Organic Chemistry-I

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Krishna's TEXT BOOK

ORGANIC

CHEMISTRY-I Second Paper

B.Sc. First Year

st (For B.Sc. I year Students of all Colleges affiliated to Universities in Uttar Pradesh)

As per U.P. Unified Syllabus (w.e.f. 2011-12) By Alok Bariyar Ph.D. (CSIR fellow), M.Sc.(IIT Delhi), GATE., NET

Ex. Scientist (Nuclear) Bhabha Atomic Research Centre, Mumbai (Maharashtra)

R.P. Singh M.Sc., Ph.D.

Babita Agrawal M.Sc., Ph.D.

Head, Dep’t of Chemistry

Head, Dep’t of Chemistry

KNI of Physical Social Science, Sultanpur (U.P.)

B.S.A. (P.G.) College, Mathura (U.P.)

Ashish Dwivedi

Prashant Singh

M.Sc., Ph.D.

M.Sc., Ph.D.

Asst. Prof., Dep’t of Chemistry

Asst. Prof., Dep’t of Chemistry

Ganpat Sahai P.G. College, Sultanpur (U.P.)

KNIPSS, Sultanpur (U.P.)

KRISHNA Prakashan Media (P) Ltd. KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India

Jai Shri Radhey Shyam

Dedicated to

Lord

Krishna Authors & Publishers

P reface

W

e are happy to present this book entitled “Organic Chemistry-I”. It has been written according to the latest U.P. Unified Syllabus to fulfil the requirement of B.Sc. Ist year students of all Colleges & Universities in Uttar Pradesh.

The book is written with the following special features: 1.

It is written in a simple language so that all the students may understand it easily.

2.

It has an extensive and intensive coverage of all topics.

3.

In each Chapter, Solved Examples are given based on different Topics.

4.

The complete Syllabus has been divided into Seven Chapters under Four Units.

5.

Sufficient Numerical Problems, Subjective Questions and Objective type questions with Hints & Solutions given at the end of each chapter will enable students to understand the concept . First of all we want to express our sincere gratitude to Purnima Sinha, Dr. S.B.P Sinha,

Prof. J.C. Ahluwalia, Prof. N.K. Jha for their invaluable guidance, immense interest and constant encouragement for the successful completion of the work. We are also thankful to Bandana Bariyar, Ashish Bariyar, Abhishek Bariyar, Archi Bariyar & Aradhyaa Bariyar for their kind help at many occasions. We are extremely grateful to our respected and beloved Parents whose incessant inspiration guided us to accomplish this work. We also express gratitude to our respective Families for their moral support. We are immensely thankful to Mr. S.K. Rastogi (Managing Director), Mr. Sugam Rastogi (Executive Director), Mrs. Kanupriya Rastogi (Director) and entire team of Krishna Prakashan Media (P) Ltd., for taking keen interest in this project and outstanding Management in getting the book published. The originality of the ideas is not claimed and criticism and suggestions are invited from the Students, Teaching community and other Readers.

July 2015

–Authors

(iv)

Syllabus ORGANIC CHEMISTRY-I B.Sc. Ist Year; IInd Paper UGC Model Curriculum (w.e.f. 2001) & U.P. UNIFIED Syllabus (w.e.f. 2011-12)

UNIT-I

M.M. 50

1.

Structure and Bonding: Hybridization, bond lengths and bond angles, bond energy, localized and delocalized chemical bonding, Van der Waals interactions, inclusion compounds, clatherates, charge transfer complexes, resonances, hyperconjugation, aromaticity, inductive and field effects, hydrogen bonding.

2.

Mechanism of Organic Reactions: Curved arrow notation, drawing electron movements with allows, half-headed and double-headed arrows, homolytic and heterolytic bond fission, Types of reagents – electrophiles and nucleophiles, Types of organic reactions, Energy considerations. Reactive intermediates – Carbocations, carbanions, free radicals, carbenes, arynes and nitrenes (with examples). Assigning formal charges on intermediates and other ionic species. Methods of determination of reaction mechanism (product analysis, intermediates, isotope effects, kinetic and stereochemical studies).

3.

Alkanes and Cycloalkanes: IUPAC nomenclature of branched and unbranched alkanes, the alkyl group, classification of carbon atom in alkanes, Isomerism in alkanes, sources methods of formation (with special reference to Wurtz reaction, Kolbe reaction, Corey-House reaction and decarboxylation of carboxylic acids), physical properties and chemical reactions of alkanes, Mechanism of free radical halogenation of alkanes: orientation, reactivity and selectivity. Cycloalkanes – Nomenclature, methods of formation, chemical reactions, Baeyer's strain theory and its limitations. Ring strain in small rings (cyclopropane and cyclobutane), theory of strain less rings. The case of cyclopropane ring, banana bonds.

UNIT-II 4.

Stereochemistry of Organic Compounds: Concept of isomerism, Types of isomerism; Optical isomerism – elements of symmetry, molecular chirality, enantiomers, stereogenic center, optical activity, properties of enantiomers, chiral and achiral molecules with two stereogenic centers, disasteromers, threo and erythro diastereomers, meso compounds, resolution of enantionmer, inversion, retention and recemization. Relative and absolute configuration, sequence rules, D & L and R & S systems of nomenclature. Geometric isomerism – determination of configuration of geometric isomers, E & Z system of nomenclature, geometric isomerism in oximes and alicyclic compounds. Conformational isomerism – conformational analysis of ethane and n-butane; conformations of cyclohexane, axial and equatorial bonds, conformation of mono substituted cyclohexane derivatives, Newman projection and Sawhorse formulae, Fischer and flying wedge formulae, Difference between configuration and conformation.

5.

Alkenes, Cycloalkenes, Dienes and Alkynes: Nomenclature of alkenes, methods of formation, mechanisms of dehydration of alcohols and dehydrohalogenation of alkyl halids, regioselectivity

UNIT-III (v)

in alcohol dehydration, The Saytzeff rule, Hoffmann elimination, physical properties and relative stabilities of alkenes. Chemical reactions of alkenes – mechanism involved in hydrogenation, electrophilic and free radical additions, Markownikoff's rule, hydroboration- oxidation, oxymercuration-reduction. Epoxidation, ozonolysis, hydration, hydroxylation and oxidation with KMnO4, Polymerization of alkenes, Substitution at the allylic and vinylic positions of alkenes, Industrial applications of ethylene and propene. Methods of formation, conformation and chemical reactions of cycloalkenes; Nomenclature and classification of dienes : isolated, conjugated and cumulated dienes, Structure of allenes and butadiene, methods of formation, polymerization, chemical reaction – 1, 2 and 1, 4 additions, Diels-Alder reaction. Nomenclature, structure and bonding in alkynes, Methods of formation, Chemical reactions of alkynes, acidity of alkynes, Mechanism of electrophilic and nucleophilic addition reactions, hydroboration-oxidation, metal-ammonia reductions, oxidation and polymerization.

UNIT-IV 6.

Arenes and Aromaticity: Nomenclature of benzene derivatives, The aryl group, Aromatic nucleus and side chain, Structure of benzene; molecular formula and kekule structure, stability and carboncarbon bond lengths of benzene, resonance structure, MO picture. Aromaticity: The Huckle rule, aromatic ions. Aromatic electrophilic substitution – general pattern of the mechanism, role of s and p complexes, Mechanism of nitration, halogenation, sulphonation, mercuration and Friedel-Crafts reaction. Energy profile diagrams. Activating and deactivating substituents, orientation and ortho/para ratio, Side chain reactions of benzene derivatives, Birch reduction; Methods of formation and chemical reactions of alkylbenzenes, alkynylbenzenes and biphenyl, naphthalene and Anthracene;

7.

Alkyl and Aryl Halides: Nomenclature and classes of alkyl halides, methods of formation, chemical reactions, Mechanisms of nucleophilic substitution reactions of alkyl halides, SN2 and SN1 reactions with energy profile diagrams; Polyhalogen compounds : Chloroform, carbon tetrachloride; Methods of formation of aryl halides, nuclear and side chain reactions; The addition-elimination and the elimination-addition mechanisms of nucleophilc aromatic substitution reactions; Relative reactivities of alkyl halides vs. allyl, vingl and aryl halides, Synthesis and uses of DDT and BHC.

(vi)

Detailed Contents Unit - I Chapter 1

Structure and Bonding

O–01-62

1. 2. 3.

Introduction 03 Localised and Delocalized Chemical Bonding Characteristics of Covalent Bond

$

Solved Examples

4. 5. 6. 7. 8.

Delocalized Chemical Bonding and Resonance Hybridization 18 Electron Displacement Effect 29 Electromeric Effect 33 Strength of Acids and Bases 42

$ $ $

Exercise 55 Answers 61 Hints & Solutions

Chapter 2

03 07

10

13

62

Mechanism of Organic Reactions

1.

Fundamental Concepts of Organic Reaction

$

Solved Examples

2. 3.

Various Reaction Intermediates 76 Methods of Determination of Reaction Mechanism

$ $ $

Exercise 99 Answers 102 Hints & Solutions

Chapter 3

O–63-102

63

66

96

102

Alkanes & Cycloalkanes

1. 2. 3.

Alkanes 103 Cycloalkanes 116 Stability of Rings and Ring Strain

$ $ $ $

Solved Examples Exercise 130 Answers 133 Hints & Solutions

O–103-134

124

126

134

Unit - II Chapter 4 1.

Stereochemistry of Organic Compounds

Introduction

135 (vii)

O–135-200

2.

Structural Isomerism

$

Solved Examples

3. 4. 5.

Stereoisomerism Geometrical Isomerism Conformational Isomerism

$ $ $

Exercise 186 Answers 194 Hints & Solutions

136 140

140 159 173

195

Unit - III Chapter 5

Alkenes, Cycloalkenes Dienes and Alkynes

1.

Alkenes

$

Solved Examples

2. 3. 4.

Cycloalkene (or Cycloolefin) Dienes 238 Alkynes 245

$ $ $

Exercise 266 Answers 273 Hints & Solutions

O–201-274

201 211

233

273

Unit - IV Chapter 6

Arenes and Aromaticity

O–275-344

1.

Arenes

$

Solved Examples

2. 3. 4. 5. 6. 7. 8.

