, five membered)

215 nm 215 nm 202 nm

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a, β-unsaturated aldehydes a, β-unsaturated carboxylic acids and esters Increments for Double bond extending conjugation Homoannular diene component Exocyclic double bond to one ring One double bond, exocyclic to two rings Nature of group –OH –OCH3 –OCOCH3 –Cl –Br R or ring residue

210 nm 195 nm

+ 30 nm + 39 nm + 5 nm + 10

Position of substituent a + 10 + 35 +6 + 15 + 25 + 10

β + 12 + 30 +6 + 12 + 30 + 12

γ + 18 + 17 +6 + 12 + 25 + 18

d + 18 + 31 +6 + 12 + 25 + 18

Notes: As already mentioned, the experimentally observed position of absorption is affected by the solvent used for dissolving the sample. Hence the final experimental value of λmax is calculated after adding the solvent correction to the value calculated by applying the Wood Ward rules. Solvent Water Ethyl/Methyl alcohol Dioxane Chloroform Cyclohexane/Hexane

Correction –8m 0 +5m + 7 nm + 11 nm

Here are a few solved examples to illustrate the application of these rules. Structure of compound

Theoretical calculated value of λmax Base value = 215 nm 2β R groups = 2 × 12 = 24 Total = 239 nm Base value (a) R (β) R Total Base value Extension of conjugation Homoannular diene component Ring residue (d) Total

= = = =

215 nm + 10 + 12 237 nm

= 215 = + 30 = + 39 = + 18 = 302 nm

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Base value a-alkyl β-alkyl Total

Base value Extension of conjugation β Ring residue d Ring residue Exocyclic double bond Total Base value β-alkyl gp Total Base value a-Ring residue β-Ring residue Exocyclic double bond Total

Spectroscopy

= 202 = 1 × 10 = 2 × 12 = 24 = 236 nm

= = = = = = = = =

215 + 30 + 12 + 18 +5 280 nm 195 nm + 12 207 nm

= 210 = +10 = 2 × 12 = +5 = 247

Questions Q.1.

Define the following terms: (a) Chromophore (b) Auxochrome (c) Hypsochromic shift (d) Hyperchromic shift

Q.2.

Name a few solvents which are used for dissolving the sample in UV spectroscopy.

Q.3.

Why quartz cells are used for sample handling while recording the UV spectrum?

Q.4.

What do you understand by the following? π* and σ molecular orbitals.

Q.5.

Why are the absorption signals broad and not sharp in the UV spectrum?

Q.6.

Which out of the following will require least energy for electronic excitation? σ→σ*, π→π* or n→π*

Q.7.

Where do the π→π* transition show an absorption in the UV of the following compounds? (i) CH2 = CH2 (ii) CH ≡ CH (iii) CH3COCH3 (iv) CH2=CH–CH=CH2 (v) CH2=CH–COCH3

Q.8.

Which out of the following pair of compounds will show an absorption at a longer wavelength and why? (i) CH2=CH–(CH2)2–CH=CH2 CH3–CH–CH=CH–CH=CH2 (ii)

2.3 UV Spectroscopy Advanced Experimental Organic Chemistry

(iii)

(iv)

(v)

Q.9. Why polyunsaturated compounds like Lycopene are coloured. Q.10. Why is nitrobenzene light and p-nitroaniline dark yellow in colour. Q.11. Calculate the λmax position for the following compounds: (i) CH3 − CH = CH − C = CH − COCH3 | CH3

(ii)

Hint: For a cross conjugated ketone, the calculations are done for a more substitued chromophore

(iii)

(iv) (v) CH 2 = C − CH = C − CH − CH 2 | CH3

(vi)

| CH3

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(vii)

(viii)

2.4

NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

2.4.1 Introduction It is a type absorption spectroscopy in which certain nuclei absorb electromagnetic radiation in the radio frequency range (60-600 MHz) when placed in a magnetic field. A plot of frequency of radiowave absorption Vs the peak intensity of absorption is called the NMR Spectrum. Nuclei which possess a spin quantum number (I) can show the nuclear magnetic resonance. The spin quantum number of a nucleus depends on the number of protons and neutrons in it. • Nuclei which have even number of protons and odd number of neutrons or vice versa possess spin quantum number which has half integral value i.e. 1/2, 3/2, 5/2 ..... • I for nuclei having odd number of protons as well as neutrons is an integer 1, 2, 3 .... . • Nuclei with even number of protons and neutrons each have a zero quantum number and do not show NMR phenomena. For example, 2H , 1H, 15N, 13C, 17O, 19F possess spin quantum number and are NMR active whereas 12C.16O etc. have I = O and are NMR inactive. A nucleus with spin quantum number I can have 2I + 1 orientations when placed in a magnetic field. For example H with I = 1/2 can orient in two different ways (2 × 1/2 + 1 = 2) or it means that such a nucleus will have two nuclear energy levels when placed in a magnetic field and 13C will have three nuclear energy levels in a externally applied magnetic field.

Fig. 2.38: Energy levels of a proton under the influence of a external magnetic field

1  Figure 2.38 shows the energy levels of a nucleus  I =  under the influence of an applied external 2  magnetic field (H0). The energy gap between the nuclear energy levels E1 and E2 is governed by the equation H 0 γh ∆E = 2π where H0 = strength of applied magnetic field

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195

h = Planck’s constant γ = Magnetogyric ratio, a constant for a given isotope ∆E = hv (v is the frequency of radiations) γ H0 v = , this value of frequency corresponds to the energy difference (∆E) between 2π the higher nuclear energy state and the lower nuclear energy state, the frequency of absorption v ∝ H0

2.4.2 Instrument It basically consist of the following: • A radio frequency source • A cylindrical tube for holding the sample solution • An electromagnet • A detector • A recorder

Fig. 2.39: Schematic diagram of an NMR spectrometer

NMR spectrum of a sample can be obtained by keeping the applied field (H0) constant and varying the frequency or varying v and keeping H0 constant. Many instruments have fixed frequency and varying magnetic field. A 10% solution of the sample (0.5 mL) is prepared in one of the solvents given below, placed in a cylindrical tube and is spun continuously while recording the spectrum so that the sample experiences a uniform magnetic field.

2.4.3 Solvents • The compound should have at least 10% solubility in the chosen solvent • The solvent should be preferably aprotic • Most of the solvents used are deutrated organic solvents in which the hydrogen has been replaced by deuterium (D).

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Some of the common solvents are CDCI3, CCI4, CS2, C6D6, D2O, (CD3)2SO and (CD3)2CO etc.

TMS (tetramethylsilane) is added in the solution of the sample, except in those cases where it is non miscible with the solvent for example D2O. In such cases either TMS is used as external standard or 3-(trimethylsilyl)-propane sodium sulphonate is used as an internal standard.

2.4.4 Uses of 1HNMR (PMR) NMR has become a very important tool for structure elucidation as it provides much more information about the structure of compound as compared to that provided by IR and UV spectroscopy. PMR spectrum provides the following information about the structure of a compound. • It gives the number of magnetically nonequivalent types of protons in an organic compound. For example the spectrum of CH4 gives one absorption signal indicating the presence of only one type of protons. The NMR of CH3CHO shows two types of protons. • It provides information of the relative number of each type of proton. For example, in the PMR spectrum of CH3CH2CH3, there are two absorption signals, the area under the absorption signal is in the ratio 3:1 indicating their relative number. • If the molecular formula of a compound is known, the exact number of each type of proton can be known. For example in the above case, the molecular formula is C3H8, it implies that there are six protons of one and two of the other type. • The position of absorption (known as the chemical shift) provides important information about type of proton. For example H of –CHO group absorbs at a different frequency as compared to an H of –CH3 group. Comparison with the correlation data available in literature can help identify the type of proton. • PMR provides information about the number of neighbouring protons. Each absorption signal splits into n + 1 peaks, where n is the number of non equivalent neighbouring protons. For example in CH3CH2Cl, the signal due to CH3 splits into three peaks called a triplet (n = 2, 2+1) and that due to CH2 into four peaks known as quartet (n = 3, 3 + 1). 2.4.5 Principle The nucleus of hydrogen atom behaves like a tiny bar magnet and spins around its axis. The spin quantum number of hydrogen is 1=1/2. When placed in an external magnetic field, the tiny magnet can take up two orientations. It can • align with the external magnetic field (lower energy orientation) • align against the external magnetic field (higher energy orientation)

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197

The population of nuclei in the lower energy state being slightly higher than those in the higher energy state (Boltzman distribution). The nucleus under the influence of an external magnetic field, spins like a top (spinning axis moves around the vertical axis). This type of motion is called precession motion. The frequency of precession motion is directly proportional to the strength of external magnetic field.

Fig. 2.40: Proton precessing in a magnetic field H0

A proton precesses 60 million times (60 MHz) when placed in an external magnetic field of strength 1.4 Telsa (T) or 14000 gauss. When a proton is irradiated with radio frequency radiations, the protons in the lower energy orientation may absorb these electromagnetic radiations and jump to higher energy orientation. The absorption occurs when the frequency of external radiation matches (resonates) with the precession frequency of the proton nucleus.

2.4.6 Chemical Shift All protons do not resonate with the radio frequency at the same wavelength. The reason is that all protons do not experience a magnetic field exactly equal to the external field. The electrons revolving round the nucleus generate their own magnetic field (induced field), the strength of which depends on the electron density around the nucleus. This induced magnetic field around each nucleus either aligns or opposes the external magnetic field. As a consequence of this a proton experiences a magnetic field which is either H0 + H1 or H0 – H1 Since H1 (induced magnetic field) is different for different types of protons, they resonate with different wavelength under the influence of an externally applied magnetic field Ho. The precession frequency of any proton is not calculated in frequency units but is represented as the difference in frequency when compared to a reference or a standard proton. Tetramethyl silane (TMS) or the sodium salt of 3-(trimethylsilyl)–propanesulphonic acid are used as standard compounds. TMS gives a sharp peak at a frequency lower than most organic compounds because it contains Si which is more electropositive than carbon. The chemical shift values are expressed in d or t scale.

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vx − vTMS v0 where vx is the frequency in Hz of the signal of proton Hx vTMS = Frequency (Hz) at which TMS absorbs v0 = Operating frequency of the instrument. Delta is a unit free number and is independent of the external magnetic field. The chemical shift is the resonant frequency of nucleus as compared (or relative) to a standard (e.g. TMS) in a mgnetic field. dx =

2.4.7  Factors Influencing the Chemical Shift 2.4.8 Electronegativity Electrons density around a hydrogen atom shields the nucleus from the external magnetic field, the greater the shielding, the lower is the precession frequency. If a hydrogen is attached to a carbon bearing an electronegative group, the electron density around such a hydrogen is reduced because of the electron withdrawing effect of the electronegative group. In such cases, the nucleus of hydrogen is shielded to a lesser extent ( the term used for such a proton is deshielding) and experience a magnetic field H0 + H1 and precess with higher frequency. For this reason, in CH3F, the hydrogens are most deshielded and CH3I least of CH3F, CH3Cl, CH3Br and CH3I. In contrast the protons attached to an electropositive element are most shielded. 2.4.9 Anisotropic Effect In compounds containing multiple bonds like C=C, C≡C, C=O and aromatic compounds, the extent of deshielding cannot be accounted for on the basis of electronegativity alone. In such cases the moving pi electrons of the multiple bond effect the electron density around protons of the molecule depending on their orientation in space or cause nonuniform magnetic field in space around the molecule. Consequently some protons are deshielded more than others. In alkenes (Fig. 2.41) for example, in the presence of an external magnetic field, the plane of the double bond is aligned at 90° to the direction of the applied field. The circulation of pi electrons produces a secondary magnetic field unequal in space, being paramagnetic in the alkene proton region and diamagnetic around the carbon atom. So the alkene proton experiences a deshielding effect.

Fig. 2.41: Allignment of an alkene in a external field and π electron movement

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In alkynes (Fig. 2.42), the molecule orients in such a way, that the C≡C plane is aligned in the direction of the external magnetic field, the protons attached to Sp carbon therefore experience less diamagnetic shielding effect.

Fig. 2.42: An alkyne in an external magnetic field

In aromatic rings (Fig. 2.43), the protons lie in the deshielded zone.

Fig. 2.43

In aldehydes (RCHO), the H of CHO group is most deshielded. There are a number of molecules in which, the anisotropic deshielding influence of the ring current of electrons is very evident. One such example is [18] annulene (Fig. 2.44). The protons lying outside the ring are highly deshielded and absorp at 8.9d and protons that lie in the cavity of the molecule are highly shielded and adsorb at –1.8d, a value lower than TMS.

Fig. 2.44

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2.4.10 Intensity of Peaks As already stated, the intensity of a peak/area under the peak gives the relative number of protons responsible for that absorption. If the molecular formula of the compound is known, the exact number of protons of a type can be known. A system of automatic integration is installed in the spectrometer and the height of integral gives the number of protons of a type.

Fig. 2.45: NMR of isopropyl benzene showing integration

For example in the spectrum of isopropyl benzene, the integration of three types of protons shows that the signals correspond to 6, 1 and 5 protons respectively.

2.4.11 Equivalent and Nonequivalent Protons Protons having the same chemical shift are known as isochronous. It is not always possible to say by simply looking at the structure of a molecules whether the protons will have the same chemical shift or not. An easy way to tell whether the protons in question are equivalent is to replace each (one at a time) with deuterium and compare the resultant structures (isomers) that are formed. • Protons which are diastereotopic have different chemical shifts. Enantiotropic protons have same chemical shift. As an example, consider the molecule CH2BrCl

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I and II are enantiomers, hence HA and HB are equivalent. In

, replacement of one H of –CH2– group by D gives two isomers which are

diastereoisomers and hence the two protons of –CH2– group in the above example are non equivalent.

Similarly, in the examples given below, the equivalence or nonequivalence of two protons can be established.

Two compounds are same and hence HA and HB in the above example are equivalent.

Now consider

replacement of HA and HB by D will give two isomers and hence HA and HB are non-equivalent. Following underlined protons in each structure are equivalent

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The underlined protons in the following compounds are non-equivalent:

2.4.12 Spin Spin Coupling Consider two protons marked HA and HB in the structure given below. These are non equivalent protons

and influence the magnetic field of each other. The nuclear spin of HA couples with the electron spin of HB and so does the nuclear spin of HB couple with the electron spin of HA. These electron coupled spin interactions operate through the bonds and their effect declines with the increasing number of bonds between the atoms. HA can align with the spin of HB (lower energy orientation) or against (higher energy orientation).

As a consequency of this, the absorption signal of HA is split into two peaks of equal intensity and so is the signal due to HB. Thus each absorption appears as a doublet (two peaks). This is called spinspin coupling. Now consider a molecule CH3–CH2–Cl. There are two sets of equivalent protons, three of one type (CH3) and two of the other type (–CH2–).

The absorption signal due to –CH3 splits into three peaks (triplet) due to –CH2–group and –CH2– signal splits into four peaks (quartet). The general rule governing the spin-spin coupling is

• An absorption signal will split into (n + 1) peaks, if there are n adjacent, non-equivalent protons. • If a hydrogen has non equivalent protons attached to two adjacent groups then the signal will split into (n + 1) (n′ + 1) peaks, where n and n′ are the number of two different types of non equivalent protons. Consider CH3–CH2–CH2Cl, the underlined –CH2– group has three protons of one type (CH3) and two (–CH2Cl) of other type on the adjacent carbon atoms.

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So the multiplicity of –CH2– will be (3 + 1) (2 + 1) = 4 × 3 = 12, which of course is normally seen as a multiplet. Predict the multiplicity of the protons that have been underlined in the following structures: (1) CH3–CHCl2 (2)

CH3 − CH 2OH

(3)

CH3 − O − CH3

(4)

(5)

(6) (7)

CH3–CH2–CH2–NO2

(8)

(9)

(10) The relative intensity of spin-spin coupled signals can be known by using the famous Pasal’s triangle

Fig. 2.46: Pascal’s triangle depicting the multiplicity and relative intensities of the first order coupling. The d value of each multiple is the centre of the multiplet

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2.4.13 Coupling Constant The measure of degree of interaction between two nuclei is called the spin-spin splitting constnat or coupling constant. Symbol used is J and the units are Hertz (Hz). It is important as two protons splitting each other will have the same J value. The coupling constants of a few systems are given in Table 2.8 below. Table 2.8 Group

J(Hz) 12–18

2–10 1–3

11–18

6–15 Cyclohexane e-e

3

a-a

5–10

a-e

3

o-

7–10

m-

2–3

p-

0–1

Aromatic

2.4.14 NMR of Some Common Functional Groups 2.4.15 Alkanes Saturated alkanes (acyclic) have three basic types of protons Type of proton

Chemical shift

CH3–R

0.7–1.3 d

RCH2R

1.2–1.4 d

R3CH

1.4–1.7 d

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In cycloalkanes, the following signals are observed Cyclohexane (–CH2–)

1.35 d

Cyclopentane (CH2)

1.42 d

Coupling constant

Fig. 2.47: NMR of n-propane

Following data is helpful in recognising the carbon skeleton in alkanes and other saturated compounds: Singlet at 0.9d (3H)

CH3–CH2–Z

CH3 appears as a triplet and CH2 as a quartet 2XCH3–overlapping doublet and CH–Septet split by 6 identical protons

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CH3 as a triplet CH3–CH2–CH2–as multiplet at (being split by CH3 and CH2 i.e. (3 + 1) × (2 + 1) = 12) CH3–CH2–CH2 as a triplet

CH3–CH2–CH2–Z

Singlet due to 9 equivalent protons

Z represents a group not containing any proton. In cyclohexane, 12 hydrogens appear as a singlet at room temperature because the flipping of chair conformation of cyclohexane can occur at room temperature there by giving a single peak.

Fig. 2.48: Flipping of the chair form of cyclohexane

At low temperature the flipping is not permited and hence cyclohexane show two absorption signal, one due to axial and other due to equitorial protons.

2.4.16 Alkenes The hydrogens attached to an sp2 hybridised carbon are deshielded by the anisotropic effect of the electrons of the double bond. Some important values are given in Table 2.9 below: Table 2.9: d values of some common types of protons in alkenes Type of proton

Chemical shift

C=C − H

4.5–6.5d

C=C − C | H

1.6–2.6d

Coupling constants in alkenes are given in Table 2.8.

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Fig. 2.48: NMR of propene

2.4.18 Aromatic Compounds A larger anisotropic effect is experienced by protons attached to an aromatic ring due to ring current generated by the π electrons. These protons are more deshielded than the protons attached to the sp2 hybridised carbon in alkenes. For the same reason, the alkyl group hydrogens are also deshielded. Type of proton C6H5–H

Chemical shift

C6H5–CH

2.3–2.7 d

6.5–8 d

Benzene shows one absorption at 7.2 d. (Fig. 2.49) The position of absorption shifts with the nature of substituents and the pattern of substitution.

Fig. 2.49: NMR of Benzene

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Fig. 2.50: NMR of toluene The electron withdrawing group deshield the other aromatic protons becuase they reduce the electron density of the nucleus. For example in nitrobenzene the o-proton absorbs at 8.1d.

In acetophenone

The electron donating groups have the opposite effect because these increase the electron density of the nucleus

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209

A moderately activating group like –CH3 shows a small effect on the chemical shift of aromatic protons

The number, nature and their relative positions also have an influence on the chemical shift value.

Fig. 2.51: NMR of p-nitrotoluene

Table 2.10 gives the coupling constants of o-, m- and p-protons in aromatic compounds. Table 2.10 Coupling constants ortho protons meta protons para protons

J = 7–10 Hz J = 2–3 Hz J = 0–1 Hz

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2.4.19 Alkyl Halides The halogen attached to the carbon in an organic compound exerts diamagnetic shielding effect. The extent of deshielding depending on the electronegativity of the halogen as shown in Table 2.11 below. Table 2.11: The deshielding effect increasing with an increase in the number of halogens Compound

Chemical shift

|

2–4 d

|

2.7–4.1 d

− C H−I − CH Br |

− CH − Cl |

− CH − F

3.1–4.1 d 4.2–4.8 d

Fig. 2.52: NMR spectrum of ethyl bromide

2.4.20 Alcohols Hydrogen attached to oxygen is deshielded, the chemical shift value of the H is variable as it depends on a number of factors, the significant ones being • Concentration of the solution. • Nature of the solvent used for dissolving the alcohol i.e. whether polar or non-polar. • Temperature. Table 2.12: Chemical shift values of alcohol Compound C–OH CH–OH

Chemical Shift 0.5–5 d 3.2–3.8 d

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211 211

Since the position of absorption is highly variable, one of the method used is to exchange the H of OH with deuterium by adding D2O to the sample solution and re-recording the spectrum. The peak due to –OH diminishes and then disappears because of the following exchange – OH + D2O —→ – OD + HOD

This exchange occurs due to the acidic nature of proton attached to oxygen. The peak due to HOD appears at 5.0d.

Fig. 2.53: NMR of isopropanol in CDCl3

Fig. 2.54: NMR of isopropanol in DMSO

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The spin-spin coupling between the H of OH group and the adjacent a-hydrogen atom is normally not seen in the NMR spectrum in CDCl3 etc. This is because the proton of the –OH group does not stay permanently on the oxygen. To prevent this exchange, the spectrum is recorded in a solvent like DMSO, which forms a very strong hydrogen bond with H of –OH group. When this is achieved the proton of –OH couples with adjacent H of –CH, if appears as a doublet and hence, it is possible to distinguish between a primary, secondary and a tertiary alcohol. The spin-spin coupling pattern in 1°, 2° and 3° alcohols in DMSO is given below: A

B

CH3OH A

B

CH3 CH 2OH

A

B

(CH3 ) C − OH

A doublet (3H)

B-quartet (1H)

A-Octet (2H)

B-triplet (1H)

A-multiplet (1H)

B-doublet (1H)

A no proton

B-singlet (1H)

2.4.21 Ethers a-hydrogens in simple ethers appear between 3.2-3.8d Table 2.13 Compound

Chemical shift

ROCH3

3.2 – 3.8d

2.4.22 Aldehydes The aldehydic proton (–CHO) is a highly deshielded proton and appears between 9-10d (Table 2.14) Table 2.14 Type of proton

Chemical shift

R–CHO

9.0–10.0d

R–CH–CHO

2.1–2.4

The aH is shielded because of anisotropic effect of the electron of C=O group. The spin-spin coupling pattern between a hydrogen and H of –CHO gives information about the number of a-Hydrogens Table 2.15 Compound (aldehyde)

Spin-spin coupling pattern of –CHO

HCHO

Doublet

CH3CHO

Quartet

CH3CH2CHO

Triplet

C6H5CHO

Singlet

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Fig. 2.55: NMR of acetaldehyde

2.4.23 Ketones shows absorption between 2.1–2.4d. A number of other compounds like aldehydes, acids, acid amides, esters and acid chlorides also show absorption in this region due to a-hydrogen to the C=O group. Ketones can be recognised by the absence of certain specific absorptions in these compounds as shown in table 2.16.

Fig. 2.56: NMR of acetone

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Table 2.16 Compound

Specific absorption in the NMR

RCHO

9.0–10.0 d

RCOOH

11.0–12.0 d

RCOOCH3

3.5–4.8 d

RCONH2

5.0–9.0 d

2.4.24 Esters The protons attached to the a-carbon of C=O group absorb in almost the same range as in other carbonyl compounds. But the protons attached to the carbon directly attached to the oxygen are more deshielded because of electron withdrawing effect of oxygen RCHCOOR 2.1–2.5 d RCOOCH 3.5–4.8 d

Fig. 2.57: NMR of ethyl acetate

2.4.25 Carboxylic Acids Hydrogen directly attached to oxygen is highly deshielded. The a-hydrogens to C=O group absorb in the same range as in other compounds RCOOH 11.0–12.0 d O 

RCH 2 − C − OH

2.1–2.5 d

The H of OH group can be exchanged with D2O as in alcohols.

2.4 Nuclear Magnetic Resonance Spectroscopy Advanced Experimental Organic Chemistry

Fig. 2.58: NMR of CH3COOH

2.4.26 Amides Amides show absorption due to following types of protons R CH CO–N 2.1–2.5 d R CO NH 5.0–9.0 d [depends on solvent, concentration and temperature] In substitued amides, the H of –N–CH group is also present RCO–N–CH 2.2–2.9 d

Fig. 2.59: NMR of CH3CONH2 in D2O

215

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2.4.27 Acid Chlorides The protons on the a-carbon appear between 2–3 d.

Fig. 2.60: NMR of Acetyl Chloride

2.4.28 Nitriles The –CH–CN appear between 2 – 3d.

Fig. 2.61: NMR of Acetonitrile

Spectroscopy

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217

2.4.29 Amino Acids Amino acids exist in equilibrium mixture of the following forms +

+

−

−     NH3 − CH − COOH    NH3 − CHCOO    NH 2 − CHCOO |

|

R

R

|

R

The NMR of glycine shows the following absorptions H 2 N − CH 2 − COOH 3.8δ

2δ

2.4.30 Nitroalkanes The a-hydrogens are deshielded by the strong electron withdrawing effect of nitro group –CHNO2 absorbs between 4.4–4.8d.

Fig. 2.62: NMR of nitromethane

2.4.31 Amines Following absorptions are typical to amines |

−C H − N R–NH Ar–NH

2.1–2.9 d 0.5–4.0 d C present in both aliphatic primary and secondary amines 3.0–5.0 d C present in both aromatic primary and secondary amines, deshielded by the aromatic ring.

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Chapter 2

Fig. 2.63: NMR of Aniline

Fig. 2.64: NMR of N-methylaniline

Spectroscopy

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219

2.4.32 Sulphonic Acids The salient feature of sulphonic acids is the absorption due to –SO3 H The position is variable, below is given the NMR of ethane sulphonic acid. –SO3H 11.0 d α

CH3 − C H 2 −

3.1 d

CH3–CH2–

1.2 d

Fig. 2.65: NMR of Ethanesulphonic acid in CDCl3

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Chapter 2

2.4.33 Sulphonamides

Fig. 2.66: N-methyl benzene sulphonamide

2.4.34 Thiols –CH–SH is seen between 2.5–3.5d.

Fig. 2.67: NMR of 2-butanethiol

Spectroscopy

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221

Table 2.17: Chemical shifts for –CH3, –CH2– and –CH– Methyl Protons

d

Methylene protons

d

Methine protons

d

0.9

1.4

1.5

1.6

2.3

2.6

2.1

2.4

2.50

CH3–NR2

2.2

–CH2–NR2

2.5

2.90

CH3–Ar

2.3

–CH2–Ar

2.7

3.0

CH3–Br

2.7

–CH2–Br

3.3

4.1

CH3–Cl

3.1

–CH2–Cl

3.4

4.1

CH3–F

4.3

–CH2–F

4.5

5.2

CH3–S

2.1

–CH2–S

2.4

3.2

CH3–N

2.3

–CH2–N

2.5

2.8

CH3OH

3.3

–CH2OH

3.3

3.8

CH3–C≡N

2.0

–CH2–C≡N

2.3

2.7

4.4

4.6

CH3–NO2

4.3

–CH2–NO2

Table 2.18: Chemical shifts for the Protons of some common Functional Groups and Unsaturated Types of Protons R–CHO Ar–CHO R–NH2 Ar–NH2 R–OH Ar–OH

d 9.4–9.7 9.7–10.0 5.0–8.0 3.4–6.0 0.5–4.0 4.4–12

Types of protons HC=C (isolated) HC=C (conjugated) H2C=C (isolated) H2C=C (conjugated) HC≡C (isolated) HC≡C (conjugated)

d 5.1–5.8 5.8–6.6 4.5–5.0 5.3–5.8 2.4–2.2 2.7–3.1 (Contd...)

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Chapter 2

R–SH Ar–SH R–COOH

1.0–2.0 3.0–4.0 10.5–13.0

Ar–H

3.7–4.1 R–O–C–H

3.3–4.0

Correlation data of common types of protons

7.2

Spectroscopy

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224

Chapter 2

Questions Q.1.

Define the following terms: (a) Chemically equivalent protons (b) Magnetically equivalent protons (c) Chemical shift.

Q.2.

How many different types of protons are present in the following: (a) CH3CH2CH2CH3 (b) CH2Cl–CH2Cl

Spectroscopy

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(c) CH2=CH2 (d) CH3CH=CH2 (e) C6H5NO2 (f) (CH3)2CH–O–CH(CH3)2 (g) (CH3)4C (h) (CH3)4Si (i) CH3CH2CHO Q.3.

(j) CH3CH2COOCH2CH3 Which compound has the most deshielded protons?

(a) (b) CH3Cl, CH2Cl2, CHCl3 Q.4.

(c) CH4, R–CH3, R–CH2–R, R3CH. What do you understand by spin-spin coupling. Indicate the multiplicity of the following absorption signals (a) CH3–CH2–CH2–NO2 (b) CH3–O CH3 (c) CH3–CH2–Cl

Q.5. Q.6. Q.7. Q.8. Q.9.

(d) (CH3)3C–CH2–Cl. Why is the position of H of –OH and H of –NH2 group variable? How will you distinguish the following compounds from each other on the basis of their NMR? CH3CHO from CH3COCH3 and CH3COOH from CH3CONH2. Name a few common solvents that are used for recording the NMR spectrum of a compound. In addition to being used as a solvent, what is the other important role of D2O? What will be the difference in the NMR spectra of ethyl alcohol in CDCl3, after the addition of D2O and in DMSO as solvent. Explain the differences in the three spectra. An organic compound (A) on reaction with iodine and sodium hydroxide gives (B), a yellow coloured solid. Enclosed are the NMR spectra of A and B. Identify the compounds and analyse the spectra systematically.

226

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Q.10. Following sequence of reactions is carried out starting with a compound A

F shows a single absorption signal in ‘HNMR (60 MHz) at d7.2. Identify A, B, C, D, E and F. Give an approximate position of absorption due to the functional group in the IR spectrum of each of the compound. Q.11.

Following are given the NMR spectra of two isomeric compounds A and B at 90 MHz. Identify A and B. [Molecular formula –C4H8O2] [Hint: Both give a +ve hydroxamic acid test]

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Q.12. Enclosed are the NMR spectra of compound A and B. Both give benzoic acid on oxidation with alkaline KMNO4. Identify A and B.

2.4 Nuclear Magnetic Resonance Spectroscopy Advanced Experimental Organic Chemistry

Q.13. Identify A and B on the basis of following: (i) The signal between 2-3d disappears when the spectrum is recorded after adding D2O. (ii) Both give aldehyde on mild oxidation.

229

230

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Spectroscopy

Q.14. A disubstituted aromatic compound (A) shows the following spectrum. Molecular formula of A is C10H13O2N. Identify the compound.

[Hint: It was being used as an antipyretic drug, it is now banned in a number of countries in the world.] Q.15. The NMR spectra of compound A and its two derivatives B and C are given below. The molecular ion peak of compound A appears at 94. The NMR spectrum of A shows an absorption at d5.25 which is not seen in the NMR spectrum of B and C (mol. wt. 198). Identify A, B and C and explain your answer.

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2.5 MASS SPECTROMETRY 2.5.1

Introduction

When a molecule in its vapour phase is exposed to an electron beam, depending on the energy of electron beam, it first loses an electron to form a molecular ion A − B− C  → [ A − B− C ] + e 70 eV e

+

The molecular ion can then fragment to produce some positively charged ions, some radicals and/or neutral fragments. The mass spectrometer records only the +ve ions produced by fragmentation.

In general, when there are many possible ways of fragmentation, more stable ions are produced. The mass spectrum of a compound is a plot of intensity of a signal and the ratio of mass/charge of the ion. Since the charge is one, it is a graph of mass vs the signal intensity. The most intense peak is known as the base peak. The intensity of the other peaks depends on the ease of formation and stability of the fragmented ion. The intensity of base peak is equated to 100%.

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233

For example, below is shown the fragmentation pattern of 2-methyl butane

The mass spectrum can show signals at 72, 71, 57 and 43. Mass spectrum of a molecule provides following important information: 1. Exact molecular mass of the compound. 2. Presence of N, S, Br, Cl etc. 3. From the intensities of peaks at higher value than the mass of the molecules i.e. M + 1, M + 2 peaks etc., it is possible to know the presence of isotopes.

2.5.2 Instrument The mass spectrometer consists of following parts: 1. Sample vapouriser. 2. The vapourised sample is bombarded with high energy electron beam in an ionizer (usually 70 eV). Lower energy electron beam (10 eV) produces only the molecular ion. −e M  → M +

3. The positively charged ions are accelerated into an analysing chamber. 4. The analysing tube is surrounded by a curved magnetic field which causes the deflection of cations. The deflection is proportional to the mass/charge ratio. 5. The ions strike the detector at different speed which then records the spectrum.

2.5.3 Molecular Formula from Molecular Mass The molecular formula of a compound can be calculated from molecular mass following the below mentioned guidelines. 1. Generate a base formula for the compound by applying rule of thirteen which assumes that the compound contains only carbon and hydrogen M r = n + , n is number of C atoms and r is the remainder, which is a smaller number than 13 13 13 (wt of one C + 1H) The base formula is CnHn + r.

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2. Then obtain the index of unsaturation as per the formula given below: ( n − r + 2) u= 2 3. Useful information can be obtained from M + 1 and M + 2 peaks, since the natural abundance of isotopes is well known Elements H C S Cl Br I

M 100% 98.9% 95% 75.5% 50.5% 100%

M+1 1.1 0.8

M+2

4.2 24.5 49.5

The presence of the hetero atoms like O, S, halogens (Cl, Br and I) can be easily assertained. 4. Presence of nitrogen can also be established. The molecular ion peak of compounds with one nitrogen is odd, all other common atoms give an even mass. 5. The molecular formula thus calculated is then used to get exact molecular mass by adding up the exact atomic masses of the atoms. If it exactly matches with M obtained from mass the formula that has been derived is correct. 6. Correlation data in literature can be used to arrive at the molecular formula of a compound.

2.5.4 Fragmentation Pattern of some Types of Compounds General guidelines are available to predict the prominent peaks that are obtained from various types of compound. These are: • Straight-chain alkane give an intense molecular ion peak. So is the case with aroamtic and heteroaromatic compounds. • Increase in branching results in lowering the intensity of molecular ion peak. • Increase in molecular weight in a homologous series also reduces the intensity of M+ •. • Cleavage occurs at the most substituted carbon, because if produces the most substituted cation, which is most stable +

+

+

+

R 3 C > R 2 CH > R CH 2 > CH3 +

C6H5CH2,

etc. are more stable than primary ions.

• Allylic cleavage is preferred in compounds containing a double bond because a resonance stabilised ion is formed. • Unsaturated cyclic enes can undergo retro Diels-Alder reaction. • Cleavage many at times occurs to eliminate small neutral molecules. • Rearrangements also take place to produce unexpected ions from molecules.

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2.5.5 Alkanes and Cyclic Alkanes In long chain alkanes, the fragmentation pattern contains a number of peaks separated by 14 mass unit corresponding to cleavage of –CH2–groups.

2.5.6 Alkenes The common fragmentation pattern is to create resonance stabilised allylic ion

2.5.7 Alkyl Halides

[CH3CH 2CH 2− CH 2− Br ] + •

| ↓ + CH3CH 2CH 2 CH 2 + Br •

But more prominent peaks are M + 2 due to the high abundance of 82Br isotope.

2.5.8 Alcohols The molecular ion peak in primary and secondary alcohols is small and undetectable for a tertiary alcohol. 1. Peak resulting by loss of a water molecule is most distinct in primary alcohols.

[CH3 − CH 2 − CH 2 − CH 2OH ] + → [CH3CH 2CH = CH 2 ] + + H 2O •

•

2. Cleavage at a-position of an alcohol is another common way of fragmentation in alcohols

2.5.9 Ethers 1. M + 2 peak is detectable. 2. Cleavage occurs between a−β carbon atoms

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3. The O–C bond also cleaves, the charge remains on the alkyl group

In aromatic ethers 1. The molecular ion peak is quite prominent. 2. Cleavage occurs at the O–R bond in ArOR

2.5.10 Aldehydes 1. H• cleaves from M•+ to give a peak at M–1. 2. –CHO group is eliminated to produce a peak at M-29 (–CHO = 29) Aromatic aldehydes M-1 peak is most prominent in aromatic aldehydes because loss of H• produces a stable cation [Ar–C≡O]+

2.5.11 Ketones

O ||

1. The C − C bond is cleaved. 2. Ketones containing a γ-hydrogen can rearrange [Mclafferty rearrangement) to form an enol ion radical. This fragmentation produces the most prominent peak.

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237 237

2.5.12 Carboxylic Acids Mclafferty rearrangement is seen 1. Loss of OH (M–17) 2. Loss of COOH (M–45) In aromatic acids (M–18) peak is obtained due to loss of a water molecule. 2.5.13 Esters Fragmentation occurs by loss of RCO O 

+ − CH3 − C − O − CH 2 − CH3 → CH3 CO + CH3CH 2 O −

→ CH3COO+ CH3CH +2

2.5.14 Amides 1. Amides can undergo Mclafferty rearrangement. 2. Primary amides give an intense peak at M = 44, due to the formation of [CONH2)+ ion.

Aromatic amides 1. Loose –NH2 to form a stable cation, so M–16 peak is prominent

2.5.15 Nitro Compounds Aliphatic nitro compounds produce peaks at M = 30 (NO+) and 46 (NO2+). Aromatic nitro compounds M-46 (loss of NO2 from ring) The loss of NO (–30) produces an (M–30) peak in aromatic nitro compounds. 2.5.16 Amines Aliphatic amines 1. The molecular ion peak is normally weak. 2. The base peak results from Ca – Cβ cleavage

In RNH2, M–1 peak is visible.

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Chapter 2

Spectroscopy

Aromatic amines 1. M+ is intense. 2. M – 1 peak (resulting from loss of H of – NH2 group) is also prominent. 3. In [Ar–NH–CH2–CH2CH3]•+, a−β cleavage in the aliphatic side chain produces a cation, with the +ve charge on the fragment containing nitrogen

Questions

Q.1. Q.2. Q.3. Q.4. Q.5. Q.6.

What important information can be obtained from the mass spectrum of a compound. What do you understand by molecular ion peak? Why are M + 1 and M + 2 peaks obtained in the mass spectrum of some compound? What do you understand by the term base peak in the mass spectrometry. Why do alkenes fragment to produce an allylic cation. Out of the carbocations given below, which will be formed in most abundance and give an intense peak for an alkane containing these structures

Q.7. Q.8.

Which class of compound/s produce an intense (M–1) peak. Disucss the probable fragmentation of the following compounds (a)

Q.9.

CH3–CH2–CH2–CH=CH2

(b)

(c) C6H5CHO

Given below are the mass spectra of two compounds A (M = 72) and B (M = 60). Both give a +ve iodoform test. Analyse the mass spectrum of each and identify the structure of A and B. Which ion is responsible for the base peak in each case?

Mass spectrum of A

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Mass spectrum of B

Miscellaneous Questions Q.1.

Suggest the best spectroscopic method for distinction between following pairs of compounds: (a) CH3CH2COCH2CH3 and CH3COCH2CH2CH3 (b) C6H5COCH3 and C6H5CH2CHO (c)

(d)

(e) (f)

C6H5NO2 and C6H5NO

Q.2.

A compound (mol. formula C2H6O) shows different pattern of absorption in the region 3000-3600 cm–1 in its IR recorded with the neat liquid and vapour phase. Provide an explanation and also predict its NMR spectrum.

Q.3.

Below are given the IR and the NMR spectra of a dichloro substitued ethane. Identify the compound. Write the structures of isomeric compounds and predict the NMR of other isomer.

240

Q.4.

Chapter 2

Spectroscopy

A compound (A) (Mol. formula C5H8) the monomer of a naturally occuring polymer shows the following: UV: λmax 221 nm IR

C–H stretching above 3050 cm–1 and a carbon–. Carbon bond stretching at ~1640 cm–1.

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NMR spectrum

Identify compound A and its Polymer Q.5.

Mass and IR of the compound are given below. Identify the compound. [Hint: It gives a positive iodoform test].

241 241

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Q.6.

Chapter 2

Spectroscopy

An organic, compound A (NMR given below) shows a prominent M + 2 peak in its mass spectrum reacts with magnesium to form B. B on reaction with (C) (NMR given below) followed by hydrolysis gives D (IR given below). The molecular ion peak of D appears at 72. Identify A–D.

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NMR of C

3 Preparation of Organic Compounds 3.1 INTRODUCTION Small scale preparation of organic compounds forms an important component of student’s training in chemistry laboratory. Small quantities of reagents should be used for these preparations and the products obtained should be subsequently used for other student as unknown compounds for other types of exercises like (i) detection of extra elements (ii) recrystallisation (iii) determination of melting points (iv) detection of functional groups etc. or used as starting compounds in the multistep preparations so as to reduce consumption of chemicals and reduction of waste produced in the laboratory. This chapter includes the procedures adopted for certain simple preparations and chemistry and mechanism of each reaction. A few terms that have been used in this chapter are Working up of a reaction mixture Washing the product with a suitable solvent Yield Yield % Atom economy. Working up of a reaction mixture The term is used to describe the procedure followed for the isolation of the reaction product from the reaction mixture. For example in the preparation of m-dinitrobenzene from nitrobenzene or benzene, the reaction mixture is poured into ice cold water. The solid m-dinitrobenzene separates from the mixture of acids (nitric and sulphuric acid), which does not happen otherwise.Water also removes the excess of acids and any other water soluble impurity. Washing the product with a suitable solvent Continuing with the above example of the preparation of m-dinitrobenzene, small amount of the concentrated acids used in the reaction remain adhering on the crude m-dinitrobenzene making it darker in colour and interfering in the crystallisation process. So the precipitate is washed with water to remove the adhering acid. There are two ways to do it. • The crude product is filtered and after the whole of the solvent has drained, some water is added on the precipitate on the filter funnel and the water removed by filtration, as water percolates through the precipitate, it washes the adhering acids.

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• The crude product is filtered and transferred to a small beaker. Small amount of water is added in the beaker and stirred with a glass rod. The solid is filtered again and the process repeated once again if required. This method is more efficient. Yield The mass of the crude product obtained by weighing after filtration and drying is known as the yield of the reaction. Yield % The percentage yield is calculated using the formula Mass of actual yield × 100 Percentage yield = Mass of theoretical yield Atom economy A new concept called atom economy has been recently introduced. It is defined as the conversion efficiency of a chemical process, if all atoms of the reactants are incorporated in the final product, the reaction has 100% atom economy. An example of this type of a reaction is the Diels-Alder reaction between ethene and butadiene.

The % of atom economy is calculated by the following formula Mass of atoms in the desired product × 100 Atom economy = Mass of atoms in the reeactants An ideal reaction is one in which both yield and the atom economy are high.

3.2 ACETYLATION Substitution of an active hydrogen attached to an electronegative element (N, O or S etc.) by an acetyl group (–COCH3) in an organic compound is known as acetylation reaction and substitution of an active hydrogen by an acyl group (–COR) is known as acylation. Acetylation is an important reaction and is used for (a) Protection of – OH, –NH2 etc. in a number of reactions and can be easily removed by hydrolysis. For example aniline cannot be nitrated directly to p-nitroaniline as it undergoes oxidation and charrs on reaction with concentrated nitric acid. p-Nitroaniline is successfully prepared as outlined below:

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(b) Acetyl derivatives of unknown compounds like carbohydrates, amines and phenols are prepared for their identification. (c) Fries migration of o-acetoxy compounds is a useful method of synthesis of o-hydroxy acetophenones

(d) Acetylation also plays an important role in human body chemical transformations like (i) protein synthesis (ii) excretion of certain drugs from the body occurs easily after acetylation (iii) regulation of DNA and other genetic elements using histone acetylation. Chemistry of acetylation As already mentioned, the term means substitution of an –H by a –COCH3 group. In compounds like primary and secondary amines, phenols and alcohols, the H group can be replaced by COCH3 group.

Acetic acid, acetic anhydride or acetyl chloride can be used as acetylating agent. Each of these reagents has certain advantages and certain disadvantages. Acetic acid is most cost effective and benign of the three reagents. The acetylation with acetic acid has high atom economy as only water is produced as a by product.   RNH + CH COOH    RNHCOCH + H O 2

3

3

2

But the main disadvantage of this reaction is that it is reversible and the yield of the product is low. Acetic anhydride is normally the reagent of choice because the reaction is not reversible and the yields are high. The atom economy is less (as compared to acetylation with acetic acid) because acetic acid is obtained as a by product. RNH2 + (CH3CO)2O ——→ RNHCOCH3 + CH3COOH

Acetyl chloride is the most reactive acetylating agent, the yields of the products are high.

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RNH2 + CH3COC1 ——→ RNHCOCH3 + HC1

The main disadvantages associated with use of acetyl chloride are (i) on storage, it undergoes slow hydrolysis (ii) HC1, that is produced as a byproduct is harmful for eyes, skin and respiratory system. The mechanism of the reaction is • With acetic anhydride

• With acetyl chloride

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• With CH3COOH and Zinc

Cl– > CH3COO– > HO– is the order of the ease of leaving group. Some organic preparations involving acetylation reaction are discussed below: Acetanilide Aspirin Paracetamol Hydroquinone diacetate Hydroxyhydroquinone triacetate b-Naphthyl acetate a-D-Glucose pentaacetate b-D-Glucose pentaacetate

3.2.1 Acetanilide Preparation of acetanilide is a typical example of acetylation of amines. It is prepared by the reaction of aniline with acetic anhydride or acetic acid. Reaction with acetic anhydride The water insoluble amine (aniline) is dissolved in dilute acid like hydrochloric acid. The solution is then treated with acetic anhydride and immediately after with an aqueous solution of sodium acetate. Sodium acetate liberates the amine from its salt which then reacts with acetic anhydride to form acetanilide. C6H5NH2 + HCI ——→ C6H5NH2.HCI

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C6H5NH2.HCI + CH3COONa ——→ C6H5NH2 + CH3COOH +NaCl C6H5NH2+ (CH3CO)2O ——→ C6H5NHCOCH3 + CH3COOH Procedure Reagents Aniline 1 mL Acetic anhydride 1.5 mL Sodium acetate 1.5 g Take freshly distilled aniline (1 mL) in a conical flask (100 mL) and suspend it in water (~5 mL). Add concentrated hydrochloric acid (few drops) to the above, suspension till aniline dissolves completely. Add crushed ice (10-20 g). Make a solution of sodium acetate (1.5 g) in water ( 5-7 mL) in a boiling tube. Add acetic anhydride (1.5 mL) to the flask containing aniline solution, shake and immediately add to it sodium acetate solution. Shake, acetanilide separates out as a white solid. Keep the reaction mixture for 5-10 minutes, filter the solid under pressure using a filter tube and a funnel. Dry, weigh the solid and recrystallise ~100 mg from water. Take the melting point of the dry crystalline sample. Yield 1.2 g, m.p. 114°C. Notes: • Freshly distilled aniline should be used in the preparation otherwise a yellowish coloured acetanilide is obtained. • Sodium acetate is used to liberate the amine, which is dissolved in an acid and is thus present as its salt and not as free amine. • A small amount of diacetyl derivative might be formed in the reaction C6H5NHCOCH3 + (CH3CO)2O ——→ C6H5N(COCH3)2 + CH3COOH

However this diacetyl derivative gets converted to acetanilide during recrystallisation from water/ dilute alcohol. C6H5N(COCH3)2 + H2O ——→ C6H5NHCOCH3 + CH3COOH

• Toludines (o–, m and p–) and anisidines (o-, m and p-) etc. can be given as unknown amines for carrying out acetylation in the class. Above mentioned procedure is adopted for these amines also. Acetanilide can also be prepared by acetylation of aniline with acetic acid ∆

 C6 H5 NH 2 + CH3COOH    C6 H5 NHCOCH3 +H 2 O Procedure Reagents Aniline 1 mL Zinc dust 50 mg Glacial acetic acid 1.5 mL Take aniline (1 mL), glacial acetic acid (l.5 mL), zinc dust (50 mg) and few pieces of pumice stones in a dry round bottomed flask fitted with a quick fit joint. Set up the apparatus for refluxing on a wire gauze/sand bath. Reflux the reaction mixture for ~30 minutes such that the temperature at the open end of the condenser is ~ 105°C during heating. Filter the hot solution so as to remove any zinc dust, allow the filtrate to cool and pour it, with stirring, into a beaker containing some crushed ice (30-40 g). Acetanilide

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separates out as a white solid. Filter, dry, weigh, recrystallise from water, determine the melting point. Yield 0.8 g, m.p 114°C. Notes: • Zinc is added to prevent aerial oxidation of aniline during refluxing. Reaction between zinc and acetic acid produces hydrogen which displaces air from the reflux apparatus. • The temperature at the open end of condenser is maintained above ~100°C so as to allow water vapour to escape and not condense back into the reaction flask. Removal of water allows forward reaction and prevents the backward reaction.

3.2.2 Aspirin (Acetylsalicylic acid) Aspirin is an analgesic, anti inflammatory, anti rheumatic drug. It is also prescribed to people suffering from heart problems. It is prepared by acetylation of salicylic acid with acetic anhydride in presence of a small amount of concentrated sulphuric acid.

Procedure Reagents Salicylic acid 1g Acetic anhydride 3 mL Conc. sulphuric acid 0.2 mL Take salicylic acid (l g), acetic anhydride (3 mL) in a dry boiling tube. Add concentrated sulphuric acid (0.2 mL ) and mix well using a glass rod. Heat the reaction mixture in a hot water bath maintained at 60°C for 15 minutes. Stir the reaction mixture during heating with a glass rod. Cool the reaction mixture and pour it with stirring into a beaker containing crushed ice (~30 g). Aspirin separates out as a white solid. Filter, wash on the funnel with water (~5 mL) and allow water to filter completely. Dry the solid on a filter paper, weigh, recrystallise with dilute alcohol and determine the melting point. Yield l g m.p 133-135°C. Notes: • Aspirin may be contaminated with some salicylic acid, which can be tested by treating the alcoholic solution of aspirin with ferric chloride. The impure sample produces a violet colouration. • Concentrated sulphuric acid is used during acetylation to break the hydrogen bond between the -OH and -COOH group of salicylic acid so that acetylation proceeds smoothly. Microwave assisted synthesis of aspirin Take salicylic acid, acetic anhydride and concentrated sulphric acid as above in a round bottomed flask and set up the reflux apparatus in a commercial microwave as shown in Fig. 1.9. Reflux the reaction mixture for 2+5 minutes at 70°C (2 minutes for attaining the temperature and 5 minutes at 70°C). Work up the reaction mixture as given above, recrytallise and determine its melting point.

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3.2.3

Preparation of Organic Compounds

Paracetamol (4-Acetamidophenol)

Paracetamol is a very commonly used antipyretic drug. It is prepared by selective acetylation of p-amino phenol with acetic anhydride in presence of phosphoric acid.

Procedure Reagents p-Aminophenol

1.5g

Acetic anhydride

1.8 mL

Phosphoric acid

20 drops

Take p-aminophenol (1.5g) in a conical flask (100 mL), add water (20 mL) and concentrated phosphoric acid (20 drops), swirl to get a clear solution. Now add acetic anhydride (1.8 mL) and heat the flask in a warm water bath for 10 minutes. Paracetamol separates out on cooling. Filter the product, wash with ~3-4 mL water, dry, weigh and recrystallise ~100 mg from water and determine the melting point. Yield 1.3 g, m.p. 169-171°C.

3.2.4

Hydroquinone Diacetate

Hydroquinone diacetate is prepared by acetylation of hydroquinone with acetic anhydride in presence conc. sulphuric acid.

Procedure Reagents Hydroquinone Acetic anhydride Conc. sulphuric acid

1g 2 mL 1-2 drops

Take hydroquinone (1 g) and acetic anhydride (2 mL) in a dry boiling tube. Add a drop of conc. sulphuric acid. Mix well with a glass rod, an exothermic reaction starts. Allow the reaction mixture to stand for ~ 5 minutes. Pour the reaction mixture into a beaker containing ice cold water (~20-30 mL) with stirring. A white solid separates. Filter the solid, wash with water and recrystallise from dilute alcohol. Determine its melting point. Yield 0.9 g, m.p.l22°C.

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3.2.5 Hydroxyhydroquinone Triacetate It is obtained by acetylation of p-benzoquinone with acetic anhydride in presence of conc. sulphuric acid. The reaction is known as Thiele’s acetylation. The reaction is initiated by 1,4-addition of acetic anhydride to p-benzoquinone as shown below.

Mechanism

Procedure Reagents p-Benzoquinone 1g Acetic anhydride 3 mL Conc. sulphuric acid 0.15 mL Take acetic anhydride (3 mL) in a dry conical flask (50 mL) and add drop wise with shaking, conc. sulphuric acid (0.15 mL) to it. Now add powdered p-benzoquinone (1g) in small lots in ~10 minutes, continuously shake or stir (using a magnetic stirrer) and maintain the temperature of the reaction mixture below 40°C. Stir the reaction mixture for another 5-10 minutes and pour it into a beaker containing ~20-30g of crushed ice. The hydroxyhydroquinone triacetate separates as a white solid. Filter the solid, wash with water and recrystalise from alcohol. Report the yield and melting point. Yield 2.1 g, m.p. 96-97°C.

3.2.6 b-Naphthyl Acetate It is prepared by the reaction of b-naphthol with acetic anhydride in presence of sodium hydroxide.

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Procedure Reagents b-Naphthol

1g

Acetic anhydride

1.2 mL

Sodium hydroxide

5 mL, 10%

Procedure Cool a solution of b-naphthol (Ig) in freshly prepared sodium hydroxide (5 mL, 10%) in a conical flask in an ice bath. Add acetic anhydride (1.2 mL) and shake the reaction mixture for 15-20 minutes. b-naphthyl acetate separates as a light pink coloured solid. Filter, wash with water, dry and weigh. Recrystallise ~100 mg of the solid from dilute alcohol and determine its melting point. Yield 1.2 g, melting point 71°C.

3.2.7 a-D-GIucose Pentaacetate It is prepared by the reaction of glucose with acetic anhydride in presence of anhydrous zinc chloride.

Procedure Reagents Glucose

1g

Acetic anhydride

10 mL

Fused zinc chloride 0.25 g Take anhydrous zinc chloride (0.25 g, see page 91 for the procedure used for its preparation), acetic anhydride (10 mL) in a small dry round bottomed flask and add glucose (1g). Shake the mixture well and set up the apparatus for refluxing in a water bath. Reflux the reaction mixture for an hour. Check for completion of reaction which is indicated by formation of a solid product on pouring with stirring,

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a drop of reaction mixture into ice cold water in a test tube. If the reaction is complete, pour the cooled reaction mixture into ice cold water (–50 mL) in a beaker. The a-D-glucose pentaacetate separates as a white solid. Filter, wash with water (5-7 mL), dry and weigh. Recrystallise from methanol and determine melting point. Yield 1.5g, m.p. 112-114°C.

3.2.8 b-D-Glucose Pentaacetate The b-D-Glucose pentaacetate is prepared by acetylation of glucose with acetic anhydride in presence of anhydrous sodium acetate.

Procedure Reagents D-Glucose 1g Acetic anhydride 8 mL Anhydrous sodium acetate 1.5 g Add acetic anhydride (8 mL) to anhydrous sodium acetate (1.5g, see page 92 for the procedure adopted for preparation of anhydrous sodium acetate) in a small, dry round bottomed flask and shake well. Add glucose (1g) to the above mixture and shake well again. Reflux the reaction mixture for 30-40 minutes in a boiling water bath. Cool and pour the reaction mixture in a beaker containing ice-cold water (~ 50 mL) and stir well with a glass rod to decompose excess acetic anhydride. A white solid separates out. Filter the solid, dry and weigh it. Recrystallise ~ 100 mg of solid from methanol and determine its melting point. Yield 1.4g, m.p. 131°C.

3.3 ANHYDRIDES Carboxylic acids can be converted to their anhydrides by elimination of a water molecule from two carboxylic acid groups of either the same molecule i.e. a dicarboxylic acid (intramolecular elimination reaction) or from two molecules of the same or different monocarboxylic acid (intermolecular elimination reaction).

3.3.1 Phthalic Anhydride It is obtained by sublimation of phthalic acid.

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Procedure Reagent Phthalic acid 1g Take phthalic acid (1g) in a dry porcelain dish (3 cm in diameter). Take a circular Whatmann filter paper No. l, slightly bigger in diameter. Punch small holes in the paper and place it on the dish. Cover the filter paper with an inverted water-jacketed glass/ordinary filter funnel and plug its stem loosely with cotton. Heat the dish on a sand bath with the help of a small bunsen burner flame. The anhydride that is formed rises through the holes in the filter paper and condenses in the hollow of the funnel and on the filter paper. Remove the burner and allow the apparatus to cool. Collect the anhydride and determine its melting point. Yield is 0.8 g, m.p. is 132°C. Notes: • Use a small flame for heating otherwise phthalic acid also volatilises and condenses under the funnel along with the anhydride. • If the anhydride is contaminated with the acid, it gives effervescence with a solution of sodium bicarbonate.

3.3.2 Succinic Anhydride It is prepared by either heating succinic acid as above or heating succinic acid with acetic anhydride

Procedure Reagents Succinic acid 2g Acetic anhydride 4 mL Gently reflux a mixture of succinic acid (2g) and redistilled acetic anhydride (4 mL) in a small, dry round bottomed flask for 15-20 minutes. Cool the flask first at room temperature and then in an ice bath. Filter the solid anhydride that separates as a white solid and wash it with ether (2 × 3 mL), keep in a dessicator over anhydrous calcium chloride and determine its melting point. Yield 1.5 g, m.p. 119-20°C.

3.3.3 Maleic Anhydride It is obtained by distilling maleic acid with an inert solvent having a high boiling point. The water distils over with the solvent leaving behind the anhydride.

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Procedure Reagents Maleic acid

l0 g

Tetrachloroethane

10 mL

Take maleic acid (10 g) and sym-tetrachloroethane (10 mL) in a small, dry distillation flask. Set up the apparatus for distillation (use water condenser). Heat the mixture slowly and collect the liquid that distils over up to 160°C. This consists of a mixture of tetrachloroethane and water. Replace the water condenser by an air condenser and place a fresh dry receiver and continue distillation. Maleic anhydride distils over between 190-200° C. Recrystallise ~200 mg of maleic anhydride from chloroform. Yield 7.5 g, m.p. 54°C.

3.4 BENZOYLATION Substitution of an active hydrogen (H of an –OH, –NH or –SH etc.) by a benzoyl (–COC6H5) group is known as benzoylation. Benzoylation reaction when carried out in presence of a base like sodium hydroxide is known as Schotten Baumann Reaction. Benzoylation reaction of phenol and aniline are given below: C6H5OH + C6H5COCl + NaOH ——→ C6H5OCOC6H5 + H2O + NaCl

Phenol

Phenyl benzoate

C6H5NH2 + C6H5COCl + NaOH ——→ C6H5NHCOC6H5 + H2O + NaCl Aniline

Benzanilide

Sodium hydroxide used in the reaction performs a number of functions: • A slight excess of benzoyl chloride is required for completion of reaction. The unreacted benzoyl chloride interferes in the separation of the reaction product after the completion of the reaction. Sodium hydroxide present in the reaction mixture hydrolyses the unreacted benzoyl chloride, forming sodium benzoate, which remains dissolved in the medium. C6H5COC1 + NaOH ——→ C6H5COONa + HC1 • Sodium hydroxide neutralises the HC1 formed during the reaction. • During the benzoylation of acidic compounds like phenols, the unreacted compound remains dissolved in the basic medium and therefore does not contaminate the final product. • Sodium hydroxide also activates the phenolic compounds for benzoylation by their conversion to more reactive species i.e. phenoxide ion. C6H5OH + NaOH ——→ C6H5O– Na+ + H2O Benzol group like the acetyl group is used as a protecting group. Benzoyl derivatives of amines, phenols and aminoacids etc. are crystalline compounds with sharp melting points, these are prepared in the laboratory for identification of unknown amines, phenols and amino acids etc. Benzoylation is an example of an electrophilic reaction in which the phenol or aniline etc. behave like a nucleophile and the C6H5CO group of benzoyl chloride as an electrophile. The step wise reaction mechanism is given below.

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Benzoylation of amine (aniline)

Benzoylation of phenol

Preparation of the following compounds is discussed in this section. Benzanilide Hippuric acid Phenyl benzoate

3.4.1 Benzanilide Benzanilide is prepared in the laboratory by the reaction of aniline with benzoyl chloride in presence of sodium hydroxide. As already mentioned, this is known as Schotten Baumann Reaction. The reaction mechanism and the role of each reactant is discussed above (section 3.4) Procedure Reagents Aniline Benzoyl chloride Sodium hydroxide

0.5 mL 0.75 mL 10 mL, 10%

Take aniline (0.5 mL) in a clean conical flask (100 mL) and add to it a freshly prepared dilute solution of sodium hydroxide (10 mL, 10%). Take benzoyl chloride (0.75 mL) in a test tube. Add 2-3

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drops of benzoyl chloride from the tube to the flask containing aniline and sodium hydroxide. Stopper the flask with a bark cork and shake the mixture vigorously. Add the remaining benzoyl chloride dropwise (2-3 drops at a time) and shake the reaction mixture vigorously after each addition. A solid product is formed (if lumps of solid are formed, break these gently into fine solid particles with the help of a glass rod). The completion of reaction is indicated by the absence of smell of benzoyl chloride. Check the alkalinity of the reaction mixture, it should be alkaline. Otherwise add ~ 5 mL of sodium hydroxide to make it alkaline. Filter the solid and wash twice with water on the funnel and filter. Dry, weigh and recrystallise ~100 mg of solid from alcohol. Determine its melting point. Yield 0.7 g. m.p. 162°C. Notes: • Use freshly distilled aniline for the reaction. • There should not be any unreacted aniline at the completion of the reaction. • Big lumps of benzanilide may contain unreacted benzoyl chloride hence these must be broken to smaller particles to release the trapped benzoyl chloride. • There should be no smell of unreacted benzoyl chloride and the reaction mixture should be alkaline at the completion of the reaction. • Many other amines can also be benzoylated following the same procedure.

3.4.2 N-Benzoylglycine (Hippuric Acid) Glycine on benzoylation with benzoyl chloride in presence of sodium hydroxide gives sodium salt of benzoylglycine from which the benzoylglycine, also known as Hippuric acid is obtained on acidification.

Hippuric acid is a natural compound of limited occurrence and is found in urine of horses and some other herbivorous animals. Traces of it are also present in human urine. Procedure Reagents Glycine 0.5 g Benzoyl chloride 0.9 mL Sodium hydroxide 10 mL, 10% Take glycine (0.5g) and dissolve it in sodium hydroxide solution (10 mL, 10 %) in a conical flask. Add benzoyl chloride (0.9 mL) in 2-3 lots to the solution of glycine. Stopper the flask with a bark cork and shake vigorously after each addition. The reaction is exothermic, cool if the reaction mixture becomes hot. After the reaction is complete, which is indicated by absence of any smell of benzoyl chloride and alkaline pH of the mixture, acidify the reaction mixture to congo red with cold dilute hydrochloric acid. A solid product is obtained which contains benzoylglycine and benzoic acid (benzoic acid is obtained by the hydrolysis of excess unreacted benzoyl chloride). Filter the solid, wash with ~5 mL water twice and

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allow it to dry. Take the solid in a dry boiling tube and heat it with ~5 mL carbon tetrachloride in a water bath, filter and repeat the treatment with carbon tetrachloride once more. Filter, dry, weigh and recrystallise with hot water. Determine the melting point of the purified product. Yield 0.9g, m.p 187-188°C. Notes: • The reaction mixture should be alkaline throughout the benzoylation reaction. More sodium hydroxide solution should be added, if required. • There should be no smell of unreacted benzoyl chloride at the completion of reaction. • Benzoic acid should be completely removed by treating with carbon tetrachloride before recrystallising benzoylglycine.

3.4.3 Phenyl benzoate Like amines, phenols also contain active hydrogen which can be substituted by a benzoyl group. Phenol on benzoylation by Schotten Baumann reaction gives phenyl benzoate. Other phenols like catechol, resorcinol, quinol, a-, b- naphthol etc. can also be benzoylated by the procedure described below. Procedure Reagents Phenol

0.5 g

Benzoyl chloride

0.75 mL

Sodium hydroxide

10 mL, 10%

Take freshly distilled phenol (0.5 g) in dilute sodium hydroxide (10 mL, 10%) in a conical flask (100 mL). Add benzoyl chloride (0.75 mL) in 2-3 lots and shake well after each addition as described above in the preparation of hippuric acid. After the reaction is complete (indicated by the disappearance of smell of benzoyl chloride) and the reaction mixture is alkaline (add ~5 mL sodium hydroxide if not so), filter the solid that separates out and wash with water (~5 mL) twice. Filter, dry, weigh and recrystallise ~100 mg of it from dilute alcohol. Yield 0.7g, m.p. 70°C.

3.5 BECKMANN REARRANGEMENT Acid catalysed rearrangement of keto oximes to substituted anilides is known as Beckmann rearrangement. It is an example of a stereospecific rearrangement reaction in which the alkyl group trans to the –OH group of keto oxime migrates from carbon to nitrogen. Phosphorus pentachloride, phosphorus pentoxide, phosphorus oxychloride, phosphoric acid, sulphuric acid etc. are commonly used as catalyst to bring about the rearrangement.

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Mechanism

3.5.1 Benzanilide from Benzophenone Oxime Benzophenone oxime gives benzanilide on treatement with phosphorus pentachloride or thionyl chloride

Benzophenone oxime

Procedure Reagents Benzophenone Hydroxylamine hydrochloride Sodium hydroxide Ethanol Conc. hydrochloric acid

2g 1.2 g 2.4 g 7.5 mL 6 mL

Dissolve Benzophenone (2 g) and hydroxylamine hydrochloride (1.2 g) in a mixture of ethanol (7.5 mL) and water (2 mL). Stir the solution continuously and add to it a solution of sodium hydroxide (2.4 g in 5 mL water) in small portions. In case the reaction becomes vigorous, cool the flask under tap water. After the addition is over, reflux the reaction mixture for 20 minutes. Cool the reaction mixture and dilute with water (20 mL). Filter the unreacted benzophenone that separates out. Now pour the filtrate with stirring into dilute hydrochloric acid (6 mL concentrated hydrochloric acid in 30 mL water). Filter the precipitated oxime and wash with cold water. Recrystallise from methanol. Yield 2g, m.p. 142°C. Store the oxime in a vacuum desiccator for the next step.

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Benzanilide (Beckmann rearrangement)

Procedure Reagents Benzophenone oxime 2g Dry ether 20 mL Phosphorus pentachloride 3g or thionyl chloride 3 mL Dissolve benzophenone oxime (2 g) in anhydrous ether (20 mL) in a conical flask. Cool the solution in an ice bath. Add finely powdered phosphorus pentachloride (3 g) or thionyl chloride (3 mL) with continuous stirring to the solution of benzophenone oxime. Remove ether (caution, ether is highly inflamable). Pour the residue into ice-water in a beaker with stirring. Filter the separated product, wash with water and crystallise from alcohol. Yield 1.7 g, m.p. 163°C. Notes: • Use dry benzophenone oxime for this reaction. • Ether is highly inflamable, do not use ether near a hot object in the laboratory.

3.5.2 Acetanilide from Acetophenone Oxime Acetanilide can be obtained by Beckmann rearrangement of acetophenone oxime with concentrated sulphuric acid or thionyl chloride as catalyst. Acetophenone oxime

Procedure Reagents Acetophenone 2 mL Hydroxylamine hydrochloride 2g Sodium acetate 2g Ethanol 6 mL Dissolve acetophenone (2 mL) in aqueous alcohol (6 mL alcohol and 20 mL water) in a conical flask. Add sodium acetate (2 g) and hydroxylamine hydrochloride (2 g) to it. Heat the mixture in a boiling water bath for 10-15 minutes and then cool it in an ice-salt bath. Filter the separated oxime and crystallise from water, scratch the sides of the tube from inside with a glass rod if necessary to induce crystal formation. Yield 2g, m.p. 59°C. Note: Use dry acetophenone oxime for the next step.

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Acetanilide

Procedure Reagents Acetophenone oxime 2g Conc. sulphuric acid 4 mL or thionyl chloride 2 mL Take concentrated sulphuric acid (4 mL) in a dry conical flask and warm it in a hot water bath. In case thionyl chloride is used, heat the flask only upto 65-70°C. Add acetophenone oxime (2 g) slowly, carefully with stirring to the warm acid or thionyl chloride. Keep the reaction mixture for 10-15 minutes, continuously stir during this. Cool and pour slowly with stirring over crushed ice in a beaker. Filter the separated product and wash with water. Recrystallise from boiling water. Yield is 1.9 g, m.p. 114°C.

3.6 BENZOIN CONDENSATION Aromatic aldehydes undergo self condensation in presence of potassium cyanide as catalyst and produce a-hydroxy ketones commonly known as benzoins and hence the reaction is known as benzoin condensation.

Mechanism Cyanide was believed to be a specific catalyst for this reaction and because of its electron withdrawing nature, it helps in increasing the acidity of H of –CHO groups

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3.6.1 Benzoin Procedure Reagents Benzaldehyde 1.8 mL Rectified spirit 4 mL Potassium cyanide 0.25 g (in 2 mL water) Reflux gently, a mixture of benzaldehyde (1.8 mL ), potassium cyanide solution (0.25 g in 2 mL water) and rectified spirit (4 mL) for 30 minutes in a small round bottomed flask. Cool the solution in an ice-bath. A white solid is formed. Filter the solid, wash with water (~5 mL) and cold rectified spirit (~5 mL). Dry, weigh, recrystallise ~100 mg of it from rectified spirit. Yield 1.6 g, m.p. 135°C. Notes: • Potassium cyanide is a deadly poisonous compound. It is fatal if inhaled, absorbed through the skin or swallowed. IMMEDIATE MEDICAL HELP SHOULD BE SOUGHT. It should be handled with extreme care using hand gloves. • Use freshly distilled benzaldehyde for this reaction. Green synthesis of Benzoin In this method, thiamine hydrochloride is used instead of cyanide as a catalyst. Mechanism

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Procedure Reagents Benzaldehyde 5g Thiamine hydrochloride 0.8 g Ethanol 6 - 7 mL Sodium hydroxide 2.5 mL, 2M Take thiamine hydrochloride (0.8 g) in a round bottomed flask (50 mL), and dissolve it in water (2-3 mL) and ethanol (6-7 mL). Cool the solution in ice-cold water. To this now add a cooled solution of sodium hydroxide (2.5 mL, 2 M) dropwise with swirling in 5-7 minutes and then freshly distilled benzaldehyde (5 g). Fix up a water condenser on the round bottomed flask and heat gently, the reaction mixture in a hot water bath for 90 minutes. Allow the reaction mixture to cool slowly to room temperature and then in an ice bath. Scratch the inner sides of the flask to induce crystal formation if solid does not form on cooling. Yield 3 g, m.p. 135°C. Notes:

• Use freshly distilled benzaldehyde. • Store thiamine hydrochloride in a refrigerator.

3.6.2 Benzil Benzil, an a-diketone is obtained by oxidation of Benzoin with concentrated nitric acid.

Mechanism of the reaction

Procedure Reagents Benzoin Conc. Nitric acid Acetic acid

2g 5 mL 10 mL

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Reflux in a fume chamber a mixture of benzoin (2g), conc. nitric acid (5 mL) and acetic acid (10 mL) in a round bottomed flask in a water bath for 2 hrs. Cool the reaction mixture and pour it into a beaker containing crushed ice (50 g). Benzil separates as a yellow coloured solid. Filter, wash with water, dry, weigh and recrystallise from rectified spirit. Yield l.8 g, m.p. 95°C.

3.6.3 Benzilic Acid It is obtained by base catalysed rearrangement of benzil

Mechanism

Procedure Reagents Benzil 1.5 g Potassium hydroxide 1.9 g Rectified spirit 4.8 mL Conc. hydrochloric acid 6.6 mL Reflux a mixture of benzil (1.5 g), potassium hydroxide (1.9 g dissolved in 3-4 mLwater) and rectified spirit (4.8 mL) in a water bath for 10-15 minutes in a round bottomed flask. Pour the reaction mixture into water (25-30 mL in a beaker). Cool the clear filtrate by adding ice and then acidify with conc. hydrochloric acid. Filter the product, wash with cold water, dry, weigh and recrystallise ~200 mg from hot water. Yield l.l g, m.p. 150°C.

3.7 BENZIDINE REARRANGEMENT The acid catalysed rearrangement of hydrazobenzene to 4,4′-diaminobiphenyl is known as benzidine rearrangement. Minor amounts of other products are also formed during the reaction.

Other products which are obtained in minor amounts are:

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Mechanism

3.7.1 Benzidine It is obtained from nitro benzene in two steps: • Reduction of nitrobenzene to hydrazobenzene • Rearrangement of hydrazobenzene to benzidine. Hydrazobenzene

Procedure Reagents Nitrobenzene 5 mL Absolute methanol 100 mL Magnesium turnings l0 g Iodine Take nitrobenzene (5 mL) and absolute methanol (50 mL) in a dry round bottomed flask fitted with a reflux condenser. Add a few crystals of iodine and magnesium turnings (6 g) to the solution of nitrobenzene. In case the reaction does not start within 2-3 minutes, warm the reaction mixture on a water bath until the reaction starts. If the reaction becomes too vigorous, cool it in cold water for 2-3 minutes. Add more absolute methanol (50 mL) and shake the reaction mixture in the flask. When the reaction has subsided, add more magnesium turnings (4g) and allow the reaction to complete as above. Now heat the

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reaction mixture on a water bath (80-85°C) until the mixture is colourless or light pale yellow. Frequently shake the reaction mixture during heating. Pour the hot reaction mixture into a beaker containing icecold water (100 mL). Acidify the reaction mixture with acetic acid till acidic to litmus. The separated magnesium hydroxide dissolves and a pale yellow solid of hydrazobenzene floats on the surface. Filter the yellow solid at the pump leaving the magnesium granules in the beaker. Wash with water and recrystallise from 90% ethanol containing a little dissolved sulphur dioxide (with minimum exposure to air to avoid oxidation of the product). Yield 2.2g, m.p. 125-26°C. Dry the product in a vacuum desiccator. Benzidine Procedure Reagents Hydrazobenzene 2g Methanol 15 mL Conc. hydrochloric acid 12 mL (5+7 mL) Stannous chloride (0.6g SnCl2 in 0.6 mL. HCl + 3 mL water) Cool a mixture of methanol (15 mL), stannous chloride solution (3 mL) and conc. hydrochloric acid (5 mL) in a conical flask to ~0°C in an ice-bath. Add this cooled solution to hydrazobenzene (2g) with shaking. Keep the resultant mixture in the ice-bath for 15 minutes with frequent shaking. Benzidine dihydrochloride separates out. Warm the solution to dissolve the solid and add (i) few drops of stannous chloride if the solution is yellow (ii) conc. hydrochloric acid (7 mL) slowly with shaking. Now again cool the reaction mixture, filter the benzidine hydrochloride, wash with dilute hydrochloric acid (3-4 mL) and methanol (2-3 mL). To get benzidine, dissolve the hydrochloride obtained above in water (~20 mL) and neutralise slowly by adding sodium bicarbonate solution (1.5g in ~ 15 mL water). Cool, filter benzidine and recrystallise from dilute alcohol, store in a vacuum desiccator yield, 1g, m.p. 128°C.

3.8 CANNIZZARO REACTION Reaction of aldehydes (which lack a-hydrogen/s) with concentrated sodium/ potassium hydroxide to produce an alcohol and the salt of the corresponding acid is known as cannizzaro reaction. For example, benzaldehyde gives benzyl alcohol and sodium benzoate

Formaldehyde gives methyl alcohol and sodium formate 2HCHO + NaOH → CH3OH + HCOONa Formaldehyde Methyl Sodium alcohol formate Intramolecular cannizzaro’s Glyoxal contains two aldehyde groups both of which lack a-hydrogen and hence it undergoes intramolecular cannizzaros’ reaction in which one –CHO group is reduced to –CH2OH and the other is

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oxidised to –COOH. Small amount of oxalic acid and glycol are obtained as a result of inter molecular reaction.

Cross or intermolecular cannizzaro’s reaction between (for example) benzaldehyde and formaldehyde gives benzyl alcohol and formic acid. C6H5CHO + HCHO + NaOH ——→ C6H5CH2OH + HCOONa Mechanism of Cannizzaro’s reaction

Mechanism of cross Cannizzaro’s reaction

3.8.1 Cannizzaro’s reaction with Benzaldehyde As already mentioned, a mixture of benzyl alcohol and benoic acid is obtained. Procedure Reagents Benzaldehyde 10 mL Potassium hydroxide 6.2g (pellets) Dissolve potassium hydroxide pellets (6.2g) in water (30 mL) in a conical flask, add it to freshly distilled benzaldehyde (10 mL) in a conical flask slowly with shaking. Allow the reaction mixture to stand overnight. Cool the mixture and add just enough water to dissolve any precipitated potassium benzoate. Extract the mixture with ether (3 × 15 mL). Cool the clear alkaline solution and acidify with cold hydrochloric acid (1:1, 60 mL). Cool the acidified solution, filter the separated benzoic acid and crystallise from hot water. Yield 4.2g, m.p. 121°C.

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Dry the combined ether extract (obtained above) over anhydrous magnesium sulphate, filter and remove ether in a rotary evaporator. Distil the residue containing benzyl alcohol using a flame. Yield is 2.6g, b.p. 205-206°C. Note: Ether is highly inflamable.

3.9 ETHYL ACETOACETATE/ ACETOACETIC ESTER (CLAISEN CONDENSATION) Base catalysed condensation of an ester containing an a-hydrogen atom with a molecule of the same ester or with a molecule of another ester to give b-keto esters is known as Claisen condensation. For example, the reaction of ethyl acetate with sodium ethoxide gives ethyl acetoacetate. Sodium ethoxide or sodium (which forms small amount of sodium hydroxide on reaction with traces of moisture present in the reaction mixture) are used as catalyst. NaOC H

2 5 → CH3COCH 2 COOC 2 H5 + C2 H5 OH 2CH3COOC2 H5 

Ethyl acetate

Ethyl acetoacetate

Mechanism

3.9.1 Ethyl Acetoacetate Procedure Reagents Ethyl acetate (dry) 50 mL Sodium metal wire 4.5g Dilute acetic acid 1:1 Take ethyl acetate (50 mL) and small pieces of finely cut sodium metal or preferably sodium wire (4.5 g) in a round bottomed flask (250 mL) fitted with a reflux condenser having a calcium chloride drying tube at the top. Warm the flask in an oil bath to start the reaction. Once the reaction starts, it becomes vigorous, cool in cold water if necessary. When the reaction has subsided, heat it gently for 1.5 hours or until all the sodium has reacted. Cool the reaction mixture. In case there is still a small amount of sodium left in the reaction mixture, add 2-3 mL of methanol to react it. Now add acetic acid (dilute

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1:1) slowly to make it just acidic to litmus. Pour the resultant acidic solution into cold saturated sodium chloride solution (50 mL) in a beaker. Transfer the solution to a separating funnel and shake. Allow the upper ester layer to separate. Take it out in a dry conical flask, dry over anhydrous magnesium sulphate. Filter and distil the ester under reduced pressure. Ethyl acetate distils over first and ethyl acetoacetate distils at 72-76°C (14 mm). Yield 10g, b.p. 181°C. Notes: • Ethyl acetate should be dried over anhydrous calcium chloride and distilled before use. • Sodium should be cut into small pieces or made into wire under dry ether in order to avoid exposure of sodium to the atmosphere. • The reaction should be performed in a fume cupboard preferably under nitrogen atmosphere.

3.10

DERIVATIVES OF ALDEHYDES AND KETONES

Aldehydes and ketones react with substituted ammonias like Hydrazine (NH2NH2), phenylhydrazine (C6H5NHNH2), hydroxylamine (NH2OH), semicarbazide (NH2CONHNH2) and 2,4-Dinitrophenyl hydrazine to form crystalline compounds with sharp melting points. Therefore these are prepared as derivatives for identification of aldehydes and ketones. The substituted ammonias are not available as free bases (except 2,4- dinitrophenyl hydrazine) but are stored as their hydrochlorides or hydrogen sulphates, because the free bases undergo slow oxidation on exposure to atmosphere.

A weak acid like acetic acid is used as a catalysts for the reaction between the ZNH2 and aldeyde or ketone.

However a strong acid interferes in the reaction by reducing the concentration of free base i.e. the nucleophile ZNH 2 + H + → ZN + H3 , which can no longer act like a nucleophile

3.10.1 2,4–Dinitrophenylhydrazone 2,4- Dinitrophenyl hydrazone of a carbonyl compound is prepared by its reaction with 2,4, dinitrophenyl hydrazine in presence of a strong acid catalyst like sulphuric acid or hydrochloric acid.

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2,4-Dinitrophenylhydrazine is insoluble in water as well as alcohol. It is dissolved in alcohol by addition of conc. sulphuric acid.

The solution of 2, 4- dinitrophenylhydrazine is then added to the solution of the carbonyl compound when H+ from the protonated 2,4- dinitrophenylhydrazine shifts to oxygen of the C = O group, liberating the free base

Procedure Reagents Acetone 0.5 mL 2,4-Dinitrophenylhydrazine 1.25 g Conc. sulphuric acid 1-2 mL Methanol 25 mL Suspend 2,4-dinitrophenylhydrazine (1.25 g) in methanol (~ 20 mL) in a conical flask. Add just enough conc. sulphuric acid (1-2 mL) dropwise, cautiously with shaking till the suspended 2,4-dinitrophenylhydrazine dissolves. (Do not add more sulphuric acid (i) if a few particles of the solid do not dissolve (ii) excess of sulphuric acid reprecipitates the dissolved solid). Filter is necessary. Take acetone (0.5 ml) in another clean, dry conical flask and dissolve it in

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minimum amount of methanol (3-4 mL). Add the solution of 2,4-dinitrophenylhydrazine with shaking to the solution of acetone. A yellow solid separates out. Filter, wash with methanol (~ 5 mL), dry and recrystallise from ethanol. Yield 1.25g, m.p. 128°C. Note: Follow the same procedure for other carbonyl compounds.

3.10.2 Semicarbazone Semicarbazones are crystalline derivative of aldehydes and ketones and are prepared by reaction of these compounds with semicarbazide hydrochloride in presence of sodium acetate. Sodium acetate liberates the amine from its salt, which then reacts with the carbonyl compounds. NH2CONHNH2 HCl + CH3COONa ——→ NH2CONHNH2 + CH3COOH + NaCl

Semicarbazide has two –NH2 groups marked 1 and 2 in the structure given below. 2

1

NH 2CONH NH 2 Only the NH2 marked 1 can react as nucleophile because the NH2 marked 2 is in resonance with the electron withdrawing C = O group and therefore is not available as free nucleophile

Mechanism The acetic acid obtained by reaction between semicarbazide hydrochloride and sodium acetate catalyse the reaction

Procedure Reagents Acetone 1 mL Semicarbazide hydrochloride 1g Sodium acetate 2g Add acetone (1 mL) dropwise to a stirred solution of semicarbazide hydrochloride and sodium acetate dissolved in minimum amount of water. A white solid separates. Filter, dry and recrystallise from water. Yield 0.5g, m.p. 187°C.

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3.10.3 Oxime Reaction between a carbonyl compound, hydroxylamine hydrochloride and sodium acetate yields the oxime of the carbonyl compound NH2OH. HCl + CH3COONa ——→ NH2OH + CH3COOH + NaCl Hydroxylamine hydrochloride

Hydroxyl amine

Procedure Reagents Cyclohexanone

1 mL

Hydroxylamine hydrochloride

1g

Sodium acetate

1.5 g

Dissolve hydroxylamine hydrochloride (1g) and sodium acetate (1.5g) in minimum amount of water in a conical flask. Add to this a solution of cyclohexanone (1 mL) in minimum amount of methanol (2-3 mL). Shake well, a white solid separates out. Filter, dry and recrystallise from dilute alcohol. Yield 0.5g, m.p. 90-91°C.

3.10.4 Phenylhydrazone These are obtained by reaction of aldehydes or ketones with phenylhydrazine hydrochloride and sodium acetate. C6H5NHNH2.HCl + CH3COONa ——→ C6H5NHNH2 + CH3COOH + NaCl Phenylhydrazine

Mechanism

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Procedure Reagents Cyclohexanone

0.5 mL

Phenylhydrazine hydrochloride

1g

Sodium acetate

1.6 g

Dissolve 0.5 mL of cyclohexanone in minimum amount of alcohol (2-3 mL) in a round bottomed flask. Add to it a solution of phenylhydrazine hydrochloride and sodium acetate in minimum amount of water. Reflux the mixture for 15-20 minutes in a water bath. Cool, filter the separated phenylhydrazone, dry and recrystallise from dilute alcohol. Yield 0.5 g, m.p. 77°C.

3.11 DIAZONIUM SALTS AND THEIR REACTIONS 3.11.1 Diazotisation Aromatic primary amines react with nitrous acid (produced in situ by reaction between sodium nitrite and a mineral acid like hydrochloric or sulphuric acid) at low temperature to yield diazonium salts. This reaction is called Diazo reaction. Since it was discovered by Johan Peter Griess (1858) it is referred to as Griess diazo reaction. The procedure of formation of diazonium salts is referred to as diazotisation. Since both nitrous acid and the formed diazonium salt are unstable, hence the reaction is carried out at a low temperature. It is important to mention here that aliphatic primary amines also form diazonium salts on reaction with nitrous acid, but these are unstable and produce the corresponding alcohol by reacting with water used in the diazotisation reaction.

Mechanism Various steps involved in the diazotisation reaction are as shown: It is believed that nitrous anhydride (dinitrogentrioxide) is first formed from nitrous acid   2 HNO2    N2O3 + H2O

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+

N2O3 thus formed dissociates to give NO (nitrosonium ion), which is the attacking electrophile

+

+

C6 H5 − N = N − OH + H → C6 H5 − N = N + H 2 O +

−

+

−

C6 H5 − N = N + Cl → C6 H5 − N = NCl or N2O3 reacts with aniline

rest of the mechanism is same as given above. The diazonium salts serve as important intermediates as these react with a number of other compounds to yield a large variety of commercially useful organic compounds. Diazonium salts can undergo the following types of reactions • Substitution: The diazo group can be replaced by a number of other groups like –H, –OH, –Cl, –Br, –I, –F, –SH, –CN etc., in such reactions, the –N=N– group is eliminated as nitrogen gas.

• Reduction: Nitrogen is not eliminated during the reduction reaction of the diazonium salt. The diazo group can be reduced to –NHNH2 or –NH2.

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• Coupling: Coupling of diazonium salts with other activated aromatic compounds gives azo compounds which contain –N=N– group. Coupling with phenols occurs under mild alkaline conditions

Coupling with amines is best carried out under mild acidic conditions

Mechanism • Coupling with phenol

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• Coupling with an amine

3.11.2 Substitution of Diazo Group by Hydrogen The amino group in an aromatic primary amine can be replaced by hydrogen (deamination) via the diazonium salt by boiling it with ethyl alcohol or better with hypophosphorus acid in presence of Cu1+ salts as catalyst. Mechanism +

+

C6 H5 N = N + Cu → C6 H5• + N 2 + Cu 2+

+

−

C6 H5 N = NC l + C2 H5 OH → C6 H 6 + CH3CHO + N 2 + HCl The deamination with ethanol proceeds via the formation of the intermediate aryl cation, which can react with ethanol in two ways • Substitution by hydrogen to form the aromatic hydrocarbon. • Substitution by the ethoxy group to form the corresponding ether. C6H5N2+Cl– + C2H5OH

→

C6H6 + CH3CHO + N2 + HCl

C6H5N2

→

C6 H5+ + N 2

C6H5 + C2H5OH

→

C6H5OC2H5 + H+

+

+

The ratio of hydrocarbon: ether generally depending on the nature of the substituents in the aromatic nucleus and the alcohol. In general, the presence of electron withdrawing groups or atoms (e.g., halogen or nitro) favours replacement of diazo group by hydrogen. Substitution of diazo group by hydrogen using hypophosphorus acid produces only the hydrocarbon.

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1, 3, 5-Tribromobenzene It is prepared from aniline in two steps: • Preparation of 2, 4, 6-Tribromoaniline • Preparation of 1, 3, 5-Tribromobenzene

2, 4, 6-Tribromoaniline See section 3.16.3. 1, 3, 5-Tribromobenzene It is obtained by deamination of 2,4,6- Tribromoaniline Procedure Reagents 2,4,6- Tribromoaniline 2g Absolute alcohol 8 mL Benzene 2 mL Conc. sulphuric acid 0.5 mL Sodium nitrite 0.8 g Take 2,4,6-Tribromoaniline (2 g) (prepared as given in section 3.16.3) in a round bottomed flask and dissolve it in absolute alcohol (8 mL) and benzene (2 mL) by heating on a water bath. Fix a reflux condenser on the flask and add concentrated sulphuric acid (0.5 mL) dropwise with stirring. Heat the reaction mixture in a water bath to obtain a clear solution. Remove the flask from the water bath, detatch the reflux condenser and add powdered sodium nitrite (0.4g) with shaking and warming in a water bath. Allow the reaction to subside and add the second lot (O.4g) of sodium nitrite and again allow the reaction to subside. Fix a reflux condenser on the flask and heat the reaction mixture on a boiling water bath for 1 hour, shaking at frequent intervals. Now allow the reaction mixture to cool to room temperature and then cool in an ice bath. Filter the separated product, wash with a small quantity of alcohol (2-3 mL) and then with water (5 × 5 mL) till the sodium sulphate is removed. Recrystallise tribromobenzene from alcohol. Yield 1.2g, m.p. 122°C.

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m-Nitrophenol

Substitution of Diazo Group by -OH group The diazo group can be replaced by an –OH group to produce phenols. This method is used to produce some phenols which generally cannot be made by other methods of synthesis of phenols. m-Nitrophenol It is obtained by diazotisation of m-nitroaniline followed by replacement of diazo group by –OH group.

Procedure Reagents m-Nitroaniline 2g Conc. sulphuric acid 9.2 mL Sodium nitrite 1.0 g Make a homogeneous paste of m-Nitroaniline (2g) by adding it to a cold solution of concentrated sulphuric acid (2.9 mL) and water (7.5 mL) and stirring the resultant mixture. Cool the paste to 0-5°C in an ice bath. Add sodium nitrite solution (l g in ~3 mL water) dropwise during 5 minutes to the stirred paste until it gives a permanent colour with potassium iodide-starch paper. Stir the mixture for 5 minutes more and allow it to stand for another 5 minutes. Carefully prepare a solution of concentrated sulphuric acid (6.3 mL) in water (6.3 mL) in a round bottomed flask (100 mL). Heat it to boiling and add the diazonium salt solution prepared above in small lots (1 mL each time) during 10 minutes. In case some solid has formed during diazotization, add it also to the solution of sulphuric acid. When all the diazonium salt has been added, boil the mixture for 5 minutes and pour it into a beaker (250 mL). Cool the solution and stir it vigorously. Filter the separated m-nitrophenol, wash with a small quantity of cold water and recrystallise from dilute hydrochloric acid (1:1). Yield 1.3 g, m.p. 96°C.

3.11.4

Iodobenzene

Substitution of diazo group by iodine Aryl iodides can be obtained by reaction of diazonium salts with aqueous solution of potassium or sodium iodide.

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Mechanism + −

C6 H5 NH 2 + HNO 2 + HCl → C6 H5 N = NCl + H 2 O +

C6 H5 N = N → C6 H5+ + N 2 ↑ lodobenzene

C6 H5+ + KI → C6 H5 I + K +

Procedure Reagents Aniline 2g Dil. hydrochloric acid (1:1) ll mL Sodium nitrite 1.6 g Potassium iodide 3.6 g Dissolve aniline (2g) in dilute hydrochloric acid (l1 mL, 1:1) in a conical flask and cool it in an ice bath to 0-5°C. Separately prepare a solution of sodium nitrite (1.6 g in 8 mL water) and add it drop wise with shaking (using a separating funnel) to the cooled solution of aniline keeping the temperature under 7-8°C. Add a small amount of ice to the reaction mixture to maintain low temperature. Continue the addition of sodium nitrite solution till a drop of the reaction mixture gives a blue colour with potassium iodide-starch paper. The reaction mixture should be acidic throughout the diazotisation. To the diazonium salt solution obtained above add a solution of potassium iodide (3.6 g in 5-6 mL water) with shaking, nitrogen is evolved during the addition. Allow the reaction mixture to stand for one hour and then heat it in a boiling water bath for 20-30 minutes till the evolution of nitrogen ceases. The iodobenzene separates as dark heavy oil. Steam distil the mixture, the distillate contains iodobenzene. Cool the steam distillate and extract with ether (3 × 10 mL), wash the combined ether extract with water and then with dilute sodium hydroxide solution till it is free from iodine. Dry the ether extract over calcium chloride, remove ether and distil the residue to obtain iodobenzene. Yield 3.6g, b.p. 188–189°C. Note: Solvent ether is highly inflamable.

3.11.5 p-Chlorotoluene and p-Bromotoluence Substitution of diazo group by Cl or Br (Sandmeyer’s reaction) The –N=N– group of diazonium salts can be replaced by chlorine or bromine by treatment with CuCl+HCl or CuBr+HBr respectively. +

−

+

−

CuCl+ HCl

C6 H5 N = N C l → C6 H5 Cl + N 2 CuBr + HBr

C6 H5 N = N C l  → C6 H5 Br + N 2 This reaction is known as Sandmeyer’s reaction. Mechanism

+

CuCl + HCl → CuCl−2 + H +

C6 H5 N = N + CuCl2− → C6 H5 + N 2 + CuCl2 C6 H5 + CuCl2 → C6 H5Cl + CuCl Chlorobenzene

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(a) p-Chlorotoluene p-Chlorotoluene is obtained from p-toluidine by diazotisation followed by replacement of diazo group by chlorine using freshly prepared CuCl in presence of HCl.

Procedure Reagents p-Toluidine Conc. hydrochloric acid Sodium nitrite Cuprous chloride Sodium sulphite Sodium chloride

7.5 g 45 + 20 mL 5.5 g obtained from 22.5g CuSO4⋅5H2O 5.5 g 6.2 g

Cuprous Chloride Copper sulphate is reduced by sodium sulphite in presence of sodium chloride to cuprous chloride as per the following reaction 2CuSO4 + 4NaCl + Na2SO3 + H2O → 2CuCl + 2HCl + 3Na2SO4 Prepare a solution of copper sulphate pentahydrate (22.2g) and pure sodium chloride (6.2g) in water (80 mL) (warm if necessary) in a beaker. Add a solution of sodium sulphite (5.5 g) in water (60 mL) to the hot solution (obtained above) in 5 minutes with constant shaking. Cool the reaction mixture in an ice-bath. A white precipitate of cuprous chloride is obtained. Allow the precipitate to settle and decant the supernatant liquid. Wash the precipitate of cuprous chloride by decantation with water containing little sulphurous acid (to prevent the oxidation of cuprous chloride). Dissolve the cuprous chloride in concentrated hydrochloric acid (45 mL). Keep the solution in an ice-salt mixture and use it immediately after preparation. Diazotisation Dissolve p-Toluidine (7.5g) in concentrated hydrochloric acid (20 mL) and water (20 mL) in a conical flask. Cool the solution to 0°C in an ice-salt mixture, p-toluidine hydrochloride separates out as a solid. Add to it a solution of sodium nitrite (5.5g) in water (12.5 mL) dropwise with continuous stirring maintaining the temperature of the reaction mixture between 0-5°C. When the addition is complete, hydrochloride of p-toluidine dissolves and a soluble diazonium salt is formed (check the completion of diazotization by potassium iodide-starch paper which gives blue colour with the test solution). Sandmeyer reaction Add the cold diazonium chloride solution (obtained above) slowly with stirring to the cold cuprous chloride solution. The mixture becomes viscous, due to the separation of an addition product (CH3-C6H4-

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N=N+Cl–CuCl). Allow the mixture to attain room temperature while shaking it occasionally. During this the temperature of the reaction mixture is ~15°C. p-Chlorotoluene separates as an oil. Allow the reaction to complete by warming it to 60°C with shaking. Steam distil the reaction mixture until no more oily drops come in the distillate. Take the steam distillate in a separating funnel, drain the aqueous layer, wash the organic layer containing p-chlorotoluene with sodium hydroxide solution (2 × 5 mL to remove p-cresol which might have formed), water (2 × 10 mL) and dry over calcium chloride. Distil and collect fraction between 160-164°C. Yield 6 g, b.p. 162°C. (b) p-Bromotoluene It is obtained by diazotisation of p-toluidine with sodium nitrite and sulphuric acid, treatment of the resultant aryl diazonium sulphate with cuprous bromide in presence of excess of hydrobromic acid.

Procedure Reagents p-Toluidine Conc. sulphuric acid Sodium nitrite Cuprous bromide Sodium bisulphite Sodium bromide Hydrobromic acid

7.7 g 7.7 mL 5.25 g obtained from 16.9 g CuSO4. 5H2O. 4.5 g 9.9 g 11 mL (48%)

Cuprous bromide Dissolve copper sulphate pentahydrate (16.9g) and sodium bromide dihydrate (9.9g) in water (55 mL) in a beaker. Heat it in a water bath and to the hot solution add a solution of sodium bisulphite (4.5g) in water (45 mL) during 5 minutes. If the blue colour is not discharged completely, add a little more sodium bisulphite solution. Cuprous bromide separates out as a solid. Cool the mixture in an ice-bath and decant the supernatant liquid. Wash the precipitate by decantation with water (2 × 10 mL) containing a little sulphurous acid to prevent oxidation of cuprous bromide. Dissolve the cuprous bromide in hydrobromic acid (48% HBr, ll mL) and cool the solution in an ice-salt mixture. Use it immediately for Sandmeyer reaction. Diazotisation Suspend p-Toluidine (7.7g) in water (60 mL) in a round bottomed flask and add concentrated sulphuric acid (7.7 mL) cautiously. Warm the mixture until p-toluidine dissolves. Cool the solution to 0-5°C in an

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ice-salt bath and add a solution of sodium nitrite (5.25 g) in water (10 mL) dropwise. Stir the solution during diazotization maintaining the temperature of the reaction mixture between 0-5°C. Check for the completion of the reaction with potassium iodide-starch paper (a blue colour is obtained). Sandmeyer reaction Take the cuprous bromide solution (prepared above) in a round bottomed flask. Heat the cuprous bromide solution to boiling and add the diazonium salt solution in small lots over a period of 10-15 minutes through a separatory funnel. Steam distil during the addition till no more oily drops distil over. Basify the steam distillate by adding 20% NaOH solution. The crude p-bromotoluene separates out. Take the entire solution in a separating funnel. Drain out the aqueous layer. Wash the organic layer with warm water (2 × 5 mL), sodium hydroxide solution (5 mL) and finally with water (2 × 10 mL). Dry the remaining organic layer in a conical flask over anhydrous magnesium sulphate, filter and distil, collect the fraction between 182-184°C. Yield 6g, b.p. 183° C; m.p. 26°C.

3.11.6

Phenylhydrazine

Reduction of diazonium salts to phenylhydrazine Diazonium salts on reduction with sodium metabisulphite produce the corresponding hydrazines.

Mechanism SO32– obtained from sodium metabisulphite acts as the reducing agent.

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Phenylhydrazine Procedure Reagents Aniline Conc. hydrochloric acid Sodium nitrite Sodium metabisulphite

10 g 25.2 mL 7.4 g 28.5 g

Diazotisation Dissolve freshly distilled aniline (l0g) in concentrated hydrochloric acid (25.2 mL) and water (20 mL) and cool the solution to ~0°C in a conical flask. Stir the aniline solution vigorously and add to it an ice cold solution (~0°C) of sodium nitrite (7.4g) in water (20 mL) dropwise through a separatory funnel during a period of 10-15 minutes. The reaction mixture should contain a slight excess of nitrous acid (test with potassium iodide-starch paper) at the completion of addition of sodium nitrite. Reduction Add sodium metabisulphite (28.5g) to a solution of sodium hydroxide (21 g) in water (210 mL) in a conical flask. Stir the mixture (manually or using a mechanical stirrier) until all the sodium metabisulphite has dissolved, cool the solution to 25°C, and a drop or two of phenolphthalein indicator. The solution aquires pink colour. Add more sodium metabisulphite with stirring until the pink colour just disappears. Add little more sodium metabisulphite (l-2g). Cool the resulting solution to ~5°C in an ice bath, add crushed ice (10 g). Now add rapidly the ice cold solution of diazonium chloride obtained above and stir the mixture vigorously. The mixture acquires a bright orange-red colour. Heat it in a water bath to 60-70°C and keep at this temperature for 1 hour until the colour becomes quite dark. Now acidify the solution to litmus with concentrated hydrochloric acid (~ 8-11 mL). Heat in a boiling water bath until the colour becomes lighter (4-5 hours). Filter the solution to remove any solid product and to the hot clear filtrate add with stirring, concentrated hydrochloric acid (110 mL) and cool the solution to 0°C in freezing mixture. Filter the separated phenyl hydrazine hydrochloride and wash with dilute hydrochloric acid (1:3). Recrystallise from water and charcolise if the solution is dark coloured. Cool the clear solution to 0°C, pure white crystals of phenylhydrazine hydrochloride are formed. The yield is 85-90%. Note: Phenyl hydrazine can be obtained by adding sodium hydroxide solution (25%, ~20 mL) to the phenylhydrazine hydrochloride. The free base is extracted with toluene (2 × 20 mL). The toluene extract is dried (sodium hydroxide pellets) and toluene distilled in vacuo. Phenylhydrazine is collected at 137-138°C/8 mm, b.p. is 243.5°C.

3.11.7 Coupling Reactions of Diazonium Salts Reaction of diazonium salts with aromatic compounds containing strong electron donating groups like aromatic amines (primary, secondary or tertiary), phenols etc. to give azo compounds having the general formula ArN=NAr is known as coupling reaction. Mechanism The coupling reaction is an electrophilic substitution in which the electrophile is the diazonium cation (Ar–N=N+). Coupling reaction occurs at the position of highest electron density of the amine or the phenol. (See page 277)

3.11.8 Phenylazo-b-naphthol (Sudan 1) It is obtained by coupling reaction between benzene diazonium chloride with b-naphthol under alkaline conditions

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Procedure Reagents Aniline 2 mL Conc. hydrochloric acid 6.5 mL Sodium nitrite 1.5 g b-Naphthol 3.2 g Sodium hydroxide solution 10%, 20 mL Add a solution of sodium nitrite (1.5 g in 7 mL water) dropwise with stirring to a cooled (0-5°C) solution of aniline (2 mL) in dilute hydrochloric acid (6.5 mL Conc. hydrochloric acid and 6.5 mL water). Maintain the temperature of the reaction mixture between 0-5°C during addition. When the addition is complete, keep the solution for 5 minutes more with occasional stirring to complete the diazotisation and then pour it slowly with stirring into a cold solution of b-naphthol (3.2 g) in sodium hydroxide solution (10%, 20 mL). Allow the reaction mixture to stand in an ice-bath for 30 minutes with occasional stirring. Filter the separated phenylazo-b-naphthol, wash with water and recrystallise from alcohol. Yield 4.2g, m.p. 131°C. Note: Sudan 1 is an intense orange-red coloured solid and is used for colouring waxes, oils, solvents and polishes etc.

3.11.9 4-(4′-Nitrobenzeneazo)-l-naphthol (Magneson II) It is obtained by coupling between diazotised p-nitroaniline and l-naphthol

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Procedure Reagents p-Nitroaniline

1.5 g

Conc. hydrochloric acid

3.8 mL

Sodium nitrite

1g

1-Naphthol

1.4 g

Sodium hydroxide solution

30%, 8 mL

Dissolve p-Nitroaniline (1.5 g) in dilute hydrochloric acid (3.8 mL concentrated acid and 3.8 mL water). Cool the solution to 0-5°C in an ice bath and add to it a cold solution of sodium nitrite (1 g) in water (3 mL) dropwise with stirring. At the completion of addition, the reaction mixture should contain slight excess of nitrous acid which is tested with potassium iodide-starch paper. Keep the reaction mixture in ice bath and maintain its temperature below 5-6°C. Now add a cooled solution of 1-naphthol (1.4 g) in sodium hydroxide solution (30%, 8 mL) slowly with stirring keeping the temperature of the reaction mixture low. Add concentrated hydrochloric acid slowly till the mixture is acidic to congo red. The colour changes from violet to dark brown. Filter the separated product, wash with water (until free from acid) and dry on filter paper. Yield 2.3 g. Notes: Magneson I is prepared by following the procedure described above by coupling the diazonium salt of p-nitroaniline with resorcinol.

3.11.10 Methyl Orange Methyl orange is an azo dye though it is used only as an indicator in acid-base titrations and changes colour between pH 3.1-4.4 being red below pH 3.1 and yellow above pH 4.4. It is prepared by coupling between diazotised sulphanilic acid with N, N-dimethylaniline.

Procedure Reagents Sulphanilic acid Sodium carbonate (anhyd.) Sodium nitrite

2g 0.6 g 0.8 g

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Conc. hydrochloric acid Dimethylaniline Acetic acid (glacial) Sodium hydroxide solution Sodium chloride

Preparation of Organic Compounds

2 mL 1.25 mL 1 mL 20%, 8 mL 5g

Diazotisation Dissolve sulphanilic acid (2g) in a solution of sodium carbonate (prepared by dissolving 0.6g of it in 20 mL water) in a conical flask. Warm to obtain a clear solution. Cool the clear solution to ~10°C and add slowly with stirring a cooled solution of sodium nitrite (0.8g in 2-3 mL water). Pour the resulting solution slowly with stirring into a beaker containing concentrated hydrochloric acid (2 mL) and crushed ice (10-15g) keeping the temperature between 0-5°C. After 15 minutes check for the presence of free nitrous acid with potassium iodide-starch paper. Fine crystals of diazobenzene sulphonate separate out. Coupling Prepare a solution of dimethylaniline (1.25 mL) in glacial acetic acid (1 mL) in a test tube and add it with vigorous stirring to the above diazotised mixture. Allow to stand for 15 minutes. The red or acid form of methyl orange separates out gradually. Add sodium hydroxide solution (20%, 8 mL) slowly with stirring to the reaction mixture. The mixture becomes orange due to the separation of sodium salt of methyl orange as fine particles. Heat the mixture almost to the boiling point when most of the dye dissolves. Add sodium chloride (~5g) to assist the subsequent separation of methyl orange. Heat the mixture to 80-90°C to dissolve the salt completely. First allow the reaction mixture to cool to room temperature (15 minutes) and then cool in ice-water. Filter the separated methyl orange, wash with a little saturated sodium chloride solution and crystallise from hot water. Yield 2.6g. The product does not have a well defined melting point since it is the sodium salt.

3.11.11 Orange II (b -Naphthol orange)

Procedure Reagents Sulphanilic acid

2g

Sodium carbonate (anhyd.)

0.6 g

Sodium nitrite

0.8 g

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Conc. hydrochloric acid

2 mL

b-Naphthol

1.45g

Sodium hydroxide solution

10%, 8 mL

Sodium chloride

5g

Diazotisation Carry out the diazotisation of sulphanilc acid (2g) as described under preparation of methyl orange. Coupling To a cold solution (5°C) of b-naphthol (1.45g) in sodium hydroxide (10%, 8 mL), add with stirring the cooled and well mixed suspension of diazolised sulphanilic acid. Coupling takes place and a solid product separates. Keep for 10-15 minutes in an ice bath and then heat till the solid dissolves. Now add sodium chloride (5 g). Heat the mixture to get a clear solution, allow it to cool in air for 1 hour and finally cool in ice bath. Filter the separated dye, wash with a little saturated sodium chloride solution. For crystallisation- dissolve the solid in minimum amount of hot water, cool to ~70°C and add spirit (~ twice the volume of water) and cool to obtain the dye. Yield 7g.

3.11.12 Methyl Red

Procedure Reagents Anthranilic acid

2g

Dimethylaniline

2.8 mL

Sodium nitrite

1.2 g

Conc. hydrochloric acid

4 mL

Sodium acetate

2.2 g

Sodium hydroxide

20%, 1.6 mL

Acetic acid

10%, 3 mL

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Diazotisation Dissolve anthranilic acid (2g) by heating with dilute hydrochloric acid (6 mL, 1:3 v/v) in a conical flask. Cool the solution to 2-3°C in an ice-bath and add more hydrochloric acid (2.4 mL) with stirring. To the cold acidic solution (2-3°C), add dropwise with stirring, a solution of sodium nitrite (1.2g in 3-4 mL water). Coupling Add dimethylaniline (2.8 mL) to the diazotised solution. Keep the resultant solution for 10-15 minutes at a temperature less than 5°C. Dissolve sodium acetate (2.2g) in water (~3-4 mL). Add the sodium acetate solution (1.5 mL) to the diazotised reaction mixture and allow it to stand in an ice-bath for 1 hour with occasional stirring. Add the remaining sodium acetate solution with stirring to the cooled mixture (ice-bath) and leave for 30 minutes with occasional stirring. Allow the reaction mixture to attain room temperature and add sodium hydroxide solution (1.6 mL, 20%) with stirring till a distinct odour of dimethylaniline is observed. Leave the mixture for 1 hour. Filter the separated product, wash with a little water and then with dilute acetic acid (10 per cent, 3 mL) to remove dimethyl aniline; and finally with water. Boil the crude product with methanol (15 mL) for 10-15 minutes (water bath) filter and wash with cold methanol (3 mL), the yield is 1.8 g (48%); m.p. is 181-182°C. Notes: • Methyl red is used as an acid-base indicator. It is red under pH 4, yellow at pH 6.2 and above. • Methyl red is used in methyl red test. This test is performed in biochemistry laboratories to know whether a bacteria is producing stable acids by mechanism of mixed acid fermentation of glucose.

3.12

SYNTHESIS OF OTHER DYES

Compounds which are coloured and can be used to colour a surface like fabric, paper, leather, etc. are known as dyes. A dye should have following properties. • It should have a colour • The dye should be able to fix permanently on the surface to be dyed • The colour should be stable and permanent. etc.

3.12.1 Fluorescein It belongs to a class of xanthene dyes and is obtained by condensation reaction between resorcinol and phthalic anhydride in presence of anhydrous zinc chloride or conc. sulphuric acid. Fluorescein is colourless in acidic medium and gives a green fluorescence in alkaline medium.

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Procedure Reagents Phthalic anhydride

1g

Resorcinol

1.5g

Zinc chloride (Anhyd.)

0.5g

Grind under dry conditions, phthalic anhydride (1 g) and dry resorcinol (1.5 g) together in a pestle mortar, transfer to a dry round bottomed flask and heat it to 180°C in an oil bath. Add in small portions, freshly fused powdered zinc chloride (0.5 g) during ~5 minutes (stirring the mixture with a thermometer) to the mixture of phthalic anhydride and resorcinol. At the end, the mass becomes so viscous that the stirring is not possible. Allow the oil bath to cool to 90°C and add dilute hydrochloric acid (10 mL water and 1 mL concentrated hydrochloric acid) to the reaction mixture. Now heat the oil bath again till the water in the round bottomed flask starts boiling, continuously stir with a glass rod. Cool, filter the solid product, grind it with water in a mortar and filter again. Yield 1.4g. Notes: • Carry out the reaction under anhydrous conditions. • Dry resorcinol by keeping it in an oven 120-130°C till it melts. Cool and powder in a motor for use. • Fused zinc chloride - Zinc chloride is hygroscopic in nature and absorbs moisture from air. Heat (1g) in a porcelein dish over a low flame using a wire guage. It melts after some time. Heat and stir the molten mass with a glass rod. When white fumes start coming, stop heating. Cool and powder the fused zinc chloride in a dry mortar and store in a desiccator or a tightly stoppered bottle. • Fluorescein is used in ophthalmology, optometry and fluorescein angiography etc.

3.12.2 Eosin (Tetrabromo fluorescein) Eosin is a xanthene dye and is obtained by bromination of fluorescein

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Procedure Reagents Fluorescein 1g Bromine 0.73 mL Suspend fluorescein (lg) in alcohol (5 mL) in a conical flask. Add bromine (0.73 mL) dropwise during 5-10 minutes to stirred suspension of fluorescein. When half of the bromine has been added, all the solid material dissolves, since the dibromo derivative of fluorescein that is formed at this stage is soluble in alcohol. Further addition of bromine gives the tetrabromo derivative, which separates out since it is sparingly soluble in alcohol. Allow the mixture to stand for one hour. Filter the separated orange coloured eosin, wash with cold alcohol and dry (100°C). Yield 1.2 g. Note: Eosin is used as a stain in histology. Cytoplasm is stained pink-orange and nucleus is stained blue or purple.

3.12.3 Malachite Green It belongs to a class of triphenyl methane dyes. It is obtained by condensation reaction between benzaldehyde and N,N-dimethyl aniline.

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Procedure Reagents Dimethylaniline 2g Benzaldehyde 0.8 g Conc. hydrochloric acid 2 mL Lead peroxide 0.83 g Sodium sulphate 2g Glacial acetic acid 2 mL Reflux a mixture of dimethylaniline (2g), benzaldehyde (0.8g) and concentrated hydrochloric acid (2 mL) for 12-14 hours in a round bottomed flask fitted with a reflux condenser. Basify the reaction mixture with sodium hydroxide solution and steam distil to remove the unreacted benzaldehyde and dimethylaniline. Pour the residue into water (100 mL) in a beaker. Filter the separated leucobase, wash with water and dry. Recrystallise from hot ethyl alcohol. The leuco base (obtained above) is converted into malachite green by oxidation. Dissolve the base by gentle warming in dilute hydrochloric acid containing 0.3g. hydrogen chloride (ascertained by titration with alkali) and glacial acetic acid (2 mL). Dilute the solution obtained above with water to 50 mL and cool by adding crushed ice. Make a paste of lead peroxide (0.83g) in water and add it to the solution of leuco base, shake the resultant mixture for about 5 minutes. Now add a saturated solution of sodium sulphate (2g) and filter the precipitate of lead sulphate and chloride that separates out. Heat the clear filtrate to boiling, add sodium hydroxide and filter the free base that is formed, wash with water and dry and convert the free base to either its oxalate (heating with a solution of calculated amount of oxalic acid and then cooling) or to its hydrochloride (by treating with hydrochloric acid). Notes: • Malachite green is used as a dye for silk, leather and paper. • Malachite green is used as a biological stain. • It is also used in forensic science.

3.12.4 Crystal Violet (Hexamethylpararosaniline) A triphenyl methane dye, is obtained by condensation reaction between N,N-dimethyl aniline and Michler’s ketone in presence of phosphorus oxychloride.

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Procedure Reagents Dimethylaniline 2g Michlers ketone 0.83 g Phosphorus oxychloride 0.9 mL Sodium chloride 10 g Heat a mixture of dimethylaniline (2g), Michler’s ketone (0.83g) and phosphorus oxychloride (0.9 mL) in a round bottomed flask in a boiling water bath for 4 hours. After 4 hours, add water (20 mL) to the reaction mixture, basify the resultant blue solution with a solution of caustic soda and then steam distil to remove unchanged dimethylaniline. Cool the remaining reaction mixture and filter the separated product. Dissolve the solid by boiling it with water (50 mL) containing a little concentrated hydrochloric acid (0.5 mL). Add some more water containing little hydrochloric acid if the solid does not dissolve completely. Now add sodium chloride (l0 g) in small lots, with stirring to precipitate the dye from the solution. Filter and recrystallise from water. Crystal violet separates as fine crystals having a bronze-like lusture. Filter and dry. Yield is 1.3 g. Notes: • Crystal violet is used in histological stains. • It has antifungal and antibacterial properties. It was also used as an antiseptic.

3.13

ESTERIFICATION

Reaction between a carboxylic acid and an alcohol or a phenol yields an ester and hence the reaction is known as esterification. The reaction is • slow • reversible • requires concentrated acid like sulphuric or hydrochloric acid as catalyst to speed it up. C6 H5 COOH + CH3OH Benzoic acid

Methyl alcohol

+

H   → C6 H5 COOCH3 + H 2 O ← 

Methyl benzoate

H+

  → CH3COOC6 H5 + H 2 O CH3COOH + C6 H5 OH ←  Acetic acid

Phenol

Phenylacetate

Since this reaction is reversible, the forward reaction is increased by • using one of the reactant in excess • or removing one of the product (normally water) as it is formed during the reaction. Esters are a very important class of organic compounds. • The triesters of glycerol with long chain fatty acids (commonly known as triglycerides) are naturally occurring organic compounds. Many of these are used as edible oils. Besides being a source of energy, many of these contain carboxylic acids (like w-3 and w-6 and other poly unsaturated acids) which perform important functions in our body. • Because of their pleasant smell, esters find extensive use in perfumery. • Esters are prepared as derivatives for identification of carbohydrates (acetates), alcohols and phenols (p- and 3,5-dinitrobenzoates) etc.

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• Esterification of a –CQOH, –OH etc. is used for protection of these groups. For example the – COOH group of an amino acid is prohibited from combining with the –NH2 of the other amino acid by its esterification.

Mechanism The mechanism of esterification in presence of an acid catalyst is

3.13.1 Ethyl Benzoate Ethyl benzoate is prepared by reaction between benzoic acid and ethyl alcohol in presence of concentrated sulphuric acid. Ethyl alcohol is used in excess to favour the forward reaction. Ethyl benzoate is used in some fragrances and as artificial fruit flavour. H+

  → C6 H5COOC 2 H5 + H 2 O C6 H5 COOH + C2 H5 OH ←  Ethyl benzoate

Procedure Reagents Benzoic acid

l0 g

Ethyl alcohol (absolute)

21 mL

Conc. sulphuric acid

1 mL

Add concentrated sulphuric acid (1 mL) dropwise cautiously to a mixture of benzoic acid (10 g), absolute ethyl alcohol (21 mL) in a dry round bottomed flask. Reflux the reaction mixture for 4 hours in a hot water bath. Distil excess alcohol in a steam-bath. Pour the residue in a beaker, add cold water (20 mL) and then a concentrated solution of sodium carbonate in small lots (~5 mL each time) till there is no more effervescence. Extract the ester with ether (carefully). Separate the ether layer from the aqueous layer, dry it by keeping with anhydrous potassium carbonate in a dry conical flask. After 15-20 minutes, filter potassium carbonate and first remove ether from filtrate and then distil the residue to collect ethyl benzoate. Yield l0g, b.p. 213°C.

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Notes: • Solvent ether is highly inflammable, handle it very carefully and away from flame and hot objects. • Concentrated sulphuric acid is very corrosive and hence should handled with great care.

3.13.2 Methyl Salicylate Methyl salicylate, also known as oil of winter green occurs naturally in plants, specially winter greens and is synthetically prepare by methylation of salicylic acid with methyl alcohol in presence of concentrated sulphuric acid. It is an important member of the salicylate group of compounds, many of which find medical applications. Methyl salicylate is used in liniments, as fragrance and in foods and beverages.

Procedure Reagents Salicylic acid 8.5 g Methyl alcohol (absolute) 15 mL Conc. sulphuric acid 2.5 mL Add concentrated sulphuric acid (2.5 mL) dropwise cautiously to a mixture of salicylic acid (8.5g), absolute methyl alcohol (15 mL) in a dry round bottomed flask and reflux the reaction mixture for 4 hours in a hot water bath. Work up the reaction mixture as described in the preparation of ethyl benzoate. Yeild 8g, b.p. 223-225°C. Note: As above in the preparation of ethyl benzoate.

3.13.3 Ethyl p-aminobenzoate Also known as Benzocaine, is prepared by saturating absolute ethyl alcohol with hydrogen chloride gas and then heating it with p-aminobenzoic acid. It is used as a local anesthetic, in ear pain relievers and cough drops.

Procedure Reagents p-Aminobenzoic acid Ethyl alcohol (absolute) Hydrogen chloride gas Sodium carbonate

2g 14 mL

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Pass dry hydrogen chloride gas into absolute ethyl alcohol (14 mL) in a dry conical flask (in a fume chamber) till the alcohol is completely saturated. Transfer the alcohol to a dry round bottomed flask and add to it p-Aminobenzoic acid (2 g). Reflux the mixture in a hot water bath for 2 hours. Pour the hot solution into excess water (50 mL) in a beaker and add a saturated solution of sodium carbonate in small lots until the solution is neutral to litmus. Filter the precipitated ethyl p-aminobenzoate, dry, weigh and recrystalise ~ l00 mg from rectified spirit. Yield 1.6g, m.p. 91°C.

3.13.4

Diethyl Adipate

Esterification of adipic acid with ethyl alcohol gives diethyl adipate.

Procedure Reagents Adipic acid 10 g Ethyl alcohol (absolute) 13 mL Dry benzene 25 mL Conc. sulphuric acid 2.2 mL To a mixture of adipic acid (10 g), absolute alcohol (13 mL) and dry benzene (25 mL) in a dry round bottomed flask, add concentrated sulphuric acid (2.2 mL) dropwise with continuous swirling. Reflux the reaction mixture in a hot water bath for 5-6 hours, cool and then pour carefully into water (50 mL) in a beaker with stirring. Separate the benzene layer and extract the aqueous layer with benzene (20 mL). Dry the combined benzene extract over anhydrous calcium chloride, filter and distil benzene in a hot water bath (preferably in a rotary evaporator) and then diethyl adipate under reduced pressure yield 13 g, b.p. 134-135°C/17 mm. Notes: • Benzene is known to be a human carcinogen. The reaction and work up procedure must be performed in a fume chamber. • In case of any skin or eye contact, wash the effected area with plenty of fresh water. • If inhaled, immediately move out in fresh air and take oxygen, if there is a difficulty in breathing. • If swallowed, seek medical advice immediately and do not induce vomiting. • as given in the preparation of ethyl benzoate.

3.14

FRIEDEL CRAFTS REACTION

Substitution of the hydrogen of an aromatic or a hetero aromatic compound by an alkyl (alkylation) or an acyl group (acylation) is known as the friedel crafts reaction. This reaction is extensively used in organic transformations, in industry and R&D sector.The reaction takes place in presence of a catalyst. Anhydrous aluminium chloride was traditionally used as a catalyst and was required in stotiometric amounts. Some examples of FC reaction are given below:

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Friedel crafts alkylation

Friedel crafts acylation

Alkyl halides, alcohols and alkenes have been used as alkylating agents. Acyl chlorides and anhydrides are used for acylation reaction. Lewis acids like anhydrous AlCl3, SnCl4 etc. are used as catalyst. The main drawbacks of the Friedel Craft’s reaction are: • Aromatic compounds with deactivating groups cannot undergo the FC reaction. For example, nitrobenzene, acetophenone and pyridine do not undergo this reaction. Presence of an appropriate activating group is required to carry out FC reaction in such cases. • Aromatic compounds with activating groups like aniline, phenol and pyrrole etc. do not undergo the FC reaction. A deactivating group, if present in such compounds helps them to undergo reaction. • Stoichiometric or super stoichiometric and not catalytic amount of the catalyst are required and which are not recovered after the reaction. • The reaction during alkylation does not stop at mono alkylation but normally proceeds to give poly alkylated product. • Dealkylation is also known to occur during FC alkylation.

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• Long n-alkyl chains cannot be introduced by direct alkylation as these tend to undergo isomerisation. For example, an attempt to prepare n-propyl benzene by reaction of benzene with n-propyl halide leads to the formation of isopropyl benzene. n-Propyl benzene is prepared in two steps:

Many modifications have been developed in the recent past, special interest has been in the direction of • Use of less hazardous alkylating agents. • Use of catalytic amounts of catalyst. General Mechanism of Friedel Craft’s Reaction Alkylation

Acylation

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3.14.1 Tert. Butylbenzene Tert. Butyl benzene is prepared by the Friedel Craft reaction between benzene and t-butyl chloride in presence of anhydrous aluminium chloride as a catalyst.

Procedure Reagents Benzene (dry) 20 mL Aluminium chloride (anhyd.) 2.4 g Tert. butyl chloride 2.6 mL Take anhydrous aluminium chloride (2.4 g) and pure dry benzene (20 mL) in a three necked round bottomed flask (50 mL) fitted with a stirrer, a reflux condenser and a dry dropping funnel. Keep the flask in an ice bath and add tert. butyl chloride (2.6 mL), dropwise (dropping funnel) during 15-20 minutes maintaining the temperature of the reaction mixture between 0-5°C during the addition. Continue stirring the reaction mixture for ~15 minutes. Now add slowly with stirring, crushed ice (~5g) to the above reaction mixture and then some ice cold water (~5 mL). Steam distil the reaction mixture till no more oily drops distil over. Take the steam distillate in a separating funnel and allow the upper organic layer to separate. Drain out the lower aqueous layer. Wash the upper organic layer with water, drain out water and transfer the organic layer into a dry boiling tube containing anhydrous magnesium sulphate (~lg). Filter the organic layer after 15-20 minutes and remove unreacted benzene by distillation (water bath) and then distil (b.p. 168.5°C) the residue. Yield: 2 g.

3.14.2 Triphenylmethane Triphenly methane is prepared by reaction between benzene and chloroform in presence of anhydrous aluminium chloride

Procedure Reagents Benzene (dry) 20 mL Chloroform (dry) 2.1 mL Aluminium chloride (anhyd.) 2.8 g Take dry benzene (20 mL) and dry chloroform (2.1 mL) in a round bottomed flask fitted with a reflux condenser. Add slowly anhydrous aluminium chloride (2.8 g) in small lots (0.5 g each) to a continuously stirred reaction mixture. A brisk reaction sets in and the liquid boils with the evolution of hydrogen

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chloride gas. After the reaction has subsided, reflux it on a sand bath for 25-30 minutes, cool and pour into ice cold water (~25 mL) in a beaker. Take the reaction mixture in a separating funnel and allow the two layers to separate. Remove the lower aqueous layer, transfer the organic layer in a boiling tube and dry over anhydrous calcium chloride. Filter after ~10 minutes, distil excess benzene on a steam bath and finally distil the residue under reduced pressure 190-215°C at 10 mm. Yield 2.4 g m.p. 92°C.

3.14.3

Acylation

Acetophenone Acetophenone is prepared by Friedel crafts acylation of benzene with acetyl chloride in presence of anhydrous aluminium chloride

Procedure Reagents Benzene (dry)

20 mL

Aluminium chloride (anhyd.)

7.5 g

Acetic anhydride

7 mL

Take dry benzene (20 mL) and finely powdered anhydrous aluminium chloride (7.5 g) in a double necked round bottomed flask (100 mL) fitted with a reflux condenser and a dry dropping funnel containing acetic anhydride (7 mL). Add acetic anhydride slowly dropwise with continuous shaking to the reaction mixture. Cool if the reaction mixture becomes hot. Now heat the reaction mixture in a hot water bath at 60°C for 40-45 minutes with occasional shaking. Cool and pour the reaction mixture with stirring into a beaker containing concentrated hydrochloric acid (20 mL) and ice-cold water (~50 mL). Take the reaction mixture in a separating funnel and allow the two layers to separate. Remove the lower aqueous layer, transfer the organic layer in a dry boiling tube and dry over anhydrous magnesium sulphate. Filter after ~10 minutes, distil excess benzene on a steam bath and finally distil the residue to obtain acetophenone. Yield 5 g, b.p. 202°C.