Structure of Benzene 278 Aromaticity and Huckel's Rule 283 Electrophilic Aromatic Substitution Reactions 287 Disubstitution in Benzene Ring and Theory of Substituent Effect Fused or Condensed Aromatic Hydrocarbons 312 Anthracene 322 Biphenyls 329

$ $ $

Exercise 331 Answers 341 Hints & Solutions

Chapter 7

275 277

299

341

Alkyl and Aryl Halide

1. 2.

Introduction Alkyl Halides

$

Solved Examples

3. 4. 5. 6. 7.

Aliphatic Nucleophilic Substitution 354 Polyhalogen Compounds 367 Aryl Halides 370 Synthesis and Uses of DDT 382 Synthesis and Uses of BHC 383

$ $ $

Exercise 391 Answers 395 Hints & Solutions

O–345-396

345 345 348

396

❍❍❍ (viii)

Book-2 Organic C hemistry-I Unit-I Chapter 1 : Structure and Bonding Chapter 2 : Mechanism of Organic Reactions Chapter 3 : Alkanes and Cycloalkanes

Unit-II Chapter 4 : Stereochemistry of Organic Compounds

Unit-III Chapter 5 : Alkenes, Cycloalkenes, Dienes and Alkynes

Unit-IV Chapter 6 : Arenes and Aromaticity Chapter 7 : Alkyl and Aryl Halide

UnitO-3 -I

C HAPTER

1

Structure and Bonding

1. Introduction Before starting any discussion on organic chemistry, first let us make it very clear that the fundamentals involving bond cleavage, various intermediates and their reactivities, types of reagents, field effects etc form the backbone of entire organic chemistry. These concepts will be useful in understanding the acidic or basic behaviour of many organic compounds, types of various organic reactions, mechanism of various reactions and their other aspects.

2. Localised and Delocalized Chemical Bonding 2.1 Types of Bond As stated a chemical bond is an attraction between atoms. This attraction may be seen as the result of different behaviors of the outermost electrons of atoms. Although all of these behaviors merge into each other seamlessly in various bonding situations so that there is no clear line to be drawn between them, customarily the chemical bonds are classified into different types. One more point has to be reiterated before discussing the classification of chemical bond. Atom may attain a stable electronic configuration in three different ways by losing electron, by gaining or by sharing electron. Moreover, elements may be divided into following three types depending upon their electronegativity as: 1.

Electropositive elements: Those elements whose atoms give up one or more electron fairly readily.

2.

Electronegative elements: Which will accept electron.

3.

Elements which have little tendency to lose or gain electrons.

Three different types of bond may be formed depending on the electropositive or electronegative character of atom involved. Electropositive element + Electropositive element → Metallic bond Electropositive element + Electronegative element → Ionic bond Electronegative element + Electronegative element → Covalent bond

O-4

2.2 Covalent Bond In the simplest view of a so-called 'covalent' bond, one or more electrons (often a pair of electrons) are drawn into the space between the two atomic nuclei. Here the negatively charged electrons are attracted to the positive charges of both nuclei, instead of just their own. This overcomes the repulsion between the two positively charged nuclei of the two atoms, and so this overwhelming attraction holds the two nuclei in a fixed configuration of equilibrium, even though they will still vibrate at equilibrium position. Thus, covalent bonding involves sharing of electrons in which the positively charged nuclei of two or more atoms simultaneously attract the negatively charged electrons that are being shared between them. These bonds exist between two particular same or different atoms, and have a direction in space, allowing them to be shown as single connecting lines between atoms in drawings. For Example, Two chlorine atoms react to form a Cl2 molecule Cl + Cl

Cl

Cl

Each chlorine atom gives a share of one of its electrons to other atom. A pair of electrons is shared equally between both atoms and each atom now has eight electrons in its outer shell (stable octet). In a similar way, a molecule of tetra chloromethane CCl4 is made up of one carbon and four chlorine atoms. C + 4C

Cl Cl C Cl Cl

The carbon atom is short off four electrons so as to have noble gas structure. Consequently, it forms four bonds with the chlorine atoms which themselves are short of one electron so they each form one bond by sharing electrons. In this way, both carbon and all four chlorine atoms attain a noble gas structure. N + 3[H ]

H N H H

A molecule of ammonia (NH3 ) is made up of one nitrogen and three hydrogen atoms. Other examples of covalent bonds include water (with two covalent bonds) and hydrogen fluoride (one covalent bond and three lone pairs). H O , H

H F

In a polar covalent bond, one or more electrons are unequally shared between two nuclei. Covalent bonds often result in the formation of small collections of better-connected atoms called molecules. These molecules in solid and liquid state are bound to other molecules by intermolecular forces that are often much weaker than the covalent bonds that hold the molecules internally together. Such weak intermolecular bonds give organic molecular substances, such as waxes and oils, their soft character, and their low melting points. When covalent bonds link long chains of atoms in large molecules, however (as in polymers such as nylon), or when covalent bonds extend in networks through solids that are not composed of discrete molecules (such as diamond or quartz or the silicate minerals in many types of rock) then the structures that result may be both strong and tough, Also, the melting points of such covalent polymers and networks increase greatly.

O-5

2.3 Types of Covalent Bonds Sigma (σ) and Pi (π) Bonds Depending upon the type of overlapping, the covalent bonds are mainly of two types. 1.

Sigma (σ) bond: When a bond is formed between two atoms by the overlap of their atomic orbitals along the internuclear axis the resulting bond is called sigma(σ) bond. Such type of overlap is also known as end to end or head on overlap. It is a strong bond and cylindrically symmetrical. The overlapping along the internuclear axis can take place in any of the following ways: (i)

s—s overlapping: This type of overlapping takes place between atoms having half filled s–orbitals in their outer most energy shell. For example, in the formation of hydrogen molecule, 1s orbital of one hydrogen atom overlaps with 1s orbital of other hydrogen atom thus forming a sigma bond.

+

+ 1s arbital of H–atom

1s arbital of H–atom

+

s–s overlap

Fig. 1: Overlap of s orbital

(ii)

s— p overlapping: In this case, half filled s–orbital of one atom overlaps with the half filled p–orbtial of another atom. A simple example of this type is the formation of hydrogen fluoride. Here 1s orbitals of hydrogen overlaps with 2pz orbital of fluorine.

2pz orbital of F–atom

1-s orbital of H–atom

+

+

+

s–p overlap

Fig. 2: Overlap of s–p orbital

(iii) p—p overlapping: This type of overlapping occurs when p–orbital of one atom overlaps with the p–orbital of the other as in case of fluorine molecule. The molecule of fluorine is produced by the overlapping of 2pz orbitals of the two fluorine atoms. +

+ 2pz orbital of F–atom

2pz orbital of F-atom

+ p–p overlap

Fig. 3: Overlap of p–p orbital

2.

Pi (π) bond: Pi (π)bond is formed by lateral or sidewise overlapping of p orbitals. Sideways overlap means overlapping of p orbitals in a direction perpendicular to the internuclear axis. A π bond is not formed between two bonded atoms unless the two are held together with a σ −bond. It is relatively a weaker bond since the electrons are not strongly attracted by the nuclei of bonding atoms. For example,

O-6 (i)

In case of oxygen molecule (each oxygen atom having electronic configuration, 1s2 2s2 2p2x 2p1y 2p1z ), the two atoms are held together by one σ-bond and one π-bond.

+

Two Half–filled p–orbitals of O–atom

Two Half–filled p–orbitals of O–atom

O2 molecule (O=O)

Fig. 4: Formation of O2 molecule

(ii)

In the molecule of nitrogen both nitrogen atoms are held together by one σ-bond and 2 π-bonds. Nitrogen atom has an electronic configuration 1s2 2s2 2p1x 2p1y 2p1z .

+

N–atom

N–atom

N2 molecule – (N=N) 2

Fig. 5: Formation of nitrogen

It is important to remember that the 's' orbitals can only form σ-bonds, whereas the p, d & f orbitals can form both σ and π-bonds.

2.4 Multiple Bonding When two atoms share a single pair of electrons, the bond is referred to as a single bond. Atoms can also share two or three pairs of electrons in the aptly named double and triple bonds. The first bond between two atoms is called the σ (sigma) bond. All subsequent bonds are referred to as π (pi) bonds. In Lewis structures, multiple bonds are depicted by two or three lines between the bonded atoms. The bond order of a covalent interaction between two atoms is the number of electron pairs that are shared between them. Single bonds have a bond order of 1, double bonds 2, and triple bonds 3. Bond order is directly related to bond strength and bond length. Higher order bonds are stronger and shorter, while lower order bonds are weaker and longer. The Lewis structures for some common molecules involving multiple covalent bonds can be found below.

O-7 H

H C

C

H

H

C

N

C

C C H acetylene

H

O

H Formaldehyde

H H Formaldimine

Ethylene (ethere) H

H

H

– + C O carbon monoxide

N C hydrogen cyanide

Fig. 6: Lewis structures of molecules with multiple bonds

2.5 Co–ordinate Bond A covalent bond results from sharing of a pair of electrons between two atoms, where each atom contributes one electron to the bond. It is also possible to have an electron pair bond where both the electrons come from one of the two binding atoms and there is no contribution from the other atom. Such bonds are called co–ordinate bonds or dative bonds. So, co–ordinate bond is a special type of covalent bond in which both the bonded electrons come from one of the two binding atoms. One common example is formation of ammonium ion. Even though the ammonia molecule has a stable electronic configuration it can react with a hydrogen ion (H+) by donating a lone pair of electrons from N atom to H+ ion forming the ammonium ion NH4+ . H H

N

+

H + [H+]

+

or H N H

H N H

H

H

H

H

Covalent bonds are usually shown as a straight line joining the two atoms, and co–ordinate bonds as arrows indicating which atom is donating the electron. Similarly, ammonia donates its lone pair to boron trifluoride and by this means the boron atom attains noble gas configuration.

H

H

F

H N

B

H

F

F

H +

N

B

F

F

H

F

In a similar way a molecule of BF3 can form a co–ordinate bond by accepting a lone pair from a F − ion. – F

+

F B F F

F F

B

– F

F

3. Characteristics of Covalent Bond Some important characteristics of covalent bonds like bond length, bond angle and bond energy are discussed below:

O-8

3.1 Bond Length In molecular geometry, bond length or bond distance is the average distance between nuclei of two bonded atoms which may be same or different in a molecule.