3.15

GRIGNARD REACTION

Grignard reagents (RMgX or ArMgX) are extensively used as intermediate products in the synthesis of a large variety of organic compounds. The –MgX can be replaced with a variety of groups to produce carboxylic acids, aldehydes, ketones, alcohols (l°, 2°and 3°), esters, thiols, nitriles, nitro compounds, amines etc. some examples of these reactions are given below:

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Triphenylcarbinol (Triphenylmethanol)

It is obtained by the reaction of phenyl magnesium bromide with methyl benzoate. Phenyl magnesium bromide is first prepared from bromobenzene and magnesium turnings in dry ether under anhydrous conditions.

Mechanism

3.16 HaLogenation Advanced Experimental Organic Chemistry

Procedure Reagents Bromobenzene Methyl benzoate Magnesium turning Iodine Dry ether

303

7.5 mL 4 mL 1.69 g Few crystals 50 mL

Preparation of phenyl magnesium bromide Take magnesium turnings (1.69 g), dry ether (8 mL) and a crystal of iodine in a dry double necked round bottomed flask (250 mL capacity) fitted with a reflux condenser and a dropping funnel. Dissolve bromo benzene (7.5 mL) in dry ether (30 mL) and pour into the dropping funnel attached to the round bottomed flask. Add the solution of bromobenzene (5 mL) to the reaction mixture and shake the flask. In case the reaction does not commence in 2-3 minutes (indicated by the disappearance of iodine colour, begning of the reaction is appearance of turbidity and boiling of ether). When the reaction has started, add the remaining bromobenzene solution dropwise at such a rate that the ether boils gently without any external heating. Reflux the reaction mixture gently in a water bath for 15 minutes and cool it in an ice bath. Preparation of triphenylcarbinol To the cold solution of phenyl magnesium bromide, add a solution of methyl benzoate (4 mL) in dry ether (10 mL) dropwise using a dropping funnel. Shake the reaction mixture occasionally. Adjust the rate of addition so that the mixture refluxes gently. Reflux the reaction mixture for 10 minutes in a water bath, cool and pour the reaction mixture cautiously into crushed ice (80 g) and dilute sulphuric acid (10%, 40 mL) in a beaker and stir for 10-15 minutes so that the magnesium complex is decomposed and free triphenylcarbinol is obtained. Separate the ether layer and extract the aqueous solution with ether (20 × 25 mL). Wash the combined ether extract with dilute hydrochloric acid (10%, 2 × 15 mL), water (20 mL) and finally with water (15 mL) containing sodium bisulphite (0.5 g) (to remove the iodine used to start the reaction). Distil the remaining ether in a water bath and add water (40 mL) to the residue. Steam distil the mixture until no oil containing unchanged reactants and diphenyl ketone (obtained as byproduct) pass over. Cool the residual solution and filter the separated product, dry and weigh. Crystallise from methylated spirit. Yield 5g, m.p. 162°C. Notes: • Ether is highly inflammable, never work with ether in a laboratory where other students are using a burner. • Perform the experiment under anhydrous conditions > Use apparatus which has been dried in an oven > Use ether which has been dried over sodium wire. > Distil ether carefully.

3.16

HALOGENATION

Reaction of a compound with a halogen is known as halogenations. Bromination, Chlorination, lodination and fluorination are four types of halogenation reactions. This section deals with the bromination reactions which can be easily performed in a chemistry laboratory.

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3.16.1 Bromination Two types of reactions basically take place on bromination • Addition of bromine • Substitution of an atom (like hydrogen) or a group (like –N=N– ⋅ –OH) by bromine.

The conditions used for bromination obviously depend on the reactivity of the compound to be brominated. For example, benzene undergoes substitution with bromine vapour only in presence of a catalyst. Whereas activated aromatic compounds (compounds containing strong electron donating groups like —NH2 or —OH etc.) react under mild experimental conditions i.e. with a dilute solution of bromine at room temperature. Some common brominating agents are: Bromine (vapour or solution) with or without a catalyst Potassium bromate and potassium bromide in presence of an acid N-Bromosuccinimide (NBS). Some common substitution bromination reactions of aromatic compounds which can be easily carried out in the laboratory are discussed in the present section. Substitution of hydrogen of an aromatic compound with bromine is an example of electrophilic substitution reaction in which bromine reacts as an electrophile. The detailed reaction mechanism is

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Bromine is a highly corrosive chemical, having strong adverse effect on skin, eyes and respiratory system and should therefore be handled very carefully. In case of an accidental spill or inhalation, medical help should is sought immediately. Some first aid measures are given on page 102.

3.16.2 p-Bromoacetanilide p-Bromoacetanilide is prepared by the reaction of acetanilide with a solution of bromine in glacial acetic acid. The –NHCOCH3 group is an electron donating group, directing the incoming electrophile (bromine) to o- and p- position. However, the reaction occurs mainly at the p-position, only a very small amount of o-substituted product is obtained because of the following • The –NHCOCH3 in acetanilide is a moderate activating group. The electron pair on nitrogen is not only donated to the benzene ring of acetanilide but is also shared with the electron withdrawing –COCH 3 group attached to it in the side chain.

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• The steric affect of –NHCOCH3 group hinders the reaction at the o-position. Procedure Reagents Acetanilide 1g Bromine 0.4 mL Acetic acid (glacial) 9 mL Take acetanilide (1g) in a dry conical flask (100 mL) and dissolve it in glacial acetic acid (5 mL). Take from burette, bromine (0.4 mL) in a dry test tube and dissolve it in glacial acetic acid (4 mL). Add carefully the solution of bromine drop wise with a dropper to the solution of acetanilide and shake well after each addition. Initially the colour of the bromine is discharged fast but this slows down near the completion of reaction. The reaction mixture attains an orange yellow colour at completion due the presence of a slight excess of unreacted bromine. Allow the reaction mixture to stand for 10-15 minutes with occasional shaking. If it is yellow in colour, proceed as described otherwise add a few more drops of bromine solution and allow the reaction mixture to stand for another 5-10 minutes. Pour the reaction mixture with continuous stirring into a beaker containing ice cold water (~30 mL), a white solid is formed and the supernatant solution is yellow in colour. Break gently using a glass rod if lumps of solid are formed. Add a solution of sodium bisulphite with stirring till the supernatant solution becomes colourless. Filter the solid, wash with cold water, dry ,weigh and crystallise (~100 mg) from dilute alcohol. Yield 1.1 g, m.p 167°C. Notes: • Bromine is highly injurious to health and skin. Handle it very carefully. • Perform the experiment and the work up in a fume chamber to avoid exposure to bromine fumes. • The completion of reaction is indicated by a permanent orange-yellow colour of the reaction mixture. Br2 + NaHSO3 + H2O ——→ NaHSO4 + 2HBr NaHSO3 + HBr ——→ NaBr + SO2 + H2O

3.16.3 2,4,6-Tribromoaniline It is obtained by the reaction of aniline with a solution of bromine. The amino group in aniline is a powerful activating group due to electron donating resonance effect of nitrogen and so increases the electron density at both ortho and para positions. Thus, the electrophillic substitution like bromination yields 2,4,6-tribromoaniline.

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Procedure Reagents Aniline

1 mL

Acetic acid (glacial)

8 mL

Bromine

1.6 mL

Add carefully a solution of bromine (1.6 mL) in glacial acetic acid (4 mL) drop wise with a dropper to a solution of redistilled aniline (1 mL) in glacial acetic acid (4 mL) in a conical flask (100 mL). Shake the reaction mixture after each addition and cool in ice cold water if necessary. A yellow coloured pasty mass is obtained. Keep the reaction mixture for five minutes and then pour with stirring into ice-cold water (~50 mL) in a beaker. A white solid with yellow supernatant liquid is obtained. Treat the mixture with a solution of sodium bisulphite till the yellow colour is removed. Filter the solid, wash it with water, dry, weigh and crystallise from rectified spirit. Yield 2g, m.p 120°C. Notes: • As in the preparation of p-bromoacetanilide. • Use freshly distilled aniline. 3.16.4 2,4,6-Tribromophenol It is prepared by reaction of phenol with a solution of bromine.

The –OH group in phenol activates the o- and p-positions in the benzene ring towards electrophilic substitution reaction.

Procedure Reagents Phenol

1g

Bromine

1.4 mL

Take freshly distilled phenol (1g) in water (~50 mL) in conical flask (250 mL). Add bromine (1.4 mL) dropwise with continuous stirring/shaking till the solution aquires a permanent

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yellow colour. Allow the reaction to stand for 5-10 minutes. Pour it into ice cold water (~ 50 mL) in a beaker, a white solid is obtained. Add a solution of sodium bisulphite to remove yellow colour of the supernatant solution. Filter the solid, wash it with water (twice with ~5 mL water each time), dry and weigh. Recrystallise l00 mg of it from dilute alcohol. Yeild 1.1g, m.p. 95°C. Note: See above in the preparation of p-bromoacetanilide.

3.16.5 l-Bromo-2-Naphthol It is prepared by bromination of 2-naphthol with a solution of bromine in acetic acid.

Procedure Reagents b-Naphthol (2-Naphthol) 1g Bromine

0.35 mL

Acetic acid (glacial)

6 mL

Add a solution of bromine (0.35 mL) in glacial acetic acid (1.5 mL) dropwise to a stirred solution of b-naphthol (1g) in glacial acetic acid (4.5 mL). During the addition, keep the reaction mixture in a cold water bath. A white solid separates, the completion of reaction is indicated by a permanent yellow colour in the supernatant liquid. Pour the reaction mixture with stirring into ice cold water (~50 mL) in a beaker and add a solution of sodium bisulphite till the yellow colour is removed. Filter the solid, wash with water, dry and weigh. Recrystallise ~100 mg from alcohol. Yield 1.4g, m.p 80°C.

3.17

HOUBEN HOESCH AND NENCKI REACTION

Synthesis of acyl substituted phenols/ethers by reaction of phenols or phenolic ethers, with organic nitriles, in the presence of lewis acid catalyst (e.g., zinc chloride, aluminium chloride etc.) or hydrogen chloride gas is known as Houben-Hoesch reaction or simply Hoesch reaction. A ketimine salt is first formed which is then hydrolysed to the corresponding acyl phenol or the acyl substituted ether. If glacial acetic is used in place of acetonitrile and treated with a phenol like resorcinol in presence of anhydrous zinc chloride, the reaction is known as Nencki reaction.

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3.17.1 2,4-Dihydroxy Acetophenone (b-Resacetophenone) b-Resacetophenone can be prepared by two reactions (1) Nencki reaction (2) Houben Hoesch reaction. Nencki reaction In this reaction resorcinol is treated under anhydrous conditions with glacial acetic acid in presence of a lewis acid like zinc chloride to give b-resacetophenone δ+ δ− CH3COOH + ZnCl2 → CH3CO ... ZnCl2 OH Mechanism

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Procedure Reagents Resorcinol

2g

Acetic acid (glacial)

4 mL

Zinc chloride (fused)

3g

Fuse zinc chloride (page 91) (3.5 g) in a china dish, powder and dissolve it (3 g) in a china dish in glacial acetic acid (4 mL) by heating on a sand bath. Continuously stir with a glass rod. Add resorcinol (2 g) slowly with stirring, maintaining the temperature at 140°C. Heat the mixture at 150°C for 15 minutes. Allow the reaction mixture to cool to room temperature. Add dil. Hydrochloric acid (1:1, l0 mL). Cool the acidified solution to ~5°C in an ice bath. Filter the resacetophenone, wash with dilute hydrochloric acid (1:3, 5 mL) and recrystallise from hot water. Yield 2.3 g, m.p 142-144°C. Houben Hoesch reaction

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Mechanism

Procedure Reagents Resorcinol (dry)

2g

Acetonitrile

1.4 mL

Zinc chloride (fused)

1g

Ether (dry)

10 mL

Cool in an ice bath, a solution of dry resorcinol, (2g) acetonitrile (1.4 mL), fused and powdered zinc chloride (1g) in dry ether (l0 mL) in a conical flask with a side arm (attach a calcium chloride tube to the side arm). Pass dry hydrogen chloride gas produced by reaction of sodium chloride with sulphuric acid for one hour into the solution of resorcinol. Stopper the flask and allow the reaction mixture to stand overnight. Decant ether and wash the separated imine salt with dry ether (2 × 10 mL). Reflux the salt with water (10 mL) for about 30 minutes in a water bath. Cool the solution and filter the separated 2,4-dihydroxyacetophenone. Dry, weigh and crystallise from boiling water. Yield 2.2g, m.p. is 145°C.

3.17.2

2,4,6- Trihydroxy Acetophenone (Phloracetophenone)

Phloroglucinol (dry)

2g

Acetonitrile

1.5 mL

Ether (Anhyd.)

8 mL

Zinc chloride (fused)

0.4g

Take dry phloroglucinol (2g), anhydrous ether (8 mL), freshly distilled acetonitrile (1.5 mL) and freshly fused and powdered zinc chloride (0.4g) in a Buchner flask (100 mL) fitted with a wide gas inlet

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tube and a calcium chloride tube to the side arm. Cool the mixture to 0°C in ice-salt mixture. Pass a rapid steam of dry hydrogen chloride gas through the solution for 2 hours with occasional shaking. Allow the flask to stand in an ice-chest for 24 hours and pass dry hydrogen chloride gas again for two hours into the cooled solution (0°C). Leave the flask in a refrigerator for 24 hours. Decant ether and wash the solid ketimine hydrochloride with dry ether (2 × 5 mL). Reflux the solid product with water (100 mL) in a round bottomed flask on a flame for 2 hours. Cool the solution and filter the separated product. Recrystallise from boiling water (10 mL). Yield 2.2g, m.p. 218-219°C.

3.18 3.18.1

HYDROLYSIS Hydrolysis of Ethyl Benzoate (Preparation of Benzoic Acid)

Ethyl benzoate on hydrolysis gives benzoic acid and ethyl alcohol and methyl benzoate gives benzoic acid and methyl alcohol. The hydrolysis can be carried out with water in presence of an acid or a base as a catalyst. The base catalysed reaction also known as saponification is a better experimental procedure as it is not reversible.

Mechanism Base catalysed hydrolysis

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Acid catalysed hydrolysis

Procedure Reagents Ethyl benzoate 1 mL Sodium hydroxide 15 mL, 10% Reflux a mixture of ethyl benzoate (1 mL) and freshly prepared sodium hydroxide (15 mL,10%) in a round bottomed flask (50 mL) on a sand bath or a wire gauze for 25-30 minutes. After the completion of reaction (indicated by the disappearance of the insoluble layer of ethyl benzoate), pour the reaction mixture into a beaker. Add some ice (15-20 g) and acidify the alkaline solution with dilute hydrochloric acid. Filter benzoic acid, wash with water, dry and recrystallise from hot water. Yield 0.6 g, m.p. 121°C. Note: Use a funnel to add sodium hydroxide to the flask.

3.18.2 Hydrolysis of Benzamide (Preparation of Benzoic Acid) Benzamide on base catalysed hydrolysis gives sodium benzoate, which on acidification gives benzoic acid. C6H5CONH2 + NaOH ——→ C6H5COONa + NH3 Mechanism

Procedure Reagents Benzamide Sodium hydroxide solution (freshly prepared)

1g 15 mL, 10%

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Take benzamide (1g) in a small round bottomed flask (50 mL) and add freshly prepared sodium hydroxide solution (15 mL, 10%) through a filter funnel (so that the alkali does not touch the joint of the flask). Now reflux the solution on a wire gauze or a sand bath for 25-30 minutes to complete the hydrolysis. Completion of the reaction is indicated by the absence of smell of ammonia. Allow the reaction mixture to cool. Transfer the reaction mixture to a beaker, add some ice (10-15g) and neutralise with dilute hydrochloric acid till acidic to litmus. Filter the white precipitate of benzoic acid, wash with water, dry and recrystallise from hot water. Yeild 0.5g, m.p. 121°C.

3.18.3 Hydrolysis of p-nitroacetanilide (Preparation of p-nitroaniline) p-Nitroacetanilide is hydrolysed under acid catalysed reaction conditions. The products of reaction are p-nitroaniline and acetic acid.

Mechanism

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Procedure Reagents p-Nitroacetanilide 2g Sulphuric acid 40 mL, 20% Take p-nitroacetanilide (2g) and sulphuric acid (40 mL, 20%) in a round bottomed flask. Reflux the mixture on a sand bath or a wire gauze for two hours. Cool the reaction mixture and filter any unhydrolysed p-nitro acetanilide or extract with ether (10 mL). Transfer the clear acidic solution containing p-nitro aniline into a beaker, add ice (~20g) and basify the solution with dilute sodium hydroxide solution. p-Nitroaniline separates as a yellow coloured solid, filter wash with water (2 × 5 mL), dry and crystallise from dilute alcohol. Yield 1.2 g, m.p.146°C.

3.18.4 Hydrolysis of p-bromoacetanilide p-Bromoaniline is obtained by hydrolysis of p-bromoacetanilide with hydrochloric/sulphuric acid

Mechanism Same as given above for p-nitroacetanilide Procedure Reagents p-Bromoacetanilide 2g Hydrochoric acid 40 mL, 5 M or H2So4 40 mL, 20% Follow the procedure given above for hydrolysis of p-nitroacetanilide. Yield 1.5g, m.p 66°C.

3.19

METHYLATION

It is a common term used for introduction of methyl group in a molecule. Various types of methylations are • C-methylation (Friedel craft reaction) • O-methylation (methylation of alcohols, phenols and carboxylic acids etc.) • S-methylation (methylation of thiols) • N-methylation (methylation of amines) etc. Some of the common methylating agents are • Dimethyl sulphate • Methyl iodide • Methanol • Diazomethane

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An example of C-methylation is the Friedel craft conversion of benzene to toluene

Methylation of –COOH group produces the methyl ester, methanol or diazomethane is used as methylating agent. +

H   → RCOOCH3 + H 2 O RCOOH + CH3OH ← 

RCOOH + CH 2 N 2 →

RCOOCH3 + N 2

Methylation of an alcohol or a phenol produces methyl ethers, the most commonly employed methylating agent being dimethyl sulphate in aqueous alkaline medium. Dimethyl sulphate is also used under anhydrous conditions in presence of anhydrous sodium carbonate for methylation of phenols.

Methyl iodide is generally used for methylation of amines

3.19.1 b-Methyl naphthyl ether (Nerolin) It is prepared by methylation of b-naphthol with dimethyl sulphate in presence of sodium hydroxide.

The mechanism of the reaction is:

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Procedure Reagents b–Naphthol 1g Sodium hydroxide 0.3 g Dimethyl sulphate 0.66 mL Add dimethyl sulphate (0.66 mL) dropwise to a stirred and cooled solution (10-15°C) of b-naphthol (1g) in sodium hydroxide solution (0.3 g in 5 mL water) in a conical flask. After the addition is over, heat the mixture for one hour at 70-80°C in a water bath and then cool it in ice cold water. Filter the separated product, wash first with dilute sodium hydroxide solution (10%) and then with water. Dry, weigh and crystallised from alcohol. Yield 0.9 g, m.p. 72°C. Notes: • Dimethyl sulphate is carcinogenic, mutagenic, poisonous and a corrosive substance. • On slow hydrolysis it produces sulphuric acid, which is highly corrosive in nature. • Dimethyl sulphate causes eye and skin burn, immediately wash for 10-15 minutes with fresh water and seek medical help. • It is fatal if inhaled, immediate medical help is a must. • Sodium hydroxide used in the reaction neutrilises the sulphuric acid and also keeps unreacted phenol in solution.

3.19.2 1,3-Dimethoxybenzene (Resorcinol dimethylether) It is prepared by methylation of resorcinol with dimethyl sulphate in presence of sodium hydroxide

Procedure Reagents Resorcinol 10 g Aqueous sodium hydroxide 35 mL 40% Sodium hydrosulphite 4g Dimethyl sulphate 22 mL Alcohol 30 mL Add freshly prepared solution of sodium hydroxide (35 mL, 40%) dropwise during ~2 hours to a well stirred mixture of resorcinol (10 g), sodium hydrosulphite (4g), dimethyl sulphate (22 mL) and alcohol (30 mL) at 18-20°C in a conical flask. Add ~100g of crushed ice to the reaction mixture and leave it overnight. Extract with ether (2 × 30 mL). Separate the ether layer and wash it with water. Remove

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water, transfer the ether in a conical flask and dry by keeping over anhydrous sodium sulphate. Filter the ether extract and remove ether in a rotary evaporator. Distil the residue, dimethyl resorcinol distils at 216-218°C. Yield 10 mL. Notes: See under nerolin.

3.19.3 1,2,4-Trimethoxybenzene (hydroxy hydroquinone trimethylether) It is obtained by methylation of 1,2,4- triacetoxy benzene with dimethyl sulphate in presence of sodium hydroxide

Procedure Reagents 1,2,4-Triacetoxybenzene 12.5 g Methanol 50 mL Dimethyl sulphate 45 mL Sodium hydroxide solution 37.5 g in 100 mL water Add sodium hydroxide solution (37.5 g in 100 mL water) dropwise during a period of 2 hours to cooled and well stirred solution of 1,2,4-triacetoxybenzene (12.5 g) in methanol (50 mL) and dimethyl sulphate (45 mL) in a conical flask, maintaining the temperature of the reaction mixture between 25~30°C throughout the addition. After the addition is complete, stir the reaction mixture for 30 minutes and leave it overnight at room temperature. Extract with ether twice, using ~50 mL ether each time. Wash the ether extract with water, dry over anhydrous sodium sulphate. Filter and remove ether in a rotary evaporator. Distil the residue, 1,2,4-trimethoxybenzene at 247°C. Yield 6 g. Notes: As given under the preparation of nerolin.

3.20

NITRATION

Substitution of a hydrogen of an aliphatic or an aromatic compound by a nitro group is known as nitration. It is an example of electrophillic substitution reaction, in which the effective elecrtophile is the nitronium ion (NO2+). The nitronium ion is produced in situ by reaction between nitric acid and sulphuric acid (the most commonly employed nitrating agent). Benzene, for example is converted to nitrobenzene by heating with a mixture of concentrated nitric acid concentrated sulphuric acid.

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The nitronium ion is produced by the following reaction HNO3 + 2H2SO4 ——→ NO2+ + 2HSO4- + H3O+ Sulphuric acid being a stronger acid donates a proton to nitric acid to produce the nitronium ion

The nitronium ion thus produced reacts as follows with benzene

If the nitration is continued further, m-dinitrobenzene is produced as the major product accompanied by a small amount of o- and p-dinitrobenzene.

The –NO2 group present in nitrobenzene is an electron withdrawing group. It withdraws electrons from the benzene nucleus and thus reduces the electron density at o- and p- positions making these least reactive to attack by an electrophile. The electron density at m- position is highest of all the unsubstituted positions in nitrobenzene and hence the incoming nitro group is directed to m-position.

Also the intermediate carbocation produced by attack at m-position is more stable as compared to that resulting from reaction at o- and p- position. Attack at o-, m- and p-position gives a carbocation, each having three rasonating structure. Out of the three structures, II (for p-attack) and V1 (for attack at o-position) are unstable because they have electron deficient carbon bearing an electron withdrawing group

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Further nitration of m-dinitrobenzene to 1,3,5- trinitrobenzene becomes difficult because the two deactivating nitro groups reduce the activity of benzene ring for further electrophillic substitution. Thus nitration of aromatic compounds which are already substituted by a deactivating group (like –COR, NO2 etc.) occurs at m- position and in compounds with groups like (–CH3, R, Cl, Br etc.) at o- and p-position.

For example in chlorobenzene, nitration at o- and p-position gives a more stable intermediate than that produced from nitration at m-position. Compounds containing powerful activating groups like amines and phenols cannot be nitrated in presence of concentrated nitric acid as these undergo oxidation rather than nitration. Special conditions are employed in such cases. Some of the common nitrating agents are: Conc./fuming nitric acid and sulphuric acid Conc. nitric acid and glacial acetic acid Conc. nitric acid Dil. Nitric acid Sodium nitrate and sulphuric acid etc. Aromatic hydrocarbons or those aromatic compounds containing deactivating groups like –NO2, – COCH3 etc. require stronger nitrating conditions and compounds which contain activating groups require milder conditions for nitration. Nitration is an important reaction as the nitro group introduced by nitration can be reduced to a variety of other functional groups like –NH2, –NHOH, –N=N–, –NH–NH– etc. Selective reduction of one nitro group in a polynitro substituted compound widens the applications of nitro compounds. Many of these are used as intermediates in the synthesis of commercially important compounds like dyes, drugs and polymers etc.

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As many of nitro compounds are sharp melting crystalline compounds, these are prepared as derivatives for characterisation of compounds like hydrocarbons, ethers etc.

3.20.1 Nitrobenzene It is prepared by nitration of benzene with a mixture of nitric and sulphuric acid.

Procedure Reagent Benzene

10 mL

Conc. nitric acid

11.5 mL

Conc. sulphuric acid

12 mL

Add cautiously concentrated sulphuric acid (12 mL) dropwise with shaking to concentrated nitric acid (11.5 mL) contained in a dry round bottomed flask (50 mL). Cool the reaction mixture by immersing the flask in a cold water bath if it becomes hot. Now add benzene (10 mL) slowly, 1 mL each time with stirring to the mixture of acids maintaining the temperature of the reaction mixture between 50-55°C. After the addition is over, fix up a condenser and heat the reaction mixture for 10-15 minutes at 60°C in a hot water bath, shaking the flask occasionally. Allow the reaction mixture to cool to room temperature and then pour the contents into cold water (100 mL) in a beaker with stirring. Take in a separating funnel and allow the two layers to separate. Take the lower layer of nitrobenzene and wash twice with cold water (15 mL, each time) in a separating funnel. Finally drain the lower layer and dry over anhydrous calcium chloride. Filter and distil. Nitrobenzene is collected at 210°C. Yield 11 g. Notes: • Maintain the reaction temperature as recommended in the procedure and do not allow to rise otherwise some disubstituted nitro compounds may be formed along with nitrobenzene. • Perform the experiment in a fume chamber. • Move out of laboratory and breath in fresh air if there is burning sensation in throat or respiratory system due to inhalation of fumes. • Wash with plenty of fresh water in case of burning in eyes or skin due to exposure to the acid fumes. • Do not allow the vapours to escape in the laboratory. Nitrobenzene is a skin poison and if it comes in contact with skin, it should be immediately washed with rectified spirit, followed by soap and water.

3.20.2 1-Nitronaphthalene (a-Nitronaphthalene) It is obtained by nitration of naphthalene with concentrated nitric acid and concentrated sulphuric acid. In naphthalene nitration can occur at 1- or 2-position. The nitro group enters position-1, since the intermediate carbonium ion formed is more resonance stabilised as compared to that obtained when the electrophile attacks at 2-position of naphthalene.

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Nitration at position-1 and resonating structures of the intermediate carbonium ion are

Nitration at position-2 and resonating structures of the intermediate carbonium ion are

Procedure Reagents Naphthalene 2g Conc. nitric acid 4 mL Conc. sulphuric acid 4 mL Add cautiously concentrated sulphuric acid (4 mL) dropwise with shaking to concentrated nitric acid (4 mL) in a small, dry round bottomed flask. Add finelly powdered naphthalene (2g) in small lots and shake well after each addition maintaining the temperature of the reaction mixture between 40-50°C. When all the naphthalene has been added, heat the reaction mixture at 55-60°C for 30-40 minutes until the odour of naphthalene has disappeared. Pour the reaction mixture into cold water (~30 mL) in a beaker. Decant the aqueous layer and boil the solid formed with water (20 mL) for 15 minutes and decant the aqueous layer again. The solid is a mixture of 1-nitronaphthalene and unreacted naphthalene from which naphthalene is removed by steam distillation. Filter the remaining solid product, wash with water, dry and weigh and crystallise ~100 mg from dilute alcohol. Yield 2g, m.p. is 61°C. Notes: As in the preparation of nitrobenzene above.

3.20.3 p-Bromonitrobenzene It is obtained by nitration of bromobenzene with concentrated nitric acid in presence of concentrated sulphuric acid, a small amount of o-isomer is also formed.

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Bromine in bromobenzene exerts a strong deactivating electron withdrawing effect on the aromatic nucleus (by the-1 effect) so the nitro group should attack the m- position. But the reaction occurs at o- and p- position because the intermediate resulting from o-, p- attack is more stable than that obtained from reaction at m- position and hence the incoming nitronium ion attacks the o- and p-positions.

Rasonating structures of I

Resonating structures of II

Procedure Reagents Bromobenzene 1.5 mL Conc. nitric acid 2.8 mL Conc. sulphuric acid 2.8 mL Add bromobenzene (1.5 mL) dropwise with shaking to the cautiously prepared nitrating mixture (2.8 mL each of nitric and sulphuric acid each) in a dry round bottomed flask. Fix up a condenser and heat the reaction mixture in a boiling water bath for 30 minutes. Cool the reaction mixture and pour it, with stirring, into cold water (30 mL) in a beaker. Filter the separated product, wash with cold water (~10 mL), dry, weigh and crystallise ~100 mg from rectified spirit. Yield 1.6 g, m.p. 124-125°C.

3.20.4 o- and p-Nitrophenol The –OH group present in phenol is a strong o-, p- directing group.

Resonating structures of phenol showing high electron density at o- and p- position.

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Phenol on nitration under mild nitrating conditions gives a mixture of o- and p-nitrophenol. Concentrated nitric acid cannot be used for nitration as phenol undergoes oxidation rather than nitration with it. Nitration of phenol with dilute nitric acid gives a mixture of o- and p-nitrophenols via the corresponding intermediate, nitroso substituted compounds which are oxidised to the nitro phenols in situ. The yield of o-nitrophenol is higher when nitration is carried out with a mixture of sodium nitrate and dil. sulphuric acid. The o- isomer is steam volatile where as the p- is not, this forms the basis of their separation. Procedure Reagents Phenol 3.1 g Conc. sulphuric acid 4.5 mL Sodium nitrate 5g Add sodium nitrate (5 g) to dilute sulphuric acid (made by cautiously adding 4.5 mL of concentrated sulphuric acid to 13 mL water) in a conical flask. Cool the solution in an ice bath and add to it dropwise, a suspension of phenol in water (3.1g phenol in 5 mL water) with shaking such that the temperature does not rise above 20°C. After the addition of phenol is complete, allow the reaction mixture to stand for one hour with frequent shaking. A solid product is formed. Decant the supernatant liquid and wash the residue with water (2-3 times) to remove any residual acid. Steam distil the residue until no more o-nitrophenol passes over. Filter the steam distillate after cooling and dry. Yield 1g, m.p. 46°C. Filter the residual solid (p-nitrophenol) in the flask, dry and weigh. Yeild 0.7g, m.p. 112°C. Notes: If the residual solid in the flask is coloured. Filter it and boil with dilute hydrochloric acid (2%) and decolourizing carbon (1g) for 5-10 minutes and filter while hot. Cool the clear solution and filter the p-nitrophenol. Dry and determine the yield.

3.20.5 p-Nitroacetanilide It is obtained by nitration of acetanilide with a mixture of conc. nitric and sulphuric acid. A small amount of o-nitroacetanilide is also formed which is separated by crystallisation of the crude product when only p-nitroacetanilide crystallises out.

Procedure Reagents Acetanilide

2.1 g

Acetic acid (glacial)

2.5 mL

Conc. sulphuric acid

4.9 mL

Conc. nitric acid

0.9 mL

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Take acetanilide (2.1g) in glacial acetic acid (2.5 mL) in a dry conical flask. Add conc. sulphuric acid (4.l mL) drop wise with shaking. Cool the mixture in an ice bath and add to it slowly a mixture of acids (causiously prepared by mixing concentrated nitric acid (0.9 mL) and concentrated sulphuric acid (0.8 mL) keeping the temperature of the reaction mixture below 10°C. Take out the reaction mixture from ice bath and allow it to stand at room temperature for 30 minutes. Pour the reaction mixture onto crushed ice (about , 20-30 g) in a beaker, allow ice to melt, filter the solid, wash with water, dry and weigh. Recrystallise ~100 mg of solid from alcohol. Yield 1.2 g, m.p. 213-14°C.

3.20.6

m-Dinitrobenzene

It is prepared by nitration of nitrobenzene with a mixture of fuming nitric acid and concentrated sulphuric acid.

Procedure Reagents Nitrobenzene

1 mL

Fuming nitric acid

2 mL (sp.gr.7.5)

Conc. sulphuric acid

2 mL

Prepare the mixed acids by carefully adding conc. sulphuric acid (2 mL) to fuming nitric acid (2 mL) in a small dry round bottomed flask. Add nitrobenzene (1 mL) slowly, drop wise, with stirring to the mixture of acids. Fix up an air condenser and heat with occasional shaking on a boiling water bath for 15-20 minutes. Allow the reaction mixture to cool to room temperature and then pour it cautiously into ice-cold water (30 mL) in a beaker. A yellowish coloured solid separates, filter, wash it with some water, dry and weigh. Recrystallise ~100 mg from alcohol. Yield l.l g, m.p. 89-90°C.

3.20.7

Picric Acid (2,4,6-Trinitrophenol)

Phenol is first converted to phenol sulphonic acid by reaction with concentrated sulphuric acid, which on reaction with concentrated nitric acid gives picric acid.

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Procedure Reagents Phenol

1g

Conc. sulphuric acid

2.5 mL

Conc. nitric acid

2.5 mL

Heat phenol (1g) with concentrated sulphuric acid (2.5 mL) in small, dry round bottomed flask for 5-7minutes in a boiling water until a clear solution of phenol sulphonic acid is obtained. Cool the solution in ice cold water and add concentrated nitric acid (2.5 mL) in small lots while shaking the mixture. A vigorous reaction sets in and red fumes of oxides of nitrogen are evolved and the reaction mixture becomes deep red in colour. After the reaction has subsided, fix up a condenser and heat it for 1.5 hours in a boiling water bath with occasional shaking. Cool the reaction mixture and pour it into cold water (20 mL). Filter the picric acid, wash with some cold water, dry and weigh. Recrystallise ~100mg from dilute alcohol. Yield 1.1g, m.p. 121-122°C Notes: Preserve picric acid under water as it is explosive when dry.

3.21

OSAZONE

Polyhydroxy aldehydes or ketones or those compounds which yield these on hydrolysis are known sugars or saccaharides or carbohydrates, (see page 29 for more information). Sugars in which the carbonyl

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group is present free or is liberated by opening of the cyclic structure (see page 29-30) react with phenyl hydrazine to give phenylhydrazone, which may react with more phenyl hydrazine to form osazone

3.21.1 Osazone of Glucose Glucose on reaction with phenylhydrazine or phenylhydrazine hydrochloride and sodium acetate gives osazone

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Mechanism

Procedure Reagents Glucose 1g Phenylhydrazine hydrochloride 2g Sodium acetate 3.0g Boil water (~200 mL) in a beaker (400 mL). Take glucose (1g), phenylhydrazine hydrochloride (2g) and sodium acetate (3.0g) in a boiling tube. Add water (7-8 mL) and dissolve the mixture by stirring with a glass rod. Filter into another clean boiling tube if a tarry mass separates. Loosely cork the tube and immerse it in boiling water in the beaker. Note the time. Heat the reaction mixture in the hot water bath for 30 minutes. Note the time when a yellow precipitate first appears. After heating in a boiling water bath, take out the tube. Allow to cool to room temperature and then cool in ice-cold water. Filter the yellow coloured osazone that separates out. Recrystallise from alcohol. Yield 0.5g, m.p 205°C (d). Notes: • Weigh all the reactants on a chemical balance otherwise a mixture of products is obtained, the osazone is not pure and hence melts over a range of temperature.

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• The osazone decomposes on melting and darkens on heating so darkening during heating should not be taken as the melting point.

3.22

OXIDATION REACTIONS

Oxidation is defined as loss of electrons, loss of hydrogen or addition of oxygen. An oxidising agent is one that can accept electrons, combine with hydrogen or give oxygen to another chemical species. • Oxidation reactions are of immense significance in our daily life. Combustion, combustion of fuel, respiration, digestion of food etc are some common examples of oxidation reactions. A number of enzymes known as oxygenases and dehydrogenases help to carry out vital biochemical oxidation reactions in a living cell. • Undesired oxidation reactions are also very common in nature for example: rusting of metals (for example iron), oxidative decay of food, metarial (eg. polymers), ageing process in humans etc. Some forms of cancer, hardening of arteries have also been liked to slow oxidation reactions in human body. • Oxidation reactions are used for the synthesis of a large number of organic compounds, which find application in industry (either as intermediates or as end product) and as laboratory reagents. For example terephthalic and adipic acid (both of which are obtained by oxidation of appropriate substrates) are used in industry for the synthesis of polyester and nylon66 respectively.

A wide variety of oxidising agents can be used for oxidation of organic compounds, the choice of oxidising agent depending on the nature of product required and the type of molecule being oxidised. For example a –CH3 in p-nitrotoluene may be oxidised to • an aldehyde • a carboxylic acid Chromium trioxide is used for oxidation to the aldehyde and potassium permanganate to get the carboxylic acid Some of the common oxidising agents used in laboratory are • Potassium permanganate • Hydrogen peroxide • Chromium trioxide • Sodium hypochlorite • Sodium/potassium dichromate • Potassium bromate • Nitric acid • Peracids etc Present section gives an account of some simple oxidation reactions.

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3.22.1 Oxidation Reactions with Potassium permanganate Potassium permanganate can be used under neutral, acidic and basic conditions for oxidation reactions in the laboratory. Manganese present as Mn7+ in KMnO4 is converted to Mn2+ (MnSO4) in acidic medium.

In alkaline or neutral medium, the Mn+7 is converted to Mn+4 (MnO2).

3.22.2 Benzoic Acid It can be obtained by the oxidation of benzaldehyde, benzyl alcohol, toluene or any other mono alkyl substituted benzene containing at least one benzylic hydrogen atom. For example n-butylbenzene is oxidised but t-butylbenzene is not oxidised by alkaline potassium permanganate. Below are given the procedures of oxidation of benzaldehyde and toluene to benzoic acid. (a) From benzaldehyde Benzaldehyde is easily oxidised by alkaline potassium permanganate to benzoic acid

Procedure Reagents Benzaldehyde

1 mL

Potassium permanganate

1.25g

Sodium carbonate

1g

Take benzaldehyde (1 mL) in a small round bottomed flask, add to it a solution of sodium carbonate (1g) and a solution of potassium permanganate (1.25g) slowly with shaking. Initially the colour of potassium permangate fades fast but slows down after some time. When the colour of permanganate persists, fix a condenser and set up the reaction for refluxing on a wire gauze. Reflux for 10-15 minutes and add more potassium permanganate slowly till its colour persists. Stop heating, allow the solution to cool down to room temperature. Filter the brown precipitate of manganese dioxide (some fine particles of MnO2 may also pass into the filtrate). Acidify the filtrate containing sodium benzoate, some dissolved unreacted potassium permanganate and manganese dioxide with dilute hydrochloric and then add a saturated solution of sodium sulphite slowly with stirring till the supernatant solution becomes colourless and a white precipitate of benzoic acid separates out. Filter the solid, wash with water, dry and weigh. Recrystallise ~100 mg solid from water. Yield 0.6g, m.p. 121°C.

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Notes: • Acidified sodium sulphite is added to remove MnO2 and KMnO4.

2KMnO4 +2HCI + 3NaHSO3 —→ 2MnO2 + 3NaHSO4 +2KCl + H2O HCl + NaHSO3 —→ H2SO3 + NaCl

MnO2 + H2SO3 —→ MnSO4 + H2O

• SO2 gas may be passed into the filtrate to remove KMnO4and MnO2. • Since SO2 is produced or used during the work up, the procedure must be performed in a fume chamber. SO2 is injurious for respiratory system. Do not inhale it. • Potassium permanganate causes irritation in the eyes and skin (also stains the skin brown). Wash the effected area with plenty of fresh water and seek medical help if necessary. (b) From toluene

Procedure Reagents Toluene 1 mL Potassium permanganate 1.5g Sodium carbonate 0.75 mL (2N) Reflux (for 3-4 hrs) on a wire gauze, a mixture of toluene (1 mL) and sodium carbonate solution (2N, 0.75 mL) in a round bottomed flask fitted with a reflux condenser. Add a solution of potassium permanganate (1.5 g dissolved in 15 mL water) in small lots to the reaction mixture in the round bottomed flask. The completion of the reaction is indicated by persistance of the colour of potassium permanganate. Allow the solution to cool down to room temperature. Filter the brown precipitate of manganese dioxide (some fine particles of MnO2 may pass into the filtrate). Acidify the filtrate containing sodium benzoate, some dissolved unreacted potassium permanganate and manganese dioxide with dilute hydrochloric and add a saturated solution of sodium sulphite slowly with stirring till the supernatant solution becomes colourless and a white precipitate of benzoic acid separates out. Filter the solid, wash with water, dry and weigh. Recrystallise ~100 mg solid from water. Yield 0.4g, m.p. 121°C. Notes: see above.

3.22.3 Veratric Acid (3,4-dimethoxybenzoic acid) It is obtained by oxidation of veratraldehyde with potassium permanganate under neutral conditions.

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Procedure Reagents 3,4-Dimethoxybenzaldehyde (veratraldehyde) 1 g Potassium permanganate 1.4 g Suspend veratraldehyde (1g) in water (6-7 mL) in a conical flask and warm it in a hot water bath maintained at ~50°C. Add a solution of potassium permanganate (1.4 g in 15 mL water) dropwise during one hour to the stirred suspension. After the addition is over, stir the mixture for one hour more. Filter the precipitated manganese dioxide and treat the filtrate with dilute hydrochloric acid and sodium sulphite solution till a white precipitate is formed with colourless supernatant liquid. Filter, dry, recrystallise from water, yield 1g, m.p. 180°C.

3.22.4 Oxidation with Chromium Trioxide Chromium trioxide is another widely used oxidising agent. The Cr6+ present in it is converted to Cr3+ in the oxidation process. The use of chromium containing oxidising agents is being curtailed because of the health hazards associated with these oxidising agents. Chromium trioxide is harmful to skin and eyes. Very hazardous, if inhaled. 3.22.5 9,10-Anthraquinone It is obtained by the oxidation of anthracene with a solution of chromium trioxide in glacial acetic acid.

Procedure Reagents Anthracene 1g Chromium trioxide 2g Acetic acid (glacial) 20 mL Dissolve anthracene (lg) in glacial acetic acid (15 mL) by gently warming it in a water bath in a round bottomed flask. Add a solution of chromium trioxide (2g) in water (5 mL) and glacial acetic acid (5 mL) dropwise to a well stirred solution (the reaction is carried out in a three necked R.B. flask fitted with sealed stirrer, a reflux condenser and a dropping funnel). Add chromium trioxide solution at such a rate that the mixture continues to reflux (10-15 minutes). When all the solution has been added, reflux the reaction mixture for 10 minutes. Cool and pour the reaction mixture into cold water (100 mL) in a beaker with stirring. Filter the separated anthraquinone, wash with water, hot sodium hydroxide solution (IN) and finally with cold water. Recrystallise from glacial acetic acid using decolourising carbon. Yield 1.1 g, m.p. 285-286°C. Note: Anthraquinone can also be purified by sublimation.

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3.22.6 p-Nitrobenzaldehyde It is obtained by the oxidation of p-nitrotoluene with chromium trioxide in acetic anhydride. p-Nitro benzylidene diacetate is formed first which is then converted to p-nitrobenzaldehyde.

Procedure Reagents p-Nitrotoluene 1.25 g Acetic anhydride 22.5 mL Conc. sulphuric acid 2 mL Chromium trioxide 2.5 g Add concentrated sulphuric acid (2 mL) dropwise to a continuously stirred, cooled solution (in an ice bath at 0°C) of p-nitrotoluene (1.25g) in acetic anhydride (10 mL). Prepare a solution of chromium trioxide (2.5g) in acetic anhydride (12.5 mL). Now add this to the solution of p-nitrotoluene such that the temperature does not exceed 10°C. Stir the mixture for 2 hours more and then pour it over crushed ice (200g). p-Nitrobenzylidene diacetate separates out. Filter and wash with cold water till the washings are colourless. Suspend the solid in cold sodium carbonate solution (~50 mL, 2%), stir for 5 minutes, filter and wash with cold water and finally with cold alcohol (~2 mL). Reflux the crude p-nitrobenzylidene diacetate with ethanol (2 mL), water (2 mL) and concentrated sulphuric acid (0.2 mL) for 30 minutes. Cool the mixture and filter the separated p-nitrobenzaldehyde, wash with cold water, dry in a vacuum dessiccator. Recrystallise ~100 mg of the product from dilute alcohol. Yield 0.75g, m.p. is 106°C. Notes: • Handle sulphuric acid carefully. • Jones and Sarett oxidation are also used for conversion of –CH3 to –CHO group.

3.22.7 Oxidation with Potassium Dichromate/Sodium Dichromate Potassium dichromate is another commonly used oxidising agent. The Cr6+ is reduced to Cr3+ during the redox reaction. The reactions occur in presence of mineral acids.

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3.22.8 Adipic Acid It is obtained by the oxidation of cyclohexene with potassium dichromate and sulphuric acid.

Procedure Reagents Cyclohexene 1g Potassium dichromate 0.9 g Conc. sulphuric acid 3 mL Add a solution of potassium dichromate (0.9 g in 3 mL water) to a solution of cyclohexene (1g) in dilute sulphuric acid (obtained by cautiously adding 3 mL conentrated sulphuric acid to 3 mL water) in a conical flask with vigorous shaking maintaining the temperature of the reaction between 40-50°C. Heat the reaction mixture to 80-90°C for few minutes and then cool to 5-10°C in an ice-salt bath and allow to stand for 10-15 minutes. The solution develops a green colour due to reduction of Cr6+ to Cr3+. Filter the separated product, wash with ice cold water and recrystallise from hot water. Yield 0.9 g, m.p. 152°C. Notes: • Adipic acid can also be obtained by the oxidation of cyclohexanol with nitric acid. • Chromium salts are hazardous and should be handled carefully.

3.22.9 p-Nitrobenzoic Acid It is obtained by the oxidation of p-nitrotoluene with sodium dichromate and dilute sulphuric acid.

Procedure Reagents p-Nitrotoluene 1.4 g Sodium dichromate 4.0 g Conc. sulphuric acid 4.9 mL Add, slowly concentrated sulphuric acid (4.9 mL) to a stirred mixture of p-nitrotoluene (1.4g), sodium dichromate (4g) and water (10 mL) in a round bottomed flask. Considerable heat is evolved, p-nitrotoluene melts and oxidation proceeds. Cool the reaction mixture in a water bath and stir it occassionally. After the addition of acid, reflux the mixture by gentle boiling on a wire guaze for 15 minutes. Cool and pour cautiously into cold water (20 mL) in a beaker. Filter the separated product and wash with water (~10 mL). Digest it by heating on a water bath with dilute sulphuric acid (5%, obtained by adding 0.3 mL concentrated sulphuric acid to 7.5 mL water) with stirring for 20 minutes to remove the chromium salts. Cool the mixture and filter. Dissolve the crude product in dilute sodium hydroxide solution (5%). Heat the solution with charcoal (~l g) for 5 minutes at 50°C. Filter to remove charcoal, add crushed ice

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to the clear alkaline filtrate and acidify with dilute sulphuric acid. Filter the precipitate of p-nitrobenzoic acid, wash with water and dry. Recrystallise from glacial acetic acid. Yield 1.2 g, m.p. 238-240°C.

3.22.10 9,10-Phenanthraquinone It is obtained by the oxidation of phenanthrene with potassium dichromate and dilute sulphuric acid.

Procedure Reagents Phenanthrene 1.5 g Potassium dichromate 9 g Conc. sulphuric acid 15 mL Heat the suspension of phenanthrene (1.5 g) in dilute sulphuric acid (obtained by cautiously adding 15 mL concentrated sulphuric acid to 30 mL water with stirring) to 90-95°C in a water bath. Add potassium dichromate (9g) in small lots (0.5-1 g) until a vigorous reaction sets in. Remove the reaction mixture from the hot water bath, the temperature of the mixture rises to approximately 110-115°C. Continue the addition of potassium dichromate such that the temperature of the reaction mixture does not fall below 85°C (heat in a hot water bath if necessary to maintain the temperature). Finally, heat the reaction mixture in a boiling water bath for 30 minutes. Cool and pour it into water (100 mL) in a beaker, filter the crude product and wash with water until it is free from chromium salts. Suspend the solid in rectified spirit (15 mL) and stir for 10 minutes with saturated sodium bisulphite solution. Dilute with water (~20 mL), filter if necessary and treat the clear solution containing the bisulphite addition product with saturated sodium carbonate solution. Phenanthraquinone separates as a yellow coloured solid. Filter, wash with water, dry and crystallise from glacial acetic acid. Yield 1g, m.p. is 206-207°C.

3.22.11 Terephthalic Acid It is obtained by oxidation of p-xylene with sodium dichromate and dilute sulphuric acid. Procedure Reagents p-Xylene 1.25 mL Sodium dichromate 7g Conc. Sulphuric acid 9.25 mL Add conc. sulphuric acid (9.25 mL) in 15-20 minutes to well stirred mixtrure of p-xylene (1.25 mL), sodium dichromte (7g) and water (30 mL) in a round bottomed flask.When all the acid has been added, the temperature begins to fall, reflux the reaction mixture gently for 30 minutes. Now pour it into water (60 mL) and allow it to stand for one hour. Filter the separated terephthalic acid, wash with cold water (~5 mL) and then with ether (5-6 mL). Dissolve the crude terephthalic acid in dilute sodium hydroxide

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solution, filter, add some crushed ice to the filtrate and acidify with dilute sulphuric acid. Filter the terephthalic acid, dry and weigh. It sublimes at 300°C without melting. Notes: Terephthalic acid is also obtained by the oxidation of p-xylene with alkaline potassium permanganate.

3.22.12 Oxidation with Hydrogen Peroxide Hydrogen peroxide is an important green oxidising agent. It does not contain any undesirable metal and also produces water a byproduct. 3.22.13 Diphenic Acid (Biphenyl-2,2′-dicarboxylic acid) It is obtained by the oxidation of phenanthrene with hydrogen peroxide in glacial acetic acid.

Procedure Reagents Phenanthrene 2g Acetic acid (glacial) 22.5 mL Hydrogen peroxide (30%) 8 mL Add hydrogen peroxide (30%, 8 mL) slowly during 15-20 minutes to a stirred solution of phenanthrene (2g) in glacial acetic acid (22.5 mL, 85°C) in a three necked flask fitted with a sealed stirrer, a reflux condenser and a thermometer. After the addition is complete, the temperature falls to ~80°C. Heat the reaction mixture in a water bath with stirring for 3-4 hours. Distil the solution under reduced pressure until its volume is reduced to half and then cool the residue. Filter the precipitated diphenic acid, evaporate the filtrate to almost dryness under reduced pressure. Extract the residue with sodium carbonate solution (10%, 20 mL) by warming on a water bath. Acidify with dilute hydrochloric acid to pH 4.5, cool the solution to 0°C, filter any tarry material that separates and acidify the clear solution with dilute hydrochloric acid. Filter the separated diphenic acid and wash with cold water, dry and weigh. Recrystallise from glacial acetic acid. Total yield 1.9 g, m.p. 229-230°C.

3.22.14 Oxidation with Potassium Bromate Potassium bromate acts as an oxidising agent in acidic medium. 3.22.15 p-Benzoquinone

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Procedure Reagents Hydroquinone 2g Potassium bromate l.l g Sulphuric acid (IN) 1 mL Heat a mixture of hydroquinone (2g), potassium bromate (l.lg), sulphuric acid (IN, 1 mL) and water (20 mL) in a conical flask to 60°C on a water bath with continuous shaking. After a few minutes, raise the temperature to 80°C and maintain for 10 minutes. Allow the reaction mixture to cool to room temperature and then in an ice bath. Filter the separated quinone, wash with cold water and dry in a vacuum desiccator over fused calcium chloride, weigh and recrystallise from petroleum ether. Yield 1.4g, m.p. 117°C. Note: Quinhydrone (see page 39), a dark green solid is formed as an intermediate, finally it dissolves and gives yellow p-benzoquinone.

3.22.16 Oxidation with Sodium Hypochlorite Sodium hypochlorite or sodium hypoiodite are both used as oxidising agents. These are obtained by reaction of chlorine/iodine with sodium hydroxide. Commercial bleach contains hypochlorite. A methyl ketone is oxidised to corresponding carboxylic acid by hypoiodite/hypochlorite. 3.22.17 2-Naphthoic Acid It is obtained by oxidation of 2-acetylnaphthalene with sodium hypochlorite.

Procedure Reagents 2-Acetylnaphthalene 2g Sodium hypochlorite 3.8g (see notes below) Take sodium hypochlorite solution (containing about 3.8 g) (see note below) in a small three necked flask fitted with a stirrer, a thermometer and a reflux condenser. Add 2-acetyl naphthalene (2g) to the stirred solution of sodium hypochlorite at 55°C. An exothermic reaction commences. Stir the reaction mixture for 30 minutes, maintaining its temperature at 60-70°C by cooling in an ice bath if necessary. Stir the mixture for 30 minutes more and decompose excess hypochlorite by adding sodium metabisulphite solution (25%, 5 mL), making sure that no unreacted hypochlorite remains (by testing one or two drops of the solution with acidified potassium iodide solution). Acidify the reaction mixture with concentrated hydrochloric acid (4.5 mL). Filter the product, wash with water, dry and weigh. Recrystallise from alcohol. Yield 1.7g, m.p. 184-185°C. Notes: Sodium hypochlorite solution is obtained by diluting 55 mL of commercially available chlorax with 12 mL water.

3.22.18 Elbs Persulphate Oxidation It is a method of oxidation of for example a mohydric phenol to a dihydric phenol with alkaline potassium persulphate. Hydroxylation normally occurs at p-position. However, if the p-position is blocked, then the reaction occurs at o-position with respect to –OH.

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Mechanism

3.22.19 2,5- Dihydroxybenzaldehyde (gentisic aldehyde) It is obtained by the oxidation of o-hydroxybenzaldehyde with alkaline potassium persulphate.

Procedure Reagents o-Hydroxybenzaldehyde

5g

Sodium hydroxide solution

16 mL 20% and 2.1g in 35 mL

Potassium persulphate

12.3g

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Add separately but simultaneously, an aqueous potassium persulphate solution (12.3 g in 120 mL water) and sodium hydroxide solution (20%, 16 mL) to a solution of o-hydroxybenzaldehyde (5 g) in sodium hydroxide solution (2.1g in 35 mL water) with vigorous stirring at 30-35°C during 1 hour. The mixture should always remain alkaline. After the addition is over, stir the mixture for one hour more and keep at room temperature for 36 hours. Cool the solution and acidify to congo red with dilute hydrochloric acid. Extract the unreacted o-hydroxybenzaldehyde with ether. Separate the ether layer and strongly acidify the clear aqueous solution with hydrochloric acid (~20 mL) and heat slowly at 70°C for one hour. Filter the solid product that separates out. Extract the filtrate with ether (3 × 15 mL). Dry the ether extract over anhydrous sodium sulphate. Remove ether in a rotary evaporator and purify the residue by column chromatography using silica gel as stationary phase and elute with benzene. Yield l.l g m.p. 97-98°C.

3.22.20 Oxidation with Peracids (Baeyer Villiger oxidation) Baeyer-Villiger oxidation is a method of oxidation of aldehydes and ketones with peracids. An aldehyde is oxidised to a formyl ester, which is hydrolysed to give the corresponding hydroxyl compound. A ketone is oxidised to an ester resulting from the insertion of an oxygen between the C=O and R group. [O]

C6 H5COCH3  → C6 H5O COCH3 CF COOOH Acetophenone

3

Phenylacetate

Peracetic, trifluro peracetic, performic, perbenzoic and m-chloroperbenzoic acid are some of the most commonly used peracids. Mechanism

The migrating group retains its configration.

3.22.21 3,4-Dimethoxyphenol It is obtained by oxidation of 3,4-dimethoxy benzaldehyde (veratraldehyde) with peracetic acid.

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Procedure Reagents Veratraldehyde 1g Acetic acid (glacial) 6 mL Peracetic acid 4 mL (obtained by adding 1 mL H2O2 to 3 mL glacial acetic acid) Add peracetic acid solution (4 mL) dropwise during 10 minutes to a well stirred solution of veratraldehyde (l g) in glacial acetic acid (6 mL). Leave the reaction mixture for ~10 hrs. Distil the mixture under vacuum to leave ~2 mL residue. Cool, extract with ether (2 × 10 mL), remove ether in a rotary evaporator and reflux the residue with potassium hydroxide (2 g) and aqueous alcohol (20 mL) for an hour. Concentrate the reaction mixture under vacuum to almost dryness. Cool, dissolve the residue in minimum volume of water, acidify the solution with dilute sulphuric acid and extract 3,4-dimethoxy phenol thus obtained with ether. Dry the ether layer over anhydrous sodium sulphate, filter and remove ether in a rotary evaporator. Purify the residue by column chromatography using benzene as a solvent. M.P. 78-80°C. Notes: Ether is highly inflammable and benzene carcinogenic.

3.23

PHOTOCHEMICAL REACTIONS

3.23.1 Introduction A chemical reaction initiated by absorption of light (a source of energy) is known as photochemical reaction. Photochemical reactions are very common in nature. Photosynthesis, synthesis of vitamin D from 7-dehydrocholesterol and ability of eye to see etc. are some examples of photochemical reactions occurring in nature. Harmful radiations of the sun are prevented from reaching the surface of earth by ozone, which absorbs these radiations undergoing decomposition to atomic oxygen, which in turn combines with atmospheric oxygen to give back ozone there by also preventing ozone layer from depletion, phosphorescence and fluorescence are also photochemical phenomena. The first photochemical reaction observed in 1866 by a Russian scientist was the dimerisation of anthracene. Chlorination of alkanes, side chain chlorination of alkyl substituted aromatic compounds are some very well known examples of photochemical reactions.

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Ultra violet, visible or infra red radiations are used to induce photochemical reactions.

3.23.2 Benzopinacol It is obtained by sun light induced photoreduction of benzophenone in propan-2-ol. Sunlight or a medium pressure mercury arc lamp can be used. The mechanism involved in this transformation is given below. Propane-2-ol acts as a hydrogen donor and itself is oxidised to acetone.

Procedure Reagents Benzophenone 1g Isopropanol 6 mL Dissolve benzophenone (1g) in isopropanol (6 mL) in a small test tube by warming and add a drop of acetic acid. Add more isopropanol to almost fill the tube. Stopper the tube with a rubber cork. Place the tube in bright sunlight. Colourless crystals of benzopinacol begin to separate within 10 hours. Allow to remain in sunlight until no further solid appears to separate (2-3 days). Filter the separated benzopinacol and wash with cold isopropanol. Recrystallise from glacial acetic acid. Yield 0.9g, m.p. 185-186°C. Notes: • Benzopinacol is also obtained by reductive dimerisation of benzophenone with zinc and acetic acid. • A medium pressure mercury arc lamp can be used to initiate the reaction,

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3.23.3 Isomerisation of trans-azobenzene to cis-azobenzene Azobenzene as ordinarily obtained, is the trans-isomer. It can be photochemically isomerised to the cis form. This conversion can be achieved by irradiation using a fluorescent lamp or even by sunlight.

 −N  − C H , which gives a mixture of cis and trans form The reaction occurs via a diradical C6 H5 − N 6 5 Procedure Reagents Azobenzene (trans) 1g Petroleum ether 50 mL Dissolve azobenzene (trans form) (1g) in petroleum ether (b.p. 40-60°C, 50 mL) in a beaker. Irradiate it for 30 minutes with ultraviolet light by supporting a Hanovoi fluorescent lamp above the surface of the liquid in the beaker. In the meanwhile, prepare a chromatographic column (approx. 20 cm × 1.8 cm) from activated alumina (grade 1, 50 g) using petroleum ether as an eluent. Place a circular well fitting filter paper at the top of the column and pour the irradiated solution on the filter paper in the column with the help of a glass rod. Do not disturb the column in this process. Run the column chromatogram with petroleum ether (b.p. 40-60°C, 100 mL). A coloured band of the cis form (about 2 cm) appears at the top of the column while a diffuse coloured region of the trans form moves down the column. Protect the upper band from light by covering with a carbon paper. This prevents the reconversion of cis into trans form. Elute the band of trans isomer with petroleum ether. Take out the alumina of column and cut the upper band containing the cis-isomer. Shake the cut out band with petroleum ether (150 mL) containing methanol (1.5 mL). Filter the solution and wash the filtrate with water (2 × 15 mL). Dry the petroleum ether-methanol extract over anhydrous sodium sulphate and remove the solvent in a rotary evaporator. The residual coloured product (m.p. 71.5°C) is pure cis-azobenzene. Notes: • The ethanolic solution of pure cis-azobenzene shows absorption at 281nm (e 5260) and transazobenzene at 320 nm (e 21300) in ultraviolet spectrum. • The spectrum should be recorded immediately after the isolation of cis form.

3.23.4 1,4-Naphthaquinone Photodimer It is obtained photochemically by irradiation of 1,4-naphthaquinone in benzene solution with UV light.

Procedure Take a solution of 1,4-naphthaquinone (3g) in thiophen free benzene (50 mL) in a conical flask (50 mL). Pass a slow stream of nitrogen through the solution to remove any air or oxygen. Stopper the flask and

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place in direct sunlight. Cover the rubber stopper with aluminium foil. A solid product slowly separates from the solution after 4-5 days. Filter the separated product and expose the filtrate again to sunlight after passing nitrogen gas (as earlier). Filter the separated product after 4-5 days again. Total yield 0.44 g, m.p. is 244-246°C.

3.24

POLYMERISATION

3.24.1 Introduction Polymers also known as macromolecules are formed by chemical combination of small molecules. The small molecules are known as monomers or building units and the process of chemical combination of monomers is known as polymerisation. Polymerisation can occur by addition of monomers to one another (addition polymerisation) or condensation (condensation polymerisation) between monomers. The reaction is generally catalysed, the nature of the catalyst being free radical or ionic. A few examples of polymerisation reactions are given below:

Polymers are of great significance in our lives, natural polymers like starch, cellulose, glycogen, proteins, DNA, RNA etc play vital role in the survival of all living beings. The man made polymers like plastics, synthetic fibers and elastomers and products made from these are indispensable in our daily life. Here is given an account of synthesis of a few polymers.

3.24.2 Nylon 66 It is one of the most widely used polyamide synthetic fibre. It is obtained by condensation polymerisation between adipoyl chloride and hexamethylenediamine in presence of sodium hydroxide. The first number in the name indicates the number of carbon atoms in the amine and the second, the number of carbon atoms in the acid (used as its chloride). Nylon 66 has excellent abrasion resistance, high strength, is stable

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to light and is not effected by sea water. It is used for making fabrics, conveyer belts, carpets, laundary bags, fishing nets, bristles for brushes and brooms etc.

Procedure Reagents Adipoyl chloride

1.5 mL

Hexamethylenediamine

1.5 g

Sodium hydroxide

1.2 g

Hexane

75 mL

Add a solution of hexamethylenediamine (1.5 g) in sodium hydroxide solution (1.2 g in 40 mL water) slowly to a solution of adipoyl chloride (1.5 mL) in hexane (75 mL) in a beaker. A polymeric film starts forming at the interface of the two liquids. Remove the polymer film with a forcepts or a spatula and stretch to form a thin rope. Collect the long rope in a beaker, wash with water and then wash with aqueous acetone (50%). Dry and place on a watch glass and melt carefully on a hot plate so that charring does not taken place. The molten polymer can then be drawn with a glass rod into a finer thread.

3.24.3

Phenol Formaldehyde Resin (Bakelite)

It is the oldest known synthetic polymer and is still extremely important and extensively used. It is obtained by condensation polymerisation between phenol and formaldehyde. The reaction is catalysed by a weak acid or a base. The nature of the polymer (whether linear or cross linked) depends on the ratio of phenol to formaldehyde and the catalyst used. A highly cross linked polymer (known as bakelite) is obtained with higher ratio of formaldehyde. Mechanism and structure of polymer Acid catalysed polymerisation

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Base catalysed polymerisation

Mono, di and trimethylol derivatves of phenol are formed. The structure of the final product depending on the ratio of phenol: formaldehyde

and finally a highly crosslinked polymer is formed

Procedure Reagents Phenol Formaldehyde solution (Formalin) Oxalic acid

5g 4.1 g (37%) 0.05 g

Reflux a mixture of phenol (5g), formalin (4.1g) and oxalic acid (0.05g) in an a round bottomed flask for one hour. The mixture becomes viscous. Add water (15 mL), cool the flask and allow the reaction

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mixture to stand for some time. Decant the supernatant liquid. The resin is obtained as a pasty mass. Remove water at the pump, the resin turns into a brittle solid. Yield 5g.

3.24.4 Thiokol Rubber It is one of the oldest known synthetic elastomer and is obtained by the reaction of 1, 2-dichloroethane with sodium polysulphide.

Thiokol is used for making hoses, gaskets, seals and as solid fuel in rocket engines. Procedure Reagents 1, 2-Dichloroethane

7.5 mL

Sulphur

3g

Sodium hydroxide

1.5g

Add powdered sulphur (3g) in small lots with constant stirring to a warm solution of sodium hydroxide (1.5g) in water (30-35 mL). A deep red solution is obtained. Filter the solution if some sulphur remains undissolved. To the clear solution add, with stirring, 1, 2-dichloroethane (7.5 mL). Stir the mixture for 15-20 minutes, a rubber like polymer separates out as a lump. Decant the supernatant liquid, wash the polymer under a tap. Dry it within the folds of a filter paper. Yield l.l g.

3.24.5

Polystyrene

It is obtained by addition polymerisation of styrene in presence of benzoyl peroxide as a free radical catalyst. It is one of the most widely used polymer. The expanded polystyrene (known as thermocol) is extensively used in packaging, polystyrene is used for making parts of refrigerators, air conditioners, washing machines, vacuum cleaners, TV’s, CD’s, DVD’s etc. polystyrene resins are extensively used in building and construction material. It finds use in medical technology etc.

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Mechanism:

Procedure Reagents Styrene 10 mL Toluene 20 mL Benzoyl peroxide 0.2 g Sodium hydroxide 5 mL, 10% Methanol 150 mL Take commercial styrene (10 mL) in a conical flask, add water (20 mL) and sodium hydroxide (5 mL, 10%). Shake the mixture well. Now take the mixture in a separating funnel, discard the aqueous layer, wash styrene layer thrice with Water (~10 mL). Dry the washed styrene over anhydrous calcium chloride, filter and use for polymerisation. Take styrene in a boiling tube, add benzoyl peroxide (0.2g), toluene (20 mL), heat the mixture for one hour in a boiling water bath, maintained at 90-95°C. Cool and pour the reaction mixture in methanol (150 mL) in a beaker. Filter the white precipitate of polystyrene, wash with water and dry it in air. Note: Styrene undergoes auto polymerisation on standing. To prevent this, a small amount of an anti oxidant is added during storage. Tri-tertiaryphenol is a common commercially used anti oxidant. It is removed by washing styrene with dilute solution of sodium hydroxide.

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REDUCTIONS

Loss of oxygen or gain of electrons or hydrogen is called reduction.

3.25.1 Reduction of Nitrobenzene to Aniline Nitrobenzene is reduced to aniline with acidic reducing agents like iron or tin and hydrochloric acid. Iron is less costly as compared to tin and that is why it is used for commercial production of aniline. Sn + 2HCl —→ SnCl2 + 2H C6H5NO2 + 6H —→ C6H5NH2 + 2H2O Mechanism

Procedure Reagents Nitrobenzene 6.3 mL Tin (granulated) 12g Conc. hydrochloric acid 30 mL Take nitrobenzene (6.3 mL), 12g of granulated tin (known as mossy) in a round bottomed flask and add hydrochloric acid (3-4 mL) to it in small lots with constant shaking. Cool if the reaction becomes vigorous and hot. After the reaction has subsided, add more hydrochloric acid (5 mL) slowly as before, and add the whole of acid over a period of 15-20 minutes. Now fix up a condenser and reflux the reaction mixture for 20-25 minutes. Take a drop of the reaction mixture and add it into water (~5 mL) in a test tube, formation of a clear solution indicates that the reduction is complete. If the reaction is complete, transfer the reaction mixture to a beaker and cool it. Some aniline chlorostanate may separate out as a solid. Basify the solution with sodium hydroxide (22 g in 35 mL water). Aniline separates out as insoluble oil. Steam distil the reaction mixture till the distillate is clear. Cool the steam distillate, saturate it with sodium chloride, take it in a separating funnel and extract aniline with ether (3 × 20 mL). Separate the ether layer, wash with water and dry over anhydrous potassium carbonate. Filter Potassium carbonate, distil ether in a rotary evaporator and finally distil the residue in a small distillation flask. Yeild 5.5g, b.p. 182°C. Note: Ether is highly inflamable, handle it very carefully.

3.25.2 Reduction of Nitrobenzene to Azobenzene It is obtained by the reduction of nitrobenzene with zinc powder in methyl alcohol.

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Procedure Reagents Nitrobenzene

2.6 mL

Methanol

35 mL

Zinc powder

3.6g

Take nitrobenzene (2.6 mL) in a round bottomed flask and dissolve in methanol (35 mL). Add Zinc powder (1.5 g) and iodine (a small crystal) to the solution of nitrobenzene. If the reaction does not commence in 2-3 minutes, warm the mixture in a water bath to start the reaction. If the reaction becomes too vigorous, cool in a cold water bath for a few seconds. When most of the zinc has reacted, add the remaining zinc (total 3.6g) and allow the reaction to proceed (as above). Fix a reflux condenser and heat the reaction mixture for 30 minutes on a water bath. Cool, pour into water (100 mL) acidify with glacial acetic acid until the mixture is acidic to congo red. Cool the mixture and filter the separated azobenzene. Recrystallise from ethanol (90%). Yield 1.5g, m.p. 67-68°C. Note: Avoid use of excess zinc as it reduces azobenzene to hydrazobenzene. Sodium hydrogen sulphide Sodium hydrogen sulphide is another important reducing agent, used for a number of reduction reactions. It is used as a reducing agent for partial reduction of polynitro compounds. Here is given the reduction of m-dinitrobenzene to m-nitroaniline.

3.25.3 Reduction of m-Dinitrobenzene to m-Nitroaniline m-Nitroaniline is obtained by selective reduction of m-dinitrobenzene with sodium or ammonium hydrogen sulphide.

Procedure Reagents m-Dinitrobenzene

2g

Sodium hydrogen sulphide

1.15 g (obtained as given below)

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Preparation of sodium hydrogen sulphide Add sodium bicarbonate (1.8g) to a solution of crystallised sodium sulphide (Na2S.9H2O, 5.4g in 15 mL water) with constant stirring. After the bicarbonate has dissolved, add methanol (15 mL) with stirring. Filter the precipitated sodium carbonate and wash the precipitate with methanol (2 mL). Use the combined filtrate and washings for reduction. Na2S + NaHCO3 —→ NaSH + Na2CO3↓ Reduction of m-dinitrobenzene Add with shaking, a solution of m-dinitrobenzene (2g) in hot methanol (15 mL) to the solution of sodium hydrogen sulphide prepared above. Reflux the mixture gently for 20-25 minutes in a hot water bath. Some more sodium carbonate may separate. Distil methanol in a water bath and pour the residual mixture with stirring into ice cold water (50 mL) in a beaker. Filter the separated yellow precipitate of m-nitroaniline, wash with water and crystallise from dilute methanol. Yield 1.1 g, m.p. 113-114°C. 3.26 S-Benzyisothiuronium chloride It is obtained by reaction of benzyl chloride with thiourea

Mechanism:

Procedure Reagents Thiourea 1.3 g Benzyl chloride 2.0 g Rectifield spirit 5 mL Take thiourea (1.3g), benzyl chloride (2g) and rectified spirit (5 mL) in a round bottomed flask. Gently warm the reaction mixture in a water bath. A vigorous reaction sets in. Take out the flask from the water bath and allow the reaction to subside. Now reflux the reaction mixture in a water bath for 30 minutes. Cool the reaction mixture first in air and then in an ice bath. Filter, dry, recrystallise from dilute hydrochloric acid. Yield 3g, m.p. 175°C.

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Questions Q.1.

Which of the following compounds can undergo acetylation reaction? Write the reactions and name the product in each case.

Q.2.

Why is acetic anhydride considered a better acetylating agent than acetic acid?

Q.3.

Mention a few advantages and a few disadvantages of using acetyl chloride for acetylation.

Q.4.

How is excess of acetic anhydride decomposed after the completion of acetylation reaction?

Q.5.

Name a few important naturally occurring acetyl compounds.

Q.6.

See the MSDS data sheet of the following compounds: Acetic acid, Acetic anhydride and Acetyl chloride.

Q.7.

What is the difference in structure of a-D and b- D-glucose pentaacetate?

Q.8.

How is 1,2,4- triacetoxy benzene prepared from hydroquinone? Give only the equations.

Q.9.

What do you understand by the term Benzoylation?

Q.10. Why is sodium hydroxide used in the Schotten Baumann reaction? Q.11.

Find out the CAS Registry number of benzoyl chloride and hydrochloric acid. List their adverse side effects on our health.

Q.12. What is the difference in the work up procedure after benzoylation ol aniline and glycine? Why? Q.13. Benzoylated product of a pure sample of p-toludine (a water insoluble solid amine) gave a lower melting point than expected. What could be the impurities in it? How will you purify it? Q.14.

Why a tribromo derivative of aniline is obtained where as a monobromo derivative of acetanilide is obtained on bromination.

Q.15. Explain why the major product of bromination of acetanilide is the para isomer and the ortho isomer is obtained only in minor amount. Q.16. Why does bromination of 2-naphthol gives 1-Bromo-2-naphthol and not 3-Bromo-2-naphthol. Q.17. Predict the product of bromination of following:

Q.18. Why should a slight excess of bromine be added for complete bromination of a given compound.

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Q.19. How is bromine obtained from potassium bromate and potassium bromide? Q.20. Which of the following will not react with a solution of bromine in acetic acid at room temperature

Q.21. Why it is better to use a mixture of conc. HNO3 and H2SO4 than nitric acid alone for nitration of compounds like benzene, nitrobenzene and chlorobenzene etc. Q.22. Why compounds like phenol, aniline etc. cannot be successfully nitrated with conc. nitric acid? Q.23. Complete the following: (a) Nitration is an _____ substitution reaction. (b) _____ is the effective _____ in nitration with mixed acids. (c) Nitro group is a ______ and a _____ directing group. (d) ______ can be obtained by nitration of nitrobenzene. It is difficult to prepare ______ by further nitration. (e) TNT is ______? Q.24. Chlorine in chlorobenzene is deactivating but o, p-directing group. Explain. Q.25. Explain why picric acid cannot be prepared by direct nitration of phenol. Explain the method used for its preparation. Q.26. Name a few (a) Catalyst used in Friedel Crafts reaction (b) Alkylating agents (c) Acylating agents. Q.27. Write the mechanism of following: (a) Reaction between C2H5I and benzene in presence of AlCl3. (b) Benzene with C2H5OH (c) Benzene and CH2 = CH2

    

in presence of sulphuric acid

Q.28. Which of the following cannot be obtained by Friedel crafts alkylation of benzene

Q.29. Why is Friedel Crafts reaction performed under anhydrous conditions? Q.30. Why should the work up in Friedel crafts reaction be carried out in a fume chamber. What are the harmful gas/es released during the reaction and work up? Q.31. Why is the use of benzene banned specially in under and post graduate laboratories? List a few harmful properties of benzene.

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Q.32. Arrange the following compunds in order of ease of oxidation with alkaline potassium permanganate

Q.33. Which of the following compounds cannot be oxidised by alkaline KMnO4

Q.34. How can acetophenone be oxidised to benzoic acid? Write the relevant reactions and the role of each reactant in the reaction. Q.35. Name the reactions and the reagents that can be used to bring about the following conversions

Q.36. Define the terms (i) Hydrolysis (ii) Saponification (iii) Alcoholysis. Q.37. Why is alkaline hydrolysis preferred over acid catalysed hydrolysis of an ester and an amide. Q.38. What will be the products of alkaline hydrolysis of the following compounds:

Q.39. An anilide (A) is hydrolysed with sulphuric acid. How will you separate the products of hydrolysis?

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Q.40. Why it is easy to hydrolyse acetanilide as compared to benzanilide. Q.41. Write step wise, the reaction between Glucose and phenylhydrazine. Q.42. Why do glucose and fructose form the same osazone. Q.43. Which of the following sugars can form an osazone? Explain your answer

Q.44. What do you understand by the term diazo group? Q.45. Why is nitrous acid not available as shelf reagent like nitric acid in the laboratoy? How is it obtained? Q.46. Why is diazotisation normally carried out at low temperature (~0-5°C)? Q.47. Why is benzene diazonium chloride soluble in water? Q.48. What product is obtained when p-toluidine is diazotised with sodium nitrite and dilute sulphuric acid? Q.49. Why should a slight excess of nitrous acid be present at the completion of diazotisation reaction? What happens if an amine is not diazotised completely before coupling? Q.50. What is Sandmeyer’s reaction? Write the mechanism of conversion of benzene diazonium chloride to bromo benzene. Q.51. Why it is better to substitute the diazo group by hydrogen with hypophosphorous acid? What happens when the same reaction is carried out with ethly alcohol.? Q.52. Why is methyl orange not used as a dye? Q.53. What are (i) azo dyes (ii) xanthene dyes (iii) Triphenyl methane dyes? Give atleast one example of each type. Q.54. Why do phenolphthalein, fluorescein and eosin show different colour in acidic and alkaline medium? Q.55. List a few uses of malachite green and crystal violet. Q.56. Define a monomer, a polymer, polymerisation, addition and condensation polymers. Q.57. Name a few free radical initiators and ionic catalyst used in polymerisation. Q.58. What are special properties of Z-N catalyst? Explain with a suitable example. Q.59. Why are inhibitors added to styrene during storing? Why and how are these removed before polymerisation? Q.60. What is Nylon 66? Why it is named so? Give one more example of a Nylon.

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Q.61. Which of the following is/are not true (a) Bakelite is a copolymer (b) Bakelite is a cross linked polymer of cresol and formaldehyde (c) Bakelite is an elastomer (d) Bakelite is a plastic (e) Bakelite was first prepared by Baekeland. Q.62. What do you understand by the term “degree of polymerisation” Q.63. Define (i) Reduction (ii) oxidation (iii) Redox reaction. Q.64. What is reductive dimerisation. Give an example. Q.65. In the reduction of nitrobenzene to aniline (a) Name two reducing agents that can be used (b) How is the completion of reduction reaction indicated (c) How will you separate pure aniline from a mixture of nitrobenzene and aniline? Q.66. How is m-dinitrobenzene converted to m-nitroaniline? Q.67. Name the reducing agent used for reduction of nitrobenzene to hydrazobenzene. Q.68. How is hydrobenzene converted to benzidine? How can benzidine be converted to 4,4’- dichlorobiphenyl. Give only the sequence of reactions. Q.69. Write the mechanism of formation of benzidine form hydrazobenzene. Q.70. Can you use dye test to distinguish between hydrazobenzene and benzidine? Q.71. Which of the following aldehydes can undergo Cannizzaro’s reaction? CCl3CHO, CH3CHO, HCHO, CHO—CHO and

Q.72. What will be the product of cross Cannizzaro’s reaction between HCHO and p-methylbenzaldehyde? Q.73. How would you separate a mixture of benzoic acid and benzyl alcohol in the laboratory? Q.74. Give a brief account of the products obtained by reduction of nitrobenzene with (a) Acidic reducing agent (b) Neutral reducing agent (c) Alkaline reducing agent. Q.75. What is benzoin condensation? Name the conventional and green catalyst used in this reaction. How is benzoin converted to benzylic acid? Give only the reactions.

4 Isolation of Natural Products 4.1 INTRODUCTION Natural products play an inevitable role in our lives and are the very reason of existence, survival and evolution of life on earth. All our needs are fulfilled by these compounds. Food (including oxygen and water), shelter, clothing—the three basic requirements of any living being are provided by nature. The medicinal plants contain biologically active compounds and have been used for treatment of diseases for thousands of years by humans are the basis of our present day pharmaceutical industry. Most of the man made medicines are developed by taking lead from the naturally occurring active compounds. Though man made compounds have improved our living standards and increased our life expectancy, natural products have an edge over them. Many of the man made products have adverse effect on us and the environment in general, the natural compounds on the other hand are by far safe. Isolation of natural products therefore is an integral component of training of students. Present chapter deals with the isolation of some natural compounds.

4.2 ISOLATION OF CAFFEINE FROM TEA LEAVES Caffeine (1,3,7-trimethylxanthine) is an alkaloid. It is a white, bitter, crystalline substance and is present in tea leaves, cola beans and coffee.

Procedure Reagents Tea leaves

20 g

Chloroform

60 mL

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Petroleum ether 40 mL Lead acetate 60 mL, 10% Boil dry tea leaves (20g) with water (~250 mL) in a beaker for 20-25 minutes. Filter the tea leaves and add a solution of lead acetate (60 mL, 10%) to the clear filtrate, leave the mixture for 2-3 days to allow the tannins to precipitate out as lead salts. Filter through glass wool and concentrate the filtrate in a rotary evaporator or on a sand bath in a beaker or china dish till the volume is reduced to ~25 mL. Cool the concentrated solution and extract with chloroform (~20 mL), thrice in a separating funnel. concentrate the combined chloroform extract in a rotary evaporator, cool add petroleum ether (40 mL). Caffeine is precipitated out. Filter, dry and recrystallise from dilute alcohol. Yield 0.2g, m.p. 235-37°C.

4.3 ISOLATION OF CAFFEINE USING SUPERCRITICAL CARBON DIOXIDE Caffeine is a bitter substance present in coffee and the coffee beans are decaffeinated before making coffee powder. Super critical carbon dioxide is now used commercially to decaffeinate coffee. Notes: • Caffeine is added to colas and some chocolates etc. • It is also added to some headache and migraine medicines. • Caffeine is a habbit forming alkaloid but is not addictive. • It is a natural pesticide. 4.4 ISOLATION OF CASEIN FROM MILK Casein is a phospho protein present in milk as a suspension of particles called micelle. It provides a slow release of amino acids to the blood stream. Procedure Reagents Milk 100 mL Acetic acid 10% Ethanol 15 mL Ether 15 mL Take milk (l00 mL) in beaker and dilute with ~300 mL water. Warm to ~40°C and add acetic acid (10%, the pH of milk is ~6.6, casein is precipitated when pH falls to ~4.5) to obtain all the casein (the remaining solution becomes almost clear). Filter using a Buckner funnel, wash the solid thrice with water, ethanol and ether (~5 mL each) to remove the fat. Dry in a desiccator and weigh.

4.5 ISOLATION OF PIPERINE FROM BLACK PEPPER Piperine is an alkaloid present to an extent of 5-9% in black pepper. Black pepper is extensively used in traditional medicine and a preseservative. It stimulate the digestive enzymes of pancreas.

Procedure Reagents Black pepper powder Alcohol

25 g 300 mL, 95%

4.6 Isolation of Lycopene from Tomatoes/WaterMelon Advanced Experimental Organic Chemistry

Acetone

10 mL

Ethanolic KOH solution

30 mL, 2N

359

Reflux finely powdered black pepper (25 g) with alcohol (300 mL, 95%) in a round bottomed flask (or a soxhlet apparatus) for ~3 hours. Filter the solid particles and concentrate the filtrate to about 20 mL in a rotary evaporator. Add warm ethanolic solution of potassium hydroxide (30 mL, 2N), stir and filter to obtain a clear filtrate. Warm on a steam bath and add water (15 mL). Keep the solution overnight. Piperine separates out as yellow solid. Recrystallise from acetone (inflamable) when piperine is obtained as yellow needles. m.p. 129-131°C.

4.6 ISOLATION OF LYCOPENE FROM TOMATOES/WATERMELON Lycopene is a carotenoid and a compound with high degree of unsaturation. It is regarded as one of the best natural antioxidant. Its role as an anticancer agent is under scientific study. It is present in guava (pink from inside), watermelon, tomatoes, papaya, purple cabbage, mango, carrot etc.

Procedure Reagents Tomatoes 100 g OR Watermelon 100 g Benzene 80 mL Ether 15 mL Mash ~100 g of ripe tomatoes/ watermelon and take in a conical flask. Warm ~40 mL benzene (40°C) in a hot water bath and add it to tomato/watermelon paste. Mix well and add another lot of warm benzene (40 mL) and shake well. Decant the benzene layer and remove all benzene by distillation in a rotary evaporator. Recrystallise lycopene from ether, m.p. 173-74°C. Notes: • Benzene is a human carcinogen, avoid exposure to it. • Both benzene and ether are highly inflammable, do not use these solvents in the laboratory where burners are being used by other students. • An emulsion is formed sometimes when the aqueous paste of tomatoes or watermelon is treated with benzene, allow it to settle before decanting benzene. • Lycopene can also be isolated from ripe papaya following the above mentioned procedure.

4.7 ISOLATION OF CHOLESTEROL FROM GALLSTONES Cholesterol is a steroid, it is an essential component of mammalian cell wall. Its function is to establish proper membrane permeability. Vitamin D, bile acids and steroidal hormones are biosynthesised from

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cholesterol. In human body, the gallbladder is a store house of bile (a digestive enzyme secreted by liver ) and cholesterol. Cholesterol precipitates out in gallbladder if present in excess and is mainly responsible for the formation of gallbladder stones. Excess cholesterol also has adverse effect on the cardiovascular system.

Procedure Gallstones

2g

Ethyl methyl ketone

25 mL

Methanol

10 mL

Crush the gallstones (2 g) in a pestle mortar and sterilise by washing with ~5 mL house hold bleach. Dry the crushed stones between the folds of filter paper and transfer to a round bottomed flask. Add ethyl methyl ketone (~ 20 mL) to the flask and gently boil for 2–3 minutes. Filter the hot solution using a filter paper, leaving the crushed stones in the flask. Add ~ 5 mL more of ethyl methyl ketone to the flask, bring to boil and filter into the previous filtrate*. Add methanol (l0 mL) to the filtrate and reheat the mixture gently. Now add water dropwise till a turbidity appears. Allow the turbid solution to cool and filter the crystals of cholesterol. m.p.148-50 °C. Notes: • Boil the filtrate* if it is highly coloured with animal charcoal (which adsorbs the coloured impurities), filter and continue with the procedure. • Cholesterol can also be isolated from egg yolk.

4.8

ISOLATION OF EUGENOL FROM CLOVES

It is an essential oil and is present in cloves. Small amount of it is also present in cinnamon, sweet basil etc. It is used by dentist because of its antiseptic and anti-inflammatory properties. It is a common ingredient in mouth wash, toothpaste and soaps etc. it is a fat soluble antioxidant.

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Procedure Reagents Cloves

50 g

Dichloromethane

250 mL

Potassium hydroxide

150 mL, 5%

Place cloves (~50 g) in a round bottomed flask and add water (~200 mL) to it. set up the apparatus for distillation. Distil ~ 150 mL of the solution. Add some more water (100 mL) to the distillation flask, distil and collect –100 mL of the distillate. Mix the two distillates and transfer to a separating funnel. Extract the distillate twice with dichloromethane (~75 mL each time). Drain out the lower organic layer. Take it in a clean separating funnel and extract thrice with 5% solution of potassium hydroxide (50 mL each time). Eugenol being a phenolic compound, dissolves in alkali. Mix the alkaline extract and acidify with cold dilute hydrochloric acid. Free eugenol is liberated. Extract the acidic solution with dichloro methane (2×50 mL). Drain the organic layer, dry it over anhydrous sodium sulphate. Filter, distil dichloro methane in a rotary evaporator. The residue is almost pure eugenol. b.p. 250°C. Notes: • Eugenol gives a yellow green colour in the ferric chloride test for phenols (see sec. 1.4.4) • It forms a brown red coloured picrate on reaction with picric acid (see sec. 1.5.10)

4.9

ISOLATION OF LIMONENE FROM ORANGE PEEL

Limonene is a colourless liquid. It belongs to a class of cyclic terpenes and is present in citrus fruits and pine trees. It is a powerful antioxidant and an antiinflammatory compound, it is added as a favouring agent to foods, beverages and chewing gums etc. it is added to some pharmaceutical creams and helps them to penetrate the skin. Limonene containing wash liquids are used to clean oil stains in parking complexes and airport runways.

Procedure In a blender, grind orange peel of two oranges for ~30 seconds. Add water (~150 mL) and blend again for 15-20 seconds. Take this in a round bottomed flask and distil. Collect ~ 25-30 mL of the distillate. Stop distillation when a clear distillate starts coming because as long as limonene is distilling over, the distillate is slightly turbid. Transfer the distillate into a small, slender separating funnel, allow the limonene layer to separate at the top. Carefully drain out as much as possible of the lower aqueous layer, now transfer the remaining liquid in a test tube, separate the limonene using a dropper into another dry tube and dry in a vacuum desiccator.

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ISOLATION OF LIMONENE USING SUPERCRITICAL CARBON DIOXIDE Carbon dioxide is a gas at normal temperature and pressure but is a liquid at and over its critical tamperature and pressure. ScCO2 is becoming a popular solvent for extraction as compared to organic solvents which are inflamable, many of these cause enviromental pollution. CO2 is non-flamable, non-toxic, environmentally benign and is cost effective. Procedure Grate the coloured part of an orange peel. Take a 15 mL plastic centrifuge tube. Make a loop from a copper wire, (see diagram below) and place a circular paper piece on the loop. Insert the loop into the centrifuge tube. Place ~2.5 g of grated orange peel on the loop and fill the tube with solid dry ice. Close the centrifuge tube with the cap and immerse it into a plastic cylinder containing warm water (40-50°C). As soon as the temperature rises, the dry ice melts and extracts limonene, which collects at the bottom of the tube and the CO2 gas starts escaping from the cap of the tube. Extract again with more dry ice if required.

Fig. 4.1: Isolation of Limonene with CO2

4.10

ISOLATION OF NICOTINE FORM TOBACCO LEAVES

Nicotine is a tobacco alkaloid, classified chemically as pyridine-piperidine alkaloid. It is present in plants belonging to solanaceae family. It is an addictive compound. It is known to have adverse effect on human body. It increases blood pressure and heart beat rate. It is known to be a human carcinogen

Procedure Take dry tobacco leaves (10 g) in a beaker and stir with sodium hydroxide (~70 mL, 25%) for 10-15 minutes. Filter the solid mass and extract the clear alkaline solution twice with ether (50 mL each time). Separate the ether extract from aqueous layer, wash with water (~5-7 mL) and remove ether in a rotary evaporator. The residue contains nicotine. Note: Boiling point of nicotine is 247°C.

4.11 Isolation of DNA (Deoxyribonucleic acid) from Onion Advanced Experimental Organic Chemistry

4.11

363

ISOLATION OF DNA (DEOXYRIBONUCLEIC ACID) FROM ONION

DNA is storehouse of genetic information and carries the code for protein synthesis. It is a poly nucleotide consisting of two long complementary chains of nucleotides which run antiparallel and are held together by strong hydrogen bonding between the base pairs. A nucleotide is a molecule formed from a sugar i.e. 2-deoxyribose, a phosphate and a base (adenine or guanine or cytosine or thymine).

Fig. 4.2: Bases present in DNA

Fig. 4.3: A Nucleotide

Fig. 4.4: Hydrogen bonding between base pairs in DNA

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Procedure Keep a clean empty beaker (250 mL) in an ice bath or a refrigerator for cooling for later use. Cool ethanol (80 mL) in another beaker Cut a white onion into small cubes (~3 mm in size) and take 60 g of it in a beaker (500 mL). Add homogenising medium* (100 mL) to it and incubate at 52°C for 15 minutes in a hot water bath. While incubating, press the onion pieces against the inner sides of the beaker with a spoon spatula. Quickly cool the solution to 10-14°C in an ice bath, not taking more than 5-6 minutes to attain this temperature. Filter the cooled homogenate through the cheese cloth (four folds) into the chilled beaker (250 mL) in the ice-bath. Add ice- cold ethanol (80 mL) into the chilled beaker (from sides) containing the DNA until a stringy DNA precipitate appears, Wind the stringy DNA by slowly rotating a clean glass rod in the solution containing precipitated DNA. The DNA collects on the glass rod. Take out the rod and either uncoil the DNA on a watch glass or store in a small vial filled with ethanol (50%). Notes: • Make the homogenising medium by diluting 6.5 mL of a liquid detergent with water (93.5 mL) and add 1.4 g of sodium chloride to it and mix well to dissolve. • Wear gloves while performing the experiment to prevent hydrolysis of DNA by the enzymes present on the hand. • Keep for cooling an empty beaker (250 mL), a measuring cylinder and ethanol before starting the experiment.

4.12

ISOLATION OF CITRIC ACID FROM LEMON JUICE

Citric acid (2-hydroxy-propane -1, 2, 3 tricarboxylic acid) occurs in lemon and citrus fruits. It is an important intermediate in citric acid/Kreb’s cycle. Its isolation from lemon juice is described below. It is commercially obtained by action of Aspergillus niger on molasses and hydrolysed corn syrup etc. Citric acid is commercially used as a food preservative and flavouring agent.

Procedure Extract the juice of two lemons. Filter to remove the suspended solid. Take the clear filtrate in a beaker and add a solution of calcium hydroxide till no more precipitation occurs. A white precipitate of calcium citrate is obtained. Filter and dry the precipitate, take it in a beaker. Add concentrated sulphuric acid carefully with stirring to the precipitate. Calcium sulphate is precipitated and citric acid is obtained in the filtrate. Filter the calcium sulphate and concentrate the clear filtrate. Cool the concentrated solution to obtain crystals of citric acid. Filter, dry, weigh and determine melting point. m.p. 153°C.

4.13

ISOLATION OF b-CAROTENE FROM CARROTS

b-carotene, also known as provitamin-A or beta, beta-carotene is a naturally occurring, water insoluble tetraterpenoid (Mol. formula C40H56)

4.13 Isolation of b-carotene from carrots Advanced Experimental Organic Chemistry

365

It is orange in colour, melts with decomposition between 176-184°C and a highly unsaturated hydrocarbon. Carrots, sweet potatoes, dark green leafy vegetables like spinach and sweet red peppers etc. are some good natural sources of b-carotene.

b-carotene is converted to retinol (vitamin A) and is therefore commonly known as provitamin A. b-Carotene helps in vision, growth and development and immune system functioning. It possesses antioxidant properties and helps in maintaining healthy skin. In addition to b-carotene, carrots also contain a-carotene and lutein. Lutein is an oxygenated a-carotene containing two hydroxyl groups.

b-Carotene is extracted from carrots either using a traditional method of solvent extraction or using supercritical fluids like supercritical carbon dioxide. Solvents like hexane, acetone, chloroform, petroleum ether and ethanol etc. can be used. Procedure Take carrots (20 g, chopped and freeze dried), add hexane: acetone (100 mL, 6 : 4 v/v) and stir vigorously for 5-10 minutes. Filter, wash the residue with acetone (2 × 25 mL) and hexane (2 × 25 mL). Mix the washings and filterate and remove the solvent in a rotary evaporator. Dry the residue in a vacuum desiccator. Residue contains mainly b-carotene, a small amount of a-carotene and no Lutein. OR Wash and clean 25 g of carrot. Cut into small, thin slices and spread these on sheets of filter paper. Allow the slices to dry for two days. Powder the slices and place in a one litre round bottomed flask. Add ~ 25 mL of carbon tetrachloride, fix a water condenser on the flask and reflux for ~5 minutes on a heating mantle or an electrically heated boiling water bath. Filter the yellow carbon tetrachloride containing b-carotene, add 10 mL more of carbon tetrachloride to the residue and reflux again for five minutes as described above. Do another extraction of the residue with ~ 10 mL of carbon tetrachloride. Combine all the three extracts (~ 45 mL) and wash with ~ 200 mL water in a separating funnel. Pour out the lower organic layer in a dry conical flask (100 mL) and add ~ 50 g of anhydrous sodium sulphate. Keep for overnight, filter and distil carbon tetrachloride in a rotary evaporator. Take the residue in a china

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dish, add 2-3 mL benzene and heat on an electrically heated water bath in a fume chamber to evaporate last traces of carbon tetrachloride and benzene. Notes: • b-Carotene can also be extracted with ethanol (Boiling 20 g of dried and chopped carrots with 2 × 100 mL boiling ethanol for 2 × 30 minutes) or supercritical carbon dioxide. • lmax 451 nm in hexane for b-carotene. • lmax 447 nm for a-carotene.

Questions Q.1.

Write a short note on importance of natural products.

Q.2.

What are the uses and adverse effects of cholesterol on human body?

Q.3.

Why do gall bladder stones contain cholesterol?

Q.4.

Why is lycopene coloured?

Q.5.

Which of the following compounds will decolourise bromine and potassium permanganate solution

Q.6.

Which of the following will give a colouration with ferric chloride

Lycopene, Limonene, caffeine and piperine? Eugenol, lycopene and cholesterol? Q.7.

Name a few natural sources of Citric acid, Nicotine and Limonene.

Q.8.

What are alkaloids? Nicotine belongs to which class of alkaloids?

Q.9.

What is supercritical carbon dioxide? Why it is preferred for decaffeination of tea and coffee?

Q.10. What methods are used for isolation of limonene from citrus peel.

5 Estimations 5.1 ESTIMATION OF AMINES Amines can be estimated by a number of methods out of which two are discussed in the following section • Acetylation method • Bromination method

5.1.1

Acetylation Method

Theory This method can be used for estimation of both aliphatic and aromatic primary and secondary amines. Tertiary amines cannot be estimated by this method. A known weight of an amine containing a replacable hydrogen on the nitrogen is treated with excess but an exact amount of acetic anhydride in presence of pyridine. The hydrogen on nitrogen is substituted by an acetyl group. After the completion of reaction, excess of acetic anhydride is hydrolysed to acetic acid which is estimated by titration with a standardised solution of sodium hydroxide. (Exp. 1)

Some diacetyl derivative may also be formed for primary amines that undergoes deacetylation during refluxing with water to the monoacetyl derivative

Excess acetic anhydride is hydrolyed to acetic acid

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The acetic acid obtained is estimated volumetrically with standerdised sodium hydroxide CH3COOH + NaOH ——→ CH3COONa + H2O A simultaneous experiment is performed in which exactly same amount of acetic anhydride that is used above is hydrolysed to acetic acid, which is also titrated (blank titration) with the standardised sodium hydroxide. (Exp.2) The difference in volume of sodium hydroxide in the second titration (Exp.2) and the first titration (Exp.1) gives the amount of acetic anhydride that reacted with a known weight of amine and hence the strength of given amine solution can be calculated. Procedure Requirements Freshly distilled amine (aniline) 0.5 g Acetic anhydride 20 mL Dry pyridine 20 mL Oxalic acid (standard solution) N/2, 100 mL Sodium hydroxide N, 200 mL Phenolphthalein • Prepare a standard solution (N/2) of oxalic acid by weighing exactly 3.1 g of oxalic acid and dissolve in 100 mL. • Take sodium hydroxide (1N) in a burette and titrate with oxalic acid using phenolphthalein as indicator. • Weigh exactly the given amine (0.5 g) in a round bottom flask (a) and reflux it with acetic anhydride and pyridine (10 mL, 3:1 accurately) for one hour. • Take acetic anhydride and pyridine (10 mL 3:1 exactly) in another round bottomed flask (b) and reflux it for one hour. • Add distilled water (20 mL) through the condenser to each of the flask, shake well and reflux the contents of each flask for five minutes. • Allow each flask to stand for ten minutes. • Titrate the contents of each flask separately with standardised sodium hydroxide. Observations and calculations Weight of the sample = wg Normality of sodium hydroxide = N Volume of NaOH used in blank = V1 mL Volume of NaOH used for titration with the sample = V2 mL (V1–V2) of N NaOH = Acetic acid used for acetylation of sample 100 mL of 1N NaOH = 1g mole of NaOH = 1g mole of acetic acid = One –NH2 group in amine V1 -V2 V1–V2 of IN NaOH = NH2 groups 1000

5.1 Estimation of Amines Advanced Experimental Organic Chemistry

wg of sample contains

V1 -V2 (–NH2) groups 1000

1 mole of Amine (M) will contain −

5.1.2

369

( V1-V2 ) × M 1000

w

(–NH2) groups

Bromination Method

Theory This method can be used only for the estimation of aromatic amines. It is based on the fact that aromatic amines contain a strong electron donating amino group, which increases the electron density of the aromatic nucleus making it prone to electrophillic attack by bromine.

Fig. 5.1: Canonical forms of aniline showing high electron density at o- and p- positions

Such amines undergo bromination reaction (an electrophillic substitution reaction) at o- and p- position when treated with excess of bromine or a brominating agent which produces bromine in situ. For example potassium bromide, on reaction with an oxidising agent like potassium bromate in presence of hydrochloric acid produces a quantitative amount of bromine. The liberated bromine then reacts at activated o- and p-position to give the corresponding polysubstituted bromo derivative. For example aniline gives 2,4,6-tribromo aniline, p-toluidine produces 2,6- dibromotoluidine on bromination.

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Estimations

Excess of unreacted bromine is treated with potassium iodide, corresponding equivalent amount of iodine that is liberated is titrated with the standardised sodium thiosulphate solution.

A blanck titration is simultaneously performed to know the total amount of bromine liberated from the potassium bromide + potassium bromate + hydrochloric acid used in the above reaction with the amine. The difference in the titre value of blank and the reaction mixture gives the amount of bromine that reacted with the given amount of the amine and hence the strength of amine can be calculated Procedure Potassium bromate = 2g   Dissolve to make 1 litre solution Potassium bromide = 13g  Potassium iodide = 10% Sodium thiosulphate

= N/20

Standard dichromate solution

= N/20

Given aniline solution Standardisation of sodium thiosulphate • Prepare a standard solution of potassium dichromate (N/20) by weighing an exact amount (equivalent weight= 49), ~ 0.21 g in 100 mL. • Take given sodium thiosulphate solution (~N/20) in the burette. • Pipette out 10 mL of the dichromate solution in a conical flask. • Add dilute sulphuric acid ( N, 10 mL) and 5 mL (5%) potassium iodide solution • Cork the flask and keep in dark for 5 minutes.The solution becomes dark yellow. • Titrate with sodium thiosulphate to get a pale yellow solution. • Add 10 drops of starch solution and continue titration to get a clear light green solution. Note down the volume of thiosulphate used (V). • Repeat to get three concordant values. Titration of brominated amine (aniline) solution • Take 20 mL of given aniline solution in an iodine flask. • Add 20 mL of potassium bromate + bromide solution, 20 mL water and 3 mL of concentrated hydrochloric acid (use a burette for measuring conc. hydrochloric acid)

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• Stopper the flask and allow it to stand for 10 minutes with occasional swirling. A white precipitate of 2,4,6-tribromoaniline separates out and the solution becomes yellow due to the presence of unreacted bromine. • After exactly 10 minutes, add 10 mL, 10% solution of potassium iodide. • Stopper the flask, allow it to stand for 10 minutes, shaking occasionally during this period. • Rinse the stopper and the neck of the flask with 5 mL water into the flask. • Titrate the solution in the flask with sodium thiosulphate till a pale yellow solution is obtained. • Add 10 drops of starch solution and titrate with shaking to a colourless solution containing white precipitate. Note down the volume of thiosulphate (V2) used.

• Repeat to get two concordant readings. Blank titration

• Take 20 mL of potassium bromate + potassium bromide, 20 mL water and 3 mL concentrated hydrochloric acid in an iodine flask. • Stopper the solution and allow it to stand for 10 minutes. • After exactly 10 minutes, add potassium iodide as above, allow to stand for 10 minutes and titrate with thiosulphate as described above. • Volume of thiosulphate (V1). Observations and Calculations Normality of standard pot. dichromale = N Volume of dichromate solution = 10 mL Volume of thiosulphate used for 10 mL dichromate = V N×10 Normality of thiosulphate (N1) = V Volume of thiosulphate used in blank titration = V1

Volume of thiosulphate used with solution containing aniline = V2

No. of bromine atoms that reacted with aniline = 6 (three molecules of Br2/molecule of aniline)

Molecular weight of aniline = 93

V1–V2 of (N1) thiosulphate ≡ V1–V2 of (N1) bromine which reacted with aniline ≡ V1–V2 of (N1) of iodine

≡ mass of aniline of Aniline 6 (Because 6 atoms of bromine reacted with one molecule of aniline) 1000 × N1 thiosulphate ≡ 1000 × N1 of I2 ≡ 1000 × N1 of Br2 ≡ V1–V2 of N1 thiosulphate ≡

93 of aniline 6

(V1 -V2 )×N1 × 93 of aniline 6 × 1000

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Estimations

Strength of aniline solution =

(V1 − V2 )×N1 × 93 1000 × g/litre 6 × 1000 20

=

(V1 -V2 )×N1 × 93 g/litre 6 × 20

Notes: • Dissolve KBrO3 first in water in a one litre standard flask and then add KBr to the same flask. Dissolve and finally make upto the mark. • The strength of any other amine in a given solution which can undergo quantitative reaction with bromine can be determined. For example for p-toluidine Mol. weight = 107 No of bromine atoms react with a mole of p-toluidine = 4 (not 6 as in aniline)

5.2 ESTIMATION OF AMIDES Percentage purity of a given sample of amide Theory A known weight of the amide is refluxed with an excess of known volume of standardised sodium hydroxide solution. The unreacted sodium hydroxide is then titrated with a standardised hydrochloric acid. A blank titration is simultaneously performed with exactly the same volume of sodium hydroxide. Thus the amount of sodium hydroxide used for hydrolysis of the given amide and hence the percentage purity is calculated. RCONH2 + NaOH —→ RCOONa + NH3 ↑ NaOH + HCl —→ NaCl + H2O Procedure Reagents Given amide 0.5 g Ethanol ~ 20 mL Sodium hydroxide solution 100 mL (2N) Hydrochloric acid 100 mL (2N) • Weigh accurately ~0.5 g of given amide and take it in a clean round bottomed flask. • Dissolve it in minimum amount of ethanol. • Add 20 mL of 2N sodium hydroxide to it. • Reflux the solution for two hours on a sand bath to completly hydrolyse the amide. • Remove the condenser, transfer the solution from the round bottomed flask to a conical flask, rinse the round bottomed flask three to four times with water (5 × 4 mL) and the condenser once with water. Collect the washings in the conical flask. • Titrate the entire alkaline solution in the conical flask with standardised hydrochloric acid using phenolphthalein as an indicator.

5.3 Estimation of Phenols Advanced Experimental Organic Chemistry

373

Blank titration • Take 20 mL of 2N sodium hydroxide in a round bottomed flask. • Reflux for two hours. • Transfer the solution to a conical flask and wash the round bottomed flask and the condenser with water as above and collect the washings in the conical flask. • Titrate the alkaline solution as above. Observations and Calculations Weight of sample = w.g Volume of hydrochloric acid used for titration with the sample = V1 mL Volume of hydrochloric acid used in blank titration = V2 mL Normality of hydrochloric acid = N M 1000 mL of 1N HCl ≡ 1000 mL of 1N NaOH ≡ g of sample (where M is molecular weight of n given amide and n is the number of –CONH2 groups present in it) (V2–V1) mL of N HCl ≡ % purity =

M V2 − V1 × ×N of sample n 1000 100 M V2 − V1 × ×N× n 1000 w

5.3 ESTIMATION OF PHENOLS Theory Phenols react with potassium bromate + potassium bromide in presence of conc. Hydrochloric acid to form poly substituted bromo derivatives. For example phenol reacts to form 2,4,6-tribromophenol.

Some 2,4,6-tirbromophenol bromide is also formed in the reaction

However, it gets converted to 2,4,6-tribromophenol in presence of hydrochloric acid and potassium iodide present in the reaction mixture

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Theory same as described for aniline in section 5.1.2. Reagents Phenol solution Same as in aniline Procedure Same as in aniline (section 5.1.2) except that after addition of phenol + bromate bromide + conc. HCl, allow the reaction mixture to stand for thirty minutes (and not ten minutes as in the case of aniline). Observation and Calculation Strength of phenol =

V1 − V2 × N1 × 94 g/litre (Molecular weight of phenol = 94) 6 × 20

5.4 ESTIMATION OF REDUCING SUGARS Theory Glucose and other reducing sugars can be estimated volumetrically by titration with a solution of fehlings solution containing an equal amount of fehling solution A and B. Fehling solution has to be first standardised with a standard glucose solution. Fehling solution A and B are stored separately and mixed just before use. This has to be done so because fehling A is a solution of copper sulphate containing a small amount of sulphuric acid (which is added to prevent hydrolysis of copper sulphate). Fehling solution B contains sodium potassium tartarate in a strong sodium hydroxide solution. If these two are mixed and stored, copper hydroxide gets precipitated, which does not react fast enough with glucose for estimation. The copper tartarate complex obtained by mixing fehling A+B keeps Cu2+ in solution but is available for reaction with glucose and is reduced to Cu1+ which gets precipitated as cuprous oxide. See chapter I, section 1.4.2 for the reaction between glucose and fehling solution. Procedure Reagents Glucose solution = 100 mL each (given and standard) Fehling solution A 100 mL Fehling solution B 100 mL • Weigh accurately ~1.25g of pure glucose (AR grade) and make its 100 mL solution in a 100 mL standard flask. • Measure using a burette 100 mL each of fehling solution A and B and mix well.

5.5 Saponification Value of a Fat or an Oil Advanced Experimental Organic Chemistry

375

• Take the mixed fehling solution in a burette and take 20 mL out of it in a conical flask and boil it. • Take standard solution of glucose in a burette, note the initial reading and add 1 mL of it to boiling fehling solution in the conical flask. • A brick red precipitate of cuprous oxide is formed and the supernatant solution is blue in colour. • Continue boiling the fehling solution and keep adding glucose solution, 1 mL each time till the supernatant solution is colourless. This titration gives an approximate volume of glucose that reacts with 20 mL of fehling solution. • Now repeat the above titration to obtain the exact volume. • Perform rough and final titration with the given glucose solution. Observations and Calculation Weight of glucose in standard solution = wg (in 100 mL) Volume of Fehling A + B used in each titration = 20 mL Volume of standard glucose solution = V1 mL Volume of given glucose solution = V2 mL 20 mL of Fehling solution ≡ V1 of standard glucose solution ≡ V2 of given glucose solution wV1 ≡ g of glucose 100 1000 mL of given glucose solution ≡

wV1×1000 100×V2

Strenght of given glucose solution ≡

w×10×V1 g/litre V2

Notes: • Traditionally the Fehling solution was heated on a wire gauze and shaken during titration. It is better to stir and heat the solution using a magnetic stirrer with a hot plate. • Fehling solution should never be sucked with the mouth as copper sulphate is highly poisonous and the solution is strongly alkaline.

5.5 SAPONIFICATION VALUE OF A FAT OR AN OIL Theory Saponification number of an oil or a fat is defined as the number of milligrams of alkali required to saponify Ig of an oil or a fat. Fats and oils are triesters of glycerol with long chain fatty acids. The fatty acids normally contain an even number of carbon atoms (ranging from 12-20) and could be either saturated or unsaturated. Edible oils like olive, almond, sunflower, ground nut, coconut etc are all triglycerides.

376

Chapter 5

Estimations

Soap is obtained by alkaline hydrolysis of triglycerides. Saponification value is determined so that just the right amount of alkali is added during the commercial production of soap in order to ensure that the end product does not contain excess of unreacted alkali.

Procedure Reagents Oil/fat 1g Alcoholic KOH 200 mL (N/2) Hydrochloric acid 100 mL (N/2) • Prepare a standard oxalic acid solution (N/2,100 mL). • Standardise the given solution of potassium hydroxide (~N/2) and then using the potassium hydroxide, Standardise the given hydrochloric acid solution (~N/2). In both titrations use phenolphthalein as the indicator. • Weigh accurately ~0.5g of given oil or fat in a clean, dry round bottomed flask. • Add 50 mL of standardised alcoholic potassium hydroxide solution to it. • Reflux the contents of the flask till the oily layer disappears (l-2hrs). • Cool and transfer quantitatively into a conical flask. • Titrate the solution with a standardised hydrochloric acid using phenolphthalein as an indicator. • Perform a blank titration using 50 mL of standerdised potassium hydroxide. Observation and Calculations Weight of oil/fat = wg Volume of hydrochloric acid used (i) Blank titration = V1 (ii) With the sample = V2 1000 mL of 1NHCl ≡ 1000 mL of in KOH ≡ 56g of KOH 56 × (V1 − V2 ) × N g of KOH V1–V2 mL of N HCl = 1000  56×(V1 − V2 )N  Saponification value (S) =   mg/g w  3×56×1000 Average molecular weight of fat/oil = S Notes: Weigh 2g of potassium hydroxide pellets and dissolve in 100 mL of 95% ethanol. Keep the solution overnight and filter any suspended particles before use.

5.6 Determination of Iodine Number of a Fat or an Oil Advanced Experimental Organic Chemistry

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5.6 DETERMINATION OF IODINE NUMBER OF A FAT OR AN OIL Theory Iodine number is defined as the number of grams of iodine that reacts with 100 gram of an oil or a fat. It is a measure of unsaturation present in an oil or a fat. Oils containing high degree of unsaturation have high iodine number. The iodine number of fats containing only saturated fatty acids is zero. The given fat or oil (wg) is treated with known volume (V) of a solution of ICl (known as Wij’s solution), it adds on to the double bonds present in oil and fat

The excess unreacted ICl is treated with KI, the equivalent amount of iodine thus liberated is then titrated with standardised sodium thiosulphate solution ICl + KI —→ KCl + I2 I2 + 2Na2S2O3 —→ Na2S4O6 + 2 NaI Simultaneously a blank titration is perfomed to know the total volume of iodine evolved from the V mL of ICl solution. The difference between the two sets of titration gives the amount of iodine that reacts with wg of fat/oil. Procedure Reagents Given sample = (2g in 100 mL of CCl4, ~2.0% solution) ICl solution (Wij’s solution) = Dissolve iodine (13 g) in one litre of glacial acetic acid. A dark brown solution is obtained, pass dry chloride gas into it till an orange yellow solution is obtained) N Standard potassium dichromate solution = 10 Sodium thiosulphate solution = N/10 Potassium iodide = 10% • Standardise sodium thiosulphate solution with standard dichromate solution. • Measure exactly 10 mL of the given fat/oil solution in a iodine flask with a stopper. • Add iodine monochloride solution (25 mL), stopper the flask and keep it for one hour away from direct light. • Rinse the stopper and neck of flask with water (~ 5 mL) into the flask. • Add potassium iodide solution (10 mL). • Titrate the solution with standardised thiosulphate solution to obtain a light yellow solution. • Shake vigorously during the titration so that all iodine passes into aqueous layer, add a few drops of starch and continue to titrate till a colourless solution is obtained. Let the volume of thiosulphate used = V1 • Perform a blank titration to get the volume of thiosulphate = V2

378

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Estimations

Observation and calculations Amount of oil in 100 mL of given solution = wg Normality of thiosulphate = N Volume of thiosulphate used for iodine used by sample = V2–V1 1000 mL of 1N Na2S2O3 ≡ 1000 mL of 1N iodine ≡ 127 g of iodine V2–V1 of N Na2S2O3 ≡ V2–V1 of N iodine ≡ Iodine number =

( V2 − V1 ) × N × 127 1000

g of I2

(V2 − V1 ) × N × 127 100 g × 1000 w

5.7 ESTIMATION OF AMINO ACIDS Theory Amino acids contain one or more basic groups (–NH2) and acidic group/s (–COOH). They exist in an equilibrium mixture of the following forms

The percentage of each form depends on the pH of the solution. At a particular pH (known as the isoelectric point which is specific to each amino acid), the amino acid exists only as a zwitter ion. The amino acid solution is estimated volumetrically, which is carried out by first reacting the amino acid with formaldehyde solution, and then titrating the –COOH group with a standard base solution. Formalin reacts with the amino group thus preventing it from participation in the acid-base titration H 2 NCH 2 COOH + HCHO → CH 2 = NCH 2 COOH + H 2 O or

H 2 NCH 2 COOH + HCHO → HOCH 2 − NHCH 2 COOH

Procedure Reagents Given glycine solution

100 mL, N/10

Sodium hydroxide solution

100 mL, N/10

Formalin solution Procedure • Prepare a standard solution of glycine (accurately weigh 0.8g of glycine and dissolve in water to make 100 mL solution) • Pipette out 20 mL of the standard glycine solution in a conical flask, add 1-2 drops of phenolphthalein to it.

5.8 Equivalent Weight of Carboxylic Acids Advanced Experimental Organic Chemistry

379

• Titrate with standardised sodium hydroxide solution till a faint pink colour is obtained. • Add 10 mL of neutralised formalin solution (the solution becomes colourless) continue to titrate with sodium hydroxide till a pink colour is obtained. • Repeat to get two concordant values. • Repeat the titration with the given glycine solution. Note: *Neutral formalin solution is prepared by adding a drop of phenolphthalein to it followed by addition of dilute sodium hydroxide with shaking to get a very light pink colour. (i) Calculations-strength of glycine Weight of glycine in 100 mL = wg Volume of sodium hydroxide used for standard solution of glycine (20 mL) = V1 Volume of sodium hydroxide used for 20 mL of given glycine solution = V2 V1 mL of NaOH solution ≡ 20 mL of standard w glycine = g 5 w V V2 mL of NaOH soultion ≡ 20 mL of unknown glycine ≡ × 2 g of glycine 5 V1 Strengh of glycine ≡

w V2 1000 × × g/litre 5 V1 20

(ii) Calculation-percentage purity of the given sample % purity =

M (V1 − V2 ) N 100 × 20 w

5.8 EQUIVALENT WEIGHT OF CARBOXYLIC ACIDS Equivalent weight of carboxylic acid is defined as the number of grams of acid required to neutralise one litre of 1N alkali. It can be measured • Volumetrically by titration with a standardised alkali solution • Gravimetrically by precipitation as silver salt of the acid. 5.8.1

Volumetric Estimation

Theory Carboxylic acids are acidic in nature, because the conjugate base resulting after the loss of the acidic proton is more stable than the corresponding acid.

Fig. 5.2: Resonating structures of carboxylate ion

A solution of a carboxylic acid in water or acid free alcohol can be estimated by titration with a standardised base. Equivalent weight can be calculated and if the basicity of the given acid is known, its molecular weight can also be calculated.

380

Chapter 5

Estimations

RCOOH + NaOH ——→RCOONa + H2O

Reagents Standard oxalic acid solution Sodium hydroxide Given acid solution

100 mL (N/2) 100 mL (~N/2) 100 mL

Procedure • Prepare a standard solution of oxalic acid (N/2) • Standardise (~N/2 ) solution of sodium hydroxide with it using phenolphthalein as an indicator. • Weigh accurately ~1.5-2g of given acid, dissolve in water or acid free alcohol and make the solution up to 100 mL in a standard flask. • Pipette 20 mL of the given acid solution, add a drop of phenolphthalein and titrate with the standard alkali till a light pink colour is obtained • Repeat the titration to get two concordant values Obsrevations and calculations Weight of carboxylic acid in 100 mL = wg Volume of the carboxylic acid = 20 mL Volume of alikali used = V1 mL Normality of alkali obtained from titration with standered oxalic acid = N w V1 mL of NaOH ≡ 20 mL of N carboxylic acid ≡ g of acid 5 w 1000 1000 mL of IN NaOH ≡ × g of acid 5 V1 N w × 1000 5 × V1 N Molecular weight = equivalent weight × basicity

Equivalent wieght of acid =

5.8.2 Gravimetric Estimation Theory A carboxylic acid as its ammonium salt (obtained by treating the carboxylic acid with ammonium hydroxide) is treated with a solution of silver nitrate, the precipitated silver salt of carboxylic acid is filtered, washed, dried and incinerated. The residue containing silver corresponding to the amount of acid used is weighed, from which the equivalent weight is calculated

Reagents Given acid Ammonium hydroxide Silver nitrate solution

1g 10%, ~ 20 mL 5%, ~ 20 mL

5.9 Estimation of Ester Group Advanced Experimental Organic Chemistry

381

Procedure • Suspend 0.5-lg (accurately weighed) of the given acid in ~20 mL of deionised water in a beaker. • Add 10% ammonium hydroxide solution slowly with stirring till the acid dissolves completely. • Warm the solution in the beaker to expel excess of ammonia. • Cool and add silver nitrate solution with continuous stirring till no more precipitation occurs. • Filter the solid through a Whatmann filter paper No. 41. • When the filtration is complete, wash the precipitate with water (3-4 times, 5 mL × 4). • Cover the dessiccator with a carbon paper. Transfer the filtered silver salt into a porous dish and keep in the dessiccator till the silver salt is completely dry. • Weigh accurately (0.5g) of the dry silver salt and incinerate in a preheated constant weight silica crucible. • Heat, cool and weigh till a constant weight of crucible with the precipitate is obtained. Observations and calculations Weight of the dry silver salt of the carboxylic acid = wg Weight of silver obtained from wg of the acid = w1g Equivalent weight of the acid = Z Eq. wt of the silver salt = Z – 1 + 107.86 107.86 of silver is present in

w × 107.86 g of Ag salt w1

hence

w × 107.86 = Z – 1 + 107.86 w1  w × 107.86  Z=  − 106.86 w  1

Mol. wt = Z × basicity

5.9

ESTIMATION OF ESTER GROUP

The method used is similar to that used for determination of saponification value of an oil or a fat. A known weight of ester is refluxed with a known volume of standardised alkali solution. The amount of unreacted alkali is then titrated with a standard acid solution. A blank titration of the total alkali added in hydrolysis of ester is simultaneously performed. The difference in the value of alkali in blank and titration of the sample gives the amount of alkali used for hydrolysis of the ester. RCOOR ′ + NaOH → RCOONa + R ′OH NaOH + HCl → NaCl + H 2 O ( Excess)

Reagents Ester Alcoholic/Aqueous potassium hydroxide Sodium carbonate Hydrochloric acid

0.5 g 100 mL (0.5 N) 100 mL (0.5 N) 100 mL (0.5 N)

382

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Estimations

Procedure • • • • • • • •

Standardise the given hydrochloric acid by titration with a standard solution of sodium carbonate. Weigh accurately ~0.5 g of ester in a clean round bottomed flask. Add 25 mL of the potassium hydroxide solution. Reflux the reaction mixture for one hour on a wire gauge if aqueous alkali has been used and a water bath if alcoholic alkali has been used. Cool and transfer the reaction mixture into a conical flask . Wash the condenser with water (5 mL) and the round bottom flask three times with water (2-3 mL each), collecting all the washings into the conical flask. Titrate the alkaline solution with standardised acid using phenolphthalein as the indicator, disappearance of the pink colour is the end point. Reflux 25 mL of alkali for one hour, transfer to a conical flask, wash as above and titrate with acid solution as described above.

Observations and calculations Weight of ester = wg Volume of HCl (i) Blank = V1 (ii) with the ester = V2 Volume of alkali used for hydrolysis of ester = V1 – V2 Normality of HCl = N V1–V2 of N HCl ≡ V1 – V2 of N KOH (V1 – V2) mL of N KOH ≡ wg of ester 1 g equivalent of KOH is required for = Saponification value =

w × 1000 g of ester (V1 − V2 ) × N w × 1000 (V1 − V2 ) × N

Notes: • Dissolve 1.25g of potassium hydroxide in water and make up to 100 mL in a standard flask. Make a fresh solution for the experiment. • Dissolve 1.25g of potassium hydroxide in 95% alcohol in a standard flask and make it up to 100 mL. Keep the solution overnight, filter if necessary.

5.10

ESTIMATION OF NITRO GROUP

The nitro group is reduced to amino group by using a known volume of stannous chloride solution and hydrochloric acid. After the reaction is complete (indicated by the disappearance of insoluble layer of the nitro compound). The excess of stannous chloride is estimated by titration with a standardised iodine solution. The same volume of stannous chloride is then titrated with a standardised iodine solution. The difference in the volume of two sets of titration gives the amount of stannous chloride used for the reduction.

5.10 Estimation of Nitro Group Advanced Experimental Organic Chemistry

383

Reagents Given nitro compound 0.5 g Stannous chloride solution; take 12.5 g of pure tin in a round bottomed flask fitted with a condenser. Add slowly 125 mL of conc. hydrochloric acid and warm to start the reaction. After the complete dissoultion of tin, transfer the solution into a 500 mL standard flask, dilute up to the mark with water and shake well. Iodine solution Procedure • Weigh accurately 0.2-0.3g of the nitro compound in a clean, dry round bottomed flask. • Add 50 mL of stannous chloride solution to it. • Reflux the mixture for one hour. • Cool and transfer into a conical flask. Wash the condenser (once) and the round bottomed flask (thrice) with 5 mL water each time and collect the washings in the conical flask. • Titrate the combined solution in the conical flask with iodine solution using starch as the indicator. • Take 50 mL stannous chloride solution in a round bottomed flask, reflux for one hour, cool wash with water as before and titrate as above. Observations and calculations Weight of sample = w Normality of iodine solution obtained by titration with standardised thiosulphate soultion = N1 Volume of iodine used in blank = V1 Volume of iodine used for solution containing the sample = V2 46 1000 mL of I2 (IN) = 1000 mL of 1N SnCl2 = g of NO2 group 6 46 (V1 − V2 ) × N g of NO2 group 6 1000 46 × (V1 − V2 ) × N × 100 100 g will contain = 6 × w × 1000 46 × M × (V1 − V2 ) = 6 × w × 10 If M is the molecular weight of given sample, the number of nitro groups M × N × (V1 − V2 ) in it = 6 × 1000 × w (V1–V2) mL of I2 (IN) =

384

5.11

Chapter 5

Estimations

DETERMINATION OF DISSOLVED OXYGEN (DO) IN A GIVEN SAMPLE OF WATER

Dissolved oxygen (DO) refers to the level of free oxygen present in water. The source of this oxygen is (i) photosynthesis activities occurring under the surface of water or (ii) absorption or diffusion. The dissolved oxygen is expressed in terms of mg/litre and its optimal value should be 4-6 mg/litre. Dissolved oxygen is essential for equatic life, too much of it is not good for industrial water use as the oxygen corrodes the machines due to slow oxidation. A few methods are used for determination of DO, one of which is as described below. (i) Iodometric method or Winkler’s method When Mn2+ salt solution (eg MnSO4) is treated with alkaline potassium iodide in water, Mn(OH)2 gets precipitated. This reacts with dissolved oxygen present in water MnSO4 + 2KOH → K 2SO 4 + Mn (OH)2 ↓ Μn (OH)2 + O → MnO (OH)2

Brown precipitate

On addition of sulphuric acid, Mn(SO4)2 is formed, which is equivalent to the dissolved oxygen that combined with Mn(OH)2. MnO(OH)2 + 2H2SO4 ——→ Mn(SO4)2 + 3H2O The Mn(SO4)2 combines with potassium iodide, liberating an equivalent amount of iodine, which is then estimated volumetrically by titration with sodium thiosulphate solution. Mn (SO 4 ) 2 + 2KI → MnSO4 + K 2SO 4 + I2 I 2 + 2 Na 2S2 O3 → Na 2S4 O6 + 2 NaI Reagents Manganous sulphate solution–obtained by dissolving 50 g of it in 100 mL distilled water. Alkaline potassium iodide solution – 25g KI + 50g KOH in 100 mL distilled water Sodium thiosulphate solution (N/40) Starch – 1% Conc. H2SO4 Procedure • Standardise sodium thiosulphate with standard dichromate solution (N/40) 100 mL. • Collect sample of water in a BOD bottle. Add 2 mL of manganous sulphate solution and 2 mL of alkaline potassium iodide solution and stopper the bottle. • Shake well, a brown precipitate is formed, allow the precipitate to settle. • Add 2 mL of sulphuric acid and shake to dissolve the precipitate. • Titration the whole solution with N/40 thiosulphate solution. • When the solution has aquired light yellow colour, add 4-5 drops of starch and titrate till the solution becomes colourless. • Repeat to get two concordant values.