It is a transferable property of a bond between atoms of fixed types, relatively independent of the rest of the molecule. Bond length is related to bond order, when more electrons participate in bond formation the bond will get shorter. Bond length is also inversely related to bond strength and the bond dissociation energy, as all other things being equal a stronger bond will be shorter. In a bond between two identical atoms half the bond distance is equal to the covalent radius. Bond lengths are measured in the solid phase by means of X-ray diffraction, or approximated in the gas phase by microwave spectroscopy. A set of two atoms sharing a bond is unique going from one molecule to the next. For example the carbon to hydrogen bond in methane is different from that in methyl chloride. It is however possible to make generalizations when the general structure is the same. The actual bond length between two atoms in a molecule depends on such factors as the orbital hybridization and the electronic and steric nature of the substituents. The carbon-carbon bond length in diamond is 154 pm which is also the largest bond length that exists for ordinary carbon covalent bonds. There are compounds in which shorter than average carbon–carbon bonds distances are also possible. Alkenes and alkynes have bond lengths of respectively 133 and 120 pm due to increased s-character of the sigma bond and presence of π bond.

153.5 pm

C2H6

133.9 pm

C2H4

120.3 pm

C2H2

Fig. 8: Bond length in different hydrocarbons

O-9 In benzene all C-C bonds have the same length, 139 pm, due to resonance. Tabel 1: Bond lengths in organic compounds C—H

Length (pm)

C—C

Length(pm)

Multiple–bonds

Length (pm)

sp3—H

110

sp3—sp3

154

Benzene

140

sp2—H

109

sp3—sp2

150

Alkene

133.9

sp—H

108

sp2—sp2

147

Allene

120.3

sp3—sp

146

sp2—sp

143

sp—sp

137

3.2 Bond Angle Molecular geometry is the three-dimensional arrangement of the atoms that constitute a molecule. It determines several properties of a substance including its reactivity, polarity, phase of matter, color, magnetism, and biological and physical activities as well. Molecular geometries can be specified in terms of bond lengths, bond angles and torsional angles. A bond angle is the angle formed between three atoms across at least two bonds. For four atoms bonded together in a chain, the torsional angle is the angle between the plane formed by the first three atoms and the plane formed by the last three atoms. As a matter of fact, the angles between bonds that an atom forms also depend on the rest of molecule, albeit weakly. Some of the molecules alongwith bond angles are shown below,

109° 47 120°

180° Linear

Trigonal planar

Tetrahedral

90° 90°

120°

Trigonal bipyramidal

90°

Octahedral

Fig. 9: Different molecular shapes along with bond angles

O-10

Solved Examples Example1: Predict all bond angles in the following molecules.

(i) CH3Cl

(ii) CH3CN

(iii) CH3COOH

Solution: (i) The Lewis structure of methyl chloride is: H H C

Cl

H

In the Lewis structure of CH3Cl carbon is surrounded by four regions of high electron density, each of which forms a single bond. Based on the VSEPR model, we predict a tetrahedral distribution of electron clouds around carbon, H - C - H and H - C - Cl bond angles of 109.5°, and a tetrahedral shape for the molecule. Note the use of doted lines to represent a bond projecting behind the plane of the paper and a solid wedge to represent a bond projecting forward from the plane of the paper.

(ii) The Lewis structure of acetonitrile, CH3CN is: H H C

C

N

H

The methyl group, CH3-, is tetrahedral. The carbon of the -CN group is in the middle of a straight line stretching from the carbon of the methyl group through the nitrogen. 109.5° H H C C N H 180°

(ii) The Lewis structure of acetic acid is: H O H C H

C

O H

O-11 Both the carbon bonded to three hydrogens and the oxygen bonded to carbon and hydrogen are centers of tetrahedral structures. The central carbon will have 120° bond angles.

The geometry around the first carbon is tetrahedral, around the second carbon atom is trigonal planar, and around the oxygen is bent.

3.3 Bond Energy In chemistry, bond energy (E) is the measure of bond strength in a chemical bond. It is the energy required to break one mole of molecules into their individual atoms. For example, the carbon-hydrogen bond energy in methane E(C-H) is the enthalpy change involved with breaking up one molecule of methane into a carbon atom and 4 hydrogen radicals divided by 4. Bond energy (E) must not be confused with bond-dissociation energy. As such, bond energy is an average of different bond dissociation energies of the same type of bonds. The same bond can appear in different molecules, but it will have a different bond energy in each molecule as the other bonds in the molecule will affect the bond energy of the specific bond. So the bond energy of C-H in methane is slightly different than the bond energy of C-H in ethane. One can calculate a more general bond energy by finding the average of the bond energies of a specific bond in different molecules to get the average bond energy. Table 2: Average bond energies (kj/mol) Single Bonds

Multiple Bonds

H—H

432

N—H

391

I–I

149

C=C

614

H—F

565

N—N

160

I–Cl

208

C≡C

839

H—Cl

427

N—F

272

I–Br

175

O=O

495

H—Br

363

N—Cl

200

S–H

347

C = O*

745

H—I

295

N—Br

243

S–F

327

C≡O

1072

C—H

413

N—O

201

S–Cl

253

N=O

607

C—C

347

O—H

467

S–Br

218

N=N

418

C—N

305

O—O

146

S–S

266

N≡N

941

C—O

358

O—F

190

Si–Si

340

C≡N

891

C—F

485

O—Cl

203

Si–H

393

C≡N

615

C—Cl

339

O—I

234

O-12 C—Br

276

F—F

154

Si–C

360

C—I

240

F—Cl

253

Si–O

452

C—S

259

F—Br

237

Cl—Cl

239

Cl—Br

218

Br—Br

193

Some of the characteristics of bond energy values are: 1.

Average bond energy values are not as accurate as a molecule specific bond-dissociation energies.

2.

Double bonds are higher energy bonds in comparison to a single bond (but not necessarily 2-fold higher).

3.

Triple bonds are even higher energy bonds than double and single bonds (but not necessarily 3-fold higher).

Example 2: (i) What is the definition of bond energy? When is energy released and absorbed?

(ii) If the bond energy for H-Cl is 431 kJ/mol. What is the overall bond energy of 2 moles of HCl? (iii) Using the bond energies given in the chart above, find the enthalpy change for: the decomposition of water 2H2O(g)→2H2+O2(g) Is the reaction written above exothermic or endothermic? Explain. (iv) Which bond in list below has the highest and lowest bond energy? H-H, H-O, H-I, H-F. Solutions (i) Bond energy is the energy required to break a bond that exists between two atoms. Energy is given off when the bond is broken, but is absorbed when a new bond is created.

(ii) Simply multiply the average bond energy of H-Cl by 2. This leaves one with 862 kJ/mol (using the table). (iii) The enthalpy change deals with breaking two mole of O-H bonds and the formation of 1 mole of O-O bonds and two moles of H-H bonds.The sum of the energies required to break the bonds on the reactants side is 4×460 kJ/mol = 1840 kJ/mol.The sum of the energies released to form the bonds on the products side is 2 moles of H-H bonds = 2 x 436.4 kJ/mol = 872.8 kJ/mol 1 moles of O–O bond – 1 x 498.7 kJ/mol = 498.7 kJ/mol The released energy = 872.8 kJ/mol + 498.7 kJ/mol = 1371.5 kJ/mol. Total energy difference is 1840 kJ/mol - 1371.5 kJ/mol = 469 kJ/mol, which indicates that the reaction is endothermic and that 469 kJ of heat is needed to be supplied to carry out this reaction.

O-13 (iv) H-F has the highest bond energy since the difference in electronegativity is the greatest. However, the H-I bond is the lowest bond energy (not due to the electronegativity difference, but due to the greater size of the I atom).

3.4 The Shape of Molecules The three dimensional shape or configuration of a molecule is an important characteristics. This shape is dependent on the preferred spatial orientation of covalent bonds to atoms having two or more bonding partners. Three dimensional configurations are best viewed with the aid of models. In order to represent such configurations on a two-dimensional surface (paper, blackboard or screen), one can normally use perspective drawings in which the direction of a bond is specified by the line connecting the bonded atoms. In most cases the focus of configuration is a carbon atom, so the lines specifying bond directions will originate there. As defined in the diagram, a simple straight line represents a bond lying approximately in the surface plane. The two bonds to substituents A in the structure below are of this kind, normal bond

A A

C

D B

wedge bond or

}hatched bond dashed bond

A wedge shaped bond is directed in front of this plane (thick end toward the viewer), as shown by the bond to substituent B; and a hatched bond is directed in back of the plane (away from the viewer), as shown by the bond to substituent D. Some texts and other sources may use a dashed bond in the same manner as we have defined the hatched bond, but this can be confusing because the dashed bond is often used to represent a covalent bond that is partially formed or partially broken.

4. Delocalized Chemical Bonding and Resonance In chemistry, resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by one single Lewis formula. So, resonance is a bonding behaviour in which a molecule or ion with such delocalized electrons is represented by several contributing structures also known as resonating structures or canonical forms. Each contributing structure can be represented by a Lewis structure, with only an integer number of covalent bonds between each pair of atoms within the structure. Several Lewis structures are used collectively to describe the actual molecular structure. However these individual contributors are not real structures. So, these cannot be observed in the actual resonance-stabilized molecule; the molecule does not oscillate back and forth between the contributing structures, as might be assumed from the word "resonance". The actual structure is an approximate intermediate between the various canonical forms, but its overall energy is lower than each of the contributors. This intermediate form between different contributing structures is called a resonance hybrid.The point of importance is that the contributing structures differ only in the position of electrons, not in the position of nuclei.

O-14

4.1 Resonance Energy Concept along with Development Resonance is a key of valence bond theory. Electron delocalization lowers the potential energy of the substance and thus makes it more stable than any of the contributing structures. The difference between the potential energy of the actual structure and that of the contributing structure with the lowest potential energy is called the resonance energy or delocalization energy. Resonance is different from tautomerism and conformational isomerism, which involve the formation of isomers, thus the rearrangement of the nuclear positions.