5.12 Determination of Chemical Oxygen Demand (COD) in a given Sample of Water Advanced Experimental Organic Chemistry

385

Calculations 5 mL of N/40 sodium thiosulphate = 1mg/litre of DO 1 × 40 × V × N

5.12 DETERMINATION OF CHEMICAL OXYGEN DEMAND (COD) IN A GIVEN SAMPLE OF WATER COD is defined as the mass of oxygen consumed (in mg/litre) for the oxidation of organic matter (including biologically degrable and non biological degrable organic matter) present in a sample of water. It is a measure of water quality. Water sample is refluxed with excess of an oxidising agent like acidified potassium dichromate. Some of the dichromate is used for oxidation of the organic matter, the unreacted excess dichromate is then estimated by titration with standard ferrous ammonium sulphate solution. The amount of dichromate used for oxidation of organic matter in water is equivalent to oxygen that would be required for the same purpose. K 2 Cr2 O7 + 6FeSO 4 .NH 4SO 4 + 7H 2SO 4 ↓ NH 4 ) 2 SO 4 + 7H 2 O Cr2 (SO 4 )3 + K 2SO 4 + 3Fe 2 (SO 4 )3 + 6(N Reagents • Potassium dichromate (0.25N)- Dissolve 3.065g of potassium dichromate in 100 mL water. • Ferrous ammonium sulphate - Dissolve 24.5g in minimum amount of water, add 20 mL dilute sulphuric acid shake and make upto 250 mL in a standard flask. • Dry powdered silver sulphate and mercuric sulphate • Ferroin • Conc. sulphuric acid Procedure • Take 20 mL of sample water in a round bottomed flask • Take 20 mL distilled water in another round bottom flask • Add a pintch of mercuric sulphate and silver sulphate to each round bottom flask. • Add 10 mL. of potassium dichromate to each flask. • 30 mL dilute sulphuric acid to each flask. • Reflux the solutions in both the flasks separately for two hours each. Cool both the flasks to room temperature. Transfer the solution to two separate conical flasks (500 mL), wash each round bottom flask thrice with distilled water and transfer the combined washings into the respective conical flask. • Add 2-3 drops of ferroin to each and titrate with standard ferrous ammonium sulphate solution.

386

Chapter 5

Estimations

• End point is change in colour from blue green to reddish brown Volume of ferrous ammonium sulphate for blank = V1 mL Volume of ferrous ammonium sulphate = V2 mL Normality of ferrous ammonium sulphate = N1 Volume of water sample = V3 mL COD =

(V1 − V2 ) N1 × 100 × 8 Volume of sample (V3 )

8 is equivalent weight of oxygen.

5.13

PLOT THE TITRATION CURVE OF GLYCINE

Glycine is the simplest amino acid, out of twenty two amino acids which occur in nature and are the building units of proteins. Glycine and most other amino acids are a-amino acids where as proline and hydroxyl proline are imino acids. The –NH2 group is the basic and –COOH, the acidic group of each amino acid. Many amino acids like glycine, alanine, phenyl alanine, tyrosine etc. contain one each of –NH2 and –COOH group

There are others like glutamic, aspartic acid, lysine, histidine which contain unequal number of these groups as shown below:

5.13 Plot the Titration Curve of Glycine Advanced Experimental Organic Chemistry

387

Glycine, like other amino acids can occur in the following forms

The pH at which an amino acid is present only as a dipolar ion (Zwitter ion) is known as the isoelectric point and is specific to each amino acid. Titration curves can be used to find out the isoelectric point (pl) of any amino acid. A typical titration curve is shown below (Fig. 5.3).

Fig. 5.3

Procedure Reagents Hydrochloric acid 0.1M Sodium hydroxide 0.1 M Glycine solution 0.1M Standard buffers pH (4, 7, 10) • Use a calibrated pH meter or if it is not calibrated use a standard buffer solution to calibrate it. Soak the pH electrode in distilled water while getting other solutions and apparatus ready for the experiment. • Prepare glycine solution (100 mL) in a beaker. Add hydrochloric acid (0.1M) to it until its pH reaches 1.5 (monitor the pH of the solution with the pH meter).

388

Chapter 5

Estimations

• Pipette glycine hydrochloride solution (10 mL) in a clean beaker (100 mL) and add de ionised water (25 mL) to it. • Place the pH electrode in the beaker such that the electrode immerses into the glycine solution and note down the pH of the solution. • Take a clean dry burette, rinse it with sodium hydroxide solution. Now clamp the burette on the stand above the beaker containing amino acid solution and add sodium hydroxide (exactly 1 mL) into the beaker while stirring it continuously. Note down the pH. • Add sodium hydroxide solution to glycine solution in 1 mL increments, mix well and record pH after each addition. • When the pH of the solution starts to change fast, add 0.5 mL sodium hydroxide instead of 1 mL and keep recording the pH after each addition. • Keep adding sodium hydroxide solution till the pH of the solution becomes 12. • Construct a titration curve by plotting volume of sodium hydroxide (x-axis) versus the pH (y-axis) • Calculate (i) pKa of acid and amino group (ii) pl (isoelectric point) of glycine. Note: Titration curves can be obtained for other amino acids following the same procedure.

5.14

EFFECT OF TEMPERATURE ON THE ACTIVITY OF AMYLASE

Enzymes are proteins present in living cells where they act as catalysts for biochemical reactions. The activity of these enzymes is affected by factors like concentration of the enzyme and substrate, pH, temperature of the reaction and presence of inhibitors. Drastic reaction conditions destroy the reactivity of enzymes because these denature the enzyme i.e. affect or destroy the tertiary structure of the enzyme. A group of enzymes that are present in saliva, pancreatic juice or plants which hydrolyse the a-glycosidic linkages of a complex carbohydrate (starch, glycogen) into simpler forms (matose) are known as amylases. There are two types of amylases (i) a-amylase, also known as ptyalin (ii) b-amylase, a principle component of diastase. a-amylase, a digestive enzyme is present in humans and many other animals. b-Amnylase is present in seeds, bacteria, molds and yeast etc. Procedure Requirements Starch weigh 1g starch and dissolve in 100 mL water by heating. Iodine solution: 2.5g of KI and 1.3g of solid iodine in 100 mL water. Amylase: (1) take a 100 mL beaker and collect your own saliva (1-2 mL) by spitting in the beaker, add water (30 mL) and mix well. (2) rinse your mouth with water 2-3 times to collect ~ 100 mL of the saliva solution. (3) make 1% solution of amylase purchased from the suppliers of enzymes: • Set up 5 water baths at different temperatures 10°C, 20°C, 35°C, 50°C and 80°C. • Take 4 mL of 1% starch solution and 3 drops of iodine solution in each in 5 clean and dry tubes.

5.14 Effect of Temperature on the Activity of Amylase Advanced Experimental Organic Chemistry

389

• Take 4 mL of amylase solution in another set of 5 clean and dry test tubes. • Place a set of two tubes one containing starch and other containing amylase in a water bath maintained at 10°C. • Similarly keep a set of two tubes (starch and amylase solution) in other water baths. • Keep the tubes in the water baths for 10 minutes. • Pour the amylase into the tube containing starch and iodine solution. • Place the tube back into the water bath and note the time taken for disappearance of the blue colour. • Repeat the above step with the tubes in each bath. • Plot a graph of time Vs the temperature.

Questions Q.1.

Which out of the following compunds can be estimated by acetylation

Q.2.

Which out of the following amines can be estimated by bromination method

and how many bromine atoms will react with those which can be estimated by this method. Q.3.

In the bromate-bromide method for estimated of amines and phenols, which of the following statements is not true (a) Potassium bromide is oxidised (b) KI is reduced to I2 (c) Starch is added as an indicator before starting the titration (d) Sodium thiosulphate is a primary standard (e) A solution of bromine can also be used for estimation.

Q.4.

Which of the following sugars cannot be estimated by titration with Fehling solution glucose, fructose, sucrose, maltose.

Q.5.

What is colour of Fehling solution A Fehling solution B Mixed Fehling solution A + B.

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Chapter 5

Estimations

Q.6.

Why are Fehling solution A and B stored separately? Why these are mixed just before use?

Q.7.

What do you understand by the term saponification value?

Q.8.

Which of the following triglyceride has the highest and which has the lowest saponification value

Q.9.

Why is the alkaline hydrolysis of ester known as saponification.

Q.10. Why should an acid free alcohol be used for making alcoholic solution of KOH. Q.11.

In the estimation of –NO2 group, what is the role of Sn2+ and what does it get converted to?

Q.12. Why is glycine or any other amino acid treated with formaldehyde solution before titration with a standardised alkali. Q.13. Strength of aniline =

(V1 − V2 ) × N1 × M g /litre ( Z × 2) × 20

M = Molecular weight of aniline Z = 3 (No of atoms of bromine that react with each molecule of aniline) What will be M and Z for the following compounds

Q.14. What do you understand by DO, BOD, and COD? Q.15. What do high values of DO, BOD and COD indicate?

6 Chromatography 6.1 INTRODUCTION This technique of separation of the components of a mixture was invented by a Russian botanist named Mikhail Tuswett in early 1900s. Since the technique was first used for the separation of coloured components of plants, it was called chromatography (chroma means colour and graph means writing). Chromatography is one of the most useful technique for separation of closely related components of a mixture of compounds. The technique is no longer restricted to separation of coloured compounds. Various types of chromatographic techniques have been developed and their main uses are: • Identification of the number of components present in a mixture of compounds. • Separation of components of a mixture of compounds • Purification of compounds • Establishing identity and non-identity of compounds. Principle The separation of components of a mixture is brought about by two immiscible substances. One of which is called the Stationary phase, as it is immobile, the other is called the mobile phase- it is the moving phase. The stationary phase is either a solid or a liquid and the mobile phase is either a liquid or a gas When the stationary phase is a solid, the mixture to be separated is adsobed on it. As the mobile phase passes through, it carries the components of the mixture along with it in the direction of its flow. The separation in this case is based on differential adsorption of the components on the stationary phase. This type of chromatography is called adsorption chromatography. More strongly adsorbed components of the mixture move slowly as compared to the less strongly adsorbed components. If the stationary phase is polar in nature, the components with higher polarity move slowly and components with lower polarity move fast. This is called normal phase chromatography when the stationary phase is non-polar, the components with higher polarity move faster along with the mobile phase as compared to components with the lower polarity. This techique is known as Reverse phase chromatography.

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When both stationary and mobile phases are liquids, the separation of the components is based on their partition coefficients into the individual liquid layers. This type of separation technique is called Partition Chromatography. The separation of components of a mixture using a liquid as stationary phases and a gas as a mobile phase is also based on the principle of partion between the two phases.

6.2

TYPES OF CHROMATOGRAPHY • Based on the prinicple — adsorption chromatography — Partition chromatography — Normal phase — Reverse phase • Based on the technique — Paper chromatography — Thin layer chromatography — Column chromatography — Gas chromatography — High performance liquid chromatography (HPLC) — Ion exchange chromatography.

6.3 PAPER CHROMATOGRAPHY It is one of the simple type of chromatographic procedure in which the separation of components is achieved on a paper, normally Whatmann fillter paper. The water held in the pores of the filter paper is the stationary phase and the mobile phase is a liquid. It is a liquid-liquid type of partition chromatography. Depending on the direction of the flow of the mobile phase and hence the components of the mixture, the paper chromatohraphy is of the following types: (i) Ascending Chromatography (ii) Descending Chromatography (iii) Horizontal/Radial/Circular Chromatography As the name suggests, the solvent moves up against gravity in ascending chromatography. The solvent moves down in Descending Chromatography. Solvent moves horizontally in Radial Chromatography.

6.3.1 Ascending Paper Chromatography Requirements • Whatmann paper strip (2 × 20 cm) • Capillary tubes • Chromatography jar • Sample • Solvent (Irrigant).

6.3 Paper Chromatography Advanced Experimental Organic Chemistry

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Procedure • Take a Whatmann filter paper strip (2 × 20 cm) • Take a clean dry chromatography jar, pour about 20-25 mL of the selected solvent and cover it with the lid. The space in jar should get saturated with the vapours of the solvent. • Take a capillary tube and make two fine jets from it as shown below.

Fig. 6.1: Capillary Jet

• Dissolve ~ 20 mg of the compound in a solvent (1-2 mL). The choice of the solvent depends on the nature of the components of the mixture. • Using the capillary jet, spot the solution on the filter paper strip (2 cm above one end) twice at the same point. Dry the paper after each application. • Now hang the spotted filter paper strip in the chromatography jar such that — The paper strip is vertical — The strip does not touch the sides of the jar — The lower tip of the strip is dipped into the solvent — The spotted portion is not submerged in the solvent • Cover the jar and leave it undisturbed. • The solvent rises up. Allow it to reach near the top end. • Take out the paper strip • Mark the point upto which the solvent has moved (solvent front) • Allow the paper to dry • If the components of the mixture are coloured, coloured spot/s are seen on the chromatogram. The number of coloured spots is the number of components in the given mixture. • If the components are not coloured, then spray the chromatogram with a suitable reagent (visualizer) and find out the number of components. Notes: • The volume of the solvent taken in the chromatography jar obviously varies with the diameter of the jar. • This techniques is used for separation or comparison of the components of a mixture with known samples. • Calculation of retention factor (Rf) and comparision of the calculated value with those given in literature can help identify the components. Though many a times, the results are not reproducible.

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Fig. 6.2: Ascending paper chromatography

The value of Rf for any substance is less than unity. Rf =

Distance moved by component A Distance moved by the solvent

6.3.2 Radial Chromatography In this type of paper chromatography, the solvent moves horizontally along the paper. Requirements • Sample • Solvent • Petridish • Circular Whatmann filter paper. Procedure • Take a petridish. Pour ~ 20-30 mL of the selected solvent, cover the dish and allow it to stand for 15-20 minutes so that the space in the dish is saturted with the vapours of the solvent • Take a circular whatmann filter paper (~ 12 cm in diameter) • Make a small hole in the centre • Spot the sample solution in the centre of the paper so that it spreads uniformally around the hole.

Fig. 6.3: Radial chromatography

6.4 Thin Layer ChromatographY (TLC) Advanced Experimental Organic Chemistry

395

• Make a cotton wick and insert its upper thread into the hole. • Place the circular filter paper on the petridish containing the solvent such that the other end of the wick dips into the solvent and cover with the lid. • The solvent rises through the wick and moves horizontally along the circular filter paper. • When the solvent has moved near the circumference. Take out the paper, mark the solvent front and remove the wick. • Allow the paper to dry. • Coloured circular rings of the components are visible, otherwise spray with the spraying agent to visuallise these. • Calculate the Rf of the components.

6.3.3 Descending Paper Chromatography It is a partition type of chromatography in which the solvent moves in the direction of gravity. The apparatus used is shown in the Fig. 6.4 below:

Fig. 6.4: Descending chromatography

Requirements • Descending chromatography jar • Paper strip • Solvent • Sample Procedure • Spot the paper as described in ascending chromatography • Take ~ 20-30 mL of the solvent in the boat shaped container placed at the top of the jar • Hang the spotted paper from top to down such that the end of the paper dips in the solvent but not spots near this end. • Allow the chromatograph to develope, visuallise spots and calculate Rf.

6.4 THIN LAYER CHROMATOGRAPHY (TLC) It is an adsorption type of chromatography in which the stationary phase is a solid and the mobile phase is a liquid. A glass plate or a thin sheet of a metal is used as a support for the thin layer of the solid stationary phase and hence the name thin layer chromatography. Silica or a alumina are normally used

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as solid support. Surface of these may be modified depending on the type of components to be separated. A fluorescent dye is added to silica gel or alumina or these may be made acidic etc. The size of the solid particles supported on the TLC plate also plays an important role in the effective separation of the components. The choice of the mobile phase is also made depending on the nature of the compounds to be separated. TLC can be used for • • • • •

Knowing the number of components in a mixture of compounds. Preparative TLC can be used for separation of components of a mixture of compounds. The progress of a reaction can be monitored on TLC. The identity of fractions collected in column chromatography can be established by TLC. TLC can be used for checking the purity of a compound.

6.4.1  Identification of Number of Components in a mixture of Compounds Requirements • TLC plate or strip cut out from merck plate • Capillaries • TLC Jar • Sample • Solvent. Procedure • Take a TLC plate • Add ~ 15-25 mL of the selected solvent, (the volume required could be less or more depending on the size of TLC jar) and cover the jar with the lid. (The air in the jar gets saturated with the solvent vapours). • Make a fine jet using a capillary as shown Fig. 6.1. • Dissolve ~ 20 mg of the sample in 1mL of a suitable solvent • Spot the sample twice near one end of the TLC plate using the capillary jet. • Allow the spot/s to dry. • Place the TLC plate in the TLC jar, making sure that the plate does not touch the sides of the jar and the spots are above the solvent layer in the jar. • Cover the jar. • Allow the solvent to rise up. • When the solvent has moved up to almost the upper end of TLC, take out the plate, mark the solvent front and allow it to dry in air. • Coloured spots are visible on the TLC if the components of the mixture are coloured. • If the components are not coloured. Spray the plate with a visualizing reagent. • or view it under a U.V. lamp to see the number of spots (only few types of compounds can be visualised under a U.V. lamp) • The colourless spots become yellow-brown if exposed to iodine vapours. (This is done by keeping the plate in a jar containing some iodine crystals, the compounds adsorb iodine and become yellow-brown).

6.4 Thin Layer ChromatographY (TLC) Advanced Experimental Organic Chemistry

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Notes: • Conventionally, the TLC plates were made by making a slurry of silica gel or alumina and spreading it uniformally on the glass plate, allowing to dry and then drying in a hot air oven. • Now these plates are available commercially. The solid stationary phase is supported on a thin aluminium sheet. The plate can be cut into any size (like a paper) and used in place of a glass coated plate. • A small beaker or a small bottle (fitted with a lid) can be used in place of a jar to hold the solvent.

Fig. 6.5: Thin layer chromatography

6.4.2 Preparative TLC TLC can also be used for separation of components of a mixture on a small scale. The technique used is known as preparative TLC. • The procedure used is same as described above (Sec. 6.4.1) except. • The preparative TLC plate is 20 × 20 cm in size. • Instead of one or two spots, a series of spots are put on the plate as shown below Fig. 6.6. • If the components are coloured, these are visible as coloured bands on the plate after the development of chromatogram. • If the components are not coloured, the plate is either

Fig. 6.6: Preparative Chromatography

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Chapter 6

Chromatography

— Viewed under a UV Lamp. — or sprayed only at one corner to know the location of the bands. • After the band/s have been located, scratch the silica gel of one band (A) with a spatula, collect it on a clean watch glass. If there is another component (B) collect the silical gel on another clean watch glass. • Elute the component of one band as shown in Fig. (6.6) with an organic solvent. Remove solvent to collect A • Similarly collect B.

6.4.3  Monitoring the Progress of a Reaction Procedure TLC can be used to check the completion or the progress of a reaction, specially in those cases where there is no visible change during the reaction. • Starting compound/s and the reaction mixture are spotted on the plate. • The chromatogram is developed as given in section 6.4.1. • The plate is viewed under a UV lamp in an iodine jar or sprayed with an appropriate reagent. • If the reaction mixture shows spots corresponding to starting metarial, the reaction under examination is not complete, otherwise it is complete.

6.4.4 Monitoring Fractions in Column Chromatography Different fraction of eluent collected in column chromatography are compared on TLC. For example if ten fractions (1-10) have been collected, for comparision lets us say fractions 1-4 are spotted on a TLC plate and chromatogram developed following the procedure described in section 6.4.1. If these show a single same spot (same Rf) , these are mixed and concentrated to obtain compound A. If fractions 5 and 6 show two spots, these are discarded 7-10 show same spot on TLC having same Rf value which is different from A, these are mixed, concentrated to obtain compund B. 6.4.5  Checking the Purity of a Compound A pure sample shows a single spot when checked on TLC. 6.5 COLUMN CHROMATOGRAPHY It is a partition type of chromatography in which the stationary phase is a solid and the mobile phase, a liquid. The stationary phase is packed in a glass tube, called the column and hence the name column chromatography. The stationary phase is also known as the adsorbent. Alumina (Al2O3, available as neutral, acidic or basic alumina), silica gel, calcium carbonate, and starch etc. are used as adsorbents. The mobile phase is a liquid and is also known as the eluting solvent. The nature of the eluting solvent plays an important role in the effective separation of the components of the mixture. In general, the eluting solvent should be less polar than the components of the mixture, but the components should be soluble in it otherwise these compounds remain adsorbed on the top of the column and do not move along with the mobile phase. Chromatographic separations are brought about by using a range of solvents. The least polar sovent is first used to elute the least strongly adsorbed components of the mixture and then solvents with increasing polarity and greater eluting power are used for elution of the more strongly adsorbed components. A mixture of solvent may also be used for elution.

6.5 Column Chromatography Advanced Experimental Organic Chemistry

399

Common solvents used in column chromatography

Requirements • A glass column with a stopper • Beaker • Conical flasks (100 mL) • Glass rod rounded at the two ends. • Stationary phase-normally silica gel/alumina. • Eluting liquid • Sample Procedure • Select a column which is suitable for the amount of components to be separated and close the stopper. A thin long column is always preferred to a small broad column. • Add ~ 10-15 mL of the eluting solvent (least polar if a variety of the solvents have to be used). Push a small ball of cotton into the lower end of the column using a long glass rod or a clean metal wire. • Clamp the column and place a clean, dry conical flask under it. • Make a slurry of the soild stationary phase in the solvent that has been already added to the column. • Keep a funnel on the top of the column and pour the slurry with continuous stirring into the column. • Gently tap the column during this to get a uniform packing of stationary phase in the column. • Keep the stopper partially open during the addition of slurry and collect the solvent in the conical flask placed under it. 3 of its length, stop adding the slurry and allow the 4 solvent to elute leaving about half an inch layer of the solvent on the top of the solid phase. (Do

• When the column has been packed to ~

not elute all the solvent. Otherwise the column will dry at the top). • Dissolve the given sample in minimum amount of a solvent in which it is freely soluble. • Pour it on the top of the column and open the stopper to allow excess liquid to elute out again, leave a thin layer of liquid above the stationary phase.

400

Chapter 6

Chromatography

Fig. 6.7: Column Chromatography

• Now add the solvent on the top (if more than one solvent is to be used, add the least polar solvent first and then the solvents with increasing polarity). Open the stopper partially and start collecting the fractions ( the volume of the fraction collected varies and depends on how fast the components are eluting out and could vary form 5 mL to 20-30 mL). • If the components are coloured, their movement can be seen easily. If the components are not coloured, collect the fractions in separate conical flasks. • Compare the fractions by performing TLC (Sec. 6.4.4) • Mix all those fractions which contain same single component, distil the solvent and the residue is one component • Simlarly separate other components.

6.6

GAS CHROMATOGRAPHY

It is based on the principle of partition between a stationary liquid phase and a gaseous moving phase. The components partitioned into gas move out at a faster rate. This technique is basically used to separate

Fig. 6.8: Schematic diagram of Gas Chromatography

6.7 High Performance Chromatography (HPLC) Advanced Experimental Liquid Organic Chemistry

401 401

volatile components of a mixture. The liquid phase is supported on a solid support. The sample is injected into the stationary phase and as the gas is allowed to pass through it at a constant rate, the components of the mixture move at different rate and thus get separated.

6.7 HIGH PERFORMANCE LIQUID CHROMATOGRAPHY (HPLC) It is a modern chromatographic technique, Most instruments use the reverse phase technique for separation and consists of the following: Column: Two types of columns are used (i) A solid column mainly consisting of silica or polymer. (ii) A liquid supported on a solid column like silica or polymer.

Fig. 6.9: HPLC System

The particle size of silica or alumina is an impotant factor in effective separation of the consituents of a mixture. The pump is used to drive the solvent and the sample through the column. A syringe or multi piston pump is used to maintain constant flow rate. Solvent The mobile pahse is a combination of polar and non polar solvents. The choice of the solvent depends on the constituents of the mixture. The solvent should be free of dissolved gases. Detector Various types of detectors are used as there is no universal detector that can be used for all types of compounds. Some common types are (a) U.V. detector (b) NMR detector (c) Refractive index detector (d) Fluorescence detector etc. Applications of HPLC • It is used for both qualitative and quantitative work.

402

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Chromatography

• The technique is effectively used for separation of a vast variety of compounds ranging from amino acids, peptides, proteins, pharmaceutically active compounds, amines, sugars, lipids to simple organic compounds which differ very minutely in structure. Limitations Reverse phase HPLC cannot be used for separation of a few types of compounds. • Inorganic ions • Strong hydrophobic compounds • Polysaccharides • Polynucleotides etc

6.8 ION EXCHANGE CHROMATOGRAPHY It is a method used to separate a mixture of ions or polar molecules. The stationary phase is a column consisting of non reactive metarial beads which have been functionalised with covalently bound functional groups depending on whether a cation exchange is required (cation exchange resin) or an anion exchange is to be carried out (anion exchange resin) The mobile phase is a bufferred aqueous solution, the pH of which is maintained or changed depending on type of compounds or ions to be separated. The cation exchange resin is functionlised with a groups like—SO3– or –COO– (X– in the equation below + + − −+ + − Resin −X − C + MB → Resin − XM + C + B + −

+

The cation of the molecule (M of M B) is held on the resin. How strongly M+ binds to the resin depends on its charge. If a mixture of two ionic compounds M+1, B– and M2+ B– is paused through an ion + exchange resin and if M1, has more affinity for resin, it will be held more strongly on the resin. M 2 will elute first. If an ion gets strongly held to the resin, it may be eluted with a strong electrolyte solution like NaCl, Na+ will displace M+, which will be eluted out. The anion exchange resin is used to separate anion + −

+ −

+ −

+

−

Resin − X A + M B → Resin − X B + M + A

6.9 APPLICATIONS 6.9.1 Chromatography of Amino Acids Amino acids are a class of naturally occurring organic compounds and are the building blocks of proteins. There are in total twenty two naturally occurring amino acids. All except two are a-amino acids (containNH2 and - COOH groups) Proline and hydroxy proline are imino acids

6.9 Applications Advanced Experimental Organic Chemistry

403

A mixture of amino acids are produced on hydrolysis of proteins. These can be separated and identified by various chromatographic techniques. TLC and paper chromatography can be used for finding out the number of amino acids in a mixture (containing may be 2-3 components) and by comparing with known samples, the identity of these components can be established. Calculation of Rf of an amino acid under specified conditions helps to identify some amino acids. Amino acids are colourless compounds and hence are visualised on a chromatogram by spraying with visualising agents. Ninhydrin (2,2-Dihydroxyindane-1,3-dione) is one of the most commonly used visualising reagent. It gives purple colour with all amino acids except proline and hydroxy proline.

Rf values of some common amino acids on a silica gel coated plate using 1- Butanol: Acetic acid: water = 3:1:1 Amino acid Rf Alanine 0.30 Aspartic acid 0.24 Leucine 0.61 Isoleucine 0.53 Glycine 0.25 Methionine 0.51 Phenylalanine 0.62 Tryptophan 0.61 Valine 0.44 Notes: • Rf values are not always exactly reproducible. • A mixture of amino acids (two or three) given for practice in the class should have amino acids with Rf quite different from each other.

404

Chapter 6

Chromatography

For example: Following mixtures may be given for practice 1. Alanine and phenylalanine 2. Glycine and leucine 3. Valine and phenylalanine etc. Procedure Follow the procedure described in sec. 6.4.1 for TLC and sec. 6.3.1 for paper chromatography • Solvent system - Butanol : Acetic acid: water = 3:1:1 (upper layer) • Spraying agent - 0.2 % solution of ninhydrin in water (warm the TLC or paper in the oven for 2-3 minutes after spraying to get the coloured spots) Other solvent • Ethanol : water = 63:37 • n- propanol : water = 64:36 Other spraying agents • Pyridine-isatin reagent

6.9.2 Chromatography of Sugars Paper chromatography/ TLC Follow the procedure described in sec. 6.3.1 or sec. 6.4.1. Solvent system Ethyl acetate: Pyridine : water = 14:6:5 Locating agent obtained by dissolving 0.5 g of m-phenylenediamine, 1.2g of stannous chloride in 20 mL acetic and 80 mL alcohol) Dry the paper after spraying the locating agent and heat it at ~100°C for 4-5 minutes in an oven. Dark yellow spots are obtained. Notes: • A mixture of glucose, xylose and Lactose may be given • Other solvent systems are • Butanol: Acetic acid: water = 3:1:1 (upper layer) • Butanol: Ethanol: water = 5:1:4 (upper layer) • Other spraying agents • 50% sulphuric acid, dry and heat in oven at 100°C. 6.9.3 Chromatography of Phenols TLC Solvent sytem – Toluene : Dioxane = 6:1 Visualising agent – UV lamp or 0.1% solution of 2,7-dichloro fluorescein in methyl alcohol Notes: A mixture of any two phenols (3% solution in ethanol) can be used. Resorcinol Phloroglucinol o-nitrophenol m-nitrophenol

6.9 Applications Advanced Experimental Organic Chemistry

405

6.9.4 Chromatography of Chlorophyl (Ascending Paper Chromatography) Procedure • Grind spinach leaves after washing (3-4 leaves) • Add 3-4 mL alcohol • Filter and concentrate in a hot water bath • Use concentrated extract for spotting Solvent system: Dilute alcohol (1:1) • Follow the procedure given in section 6.3.1. Result: Three coloured spots are seen • Red (carotene) • Green (chlorophyl) • Orange (xanthophyl)

6.9.5 Chromatography of Dyes Separation of methylene blue and malachite green by column chromatography Requirements Malachite green

0.1g

Methylene blue

0.1g

Column

Long thin column

Stationary phase

Alumina

Solvent

1. water 2. Alcohol

• Prepare the column of alumina with water as described in section 6.5. • Pour the solution of malachite green + methylene blue in water on the column • Elute malachite green completely with water • After the complete elution of malachite green, elute methylene blue with alcohol

6.9.6  Separation of a mixture of o-Nitroaniline and p-Nitroanline by Column  Chromatography Requirements o-Nitroaniline

0.1g

p-Nitroaniline

0.1g

Alumina Benzene Procedure As given in section 6.5.

406

Chapter 6

Chromatography

o-Nitro aniline elutes out first. The movement of both the components is visible as these are yellow coloured solids. Note: Benzene is known to be a human carcinogen, so work in a fume chamber and avoid exposure to benzene vapours.

6.9.7  Separation of Syn and Anti-azobenzene by Column Chromatography Requirements Syn-azobenzene

0.1g

Anti-azobenzene

0.1g

Alumina Petroleum ether Methanol Procedure Same as given in section 6.5. Orange coloured anti-azobenzene is eluted first using petroleum ether. Yellow coloured syn-azobenzene is then eluted using petroleum ether - methanol (95:5)

Questions Q.1.

Explain the following terms: Stationary phase, Mobile phase, Solvent front, Retention factor.

Q.2.

What do you understand by Adsorption, Partition, Normal and Reverse phase chromatography?

Q.3.

Why should the spotting be done by • Using a capillary jet • Using 10-20% solution of the compound?

Q.4.

What are the major differences between ascending paper and thin layer chromatography?

Q.5.

Mention a few techniques used for visualizing the spots on a paper or a TLC.

Q.6.

Can the chromatography (specially TLC) be used to check the completion of a reaction? How? Explain with a suitable example.

Q.7.

Why o-nitrophenol and p-nitrophenol move at different rates in column chromatography?

Q.8.

Which of the following statement is not true (a) The spots of amino acids can be visulalised by spraying ninhydrin on the chromatogram. (b) All amino acids have same Rf. (c) Proline and hydroxy proline are imino acids.

Q.9.

How is the elution of components monitored in HPLC?

Q.10. List a few applications of ion-exchange chromatography.

7 Tables of Organic Compounds and their Derivatives Table 7.1 M.P.a

Name and structure

(°C)

D-Ribose HOCH2

Osazone M.P. Time of (°C) formation (min)

87 O

Carbohydrates

163

—

Acetate M.P. (°C) —

H

H

H

H

OH OH

Specific tests and derivatives [[a]D20 in H2O] – Reduces Fehling’s solution and Barfoed’s reagent. [– 23° → – 24°] (add trace of NH3) (C = 4 in H2O)e

OH

2-Deoxy D-ribose HOCH2 O H H

H

OH

H

H

OH

90













– Reduces Fehling’s solution and Barfoed’s reagent. – p-Nitrophenyl hydrazone, m.p. 160°C. [– 56° (final) (C = 1 in H2O)e]

D-Fructose (Laevulose) Fruit sugar)

102

205

2

70 (α)c 109 (β)d

– Reduces Fehling’s solution in cold and Barfoed’s reagent on warming. – Gives Seliwanoff test. – Blue colour on warming with ammonium molybdate (specific test). – p-Nitrophenylhydrazone m.p. 180°C. [– 92° (30 min. after dissolving) (C = 4 in H2O)5]

Maltose C12H22O11

102 (mono-hydrate) 160 – 65 (anhydrous)

208

20–25 (separates on slow cooling) in 5 min. after hydrolysis, see page 94)

125(α)c 159 (β)d

– Reduces Fehling’s solution but not Barfoed’s reagent. +111° → +130° (C = 4, hydrate in H2O)e

408

Chapter 7 Tables of Organic Compounds and their Derivatives M.P.a (°C)

Name and structure

Raffinose C18H32O16

Osazone M.P. Time of (°C) formation (min)

118 (anhydrous) 78 (pentahydrate









Acetate M.P. (°C)

Specific tests and derivatives [[a]D20 in H2O]

99 (β)4 (Hendecaacetate)

– Does not reduce Fehling’s solution and does not react with phenylhydrazine. On hydrolysis yields galactose, glucose and fructose. [+105° (C = 4)e]

CH2OH O

HO H OH

H

H

OH

CH2

H

O

O H H OH

H

H

D-Mannose CH2OH O

H H OH

H

HO

HOCH2

H

H

O

O

H OH CH2OH

OH

OH

H

132

205

0.5

74(α)c 115(β)d

– Reduces Fehling’s solution and Barfoed’s reagent. – Phenylhydrazone, m.p. 199°C. [+29.3 → + 14.2]

145

163

8–10

Tetraacetate 59 (α)c 126 (β)d

– Reduces Fehling’s solution and Barfoeds reagent. – Gives rapid furfural test (see D-fructose). – On warming with phloroglucinol + dil. HCl gives red colour. + 93° → 18.5 (add trace of NH3; (C = 10 in H2O)e

146 (anhydrous) 85 (monohydrate)

205

5

Pentaacetate 112 (α)c 134 (β)d

– Reduces Fehling’s solution and Barfoed’s reagent. – On oxidation (HNO3) gives saccharic acid, m.p. 125° C. [+112° → + 52.8°] (add trace of NH3; C = 3.9 anhydrous in H2O)e

160

166

10

Tetraacetate 96 (α)c 86 (β)d

Reduces Fehling’s solution and Barfoeds reagent. – Gives rapid furfural test (see D-fructose). – On warming with phloroglucinol + dil. HCl gives red colour. [+ 180° → + 105°] (add trace of NH3; C = 1 in H2O)e

H

HO OH

HO H

H

α-D-Xylose (wood sugar) H O

H H OH

H

H OH

HO H

OH

α-D-Glucose (dextrose, grape sugar) CH2OH O

H H OH

H

H OH

HO H

OH

β-L-Arabinose H O

HO H OH

H

H OH

H H

OH

Table 7.1 Experimental Carbohydrates Advanced Organic Chemistry M.P.a (°C)

Name and structure

D-Galactose CH2OH O

HO H OH

H

H

409

Osazone M.P. Time of (°C) formation (min)

Acetate M.P. (°C)

Specific tests and derivatives [[a]D20 in H2O]

168 (anhydrous) 119 (monohydrate)

196

15-20

Pentaacetate 96(α)c 142(β)d

– Reduces Fehiling’s solution and Barfoed’s reagent [+150°→+80°] (add trace of NH3; C = 4 in H2O)e

169 (185)

205

30 (after hydrolysis in 5–10 min.)

Octaaceate 82(β)d

– Non-reducing sugar, no reaction with Fehlings solution or phenylhydrazine. – With CuSO4 + NaOH solution gives blue solution. – On oxidation (HNO3) gives saccharic acid m.p. 125°C. – On warming with very dil. HCl gives invert sugar (mixture of glucose + fructose), [α]D20 –37, which reduces Fehling’s solution and gives osazone, m.p. 205°C. [+ 66.5° (C = 26)e]

20–25

Octaacetate 152(α)c 100(β)d

OH

H H

OH

Sucrose (cane sugar) C12H22O11 CH2OH O

H H OH

HOCH2

H

H

H

CH2OH

OH

OH

Lactose (milk sugar)

O OH

H

O

H

H

H OH

H H

H H

200

CH2OH O

H OH

H

203 (hydrate) 223 (anhydrous)

CH2OH HO

H OH

H

O

HO

O

OH

H

Loses H2O at approx. 130°C and turns yellow at approx. 160°. Reduces Fehlings solution but not Barfoeds reagent. With warm dil. HCl gives glucose + galactose. [+85°→+52.5°] (add trace of NH3; C = 7.6, hydrate in H2O)e

OH

a

The melting points of most carbohydrates vary with the rate of heating.

b

The melting points of the osazones given are really decomposition temperatures; it is advisable to determine the decomposition temperature by heating rapidly.

c

The α-acetate is prepared by heating with fused zinc chloride and acetic anhydride. (see page 91).

d

The β-acetate is prepared by heating with fused sodium acetate and acetic anhydride. (see page 92).

e

For giving specific rotation [α]D10 in water, the concentration is specified. For example, C = 4 in water means 4 g of the carbohydrate is dissolved in water so that the volume of the solution is 10 ml. The concentration is expressed in g/dm3.

410

Chapter 7 Tables of Organic Compounds and their Derivatives

Table 7.2 Name

B.P. (°C)

Structure

Carboxylic Acids Derivatives, M.P. (°C) Anilide

Formic acid Methanoic acid

101

p-Tolui- Amide dide

HCOOH

47

53

Specific tests and derivatives

S-Benzylisothiouronium salt 146

– Reduces Fehling’s and Tollens reagent. – Gives yellow colour with conc. NaHSO3 solution. – Decolourises KMnO4 solution.

Acetic acid

118 (m.p. 16)

CH3COOH

114

148

82

134

– Gives fruity smell of ethyl acetate on heating with ethyl alcohol and a few drops of conc. H2SO4

Acrylic acid

140 (m.p. 8)

CH2 = CHCOOH

104

141

84

—

– Decolourises KMnO4 solution and Br2 water.

Propionic acid

140

CH3CH2COOH

104

124

79

148

Isobutyric acid

155

(CH3)2CHCOOH

105

104

129

143

n-Butyric acid

163

CH3(CH2)2COOH

96

72

114

146

– Gives fruity odour of ethyl butyrate on heating with absolute alcohol and 2 drops of conc. H2SO4 and pouring into water.

CH3COCOOH

104

109

125

158(d)

– Decolourises Br2 water and KMnO4 solution.

Pyruvic acid

165(d) (m.p. 13)

– Reduces ammoniacal AgNO3 solution. – Yields iodoform. Crotonic acid (cis)

165(d)

CH3CH= CHCOOH

102

132

—

—

Isovaleric acid

176

(CH3)2CHCH2COOH

110

109

135

153

α-Chloropropionic acid 186

CH3CHClCOOH

92

124

80

—

n-Valeric acid

186

CH3(CH2)3COOH

63

70 58

104 (114)

—

Methoxyacetic acid

203

CH3OCH2COOH

96

α-Bromopropionic acid 205 (m.p. 24)

CH3CHBrCOOH

99

125

123

—

n-caproic acid (n-Hexanoic acid)

205

CH3(CH2)4COOH

95

75

100

159

Ethoxyacetic acid

206

C2H5OCH2COOH

92

n-Heptoic acid

223

CH3(CH2)5COOH

71

80

94

81

Octanoic acid

237

CH3(CH2)6COOH

57

70

105

157

80

– Decolourises Br2 water and KMnO4 solution.

Table 7.2 Experimental Carboxylic Acids Advanced Organic Chemistry Name

M.P. (°C)

411

Structure

Derivatives, M.P. (°C) Anilide

(±) Lactic acid

18 b.p. 119 (12 mm)

Thiobenzoic acid

CH3CH(OH)COOH

24 (b.p.)

C

122/30 mm

O

o-Ethoxybenzoic acid 25 b.p. 300 (d)

SH

p-Tolui- Amide dide

Specific tests and derivatives

S-Benzylisothiouronium salt

59

107

74

153

– Decolourises KMnO4 solution on warming.

102

128

116

—

– On oxidation gives benzoyl disulphide, m.p. 130°C. – On boiling with dil, NaOH solution gives H2S and benzoic acid, m.p. 121°C.

OC2H5

132

COOH

CH3CO(CH2)2COOH

102

109

108

—

(CH3)3C COOH

128

120

155

—

β-Chloropropionic acid 42

ClCH2CH2COOH

—

—

101

—

Bromoacetic acid

50

BrCH2COOH

130

91

91

4-Phenylbutyric acid

52

C6H5CH2CH2CH2 COOH

Myrisitic acid

54

CH3(CH2)12COOH

84

93

102

Trichloroacetic acid

58

Cl3CCOOH

95

113

141

149

Palmitic acid

62

CH3(CH2)14COOH

90

98

106

141

Chloroacetic acid

63

ClCH2COOH

134

162

120

160

Phenylglyoxalic acid

66

Levulinic acid

Pivalic acid (Trimethyl acetic acid

33 (b.p. 245) (d)

35 (b.p. 163)

84

91 C O

COOH

– Gives iodoform, m.p. 119°C in cold with iodine in KI and NaOH solution. – Phenylhydrazone, m.p. 108° C. – Oxime, m.p. 95°C. – Semicarbazone, m.p. 184°C (d)

– Oxidation (CrO3) gives benzoic acid, m.p. 121°C.

– Solution of sodium salt on prolonged boiling yields chloroform.

– With β-naphthol and NaOH yields β-naphthoxyacetic acid, m.p. 156°C. – 2, 4-Dinitrophenylhydrazone, m.p. 196 (d). – On heating with aniline gives benzylidene aniline m.p. 54°C.

412 Name

Chapter 7 Tables of Organic Compounds and their Derivatives M.P. (°C)

Structure

Derivatives, M.P. (°C) Anilide

Crotonic acid (trans)

72

Phenylacetic acid

76

Glycolic acid

80

Iodoacetic acid

83

p-Sulphobenzoic acid

Phenoxyacetic acid Glutaric acid

Citric acid

94 (hydrated) 259 (anhydrous) 98 98 (b.p 303) 100 (hydrated)

CH3CH=CHCOOH

p-Tolui- Amide dide

Specific tests and derivatives

S-Benzylisothiouronium salt

116

132

158

118

136

154

160

CH2OHCOOH

96

143

120

142

ICH2COOH

143

—

95

—

253 COOH (di)

—

236 (di)

CH2COOH

HO3S

C6H5OCH2COOH HOOC(CH2)3COOH

99 224

— 217

101 174

CH2COOH

199

189

210

131

128

—

197

207

157 (di)

– Reduces KMnO4 solution and ammoniacal AgNO3 solution on warming. – With bromine gives α, β− dibromobutyric acid, m.p. 87°C. – With HI yields β-idobutyric acid, m.p. 110°C.

– Nitro derivative m.p. 130°C – On fusion with KOH yields p-hydroxybenzoic acid, m.p. 213°C.

—

C(OH)COOH

COOH

– Gives fluorescein test. – Amide on heating gives imide, m.p. 154°C. – With warm conc. H2SO4 gives a yellow solution; the solution on neutralising with alkali and addition of sod. nitroprusside gives wine red colour. – Gives blue colour with β-naphthol in conc. H2SO4

CH2COOH

o-Methoxybenzoic acid 100

– With acidic KMnO4 gives benzaldehyde. – With alkaline KMnO4 gives benzoic acid, m.p. 121°C.

OCH3

(–) Malic acid

100

CH2COOH CH(OH)COOH

124

– With cobalt nitrate solution and excess NaOH gives blue colour. – With β-naphthol in conc. H2SO4 gives greenish-yellow colour, clear yellow on warming and pale yellow on dilution. – With PCl 5 in cold and on subsequent treatment with H2O yields chlorosuccinic acid, m.p. 176°C (decom.)

Table 7.2 Experimental Carboxylic Acids Advanced Organic Chemistry Name

M.P. (°C)

Structure

413 Derivatives, M.P. (°C) Anilide

Oxalic acid

101

o-Toluic acid

104

(COOH)2 . 2H2O

p-Tolui- Amide dide

246 (di)

267 (di)

419 (di)

125

144

142

126

118

95

Specific tests and derivatives

S-Benzylisothiouronium salt 196

– With urea gives urea oxalate. m.p. 171°C. – On oxidation (KMnO4) gives phthalic acid, m.p. 195°C.

COOH CH3

m-Toluic acid

110

COOH

164 (140)

– On oxidation (KMnO4) gives isophthalic acid, m.p. > 300°C (sublimes).

166

– Gives smell of bitter almonds on heating with KMnO4 and dil. H2SO4. – On heating with β-naphthol and conc. H2SO4 (1 min.) and cooling gives wine red colour.

204

– Gives red colour with FeCl3.

CH3

Ethyl malonic acid

111

Methyl succinic acid

115

(±) Mandelic acid

118

C2H5CH(COOH)2

150 200

CHOHCOOH

5-Sulphosalicylic acid 120 (anhydrous)

151

214 164 165 (mono) 225(di)

172

133

COOH OH

HO3S

Benzoic acid

3-Nitrosalicylic acid

122

C6H5COOH

164

158

128

167

– Neutral solution of the compound with FeCl3 gives buff coloured precipitate.

125 (hydrated)

COOH

—

—

145

—

– Gives red colour with aqueous FeCl3.

—

—

172

—

195

—

165

—

OH

NO2

dl-Saccharic acid

125

o-Benzoylbenzoic acid 126

HOOC(CHOH)4COOH COOH COC6H5

414 Name

Chapter 7 Tables of Organic Compounds and their Derivatives M.P. (°C)

Structure

Derivatives, M.P. (°C) Anilide

Maleic acid (cis)

130 to 140 (varies with rate of heating)

Cinnamic acid

133

Furoic acid (Pyromucic acid)

133

Malonic acid

133 (d)

CH

CHCOOH

S-Benzylisothiouronium salt

187 (di) 198 (mono)

142

153

175 163

– Decolourises bromine water only on warming. – Gives dibromide, m.p. 160°C.

151

168

147

175

– On oxidation (acidic KMnO4) gives smell of bitter almonds – On oxidation (alkaline KMnO4) yields benzoic acid, m.p. 122°C. – With fuming HNO3 (cold) yields p-nitro derivative, m.p. 285°C.

123

108

142

211

225

252

170

147

COOH O



CH2(COOH)2

o-Sulphobenzoic acid

134 (anhydrous) 94 (hydrated)

Acetyl salicylic acid (aspirin)

p-Tolui- Amide dide

Specific tests and derivatives

COOH









190

—

—

206

– On KOH fusion forms salicyclic acid m.p. 159°C. – On warming with phenol and conc. H2SO4 and pouring into NaOH solution gives red colour which turns yellow in acidic medium.

136

—

138

144

– Gives smell of oil of winter green on heating in methanol with conc. H2SO4 and pouring in H2O.

—

—

—

—

– N-oxide, m.p. 161°C. – Hydrochloride, m.p. = 210 –12°C. – Gives smell of pyridine on heating with sodalime

182

—

217

—

153

162

142

163

– With Sn + HCl gives m-aminobenzoic acid, m.p. 174°C.

—

—

170

163

– On KOH fusion, gives m-hydroxybenzoic acid, m.p. 201°C.

SO3H

135

COOH

OCOCH3

Picolinic acid

136 N

Methyl malonic acid

137

m-Nitrobenzoic acid

141

COOH

CH3CH(COOH)2 COOH

– On warming with Ac2O gives orange colour with green fluorescence.

NO2

m-Sulphobenzoic acid 141

Table 7.2 Experimental Carboxylic Acids Advanced Organic Chemistry Name

M.P. (°C)

415

Structure

Derivatives, M.P. (°C) Anilide

o-Chlorobenzoic acid

142

COOH

p-Tolui- Amide dide

Specific tests and derivatives

S-Benzylisothiouronium salt

118

131

142

—

– With sodium amalgam yields benzoic acid, m.p. 122°C. – On KOH fusion gives o-hydroxybenzoic acid m.p. 159°C.

155

203

174

159

– On reduction (Sn + HCl) gives anthranilic acid, m.p. 144°C.

180

173

168

145

175

190

154

—

– Dissolves in conc. H2SO4 giving red colour. – On oxidation (CrO 3 ) gives benzophenone, m.p. 48°C. – Acetyl derivative, m.p. 98°C.

141

—

155

171

– With conc. HNO3 and H2SO4 at 130° gives dinitroderivative m.p. 213°C.

239

241

220

159

212

208

198

—

—

—

233

—

—

—

155

—

146

—

155

168

– On KOH fusion gives m-hydroxybenzoic acid, m.p. 201°C.

124

—

134

155

– With sodium amalgam yields benzoic acid, m.p. 122° C. – On KOH fusion gives m-hydro xybenzoic acid, m.p. 201°C.

Cl

o-Nitrobenzoic acid

146

COOH

NO2

Diphenylacetic acid

148

3-Nitrosalicylic acid

148 (anhyd)

(C6H5)2CH COOH COOH OH

NO2

Benzilic acid

150

o-Bromobenzoic acid

150

Adipic acid

151

p-Nitrophenylacetic acid

152

Phenylmalonic acid

153

2, 5-Dichlorobenzoic acid

153

(C6H5)2C(OH)COOH

COOH Br

HOOC (CH2)4COOH O 2N

CH2COOH

C6H5CH(COOH)2

m-Bromobenzoic acid 155

COOH Br

m-Chlorobenzoic acid 158

COOH

Cl

416 Name

Chapter 7 Tables of Organic Compounds and their Derivatives M.P. (°C)

Structure

Derivatives, M.P. (°C) Anilide

Salicyclic acid

159

134

COOH

p-Tolui- Amide dide

156

139

Specific tests and derivatives

S-Benzylisothiouronium salt 148

– Gives deep violet colour with – aqueous FeCl3. – Gives smell of oil of wintergreen on heating with CH3OH and conc. H2SO4 and pouring into water.

OH

– Acetyl derivative m.p. 135°C. o-Iodobenzoic acid

COOH

162

142

—

184

—

I

α-Naphthoic acid

162

COOH

160

—

205

—

– On heating with sodalime. gives naphthalene, m.p. 80°C.

Thiosalicylic acid

165

COOH

—

—

—

—

– Acetyl derivative m.p. 125°C.

192

172

200

—

SH

4-Nitrophthalic acid

165

O 2N

COOH COOH

α, β-Dibromosuccinic 167 acid (±)

HOOCCHBrCHBrCOOH

Tartaric acid (+) or (–) 170 140 (meso) 206 (±)

COOH

3,5-Dinitrosalicylic acid

– p-Nitrobenzyl ester, m.p. 168°C. 180

264

195

—

– On heating with conc. H2SO4 it chars and gives smell of burnt sugar. – With cobalt nitrate and excess NaOH solution gives blue solution in cold which becomes colourless on warming. – On adding successively FeSO4, H2O2 and NaOH solution gives violet colour.

181

—

197

—

– Gives red colour with FeCl3 solution.

CHOH CHOH COOH

173

COOH OH

O 2N

NO2

Table 7.2 Experimental Carboxylic Acids Advanced Organic Chemistry Name

417

M.P. (°C) Structure

Derivatives, M.P. (°C) Anilide

p-Toluic acid

178

Veratric acid

H 3C

140

COOH

180 H3CO (anhyd)

p-Tolui- Amide dide

160

166

COOH

159

Specific tests and derivatives

S-Benzylisothiouronium salt 190

– On oxidation (alkaline KMnO4) gives terephthalic acid m.p. > 300°C.

164

OCH3

203

2,4-Dinitrobenzoic acid 183

p-Fluorobenzoic acid

183

Anisic acid

184

β-Naphthoic acid

184

COOH

Succinic acid

185

HOOCCH2CH2COOH





(+) Camphoric acid

187

m-Iodobenzoic acid

187

Oxalic acid Anhydrous

189

F

154

COOH

H3CO

168

186

162

185

– On boiling with 48% HBr or on KOH fusion yields p-hydroxybenzoic acid, m.p. 213°C.

170

191

192

—

– On heating with sodalime yields naphthalene, m.p. 80°C.

228(di) 145 (mono)

255 (di)

260 (di)

149

– Gives brown ppt. with neutral aq. FeCl3 solution. – Gives fluorescein on heating with resorcinol and conc. H 2SO 4 and pouring into excess NaOH solution.

226

—

192

—

—

—

186

—

250 (di) 169 (mono)

201 (di)

220 (di) 149 (mono)

158

COOH



COOH

– On reduction (Sn + HCl) gives m-phenylenediamine, m.p. 63°C.



– On warming with Ac2O yields anhydride, m.p. 221°C.

I

Phthalic acid

195-213 (Varies with rate of heating)

See oxalic acid hydrated m.p. 101 COOH

COOH

– On heating above m.p. gives anhdride, m.p. 132°C. – On heating with resorcinol and conc. H2SO4 and pouring in dilute aq. NaOH solution (excess) gives orange or reddish solution with intense green flourescence.

418 Name

Chapter 7 Tables of Organic Compounds and their Derivatives M.P. (°C) Structure

Derivatives, M.P. (°C) Anilide

m-Nitrocinnamic acid (trans)

200

CH

—

CHCOOH

p-Tolui- Amide dide

—

196

Specific tests and derivatives

S-Benzylisothiouronium salt —

– Decolourises KMnO4 solution and Br2 water. – On oxidation (KMnO 4 ) it gives m-nitrobenzoic acid, m.p. 140°C.

NO2

– On reduction (SnCl2 + HCl) gives m-aminocinnamic acid, m.p. 191°C. Protocatechuic acid (3, 4-dihydroxybenzoic acid)

200 (d)

166

—

212

—

– Gives blue green colour with neutral FeCl3 solution. – Reduces Tollen’s reagent. – With dimethyl sulphate + NaOH gives veratric acid, m.p. 180°C. – Diacetyl derivative, m.p. 162°C.

Fumaric acid (trans) m-Hydroxybenzoic acid

200 sublimes 201

COOH

314

—

266 (di)

195

– Decolourises Br2 water.

155

163

170

—

– No colour with FeCl3. – Acetate, m.p. 127°C.

185

—

176

234

—

183

HO

Mesaconic acid Methyl fumaric acid

204 (sublimes)

H3C

COOH C C

HOOC

3,5, Dinitrobenzoic acid

205

H

COOH

O 2N

o-Coumaric acid

207

NO2

209 (d)

OH CH CH

Vanillic acid

207

COOH

OCH3 HO

– On reduction (Sn + HCl) gives 3, 5-diaminobenzoic acid m.p. 240°C.

COOH

– With HBr gives coumarin, m.p. 67°C. – Acetyl derivative m.p. 146°C.

– Gives no colour with FeCl3. – On methylation (dimethyl sulphate and alkali) gives veratric acid, m.p. 180°C. – Acetyl derivative m.p. 146°C.

Table 7.2 Experimental Carboxylic Acids Advanced Organic Chemistry Name

419

M.P. (°C) Structure

Derivatives, M.P. (°C) Anilide

p-Hydroxybenzoic acid 213

p-Tolui- Amide dide

Specific tests and derivatives

S-Benzylisothiouronium salt

202

—

162

—

127

—

222

—

234 (di) 181 (mono)

224

201

—

– On heating above m.p. yields the anhydride, m.p. 162°C.

—

—

264 (d)

—

– With NaOH solution gives red colour, which is discharged on acidification. – On boiling with water gives trinitrobenzene, m.p. 122°C.

243

222

217

—

– Gives blue colour with FeCl3 solution. – Acetyle derivative m.p. 185°C. – Benzoyl derivative m.p. 208°C.

229

—

212

—

COOH OH

224

—

225

—

COOH

85

—

122

—

COOH

OH

β-Resorcylic acid

213

COOH

HO

– Gives red colour with FeCl3 solution. – Acetyl derivative, m.p. 185°C. – On methylation (Me 2 SO 4 + alkali) gives anisic acid, m.p. 184°C.

OH

3-Nitrophthalic acid

218

COOH COOH

NO2

2,4,6-Trinitrobenzoic acid

220 (d)

O 2N

COOH NO2

NO2

3-Hydroxy-2-naphthoic 222 acid

OH

COOH

Diphenic acid

228

HOOC COOH

5-Nitrosalicylic acid

230

O 2N

Nicotinic acid

236 N

– Gives red colour with alcoholic FeCl3 solution.

420 Name

Chapter 7 Tables of Organic Compounds and their Derivatives M.P. (°C) Structure

Derivatives, M.P. (°C) Anilide

o-Nitrocinnamic acid

240

CH

CHCOOH

—

p-Tolui- Amide dide

Specific tests and derivatives

S-Benzylisothiouronium salt

—

185

—

– Decolourises KMnO4 solution. – On oxidation (KMnO4) gives o-nitrobenzoic acid, m.p. 146°C.

201

201

182

– On reduction (Sn + HCl) gives p-aminobenzoic acid, m.p. 186°C.

NO2

p-Nitrobenzoic acid

240

217 COOH (204)

O2N

Gallic acid

204 (d)

207

COOH

HO

244

– Gives blue-black ppt. with neutral FeCl3 solution. – Wi t h M e 2 S O 4 + N a O H solution gives trimethyl ether m p. 170°C.

OH HO

p-Chlorobenzoic acid

243

p-Bromobenzoic acid

251

p-Iodobenzoic acid

270

Cl

COOH

Br

COOH

COOH

194

—

179

186

197

—

189

—

210

—

217

—

—

204

—

– Decolourises KMnO4 soloution and Br2 solution. – On oxidation (alkaline KMnO4)-p-nitrobenzoic acid. m.p. 240°C.

250

—

280

215

– Me ester, m.p. 67°C.

337

—

—

—

– Methyl ester, m.p. 140°C.

156

—

—

—

– N-oxide, m.p. 83°C.

I

p-Nitrocinnamic acid

285

CH

CHCO2H

NO2

Isophthalic acid

>300

COOH

COOH

Terephthalic acid

>300 HOOC

Isonicotinic acid

319

COOH

Table 7.3 Experimental Carbonyl Compounds (Aldehydes and Ketones) Advanced Organic Chemistry

Table 7.3 Name

421

Carbonyl Compounds (Aldehydes and Ketones)

B.P. (°C)

Structure

2,4-dini- trophenylhydrazone

Semi- carbazone

Oxime

Specific tests and derivatives

Formaldehyde

Gas

HCHO

168

169(d)

—

– Reduces Fehling’s and Tollen’s reagents. – Aq. solution + resorcinol + conc. H2SO4–red layer at the junction of two liquids. – p-Nitrophenyl hydrazone, m.p. 182°C.

Acetaldehyde

21

CH3CHO

168

163

47

– – – –

Propionaldehyde

49

CH3CH2CHO

155

154 (89)

40

– On warming with ZnCl2 gives intense odour like that of skatole p-Nitrophenylhydrazone m.p. 123°C.

Glyoxal

50

OHC–CHO

328

270

178

Acrolein

52

CH2=CHCHO

165

171

Acetone

56

CH3COCH3

126

188

59

– Gives red colour with freshly prepared sodium nitroprusside and excess NaOH solution. – Gives idoform test. – p-Nitrophenylhydrazone, m.p. 148°C.

iso-Butyraldehyde

63

(CH3)2CHCHO

187 (182)

125

—

– p-Nitrophenylhydrazone, m.p. 131°C.

n-Butyraldehyde

74

CH3CH2CH2CHO

123

104

—

– Gives Fehling’s and Tollen’s test. – p-Nitrophenylhydrazone, m.p. 119°C.

(CH3)3CCHO

210

190

—

– p-Nitrophenylhydrazone m.p. 119°C.

115

147

—

– Given iodoform test. – Gives sodium nitroprusside test

Trimethylacetaldehyde 75 (Pivaldehyde)

Reduces Tollen’s reagent Reduces Fehling’s solution. Gives iodoform, m.p. 119°C. With sodium nitroprusside and excess NaOH gives wine red colour.

– Phenylhydrazone, m.p. 180°C. – Gives tests for unsaturation (KMnO4 and bromine water). – Reduces Tollen’s reagent.

Ethyl methyl ketone

80

CH3COCH2CH3

Methyl vinyl ketone

80

CH2 = CHCOCH3

Diacetyl

88

CH3COCOCH3

315 (di)

278 (di)

234 (di)

Isovaleraldehyde

92

(CH3)2CHCH2CHO

123

132 (107)

48

– p-Nitrophenylhydrazone, m.p. 109°C.

Isopropylmethyl ketone

94

CH3COCH(CH3)2

117

112

109

– Gives iodoform test. – With sodium nitroprusside gives red colour.

141

– (see acetone above)

.

422

Name

Chapter 7 Tables of Organic Compounds and their Derivatives

B.P. (°C)

Structure

2,4-dini- trophenylhydrazone

Semi- carbazone

Oxime

Specific tests and derivatives

Chloral 98 (Trichloroacetaldehyde)

CCl3CHO

131

—

56

– On shaking with conc. NaOH solution it gives smell of chloroform. – Reduces Fehling’s and Tollen’s regent.

Diethyl ketone

102

(C2H5)2CO

156

139

69

– Does not give iodoform test – p-Nitrophenylhydrazone. m.p. 144°C.

Methyl n-propyl ketone (Pentanone-2)

102

CH3COC3H7(n)

142

110

58

– Gives iodoform test.

Crotonaldehyde

103

CH3CH=CHCHO

189

201

119

– Reduces Tollens reagent. – With sodium nitroprusside and NaOH solution gives wine red colour.

n-Valeraldehyde

103

CH3(CH2)3CHO

107

—

52

Pinacolone (Methyl t-butyl ketone)

106

CH3COC(CH3)3

125

157

75

– Gives iodoform reaction. – Gives red colour with sodium nitroprusside. – Gives iodoform test.

Isobutyl methyl ketone 117 Di-isopropyl ketone

124

Paraldehyde

124

CH3COCH2CH(CH3)2

95

135

58

[(CH3)2CH]2CO

96

160

34

CH3 H

n-Butyl methyl ketone 129 (Hexanone-2)

– Gives acetaldehyde on warming with dil. H2SO4.

CH3

O

H O

O

H

CH3

CH3COC4H9(n)

106

122

49

– Gives iodoform test.

Cyclopentanone

130

O

142

206

56

– Phenylhydrazone, m.p. 50°C.

Mesityl oxide

130

CH3COCH = C(CH3)2

203

164

49

– Decolourises bromine water. – On boiling with dil H2SO4 yields acetone. – p-Nitrophenylhydrazone.m.p.133°C.

Caproaldehyde (n-Hexaldehyde)

131

CH3(CH2)4CHO

104

106

51

Methyl pyruvate

135

CH3COCOOCH3

187

208

69

Acetylacetone

139

CH3COCH2COCH3

209(di)

– Gives iodoform test. – It is a keto-enolic compound. – Gives red to violet colour with neutral aq. FeCl3. – Gives light blue ppt. of Cu salt with Copper acetate solution.

Tetrahydrofurfural

143

134 CHO O

166

—

– On warming with conc. HCl gives a bright red colour.

Table 7.3 Experimental Carbonyl Compounds (Aldehydes and Ketones) Advanced Organic Chemistry

Name

4-Heptanone (Di-n-propyl ketone) Acetoin

423

B.P. (°C)

Structure

2,4-dini- trophenylhydrazone

Semi- carbazone

Oxime

145

C3H7COC3H7

75

133

—

148 (m.p. 15°)

CH3CHOHCOCH3

318

185

—

CH3CO(CH2)4CH3

89

127

—

n-Amyl methyl ketone 151 (2-Heptanone)

Specific tests and derivatives

Cyclohexanone

155

O

162

166

90

– On oxidation with nitric acid yields adipic acid, m.p. 151°C. – Phenylhydrazone, m.p. 77°C.

Ethyl pyruvate

155

CH3COCOOC2H5

155

206(d)

94

– Gives iodoform test.

n-Heptaldehyde

156

CH3(CH2)5CHO

108

109

57

Furfural

161

229

203

89 (74)

– With alcoholic KOH gives Cannizzaro reaction. – On oxidation (KMnO4) gives Furoic acid, m.p. 133°C.

202

—

56

– Yields acetone on distillation with dil. NaOH solution.

136

197

43

92

124

155

190

—

280

—

172(di)

134

199

38

CH3COCH2COOCH3

—

152

—

CH3CO(CH2)5CH3

58

123

—

CHO O

Diacetone alcohol

164

2-Methylcyclohexanone

165

Di-isobutyl ketone

168

3-Methylcyclohexanone

169

(CH3)2C(OH)CH2COCH3 O CH3

[(CH3)2CH CH2]2CO O

CH3

Succinaldehyde

169

4-Methylcyclohexanone

170

Methyl acetoacetate

170

n-Hexyl methyl ketone 172

OHCCH2CH2CHO O

CH3

– It is a keto-enolic compound. – Gives red colour with aq. FeCl3 solution.

424

Name

Chapter 7 Tables of Organic Compounds and their Derivatives

B.P. (°C)

Structure

Benzaldehyde

179

CHO

Ethyl acetoacetate

181

Cycloheptanone

181

Di-n-butyl ketone

187

Phenylacetaldehyde

193 (m.p. 34°)

Acetonyl acetone (hexane-2, 5-dione)

194

Salicyladehyde

196

CH3COCH2COOC2H5

(C4H9)2CO CH2CHO

CH3COCH2CH2COCH3

CHO

2,4-dini- trophenylhydrazone

Semi- carbazone

Oxime

Specific tests and derivatives

237

222

35

– On oxidation (KMnO4) gives benzoic acid, m.p. 121°C. – On warming with aniline gives benzalaniline, m.p. 54°C. – Phenylhydrazone, m.p. 158°C. – p-Nitrophenylhydrazone m.p. 192°C.

93

130

—

– It is a keto enolic compound. – Gives red colour with aq. FeCl3 solution. – On warming with phenylhydrazine it yields methyl phenyl pyrazolone, m.p. 127°C.

148

163

—

—

90

—

121

156

103

255

220 (di)

138 (di)

252

232

57

112

221

48

211

213

60

– Phenylhydrazone, m.p. 84°C.

194

212

49

– Phenylhydrazone, m.p. 105°C.

250

198

59

– On oxidation (alkaline KMnO4) gives benzoic acid, m.p. 121°C – Phenylhydrazone, m.p. 105°C. – Gives iodoform test.

239

234

80

OH

Phorone

198

m-Tolualdehyde

199

[(CH3)2C==CH]2CO

CH3

– Gives violet colour with FeCl3 solution. – On mild oxidation gives salicylic acid, m.p. 159°C. – On shaking with Ac2O and a drop of conc. H2SO4 yields triacetyl derivative m.p. 100°C.

CHO o-Tolualdehyde

200

CH3 CHO

Acetophenone

202 (m.p. 20)

p-Tolualdehyde

205

COCH3

H 3C

CHO

Table 7.3 Experimental Carbonyl Compounds (Aldehydes and Ketones) Advanced Organic Chemistry

Name

B.P. (°C)

Structure

425

2,4-dini- trophenylhydrazone

Semi- carbazone

Oxime

Specific tests and derivatives

Ethyl levulinate

206 CH3COCH2CH2COOC2H5

101

148

—

– On boiling with NaOH solution yields sodium levulinate (free acid, m.p. 33°, b.p. 245°C)

(–) Menthone

207

146

189

59

– Phexylhydrazone, m.p. 104°C.

78

84

—

– Decolourises Br2 water.

209

226

75

– On oxidation (KMnO4) gives o-chlorobenzoic acid, m.p. 142°C. – Phenyl hydrazone, m.p. 86°C.

255

228

70

– On oxidation (KMnO4) gives m-chlorobenzoic acid, m.p. 158°C.

130

200

76

—

209

117

– Gives red to violet colour with neutral FeCl3 solution. – Phenylhydrazone, m.p. 110°C. – Acetate, m.p. 89°C. – Phenylhydrazone, m.p. 87°C.

O

Citronellal (Rhodinal)

207

CHO

o-Chlorobenzaldehyde 208 (m.p.11)

Cl

m-Chlorobenzaldehyde 213

CHO Cl

Isophorone

215

o-Hydroxyacetophenone

215

O

COCH3

OH

Benzyl methyl ketone 216 (m.p. 27°).

CH3COCH2C6H5

156

198

69

Propiophenone

218

C6H5COCH2CH3

191

174

53

Hydrocinnamal– dehyde

224

C6H5CH2CH2CHO

149

127

94

n-Butyrophenone

228

C6H5CO(CH2)2CH3

190

186

50

Citral

228

110

164

—

– On oxidation gives hydrocinamic acid, m.p. 48°C.

– Decolourises Br2 water and KMnO4 solution.

426

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

B.P. (°C)

d-Carvone

230

p-Chloroacetophenone

Structure

O

236 (m.p. 20°) Cl

Anisaldehyde

247

Cinnamaldehyde

252

COCH3

H3CO

CHO

CH

269 (d)

α-Ionone

130/ 13 mm

β-Ionone

140/ 18 mm

Oxime

191

162 (142)

72

231

203

95

253

210

—

Specific tests and derivatives

– p-Nitrophenylhydrazone m.p. 175°C. – Decolourises Br2 water.

– On oxidation with KMnO4 gives anisic acid, m.p. 184°C.

216

138

(d)

(208)

(69)

—

125 (d)

152

– Keto-enolic compound. Gives red violet colour with nautral aq. FeCl3. – With phenylhydrazine, gives diphenylpyrazolone, m.p. 137°C.

151

143 108

90

– p-Nitrophenylhydrazone. m.p. 113°C.

128

148

—

– p-Nitrophenylhydrazone, m.p. 173°C.

COCH3 258

205

88

100

146

124

– On warming with benzil in alcoholic KOH solution gives deep violet colour.

C6H5COCH2COOC2H5

CH

Semi- carbazone

255

CHCHO

(d)

Ethylbenzoyl acetate

2,4-dini- trophenylhydrazone

CHCOCH3

p-Methylacetophenone

m.p. 28 (b.p. 226)

Dibenzyl ketone

m.p. 34 (b.p.) 331)

Methyl α-naphthyl ketone

m.p. 34

COCH3

—

229

137

– Picrate, m.p. 116°C. – Phenyl hydrazone, m.p. 146°C.

m.p. 34 (b.p. 292)

CHO

—

221

90

– On oxidation (KMnO4) gives I-naphthoic acid, m.p. 162°C.

1-Naphthyl aldehyde

H 3C

(C6H5CH2)2 C = O

Table 7.3 Experimental Carbonyl Compounds (Aldehydes and Ketones) Advanced Organic Chemistry

Name

p-Methoxyacetophenone Piperonal

Benzal acetone (Benzylidene acetone)

M.P. (°C)

38

38 (b.p.) 263) 42 (b.p. 262)

Structure

H3CO

COCH3

O CH2 O

427

2,4-dini- trophenylhydrazone

Semi- carbazone

Oxime

Specific tests and derivatives

220

198

87

– p-Nitrophenylhydrazone, m.p. 195°C

265

230

110

– On oxidation (KMnO4) gives piperonylic acid, m.p. 228°C.

223

186

115

– Decolourises bromine colour. – With Br2 in CCl4–di-bromide, m.p. 125°C. – Phenylhydrazone, m.p. 156°C.

258

233

145

250

256

103

– On heating with aq. NaOH gives reddish brown solution. – On oxidation (KMnO4) gives o-nitro-benzoic acid, m.p. 146°C.

270

231

110

– On oxidation (KMnO4)— p-chlorobenzoic acid, m.p.

CHO

C6H5CH=CHCOCH3

O

1-Indanone

42

o-Nitrobenzaldehyde

44

CHO

NO2

p-Chlorobenzaldehyde 47 (b.p.

Cl

CHO

214) Benzophenone

243°C.

48

C6H5COC6H5

ω-Bromoacetophenone 50 (Phenacyl bromide)

COCH2Br

p-Bromoacetophenone 51 Br

Methyl β-naphthylketone (2-Acetylnaphthalene

53

Benzal acetophenone (chalcone)

58

238

164

142

—

146

97 (89)

230

208

128

262(d)

236

145

244(d)

168

73

– On heating with freshly cut small piece of Na gives deep blue colouration (THF is used as solvent).

COCH3

COCH3

C6H5CH=CHCOC6H5

– Phenylhydrazone m.p. 171°C.

428

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

M.P. (°C)

Structure

CHO

m-Nitrobenzaldehyde 58

2,4-dini- trophenylhydrazone

Semi- carbazone

Oxime

Specific tests and derivatives

293(d)

246

118

– On oxidation (KMnO4)– m-nitrobenzoic acid, m.p. 140°C. – On heating with AC2O + fused NaOAc gives m-nitrocinnamic acid, m.p. 200°C. – Phenylhydrazone, m.p. 120°C.

204

148

98

245

156

NO2

Desoxybenzoin 60 (Phenyl benzyl ketone) β-Naphthaldehyde

Benzoyl acetone

C6H5CH2 COC6H5

60

CHO

270

61

C6H5COCH2 COCH3

151

– Exhibits keto-enol properties.

(b.p. 261) p-Chlorobenzophenone

78

p-Nitroacetophenone

80

– On oxidation (KMnO 4 ) gives β-naphthoic acid, m.p. 184°C.

– Positive ferric chloride reaction.

Cl

COC6H5

O 2N

COCH3

m-Nitroacetophenone 81

COCH3

184

—

163

– Phenylhydrazone, m.p. 106°C.

—

—

174

– Phenylhydrazone, m.p. 132°C.

233

261

132

– Phenlhydrazone, m.p. 128°C.

271

240(d)

117

– Gives blue colour with FeCl3. – On oxidation (alkaline KMnO4) gives vanillic acid, m.p. 207°C. – With bromine water gives bromo derivative m.p. 160°C.

283

234

196

– Phenylhydrazone, m.p. 152°C.

O 2N

Vanillin (3-Methoxy4-hydroxybenzaldehyde)

81

CHO

OCH3 OH

Fluorenone

83

Benzil

95

Glycoladehyde 96 (hydroxyacetaldehyde)

C

C

O

O

HOCH2CHO

189

—

224(di) 237(di) 182 (mono) 137 (mono)

—

—

– On boiling with alcoholic KOH solution yields benzilic acid, m.p. 150°C. – Quinoxaline derivative m.p. 126°C. – Phenylhydrazone, m.p. 162°C. – With Br2/H2O gives glycollic acid, m.p. 80°C.

Table 7.3 Experimental Carbonyl Compounds (Aldehydes and Ketones) Advanced Organic Chemistry

Name

m-Hydroxy acetophenone

M.P. (°C)

Structure

96

COCH3

429

2,4-dini- trophenylhydrazone

Semi- carbazone

Oxime

Specific tests and derivatives

—

195

—

260 (d)

199

88

– Negative ferric chloride reaction. – Methyl ether, b.p. 230°C. – Ethyl ether, b.p. 245°C.

320

220

129

– On oxidation (K2Cr2O7 + dil H2SO4)–p-nitrobenzoic acid m.p. 240°C. – On heating with Ac 2O + fused NaOAc p-nitrocinnamic acid, m.p. 285°C.

—

—

115

– On oxidation (alkaline KMnO4) gives p-bromobenzoic acid acid, m.p. 251°C.

261

199

145

– Phenylhydrazone, m.p. 151°C.

180

190

143

—

185

—

—

184

162

280(d)

223

73

– Gives faint violet colour with aq. FeCl3 solution.

234 (245)

206 (d)

151

– On heating wtih Fehling’s solution gives red ppt. – benzoate, m.p. 125°C. – On oxidation (conc. HNO3) gives benzil, m.p. 95°C.

216

—

161

– On mild oxidation with cupric acetate and NH4NO3 in aq. acetic acid (reflux) gives furil (see p. 430) m.p. 162°C.

OH

m-Hydroxybenzaldehyde

104

CHO HO

p-Nitrobenzaldehyde

106

p-Bromophenacyl bromide

109

p-Hydroxyacetophenone

109

Dibenzalacetone

113

Anisoin (pp′-

113

O 2N

CHO

Br

COCH2Br

HO

COCH3

(C6H5CH=CH)2CO

dimethoxy benzoin) H3CO

CH

C

– Acetate, m.p. 95°C.

OCH3

OH O O O

1,2-Naphthquinone (see quinones)

115 (d)

p-Hydroxybenzaldehyde

117

Benzoin (dl)

137

Furoin

139

HO

O

CHO

CH

C

OH O

O

430

Name

Chapter 7 Tables of Organic Compounds and their Derivatives

M.P. (°C)

β-Reseacetophenone 147 (2,4-Dihydroxyacetophenone)

Furil

162

(±) Camphor

179

Structure

HO

COCH3

2,4-dini- trophenylhydrazone

Semi- carbazone

Oxime

—

218

199

215

—

177

237

Specific tests and derivatives

– With FeCl3 gives red to violet colour. – Dibenzoate, m.p. 81°C.

OH

– p-Nitrophenylhydrazone,m.p.199°C. – Quinoxaline derivative, m.p. 134°C.

O

118

Oxidation of furoin to furil Take copper acetate (0.025 g), ammonium nitrate (1 g), furoin (1.9 g) in acetic acid (35 mL, aq. CH3COOH) in a round bottomed flask. Heat to make a clear solution (nitrogen gas is vigorously evolved during dissolution). Now reflux for about 1.5 hours, cool, filter the furil and recrystallise from methanol. Table 7.4

Phenols Derivatives

Name

o-Chlorophenol

B.P. (°C)

175

Structure

OH

Benzo- 3,5- P-Tolu- Bromo ate Dinitro ene derivabenzo- sulpho- tive ate nate —

143

74

95

181 m.p.

OH

– p-Nitrobenzoate, m.p. 115°C. – Nitration (HNO3 in acetic acid)—dinitro derivative, m.p. 115°C. – Picrate, m.p. 81°C.

Cl

Phenol

Specific tests and derivatives

69

145

96

95(tri)

– p-Nitrobenzoate, m.p. 126°C – Violet colour in FeCl3 test. – On warming with phthalic anhydride + one drop of conc. H2SO4 and pouring in alkaline solution gives red colour (phenolphthalein). – Picrate m.p. 83°C.

o-Cresol

190 m.p. 31°

CH3

OH

—

138

55

56(di)

– Gives violet colour with FeCl3 (aq.). – Picrate, m.p. 88°C.

Table 7.4 Experimental Phenols Advanced Organic Chemistry

431 Derivatives

Name

o-Bromophenol

B.P. (°C)

Structure

195

Benzo- 3,5- P-Tolu- Bromo ate Dini-tro ene derivabenzo- sulpho- tive ate nate

OH

—

—

78

95

70

188

70

199 (tetra) 49 (di)

55

165

51

84 (tri)

– Gives bluish violet colour with aq. FeCl3. – Phthalein test-bluish purple colour. – Nitration (conc. HNO3 + conc. H 2SO 4)—trinitro derivative m.p. 109°C.

57

141

85

116 (tri)

– Gives greenish blue colour with alcoholic FeCl3 solution. – p-Nitrobenzoate, m.p. 93°C.

71

156

—

—

– Gives violet colour with aq. FeCl3 solution

88

186

71

—

– Nitration (conc. HNO3)— dinitro derivative m.p. 81°C.

92

—

—

—

79

—

—

—

86

—

53

—

Br

p-Cresol

201 m.p. 36°

m-Cresol

203 m.p. 12°

o-Methoxyphenol (Guaicol)

m-Chlorophenol

OH

H3C

OH CH3

205 m.p. 28°

OH

OCH3

214 m.p. 32°

OH

Specific tests and derivatives

– On nitration (HNO3 in CH3COOH) gives dinitro derivative, m.p. 118°C. – Gives blue colour with aq. FeCl3 solution.

Cl

p-Chlorophenol

Methylsalicylate

217 m.p. 38

224

OH

Cl

COOCH3

OH

Ethyl salicylate (see esters)

234

COOC2H5

OH

m-Bromophenol

236 m.p. 32°

OH

Br

– Gives violet colour with aq. FeCl3 solution.

432

Chapter 7 Tables of Organic Compounds and their Derivatives Derivatives

Name

B.P. (°C)

Structure

Benzo- 3,5- P-Tolu- Bromo ate Dini-tro ene derivabenzo- sulpho- tive ate nate

Specific tests and derivatives

CH3 OH

Carvacrol

237

m-Methoxyphenol

243

OH

—

78

—

46

– p-Nitrobenzoate, m.p. 51°C.

—

—

—

104 (tri)

– Gives faint violet colour with FeCl3 solution.

OCH3

Eugenol (4-allyl-2-methoxy phenol)

255

70

131

85

118

– Gives blue colour in ferric chloride reaction. – Picrate, m.p. 81°C.

Isoeugenol

266

103

158

—

94

– Gives green colour with neutral alcoholic FeCl3. – Picrate, m.p. 109°C.

2, 4-Dibromophenol

m.p. 40

97

—

120

95

– Gives violet colour with FeCl3 solution.

73

183

—

—

80

—

—

—

97

—

125

68

59

155

83

117(di)

Br

OH Br OH

m-Iodophenol

Phenylsalicylate (Salol)

m.p. 40

I

m.p. 42

COOC6H5 OH

2,4-Dichlorophenol

m.p. 45 b.p.

OH

Cl Cl

209° o-Nitrophenol

m.p. 45 b.p. 215

OH

NO2

– Dissolves in dil NaOH to give orange solution. – With Zn dust + dil. CaCl2 gives o-aminophenol, m.p. 174°C. – On warming with 60% HNO3 gives 2, 4-dinitrophenol, m.p. 114°C

Table 7.4 Experimental Phenols Advanced Organic Chemistry

433 Derivatives

Name

M.P. (°C)

4-Ethyl phenol

46

2,6-Dimethylphenol

49

Structure

Benzo- 3,5- P-Tolu- Bromo ate Dini-tro ene derivabenzo- sulpho- tive ate nate

Specific tests and derivatives

60

133

—

—

—

159

—

79(di)

32

103

71

55

– With alcoholic FeCl3 gives transient green colour. – On boiling with MnO2 + dil H2SO4 gives thymoquinone, m.p. 45°C

87

—

—

—

– With ammonical AgNO3 on heating gives grey or black ppt.

88(di)

190

—

104(tri)

– With aq. FeCl3 gives blue violet colour. – Reduces ammonical AgNO3 on heating – Solution in NH4OH turns red on standing in air.

76

—

65

—

OH

58

181

—

171 (tri)

OH

102

191

94

95

– Gives violet colour with aq. FeCl3.

OH

75

136

—

—

– Gives no colour with FeCl3.

H5C2

OH

CH3 OH

CH3

Thymol 2-Isopropyl5-methyl phenol

CH(CH3)2 OH

50 b.p. 230

CH3

p-Methoxyphenol

Orcinol

56 b.p. 243

H3CO

58 (hydrated)

OH

OH

OH

CH3

2-Hydroxybiphenyl (o-Phenyl phenol)

58

3,4-Dimethylphenol

62

OH

CH3

– Acetate, m.p. 62°C.

CH3

p-Bromophenol

64

Br

Cl

2,4,6-Trichlorophenol

68

Cl

– p-Nitrobenzoate, m.p. 106°C.

Cl

1,3, 5-Xylenol (3,5-dimethyl phenol)

68

—

195

83

166(tri)

434

Chapter 7 Tables of Organic Compounds and their Derivatives Derivatives

Name

Pseudocumenol (2, 4, 5-trimethylphenol

M.P. (°C)

71 (b.p. 230)

Structure

Benzo- 3,5- P-Tolu- Bromo ate Dini-tro ene derivabenzo- sulpho- tive ate nate

H 3C

63

—

—

35 (mono)

61

137

—

178(Tri)

—

—

—

—

– p-Nitrobenzoate, m.p. 126°C.

158

—

130

—

– With aq. FeCl3 no colour but smells like quinonl on warming – With NaOH gives dark brown solution with unplesant odour. – On treatment with NaOAc + Ac2O in HCl at room temperature gives acetate, m.p. 78°C.

118

—

115

—

– Picrate, m.p. 204°C.

231 — (tribenzoate)

—

—

– Gives intense red colour with aq. FeCl3 solution. – On treatment with NaOAc + Ac2O in HCl at room temperature gives acetate, m.p. 180°C.

—

—

– With aq. FeCl3 gives purple colour which on warming gives yellow colour of quinone. – On warming with NaOH solution gives dark brown colour with unpleasant odour. – On treatment with NaOAc + Ac2O in presence of HCl at room temperature gives monoacetate, m.p. 240°C.

OH

H 3C

Specific tests and derivatives

– Negative ferric reaction. – With fuming nitric acid gives mono nitroderivative, m.p. 48°C.

CH3 OH

2,5-Dimethylphenol

CH3

75 H3C OH

CH3

2,3-Dimethylphenol

75 CH3

p-Dimethylaminophenol 75

8-Hydroxyquinoline

(CH3)2N

OH

76 N OH

2,4-Diaminophenol

79 (d)

OH NH2

NH2

p-Methylaminophenol 85

175

—

Table 7.4 Experimental Phenols Advanced Organic Chemistry

435 Derivatives

Name

M.P. (°C)

α-Naphthol

94

p-Iodophenol

94

2,4,6-Tribromophenol

95

Structure

Benzo- 3,5- P-Tolu- Bromo ate Dini-tro ene derivabenzo- sulpho- tive ate nate

OH

I

OH

OH Br

Br

Specific tests and derivatives

56

217

88

105(di)

– Give white ppt with aq. FeCl3 solution. – On warming with dil. NaOH and pure CCl4 + Cu powder gives blue colour. – On adding dil. HNO 3 to a solution of conc. H2SO 4 a n d w a r m i n g Yi e l d s 2 , 4-dintroderivative, m.p. 138°C. – Picrate, m.p. 189°C

119

—

99

—

81

174

113

120

– Acetate, m.p. 82°C.

150 (di)

—

—

—

– Purple colour will FeCl3. – Acetyl derivative, m.p. 150°C (mono).

95

159

113

91 (di)

– On heating with aq. FeCl3 solution gives violet red colour. – With NaOH solution gives orange red colour.

83

156

—

50

84 (di)

152

—

192 (tetra)

130 (di)

—

—

—

Br

o-Methylaminophenol

96

OH NHCH3

m-Nitrophenol

97

p-tert.-Butylphenol

98

Catechol

104

(CH3)3C

OH

OH OH

Chlorohydroquinone

106

HO

OH Cl

– Gives red colour on heating with Fehling solution. – Reduces Tollens reagent. – Diacetate, m.p. 63°C. – With aq. FeCl3 gives green colour, which changes to deep red on addition of NH 4 OH solution. – On oxidation (K2Cr2O7 + H2SO4) in cold yields chloroquinone, m.p. 57°C. – Diacetate, m.p. 72°C.

436

Chapter 7 Tables of Organic Compounds and their Derivatives Derivatives

Name

Resorcinol

M.P. (°C)

Structure

110

Benzo- 3,5- P-Tolu- Bromo ate Dini-tro ene derivabenzo- sulpho- tive ate nate

OH

117(di)

201

80

112(di)

– With aq. FeCl3 gives blue violet colour. – Phthalein test-yellowish green fluorescence. – On heating with aq. NaOH and chloroform gives red or violetred colour.

—

—

—

186 (di)

– With aq. FeCl3 gives bromoquinone, m.p. 55°C – Diacetate, m.p. 72°C.

142

188

97

142(di)

– With aq. NaOH gives deep yellow colour. – With aq. FeCl 3 solution on warming gives violet-red colour. – On nitration (conc. HNO3 + conc. H2SO4) yields picric acid, m.p. 122°C. – Acetate, m.p. 81°C.

—

—

—

118

OH

Bromohydroquinone

110

p-Nitrophenol

114

OH

NO2

2,4-Dinitrophenol

114

OH

Specific tests and derivatives

– Picric acid, m.p. 122°C. (On nitration with conc. HNO3+ conc. H2SO4)

NO2

NO2

S-Trinitrophenol (Picric acid)

122

OH O 2N

—

—

—

—

– With aq. NaOH (heated just to boil) gives deep yellow colour which changes to deep red on addition of 1 drop of ammonium sulphide solution. – Acetate, m.p. 76°C.

153 (di)

—

157

—

– On treatment with NaOAc + HOAc in presence of HCl at room temperature gives mono acetate, m.p. 148°C. – Diacetate, m.p. 101°C. – On heating with aq. FeCl3 gives dark solution.

107

210

125

84 (1-bromo)

NO2

NO2

m-Aminophenol

122

OH

NH2

β-Naphthol

122

– With FeCl3 solution gives white opalescence.

Table 7.4 Experimental Phenols Advanced Organic Chemistry

437 Derivatives

Name

M.P. (°C)

Structure

Benzo- 3,5- P-Tolu- Bromo ate Dini-tro ene derivabenzo- sulpho- tive ate nate

Specific tests and derivatives

– On warming in conc. KOH solution with CHCl3 blue colour is obtained. – Acetate, m.p. 70°C. – Picrate, m.p. 156°C. Toluhydroquinone

124

OH CH3

119 (di)

—

—

—

– Gives blue colour with aq. FeCl3 – On oxidation gives toluqinone, m.p. 68°C. – Diacetate, m.p. 52°C.

90 (tri)

205

—

158

– With aq. FeCl3 gives dull yellow colour. – Soluble in NaOH, turns brown in air. – with FeSO4 solution gives blue ppt. – With Tollen’s reagent grey or black ppt. – Triacetate, m.p. 165°C.

174

—

—

—

– On reduction (Sn + HCl)— 2,4-diamino-α-naphthol [triacetyl derive, of reduction product m.p. 280°C (d)].

120 (tri)

—

—

—

– Triacetate, m.p. 96°C.

137

—

—

—

– Acetate, m.p. 156°C.

149

—

179

—

OH

Pyrogallol

133

OH OH

2,4-Dinitro-1-naphthol 138

OH

OH NO2

NO2

1,2, 4-Trihydroxybenzene (Hydroxy hydroquinone)

140

OH OH

OH

2,4,6-Triiodophenol

OH

158 I

I

I

4-Hydroxybiphenyl (p-Phenyl phenol)

165

OH

438

Chapter 7 Tables of Organic Compounds and their Derivatives Derivatives

Name

M.P. (°C)

Hydroquinone (Quinol)

169

o-Aminophenol

174 (d)

Structure

Benzo- 3,5- P-Tolu- Bromo ate Dini-tro ene derivabenzo- sulpho- tive ate nate

HO

OH

OH

205 (di)

317

159

186 (di)

– Gives transient blue colour with aq. FeCl3. – Yields benzoquinone, m.p. 116°C, on warming with excess conc. FeCl3 solution. – Diacetate, m.p. 123°C.

183 (di)

—

146

—

– Give dark brown ppt. with aq. FeCl3. – On treatment with NaOAc + acetic acid in HCl at room temperature gives monoacetate, m.p. 201°C.

194 (di)

—

109

—

– Acetate, m.p. 103° (di). – (Also see primary amines) Table (7.15)

NH2

4-Amino-2-methylphenol

175

H 2N

OH CH3

2,4,6-Trinitroesorcinol 176 (Styphinic acid)

Specific tests and derivatives

– Dimethyl ether, m.p. 125°C.

OH NO2

O2N

OH NO2

p-Aminophenol

184 (d)

Pentachlorophenol

190

H2 N

OH

OH

234 (di)

—

—

—

– HCl salt, m.p. 306°C. – Diacetate, m.p. 150°C. – On treatment with NaOAc + HOAc in HCl at room temperature gives monoacetate m.p. 168°C.

159

—

145

—

– Acetate, m.p. 150°C.

—

—

82

—

– On Nitration (conc. HNO3 + conc. H 2SO 4) gives 6-nitro derivative, m.p. 280°C.

215

—

—

—

– Diacetate, m.p. 158°C.

Cl

Cl

Cl

Cl Cl

Carbostyril 199 (2-Hydroxyquinoline) N

4-Amino-1-naphthol

d

OH OH

(di)

– Hydrochloride salt m.p. 280°C.

NH2

1-Amino-2-naphthol

d

NH2 OH

235 (di)

—

—

—

– Diacetate, m.p. 206°C. – Picrate, m.p. 110°C.

Table 7.5 Experimental Quinones (Solids) Advanced Organic Chemistry

439 Derivatives

Name

Phloroglucinol

M.P. (°C)

218

Structure

HO

Benzo- 3,5- P-Tolu- Bromo ate Dini-tro ene derivabenzo- sulpho- tive ate nate OH

174 (tri)

162

—

151 (tri)

– With FeCl3 solution gives transient violet colour. – On nitration–trinitro derivative m.p. 165°C. – Triacetate, m.p. 105°C

169 (di)

—

—

—

– With NaOH solution gives red colour. The colour is discharged by addion of excess conc. NaOH solution. – Diacetyl derivative, m.p. 143°C.

OH

Phenolphthalein

254 (261)

Table 7.5

Specific tests and derivatives

Quinones (Solids) Derivatives

Name

Thymoquinone

M.P. (°C)

Structure

45 b.p. 232

O

2,4-dinitro- phenylhydrazone

Semi- carbazone

Oxime

179 (mono)

204 (mono)

161 (mono)

—

—

142 (di)

– It gives deep brown colour on shaking with KI solution containing a drop of dil. H2SO4.

269 (di)

179

135

– With AC2O + drop of conc. H2SO4–2, 4, 5-triacetoxytoluene m.p. 114°C.

—

247

169

– On reduction with SO2 (aqueous) gives the corresponding quinol m.p. 170°C.

O

o-Benzoquinone

Toluquinone

60-65

68

O CH3

Specific tests and derivatives

– On reduction with aqueous SO2. Gives hydrothymoquinore, m.p. 139°C. – Forms adduct with hydroquinone, m.p. 137°C.

O

2-Methyl-1, 4naphthaquinone

106

O CH3

O

440

Chapter 7 Tables of Organic Compounds and their Derivatives Derivatives

Name

p-Benzoquinone

M.P. (°C)

115

Structure

2,4-dinitro- phenylhydrazone

O

Semi- carbazone

Oxime

—

243 (di)

240 (di)

– Reduces ammonical AgNO3 solution in cold – With FeSO4+ dil. H2SO4 gives quinhydrone. m.p. 171°C. – On warming with Ac2O + 1 drop conc. H 2 SO 4 , gives hydroxyhydroquinone triacetate, m.p. 96°C.

—

184

162

– Gives green colour with conc. H2SO4. – On oxidation with KMnO4 gives phthalic acid, m.p. 195°C. – With Ac2O + conc. H2SO4 Warm 1, 2, 4-triacetoxynaphthalene, m.p. 135°C. – On warming with NH2OH, HCl in alcohol gives β-nitroso-αnaphthol, m.p. 103°(d).

—

247

198

– With Ac2O + conc. H2SO4 warm 1, 2, 4-triacetoxynaphthalene, m.p. 135°C.

—

—

—

– Reduces ammonical AgNO3 sol. – On reduction-Zn + HCl hydroquinone, m.p. 169. – On oxidn (K2Cr2O7 + dil H2SO4), p-benzoquinone, m.p. 115.

190

236

170 (di)

– On reduction (Zn + HCl) – α hydroxy camphor, m.p. 203. – On long boiling with alc. KOHcomphoric acid m.p. 187. – Monophenylhydrazone, m.p. 170.

—

220 (di)

162

– On reduction (SO2 in warmalcohol) hydrophenanthraquinone, m.p. 147. – On oxidation (CrO3 or KMnO4) diphenic acid, m.p. 228. – Di-p-nitrophenyhydrazone m.p. 245.

—

—

224

– On heating with Zn dust yields anthracene. m.p. 216°C. – On heating in alcohol with sodium hydrosulphite gives oxanthrol, m.p. 205˜C (d).

O

1, 2-Naphaquinone 115-120 (β-Naphthoquinone)

O O

1, 4-Naphthaquinone 125 (α-Naphthaquinone)

O

Specific tests and derivatives

O

Quinhydrone

171

Camphorquinone

198

O

O

Phenanthraquinone

206

O O

Anthraquinone

286

O

O

Table 7.6 Experimental Alcohols Advanced Organic Chemistry

441 Derivatives

Name

M.P. (°C)

Structure

2,4-dinitro- phenylhydrazone

Semi- carbazone

Oxime

Specific tests and derivatives

– On warming with Zn dust NH4OH gives dihydroanthranol, m.p. 76°C. – On heating with Zn dust and alkali a bright red colour is produced due to formation or anhydroanthraquinone; the colour disappears on shaking in presence of air due to regeneration of anthraquinone. – Soluble in conc. H2SO4 giving purple colour. Alizarin (1, 2 Dihydroxy anthraquinone)

289

O

– Ac2O + NaOAc yields mono acetyl, derivative, m.p. 201°C – Diacetate, m.p. 182°C.

OH OH

O

Table 7.6 Name

B.P. (°C)

Structure

Alcohols 3,5-dinitro benzoate

P-nitro- benzoate

Specific tests and derivatives

Methyl alcohol

65

CH3OH

109

96

– On oxidation (by a hot Cu wire) gives characteristics odour of formaldehyde.

Ethyl alcohol

78

C2H5OH

93

56

– On oxidation (alkaline KMnO4) gives acetic acid. – With iodine and dil. NaOH gives iodoform, m.p. 119°C.

Isopropyl alcohol

82

(CH3)2CHOH

122

110

– On oxidation (K2Cr2O7 + H2SO4 gives acetone. – With iodine and dil. NaOH gives iodoform, m.p. 119°C.

tert. Butyl alcohol

83 m.p. 25

(CH3)3COH

142

116

Allyl alcohol

97

CH2=CHCH2OH

50

28

n-Propyl alcohol

97

CH3CH2CH2OH

74

35

sec. Butyl alcohol

99

CH3CH(OH)CH2CH3

76

26

– Decolourises Br2 water. – On oxidation (K 2 Cr 2 O 7 + dil. H2SO4) gives pungent irritating odour of acrolein. – On oxidation (K2Cr2O7 + dil H2SO4) Gives ethyl methyl ketone (detected with sodium nitroprusside –NaOH solution-red colour) and iodoform test

442 Name

Chapter 7 Tables of Organic Compounds and their Derivatives B.P. (°C)

Structure

3,5-dinitro benzoate

P-nitro- benzoate

tert. Amyl alcohol (2-Methylbutane-2-ol)

102

(CH3)2C(OH)CH2CH3

117

85

iso-Butyl alcohol

108

(CH3)2CHCH2OH

88

69

3-Methylbutan-2-ol

112

(CH3)2CHCH (OH)CH3

76

—

Pentan-3-ol

116

CH3CH2CH(OH) CH2CH3

100

—

n-Butyl alcohol

118

CH3CH2CH2CH2OH

63

35

2, 3-Dimethyl-2-butanol

118

(CH3)2C(OH)CH(CH3)2

111

82

sec-Amyl alcohol (Pentan-2-ol)

119

CH3(CH2)2CHOHCH3

62

—

2-Methyl-2-pentanol

121

CH3CH2CH2C(OH) (CH3)2

72

70

2-Methoxyethanol (Ethylene glycol monomethyl ether)

124

CH3OCH2CH2OH

—

51

3-Methyl-3-pentanol

128

62

—

Specific tests and derivatives

– On oxidation (dil. K 2 Cr 2 O 7 + H2SO4) yields n-butyraldehyde.

(79) 2-Methylbutan-1-ol (Active amyl alcohol)

129

CH3CH2CH(CH3)CH2OH

70

—

Ethylene chlorohydrin

129

ClCH2CH2OH

92

—

– On oxidation (alkaline KMnO4) gives oxalic acid (identified by acidification and adding CaCl2 solution-white ppt. of calcium oxalate is obtained.

iso-Amyl alcohol (3-Methylbutan-1-o1)

132

(CH3)2CH CH2CH2OH

62

—

– On oxidation (dil K2Cr2O7 +H2SO4) gives iso-valeraldehyde.

2-Ethoxy ethanol

135

C2H5OCH2CH2OH

76

—

n-Amyl alcohol (1-pentanol)

138

CH3CH2CH2CH2CH2OH

46

2-Hexanol

139

CH3CH2CH2CH2CH(OH)CH3

39

—

Cyclopentanol

140

115

62

Ethylene bromohydrin

149

BrCH2CH2OH

86

—

4-Heptanol

156

CH3CH2CH2CH(OH)CH2 CH2CH3

64

35

n-Hexyl alcohol (1 Hexanol)

157

CH3(CH2)4CH2OH

58

—

OH

Table 7.6 Experimental Alcohols Advanced Organic Chemistry Name

Cyclohexanol

Furfuryl alcohol

B.P. (°C) 160 (m.p. 20)

Structure

OH

170 O

n-Heptanol (1-Heptanol)

443

176

P-nitro- benzoate

Specific tests and derivatives

113

50

– On oxidation (dil, K2Cr2O7 + H2SO4) gives cyclohexanone\ b.p. 155°C. – On oxidation (hot conc. HNO3) gives adipic acid, m.p. 151°C.

80

76

– On oxidation (KMnO4) gives furoic acid, m.p. 133°C. – GIves green colour with conc. HCl.

48

—

84

47

CH2OH

CH3(CH2)5CH2OH

Tetrahydrofurfuryl alcohol 177 O

3,5-dinitro benzoate

CH2OH

Octan-2-ol (capryl alcohol)

179

CH3(CH2)5 CHOH⋅CH3

32

28

Propylene glycol (propan-1, 2 diol)

188

CH3CH(OH)CH2OH

147 (di)

127

n-Octanol (1-Octanol)

194

CH3(CH2)6CH2OH

62

—

Ethylene glycol

197

HOCH2CH2OH

169

140

– On oxidation (conc. HNO3) gives oxalic acid (tested by CaCl2 solution white ppt.)

l-Linalool

198

—

70

– Decolourises KMnO4 solution.

Benzyl alcohol

206

112

86

– On oxidation (KMnO4) gives benzoic acid, m.p. 122°C. – On heating with dil. HNO3, water bath, 2 min) gives smell of bitter almonds (benzaldehyde).

Trimethylene glycol (propan 1, 3-diol)

216

HOCH2CH2CH2OH

178 (di)

119

– Dibenzoate, m.p. 59°C.

β-Phenyl ethyl alcohol

219

C6H5CH2CH2OH

108

62

– On oxidation (K2Cr2O7 + dil. H2SO4) gives phenyl acetic acid, m.p. 76°C.

(±) α−Terpineol

220

79

139

– Decolourises KMnO4 solution.

CH2OH

(m.p. 35°)

OH

– On oxidation (dil K 2 Cr 2 O 7 + H2SO4) gives hexyl methyl ketone, b.p. 172°C.

444

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

B.P. (°C)

Structure

P-nitro- benzoate

Specific tests and derivatives

—

—

– On oxidation first with alkaline KMnO4 and then with K2Cr2O7 + dil. H2SO4 yields β-methyladipic acid, m.p. 89°C. – Decolourises KMnO4 solution.

63

35

– Decolourises KMnO4 solution. – Reduces ammoniacal AgNO3 solution. – On heating 2 drops with 2 drops of phenol and 2 drops of conc. H 2 SO 4 at 120°, diluting and adding NH4OH-red colour is formed.

(±) Citronellol

222

Geraniol

229

n-Decanol

230

CH3(CH2)8CH2OH

57

30

γ-Phenyl propyl alcohol

237

C6H5CH2CH2CH2OH

92

46

Diethylene glycol

245

(HOCH2⋅CH2)2O

149

—

Cinnamyl alcohol

257 (m.p. 33)

C6H5CH= CHCH2OH

121

78

– Decolourlise KMnO4 solution

CH2OH

—

188

– On heating with KHSO4 gives pungent irritating odour of acrolein. – On heating with phenol and conc. H2SO4 (120°C) and then pouring in excess dil. NH4OH gives red colour. – Reduces ammonicalAgNO3 solution

94

– On oxidation (KMnO4) gives anisic acid m.p. 184°C.

—

—

– Benzoate, m.p. 72°C. – Oxidation (KMnO4) gives m-nitrobenzoic acid, m.p. 141°C. – On reduction (Zn dust + CaCl2) solution gives m-aminobenzyl alcohol, m.p. 97°C.

—

—

153

62

Glycerol

290 (d) (m.p. 18)



3,5-dinitro benzoate

CHOH CH2OH

Anisyl alcohol

H3CO m.p. 25 (b.p. 259)

m-Nitrobenzyl alcohol

m.p. 30 (b.p. 175-80/ 30 mm)

CH2OH

CH2OH

NO2

Pinacol

m.p. 40 (b.p. 174)

(–) Menthol

m.p. 44 (b.p. 216)

(CH3)2C(OH)C(OH)(CH3)2

OH

– On warming with iodine and dil. NaOH solution yields iodofrom, m.p. 119°C. – On oxidation (Cr2O3) gives acetone, b.p. 56°C. – Benzoate, m.p. 54°C.

Table 7.6 Experimental Alcohols Advanced Organic Chemistry Name

M.P. (°C)

Cetyl alcohol

50 (b.p. 189 15 mm)

Diphenyl carbinol (Benzlydrol)

69

o-Nitrobenzyl alcohol

74

445

Structure

3,5-dinitro benzoate

P-nitro- benzoate

CH3(CH2)14CH2OH

66

52

(C6H5)2CHOH

141

131

—

—

– Benzoate, m.p. 101°C. – On oxidation (K2Cr2O7 + acid) gives o-nitrobenzoic acid, m.p. 144°C. – On reduction (Zn-dust, CaCl 2 solution) gives o-amino benzyl alcohol m.p. 83°C.

—

—

– Benzoate, m.p. 94°C.

CH2OH NO2

p-Nitrobenzyl alcohol

93

O 2N

CH2OH

Specific tests and derivatives

– On KOH fusion or on oxidation (Cr2O3 in acetic acid) yields palmitic acid, m.p. 62°C.

– Acetate, m.p. 78°C. – On oxidation (KMnO 4 ) gives p-nitrobezoic acid, m.p. 240. – On reduction (Zn dust + CaCl2 solution) gives p-aminobenzyl alcohol, m.p. 65°C.

(–) Cholesterol

148

—

190

HO

Triphenyl carbinol

Ergosterol

162 (b.p. 380)

(C6H5)3COH

—

165

202

—

– Acetate, m.p. 88°C – With conc. HCl gives chloride, m.p. 108°C. – With conc. H2SO4 forms intense yellow solution.

—

– Acetate, m.p. 180°C. – Benzoate, m.p. 168°C.

H HO

D(+) Mannitol

166 (d)

(±) Borneol

208

CH2OH(CHOH)4 CH2OH

– Hexaacetate, m.p. 120°C. – Hesxabenzoate, m.p. 149°C. 154

137

– On heating with HNO3 gives camphor, m.p. 179°C.

446

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

Inositol

M.P. (°C)

Structure

225

OH

OH

H OH

H H

H

—

—

—

—

Specific tests and derivatives

– Hexaacetate, m.p. 216°C. – Hexabenzoate m.p. 258°C. – On heating with HNO3 gives oxalic acid.

H H

269

P-nitro- benzoate

OH

OH

Pentaerythritol

3,5-dinitro benzoate

OH

C(CH2OH)4

– Tetraacetate, m.p. 83°C. – Tetrabenzoate, m.p. 99°C.

Table 7.7

Name

B.P. (°C)



Esters (Liquids)

Structure

Acid obtained by Saponification b.p. (°C)

m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C) 109

Methyl formate

32

HCOOCH3

101

—

Ethyl formate

54

HCOOC2H5

101

—

93

Methyl acetate

57

CH3COOCH3

118

—

109

Isopropylformate

68

HCOOCH(CH3)2

101

—

112

Vinyl acetate

72

CH3COOCH=CH2

118

—

—

Methyl chloroformate

75

ClCOOCH3

—

—

109

Ethyl acetate

77

CH3COOC2H5

118

—

93

Methyl propionate

79

CH3CH2COOCH3

140

—

109

n-Propyl formate

81

HCOOC3H7(n)

101

—

74

Allyl formate

83

HCOOCH2CH=CH2

101

—

50

Methyl acrylate

85

CH2=CHCOOCH3

140

—

109

Isopropyl acetate

91

CH3COOCH(CH3)2

118

—

122

Methyl isobuyrate

92

(CH3)2CHCOOCH3

155

—

109

Ethyl chloroformate

93

ClCOOC2H5

—

—

93

t-Butyl acetate

98

CH3COOC(CH3)3

118

—

142

iso-Butylformate

98

HCOOCH2CH(CH3)2

101

—

88

Ethylpropionate

98

CH3CH2COOC2H5

140

—

93

Ethyl acrylate

101

CH2=CHCOOC2H5

140

—

93

n-Propyl acetate

101

CH3COOC3H7(n)

118

—

122

Methyl n-butyrate

102

CH3CH2CH2COOCH3

163

—

109

Allyl acetate

103

CH3COOCH2CH=CH2

118

—

50

n-Butyl formate

107

HCOOC4H9(n)

101

—

63

Ethyl isobutyrate

110

(CH3)2CHCOOC2H5

155

—

93

See-Butyl acetate

111

CH3COOCH(CH3)C2H5

118

—

76

Table 7.7 Experimental Esters (Liquids) Advanced Organic Chemistry



Name

B.P. (°C)



Structure

447 Acid obtained by Saponification b.p. (°C)

iso-Propyl priopionate

140

m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

—

122

111

CH3CH2COOCH(CH3)2

iso-Butyl acetate

116

CH3COOCH2CH(CH3)2

118

—

88

Ethyl n-butyrate

120

CH3CH2CH2COOC2H5

163

—

93

n-Propyl propionate

122

CH3CH2COOC3H7(n)

140

—

74

iso-Amyl formate

123

HCOO(CH2)2CH(CH3)2

101

—

62

n-Butyl acetate

125

CH3COOC4H9(n)

118

—

63

Methyl n-valerate

128

CH3CH2CH2CH2COOCH3

186

—

109

iso-Propyl-n-butyrate

128

CH3CH2CH2COOCH(CH3)2

163

—

122

Ethyl glyoxalate

130

OHC COOC2H5

—

70–75 (Hemithydrate)

93

Methyl chloracetate

130

CICH2COOCH3

—

63

109

Ethyl isovalerate

134

(CH3)2CHCH2COOC2H5

176

—

93

Methyl pyruvate

136

CH3CO⋅COOCH3

165

—

109

Iso-Butyl propionate

137

CH3CH2COOCH2CH(CH3)2

140

—

88

Ethyl crotonate

138

CH3CH=CHCOOC2H5

165(d)

—

93

iso-Amyl acetate

142

CH3COOCH2CH2CH(CH3)2

118

—

62

Ethyl chloroacetate

144

CICH2COOC2H5

—

63

93

Methyl bromoacetate

144

BrCH2COOCH3

—

50

109

n-Propyl n-butyrate

144

CH3CH2CH2COOC3H7(n)

163

—

122

n-Butyl propionate

144

CH3CH2COO(CH2)3CH3

140

—

63

Ethyl n-valerate

145

CH3CH2CH2CH2COOC2H5

186

—

93

Methyl lactate

145

CH3CH(OH)COOCH3

119/12 mm

—

109

n-Amyl acetate

146

CH3COO(CH2)4CH3

118

—

46

Ethyl-a-chloropropionate

146

CH3CH(CI)COOC2H5

186

—

93

iso-Butyl isobutyrate

147

(CH3)2CHCOOCH2CH(CH3)2

155

—

88

Ethyl lactate

152

CH3CH(OH)COOC2H5

119/12 mm

—

93

Ethyl pyruvate

155

CH3CO⋅COOC2H5

165

—

93

iso-Butyl n-butyrate

157

CH3CH2CH2COOCH2CH(CH3)2

163

—

88

Ethyl bromoacetate

159

BrCH2COOC2H5

—

50

93

iso-Amyl propionate

160

CH3CH2COO(CH2)2CH(CH3)2

140

—

62

Ethyl glycolate

160

HOCH2COOC2H5

—

80

93

Cyclohexylformate

162

HCOO

101

—

113

Ethyl-α-bromopropionate

162

CH3CH(Br)COOC2H5

205

—

93

n-Butyl n-butyrate

165

CH3CH2CH2COO(CH2)3CH3

163

—

63

Ethyl n-caproate

166

CH3(CH2)4COOC2H5

205

—

93

448



Chapter 7 Tables of Organic Compounds and their Derivatives Name

B.P. (°C)



Structure

Acid obtained by Saponification m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

—

58

93

119/12 mm

—

122

b.p. (°C) Ethyl trichloroacetate

167

Cl3CCOOC2H5

Isopropyl lactate

168

CH3CH(OH)COOCH(CH3)2

Methyl acetoacetate

169

CH3COCH2COOCH3

—

—

109

Cyclohexylacetate

175

CH3COOC6H11

118

—

113

Isoamyl-n-butyrate

178

CH3CH2CH2COOCH2CH2CH(CH3)2

163

—

62

Ethyl-β-bromopropionate

179

BrCH2CH2COOC2H5

—

62

93

Dimethyl malonate

181

CH2(COOCH3)2

—

113(d)

109

Ethyl acetoacetate

181

CH3COCH2COOC2H5

—

—

93

Ethylene glycol monoacetate 183

HOCH2CH2OOCCH3

118

—

169

Diethyl oxalate

186

(COOC2H5)2

—

101

93

Ethyl methyl acetoacetate

187

CH3COCH(CH3)COOC2H5

—

—

93

Di-isopropyl oxalate

190

[COOCH(CH3)2]2

—

101

122

Ethylene glycol diacetate

190

(CH2OOCCH3)2

118

—

169

Dimethyl succinate

195

(CH2COOCH3)2

—

185

109

Phenyl acetate

196

CH3COOC6H5

118

—

145

Diethyl malonate

198

CH2(COOC2H5)2

—

133(d)

93

Methyl benzoate

198

C6H5COOCH3

—

122

109

Benzyl formate

203

HCOOCH2C6H5

101

—

112

Ethyl levulinate

205

CH3COCH2CH2COOC2H5

245

—

93

o-Cresyl acetate

208

118

—

138

140

—

145

118

—

165

118

—

188

—

101

74

CH3 OCOCH3

Phenyl propionate

211

CH3CH2COOC6H5 CH3

m-Cresyl acetate

212 OCOCH3

p-Cresyl acetate

212

H3C

Di-n-propyl oxalate

213

[COOC3H7(n)]2

Ethyl benzoate

213

COOC2H5

—

122

93

Methyl pelargonate

213

CH3(CH2)7COOCH3

253

—

109

OCOCH3

Table 7.7 Experimental Esters (Liquids) Advanced Organic Chemistry



Name

B.P. (°C)



449

Structure

Acid obtained by Saponification m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

—

104

109

—

110

109

CH3COOCH2C6H5

118

—

112

b.p. (°C) CH3 COOCH3

Methyl-o-toluate

213

CH3

Methyl m-toluate

215 COOCH3

Benzyl acetate

216

Diethyl succinate

216

(CH2COOC2H5)2

—

185

93

Methyl phenyl acetate

218

C6H5CH2COOCH3

—

76

109

Isopropyl benzoate

218

C6H5COOCH(CH3)2

—

121

122

Benzyl propionate

222

CH3CH2COOCH2C6H5

140

—

112

—

130

93

—

159

109

Diethyl maleate

224

Methyl salicylate

224

CHCOOC2H5 CHCOOC2H5

Phenyl n-butyrate

227

CH3CH2CH2COOC6H5

163

—

145

Ethyl pelargonate

227

CH3(CH2)7COOC2H5

253

—

93 93

Ethyl phenyl acetate

229

C6H5CH2COOC2H5

—

76

n-Propyl benzoate

230

C6H5COOCH2CH2CH3

—

122

74

Methyl n-caprate

232

CH3(CH2)8COOCH3

269

—

109

Ethyl salicylate

234

—

159

93

Ethyl glutarate

234

—

98

93

Methyl o-chlorobenzoate

234

—

142

109

—

159

122

163

—

112

CH2(CH2COOC2H5)2

OH

Isopropyl salicylate

237 COOCH(CH3)2

Benzyl n-butyrate

238

CH3CH2CH2COOCH2C6H5

450



Chapter 7 Tables of Organic Compounds and their Derivatives Name

B.P. (°C)

Structure



Acid obtained by Saponification m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

—

243

93

—

159

74

b.p. (°C) Ethyl p-chlorobenzoate

238

Cl

COOC2H5 OH

n-Propyl salicylate

238 COOC3H7

Di-n-butyl oxalate

242

[COOC4H9(n)]2

—

101

63

Isobutyl benzoate

242

C6H5COOCH2CH(CH3)2

—

122

88

—

142

93

—

150

109

Cl

Ethyl o-Chlorobenzoate

243 COOC2H5 Br

Methyl o-bromobenzoate

244 COOCH3

Diethyl adipate

245

[CH2CH2COOC2H5]2

—

151

93

Ethyl n-caprate

245

CH3(CH2)8COOC2H5

269

—

93

—

158

93

COOC2H5

Ethyl m-chlorobenzoate

245 Cl

Di-n-propylsuccinate

246

(CH2COOC3H7(n))2

—

185

74

n-Butyl benzoate

248

C6H5COO(CH2)3CH3

—

122

63

Diethyl acetone diacrboxylate 250

CO(CH2COOC2H5)2

—

135(d)

93

Diethyl (–) maliate

CH2COOC2H5

—

100

93

—

150

93

—

144

109

118

—

—

—

155

93

253

CH(OH)COOC2H5 Br

Ethyl o-bromobenzoate

254 COOC2H5 COOCH3

Methyl anthranilate

255 NH2

Glyceryl triacetate (triacetin)

258

CH3COOCH2CH(OCOCH3)CH2OCOCH3 Br

Ethyl m-bromobenzoate

259 COOC2H5

Table 7.7 Experimental Esters (Liquids) Advanced Organic Chemistry



Name

B.P. (°C)

451

Structure



Acid obtained by Saponification b.p. (°C)

m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

Isoamyl benzoate

262

C6H5COO (CH2)2CH(CH3)2

—

122

62

Ethyl p-bromobenzoate

263

Br

—

251

93

Ethyl anthranilate

265

—

144

93

—

159

63

—

184

93

—

133

93

—

146

109

118

—

201

—

206

93

—

195

109

—

100

93

—

174

93

COOC2H5 COOC2H5 NH2 OH

n-Butyl salicylate

268 COOC4H9(n)

Ethyl-p-anisate

269

Ethyl cinnamate

271

CH

CHCOOC2H5

COOCH3

Methyl o-nitrobenzoate

275 NO2

Resorcinol diacetate

277

Diethyl (±) tartrate

280

Dimethyl phthalate

282

C2H5OOC(CHOH)2COOC2H5 COOCH3 COOCH3 CH2COOC2H5

Triethyl citrate

294

C(OH)COOC2H5 CH2COOC2H5 COOC2H5

Ethyl m-aminobenzoate

294 NH2

452

Chapter 7 Tables of Organic Compounds and their Derivatives

m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

—

195

93

—

122

138

—

162

93

—

76

112

—

159

112

—

122

112

338

—

195

63

Methyl m-chlorobenzoate

m.p. 21 (b.p. 231)

—

158

109

Cetyl acetate

22 (m.p.)



Name

B.P. (°C)



Structure

Acid obtained by Saponification b.p. (°C)

COOC2H5

Diethylphthalate

298 COOC2H5 CH3

o-Cresyl benzoate

307 OCOC6H5 COOC2H5

Ethyl 1-naphthoate

309

Benzyl phenyl acetate

317

C6H5CH2COOCH2C6H5 OH

Benzyl salicylate

320 COOCH2C6H5

Benzyl benzoate

Di-n-butylphthalate

Ethyl palmitate Methyl anthranilate

323 m.p. 21

m.p. 24 b.p. 185/10 mm

C6H5COOCH2C6H

CH3COOCH2(CH2)14CH3

118

—

66

CH3(CH2)14COOC2H5

—

62

93

—

144

109

—

155

109

118

—

154

COOCH3

m.p. 25 (b.p. 255)

NH2

Methyl m-bromobenzoate

COOCH3

m.p. 29 Br

Bornyl acetate

m.p. 30 b.p. 221

Table 7.7 Experimental Esters (Liquids) Advanced Organic Chemistry



Name

M.P. (°C)

453 Structure



Acid obtained by Saponification m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

—

146

93

—

184

93

—

69

93

—

178

109

b.p. (°C) NO2

Ethyl o-nitrobenzoate

30 COOC2H5 COOC2H5

Ethyl 2-naphthoate

32

Ethyl stearate

33

CH3(CH2)16COOC2H5

Methyl p-toluate

33 (b.p. 217)

Methyl cinnamate

36 (b.p. 262)

C6H5CH=CHCOOCH3

—

133

109

Ethyl mandalate (±)

37

C6H5CH(OH)COOC2H5

—

118

93

Benzyl cinnamate

39

C6H5CH=CHCOOCH2C6H5

—

133

112

Ethyl m-nitrobenzoate

41

—

141

93

—

195

112

—

185

112

—

159

145

—

243

109

H3COOC

CH3

COOC2H5 NO2 COOCH2C6H5

Dibenzyl phthalate

42 COOCH2C6H5

Dibenzyl succinate Phenyl salicylate (salol)

42

(CH2COOCH2C6H5)2 COOC6H5

42 (b.p. 173/ 12 mm)

OH

Methyl p-chlorobenzoate

44

Cl

Cinnamyl cinnamate

44

C6H5CH=CHCOOCH2CH=CHC6H5

—

133

121

Diethyl terephthalate

44

H5C2OOC

—

>300

93

Methyl-p-anisate

45

H3CO

—

184

109

Dimethyl (±) tartrate

48

CH3OOC[CH(OH)]2COOCH3

—

206

109

COOCH3

COOC2H5

COOCH3

454



Chapter 7 Tables of Organic Compounds and their Derivatives Name

M.P. (°C)

Structure

m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

118

—

217

NH2COOC2H5

—

—

93



Acid obtained by Saponification b.p. (°C)

OCOCH3

1-Naphthyl acetate

49

Ethyl carbamate (Urethane)

50

Methyl carbamate

52

H2NCOOCH3

—

—

109

Dimethyl oxalate

53

(COOCH3)2

—

101

109

m-Cresyl benzoate

54

—

122

165

—

240

93

—

122

217

—

118

109

OCOC6H5 H 3C

Ethyl p-nitrobenzoate

56

1-Napthyl benzoate

56

(±) Methyl mandelate

58

O2N

COOC2H5

C6H5CH(OH)COOCH3

n-Propyl carbamate

60

NH2COOC3H7(n)

—

—

74

Phenyl benzoate

68

C6H5COOC6H5

—

122

145

—

201

109

118

—

210

—

195

145

—

122

188

—

201

93

OH

Methyl m-hydroxybenzoate

70 COOCH3

2-Naphthyl acetate

70

COOC6H5

Di-Phenyl phthalate

70 COOC6H5

p-Cresyl benzoate

71

Ethyl m-hydroxybenzoate

72

COOC2H5 HO

Table 7.7 Experimental Esters (Liquids) Advanced Organic Chemistry



Name

M.P. (°C)

455 Structure



Acid obtained by Saponification m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

—

240

109

—

133

145

122

169

—

184

109

—

—

145

—

200

93

b.p. (°C) CH

Methyl o-nitrocinnamate

CHCOOCH3

72 NO2

Phenyl cinnamate

72

Ethylene glycol dibenzoate

73

CH

CHCOOC6H5

(CH2OOCC6H5)2 COOCH3

Methyl 2-naphthoate

77

Diphenyl carbonate

78

Ethyl m-nitrocinnamate

78

C6H5OCOOC6H5 CH

CHCOOC2H5

NO2

Methyl m-nitrobenzoate

78

—

141

109

Trimethyl citrate

78

—

100

109

Dibenzyl oxalate

80

C6H5CH2OOC–COOCH2C6H5

—

101

112

Methyl p-bromobenzoate

81

Br

—

251

109

Ethyl p-aminobenzoate

92

C2H5OOC

—

186

93

—

205

93

—

159

210

—

240

109

COOCH3

NH2

O2N

Ethyl 3, 5-dinitrobenzoate

93

COOC2H5 O2N OOC

β-Naphthyl salicylate

95

Methyl p-nitrobenzoate

96

HO

O2 N

COOCH3

456



Chapter 7 Tables of Organic Compounds and their Derivatives Name

M.P. (°C)

Structure



Acid obtained by Saponification m.p. (°C)

3, 5 dinitrobenzoate m.p. (°C)

—

122

210

—

205

109

b.p. (°C)

β-Naphthyl benzoate

107

O 2N

Methyl 3,5-dinitrobenzoate

108

COOCH3 O 2N

Ethyl oxamate

114

NH2COCOOC2H5

—

Ethyl p-hydroxybenzoate

116

HO

—

213

93

Resorcinol dibenzoate

117

—

122

201

Diphenyl succinate

121

—

185

145

Hydroquinone diacetate

123

118

—

317

Methyl m-nitrocinnamate

124

—

200

109

Methyl p-hydroxybenzoate

131

—

213

109

Diphenyl oxalate

136

—

101

145

Ethyl p-nitrocinnamate

137

—

285

93

Dimethyl terephthalate

140

H3COOC

COOCH3

—

>300

109

Methyl p-nitrocinnamate

161

O 2N

CHCOOCH3

—

285

109

118

—

205

COOC2H5

(CH2COOC6H5)2

(COOC6H5)2

CH

93

OCOCH3 OCOCH3

Pyrogallol triacetate

161 OCOCH3

Table 7.9 Experimental Acid Anhydrides Advanced Organic Chemistry

457

Table 7.8 Name

M.P. (°C)

Coumarin

67



Structure

Specific tests and derivaties

Decolourises Br2water. Dibromide, m.p. 105°C. On methylation with dimethyl sulphate and NaOH solution (shaking 20 min) and acidification gives cis o-methoxycinnamic acid, m.p. 88°C: on long boiling with conc. KOH solution (20 min) followed by acidification gives coumaric acid, m.p. 207°C.

O

O



Phthalide

Lactones



73

On boiling with alkali yields o-hydroxymethyl benzoic acid, m.p. 120°C. On oxidation (KMnO4) gives phthalic acid, m.p. 195°C.

CO O CH2

Glycollide

84

Lactide

124

L-Ascorbic acid

Glycolamide, m.p. 120°C Glycol anilide, m.p. 96°C On boiling with water gives glycollic acid. m.p. 80°C

CH

CH3

O

O

CO

CH

CH3

190 (d)

With neutral ferric chloride gives intense violet colour. Reduces Tollen’s reagent. Decolourises Br2 water.

Table 7.9 Name

Lactamide, m.p. 74°C, Lactanilide, m.p. 59°C On boiling with H2O gives lactic acid.

CO

Acid Anhydrides

M.P. (°C)

Structure

Acetic anhydride

138

(CH3CO)2O

114

148

82

– On boiling with H 2 O gives acetic acid. The aqueous solution is made neutral by NH4OH and boiling off excess NH3. On adding FeCl3 gives red colour.

Propionic anhydide

168

(CH3CH2CO)2O

104

124

79

n-Butyric anhydride

191

(CH3CH2H2CO)2O

96

72

114

– On boiling with water gives propionic acid.