4.2 Characteristics of Resonance Molecules and ions with resonance have certain basic characteristics. O–

O –O

C O

–O

–O

C

C O

O

O–

Fig. 10: Contributing structures of the carbonate ion

In diagrams, contributing structures are typically separated by double-headed arrows (←→ ). The arrow should not be confused with the right and left pointing equilibrium arrow ( ). [S=C=N

S–C N]

Fig. 11: Contributing structures of the thiocyanate ion

All structures together may be enclosed in large square brackets, to indicate they picture one single molecule or ion, not different species in a chemical equilibrium. Alternatively to the use of resonance structures in diagrams, a hybrid diagram can be used. In a hybrid diagram, π bond that are involved in resonance are usually pictured as curves or dashed lines, indicating that these are partial rather than normal complete π bonds. In benzene and other aromatic rings, the delocalized π-electrons are sometimes pictured as a solid circle. Some of the important characteristics are : 1.

They can be represented by several Lewis formulas, called "contributing structures", "resonance structures" or "canonical forms". However, the real structure is not a rapid interconversion of contributing structures. To represent the intermediate, a resonance hybrid is used instead.

2.

The contributing structures are not isomers. They differ only in the position of electrons, not in the position of nuclei.

3.

Each resonating form must have the same number of valence electrons and thus the same total charge, and the same number of unpaired electrons, if any.

4.

Bonds that have different bond orders in different contributing structures do not have typical bond lengths. Measurements reveal intermediate bond lengths.

5.

The real structure has a lower total potential energy than each of the contributing structures would have. This means that it is more stable than each separate contributing structure would be.The gap of the potential energies of the resonance hybrid and the most stable resonating structure is known as resonance energy, as stated earlier.

O-15

4.3 Resonance Hybrids The actual structure of a molecule in the normal quantum state has the lowest possible value of total energy. This structure is called the "resonance hybrid" of that molecule. The resonance hybrid is the approximate intermediate of the contributing structures, but the overall energy is lower than each of the contributors, due to the resonance energy. Any molecule or ion exists in only one form - the resonance hybrid. It does not jump back and forth between its resonance contributors- looking like one this moment and like another the next moment.

4.4 Stability of Resonating Forms One contributing structure may resemble the actual molecule more than another in the sense of energy and stability. Structures with a low value of potential energy are more stable than those with high values and resemble the actual structure more. The most stable contributing structures are called major contributors, so more is the stability of a canonical form, more is its contribution towards the resonance hybrid. Energetically unfavourable and therefore less probable structures are minor contributors. The stability of a canonical form depends upon the following factors, 1.

The atoms of the structure must obey as much as possible the octet rule, so that 8 valence electrons must be present around each atom rather than having deficiencies or surplus.

2.

More is the number of covalent bonds, higher is the stability.

3.

A major contributer is the one that carry a minimum of charged atoms. If unlike charges are present their separation must be least while for like charges the separation must be maximum.

4.

In case of presence of negative charge, if any, it must be present on the more electronegative atoms and positive charge, if any, on the most electropositive. So, in the following case, structure II is more stable than I because the negative charge is placed on more electronegative O atom. Similarly positive charges, if present, are best occupied on atoms of low electronegativity.

CH2–C–CH3

CH2

C–CH3

O

O

I

II

5.

The greater the number of contributing structures, the more stable the molecule. This is because the more states at lower energy are available to the electrons in a particular molecule, the more stable the electrons are. Also the more volume electrons can occupy, the more stable the molecule is. It can be understood by borrowing a concept of physics, which states that charge dispersed is directly proportional to stability. Here, electrons can be termed as charged bodies and the more volume they occupy, more the charge gets dispersed ultimately leading to stability.

6.

Equivalent contributors contribute equally to the actual structure; those with low potential energy (the major contributors) contribute more to the resonance hybrid than the less stable minor contributors. Especially when there is more than one major contributor, the resonance stabilization is high.

4.5 Van der Waals Interactions In chemistry, the Van der Waals force or Van der Waals interaction is the sum of the attractive or repulsive forces between molecules or between parts of the same molecule other than those due to covalent bonds, or the electrostatic interaction of ions with one another, with neutral molecules, or with charged molecules.

O-16 Van der Waals forces include attractions and repulsions between atoms, molecules, and surfaces, as well as other intermolecular forces. As such, Van der Waals forces define many properties of organic compounds, including their solubility in polar and non-polar solvents etc.For example,in low molecular weight alcohols, the hydrogen-bonding properties of the polar hydroxyl group dominate the weaker Van der Waals interactions. In higher molecular weight alcohols, the properties of the nonpolar hydrocarbon chain(s) dominate and define the solubility. Van der Waals forces quickly vanish at longer distances between interacting molecules. Van der Waals forces are relatively weak compared to covalent bonds which is quite expected of them being the intermolecular forces and not a type of chemical bond. However, these forces play a fundamental role in chemistry and its different offshoots like supramolecular chemistry, polymer science, nanotechnology, and surface science. All Van der Waals forces are anisotropic except those between two noble gas atoms. It means that the magnitude of these forces depend on the relative orientation of the molecules. The induction and dispersion interactions are always attractive, irrespective of orientation, but the electrostatic interaction changes sign upon rotation of the molecules. That is, the electrostatic force can be attractive or repulsive, depending on the mutual orientation of the molecules. The main characteristics are: 1.

They are weaker than normal covalent or ionic bonds.

2.

Van der Waals forces are additive and cannot be saturated.

3.

They have no directional characteristics.

4.

They are all short - range forces and hence only interactions between nearest need to be considered instead of all the particles. The greater is the attraction if the molecules are closer due to Van der Waals forces.

5.

Van der Waals forces are independent of temperature except dipole - dipole interactions.

4.5.1 Types of Van Der Waals Forces Van der Waals forces include a number of interactions. These are discussed below, 1.

Dipole-dipole interaction: A force between two permanent dipoles is known as dipole-dipole interaction or Keesom force. It can be diagrammatically shown below, δ+

δ–

δ–

δ+

attraction Fig. 12: Dipole–dipole interaction

Dipole-Dipole interactions result when two polar molecules approach each other in space. When this occurs, the partially negative portion of one of the polar molecules is attracted to the partially positive portion of the second polar molecule. This type of interaction between molecules accounts for many physically and biologically significant phenomena. An example is shown below: δ+ δ– H—Cl

–

δ+ δ H—Cl

Fig. 13:Dipole–dipole interaction in HCl

O-17 2.

Dipole - induced dipole interaction: A force between a permanent dipole and a corresponding induced dipole is also known as dipole - induced dipole or Debye force. This type of attractive interaction also depends on the presence of a polar molecule. However, the second participating molecule need not be polar as shown below:

+

+

H2O

Xenon

Fig. 14: Example of dipole induced dipole interaction

In the dipole-induced-dipole interaction, the presence of the partial charges of the polar molecule causes a polarization, or distortion, of the electron distribution of the other molecule. As a result of this distortion, the second molecule acquires regions of partial positive and negative charge, and thus it becomes polar. The partial charges so formed behave just like those of a permanently polar molecule and interact favourably with their counterparts in the polar molecule that originally induced them. Hence, the two molecules attract as shown below:

δ–

δ+

+

Spherical atom with no dipole, The dot indicates the location of the nucleus

Upon approach of a charged ion, electrons in the atom respond and the atom develops a dipole.

Fig. 15: Dipole–induced dipole interaction

This interaction also contributes to the intermolecular forces that are responsible for the condensation of hydrogen chloride gas. 3.

Induced dipole-induced dipole interaction : It is a force between two instantaneously induced dipoles also known as London dispersion force. This type of interaction acts between all types of molecule, polar or not. It is the principal force responsible for the existence of the condensed phases of certain molecular substances, such as benzene, other hydrocarbons, bromine, and the solid elements phosphorus (which consists of tetrahedral P4 molecules) and sulfur (which consists of crown-shaped S8 molecules). The interaction is called the dispersion interaction or, less commonly, the induced-dipole-induced-dipole interaction. Two nonpolar molecules of argon are considered near each other as shown below,

O-18

Ar

Ar

δ– δ+

δ– δ+

Fig. 16: Example of induced–dipole induced dipole interaction

Although there are no permanent partial charges on either molecule, the electron density can be thought of as ceaselessly fluctuating. As a result of these fluctuations, regions of equal and opposite partial charge arise in one of the molecules and give rise to a transient dipole. This transient dipole can induce a dipole in the neighbouring molecule, which then interacts with the original transient dipole as shown here, –

+

–

+

–

+ + Repulse

–

Attract

Fig. 17: Induced dipole induced dipole interaction

Although the latter continuously flickers from one direction to another (with an average of zero dipole overall), the induced dipole follows it, and the two correlated dipoles either attract or repel with one another.

5. Hybridization 120°

109.5° A

180° A

A

In chemistry, hybridization is one of the landmark concept of chemical bonding explaining a number of properties of covalent compounds. It refers to the concept of mixing atomic orbitals into new hybrid orbitals with different energies, shapes, etc., than the component atomic orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory. Hybrid orbitals are very useful in the explanation of molecular geometry and atomic bonding properties. Although sometimes taught together with the valence shell electron-pair repulsion (VSEPR) theory, valence bond and hybridization are in fact not related to the VSEPR model. It is the process of intermixing of atomic orbitals of same atom, having almost similar energies, followed by redistribution of energies to form new orbitals of identical energies and sizes. The new orbitals formed are called hybrid orbitals. The number of hybrid orbitals formed is always equal to the number of pure atomic orbitals employed for hybridization.

O-19 It is not known whether hybridization actually takes place or not but it is a concept which is used to explain certain observed properties of molecules. The following points are to be remembered about hybridization, 1.

Orbitals belonging to the same atom or ion having almost similar energies get hybridized.

2.

Number of hybrid orbitals is equal to the number of pure atomic orbitals taking part in hybridization.

3.

The hybridization takes place to produce equivalent hybrid orbitals which are degenerate and which give maximum symmetry.

4.

Hybrid orbitals are always involved in head on overlap, so the type of bonding resulted is always sigma (σ) bond.

5.1 Important Types of Hybridization In organic chemistry the various types of hybridization encountered are - sp3, sp2 and sp. These are discussed in detail.

Fig. 18: Shape of sp3 hybrid lobes

1.

sp3 hybridization: In this type of hybridization, one 's' and three 'p' orbitals of the same value of n (principle quantum number) mix up to form four sp3 hybridized orbitals. The mixing of orbitals is shown below,

O-20

The sp3 hybrid orbitals have 25% 's' character and 75% 'p' character. These orbitals orient themselves towards the corners of a regular tetrahedron. The angle between the orbitals is 109°28'. The commonest example is CH4 as shown below, 1s

H

H

σ 9°–

28 '

sp3

H

10

C sp3

sp3 sp3

H

C

σ

H

H

σ

H Fig. 20: Shape of methane

H

As such, in any organic compound if the atom is containing 4σ bonds or 3σ bonds plus one lone pair of electrons or 2σ bonds plus 2 lone pairs of electrons, then the the type of hybridization undergone by the atom is sp3.

Fig. 21: Lobes of sp2 hybrid orbitals

2.

sp2 hybridization: This is a second type of hybridization involving pure one 's' and two 'p' atomic orbitals of the same principle quantum number. These pure atomic orbitals mix together to form three sp2 hybridized orbitals. The mixing of orbitals can be shown as,

O-21 y

z

x

2p

2s

2py

z

2px Unhybridised unhybridised p–orbitals

sp2 hybridization

Three sp2 hybrid orbitals

120° Three sp2 hybrid orbitals in one plane

Fig. 22: Formation of sp2 hybrid orbitals

The sp2 hybrid orbitals have 33.3% 's' character and 66.6% 'p' character. These orbitals orient themselves towards the corners of an equilateral triangle, as shown above. It is important to remember that one p orbital perpendicular to the molecular plane is unhybridised. An inorganic example is BF3,where the central boron atom is sp2 hybridized. In cases like ethene (C2H4), the bonding can be shown as, H

H π σ

C

F

F B

F Fig. 23: Formation of BF3

C

H

H

Fig. 24: Bond line notation of C 2H4

Upon exhaustive analysis , it is found that the two carbon atoms have one p–orbital each with an unpaired electron. p

p

H

sp2 sp2

H

sp2

C sp2

sp2 C 2 sp

H p

p

H

Fig. 25: Hybrid orbital arrangement of C2H4

O-22 These two p-orbitals overlap in sideways manner to give rise to a double bond. Both and framework bonding is shown. H1s H1s 2pz

2pz

C

C

z sp2 C

H1s

C sp

H

2

H

H H

H1s

(a)C2H4 σ–bonded framework

(b)C2H4 π bonding

Fig. 26: σ and π bond arrangement of C2H4

It is always important to remember that when a bond is formed with head-to-head overlap, it is called a sigma (σ) bond. Similarly,when a bond is formed with sideways overlap, it is called a pi (π) bond. As such, in any organic compound if the atom is containing 3σ bonds or 2σ bonds plus one lone pair of electrons, then the the type of hybridization undergone by the atom is sp2.

Fig. 27: Shape of sp hybrid lobes

3.

sp hybridization: In this type, one 's' and one 'p' orbital of the same principle quantum number mix to form two sp hybridized orbitals, for example in BeCl2, Be is sp hybridized. The mixing of s and p orbitals can be shown as,

O-23 They possess 50% 's' and 50% 'p' character. In cases like ethyne C2H2, the central carbon atom is sp hybridized. Ethyne has the following Lewis structure. H

C

C

H

Hybridization wise, each carbon atom is sp hybridized having two perpendicular p orbitals with two unpaired electrons present on each carbon atom. These overlap in sideways fashion and produce a triple bond. All the overlaps are shown below, pz- Orbital py- Orbital σ-bond

H

C

C

H

π-bond

π-bond Fig. 29: Formation of ethyne

As such, in any organic compound if the atom is containing 2σ bonds or 1σ bonds plus one lone pair of electrons, then the type of hybridization undergone by the atom is sp.

5.2 Inclusion Compounds In host-guest chemistry, the definition of inclusion compounds is very broad, extending to channels formed between molecules in a crystal lattice in which guest molecules can fit. As such, an inclusion compound is a complex comprised of two chemical species, one chemical compound is the "host" which generally forms a cavity. Another is the "guest" compound which are located in the cavities of the host. If the spaces in the host lattice are enclosed on all sides so that the guest species is 'trapped' as in a cage, the compound is known as a clathrate. In this type of molecular encapsulation a guest molecule is actually trapped inside another molecule.

Fig. 30: Guest molecule is trapped inside the host

O-24

5.3 Clathrate Compounds A clathrate is a chemical substance consisting of a lattice that traps or contains molecules. The word clathrate is derived from the Latin clatratus meaning with a lattice. Traditionally, clathrate compounds are polymeric and completely envelop the guest molecule, but in modern usage clathrates also include host-guest complexes and inclusion compounds. According to IUPAC convention, clathrates are Inclusion compounds in which the guest molecule is in a cage formed by the host molecule or by a lattice of host molecules. Traditionally clathrate compounds refer to polymeric hosts containing molecular guests. More recently, the term refers to many molecular hosts, including some inorganic polymers such as zeolites. Many clathrates are derived from an organic hydrogen-bonded frameworks. These frameworks are prepared from molecules that "self-associate" by multiple hydrogen-bonding interactions. The most famous clathrates are methane clathrates where the hydrogen-bonded framework is contributed by water and the guest molecules are methane. Large amounts of methane naturally frozen in this form exist both in permafrost formations and under the ocean sea-bed. Other hydrogen-bonded networks are derived from hydroquinone, urea, and thiourea. A much studied host molecule is Dianin's compound which is shown below,

Fig. 31: Dianin's compound

5.4 Charge Transfer Complexes Charge-transfer complexes exist in many types of molecules, inorganic as well as organic, and in solids, liquids, and solutions. A well-known example is the iodine starch complex formed by iodine when combined with starch, which exhibits an intense blue charge-transfer band. A charge-transfer complex or CT complex or electron-donor-acceptor complex is an association of two or more molecules, or of different parts of one large molecule, in which a fraction of electronic charge is transferred between the molecular entities. The resulting electrostatic attraction provides a stabilizing force for the molecular complex. The source molecule from which the charge is transferred is called the electron donor and the receiving species is called the electron acceptor. The nature of the attraction in a charge-transfer complex is not a stable chemical bond, and is thus much weaker than covalent forces. Many such complexes can undergo an electronic transition into an excited electronic state. The excitation energy of this transition occurs very frequently in the visible region of the electro-magnetic spectrum, which produces the characteristics intense color for these complexes. These optical absorption bands are often referred to as charge-transfer bands (CT bands). Optical spectroscopy is a powerful technique to characterize charge-transfer bands.

O-25 nitrogen atoms have a higher electron density than the carbonyl groups and lend it to the acceptor

acceptor

O N O

O N O

O

O

O

N

N

N

N

O

O

O

O

O

donar carbonyl groups suck electron density away from the acceptor unit

5.5 Hydrogen Bonding Hydrogen bond is a weak force of attraction formed between an H atom and highly electronegative atom like O, N or F either of the same molecule or of a different molecule but to which it is not directly attached. Moreover the H atom itself should be attached to a highly electronegative atom like O, N or F. HF is considered as an example. The hydrogen fluorine bond in hydrogen fluoride is polar covalent bond as fluorine is a strongly electronegative element. As a result fluorine acquires a partial negative charge and hydrogen acquires a partial positive charge. δ+

δ−

H− F The lone pair on the fluorine atom in another molecule of hydrogen fluoride will attract the positive charge on hydrogen in a molecule of hydrogen fluoride electrostatically. This bond between hydrogen and fluorine of different molecules is known as hydrogen bond. This type of linkage is represented by dotted lines. δ+ δ – δ+ δ – δ+ δ – H – F .......... H – F............ H – F Hydrogen Bond

Here, it is important to remember that hydrogen bonding is not a type of chemical bond rather it is an intermolecular force of attraction.

5.5.1 Conditions for Hydrogen Bonding 1.

The molecule must contain a highly electronegative atom linked to hydrogen atom.

2.

The size of electronegative atom should be small. These conditions are met only by F, O and N atoms.

Although Cl has the same electronegativity as nitrogen, it does not frequently form effective hydrogen bond. This is because of its larger size than that of N with the result that its electrostatic attractions are weak.

5.5.2 Types of Hydrogen Bonding Generally the hydrogen bonds are classified into two types -

O-26 1.

Intermolecular hydrogen bonding

2.

Intramolecular hydrogen bonding

1.

Intermolecular hydrogen bonding: In this type of hydrogen bonding, the two or more than two molecules of same or different compounds combine together to give a polymeric aggregate. Some of the examples are,

δ+ δ– O – H..........O (iii) H – C

C–H O..........H – O δ+ δ–

(iv) Cl

2.

O–H

O – H......Cl

Intramolecular hydrogen bonding: In this type, hydrogen bonding occurs within two atoms of the same molecule. This type of hydrogen bonding results into ring formation commonly known as chelation. Organic compounds frequently show this type of H bonding. Some of the examples are, +

O – Hδ

Cl

O–H

O

δ–

N O

o–chlorophenol

o–nitrophenol

5.5.3 Effects of Hydrogen Bonding 1.

Boiling point and melting point: The high boiling point of water can be explained on the basis of hydrogen bonding.Water molecules associate through extensive network of hydrogen bonding and thus require more energy to break the intermolecular forces. Similarly we can explain the higher boiling point of alcohols, amines, amides etc. Moreover, H2O is a liquid but H2S is a gas because H2O can form hydrogen bond but H2S cannot.

O-27 Up till now those examples are considered where boiling point is increased due to hydrogen bonding. Normally, this happens in the case of intermolecular hydrogen bonding. However, there are certain cases where the boiling point is decreased due to hydrogen bonding. It happens in the case of intramolecular hydrogen bonding. For example o–hydroxy, nitro, carbonyl, carboxylic or chloro compounds have lower melting and boiling points than the respective meta or para isomers. The explanation is intramolecular hydrogen bonding taking place between the two groups placed ortho to each other. It prevents the association of the molecule with other neighbouring molecules. As a result, the intermolecular forces of attraction decreases which eventually leads to decreased melting and boiling points. O

O

O

H

H

Cl

O

H

Intramolecular Hydrogen bonding

O

N

N o–chlorophenol

O o–nitrophenol

O–H o–hydroxy benzoic acid

On the other hand, in case of meta– and para–isomers, intramoelcular hydrogen bonding is not possible because of the bigger size of the ring formed. Rather, these molecules indulge into intermolecular hydrogen bonding which causes association. This association of molecules eventually leads to higher mp and bp for meta and para isomers. 2.