Citraconic anhydride

213

175

–

187

Anilide

CH

CO

C

CO

O CH3

Derivatives p-ToluiAmide dide

Specific tests and derivatives

458

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

Benzoic anhydride

M.P. (°C)

Structure

42 (C6H5CO)2O (b.p. 360°)

Glutaric anhydride

56

CH 56 (b.p. 202°) CH

Succinic anhydride

CH2CO 119 O (b.p. 250°) CH2CO

Phthalic anhydride

131 (b.p. 284°)

158

128

– On heating with NaOH solution and acidification gives benzoic acid, m.p. 122°C.

224

217

174

– On heating with NaOH solution and acidification gives glutaric acid, m.p. 98°C.

198 (mono) 187 (di)

142

153

– On heating with NaOH and acidification gives maleic acid, m.p. 130°C. – Decolourises Br2 water on warming.

228 (di) 145 (mono)

255 (di)

260 (di)

– On heating with NaOH and acidification gives succinic acid, m.p. 185°C.

O

250 (di) 170 (mono)

201 (di)

149 (mono) 220 (di)

– On heating with NaOH solution and acidification gives phthalic acid, m.p. 195°C.

O

234(di) 181 (mono)

224

201

– On heating with NaOH solution and acidification gives 3-nitrophthalic acid, m.p. 218°C.

226

—

192

– On boiling with water gives cis camphoric acid, m.p. 187°C.

Picrate

Nitro derivative

CO O CO

CO

CO

3-Nitrophthalic anhydride 162

CO



Specific tests and derivatives

164

CH2 CO CH2 O CH2 CO

Maleic anhydride

Derivatives p-ToluiAmide dide

Anilide

CO NO2 O

(±) Camphoric anhydride

221

O O

Table 7.10 Name

B.P. (°C)

Diethyl ether

35

Propylene oxide

35

Ethylvinyl ether

36

Chloromethyl methyl ether 59 Tetrahydrofuran

Structure

Ethers

3,5-dinitro benzoate

C2H5OC2H5

Specific tests and derivatives

– Soluble in cold conc. H2SO4 and is recovered unchanged on dilution. – With hot water gives propylene glycol.

CH2=CH OCH2CH3 CH3OCH2Cl

65 O

– Decolourises Br2 water. —

163

—

– On shaking with H2O gives CH3OH + HCHO.

Table 7.10 Experimental Ethers Advanced Organic Chemistry Name

B.P. (°C)

459

Structure

Di-isopropyl ether

68

(CH3)2CHOCH(CH3)2

Chloromethyl ethyl ether

83

C2H5OCH2Cl

Ethylene glycol dimethyl ether

85

CH3OCH2CH2OCH3

Tetrahydropyran

85

3,5-dinitro benzoate

Picrate

Nitro derivative

Specific tests and derivatives

121 – On shaking with H2O gives C2H5OH + HCHO.

O

Di-n-propyl ether

90

(n)C3H7OC3H7

α-Chloracthyl ethyl ether

98

CH3CHCl⋅O⋅C2H5

Dioxan

102

– On shaking with H2O gives C2H5OH + CH3CHO.

O

– After heating with alkaline KMnO4, the filtrate is treated with H 2 O 2 (to destroy KMnO 4 ). To the residual solution is added acetic acid and CaCl2 solution and heated a white ppt. of Ca oxalate is obtained.

O

Ethylene glycol diethyl ether

121

C2H5OCH2CH2OC2H5

Ethylene glycol monoethyl 135 ether

HO CH2CH2OC2H5

75

Di-n-butyl ether

140

n-C4H9OC4H9(n)

63

Anisole (Phenyl methyl ether)

153

Diethylene glycol dimethyl ether (Diglyme)

162

CH3OCH2CH2OCH2 CH2OCH3

Methyl benzyl ether

169

C6H5CH2OCH3

Di-isoamyl ether

171

[(CH3)2CHCH2CH2]2O

o-Cresyl methyl ether (o-Methoxytoluene)

171

OCH3

CH3 OCH3

—

81

87(di)

116

– The compound in conc. H2SO4 is treated with conc. HNO3 + conc. H2SO4 (1:1) at room temperature (30°). After 10 min, the mixture is poured on ice to give dinitro derivative, m.p. 87°C.

– On oxidation (alkaline KMnO4) gives benzoic acid.

62

—

—

—

119 (mono)

64

– Bromo derivative, m.p. 63°C. – On nitration with conc. H2SO4 + conc. HNO3 gives nitro derivative, m.p. 64°C. – On oxidation (alkaline KMnO4) gives o-methoxybenzoic acid, m.p. 100°C.

460 Name

Chapter 7 Tables of Organic Compounds and their Derivatives B.P. (°C)

Phenetole (Phenyl ethyl ether)

172

p-Cresyl methyl ether (p-Methoxytoluene)

176

Cineole

176

Structure

3,5-dinitro benzoate

OC2H5

H 3C

OCH3

—

Picrate

Nitro derivative

92

58

89

122(di)

177

H3C

—

114

91 (tri)

—

118

51(di)

—

111

—

—

115

—

—

57

95 (mono)

– With fuming HNO 3 gives mono nitro derivative, m.p. 95°C.

—

—

98 (mono)

– With fuming HNO3 gives mono nitro derivative, m.p. 98°C

—

57

95(mono)

– Dibromoo derivative, m.p. 92°C. – With fuming HNO3 gives mononitro derivative, m.p. 95°C.

OCH3

o-Cresylethyl ether (o-Ethoxytoluene)

CH3

184

– Oxidation (KMnO4) gives anisic acid, m.p. 184°C. – With dry HCl at 40-50° yields dipentene dihydrochloride, m.p. 50°C.

O

m-Cresyl methyl ether (m-Methoxytoluene)

Specific tests and derivatives

– On oxidn. (KMnO4) gives m-methoxybenzoic acid, m.p. 100°C.

OC2H5

Diethyleneglycol diethyl ether

188

p-Cresylethyl ether (p-Ethoxytoluene)

189

Benzyl ethyl ether

189

m-Ethoxytoluene m-Cresylethyl ether

192

(C2H5OCH2CH2)2O

CH3

OC2H5

CH2OC2H5 CH3

OC2H5

o-Chloroanisole

195

OCH3

Cl

p-Chloroanisole

Veratrole (Catechol dimethyl ether)

200

206 (m.p. 22)

Cl

OCH3

OCH3

OCH3

Table 7.10 Experimental Ethers Advanced Organic Chemistry Name

B.P. (°C)

461

Structure

3,5-dinitro benzoate

Picrate

Nitro derivative

Specific tests and derivatives

OC2H5

o-Chlorophenetole

208

—

82 (mono)

—

61 (mono) 54(di)

—

92 (trinitro)

– On boiling with 48% HBr yields thymol, m.p. 50°C.

124(tri)

– Dibromo derivative, m.p. 140.

106 (mono)

– With fuming HNO3 + glacial acetic acid gives 4-nitro derivative m.p. 106°C.

Cl

p-Chlorophenetole

Thymol methyl ether

212 (m.p. 21)

212

Cl

OC2H3

CH(CH3)2

H 3C

– With fuming HNO3 gives 4-nitro derivative, m.p. 82°C.

OCH3

Resorcinol dimethyl ether 214

OCH3

—

58

OCH3

o-Bromoanisole

218

OCH3

Br

p-Bromoanisole

o-Bromophenetole

222 (m.p. 13)

Br

224

OCH3

OC2H5

—

—

88 (mono)

– On nitration (fuming HNO3 + glacial acetic acid) gives 2nitro derivative, m.p. 88°C.

—

—

98 (mono)

– With fuming HNO3 gives 4-nitro derivative m.p. 98°C.

Br

Safrole

Anethole (p-propenylanisole)

Resorcinol diethyl ether

– Decolourises Br2 water. – Penta bromo derivative, m.p. 169°C. – Oxidation (alkaline KMnO4) gives piperonylic acid, m.p. 228°C.

232

232 (m.p. 22)

235

OC2H5 H5C2O

—

70

—

—

58

—

– Tribromo derivative m.p. 108°C. – On oxidation (K 2Cr2O7 + H2SO4) yields anisic acid, m.p. 184°C.

462

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

B.P. (°C)

Eugenol methyl ether

244

Structure

3,5-dinitro Picrate Nitro Specific tests and derivatives benzoate derivative OCH2CH

H3CO

CH2

—

114

—

– Decolourises Br2 water. – Dibromo derivative, m.p. 77°C.

OCH3

– On oxidation (K2Cr2O7 in HOAc) gives veratric acid, m.p. 180°C. Isosafrole

246

—

75

—

– Decolourises Br2 water. – Tribromo derive., m.p. 109°C. – On oxidation (alkaline KMnO4) gives piperonylic acid, m.p. 228°C.

Diphenyl ether

252 m.p.

O

—

110

135 (di)

– Dibromo derivative. m.p. 58°C.

—

113

—

– Bromo derivative, m.p. 46°C.

—

—

95

– On reduction (Sn HCl) o-anisidine, m.p. 225°C.

—

—

95

—

119

—

112

78

—

—

44

—

28 Iso-eugenol methyl ether

264

CH

CH3O

CH

CH3

OCH3 OCH3

Methyl-l-naphthyl ether

o-Nitroanisole

265

272 (m.p. 10)

OCH3

NO2 OCH3

o-Methoxydiphenyl

274

OC2H5

Ethyl α-naphthyl ether

276

Dibenzyl ether

298

Ethyl β-naphthyl ether

m.p. 37 (b.p. 282)

C6H5CH2OCH2C6H5 OC2H5

– Bromo derivative m.p. 48°C.

– Bromo derivative, m.p. 66°C.

Table 7.10 Experimental Ethers Advanced Organic Chemistry Name m-Nitroanisole

B.P. (°C)

463

Structure

39 (b.p. 258)

3,5-dinitro Picrate Nitro Specific tests and derivatives benzoate derivative

OCH3

—

—

—

—

—

—

—

71

122 (tri)

—

81

—

—

119

—

—

—

95(di)

—

—

—

– On heating with aq. NaOH gives picric acid, m.p. 122°C.

—

108

—

– Bromo derivative, m.p. 86°C.

—

94

182

NO2

– On reduction (Sn + HCl) yields m-anisidine, b.p. 251°C. (Picrate, m.p. 169°C (d)); – On heating with 48% HBr gives m-nitrophenol, m.p. 97°C.

OC2H5 OC2H5

Pyrogallol triethyl ether

39 OC2H5 OC2H5

Catechol diethyl ether

43 OC2H5 OCH3 OCH3

Pyrogallol trimethyl ether

47 OCH3

Hydroquinone dimethyl ether

56

H3CO

OCH3

– Bromo derivative, m.p. 142°C.

Cl

Symm. trichloroanisole

62

Cl

OCH3 Cl NO2

Symm. trinitroanisole

68

O 2N

OCH3 NO2 OCH3

Methyl-β-naphthyl ether

72

p-Methoxydiphenyl

85

Dibenzofuran (Diphenylene oxide)

86

OCH3

O

464 Name

Symm. trbromoanisole

Chapter 7 Tables of Organic Compounds and their Derivatives B.P. (°C)

Structure

3,5-dinitro Picrate Nitro Specific tests and derivatives benzoate derivative —

88

Table 7.11 Name

Isoperne

B.P. (°C)

34

Structure

CH2

C

CH

—

—

– On heating with 48% BHr yields symm. tribromophenol, m.p. 95°C.

Hydrocarbons Pic- rate

Nitro Derivative Pro- (Position cedure of NO2 (See group) page 468)

Specific test and derivativs

– Decolourises Br2 water.

CH2

CH3

n-Pentane

36

CH3(CH2)3CH3

Trimethylethylene (Amylene)

38

(CH3)2 C = CHCH3

n-Hexane

68

CH3(CH2)4CH3

Benzene

80

– Decolourless Br2 water – On dilution and distillation yields tert. amyl alcohol, b.p. 102°C.

84

Cyclohexane

80 (m.p. 4°)

Cyclohexene

83

n-Heptane

98

CH3(CH2)5CH3

Toluene

110

CH3

n-Octane

125

CH3(CH2)6CH3

Ethyl benzene

136

90

(1, 3)

1 – On oxidation (on boiling with fuming HNO3) yields adipic acid, m.p. 151°C. – Decolourises Br2water. – On oxidation (HNO3) yields adipic acid m.p. 151°C (See procedure A). Page 468).

CH2CH3

88

70

(2, 4)

2

– On oxidation (alkaline KMnO4) gives benzoic acid m.p. 122°C.

96

37

(2, 4, 6)

1

– On oxidation (KMnO4) benzoic acid m.p. 122°C.

p–Xylene

137 (m.p. 15)

90

137

(2, 3, 5)

1

– On oxidation (KMnO4)– terephthalic acid, m.p. > 300.

m-Xylene

139

91

182 83

(2, 4, 6) (2, 4)

1

– On oxidation (KMnO4)– isophthalic acid. m.p. > 300.

Table 7.11 Experimental Hydrocarbons Advanced Organic Chemistry Name

o-Xylene

B.P. (°C)

465

Structure

142

Pic- rate

88

CH3

Nitro Derivative Pro- (Position cedure of NO2 group) 71 178

(4, 5) (tri)

– On oxdn. (KMnO4)– phthalic acid, m.p. 195. – With Br2 (130°) yields dibromide, m.p. 93.

CH3

Styrene

146

Cumene (Isopropylbenzene)

152

α-Pinene (d or 1)

156

n-Propylbenzene

159

CH (CH3)2

—

109

Specific test and derivativs

– Decolourises Br2 water\ – On oxidation (alk. KMnO4)Benzoic acid m.p. 122. – Dibromide, m.p. 73. – On oxidation (acidic KMnO4)— Benzaldehyde, b.p. 179. – On oxidation (KMnO4)— benzoic acid, m.p. 122

(2, 4, 6)

– Decolourises Br2 water. – Dibromide derivative, m.p. 164°C.

CH2CH2CH3

103

—

—

—

97

235 86

(2, 4, 6) (2, 4)

2

—

62

(2, 4)

– On oxidation (KMnO4)— benzoic acid, m.p. 122°C.

CH3

Mesitylene

164

H 3C CH3

tert-Butyl benzene

169

pseudo cumene

169

97

185

(3, 5, 6)

p-Cymene

176

— —

118 54

(2, 3, 6) (2, 6)

Limonene

176

– Decolourises Br2 water. – Tetrabromide, m.p. 104°C.

Dipentene (dl-limonene)

176

– Decolourises Br2 water. – Tetrabromide, m.p. 104°C.

C(CH2)3

1

466

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

B.P. (°C)

Structure

Pic- rate

Nitro Derivative Pro- (Position cedure of NO2 group)

Specific test and derivativs

CH

Indene

180

CH CH2

98 — (explosive)

—

—

– Decolourises Br2 water. – Dibromide, m.p. 32°C. – Gradually, polymerises on heating. – Oxidation (HNO3)— phthalic acid, m.p. 195°C.

Decalin (Trans)

187

cis (Decahydronaphthalene)

195

—

164

—

—

– On boiling with 35% HNO3 gives a nitro derivative (insoluble in alcoholic KOH), b.p. 96/2 mm and a dinitro derivative, m.p. 164°C.

Tetralin (Tetrahydronaphthalene)

206

—

95

(5, 7)

2

– On oxidation (CrO3)— phthalic acid, m.p. 195°C. – Bromination (2 moles Br2) yields dibromo derivative, m.p. 70°C.

α-Methylnaphthalene

241

141

71

(4)

3

– On oxidation (dil. HNO3) gives α-naphthoic acid m.p. 162°C.

β-Methylnaphthalene

241 (m.p. 32)

115

81

(1)

3

Diphenylmenthane

264 (m.p. 26)

—

172 (2, 2′, 4, 4′) 1

(–) Camphene

CH3

CH3

– On oxidation (conc. HNO3) gives b-Naphthoic acid, m.p. 184°C.

CH2

– Decolourises Br2 water. – Dibromide, m.p. 89°C. – Hydrochloride, m.p. 149°C.

m.p. 52 (b.p. 158) CH2

—

180 (4, 4′)

1

– On oxidation (KMnO4)— benzoic acid, m.p. 122°C. – On boiling with KCIO3 in HCl yields stilbene, m.p. 125°C.

m.p. 70

—

234 (4, 4′)

5

– On oxidation (CrO3)— benzoic acid, m.p. 122°C. – With Br2 in presence of Fe yields p, p’-dibromo derivative m.p. 169°C.

m.p. 80

150

61(1)

3

– Addition of anhydrous AlCl3 to chloroform solution of the compound gives green colour. – On oxidation (KMnO4) gives phthalic acid, m.p. 195°C.

Dibenzyl

m.p. 53

Diphenyl

Napthalene

(C6H5CH2)2

Table 7.11 Experimental Hydrocarbons Advanced Organic Chemistry Name

M.P. (°C)

Triphenylmethane

92

Acenaphthene

95

Retene (8-Methyl-2-isopropyl phenanthrene)

98

467

Structure

(C6H5)3CH

CH3 CH CH3

Pic- rate

Nitro Derivative Pro- (Position cedure of NO2 group)

—

206 (4, 4′, 4″) 4

Specific test and derivativs

– Reduction of trinitro derivative (Zn dust + CH3COOH) gives triamino derivative, m.p. 208°C, which on oxidation (PbO2) gives intense red colour of rosaniline. – Bromination (Br2 at 130°) gives triphenyl bromoethane, m.p. 152°C.

161

101(5)

6

– Gives addition compound with 2,4-dinitrotoluene m.p. 60°C. – Gives addition compound with 2, 4, 6-trinitrotoluene, m.p. 109°C.

124 (ustable)

—

—

—

– On oxidation (K2Cr2O7 + H2SO4) yields retenequinone m.p. 197°C. – Wi t h B r 2 a t 9 0 – 1 0 0 ° yields dibromo derivative, m.p. 180°C.

—

—

—

– On oxidation gives phenanthraquinone, m.p. 206° and diphinic acid, m.p. 228°C. – Gives compound with 2, 4-dinitrochlorobenzene, m.p. 44°C. – Gives compound with 2, 4, 6-trinitrotoluene, m.p. 87°C. – On warming with conc. H2SO4 gives blue colour. – On oxidation (CrO3 + HOAc) gives fluorenone, m.p. 83°C. (procedure B, See page 468). – On oxidation (alkaline KMnO4) gives phthalic acid, m.p. 195°C.

CH3

Azulene

99

Phenanthrene

100

143

Fluorene

114

79 199(2, 7)

—

—

—

—

—

Stilbene (trans) (cis)

125 (141/12)

CH

CH

—

– Decolourises Br2 water on warming. – Dibromide, m.p. 237°C. – On oxidation (KMnO4 + dil H2SO4) gives smell of bitter almonds.

468

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

M.P. (°C)

Structure

Pic- rate

Anthracene

216

138

Chrysene

255

273 (unstable)

Nitro Derivative Pro- (Position cedure of NO2 group) —

—

—

Specific test and derivativs

– On bromination (Br2 in CCl4) yields dibromo derivative, m.p. 221°C. – On oxidation (CrO3 in CH3COOH) gives anthraquinone, m.p. 280°C. (procedure B, see below). – On oxidation (CrO3 in HOAc) gives chrysaquinone, m.p. 240°C. (procedure B. see below).

Procedure for nitration: 1. Fuming HNO3 with conc. H2SO4 2. Mixed conc. HNO3 and H2SO4 3. Cold conc. HNO3 10 min. 4. Fuming HNO3 at room temperature 10 min. 5. Conc. HNO3 at 100°, 10 min. 6. Add solution of fuming HNO3 in glacial acetic acid dropwise to ice-cold solution of hydrocarbon in glacial acetic acid. Keep at room temperature for 10 min. Do not heat. Procedure for Oxidation Procedure A: Add 1 g. cyclohexene in small portions to 5 ml of warm conc. HNO3 (caution). When the addition is complete heat under reflux for 15 min. Cool, filter the separated adipic acid and recrysallise from small volume of water. Procedure B: Add a solution of 1 g of CrO3 in water (2 ml) to a solution of the hydrocarbon (0.5 g) in glacial acetic acid. Reflux for 30 min. cool and add water. Filter the formed quinone and crystallise from glacial acetic acid. Note: See MSDS (material safety data sheet) for polysubstituted nitro compounds as some of these are explosive in nature and should not be prepared as derivatives by students.

Table 7.12 Name Ethyl chloride

Halohydrocarbons (Alkyl and Aryl Halides)

Structure

B.P. (°C)

Derivatives

C2H5Cl

12

– S-Alkylisothiouronium picrate, m.p. 188°C

(CH3)2CHCl

36

– Alkyl-β-naphthyl ether, m.p. 41°C. – S-Alkylisothiouranium piocate, m.p., 196°C (2 hr heating)

Ethyl bromide

C2H5Br

38

– S-Alklsothiouronium picrate, m.p. 188°C.

Methylene chloride (Dichloromethane)

CH2Cl2

40

CH3l

43

– Gives yellow ppt. with alcoholic AgNO3 in cold. – β-Naphthyl ether m.p. 70°C. – Quatenary salt (with quinoline, m.p. 72°C), (with pyridine m.p. 117°C).

n-Propyl chloride

CH3CH2CH2Cl

46

– β-Naphthyl ether, m.p. 39°C. – S-Alkylisothiouronium picrate, m.p. 177°C (unstable; 3 hr heating).

Allyl chloride

CH2=CHCH2Cl

46

– Decolourises Br2 water. – S-Alkylisothiouronium picrate, m.p. 154°C.

Isopropyl chloride

Methyl iodide

β-Naphthyl ether, m.p. 133°C.

Table 7.12 Experimental Halohydrocarbons and Aryl Halides) Advanced Organic(Alkyl Chemistry Name

Structure

469 B.P. (°C)

Derivatives

n-Amyl chloride

CH3(CH2)4 Cl

106

– Picrate, of Alkyl-β-naphthyl ether m.p. 67°C. – S-Alkylisothiouronium picrate, m.p. 154°C.

Ethylene bromide

CH2BrCH2Br

132 (m.p. 10°)

– Addition product with pyridine, m.p. 295°C (decompose). – S-Alkylthiouronium picrate, m.p. 260°C.

Chlorobenzene

n-Hexyl chloride

Cl

CH3(CH2)4CH2Cl

Chlorocyclohexane (Cyclohexyl chloride) sym. tetrachloroethane Bromoform

Cl

132

134 142

Cl2CH–CHCl2

147

CHBr3

151 (m.p. 9)

Bromobenzene

p-Chlorotoluene

CH3

– Alkyl β-naphthyl ether, m.p. 116°C.

– Similar to chloroform.

156

– On nitration (conc. HNO3 + conc. H2SO4 at 80–90°) yields 2,4-dinitro derivative, m.p. 75°C. – p-nitro bromobenzene m.p. 124°C. – On adding to large excess of chlorosulphonic acid yields p-bromobenzenesulphonyl chloride, m.p. 75°C, which on treatment with NH3 gives the sulphonomaide, m.p. 160°C.

159

– On oxidation (alkaline KMnO 4 ) gives o-chlorobenzoic acid, m.p. 142°C. – On nitration (conc. HNO3 + conc. H2SO4 at 80–90°) yields, 3, 5-dinitro derivative, m.p. 63°C.

162

– Oxidation (alkaline KMnO4) gives p-chlorobenzoic acid, m.p. 243°C.

162

– On oxidation (alkaline KMnO 4 ) gives m-chlorobenzoic acid, m.p. 158°C. – On nitration (conc. HNO3 + conc. H2SO4) gives 4, 6-dinitroderivative, m.p. 91°C.

164

– β-Naphthyl ether, m.p. 116°C.

Br

o-Chlorotoluene

– Nitration (conc. HNO3 + conc. H2SO4 warming 80–90°) yields 2,4-dinitro derivative m.p. 51°C. – Mono nitro (p-) m.p. 84°C. – On adding to large excess of chlorosulphonic acid gives p-chlorobenzene sulphonyl chloride m.p. 53°, which on treatment with NH4OH yields the sulphonamide, m.p. 144°C.

Cl

m-Chlorotolueue

CH3

Cl

Bromocyclohexane (Cyclohexyl bromide)

Br

470

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

Structure

m-Dichlorobenzene

B.P. (°C)

Derivatives

172

– On nitration (fuming HNO3) gives 4, 6-dinitro derivative, m.p. 103°C.

179

– On boiling with lead nitrate solution gives smell of bitter almonds (benzaldehyde, b.p. 179°C). – On oxidation (K2Cr2O7 + H2SO4) gives benzoic acid, m.p. 122°C. – On nitration (fuming HNO3 in cold) gives p-nitro derivative, m.p. 71°C. – Forms quartenary salt with with quinoline, m.p. 170°C. − β-Napththyl ether, m.p. 101°C (Picrate m.p. 123°C).

179

– On nitration gives 4, 5-dinitro derivative, m.p. 110°C.

181

– On oxidation (alkaline KMnO 4 ) gives o-bromobenzoic acid, m.p. 150°C. – On nitration (conc. HNO3 + conc. H2SO4-warm) gives dinitro derivative, m.p. 82°C.

183

– On oxidation (alkaline KMnO 4 ) gives m-bromobenzoic acid, m.p. 155°C. – On nitration (conc. HNO3 + conc. H2SO4) gives 4, 6-dinitro derivative, m.p. 103°C (temperature 40–50° for 20 min).

185 m.p. 27

– On oxidation (alkaline KMnO4) gives p-bromobenzoic acid, m.p. 251°C. – On nitration (conc. HNO3 + conc. H2SO4) gives 2-nitro derivative m.p. 47°C.

I

188

– On nitration (conc. HNO3 + conc. H2SO4) (40–50° for 20 min) gives 4-nitro derivative, m.p. 171°C. – On bromination with bromine in glacial acetic acid gives p-bromoiodobenzene, m.p. 91°C.

CH2CH2Cl

190

– β-Napthyl ether m.p. 70°C. – Anilide m.p. 97°C.

CH2Br

198

– On oxidation gives benzoic acid, m.p. 122°C.

204

– On oxidation gives m-iodobenzoic acid, m.p. 186°C. – On nitration (conc. HNO3 + conc. H2SO4) gives nitro derivative, m.p. 108°C.

Cl

Cl

Benzyl chloride

CH2Cl

o-Dichlorobenzene

Cl

Cl

o-Bromotoluene

CH3

Br

m-Bromotoluene

CH3

Br

p-Bromotoluene

Br

Iodobenzene

β-Phenylethyl chloride

Benzyl bromide m-Iodotoluene

CH3

CH3 I

Table 7.12Experimental Halohydrocarbons and Aryl Halides) Advanced Organic(Alkyl Chemistry Name

Structure

o-Iodotoluene

471 B.P. (°C)

CH3

211

– On oxidation (dil. HNO3) gives o-iodobenzoic acid, m.p. 162°C. – On nitration (furming HNO3 in cold) gives 6-nitro derivative, m.p. 103°C.

211 (m.p. 35)

– On oxidation (CrO3) gives p-iodobenzoic acid, m.p. 265°C. – On heating (210–250°) with copper powder yields di-p-tyol, m.p. 125°C.

212

– On refluxing with excess of phenyl hydrazine in ethanol yields benzaldehyde phenylhydrazone, m.p. 158°C. – On boiling with Na 2 CO 3 solution yields benzaldehyde, b.p. 179°C.

219

– On nitration (conc. HNO3 + conc. H2SO4) gives 4-nitroderivative, m.p. 61°C.

I

p-Iodotoluene

CH3

I

Benzal chloride

m-Dibromobenzene

CHCl2

Br



– With excess chlorosulphonic acid yields sulphonyl chloride, m.p. 79°, which with NH4OH yields the sulphonamide, m.p. 190°C.

Br

Benzotrichloride

o-Dibromobenzene

Derivatives

C6H5CCl3

220

– On heating with Na2CO3 solution gives benzoic acid, m.p. 122°C. – On boiling with alcohol yields HCl and ethyl benzoate, b.p. 213°C.

Br

224

– On nitration (conc. HNO3 + conc. H2SO4) gives 4,5-dinitro derivative, m.p. 114°C. – With excess chlorosoluphonic acid yields sulphonyl chloride, m.p. 34°, which with NH4OH gives the sulphonamide, m.p. 175°C.

Br

α-Chloronaphthalene

Cl

259

– On nitration (fuming HNO3) yields 4-5-dinitro derivative, m.p. 180°C. – On heating with conc. H2SO4 gives 4-sulphonic acid, m.p. 130°C. – Picrate, m.p. 137°C.

α-Bromonaphthalene

Br

281

– On oxidation (Cr2O3-acetic acid) yields phthalic acid, m.p. 195°C. – On nitration (conc. HNO 3 ) gives 4-nitro derivative, m.p. 85°C. – Pictrate, m.p. 134°C

305

– Picrate, m.p. 128°C.

α-Iodonaphthalene

Benzyl iodide

CH2I

m.p. 24

– Forms quaternary salt with dimethylaniline, m.p. 165°C.

472

Chapter 7 Tables of Organic Compounds and their Derivatives

Name

M.P. (°C)

Derivatives

Cl

53

– On nitration (fuming HNO3, the mixture is initially cooled in ice and then allowed to stand at room temperature for 10 min) gives 2-nitro derivative, m.p. 54°C.

I

55

– Picrate, m.p. 95°C. – Anilide m.p. 170°C.

β-Chloronaphthalene

Cl

56

– On nitration (conc. HNO3 + conc. H2SO4; warming for 6-8 hrs) gives 1, 8-dinitro derivative, m.p. 175°C. – Picrate m.p. 81°C.

β-Bromonaphthalene

Br

59

– Picrate m.p. 86°C. – Anilide m.p. 171°C.

66

– On nitration (conc. HNO3 + conc. H2SO4) gives 2-nitro derivative, m.p. 72°C.

77

– On oxidation (alkaline KMnO 4 ) gives p-chlorobenzoic acid 240°C.

81

– With Br2 yields ethylene bromide, b.p. 132°C.

89

– On nitration (fuming HNO3; the mixture initially cooled in ice and then left for 10 min. at room temperature) gives 2, 5-dinitro derivative, m.p. 84°C.

89

– On oxidation (alkaline KMnO 4 ) gives p-bromobenzoic acid m.p. 251°C.

CBr4

92

– With alcoholic AgNO3 gives yellow precipitate of AgBr.

(C6H5)3CCl

113

– On boiling with H2O for 5 min gives triphenyl carbinol, m.p. 162°C.

p-Dichlorobenzene

Structure Cl

β-Iodonaphthalene

p-Bromochlorobenzene

Cl

Br

p-Chlorodiphenyl Cl

Ethylene iodide

CH2ICH2I

p-Dibromobenzene

p-Bromo biphenyl

Carbon tetrabromide Triphenyl methyl chloride

Br

p-Iodo biphenyl Iodoform

I

CHl3

114 119

– With alcoholic AgNO3 gives yellow ppt. of Agl. – On warming with small quantity of phenol or resorcinol in dil. alcoholic NaOH gives red colour. – Forms addition product (in dry ether) with quinoline, m.p. 65°C.

Table 7.13 Experimental Carboxylic Acid Halides Advanced Organic Chemistry Name

473

Structure

1, 3, 5-Tribromobenzene (sym. Tribromobenzene)

M.P. B.P. (°C) (°C) 120

Br

Br

p-Diiodobenzene

Derivatives

Br I

I

129

– 2, 5-Dinitro derivative, m.p. 171°C.

139

– 3-Nitro derivative, m.p. 99°C.

149

– On oxidation (CrO3) gives p-chlorobenzoic acid, m.p. 240°C.

164

– On oxidation (CrO3) gives p-bromobenzoic acid, m.p. 251°C.

181

– 3-Nitro derivative, m.p. 168°C.

187

– With Zn dust + dil. H2SO4 yields tetrachloro ethylene, b.p. 121°C.

228

– On boiling with fuming HNO3 yields chloranil, m.p. 290°C.

Cl Cl

1,2,4,5-Tetrachlorobenzene Cl Cl

p.p′-Dichlorodiphenyl

p-p′-Dibromodiphenyl

Cl

Cl

Br

Br

Br Br

1, 2, 4,5-Tetrabromobenzene Br Br

Hexachloroethane

Cl3C—CCl3 Cl Cl

Cl

Cl

Cl

Hexachlorobenzene Cl

Table 7.13 Name

B.P. (°C)

Structure

Carboxylic Acid Halides Anilide

Derivatives Specific tests and derivatives p-ToluiAmide dide

Acetyl chloride

55

CH3COCl

114

148

82

Oxalyl chloride

64

(COCl)2

246 (di)

267 (di)

419 (di)

– The solution in H2O (Obtained on heating) is made alkaline line (NH 4 OH) and excess NH3 boiled off. Addition of FeCl3 solution gives wine red colour.

474 Name

Chapter 7 Tables of Organic Compounds and their Derivatives B.P. (°C)

Structure

Anilide

Derivatives Specific tests and derivatives p-ToluiAmide dide

Propionyl chloride

80

CH3CH2COCl

104

124

79

– On hydrolysis (heating with water) gives propionic acid, b.p. 140°C.

Acetyl bromide

80

CH3COBr

114

148

82

– Same test as in case of acetyl chloride. On hydrolysis gives acetic acid, b.p. 118°C.

Iso-Butyryl chloride

92

(CH3)2CHCOCl

105

104

129

– On hydrolysis (NaOH solution) gives isobutyric acid, b.p. 155°C.

n-Butyryl chloride

100

CH3CH2CH2COCl

96

72

114

– On hydrolysis gives n-butyric acid, b.p. 163°C.

Chloroacetyl chloride

105

CICH2COCl

137

162

120

– On hydrolysis gives chloroacetic acid, m.p. 63°C.

iso-Valeryl chloride

115

(CH3)2 CHCH2COCl

110

109

135

– On hydrolysis gives isovaleric acid, b.p. 176°C.

n-Valeryl chloride

127

CH3CH2CH2CH2COCl

63

70

104 (114)

Bromoacetyl bromide

149

BrCH2COBr

130

91

91

– On hydrolysis gives bromoacetic acid, m.p. 50°C.

α-Bromopropionyl bromide

153

CH3CHBrCOBr

99

125

123

– On hydrolysis gives αbromopropionic acid, b.p. 205°C.

Succinoyl chloride

190 (d)

CH2COCl

228 (di) (148 mono)

255 (di)

260 (di)

– On hydrolysis gives succinic acid m.p. 185°C.

164

158

128

– On heating with NaOH solution and acidification gives benzoic acid, m.p. 122°C. – Benzoyl glycine, m.p. 187°C.

118

136

154

– Onhydrolysis gives phenyl acetic acid, m.p. 76°C.

168

186

162

– On hydrolysis yields anisic acid, m.p. 184°C.

201

COCl

250(di) 169 (mono)

CH COCl

151

168

Benzyl chloride

197

Phenylacetyl chloride

210

Anisoyl chloride

262

Phthaloyl chloride

276

Cinnamoyl chloride

257 (trans) (m.p. 35)

CH2COCl

CH2COCl

COCl

H3CO

COCl

CH

– On hydrolysis given n-valeric acid, b.p. 186°C.

220(di) – On hydrolysis yields phthalic 149 (mono) acid, m.p. 195°C.

147

– On hydrolysis yields cinnamic acid, m.p. 133°C.

Table 7.14 Experimental Nitro Compounds Advanced Organic Chemistry Name

p-Bromobenzoyl chloride

M.P. (°C)

42 (b.p. 245)

2-Naphthoyl chloride

51

3, 5-Dinitrobenzoyl chloride

74

Structure

75

Terephthaloyl chloride

83 (b.p. 263)

Anilide

—

189

– On hydrolysis (dil. NaOH solution) yields p-bromobenzoic acid, m.p. 251°C.

170

191

192

– On hydrolysis (dil. NaOH) yields 2-naphthoic acid, m.p. 184°C.

234

—

183

– On hydrolysis (dil. NaOH solution) gives 3, 5-dinitrobenzoic acid, m.p. 205°C.

204 (211)

203

201

– On hydrolysis (dil. NaOH) yields p-nitrobenzoic acid, m.p. 240°C.

337

—

—

– On hydrolysis (dil. NaOH solution) yields terephthalic acid, m.p>300°.

COCl

ClOC

NO2

COCl

COCl

Table 7.14 Name

197

COCl

O2N

B.P. (°C)

Nitromethane

101

Nitrobenzene

209

Nitro Compounds

Structure

224

NO2

CH3

Phenylnitromethane

226 (d)

Derivatives

CH3NO2 NO2

o-Nitrotoluene

Derivatives Specific tests and derivatives p-ToluiAmide dide

COCl

Br

O2N

p-Nitrobenzoyl chloride

475

CH2NO2

– On reduction (Sn + HCl, heat) gives aniline, b.p. 183°C (characterised by dye test). – On boiling with dil. sodium arsenite yields azoxybenzene, m.p. 36°C. – On warming with fuming HNO3 + H2SO4 yields m-dinitrobenzene, m.p.90°C. – On warming with Br2 + FeCl3 (or FeBr3) yields m-bromo derivative, m.p. 54°C. – On reduction (Sn + HCl) gives o-toloudine, b.p. 199°C. – On oxidation (KMnO 4 , heat) gives o-nitrobenzoic acid, m.p. 146° – On nitration (conc. HNO3 + H2SO4) at 100° for 3 min. yields 2,4-dinitrotoluene, m.p. 70°C. – Reduction (Sn + HCl) gives benzylamine, b.p. 184°C

476 Name m-Nitrotoluene

Chapter 7 Tables of Organic Compounds and their Derivatives B.P. (°C) 230 (m.p. 16)

Structure

Derivatives – On reduction (Sn + HCl) gives m-toluidine, b.p. 202°C. – On oxidation (KMnO4 or K2Cr2O7 + dil. H2SO4 boiling) gives m-nitrobenzoic acid, m.p. 141°C.

CH3

NO2

p-Nitroethylbenzene

243

p-Nitro-m-xylene

244

C2H5

O 2N

CH3

CH3 NO2

2-Nitro-p-Cymene

264

o-Nitrophenetole

267

OC2H5

272

OCH3

NO2

4-Nitro-o-xylene

30 m.p.

– On oxidation (KMnO4) gives 4-nitroisophthalic acid, m.p. 258°C. – On reduction (Sn + HCl) gives 4-m-xylidine, b.p. 212°C. – With fuming HNO 3 in cold yields 4, 6-dinitro-m-xylene, m.p. 93°C. – On reduction (Sn + HCl) gives 2-aminop-cymene, b.p. 241 (acetyl derivative, m.p. 115°C; benzoyl derivative, m.p. 102°). – On oxdiation (alkaline KMnO4) yields 2-nitro-4-isopropylbenzoic acid, m.p. 168°C. – On warming with conc. HNO3 + H2SO4 yields 2, 6-dinitro-p-cymene, m.p. 54°C.

NO2

o-Nitroanisole

– On reduction (Sn + HCl) gives p-ethyl aniline, b.p. 214, (acetyl derivative m.p. 94°).

CH3 CH3

– On reduction (Sn + HCl) yields o-phenetidine b.p. 229°C. – With cold HNO 3 + H 2 SO 4 yields 2, 4-dinitrophenetole, m.p. 86°C. – With HNO3 + H2SO4 on warming yields trinitrophentole, m.p. 78°C. – On boiling with 48% BHr yields o-nitrophenol, m.p. 45°C. – On nitration (conc. HNO3 + conc. H2SO4 at 0°C for 15 min.) gives nitro-derivatives, m.p. 95°C.

– On reduction (Sn + HCl) gives 3, 4Dimethyl anline m.p. 50 (acetyl derivative, m.p. 99°) – On nitration–dinitro derivative, m.p. 82°C.

NO2

o-Chloronitrobenzene

32 m.p.

NO2 Cl

– On nitration (conc. HNO3 + conc. H2SO4) (30–50°, 10–15 min. or at 100° for 2 min, gives 2, 4-dinitro chlorobenzene, m.p. 51°C.

Table 7.14 Experimental Nitro Compounds Advanced Organic Chemistry Name o-Nitrobiphenyl

477

M.P. (°C)

Structure

37

Derivatives – On reduction gives o-aminobiphenyl, m.p. 49°C (acetyl derivative, m.p. 121°C).

NO2

o-Bromonitrobenzene

41 (b.p. 261)

NO2

Br

Nitromesitylene

44 (b.P. 255)

CH3 NO 2

CH 3

CH3

m-Chloronitrobenzene

45 (b.p. 236)

– With conc. HNO3 and conc. H2SO4 at 30–50° for 10-15 min gives 2, 4-dinitro derivative, m.p. 74°C.

– On reduction (Sn + HCl) gives mesidine, b.p. 232°C (acetyl derivative, m.p. 216°C; benzoyl derivative, m.p. 204°C) – On warming with fuming HNO3 yields dinitromesitylene, m.p. 86°C. – On boiling with KOH and Na2S in methanol yields 3, 3′-dichloroazoxybenzene m.p. 97°C.

NO2

Cl

o-Nitrobenzyl chloride

48

CH2Cl

2,4-Dinitrochlorobenzene

51

Cl

NO2

NO2

NO2

p-Nitrotoluene

53

CH3

NO2

p-Nitroanisole

54

OCH3

NO2

– On adding SnCl2 + HCl at 40–50° and then adding Zn dust yields o-toluidine, b.p. 199°C. – With KMnO4 solution yields o-nitrobenzoic acid, m.p. 146°C. – On boiling with KCNS in 90% alcohol yields the thiocyanate, m.p. 75°C. – With Sn + HCl yields chloro-m-phenylene diamine, m.p. 91 (diacetyl, m.p. 242°C; dibenzoyl derivative, m.p. 178°C). – On warming with fuming HNO3 + fuming H2SO4 yields picryl chloride, m.p. 83°C. – On boiling with dil. NaOH solution yields 2, 4-dinitrophenol m.p. 114°C. – On oxidation (KMnO4) gives p-nitrobenzoic acid, m.p. 240°C. – With conc. HNO3 + conc. H2SO4 at 30–50° for 10–15 min gives nitro derivative, m.p. 70°C. – On reduction (Sn + HCl, heat) gives p-toluidine, m.p. 44°C. – With conc. HNO3 + conc. H2SO4 at 30–50° for 10–15 min. gives nitro derivative m.p. 88°C. – On reduction (Sn + HCl) gives p-anisidine, m.p. 57°C. – On boiling with 48% HBr gives p-nitrophenol, m.p. 114°C.

478 Name 2-Nitro-1, 4-dichlorobenzene

Chapter 7 Tables of Organic Compounds and their Derivatives M.P. (°C) 55

Structure Cl

Derivatives – With Sn + HCl yields 2, 5-dichloroaniline, m.p. 50°C b.p. 251°C (Acetyl derivative, m.p. 133°C, benzoyl derivative m.p. 120°C). – On boiling with 1 mole KOH in methanol yields 2-nitro-4-chloroanisole, m.p. 98°C. – On boiling with fuming HNO3 + H2SO4 (caution) yields, nitro derivative, m.p. 104°C.

Cl NO2

m-Nitrobromobenzene

56

– With Sn + HCl yields m-bromoaniline b.p. 251°C. – On heating with excess conc. HNO3 + H2SO4 yields nitro derivative, m.p. 59°C. – On boiling with methanolic KOH yields 3,3′-dibromoazoxy benzene, m.p. 111°C.

Br

NO2

p-Nitropheneole

59

1-Nitronaphthalene

60

2, 6-Dinitrotoluene

66

H5C2O

NO2

2, 4-Dinitrotoluene

70

– On nitration with excess conc. HNO3 + H2SO4 gives 1, 3, 8-trinitro naphhalene, m.p. 218°C. – On oxidation (CrO3—CH3COOH) gives 3-nitrophthalic acid, m.p. 218°C.

NO2

CH3 O 2N

NO2

NO2

71 O2 N

2, 4-Dinirobromobenzene

74

CH2Cl

Br NO2

NO2

– On reduction (Sn + HCl) gives 2, 6-diaminotoluene, m.p. 105°C. – On oxidation (KMnO 4) gives 2, 6-dinitrobenzoic acid m.p. 202°C. – On reduction with excess Sn + HCl gives 2,4-diamino toluene, m.p. 99°C. – On oxidation (KMnO 4 ) gives 2, 4, dinitrobenzoic acid, m.p. 183°C. – With acetone + KOH gives blue colour changing to violet by addition of acetic acid. – With benzaldehyde + little piperidine yields dinitrostilbene, m.p. 139°C.

CH3 NO2

p-Nitrobenzyl chloride

– With conc. HNO3 + conc. H2SO4, room temperature (ice cooling if necessary) for 10 min. yields nitro derivative, m.p. 86°C. – With Sn + HCl gives p-phenetidine, b.p. 254°C. (acetyl derivative, m.p. 134°C).

– On oxidation (KMnO4) yields p-nitrobenzoic acid, m.p. 240°C. – On heating with Na2CO3 solution gives p-Nitrobenzyl alcohol, m.p. 93°C. – With Sn + HCl yields m-phenylenedimine, m.p. 63°C. – On heating with Na2CO3 solution yields 2, 4-dinitrophenol, m.p. 114°C.

Table 7.14 Experimental Nitro Compounds Advanced Organic Chemistry Name

479

M.P. (°C)

2, 4-Dimethyl-6-nitroanliine

76

2, 4, 6-Trinitrophenetole

78

Structure

Derivatives – Acetyl, m.p. 176°C. – Benzoyl, m.p. 185°C.

OC2H5 O 2N

NO2

– On boiling with aq. KOH yields picric acid, m.p. 122°C.

NO2

2,4,6-Trinitrotoluene

82

CH3 O 2N

NO2

– On oxidation (CrO 3 + conc. H 2 SO 4 ) yields 2, 4, 6-trinitrobenzoic acid, m.p. 220°C.

NO2

p-Chloronitrobenzene

83

Picryl chloride (2, 4, 6-Trinitrochlorobenzene)

83

Cl

NO2

Cl O2 N

NO2

NO2

2-Nitro-p-dibromobenzene

85 Br

Br

– With conc. HNO3 + conc. H2SO4 at 30–50° for 10-15min. yields 2, 4-dinitro derivative m.p. 51°C. – On boiling with conc. NaOH solution yields p-nitrophenol, m.p. 114°C. – On boiling with dil. NaOH solution yields picric acid, m.p. 122°C. – On boiling with KOH in methanol yields trinitroanisole, m.p. 68°C. – With KOH in ethanol yields trinitrophenetole, m.p.78°C. – With NH3 in alcohol yields picramide, m.p. 189°C. – On warming with aniline yields trinitrodiphenylamine, m.p. 177°C. – On boiling with KOH in methanol yields 4-bromo-2-nitroanisole, m.p. 86°C.

NO2

m-Dinitrobenzene

90

NO2

NO2

– On reduction with NH4SH in alcohol yields m-nitroaniline, m.p. 114°C. – On reduction with Sn + HCl yields m-phenylene diamine, m.p. 63°C. – On boiling with alkaline K3Fe(CN)6 yields 2, 4-dinitrophenol, m.p. 114°C. – Gives violet colour on adding a trace of glucose to boiling solution of compound in very dil. NaOH solution.

480 Name 4,6-Dinitro-m-xylene

Chapter 7 Tables of Organic Compounds and their Derivatives M.P. (°C)

Structure

93

Derivatives – On reduction with Sn + HCl yields diamino-m-xylene, m.p. 104°C. – On warming with conc. HNO3 + H2SO4 yields trinitroxylene, m.p. 125°C.

CH3 O2N CH3 NO2

2-4-Dinitroanisole

95

– With conc. HNO3 + H2SO4 at 30–50° for 10–15 min, yields 2, 4, 6-trinitro anisole, m.p. 68°C. – On heating with conc. NaOH solution yields 2, 4-dinitrophenol, m.p. 114°C.

p-Nitro benzyl bromide

99

– On oxidation (KMnO4) yields p-nitrobenzoic acid m.p. 240°C. – On boiling with Na2CO3 solution yields p-nitrobenzyl alcohol, m.p. 93°C.

4-Nitrodiphenyl

114

o-Dinitrobenzene

117

NO2

– With NH4SH in alcohol yields o-nitro aniline, m.p. 71°C. – On reduction (Sn + HCl) yields o-phenylene diamine, m.p. 102°C.

NO2 NO2

5-Nitro-1-naphthylamine

119

1,3,5-Trinitrobenzene

122

– Acetyl derivative, m.p. 220°C.

O2N

1-Nitro-2-napthyl amine

126

p-Bromonitrobenzene

126

– On oxidation with CrO3 in acetic acid gives p-nitrobenzoic acid, m.p. 240°C. – On reduction gives p-aminobiphenyl, m.p. 53°C.

– With dil. NaOH solution gives red colour, which disappears on acidification. – With Na2CO3 solution + K3Fe(CN)6 yields picric acid. m.p.122°C.

NO2

NO2

– Acetyl derivative, m.p. 125°C. – Benzoyl derivative, m.p. 168°C. – p-Tosyl derivative, m.p. 160°C.

O 2N

Br

– With conc. HNO3 + H2SO4 at 30–50 for 10-15 min gives 2, 4-dinitro bromobenzene, m.p. 74°C. – On reduction (Sn + HCl) yields p-bromoaniline, m.p. 66°C.

Table 7.14 Experimental Nitro Compounds Advanced Organic Chemistry Name

481

M.P. (°C)

3-Nitro-4-acetamido-toluene (2-Nitro-4-methylacetanilide)

144

m-Nitrobenzanilide

154

Structure

Derivatives

NHCOCH3

H 3C

NO2

– On oxidation (KMnO4) yields 3-nitro-4-acetamidobenzoic acid m.p. 220°C. – On hydrolysis yields 3-nitro-p-toluidine, m.p. 116°C. – On hydrolysis with conc. HCl or 50% H2SO4 yields m-introbenzoic acid m.p. 141°C.

NO2

CONHC6H5

p-Nitrophenylacetic acid

154

o-Nitrobenzanilide

155

1, 8-Dinitronaphthalene

170

p-Dinitrobenzene

172

4-Nitro-2-methylacetanilide

196

CH2COOH

O 2N

– Anilide, m.p. 212°C. – Amide, m.p. 198°C. – On hydrolysis gives o-nitrobenzoic acid, m.p. 146°C.

– On boiling with fuming HNO3 + conc. H2SO4 (caution) yields, 1, 3, 8trinitronaphthalane, m.p. 218°C. – On reduction (Sn + HCl) in alcohol yields 1, 8-diaminonaphthalene, m.p. 66°C.

NO2 NO2

– With alcoholic NH4SH yields p-nitroaniline, m.p. 147°C. – On boiling with 5% NaOH solution gives p-Nitrophenol, m.p. 114°C. – On reduction (Sn + HCl) yields p-phenylenediamine, m.p. 140°C.

NO2

O 2N

– On hydrolysis with 50% H2SO4(refluxing) gives 5-nitro-o-toluidine, m.p. 130°C. – On oxidation (KMnO4) yields 5-nitro-2acetamino-benzoic acid, m.p. 215°C.

NHCOCH3 CH3

NO2

p-Nitrobenzanilide

211

1, 5-Dinitronaphthalene

214

CONHC6H5

O 2N

– On reduction (Sn + HCl) yields 1, 5-diaminonaphthalene, m.p. 190°C. – On boiling for 5 min with fuming HNO3 + conc. H2SO4 yields 1, 4, 5-trinitronapthalene, m.p. 154°C.

NO2

NO2

4,4′-Dinitrodiphenyl

233

O 2N

– On hydrolysis with NaOH solution yields p-nitrobenzoic acid, m.p. 240°C.

NO2

– On reduction (Sn + HCl) gives benzidine m.p. 127°C.

482

Chapter 7 Tables of Organic Compounds and their Derivatives

Table 7.15 Name

B.P. (°C)

Structure

Primary amines

Derivatives (M.P.°C) Acebenzp-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

Picrate

Specific tests and derivatives

Methylamine

Gas

CH3NH2

—

80

75

178

215

– Gives yellow ppt. with Nessler’s reagent.

Ethylamine

18

C2H5NH2

—

71

63

114

166

– Gives white ppt. with Nesslers reagent.

Isopropylamine

32

(CH3)2CHNH2

—

100

51

94

150

– Hydrochloride m.p. 139°C. (Hygroscopic).

tert. Butylamine

46

(CH3)3CNH2

—

134

—

—

198

– Hydrochloride m.p. 310

n-Propylamine

49

CH3CH2CH2NH2

—

85

52

96

135

– Gives brown ppt with Nessler’reagent. – Hydrochloride, m.p 157°C.

Cyclopropylamine

50

—

99

—

—

149

– Hydrochloride m.p. 100°C. – Decolourises KMnO4 solution. – Dibromide, m.p. 135°C. – Hydrochloride, m.p. 105–110°C.

NH2

Allylamine

56

CH2==CHCH2NH2

—

—

64

76

140

sec. Bytylamine

63

CH3CH2CH

NH2

—

76

55

—

140

CH3

Iso-butylamine

68

(CH3)2CHCH2NH2

—

57

78

80

151

– Hydrochloride, m.p. 174°C.

n-Butylamine

77

CH3CH2CH2CH2NH2

—

42

—

—

151

– Hydrochloride m.p. 195°C.

iso-Amylamine

95

(CH3)2CHCH2CH2NH2

—

—

65

91

138

– Hydrochloride, m.p. m.p. 215°C.

n-Amylamine

104

CH3(CH2)4NH2

—

—

—

81

139

Ethylenediamine

116

(CH2NH2)2

172

249

160

302

233

Cyclohexylamine

134

104

148

87

156

158

—

—

—

—

160

114

160

103

156

—

Ethanolamine 171 (2-hydroxy ethylamine) Aniline

184

NH2

HOCH2CH2NH2

– With bromine water. gives tribromo derivative m.p. 119°C. – Hydrochloride, m.p. 198°C. – Azo b-naphthol derivative m.p. 134°C.

Table 7.15 Experimental Primary amines Advanced Organic Chemistry

Name

Benzylamine

B.P. (°C)

483

Structure

Derivatives (M.P.°C) Acebenzp-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

184

60

106

116

116

Picrate

199

Specific tests and derivatives

– Hydrochloride, m.p. 253°C. – Oxalate, m.p. 201°C (d).

1-Phenylethylamine

185

2-Phenylethylamine

198

o-Toluidine

199

C6H5CH2CH2NH2 CH3

57

120

—

118

189

—

116

—

154

167

112

144

109

126

213

– Gives brown colour on shaking ethereal solution with dil. bleaching powder solution – Hydrochloride, m.p. 215°C. – With bromine water gives dibromo derivative m.p. 50°C. – Azo b-naphthol derivative, m.p. 128°C.

65

125

114

160

200

– Hydrochloride 228°C. – On shaking ethereal solution with dil. bleaching powder solution water layer becomes yellow brown and ether layer red. – Solution in 50% H2SO4 gives yellowish colour. – Brown colour with K2Cr2O7. – Red colour with HNO3 – Azo b-naphthol derivative, m.p. 140°C.

87

99

105

150

134

– Yields 5-Nitro derivative, on adding (1 mole) KNO3 to solution in conc. H2SO4— m..p. 117°C.

113

192

—

156

209

– Hydrochloride, m.p. 235°C. – Azo-b-naphthol derivative m.p. 166°C.

139

140

119

150

171

– Hydrochloride, m.p. 228°C. – Azo-b-naphthol derivative m.p. 150°C.

NH2

m-Toluidine

202 CH3 H 2N

o-Chloroaniline

209

NH2

Cl

4-m-Xylidine (2, 4-dimethylaniline)

212

p-Xylidine 214 (2,5-Dimethylaniline) (m.p. 15)

H 3C

NH2

CH3

484

Name

o-Ethylaniline

Chapter 7 Tables of Organic Compounds and their Derivatives

B.P. (°C)

Structure

216

Derivatives (M.P.°C) Acebenzp-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

NH2

Picrate

Specific tests and derivatives

111

147

—

—

95

177

168

212

—

180

144

136

—

—

209

85

60

127

151

200

– On boiling with 48% HBr gives o-aminophenol, m.p. 174°C. – Azo-b-naphthol derive. m.p. 180°C.

92

152

—

—

—

– Reduces Fehling‘s solution on warming.

99

116

90

161

129

79

104

164

164

119

72

121

138

184

177

117

149

—

—

—

– Hydrochloride, m.p. 221°C.

128

168

—

—

—

– Reduces Fehling‘s solution in cold. – Hydrochloride, m.p. 240°C. – With HCl and CuSO4 (warming) yields chlorobenzene, B.P. 132°C.

C2H5

2-m-Xylidine (2, 6-Dimethylaniline)

216

CH3 NH2 CH3

5-m-Xylidine (3, 5-Dimethylaniline)

220

CH3

H2N

o-Anisidine

225

N-Methyl-Nphenylhydrazine

227

o-Bromoaniline

229 (m.p. 31°)

CH3

C6H5N(CH3)NH2 NH2

Br

o-Phenetidine

229

NH2

OC2H5

m-Chloroaniline

230

3-Bromo-4-aminotoluene 240 (m.p. 26)

H2N

CH3

– Hydrochloride, m.p. 214°C. – On boiling with 48% HBr gives o-aminophenol, m.p. 174°C.

Br

Phenylhydrazine

243 (m.p. 19)

NHNH2

Table 7.15 Experimental Primary amines Advanced Organic Chemistry

Name

B.P. (°C)

m-Phenetidine

248

m-Anisidine

251

485

Structure

Derivatives (M.P.°C) Acebenzp-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

Picrate

Specific tests and derivatives

97

103

157

—

158

81

–

68

138

169

87

136

—

178

180

134

173

106

118

69

– Hydrochloride, m.p. 234°C. – Gives red colour with FeCl3. – On boiling with 48% HBr gives p-aminophenol, m.p. 184°C (d). – Azo-b-naphthol derivative m.p. 139°C.

101

100

—

—

106

– On heating with dil. HCl gives anthranilic acid, m.p. 144°C. – Hydrochloride, m.p. 178°C.

NH2 104

172

112

—

—

61

98

112

—

—

– Hydrochloride, m.p. 170°C. – On boiling with dil HCl gives anthranilic acid, m.p. 144°C.

110

114

—

—

—

– Hydrochloride, m.p. 185°C.

76

98

148

—

—

NH2

OCH3

m-Bromoaniline

251 (m.p. 18)

p-Phenetidine

254 H5C2O

Methyl anthranilate

260 (d) (m.p. 25°)

p-Aminodiethylaniline

Ethyl anthranilate

NH2

262

(C2H5)2N

265 (d) (m.p. 13)

Ethyl m-aminobenzoate 294

o-Aminoacetophenone m.p. 20 (b.p. 252)

COCH3

NH2

486

Name

Chapter 7 Tables of Organic Compounds and their Derivatives

M.P. Structure (°C)

p-Amino-N, N-dimethyl aniline

41 (b.p. 262)

Hexamethylenediamine

42 (b.p. 205)

p-Toluidine

44 (b.p. 200)

4-o-Xylidine (3,4-Dimethylaniline)

48

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

(CH2CH2CH2NH2)2

H 3C

NH2

H 3C

NH2

Picrate

132

228

—

—

186

126

155

—

—

220

153

158

117

136

181

99

118

—

141

—

119

86

—

—

—

160

161

157

190

161

133

120

—

—

—

172

230

255

—

—

Specific tests and derivatives

– Hydrochloride, m.p. 220°C – Solution of dihydrochloride gives purple colour with excess AgNO3 in cold and odour of p-benzoquinone on warming.

– Hydrochloride, m.p. 240°C. – With Br2 water gives dibromo derivative, m.p. 73°C. – Azo-β-napthol derivative m.p. 134°C. – Gives yellow colour with K2Cr2O7; blue with HNO 3 , changing gradually, through violet and red to brown.

H 3C

o-Phenylaniline (2-aminobiphenyl)

1-Napthylamine

2,5-Dichloroaniline

49

NH2 C6H5

50 (b.p. 300)

50

NH2

Cl

NH2

Cl

p-Phenylaniline (4-Aminobiphenyl)

53

C6H5

NH2

– Hydrochloride, m.p. 286°C. – Solution of hydrochloride gives a blue ppt. with FeCl3. – Azo-β-naphthol derivative m.p. 274°C.

Table 7.15 Experimental Primary amines Advanced Organic Chemistry

Name

487

M.P. Structure (°C)

2-Aminopyridine

57

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

NH2

Picrate

Specific tests and derivatives

71

165

—

—

217

127

154

114

141

117

156

115

—

—

—

184

222

—

—

—

191

240

172

—

184

– Hydrochloride, m.p. 348°C. – Solution of hydrochloride with NaNO2 gives brown colour. – With Br2 in dil HCl gives tribromo derivative m.p. 158°C.

146

117

126

116

106

– Azo-β-naphthol derivative m.p. 183°C

167

204

101

158

180

178

192

95

167

178

– Azo-β-napthol derivative m.p. 160°C.

93

98

—

—

73

– On reduction (Zn dust + NaOH) gives o-phenylenediamine, m.p. 102°C. – Azo-β-naphthol derivative, m.p. 212°C.

148

172

164

—

—

N

p-Anisidine

4-Bromo-2-methylaniline

57 (b.p. 246) H3CO

59

NH2

Br

NH2

– Hydrochloride, m.p. 216°C. – Solution of hydrochloride gives violet colour with FeCl3. – On boiling with 48% HBr yields p-aminophenol, m.p. 184°C. – Azo-β-naphthol derivative m.p. 139°C.

CH3

p-Iodoaniline

62

m-Phenylenediamine

63

NH2

I

NH2 H 2N

2,4-Dichloroaniline

63 (b.p. 245)

Cl

NH2 Cl

p-Bromoaniline

66 (b.p. 245)

Br

NH2

p-Chloroaniline

70 (b.p. 230)

Cl

NH2

o-Nitroaniline

71

NH2

NO2

2-Nitro-p-toluidine (4-Amino-2-nitrotoluene)

77

CH3

NH2 NO2

488

Name

Chapter 7 Tables of Organic Compounds and their Derivatives

M.P. Structure (°C)

2, 4, 6-Trichloroaniline

77

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

NH2 Cl

Picrate

204

174

—

—

83

Cl

Specific tests and derivatives

– On diazoatizing in alcohol + H2SO4 and then warming yields 1, 3, 5-tricholo robenzene, m.p. 63°C.

Cl

2, 4-Dibromoaniline

79

146

134

—

—

124

2, 4-Diaminochlorobenzene

88

243

178

215

—

—

m-Aminoacetanilide

89

191

—

241

—

—

– Hydrochloride, m.p. 280°C.

NH2 110

149

—

—

131

– Hydrochloride, m.p. 207°C. – On boiling with alkali yields p-aminobenzoic acid, m.p. 186°C.

157

167

—

—

—

– On reduction (Sn + HCl) yields 2, 6-diaminotoluene 105°C. – Azo-β-naphhol derivative, m.p. 215°C.

158

—

—

—

—

– On reduction (Sn + HCl) yields 2, 3-diaminotoluene, m.p. 61°C. – Azo-β-naphthol derivative, m.p. 168°C.

224

224

192

184

170

Hydrochloride m.p. 305°C.

185

301

201

—

208

– Hydrochloride, m.p. 260°C. – On heating with Na acetate + benzaldehyde yields dibenzal derivative m.p. 106°C.

NHCOCH3 H 2N

Ethyl p-aminobenzoate (Benzocaine)

92 H5C2OOC

6-Nitro-o-toluidine

92

NO2 CH3 NH2

3-Nitro-o-toluidine

95

CH3 NH2 NO2

2, 4-Diaminotoluene (m-Toluylenediamine)

99 H 2N

CH3 NH2

o-Phenylenediamine

102 (b.p. 257)

NH2

NH2

Table 7.15Experimental Primary amines Advanced Organic Chemistry

Name

489

M.P. Structure (°C)

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

Picrate

Specific tests and derivatives

O

p-Aminoacetophenone

106

4-Nitro-o-toluidine (2-Amino-4-nitrotoluene)

107

H 3C

C

NH2 167

205

203

—

—

151

183

—

—

—

– On reduction (Sn + HCl) 2, 4-diamotaluene m.p. 99°C.