Water solubility: A substance is said to be soluble in water if it is capable of forming hydrogen bonding with water molecules. For example alcohols are soluble in water due to hydrogen bonding. H – O ...........H – O .......... H – O .......... H – O H

R

H

R

It is important to note that when alkyl groups in alcohols are bulkier having four or more carbon atoms, the alkyl group predominates over OH group. So, hydrogen bond forming ability is subdued with the result that the solubility of such alcohols in water decreases. On the contrary side, when a compound has a large ratio of –OH groups to hydrocarbon group, it will be highly water soluble. Thus sugar, certain starches and polyvinyl alcohol are quite soluble in water. Solubility of other compounds (organic or inorganic) having one or more than one –OH group can be explained on the basis of hydrogen bonding. Detergents are also soluble in water due to solubilizing effect of hydroxyl groups. It is important to note that while the intermolecular hydrogen bonding increases solubility of the compound in water, the intramolecular hydrogen bonding decreases. This is due to fact that the formation of intramolecular hydrogen bond prevents hydrogen bonding of the compound with water. It eventually reduces solubility of the compound in water. For Example: o–nitrophenol is less soluble in water then p–nitrophenol because due to intramolecular hydrogen bonding –OH group is not available to form hydrogen bond with water.

O-28 O–H

O

δ–

N O

H Hydrogen bonding with H2O molecule is not possible because – OH group of o–nitrophenol is not available for intermolecular bonding.

o–Nitrophenol

3.

δ+

O–H

Volatile nature: A compound having high boiling point is less volatile and vice versa. As intermolecular hydrogen bonding increases the boiling point so it can be concluded that volatile nature of a compound decreases if intermolecular hydrogen bonding is present. On the other side, the volatility increases if intramolecular hydrogen bonding is present.

5.6 Aromaticity In organic chemistry, aromaticity is formally used to describe an unusually stable nature of some planar or flat rings of atoms such that the ring may be homonuclear as in case of benzene or hetronuclear as in case of pyridine. H H

H

H

C C

C

C

C C

H

H

H

H

H

C C

C

C

C N

H H

These structures contain a number of double bonds that interact with each other according to certain rules. As a result of their being highly stable, such rings tend to form easily, and once formed, tend to be difficult to break in chemical reactions. Since one of the most commonly-encountered aromatic system of compounds in organic chemistry is based on derivatives of benzene which is commonly found in petroleum, the word "aromatic" is occasionally used to refer informally to benzene derivatives, and this is how it was first defined. Nevertheless, many non-benzene aromatic compounds do exist as well. In living organisms, for example, the most common aromatic rings are the double-ringed bases in RNA and DNA, and interestingly benzene and its derivatives are too rare in biology. Huckel's rule helps us to decide that a given compound would be aromatic or not by observing its structure. According to this rule, for a compound to be aromatic, it must satisfy the following conditions, 1.

There must be cyclic delocalization in the compound.

2.

The closed loop formed due to cyclic delocalization must have a total of (4n + 2)π electrons where n is an integer having values 0, 1, 2, 3…… . The (4n + 2)π electrons is called Hukel number.

The first condition of cyclic delocalization takes into account the conditions required to show resonance so that the compound must have conjugation and the compound must be planar or nearly planar. If both the above conditions are fulfilled, then the compound would be aromatic and if any of the above condition fails, then it is non-aromatic.

O-29 First of all, let us take the simplest example of benzene. As it is evident from the structure that there is conjugation in the ring leading to cyclic delocalization. Moreover,the closed loop thus formed would have 6π electrons which corresponds to n=1 in the Huckel's rule. Thus, benzene is said to be aromatic compound. The same would be true for phenol and nitrobenzene also. OH

NO2

The rule is discussed in details in chapter VI.

6. Electron Displacement Effect In organic chemistry, there are some important mechanisms of electron displacement like field and inductive effect, electromeric effect, mesomeric effect and hyperconjugation. These effects are discussed below one by one.

6.1 Field and Inductive Effect Inductive Effect is an experimentally observed effect of the transfer of charge - negative or positive, through a chain of atoms in a molecule. The permanent dipole induced in one bond by another is called inductive effect. The electron cloud in a σ-bond between two unlike atoms is not uniform and is slightly displaced towards the more electronegative of the two atoms. This causes a permanent state of bond polarization, where the more electronegative atom has a slight negative charge (δ-) and the other atom has a slight positive charge (δ+). If the electronegative atom is then joined to a chain of atoms, usually carbon, the positive charge is relayed to the other atoms in the chain. This is the electron-withdrawing inductive effect, also known as the -I effect. Some groups, such as the alkyl group, are less electron-withdrawing than hydrogen and are therefore considered as electron-releasing. This is electron releasing character and is indicated by the +I effect. In short, alkyl groups tend to give electrons, leading to induction effect. Explanation First let us look at the C–C bond in ethane. This C–C bond has no polarity because it connects two similar atoms. However, the C–C bond in ethyl chloride is polarized by the presence of the electronegative chlorine atom. This polarization is actually the sum of three effects. In the first of these, the Cl atom have been deprived of some of its electron density by the higher electronegativity of Cl. Secondly, the electron deficiency of C1 is partially compensated by drawing the C–C electrons closer to itself, resulting in polarization of this bond and a slight positive charge on the C2 atom. Thirdly, the polarization of the C–C bond causes a (slight) polarization of the three methyl C–H bonds. The effect of C1 on C2 is less than the effect of Cl on C1 atom.

O-30

6.2 Field Effect In addition to such inductive effect operating through the σ–bonds in a compound, an analogous effect can also operate either through the space surrounding the molecule or in solution via the molecules of solvent that surround it which is called field effect. A point of distinction between the two effects is that the inductive effect depends only on the nature of bonds while the field effect depends on the geometry of the molecule. However, in many cases, it is not possible to distinguish inductive effect with this field effect. But in general, reference to an inductive effect is assumed to include any such field effect.

6.3 Characteristics of I Effect Some important characteristics of inductive effect are; 1.

It is a permanent electron displacement effect. It is because no attacking reagent (electrophile or nucleophile)is required to cause the electron displacement.

2.

It always causes partial charge separation as shown below,

3.

It decreases with increases of distance

4.

It does not involve delocalization of π-electrons. Rather,it involves displacement of σ-electrons only.

6.4 Types of Inductive Effect The inductive effect is divided into two types depending on their strength of electron withdrawing or electron releasing nature with respect to hydrogen. 1.

Negative inductive effect (-I): The electron withdrawing nature of groups or atoms is called as negative inductive effect. It is indicated by -I. Following are the examples of groups in the decreasing order of their -I effect:

NH +3 > NO2 > CN > SO3H > CHO > CO > COOH > COCl > CONH2 > F > Cl > Br > I > OH > OR > NH2 > C6H5 > H 2.

Positive inductive effect (+I): It refers to the electron releasing nature of the groups or atoms and is denoted by +I. Following are the examples of groups in the decreasing order of their +I effect. C(CH3)3 > CH(CH3)2 > CH2CH3 > CH3 > H

For measurements of relative inductive effects, hydrogen is chosen as reference point in the molecule CR3 H. If the H atom in this molecule is replaced by X (an atom or group of atoms), the electron density in the CR3 part of the molecule is less than in CR3 - H, then X is said to have a -I effect or electron withdrawing or electron attracting inductive effect. If the electron density in the CR3 part is greater than in CR3 - H, then X is said to have a +I effect or electron releasing or electron repelling inductive effect. Deuterium is also electron donating with respect to hydrogen. Generally speaking, all alkyl groups exhibit +I effect when they are attached to an unsaturated or trivalent carbon or other atom. Keeping other things equal, atoms with sp bonding generally have a greater electron withdrawing ability than those with sp2 bonding, which in turn

O-31 have more electron-withdrawing power than with sp3 bonding. This reasonably accounts for the fact that arylic, vinylic and alkynyl groups exhibit -I effect. Example 3: Why alkyl groups are showing positive inductive effect? Solution: Though the C-H bond is practically considered as non-polar, there is partial positive charge on hydrogen atom and partial negative charge on carbon atom. Therefore each hydrogen atom acts as electron donating group. This cumulative donation turns the alkyl moiety into an electron donating group.

6.5 Applications of Inductive Effect Some of the prominent applications of inductive effect are discussed below. 1.

Stability of carbonium ions: The stability of carbonium ions increases with increase in number of alkyl groups due to their +I effect. The alkyl groups release electrons to carbon, bearing positive charge and thus stabilize the ion. The order of stability of carbonium ions is: R

R R

most stable

C+ >

R

C+

> R

H R a secondary a tertiary carbocation carbocation

2.

H

H

C+ > H

C+

least stable

H H a primary methyl cartion carbocation

Stability of free radicals: In the same way the stability of free radicals increases with increase in the number of alkyl groups. Thus the stability of different free radicals is: H

CH3


3°

O-32 4.

Acidic strength of carboxylic acids and phenols: The electron withdrawing groups (-I) decrease the negative charge on the carboxylate ion and thus by stabilizing it. Hence the acidic strength increases when -I groups are present. However the +I groups decrease the acidic strength. For example, (i)

The acidic strength increases with increase in the number of electron withdrawing Fluorine atoms as shown below. CH3COOH < CH2FCOOH < CHF2COOH < CF3COOH

5.

(ii)

Formic acid is stronger acid than acetic acid since the -CH3 group destabilizes the carboxylate ion. Generally in gaseous medium, the order of basic strength is,

(iii)

On the same lines, the acidic strength of phenols increases when-I groups are present on the ring. e.g. The p-nitrophenol is stronger acid than phenol since the -NO2 group is a -I group and withdraws electron density. Whereas the para-cresol is weaker acid than phenol since the -CH3 group shows positive (+I) inductive effect.

Basic strength of amines: The electron donating groups like alkyl groups increase the basic strength of amines whereas the electron withdrawing groups like aryl groups decrease the basic nature. Therefore alkyl amines are stronger Lewis bases than ammonia, whereas aryl amines are weaker than ammonia. Thus the order of basic strength of alkyl and aryl amines with respect to ammonia is :CH3NH2 > NH3 > C6H5NH2 . Generally in gaseous medium, the order of basic strength for primary, secondary and tertiary amines with respect to ammonia is, CH3 CH3

N CH3

CH3 >

NH > CH3

NH2 > NH3

CH3

However, the same order changes in aq. medium depending upon the fact that ammonia is ethylated or methylated.So, the order of basic strength are given below, (CH3)2NH > CH3NH2 > (CH3)3N 3° 1° 2° (C2H5)2NH > (C2H5)3N > C2H5NH2 1° 3° 2° Basic character decreases

6.

Reactivity of carbonyl compounds: The +I groups increase the electron density at carbonyl carbon. Hence their reactivity towards nucleophiles decreases. Thus formaldehyde is more reactive than acetaldehyde and acetone towards nucleophilic addition reactions.

Example 4: The inductive effect can be best described as: (i)

The conjugation of σ-bonding orbital with the adjacent π-orbital.

(ii) The ability of atom or group to cause bond polarization. (iii) The transfer of lone pair of electrons from more electronegative atom to lesser electronegative atom in a molecule.

O-33 (iv) All of the above. Solution. (iii) Example 5: Which of the following statement is incorrect about the inductive effect ? (i)

It is a permanent effect.

(ii) It decreases with increases of distance. (iii) It involves delocalization of π-electrons. (iv) It involves displacement of σ-electrons. Solution. (iii) Example 6: Which of the following shows positive (+I) inductive effect? (i)

- NO2

(ii) -OCH3 (iii) -COOH (iv) -CH3 Solution. (iv) Example 7: Which of the following group shows negative (-I) inductive effect ? (i)

-CH2CH3

(ii) -C(CH3)3 (iii) -C6H5 (iv) -CH3 Solution. (iii)

7. Electromeric Effect Electromeric effect is a molecular polarizability effect taking place by an intramolecular electron displacement. It is characterized by the substitution of one electron pair for another within the same atomic octet of electrons. So, the simplified version is that electromeric effect is the movement of electrons from one atom to another as a reagent attacks a π–bond. This effect is shown by those compounds containing multiple bonds. When a double bond or triple bond is exposed to an attack by a reagent, a pair of bonding electrons involved in the π bond is transferred completely from one atom to another. This is referred as a temporary electron displacement effect because this effect will remain as long as the attacking reagent is present. As soon as the reagent is removed, the polarized molecule will come back to the original state as shown below, In the presence of reagent + A —B– A=B In absence of reagent

O-34 Since A has lost its shared pair of electrons, it acquires a positive charge while B has gained the electron pair, thus it acquires a negative charge. The most common example illustrating the electromeric effect is the reaction of an alkene with Br2 in CCl4. R C=C 1

2

Br2 in CCl4

R

+ –

C—C 1

–

Br

–Br–

2

Br

R

–

Br

C—C +

Br

+

Br

R

C—C Br

cyclic bromonium ion vicinal dibromide

In this reaction, when the attacking reagent (bromine) approaches alkene, the temporary polarization develops on the alkene with C2 atom gaining a negative charge and Cl atom acquiring positive charge as it can be compensated by the +I effect of R group. The alkenes being electron rich compounds, due to the presence of π electron cloud, are attacked by the electrophile (Br+) to give a cyclic bromonium ion. Here, the formation of cyclic bromonium ion as intermediate is possible because bromine is of considerably large size having lone pairs to be bonded to both the carbons simultaneously. The cyclic bromonium ion is then attacked by Br − from the top (as lower side is already blocked) whereby the three membered ring is cleaved by trans opening giving vicinal dibromide as the product.

7.1 Types of Electromeric Effect Like inductive effect, electromeric effect is also of two types - +E and -E effect, as discussed below in details. +E Effect: If the attacking species is an electrophile, the π electrons are transferred towards the positively charged atom. This is the +E effect.

+ C = C +H

+ C – C H +E effect

An example is the protonation of ethene. When the H+ comes near the double bond, the bond is polarized towards the proton.

H2C=CH2

+

+

H

δ+ δ– H2C–CH2 +

+

H

+

H2C–CH3

-E Effect: If the attacking reagent is a nucleophile, the electrons are transferred away from the attacking reagent and into the π system. This is the -E effect and it is being shown below,

O-35

C

Z

δ–

δ+ O

+

C

Z

– O

Attacking species

An example is attack on a carbonyl carbon by cyanide ion which can be represented as,

7.2 Resonance Effect or Mesomerism The mesomeric effect or resonance effect in chemistry is a property of compounds having a characteristic set of substituents or functional groups present. The effect is used in a qualitative way and describes the electron withdrawing or releasing properties of substituents based on relevant resonance structures and is symbolized by the letter M or R. As such, it is another type of electron displacement effect shown by various compounds. Definitionwise, it refers to the electron withdrawing or releasing effect attributed to a substituent through delocalization of p or π electrons. It can be visualized by drawing various canonical forms. It is important to know about the conditions necessary for a compound to show resonance effect. The two essential conditions are, 1.

There must be conjugation in the molecule. The conjugation is defined as the presence of alternate double and single bonds in the compound like. –C=C–C=C–

2.

The part of the molecule having conjugation must be essentially planar or nearly planar.

The first condition of conjugation is not only confined to the presence of alternate double and single bonds as mentioned above. There are some other ways available to achieve conjugation. These are +

(i) – C = C – C –

(ii) – C = C – C –

(iii) – C = C – C –

(iv) – C = C – O –

O (v) – C = C – N O

So, any molecule satisfying both the conditions will show resonance and hence resonance effect. For example, phenol is considered to show mesomeric effect. The structure of phenol is

O-36

O–H

By looking at the structure, it must be clear that the compound possess conjugation of the type − C = C − C = C − as well as the category (iv) because the lone pairs on oxygen are in conjugation |

|

|

|

with unsaturated (sp2 hybridised) carbon of the ring. Since, oxygen atom is sp3 hybridised in phenol, the lone pairs on oxygen are nearly planar with respect to the p z orbital of carbon linked to oxygen. Thus, both the conditions are fulfilled by phenol. So the compound shows resonance and hence mesomeric effect is operational. Its various resonating structures are represented as O–H

⊕O

⊕O

–H

–H

⊕O

–H

O–H

This has to be borne in mind that resonance and henceforth the subsequent the electron displacement always results in different distribution of electron density. It is a permanent effect because no attacking reagent is required to carry out the displacement. Some more examples of electron displacement are shown below. In, 1, 3-Butadiene the different resonance structures and the electron displacement therein is shown.

CH2

CH—CH

CH2

⊕

CH2 —CH

CH —CH2

CH2—CH

⊕

CH—CH2

Similarly, electron displacement can be shown in vinyl chloride, CH2

CH—Cl

CH2 —CH

⊕

Cl

7.3 Types of Resonance Effect Resonance (mesomeric) effect is of two types 1.

If the atom or group of atom is electron releasing through resonance, it is called +R or +M effect. For example,

Other groups that show +M effect are –NHR, –NR2, –OH, –OR, –NHCOR, –Cl, –Br, –I etc. 2.

If the atom or group of atom is withdrawing electrons through resonance, it is called –R or –M effect. For example,

O-37

Other groups showing M effect are –CN, –CHO, –COR, –CO2H, –CO2R, –CONH2, –SO3H, –COCl etc. In general, if any atom (of the group) attached to the carbon of benzene ring bears atleast one lone pair, then the group shows +M effect while if the atom (of the group) linked to the benzene carbon bears either a partial or full positive charge, then the group exhibits –M effect.

7.4 Illustrations & Applications of Resonance Effect or Mesomeric Effect 1.

The negative resonance effect (–R or –M) of carbonyl group is shown below. It withdraws electrons by delocalization of π electrons and reduces the electron density particularly at ortho and para positions. H3C

O C

– O

H3C

H3C

C

–

C + etc.

2.

Oδ

δ+

δ+

δ+

The nitro group, -NO2, in nitrobenzene shows -M effect due to delocalization of conjugated π electrons as shown below. Note that the electron density on benzene ring is decreased especially on ortho and para positions.

This is the reason for why nitro group deactivates the benzene ring towards electrophilic substitution reaction. 3.

The -NH2 group in aniline also exhibits +R effect. It releases electrons towards benzene ring through delocalization. As a result, the electron density on benzene ring increases particularly at ortho and para positions. Thus aniline activates the ring towards electrophilic substitution.

O-38

It is also worth mentioning that the electron density on nitrogen in aniline decreases due to delocalization which is the reason for its less basic strength when compared to ammonia and alkyl amines. 4.

The positive resonance effect (+R or +M) of methoxy group is shown in the compound below

7.5 Inductive Effect vs. Resonance Effect In most of the cases, resonance effect is stronger and outweighs inductive effect. For example, the -OH and -NH2 groups withdraw electrons by inductive effect (-I). However they also release electrons by delocalization of lone pairs (+R effect). Since the resonance effect is stronger than inductive effect the net result is electron releasing to rest of the molecule. This is clearly observed in phenol and aniline, which are more reacting than benzene towards electrophilic substitution reactions. However,the inductive effect is stronger than the resonance effect in case of halogen atoms attached to benzene ring. Halogens are electronegative and hence exhibit -I effect. However, at the same time they also release electrons due to delocalization (+R effect) of lone pair. This is evident in case of reactivity of halobenzenes, which are less reactive than benzene towards electrophilic substitution due to -I effect of halogens. But it interesting to note that the substitution is directed at ortho and para positions rather than meta position. It can be ascribed to the fact that the electron density is increased at ortho and para positions due to +R effect of halogens as shown below, Cl

Cl

Cl+

Cl+ –

II

III

–

–

I

Cl+

IV

V

7.6 Hyperconjugation (Baker Nathan effect) As it has been observed that the general inductive effect of alkyl groups is R3C > R2CH > RCH2 > CH3. This inductive order has been used satisfactorily to explain many phenomenon like the dipole moments in the gas phase of PhCH3, PhC2H5, PhCH(CH3)2 and PhC(CH3)3 are 0.37, 0.58, 0.65 and 0.70 D respectively. However, in some reactions, this order is reversed. For example, the rate of reaction of p-substituted benzyl bromide with pyridine in acetone (follows SN2 pathway) is just opposite than that expected from the electron

O-39 release by the inductive effect. That is, the methyl substituted compound reacted fastest and the t-butyl substituted compound reacted slowest. The rate order for the R groups was Me > Et > iso Pr > t-Bu.

R

CH2

N

R

CH2

Br + N

Thus, the order of electron release is exactly the reverse of inductive effect of alkyl groups. Therefore, alkyl groups must also possess some other mechanism of electron release in which the order is Me > Et > isoPr > t- Bu. This mechanism of electron release is termed as hyperconjugation effect or Baker Nathan effect after its discoverer. Thus, when the C—H bond is attached to an unsaturated carbon atom, the σ electrons of the C—H bond becomes less localized by entering into partial conjugation with the attached unsaturated system i.e., σ, π- conjugation.