205 203 (N-acetyl)

—

—

—

– – – –

CH3

O2 N

NH2

β-Glucosamine

110

β-Naphthylamine

112 (b.p. 294)

m-Nitroaniline

114 (b.p. 285)

132

162

133

179

195

Hydrochloride, m.p. 260°C. – Azo-β-napthol deriv. m.p. 174°C.

154

155

138

—

143

On reduction (Sn + HCl) yields m-phenylenediamine, m.p. 63°C – Azo-β-naphthol deriv., m.p. 194°C.

96

148

145

232

198

—

—

—

203 (di)

176 (di)

—

—

240

NH2 153

152

—

—

—

NH2

NH2

NO2

3-Nitro-p-toluidine (4-Amino-3-nitrotolueune)

116

CH3

H2N

– On reduction (Zn dust+ NaOH in dil. alcohol) yields 3, 4-diaminotoluene, m.p. 89°C.

O 2N

2, 4, 6-Tribromoaniline

119

Br

, Br

NH2 Br

2,6-Diaminopyridine

121 H 2N

p-Aminobenzophenone

124

C6H5CO

N

Reduces Fehling’s solution Hydrochloride, m.p. 200°C. Oxime, m.p. 127°C(d). Semicarbazone, m.p., 165°C.

NH2

– On diazotising in alcohol + H2SO4 and then warming yields 1,3,5-tribromobenzene, m.p. 120°C.

490 Name

p-Aminoazobenzene

Chapter 7 Tables of Organic Compounds and their Derivatives M.P. Structure (°C)

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

125

144 N=N

Benzidine

317

Specific tests and derivatives

211

—

—

—

– On reduction (Sn + HCl) gives p-phenylenediamine, m.p. 140°C. – Azo-β-naphthol derivative, m.p. 202°C.

352

243

—

—

– On boiling with dil. H2SO4

NH2

127 H2N

Picrate

NH 2

–

– – –

o-Tolidine

129

314 CH3

5-Nitro-o-toluidine

130

o-Dianisidine

135

138

—

—

185

On boiling with dil. H2SO4 + MnO2 yields toluquinone, m.p. 68°C. – Dihydrochloride, m.p. 355°C – Diazo-β-naphhol derivative m.p. 297°C.

174

—

—

—

– With Sn + HCl gives 2, 5-diaminotoluene m.p. 64°C. – Azo-β-naphtol derivative, m.p. 248°C.

OCH3 243 236

—

—

225

– Dihydrochloride, m.p. 312°C. – Diazo-β-nasphthol derivative m.p. 298°C.

—

—

—

– With Sn + HCl yields 1, 2 3-triaminobenzene, m.p. 103°C.

NH 2

200

OCH3 H2N

2,6-Dinitroaniline

265

CH3

H2N

NH2

NO2 NH2 NO2

+ MnO2 yields bonzoquinone, m.p. 115°C. On gradually adding Br2 to a solution in CS2 under H2O, the aq. layer becomes first blue then green and finally colourless; the CS2 layer becomes red. Hydrochloride, m.p. 385°C. Diazo-β-naphthol derivative m.p. 302°C. On diazotization in dil H2SO4 and boiling yields p, p′-dihydroxy diphenyl, m.p. 272°C.

197

—

Table 7.15 Experimental Primary amines Advanced Organic Chemistry Name

491

M.P. Structure (°C)

p-Phenylenediamine

140

p-Nitroaniline

147

p-Nitrophenylhydrazine

157(d)

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

H 2N

NH2

O 2N

NH2

O2N

NHNH2

303

> 300 266

Picrate

Specific tests and derivatives

177

202

– Dihydrochloride, m.p. > 390°C. – Solution of dihydrochloride with excess AgNO3 solution in cold gives green colour rapidly changing to brown through red. – On boiling with dil. FeCl3 or H2SO4 + MnO2 gives benzoquinone, m.p. 115°C.

215

199

191

—

100

– With Sn + HCl yields pphenylenediamine, m.p. 140°C. – On boiling with conc. NaOH solution yields p-nirophenol, m.p. 114°C. – Azo-β-naphthyl derivative, m.p. 250°C.

205

193

—

—

119

– With Sn + HCl yields pPhenylenediamine, m.p.

303

—

—

—

—

– On boiling with HCl, yields p-phenylenediamine HCl. – Azo-β-napthol derivative, m.p. 261°C.

201 220 — (N(N. acety) benzoyl)

—

—

– Orange colour in dil. NaOH solution

140°C. p-Aminoacetanilide

163 H 2N

Picramic acid

NHCOCH3

168

OH H 2N

NO2

NO2

4-Amino-2-methylphenol

175

2,4-Dinitroaniline

180

NH2 NO2

NO2

103 (di)

194 (di)

109

—

—

120

200

219

—

—

– On boiling with conc. NaOH solution yields 2, 4-dinitrophenol, m.p. 114°C. – Azo-β-naphthol derivative. m.p. 302°C.

492 Name

Chapter 7 Tables of Organic Compounds and their Derivatives M.P. Structure (°C)

2, 4, 6-Trinitroaniline (Picramide)

189

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4tyl oyl uene dinisulph- trophonyl enyl

NH2 O 2N

Picrate

Specific tests and derivatives

230

196

—

—

—

– On boiling with dil. NaOH solution yields picric acid. m.p. 122°C.

197

206

—

—

—

– On boiling with conc. NaOH solution gives 2, 4– dinitrophenol, m.p. 114°C. – On heating with CuSO4 solution gives m-dinitrobenzene, m.p. 90°C.

257

228

304

—

—

NO2

NO2

2,4-Dinitrophenyl hydrazine

198 (d)

NHNH2 NO2

NO2 O

2-Amino anthraquinone

NH2

302

O

Table 7.16 Name

B.P. (°C)

Structure

Secondary amines Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4- Pictyl oyl uene dini- rate sulph- trophonyl enyl

Specific tests and derivatives

Dimethylamine

7 gas

(CH3)2NH

—

42

79

87

158

– Gives white ppt. with Nessler’s solution on dilution with H2O. – Hydrochloride, m.p. 170°C (Hygroscopic).

Diethylamine

55

(C2H5)2NH

—

42

60

80

155

– Gives white ppt. with Nessler’s solution. – Gives yellow ppt. with a solution of HgI2 in KI. – Hydrochloride, m.p. 216°C (hygroscopic)

Di-isopropylamine

84

[(CH3)2CH]2NH

—

—

—

—

140

Pyrrolidine

89

—

—

123

—

112

N H

Table 7.16 Experimental Secondary amines Advanced Organic Chemistry

Name

B.P. (°C)

Piperidine

107

493

Structure

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4- Pictyl oyl uene dini- rate sulph- trophonyl enyl —

48

103 (96)

93

152

—

—

—

40

75

—

—

—

—

69

Specific tests and derivatives

– Hydrochloride, m.p. 247°C. – Nitroso deriv. b.p. 218°C.

N H

Di-n-propylamine

110

Pyrrole

131

(CH3CH2CH2)2NH

N H

Di-isobutylamine

137

[(CH3)2CHCH2]2NH

86

—

—

112

119

– Hydrochloride, m.p. 262°C.

Di-n-butylamine

159

(CH3CH2CH2CH2)2NH

—

—

—

—

60

– Nitroso derivative, m.p. 61°C.

N-Methylbenzylamine

181

C6H5CH2NHCH3

—

—

95

—

—

Di-isoamylamine

186

[(CH3)2CHCH2CH2]2 NH

—

—

—

—

—

– Hydrochloride, m.p. 276°C.

N-Methylaniline

194

C6H5NHCH3

102

63

95

167

147

– Hydrochloride, m.p. 121°C.

N-Ethylaniline

205

C6H5NHC2H5

54

60

88

95

138

– Hydrochloride, m.p. 176°C.

N-Methyl-m-toluidine

206

66

—

—

—

—

55

66

120

155

90

NHCH3 H3 C NHCH3

N-Methyl-o-toluidine

207 CH3

N-Methyl-p-toluidine

210 H3C

NHCH3

83

53

60

—

131

N-Ethyl-p-toluidine

217 H3C

NHC2H3

—

40

71

120

—

N-Ethyl-o-toluidine

218

—

72

75

114

—

—

72

—

—

132

CH3

NHC2H5

N-Ethyl-m-toluidine

221

– Hydrochloride, m.p. 159°C.

494

Name

Chapter 7 Tables of Organic Compounds and their Derivatives

B.P. (°C)

N-Propylaniline

222

1, 2, 3, 4-Tetrahydroisoquinoline

233

N-Butylaniline 1, 2, 3, 4-Tetrahydroquinoline

Structure

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4- Pictyl oyl uene dini- rate sulph- trophonyl enyl

Specific tests and derivatives

48

—

56

—

—

46

129

—

—

200

236

C6H5NHCH2CH2CH2CH3 —

56

54

—

—

250

—

75

—

—

141

103

153

119

—

173

—

112

81

105

—

58

107

140

168

—

—

68

—

—

187

– N-Nitroso derivative, m.p. 171°C.

101

180

142

—

182

– N-Nitroso derivative, m.p. 66°C. – Sulphate salt 123–25°C.

115

152

—

—

—

85

125

—

—

—

– Gives yellow colour in conc. H2SO4 with trace of HNO3. – N-nitroso derivative, m.p. 101°C.

—

—

—

170

– Hydrochloride, m.p. 167°C.

NHCH2CH2CH3

– Reduces Tollen’s reagent.

NH

N H

Dicyclohexylamine

254 (C6H11)2NH (m.p. 20)

Dibenzylamine

300 (d)

N-Benzylaniline

m.p. 37 (b.p. 306)

Indole

(C6H5CH2)2NH

– Hydrochloride, m.p. 256°C.

NHCH2C6H5

m.p. 52 N H

Diphenylamine

m.p. 54 (b.p. 310)

N-Phenyl-I-naphthylamine

m.p. 62

Di-p-tolylamine

m.p. 79

NHC6H5

H 3C

3-Methyl indole

NH

95

CH3

CH3 N H

68

Table 7.17 Experimental Tertiary amines Advanced Organic Chemistry

Name

M.P. (°C)

495

Structure

Derivatives (M.P.°C) Ace- benz- p-Tol- 2, 4- Pictyl oyl uene dini- rate sulph- trophonyl enyl

N-Phenyl-2-naphthylamine

108

NHC6H5 93

136

—

—

—

1-1′−Dinaphthylamine

115

217

—

—

—

169

69

98

—

—

186

Specific tests and derivatives

NH

245

Carbazole

N H

Table 7.17 Name

– Picrate obtained in toluene solution. – To solution of the compound in conc. H2SO4 add aq. NaNO2 solution (2–3%)— deep green colour. – Compound + isatin solution in conc. H2SO4 gives blue colour.

Tertiary amines

B.P. (°C)

Structure

Derivatives

Trimethylamine

gas

(CH3)3N

230(d)

216

– Hydrochloride, m.p. 278°C. – Gives reddish yellow ppt. with Nessler’s solution.

Triethylamine

89

(C2H5)3N

280

173

– Hydrochloride, 253°C.

Pyridine

116

117

167

—

156

230

169

– On boiling with dil.KMnO4 yields picolinic acid, m.p. 137°C.

92

150

– On oxidation (KMnO4) gives nicotinic acid, m.p. 228°C.

Meth iodide

Picrate

Specific tests and derivatives

N

Pyrimidine

124

N N

α-Picoline (2-Methylpyridine)

129 N

β-Picoline (3-Methylpyridine)

CH3 CH3

143 N

496

Chapter 7 Tables of Organic Compounds and their Derivatives Name

γ-Picoline (4-Methylpyridine)

B.P. (°C)

Structure

143

CH3

Derivatives Meth iodide

Picrate

Specific tests and derivatives

152

167

– On oxidation (KMnO4) gives isonicotinic acid acid, m.p. 308°C.

238

163 (168)

208

116

–

156

210

122

– Hydrochloride, m.p. 156°C.

228 (d)

162

– Gives green colour on treating the solution of the compound in dil. HCl with a crystal of NaNO2. – Gives deep dichromate colour on treating the solution of the compound in conc. H2SO4 with very dil. HNO3. – p-Nitroso compound, m.p. 87°C.

N

2, 6-Lutidine (2,6-Dimethylpyridine)

143 H 3C

Tri-n-propylamine

156

Collidine 2, 4, 6-Trimethyl pyridine

171

185

CH3

(n-C3H7)3N CH3

H3C

N, N-Dimethyl-o-toluidine

N

N

CH3

N(CH3)2

CH3

N, N-Dimethylaniline

193

N-Ethyl-N-methyl aniline

201

125

134

– p-Nitroso compound, m.p. 66°C. – Hydrochloride, m.p. 114°C.

N-Methyl-2-pyrrolidone

202

—

—

– Hydrochloride, m.p. 80°C.

N, N-Diethyl-o-toluidine

210

224

180

N(CH3)2

CH3

N(C2H5)2

Table 7.17 Experimental Tertiary amines Advanced Organic Chemistry Name

B.P. (°C)

497 Structure

Derivatives Meth iodide

N, N-Dimethyl-p-toluidine

210

(CH3)2N

Tri-n-butylamine

211

(n-C4H9)3N

N, N-Dimethyl-m-toluidine

212

CH3

CH3

Picrate

220

129

186

106

177

131

102

142

184

110

—

97

Specific tests and derivatives

N(CH3)2

N, N-Diethylaniline

218

C6H5N(C2H5)2

N, N-Diethyl-p-toluidine

229

H 3C

N, N-Diethyl-m-toluidine

231

N(C2H5)2

N(C2H5)2

– Hydrochloride, m.p. 157°C.

CH3

Quinoline

238

72 203 (hydrated form) 133 (anhydrous (form)

N

Isoquinoline

240 (m.p. 24)

Di-n-Propylaniline

245

2-Methylquinoline (Quinaldine)

247

263

159

222

– Acid sulphate, m.p. 205°C.

156

—

– p-Nitroso derivative, m.p. 42°C.

195

191

– On boiling with conc. HNO3 gives nitroquinaldinic acid, m.p. 219°C.

174

212

N N(CH2CH2CH3)2

N

4-Methylquinoline (Lepidine)

– Hydrochloride, m.p. 143°, (anhydrous) m.p. 93°C. (hydrated) – On adding bleaching powder solution to the solution in boric acids yields chloroquinoline, m.p. 112°C, which with NaOH solution yields carbostyril, m.p. 199°C.

CH3

N

CH3

498

Chapter 7 Tables of Organic Compounds and their Derivatives Name

B.P. (°C)

Di-n-butylaniline

271

N-Methyldiphenylamine

296

2, 2′-Bipyridyl

Structure

N(CH2CH2CH2CH3)2

C6H5N(CH3)C6H5

m.p. 69 N

Dibenzylaniline

m.p. 70

Tribenzylamine

m.p. 91

Acridine

m.p. 127

Hexamethylene tetramine (Hexamine)

m.p. 280

N(CH2C6H5)2

(C6H5CH2)3N

(C6H5)3N

N N

Table 7.18

Formamide

Specific tests and derivatives

—

125

—

—

—

158

135

131(d)

184

190

224

208

—

—

– With conc. H2SO4 and trace of HNO3 gives blue colour. – Nitration (Fuming HNO3)– Trinitro derivative, m.p. 280°C.

190

157 (d)

– Dibromide, m.p.196–200°C. – Adds Br2 to form a red tetrabormide which on keeping in air loses bromine to form yellow dibromide. – Solution in HCl gives formaldehyde. – Aq. solution with conc. HNO3 at 0° gives nitrate, m.p. 165°C.

– p-Nitroso derivative, m.p. 44°C. – With HNO3 gives violet colour.

– p-Nitroso compound, m.p. 91°C.

– Sulphate, m.p. 106°C.

N

N

Name

Picrate

N

m.p. 106 and m.p. 108 (two stable forms)

Triphenylamine

Derivatives Meth iodide

N

Amides, Imides, Anilides, Guanidines

B.P./M.P (°C)

Structure

B.P. 195(d) 110/11 mm

HCONH2

Specific test and derivatives – Gives white ppt. with HgCl2 solution. – With aq. FeCl3 gives wine red colour changing to brown ppt. on boiling. – Anilide, m.p. 46°C, I-naphthylamine derivative, m.p. 138°C.

Table 7.18 Experimental Amides, Imides, Anilides, Guanidines Advanced Organic Chemistry Name Guanidine

Formanilide

499

B.P./M.P (°C)

Structure

Specific test and derivatives

(Available as nitrate salt)

(NH2)2C = NH

– On boiling with Ba(OH)2 yields urea + NH3. – With NaOCl gives yellow colour changing to orange red. – With conc. HNO3 + H2SO4 in cold yields nitroguanidine m.p. 230°(d). – Carbonate, 197°C. Nitrate, 214°C. Acetate, 229°C. Hydrochloride, 172°C

m.p. 46 (b.p. 216/ 120 mm)

C6H5NCHO

– On boiling with dil. HCl yields formic acid + C6H5NH2HCl. – On warming with CHCl 3 + KOH in alcohol gives odour of phenyl isocyanide. – Benzyl formanilide, m.p. 48°C obtained on warming with benzyl chloride

Methyl carbanilate

m.p. 47

NHCOOCH3

– On boiling with aniline, gives carbanilide, m.p. 238°C.

Ethyl carbanilate

m.p. 52

NHCOOC2H3

– On boiling with aniline gives carbanilide, m.p. 238°C.

N-Ethyl acetanilide

m.p. 54 (b.p. 249)

N

COCH3

C2H5

– On boiling with HCl yields acetic acid, b.p. 118°C. and ethyl aniline hydrochloride. – With conc. HNO3 + H2SO4 at 40° yields p-nitro derivative, m.p. 118°C which on hydrolysis yields p-nitroethylaniline. m.p. 95°C.

Allyl thiourea

m.p. 74

CH2=CHCH2NHC SNH2

– On boiling with conc. NaOH yields allylamine, b.p. 53°C. – Yields Ag2S with ammonical AgNO3 solution. – With ethyl iodide in alcohol forms addition product, m.p. 72°C. – With CH3COCl in acetone yields acetyl derivative mp. 103°(d) which with NaOH yields N-acyl derivative, m.p. 95°C. – With Br2 in alcohol yields bromopropylene-isothiourea HBr, m.p. 140°C.

500

Chapter 7 Tables of Organic Compounds and their Derivatives Name

M.P (°C)

Structure

Specific test and derivatives

Propionamide

79 (b.p. 213°)

CH3CH2CONH2

– With 75% H2SO4 at 120° yields propionic acid, b.p. 140°C. – Propionyl-α-naphth-ylamine, 116°C

Acetamide

82 (b.p. 222°)

CH3CONH2

– On distilling with conc. H2SO4, yields acetic acid b.p. 118°C (smell of vinegar) – Picrate, 107°C, Anilide, 114, α-Naphthylamine derivative 159°C

CH3COCH2CO.NHC6H5

– Gives violet colour with FeCl3 solution. – On boiling with HCl yields acetic acid, b.p. 118°C (smell of vinegar) and aniline hydrochloride. – On heating with aniline yields acetone and carbanilide, m.p. 238°C. – On warming with conc. H2SO4 yields lepidone (4-methyl carbostyril), m.p. 224°C.

Acetoacetanilide

85

o-Nitroacetanilide

94

– On heating with NaOH solution yields o-nitroani-line, m.p. 71°C. – On heating with conc. HCl gives smell of vinegar.

NHCOCH3 NO2

n-Butylanilide

95

CH3CH2CH2CONHC6H5

Acetoacet-p-toluidide

95

CH3COCH2CONH

m-Toluamide

95

CONH2

– On boiling with HCl yields n-butyric acid, b.p. 163°C and aniline hydrochloride. CH3

– Gives violet colour with FeCl3. – On hydrolysis gives m-toluic acid, m.p. 110°C. – Anilide, m.p. 126°C

CH3

Semicarbazide

96

NH2CONHNH2

– Reduces Fehling and Tollen’s reagents. – On heating with benzal dehyde in alcohol gives benzal derivative, m.p. 227°C (d) – Hydrocloride, 173° (d)

Methyl urea

101

CH3NHCONH2

– Xanthyl derivative, 230

Propionanilide

104

CH3CH2CONHC6H5

– On boiling with HCl yields propionic acid, b.p. 140°C and aniline hydrochloride. – With conc. HNO3 + H2SO4 at 0° yields p-nitro derivative, m.p. 182°C which on hydrolysis yields p-nitro- aniline, m.p. 147°C.

Table 7.18 Experimental Amides, Imides, Anilides, Guanidines Advanced Organic Chemistry Name

Acetoacet-o-toluidine

M.P (°C) 104

501 Structure

Specific test and derivatives – Gives violet colour with FeCl3 solution. – On warming with conc. H2SO4 yields 4, 8-dimethyl-earbostyril, m.p. 217°C.

CH3COCH2CONH CH3 Cl

Acetoacet-o-chloroaniline

105

– Gives violet colour with FeCl3 solution.

CH3COCH2CONH

Dicyanidamidine

105

H2NC(=NH)NHCONH2

Iso-Valeranilide

110

(CH3)2CHCH2CONHC6H5

Acetanilide

114

n-Butyramide

114

CH3CH2CH2CONH2

– On heating with aniline gives n-butyranilide, m.p. 95°C. – Xanthylamide, 187

Phenylacetanilide

118

C6H5CH2CONHC6H5

– On boiling with HCl yields phenylacetic acid, m.p. 76°C. and aniline hydro-chloride. – With conc. HNO3 in cold yields 2, 4-dinitrophenylacetic acid, m.p. 189°C.

Acetoacet-p-nitroanilide

119

Chloracetamide

120

ClCH2CONH2

Xanthyl, m.p. 208

Cyanoacetamide

120

CNCH2CONH2

Xanthyl, m.p. 223

– On heating to 160° evolves NH3. – On warming with Ba(OH)2 solution yields NH3, CO2 and urea. – With CuSO4 + NaOH yields a sparingly soluble red Cu salt. – Hydrochloride 173 Sulphate, 194 Picrate, 265 – On boiling with HCl yields iso-valeric acid, b.p. 176°C and aniline hydrochloride. – On boiling with dil. HCl yields CH3COOH and aniline hydrochloride. – p-Bromo, 167 – p-Nitro, 210 – p-Bromo derivative, on hydrolysis yields p-bromoaniline, m.p. 66°C. – With NaOCl solution yields N-chloro derivative, m.p. 91°C which with acetic acid and a trace of HCl yields p-chloro derivative, m.p. 179°C. – p-Nitro derivative, on hydrolysis yields p-nitroaniline, m.p. 147°C.

NHCOCH3

CH3COCH2CONH

NO2

– Violet colour with FeCl3.

502

Chapter 7 Tables of Organic Compounds and their Derivatives Name

M.P (°C)

Structure

Specific test and derivatives

Dimethylcarbanilide

120

C6H5N(CH3), CO.N(CH3)C6H5

Nicotinamide

122

CONH2

– On boiling with HCl yields methylaniline hydrochloride. – On hydrolysis gives nicotinic acid, m.p. 236°C.

N

Succinimide

126

CH2

CO NH

CH2



CO

– Gives fluorescein reaction (see succinic acid) – On boiling with aniline gives succinanil. m.p. 156°C. – Succinic acid, 185 – Xanthyl, 246

Benzamide

128

Iso-butyramide (2-Methylpropionamide)

128

(CH3)2CHCONH2

Acetylphenylhydrazine

128

NHNHCOCH3

– On boiling with HCl gives phenylhydrazine hydro-chloride – Reduces Fehlings solution. – With 2 moles of Br2 in conc. HCl yields dibromo derivative, m.p. 146°C

Urea (carbamide)

132

NH2CONH2

– On heating to 150° yields NH3 and biuret, m.p. 192°C.

– On hydrolysis gives benzoic acid, m.p. 122°C. – Xanthyl, 224 – Benzanilide, 164 – On hydrolysis gives isobutyric acid, b.p. 155°C. – Anilide, 105 – Xanthyl, 211

– Evolves N2 with NaOBr solution. – Evolves NH3 on boiling with dil. NaOH solution. – On warming with aniline hydrochloride solution yields phenyl urea. m.p. 147°C and carbonilide, m.p. 238°C. – Biuret (obtd. above) on dissolving in dil. NaOH solution and treatment with a drop of dil. CuSO4 solution gives purple colour. – Nitrate, 163 – Oxalate, 171

Table 7.18 Experimental Amides, Imides, Anilides, Guanidines Advanced Organic Chemistry Name

4-Aceto-m-xylidine

503

M.P (°C)

Structure

133

NHCOCH3 CH3

CH3

Specific test and derivatives – On boiling with HCl yields acetic acid and m-xylidine hydrochloride. – With fuming nitric acid yields 5-nitro derivative. m.p. 172°C. – On adding slowly KNO3 to solution in conc. H2SO4 yields 6-nitro derivative, m.p. 159°C.

b-Acetnaphthyamine

134

Acet-p-phenetidide (Phenacetin)

134

Salicylamide

139

CONH2 OH

– Gives violet colour with FeCl3. – On boiling with NaOH solution yields salicylic acid, m.p. 159°C. – Acetyl, 143 – Benzoyl, 200

m-Tolyurea

142

NHCONH2

– On alkaline hydrolyine yields m-toluidine b.p. 202°C.

NHCOCH3

– On boiling with HCl yields acetic acid and β-naphthyl amine hydrochloride. – With Br2 in AcOH—1-bromo derivative, m.p. 140°C – With fuming HNO3 yields two isomeric dinitro derives. One soluble in alcohol, m.p. 185°C. and the other insoluble in alcohol, m.p. 235°C. – On boiling with HCl yields acetic acid and p-phenetidine hydrochloride. – On warming with 10% HNO3 yields 3-nitro derivative. m.p. 103°C which on hydrolysis yieldsnitraphenetidine, m.p. 113°C.

CH3

m-Nitrobenzamide

142

CONH2

NO2

– On boiling with NaOH solution m-nitrobenzoic acid, m.p. 141°C.

504

Chapter 7 Tables of Organic Compounds and their Derivatives Name

o-Toluamide

M.P (°C)

Structure

142

CONH2

CH3

o-Benzotoluidide

144

Specific test and derivatives – On hydrolysis yields o-toluic acid, m.p. 104°C. – Benzoyl, 158 – Anilide, 125 – On boiling with HCl yields benzoic acid m.p. 122°C and o-toluidine hydro-chloride. – With hot KMnO4 soln. yields – benzoyl anthranilic acid, m.p. 185°C.

Triphenylguanidine

145

C6H5N=C(NHC6H5)2

– On heating yields aniline, b.p. 183°C and carbodi phenylimide, h.p. 330°C – On boiling with conc. KOH solutions yields aniline. – Hydrochloride, 241 – Picrate, 180

Diphenylguanidine

147

Phenyl urea

147

N-Benzylurea

149

Cinnamanilide

151

(C6H5NH)2C=NH

NHCONH2

C6H5CH2NHCONH2 CH

CHCONHC6H5

– On boiling with HCl gives aniline hydrochloride, m.p. 198°C. – With Ac2O at 100° yields acetanilide, m.p. 114°C – Perchlorate, 162 – On heating above m.p. yields NH3, CO2 and carbanilide m.p. 238°C – With NaOH yields aniline and NH3. – Xanthyl deriv. 225 – On heating above 200° gives CO2, NH3 and dibenzylurea, m.p. 167°C. – On heating with HCl, yields cinnamic acid, m.p. 133°C and aniline hydrochloride. – On heating with acidic KMnO4 yields benzaldehyde and benzoic acid, m.p. 122°C.

Sym. Diphenyl thiourea (Thiocarbanilide)

153

C6H5NHCSNHC6H5

– With yellow mercuric oxide in warm alcohol gives H2S and carbanilide, m.p. 238°C.

Table 7.18 Experimental Amides, Imides, Anilides, Guanidines Advanced Organic Chemistry Name p-Acetotoluidide

505

M.P (°C)

Structure

Specific test and derivatives

153

NHCOCH3

– On nitration (conc. HNO3 + H2SO4) at 30-40° gives 3-nitro derivative, m.p. 144°C.

CH3

Phenylthiourea

154

m-Nitroacetanilide

154

C6H5NHCSNH2 NHCOCH3

NO2

Phenylacetamide

154

Succianil

156

CH2CONH2

CH2

CO NC6H5

CH2

N-Benzoyl-p-toluidine

158

p-Toluamide

159

CH3

H 3C

α-Acetylnaphthyl amide

160

CO

NHCOC6H5

CONH2

NHCOCH3

– On boiling with aniline gives thiocarbanilide, m.p. 153°C. – On reduction with Sn + HCl gives m-phenylenediamine. m.p. 63°C. – On heating with NaOH solution gives m-nitroaniline, m.p. 114°C. – On heating with dil. HCl— smell of vinegar. – On boiling with aniline yields phenylacetanilide. m.p. 118°C. – On hydrolysis (NaOH solution) yields pheny-lacetic acid, m.p. 76°C. – Xanthylamide, 196 – N-acetate, 129 – N-benzoate, 130 – On boiling with HCl gives succinic acid, m.p. 185°C and aniline hydrochloride. – On boiling with Ba(OH)2 solution yields succinanilic acid, m.p. 148°C. – On boiling with HCl yields benzoic acid, m.p. 122°C and p-toluidine hydro-chloride. – On nitration (fuming HNO3 at 0°) yields dinitro derivative, m.p. 186°C. – With CrO3 in acetic acid yields p-benzoylaminobenzoic acid, m.p. 278°C. – On hydrolysis (NaOH solution) gives p-toluic acid m.p. 178°C – N-acetyl, 147 – Xanthyl, 224 – On boiling with HCl yields acetic acid and α-naphythyl-amine hydrochloride. – On nitration (fuming HNO3 in acetic acid) yields dinitro derivative, m.p. 250°C. – With Br2 in acetic acid (1 mol Br2) yields 4-bromo derivative, m.p. 193°C.

506

Chapter 7 Tables of Organic Compounds and their Derivatives Name

M.P (°C)

Structure

Benzanilide

164

C6H5NHCOC6H5

p-Bromoacetanilide

167

sym. Dibenzylurea

167

CO(NHCH2C6H5)2

Benzoyl phenylhydrazine

168

NHNHCOC6H5

Malonamide

170

Alloxan

170 (d)

N-Bromosuccinimide

p-Phenetyl urea

– On bromination (Br2 in acetic acid) gives p-bromo derive, m.p. 204. – On boiling with HCl yields benzoic acid, m.p. 122. – With fuming HNO3 (50°) in acetic acid gives 2-nitro derivative, m.p. 104°C. – On acid hydrolysis gives-pbromo aniline m.p. 67°C.

CH2(CONH2)2

– On heating with HCl yields benzylamine hydrochloride. – On oxidation with HgO yields red benzoylazobenzene. m.p. 80°C. – On boiling with conc. HCl yields phenylhydrazine hydrochloride and benzoic acid, m.p. 122°C. – On heating with aniline yields malonanlide, m.p. 225°C. – On heating with NaOH solution gives malonic acid, m.p. 133°C. – On shaking the compound in aq. NaOH with 2 drops of aq. CuSO4 gives red or violet-red colour. – Gives blue colour with ferrous salt solution. – On warming with NH2OH HCl solution and then adding Na2CO3 solution gives purple colour.

173 (180)

174

Specific test and derivatives

CH2 CO NBr CH2 CO

H5C2O

NHCONH2

– Fuming HNO3 in acetic acid at 50° gives nitro derivative m.p. 104°C.

– On heating above m.p. yields NH3, CO2 and di-p-phenetyl urea, m.p. 235°C. – On boiling with 48% HBr yields p-aminophenol HBr salt.

Table 7.18Experimental Amides, Imides, Anilides, Guanidines Advanced Organic Chemistry Name

M.P (°C)

o-Nitrobenzamide

174

p-Chloroacetanilide

179

507 Structure

– On heating with NaOH solution gives o-nitrobenzoic acid m.p. 146°C. – On reduction (Sn + HCl, boil) gives anthranilic acid, m.p. 144°C. – On heating with P2O5 gives o-nitrobenzonitrile. m.p. 110°C. NHCOCH3

180

– With fuming HNO3 at 50° gives 2-nitro derivative, m.p. 104°C. – On heating with NaOH solution yields p-chloro-aniline, m.p. 70°C.

Cl

Thiourea

Specific test and derivatives

NH2CSNH2

– With K4Fe(CN)6 and dil. acetic acid gives green colour changing to blue. – On heating with benzyl chloride in dil. alcohol yields S-benzyl iso thiouronium chloride, m.p. 175°C. – Hydrochloride m.p. 136. – On warming with ethyl sodio malonate in absolute alcohol yields thiobarbituric acid. m.p. 235°C.

p-Tolylurea

181

Thiosemicarbazide

182

Diacetyl-o-phenylenediamine 185

– On heating above m.p. yields NH3, CO2 and di-ptolylurea, m.p. 268°C. – On boiling with HCl yields p-toluidine hydrochloride. – N-acetyl, 200 – N-benzoyl, 223 NH2CSNHNH2

NHCOCH3

NHCOCH3

Hippuric acid (Benzoyl glycine)

187

C6H5CONHCH2COOH

– – – –

Acetate, 165 Isopropylidine, 179 Hydrochloride, 188 Benzal, 159

– On boiling with HCl yields acetic acid and o-phenylenediamine hydrochloride. – p-Nitrobenzyl ester, 136 – p-Bromophen-acyl ester, 151 – Anilide, 208

508

Chapter 7 Tables of Organic Compounds and their Derivatives Name

5, 5-Diethyl barbituric acid

M.P (°C) 188

Structure

O

H N

Specific test and derivatives

O

(d)

– Xanthyl derivative 246.

C2H5

HN

C2H5 O

o-Tolyl urea

191

Diacetyl m-phenylenediamine 191

– On heating above its m.p. yields NH3, CO2 and di-otolyl urea, m.p. 250°C. – On boiling with HCl, yields o-toluidine hydrochloride. – N-acetyl, 168 – N-benzoyl, 210 NHCOCH3

NHCOCH3

Biuret

192

p-Nitrobenzamide

201

Isatin

203

NH2CONHCONH2

210

– Addition of one drop of aq. CuSO4 solution to an alkaline solution of the compound gives red violed colour. – On heating with aniline yields diphenylbiuret, m.p. 210°C. – Amide, 260 – On heating with NaOH solution yields p-nitrobenzoic acid, m.p. 240°C. – On reduction (Sn + HCl, boiling) gives p-amino-benzoic acid, m.p. 186°C. – On heating with P2O5 gives p-nitrobenzonitrile, 148°C.

CO N H

Phthalanil

– On boiling with HCl yields acetic acid and m-phenylene diamine hydro-chloride. – With conc. HNO3 at 0° in presence of urea, yields dinitro derivative, m.p. 228°C.

CO

– N-Acetyl, 141 3-Oxime, 225(d) – 3-Semicarba-zone, 266(d)

– On boiling with HCl yields phthalic acid, m.p. 195°C and aniline hydrochloride. – On boiling with Ba(OH)2 solution and acidifying yields phthalanilic acid, m.p. 169°C.

Table 7.18 Experimental Amides, Imides, Anilides, Guanidines Advanced Organic Chemistry Name

p-Nitroacetanilide

509

M.P (°C)

Structure

Specific test and derivatives

210

NHCOCH3

– On reduction (Sn + HCl) heat, yields p-phenylenediamine, m.p. 140°C.

NO2

Phthalamide

220 (d)

CONH2

CONH2







o-Sulphobenzimide (Saccharin)

224



Succinanilide

228

Phthalimide

238

CO NH SO2

C6H5NHCOCH2CH2 CONHC6H5

CO NH CO

sym. Diphenylurea (Carbanilide)

238

(C6H5NH)2CO

Succinamide

260

CH2CONH2 CH2CONH2







– On heating above m.p. yields phthalimide, m.p. 238°C. – On heating with aniline yields anil, m.p. 205°C. – Gives fluorescein test (See phthalic acid) – On heating with NaOH solution gives phthalic acid, m.p. 195°C – Gives red solution having green fluorescence in the fluorescein test. – Heat the compound with KOH pellets until melting, cool, dissolve in dil. HCl and add H2O then a drop of FeCl3— violet colour. – Banzyl, 118 – Xanthyl dervative, 198 – On boiling with HCl yields succinic acid, m.p. 185°C, and aniline hydrochloride. – On heating above m.p. yields aniline and succinanil. m.p. 156°C. – Gives Fluorescein test (see phthalic acid). – Xanthyl-amide, 177 – Anil 210 – N-acetate, 133 – N-Benzoate, 115 – On heating with HCl gives aniline. – On boiling (15 min) with Ac2O-acetic acid gives acetanilide, m.p. 114°C. – With conc. HNO3 (room temperature)–m-Nitro derivative. m.p. 248(d) – Acetyl derivative, 106 – On heating above m.p. gives succinimide, m.p. 126°C. – On hydrolysis (heating with NaOH solution) gives succinic acid, m.p. 185°C. – Gives fluorescein test – Anil, 156

510

Chapter 7 Tables of Organic Compounds and their Derivatives Name

M.P (°C)

Barbituric acid (malonyl urea)

245 (d)

Structure

Specific test and derivatives – With cold NaNO2 solution gives purple colour. – On treatment with Br2 water gives dibromo derivative m.p. 235(d).

Oxanilide

246

C6H5NHCOCONHC6H5

– On boiling with HCl yields Oxalic acid and aniline hydrochloride. – With fuming HNO3 in acetic acid, yields di-p-nitro derivation, m.p. 260°C, which on hydrolysis yields p-nitroaniline, m.p. 147°C.

sym. Di-o-tolyl urea

250

CH3

sym. Di-p-tolyl urea

– On boiling with HCl yields o-toludine hydrochloride.

NHCONH

– On boiling with Ac2O + Na acetate yields o-acetotoluidide, m.p. 112°C.

CH3

268 H3C

CH3

NHCONH

Diacetyl-p-phenylene diamine 305 CH3OCHN

NHCOCH3

– On boiling with Ac2O + Na acetate yields N-acetyl p-toluidine, m.p. 153°C. – On boiling with HCl yields acetic acid and p-phenylene diamine hydrochloride. – With fuming HNO3 at 0° yields dinitro derivative, m.p. 258°C.

Creatinine

315 (d)

– On oxidation (KMnO4) yields oxalic acid and methyl guanidine. – Reduces Fehlings solution – With nitroprusside + NaOH gives red colour changing to yellow. – Picrate, 220

Oxamide

419 (sealed tube)

NH2COCONH2

– Gives red colour with CuSO4 + dil. NaOH solution – On boiling with NaOH solution gives oxalic acid and NH3. – On heating with aniline gives oxanilide, m.p. 246°C.

Table 7.19 Experimental Nitriles Advanced Organic Chemistry

511

Table 7.19 Name

Nitriles

B.P.(°C)

Structure

Acrylonitrile

78

CH2=CHCN

– β-Naphthol adduct, m.p. 142°C.

Acetonitrile

81

CH3CN

– On heating with HCl gives acetic acid (smell of vinegar). – With alcohol (1 mole) + HCl gas yields acetiminoethyl ether HCl, m.p. 98°(d)

Propionitrile

97

CH3CH2CN

– On boiling with dil. acid or alkali yields propionic acid, b.p. 140°C, – With alcohol (1 mole) + HCl gas yields propioiminoethyl ether HCl, m.p. 92(d) – On boiling with alcohol + H2SO4 yields ethyl propionate, b.p. 98°C,

n-Butyronitrile

118

CH3CH2CH2CN

– On boiling with dil. acid or alkali, yields butyric acid, B.P. 163°C. – On boiling with alcohol + H2SO4 yields ethyl n-butyrate, b.p. 120°C.

n-Valeronitrile

141

CH3CH2CH2CH2CN

– On boiling with dil. acid or alkali yields n-valeric acid. b.p. 186°C. – On boiling with alcohol + H2SO4 yields ethyl n-valerate, b.p. 145°C.

Benzonitrile

190

o-Tolunitrile

205

– With conc. H2SO4 at 100° or on shaking with dil. NaOH + H2O in cold yields benzamide, m.p. 128°C. – On boiling with 75% H2SO4 yields benzoic acid, m.p. 122°C. – On boiling with CH3OH + conc. H2SO4 yields methyl benzoate, b.p. 198°C. – With conc. HNO3 + conc. H2SO4 yields m-nitro derivative, m.p. 118°C

CN

CH3

CN

Ethylcyanoacetate

207

m-Tolunitrile

212

Specific tests and derivatives

CNCH2COOC2H5

CH3

CN

– With conc. H2SO4 at 100° or on shaking with dil. NaOH + H2O2 in cold yields o-toluamide, m.p. 142°C. – On boiling with 75% H2SO4 yields o-toluic acid, m.p. 104°C. – On boiling with CH3OH + H2SO4 yields methyl o-toluate, b.p. 213°C. – With conc. HNO3 + conc. H2SO4 yields nitro derivative m.p. 105°C. – On evaporation with NH4OH yields cyanoacetamide, m.p. 118°C. – On boiling with alcohol + H2SO4 yields ethyl malonate, b.p. 198°C. – On hydrolysis with 75% H2SO4 yields m-toluic acid. m.p. 110°C. – With conc. H2SO4 at 100° or on shaking with dil. NaOH + H2O2 in cold yields the amide, m.p. 95°C.

512

Chapter 7 Tables of Organic Compounds and their Derivatives Name

B.P.(°C)

p-Tolunitrile

217

Phenylacetonitrile (benzyl cyanide)

234

Cinnamonitrile

α-Naphthonitrile

Succinonitrile

Structure NC

– With conc. H2SO4 at 100° yields phenyl acetamide, m.p. 154°C. – On boiling with 75% H2SO4 yields phenyl acetic acid, m.p. 76°C. – With CH3OH + H2SO4 (boiling) yields methyl phenyl acetate, b.p. 218°C. – With conc. HNO3 + H2SO4 yields p-nitro derivative, m.p. 116°C. CH

297

CH

m.p. 66 b.p. 305

Phthalonitrile

m.p. 141

m.p. 148

CN

– On heating rapidly with 75% H2SO4 yields β-naphthoamide m.p. 192°C, which on prolonged boiling is converted into β-naphthoic acid, m.p. 184°C. – On hydrolysis with 75% H2SO4 (refluxing) yields phthalic acid, m.p. 195°(d).

CN

O 2N

– On boiling with 75% H2SO4 yields cinnamic acid, m.p. 133°C. – On shaking with dil. NaOH + H2O2 in cold yields cinnamoamide, m.p. 147°C.

– On hydrolysis (75% H2SO4, refluxing) gives succinic acid. m.p. 185.

CH2CN

CN

p-Nitrobenzonitrile

CN

– On heating rapidly with 75% H2SO4 yields α-naphthoamide, m.p. 202°C, this on prolonged boiling is converted into a-naphthoic acid, m.p. 162°C.

CN

CH2CN

b-Naphthonitrile

– With conc. H2SO4 at 100° or on shaking with dil. NaOH + H2O2 in cold yield p-toluamide, m.p. 159°C. – On boiling with 75% H2SO4 yields p-toluic acid, m.p. 178°C. – On boiling with CH3OH + H2SO4 yields methyl p-toluate, m.p. 33°C. b.p. 217°C. – With conc. HNO3 + H2SO4 yields nitro-derivative, m.p. 107°C.

CH3

CH2CN

254 (m.p. 20°)

m.p. 55

Specific tests and derivatives

CN

– On boiling with 75% H2SO4 yields p-nitrobenzoic acid, m.p. 240°C. – Amide, m.p. 201°C.

Table 7.20 Experimental Amino Acids, Esters Chemistry Advanced Organic

513

Table 7.20 Name

Methylanthraniliate

M.P. (°C) 25

Amino Acids, Esters

Structure

Derivatives Acetyl

COOCH3



Benzoyl

101

100

– Solution in ethanol gives blue fluorescence. – Picrate, m.p. 106°C.

194

63

– On heating with a pellet of NaOH gives blue colour.

182

– On heating with solid CaCl2 and dissolving the mass in alcohol gives red solution. (violet fluorescence on standing). – Anilide, m.p. 126°C. – p-Toluidide, m.p. 151°C.

250

248

– Amide, m.p. 75°C. – Anilide, m.p. 129°C.

252

278

NH2

N-Phenylglycine

126

Anthranilic acid (o-Aminobenzoic acid)

144

C6H5NHCH2COOH

185

COOH

Specific tests and derivatives

NH2

m-Aminobenzoic acid

174

p-Aminobenzoic acid

186

H2N

Hippuric acid (benzoyl glycine)

187

C6H5CONHCH2COOH

—

—

– Anilide, m.p. 208°C. – Amide, m.p. 183°C. – On boiling with HCl gives benzoic acid, m.p. 122°C. – On treatment with conc. H 2 SO 4 + conc. HNO 3 gives m-nitro derivative, m.p. 162°C.

β-Alanine

196

NH2CH2CH2COOH

—

165

– 3, 5-Dinitrobenzoyl, m.p. 202°C.

(D-or L Glutamic acid) (a-aminoglutaric acid)

198 (d)

HOOC(CH2)2CH(NH2)COOH

—

138

– Gives deep blue colour on adding CuSO 4 solution to aq. solution of the compound. – 3, 5-dinitrobenzoyl, derivative, m.p. 217°C. – Phthalyl derivative, 189°C.

H2N

168

198

—

—

p-Aminophenyl acetic acid dl-Proline

200(d)

COOH

CH2COOH

203 COOH N H

– Picrate m.p. 137°C.

514

Chapter 7 Tables of Organic Compounds and their Derivatives Name

M.P. (°C)

Structure

Sarcosine

210 (d)

CH3NHCH2COOH

L(–) proline

222 (d)

Derivatives Acetyl

Benzoyl

Specific tests and derivatives

135

104

– Hydrochloride, m.p. 169°C. – p-tosyl, m.p. 102°C. – Anhydride, m.p. 149°C.

—

156

– Picrate, m.p. 154°C. – p-Tosyl, m.p. 132°C.

COOH N H

D or L-Lysine D or L-Asparagine

224 (d)

NH2(CH2)4CH(NH2)COOH

—

150 (di)

– 3, 5-Dinitrobenzoyl, m.p. 169°C.

226 (sealed tube)

NH2COCH2CH(NH2)COOH

—

189

– 3, 5-Dintrobenzoyl, 196°C. – Picrate m.p. 180°C. – Gives blue colour with dil. NaOH + trace of CuSO4 solution. – Gives deep blue colour on addition of CuSO 4 solution to an aq. solution of the compound. – Gives red colour with aq. FeCl3 solution. – 3, 5-Dinitrobenzoyl, m.p. 179°C.

Glycine

232

NH2CH2COOH

206

187

dl-Threonine

235 (d)

CH3CH(OH)CH(NH2)COOH

—

148 (mono)

N-(p-Hydroxyphenyl) glycine 248 (d)

HO

175 (di)

117

DL Serine

246 (d)

HOCH2CH(NH2)COOH

—

171

– 3, 5-Dintrobenzoyl, m.p. 183°C.

L-Cystine

260

180

– Forms a blue insoluble Cu salt. – With Sn + HCl yields cy steine (soluble in H 2 O, gives blue colour with FeCl3)

—

—

– p - To l u e n e s u l p h o n y l derivative m.p. 153°C.

—

185

– Gives blue colour by adding CuSO 4 solution to aq. solution of the compound

NHCH2COOH

NH2 CH2

CH COOH

S S CH2

With AgNO3 solution gives black ppt. in cold; solution becomes purple on heating.

CH COOH NH2

L-Hydroxylproline

270

HO

N

COOH

H

D or L Aspartic acid

270 (sealed tube

HOOCCH2CH(NH2)COOH

Table 7.20 Experimental Amino Acids, Esters Chemistry Advanced Organic Name

dl-Methionine

M.P. (°C) 272

515 Structure

Derivatives Acetyl

Benzoyl

Specific tests and derivatives

114

151

—

188

– 3, 5-Dinitrobenzoyl, m.p. 93°C. – On oxidation (KMnO4)— benzoic acid, m.p. 122°C.

—

230

– 3, 5-Dinitrobenzoyl, m.p. 189°C. – Picrate, m.p. 86°C.

(CH3)2C(NH2)COOH

—

198

– On oxidation with NaOCl yields acetone.

—

150

– 3, 5-Dinitrobenzoyl, m.p. 95°C.

218 (mono) 184 (di)

252

– With FeCl 3 gives red colour changing to brown ppt. – Azo-β-naphthol derivative m.p. 201°C.

189

104

– Picrate, m.p. 195°C. – 3, 5-Dinitrobenzoate, m.p. 233°C.

CH3S(CH2)2CH

COOH

NH2

dl-Phenylalanine

273 (d)

L-Histidine

277 (d)

α-Amino isobutyric acid

280 (Subl)

C6H5CH2CH(NH2)COOH

L-Methionine

283 (d)

CH3S(CH2)2CH(NH2)COOH

5-Aminosalicylic acid

283

H 2N

COOH

OH

L-Tryptophan

289

CH2CH(NH2)COOH N H

dI-Isoleucine

292 (d)

CH3CH2CH(CH3) CH(NH2)COOH

—

118

L-Leucine

294 (d)

(CH3)2CHCH2CH(NH2)COOH

—

107

– 3, 5-Dinitrobenzoate, m.p. 187°C.

dI-Alanine

295 (d)

CH3CH(NH2)COOH

—

166

– 3, 5-Dinitrobenzoate, m.p. 177°C.

dI-Valine

298 (d)

(CH3)2CHCH(NH2)COOH

148

132

– Hydrochloride, m.p. 189°C.

300 (d) (sealed tube)

CH3(CH2)3CH(NH2)COOH

—

—

CH3CH2CH(NH2)COOH

—

147

—

197

L–(+) Norleucine

dl α-Amino-n-butyric acid

307 (d)

dI Tyrosine

318

– On KOH fusion gives p-hydroxybenzoic acid, m.p. 213°C. – Gives red colour with dil. solution of CH3CHO in conc. H2SO4.

516

Chapter 7 Tables of Organic Compounds and their Derivatives Name

M.P. (°C)

dI-Norleucine

327 (d)

dI-Leucin L(–) Tyrosine

Structure

Derivatives Acetyl

Benzoyl

Specific tests and derivatives

CH3(CH2)3CH(NH2)COOH

332 344(d)

Table 7.21 Name

Sulphonic Acids

M.P. (°C)

Structure

Amide

Anilide

S-Benzyl isothiouronium Salt

Methanesulphonic acid

20

CH3SO3H

90

99

—

Benzenesulphonic acid

43 (anhydrous 65)

SO3H

153

110

148

167

126

146

156

136

170

137

110

146

144

104

175

193

115

—

SO3H

m-Nitrobenzenesulphonic acid

48 NO2

CH3

o-Toluenesulphonic acid

57 SO3H

SO3H

m-Xylene-4-sulphonic acid

CH3

60

CH3

p-Chlorobenzenesulphonic acid

68

o-Nitrobenzenesulphonic acid

70

SO3H

Cl

SO3H

NO2

Table 7.21Experimental Sulphonic Acids Advanced Organic Chemistry Name

M.P. (°C)

p-Nitrobenzenesulphonic acid

70

Naphthalene-1-sulphonic acid

88 (anhydrous 90)

p-Toluenesulphonic acid

92 (anhydrous) 105 (hydrated)

p-Bromobenzenesulphonic acid

517

103

Structure

SO3H

O 2N

SO3H

H 3C

Br

SO3H

Amide

Anilide

S-Benzyl isothiouronium Salt

179

136

—

150

112

137

137 103 (anhydrous) 105 (hydrated)

182

166

119

170

—

—

204

COOH

5-Sulphosalicylic acid

OH

120 HO3S

2-Naphthol-6-sulphonic acid (Schaffer’s acid)

122 (d)

238

—

—

Naphthalene-2-sulphonic acid

124 (anhydrous 91)

212 (217)

132

191

—

—

—

132

119

210

OH

1-Napththol-4-sulphonic acid

170 SO3H

(+)-Camphor 10-sulphonic acid

193

518

Chapter 7 Tables of Organic Compounds and their Derivatives Name

Amide

Anilide

245

310

249

254

259 (anhyd.) 94 (hydrated)

236 (di)

253 (di)

213

Metanilic acid

> 300

142

—

148

Sulphanilic acid

> 300

165

200

182

229

114

214

60

58

115

305

—

256

242

—

205

—

202

213

Napthalene-1, 5-disulphonic acid

p-Sulphobenzoic acid

M.P. (°C)

Structure

S-Benzyl isothiouronium Salt

SO3H

Benzene m-disulphonic acid

— SO3H

Ethanesulphonic acid

—

CH3CH2SO3H SO3H

Naphthalen-2, 6-disulphonic acid

— HO3S

Napthalene acid 2, 7-disulphonic acid

—

2-Napththol-3, 6-disulphonic acid (R-acid)

—

HO3S

SO3H

Table 7.22 Experimental Thiols, Mercaptans Advanced Organic(Thioalcohols Chemistry and Thiophenols) Name

M.P. (°C)

2-Naphthol 6, 8-disulphonic acid

519

Structure

Amide

—

Anilide

S-Benzyl isothiouronium Salt

—

195

228

177

141

169

261

—

195(d)

142

—

148

SO3H

Phenol p-sulphonic acid

— OH NH2

Naphthionic acid

300 SO3H

Orthanilic acid

> 300 (d)

Table 7.22 Name

B.P. (°C)

Thiols, Mercaptans (Thioalcohols and Thiophenols) Structure

2, 4-dinitro phenyl thioether

3, 5-dinitro thiobenzoate

Methylmercaptan

6

CH3SH

—

—

Ethylmercaptan

36

C2H5SH

115

62

Isopropylmercaptan

56

(CH3)2CHSH

95

84

n-Propyl mercaptan

67

CH3CH2CH2SH

81

—

Isobutyl mercaptan

88

(CH3)2CHCH2SH

76

64

Allyl mercaptan

90

CH2=CHCH2SH

72

—

n-Butylmercaptan

97

CH3CH2CH2CH2SH

66

—

Isoamyl mercaptan

117

(CH3)2CHCH2CH2SH

59

—

n-Amylmercaptan

127

CH3CH2CH2CH2CH2SH

80

—

Cyclohexyl mercaptan

159

SH

148

—

520

Chapter 7 Tables of Organic Compounds and their Derivatives Name

Thiophenol (Phenyl mercaptan)

B.P. (°C) 169

Structure

2, 4-dinitro phenyl thioether

3, 5-dinitro thiobenzoate

121

—

194

130

—

194 (m.p. 15°)

101

—

91

—

103

—

SH

CH2SH

Benzyl mercaptan

o-Tolmercaptan (Thio-o-cersol)

m-Tolyl mercaptan (Thio m-cresol)

200

CH3

SH

p-Tolyl mercaptan (Thio p-cersol)

m.p. 43 (b.p. 195°)

SH

CH3

SH

β-Thionaphthol

–2, 4-Dinitrophenyl-2-naphthyl sulphide, m.p. 145°C.

81

Table 7.23 Name

M.P.(°C)

p-Toluenesulphonanilide

103

Benzenesulphonanilide

110

(+) Camphor 10-sulphonamide

132

Sulphonamides

Structure

H3C

SO2NHC6H5

C6H5SO2NHC6H5

Specific tests and derivatives – On boiling with 20% HCl gives p-toluenesulphonic acid m.p. 92 and aniline, b.p. 184. – On boiling with 20% HCl gives benzenesulphonic acid, m.p. 43 and aniline b.p. 184°C. – On heating with benzyl chloride in alcoholic sodium hydroxide gives benzyl derivative, m.p. 119°C.

Table 7.23Experimental Sulphonamides Advanced Organic Chemistry Name p-Toluenesulphonamide

521

M.P.(°C)

Structure

137 H3C

Naphthalene 1-sulphonamide

150

Benzenesulphonamide

153

o-Toluenesulphonamide

153

SO2NH2

SO2NH2

SO2NH2

CH3

SO2NH2

m-Nitrobenzene sulphonamide

162

Specific tests and derivatives – On methylation (Me2SO4 + dil. NaOH) yields dimethyl derivative, m.p. 80. – Xanthyl derivative, m.p. 197. – On methylation (Me2SO4 + dil. NaOH) yields dimethyl derivative, m.p. 137°C. – N-Acetyl, m.p. 185°C. – N-Benzoyl, m.p. 194°C.

– On methylation (Me2SO4 + dil. NaOH) yields dimethyl derivative, m.p. 47°C. – N-Acetyl, m-p. 125°C. – N-Benzoyl, m.p. 147°C. – Xanthyl derivative, m.p. 206°C. – With K3Fe(CN)6 solution yields o-sulphonamidobenzoic acid, m.p. 154°C, which on heating gives the imide (Saccharin), m.p. 224°C. – On heating with strong H2SO4 yields toluene, b.p. 110. – Xanthyl derivative, m.p. 183°C. – On methylation (Me2SO4 + dil. NaOH) yields dimethyl derivative, m.p. 122°C. – N-acetyl, m.p. 189°C

SO2NH2

NO2

Sulphanilamide

Naphthalene-2-sulphonamide

Sulphobenzoic imide (Saccharin)

165

– Diacetyl, m.p. 217°C. – Dibenzoyl, m.p. 268°C. – Xanthyl derivative, m.p. 208°C.

212 (217)

224

SO2NH2

CO NH SO2

m-Benzenedisulphonamide

229

SO2NH2

SO2NH2

– On methylation (Me2SO4 + dil. NaOH) yields dimethyl derivative, m.p. 94°C. – N-Acetyl, m.p. 146°C.

– On methylation (Me2SO4 + dil. NaOH) gives N-methyl derivative, m.p. 131°C. – Xanthyl derivative, m.p. 198°C. – On warming with resorcinol + trace of conc. H2SO4 gives sulphonfluorescein (red solution in dil. NaOH with green fluorescence). – Xanthyl derivative, m.p. 170°C.

APPENDIX

I Preparation of Reagents Barfoeds’ reagent: Dissolve 6.5 g of neutral copper acetate in 100 mL water containing 1 mL acetic acid. Filter if necessary. Baeyers reagent: Dissolve 2 g of potassium permanganate in 100 mL water. Benedict reagent: Dissolve 17.3 g sodium citrate and 10 g of anhydrous sodium carbonate in 70 mL water in a 100 mL standard flask. Add a solution of 1.75 g of copper sulphate pentahydrate dissolved in 10 mL water to the above solution. shake to mix and add water to make up to the mark. Bials reagent: Dissolve 0.3 g of orcinol in 100 mL con RC, hydrochloric acid and then add to it 3–4 drops of ferric chloride solution. Bromine water: Take 5 mL of liquid bromine (bromine is highly corrosive, handle it cautiously) in a reagent bottle and add 100 mL water to it. Swirl to mix. Some liquid bromine settles at the bottom of the reagent bottle. Use the supernatant aqueous solution for the test. Bromine in acetic acid: 10 mL liquid bromine in 100 mL glacial acetic acid. Bromine in carbon tetrachloride: 10 mL liquid bromine in 100 mL carbon tetrachloride. Ceric ammonium nitrate: Warm 100 mL of 2N nitric acid and dissolve 40g of ceric ammonium nitrate in it. Chlorine water: Pass chlorine gas (generated by reaction between conc. hydrochloric acid and potassium permanganate) in to 100 mL water till a saturated solution is obtained. 2,4-dinitrophenyl hydrazine In aqueous alcohol (Brady’s reagent): Suspend 2 g of 2,4-dinitro phenylhydrazine in 4 mL of concentrated sulphuric acid in a conical flask and add cautiously 35 mL of methanol with stirring. Cool if the mixture becomes hot. Warm and then add 10 mL of water to it. Allow the solution to stand for 5-6 hrs. and filter to obtain a clear solution.

Fehling solution A: Dissolve 6.9 g of copper sulphate pentahydrate in ~ 50–60 mL water containing few drops of dil. Sulphuric acid and add water to get 100 mL of the solution. B: Dissolve 35 g of sodium potassium tartrate and 14 g of sodium hydroxide in ~50–60 mL water. Make up the solution to 100 mL with water.

526

Appendix

Ferric chloride solution: dissolve 13.5g of ferric chloride in a mixture of water and conc, hydrochloric acid (50:2) and dilute to obtain 100 mL solution. Neutral ferric chloride: take ~5–6 mL of the above ferric chloride solution and add dropwise with shaking a dilute solution of ammonium hydroxide till a faint permanent turbidity is obtained. Filter and use for the test. Iodine-potassium iodide solution: Dissolve 20 g of potassium iodide in 100 mL water and then add to it 5g of solid iodine with shaking. Add some more iodine if all the previously added iodine has dissolved completely. Lucas reagent: Dissolve 16 g of anhydrous (fused, see page 91) zinc chloride in 10 mL of concentrated hydrochloric acid. Millon’s reagent: Dissolve 5 g of mercury (handle very carefully) in 5 mL conc, nitric acid and dilute the solution with 10 mL water. Molisch’s reagent: Dissolve 10 g of a-naphthol in 100 mL alcohol. Prepare only 1–2 mL fresh solution for use. Nessler’s reagent: Dissolve 5 g of potassium iodide in minimum amount of water and add to it a saturated solution of mercuric chloride until a faint permanent precipitate is obtained. Add 40 mL of 50% potassium hydroxide solution to the above mixture. Dilute to 100 mL with water, allow to stand and decant the supernatant solution for use. Ninhydrin reagent: Dissolve 0.2g of ninhydrin in 100 mL of rectified spirit. Schiff’s reagent: Dissolve 0.1 g of rosaniline hydrochloride in 50 mL warm water, cool and saturate the solution by passing sulphur dioxide in it. Dilute the pale yellow solution so obtained with water to 100 mL. Selewonoff’s reagent: Dissolve 0.05 g of resorcinol in 100 mL of dilute hydrochloric acid (4N). Sodium nitroprusside reagent: Dissolve 0.1 g of sodium nitroprusside in 20 mL water. Use preferably freshly prepared solution. Modified sodium nitroprusside reagent: Dissolve 2.9g of sodium nitroprusside in a solution containing 80 mL of dimethyl sulphoxide and 20 mL water. Store it in a refrigerator. ((Modified Remini and Simon reagent). Sodium iodide solution: Dissolve 15g of sodium iodide in 100 mL acetone. Tollens’ reagent: (i) Dissolve 0.25g of silver nitrate in 5 mL water and add to it dil. Sodium hydroxide until a faint but permanent precipitate is obtained. Now add dilute solution of ammonium hydroxide with shaking to just dissolve the precipitate. Add l–2 mL more of ammonium hydroxide to make it ammonical, use only freshly prepared solution and do not store. (ii) Dissolve 0.25g of silver nitrate in 5 mL water and add to it dil. Sodium hydroxide until precipitation occurs. Decant the supernatant solution, wash the precipitate with 2–3 mL water and decant water. Now add dilute solution of ammonium hydroxide to the precipitate with shaking to just dissolve it. Add 1–2 mL more of ammonium hydroxide to make it ammonical, use only freshly prepared solution and do not store the reagent.

II Normality of Conc. Acids Name of the acid

Normality of the concentrated acid

Glacial acetic acid

17 N

Hydrochloric acid

12 N

Nitric acid

16 N

Sulphuric acid

36 N

Always add concentrated acid to water for preparing a dilute acid from concentrated acid as the reaction between conc. acid and water is an exothermic reaction.

III Boiling points of some Common Organic Solvents Solvent acetic acid acetone acetonitrile benzene 1-butanol 2-butanol 2-butanone t-butyl alcohol carbon tetrachloride chlorobenzene chloroform cyclohexane 1.2-dichloroethane diethylene glycol diethyl ether diglyme (diethylene glycol dimethyl ether) 1,2-dimethoxy ethane (glyme. DME) dimethyl-formamide (DMF) dimethyl sulfoxide (DMSO) 1,4-dioxane ethanol ethyl acetate

Solubility in water (g/100 g) Miscible Miscible Miscible 0.18 6.3 15 25.6 Miscible 0.08 0.05 0.795