This type of conjugation between the σ–electrons of single bond and π–electrons of multiple bond is known as hyperconjugation. The molecular picture of the σ, π- hyperconjugation is shown below, H

π bond C

σ bond

C C H

H

Fig. 32: σ – π hyperconjugation

Definitionwise, hyperconjugation is the interaction of the electrons in a σ bond (usually C–H) with an adjacent empty (or partially filled) non-bonding p-orbital, antibonding ( ∏ ) orbital, or filled ( ∏ ) orbital, to give an extended molecular orbital that increases the stability of the system. It is a permanent effect which stabilizes the molecule. When CH 3 group is attached to an unsaturated atom or one with an unshared orbital, the canonical forms are drawn as

O-40 H⊕

H H – C – CH = CR2

H – C = CH – CR2

H ⊕ C = CH – CR2

H – C = CH – CR2 H⊕ IV

H III

H II

H I

H

H

In such canonical forms there is no bond at all between the carbon and hydrogen but the hydrogen is not free as H+ because it will then contradict the condition of resonance, in which the relative position of the atoms in the molecule should not change. Thus, hyperconjugable hydrogens are not acidic in nature. It is due to this fact, hyperconjugation is also referred as no bond resonance. The effect of II, III and IV on the actual molecule is that the electrons in the C–H bond are closer to the carbon than they would be if II, III and IV did not contribute at all. For the other alkyl groups sequence -C2H5, -CH(CH3)2, -C(CH3)3, the hyperconjugation is further diminished because the number of C–H bonds decreases. Hence with respect to hyperconjugation effect, methyl is the strongest electron donor and t-butyl the weakest. Hyperconjugation is also operational in case of carbcations and free radicals. In these cases, electrons in bonds that are β to the positively charged carbon can stabilize a carbocation or even a free radical by hyperconjugation. The molecular orbital view is shown below, H Hyperconjugation sp3–(–C–H) σ bond

Empty 2p orbital of C H

C

+

C

H H H

Fig. 33: Hyperconjugation in case of carbocation stability

7.7 Applications of Hyperconjugation The significance of this effect is to explain certain facts like 1. o,p-directing influence of CH3 group: In toluene, as CH3 is linked to an sp2 hybridized carbon, it does show hyperconjugation.

O-41 H H—C—H

H H—C

H H

⊕

H—C

H H

⊕

H—C

H

⊕

As it is evident from the hyperconjugable structures that the electron density will be high at ortho and para positions, so electrophile attack is favoured at these two positions in preference to meta attack. Thus, the directive influence of CH3 group is ortho & para. 2. m-directing influence of -CCl3 group: Cl Cl — C — Cl

Cl Cl — C

Cl Cl

Cl — C

⊕

Cl Cl

Cl — C

Cl ⊕

⊕

From these structures it is evident that the electron density being greater at meta position than at ortho and para, the electrophilic attack will be favoured at meta position. Thus, the directive influence of -CCl3 group is meta. 3. Relative stability of alkenes: The stability of alkenes can be decided by looking at the structure of alkenes. The alkene which has more hyperconjugable hydrogens will be more stable than the other with lesser hyperconjugable hydrogens. For example,

In 1-butene, the hyperconjugable structures possible are only 2 while with 2 butene, it will be 6. Thus, 2-butene is more stable than 1-butene. In a chemical reaction leading to the formation of both these alkenes, the 2-butene will be the major product while 1-butene will be the minor product. This is also decided by Saytzeff's rule which is a result of the hyperconjugation only. Saytzeff rule states that the more substituted alkene will be the major product in a reaction which can give two or more alkenes. Thus, 2 butene being disubstituted would be the major product while 1 butene would be the minor product as it is monosubstituted.

7.8 Application of Inductive and Resonance Effect The inductive and resonance effect can be effectively employed to explain the stability of carbocations, free radicals and carbanions and the acidic and basic strengths of various organic compounds. Let us take them separately.

O-42

8. Strength of Acids and Bases There are various theories regarding acids & bases. 1.

Arrhenius theory: According to Arrhenius theory, substances producing H + ions in solution are acids and those producing OH − ions in solution are bases. Therefore, substances like H2O, HCl, H2SO4, CH3COOH etc. are acids and the ones like NH4OH, NaOH, KOH, H2O etc. are bases.

2.

Bronsted Lowry theory: In 1923, Bronsted and Lowry independently defined acids as proton donors, and bases as proton acceptors. For aqueous solutions the definition does not vary much for acids from the Arrhenius theory but it widens the scope of bases. In this, the bases need not contain OH − ions and simply have to accept protons. So ions like Cl − , CH 3 COO– , Br − etc. which do not contain OH − ions can be considered as bases under this definition.

3.

Levelling solvents: Whenever an acid is dissolved in water, it acts as an acid only if the solvent acts as a base. That is, if we dissolve acids like HCl, HNO3 etc in water, their acidic strength is almost the same since water acts as a base for both these acids. In fact, it is known that all strong acids show equal acidic strength when dissolved in water. This is because, water acts as a base to all these acids and thus forces them to donate almost the same amount of protons irrespective of their chemical nature. Since water levels the acidic strength of strong acids, it is referred to as a levelling solvent. In order to measure the strength of strong acids, they are dissolved in glacial acetic acid and the amount of protons is measured by conductometry. It is found that the strength of acids varies as: HClO4 > HBr > H2SO4 > HCl > HNO3

4.

Lewis theory: Lewis developed a definition of acids and bases that did not depend on the presence of protons nor involve reactions with the solvent. He defined acids as materials which accept electron pairs, and bases as substances which donate electron pairs. Thus a proton is Lewis acid and ammonia is Lewis base since, the lone pair of electrons on the nitrogen atom can be donated to a proton: H+

+

:NH3 → [H←:NH3]+

Ag +

+

2 NH3 → [H3N: → Ag ← :NH3]+

Lewis acid →

Lewis base →

Acid-base adduct

Conditions to be a Lewis Acid: (i)

Compounds whose central atoms have an incomplete octet e.g. BF3, AlCl3, GaCl3 etc.

(ii)

Compounds in which the central atom have empty d-orbitals and may acquire more than an octet of valence electrons.e.g. SiF4 + 2F − → SiF62– Other examples are : PF3, SF4, SeF4, TeCl4.

(iii) All simple cations : Na+, Ag+, Cu2+, Al3+, Fe3+ , Mg2+, Ca2+ etc.

Conditions to be a Lewis Base: −

−

(a)

All simple negative ions e.g. Cl , F , O −2 , SO4–2 etc.

(b)

Molecules with unshared pair of electrons: H2O, NH3 etc.

O-43 (c)

Multiple bonded compounds which form co-ordination compounds with transition-metals,e.g., CO, NO, ethylene, acetylene etc. 5. Amphiprotic species: Water can either gain or lose a proton and thus can behave as an acid as well as a base. Such species are called amphiprotic species. Similarly many other molecules and ions can either gain or lose a proton. Examples: Acid

Base

Acid

HS–

+

OH–

Base

H2O

+

S2

HBr

+

HS–

H2S

+

Br–

HSO4– +

OH–

H2O

+

SO42–

HClO4 +

HSO4–

H2SO4 +

ClO4–

–

The oxides and hydroxides of metals near the boundary between metals and non-metals are amphiprotic and can react either as acids or as bases.

8.1 Hard and Soft Acids and Bases (HSAB Principle) Hardness is measured as the property of retaining valence electrons very strongly. Lewis acids and bases can be classified as hard and soft acids and bases.Hard Acid is that in which the atom, which is accepting electrons, is smaller in size and has a high positive charge. Soft acid is that in which the atom, which is accepting electrons, is bigger in size, has low positive charge and the electrons in it can be easily polarised.Hard Base is that in which the electron donating atom is small and has high electronegativity e.g. F-, NH3, H2O, OH − etc. Soft Base is that in which the electron donating atom is bigger and has low electronegativity e.g. I − , PH3, (CH3)3P etc. A hard acid prefer to bind to hard bases to form ionic bond and the soft acids prefer to bind to soft bases to form mainly covalent bonds.

8.2 Factors Responsible for Influencing the Acidic and Basic Strength The factors responsible for influencing the acidity of an organic compound, H–X are: 1. The strength of the H–X bond 2. The electronegativity of X 3. Factors stabilizing X– as compared to HX. 4. The nature of the solvent. The most important factors among all is (3). These factors are used in tandem to predict the order of acidic strength for organic acids. (i)

Hydrocarbons: Among hydrocarbons, acidic strength increases with increase in the % 's' character. This is because, higher the 's' character, higher is the effective electro negativity of carbon atom. As a result, electrons are closer to the nucleus of carbon and moved farther away from H. So H can be easily removed as H+.Therefore, the order of acidic strength is CH3 - CH3 < CH2 = CH2 < CH ≡ CH.

(ii) Alcohols: Alcohols are more acidic than CH4 because oxygen is more electronegative than carbon and consequently O–H bond easily give H+ than C–H bond. Therefore, the order of acidic strength is H3C – H < CH3, O, H.If one has to predict the order of acidic strength of ethanol, isopropanol and tertiary butanol, one can proceed as

O-44

Me

CH2

Me

δ– O

CH

H

2δ– O

Me C

H

3δ– O

H

Me

Me

Me

As alkyl groups (methyl group) have a +I effect, the electron density at the oxygen atom will increase. More the number of alkyl groups, more will be the intensity of negative charge on oxygen. The greater the negative charge on the oxygen atom, the closer is the electron pair in the O–H bond driven to the hydrogen atom and consequently separation of a proton becomes increasingly difficult. Thus, the acidic strength of given alcohols will be in the order Ethanol > isopropanol > t–butanol Example 8: Out of ethanol and trifluoro ethanol, which is more acidic? Solution : As the methyl group is weakly +I while the –CF3 group is strongly –I. Thus, due to –I effect of –CF3 group, the electron pair in O–H bond will be drawn in towards the oxygen atom, thereby facilitating the release of the hydrogen as a proton.

(iii) Phenols and substituted phenols: Phenols are stronger acid than alcohols. This is attributed to the fact that in phenoxide (anion obtained by the loss of proton from phenol), there is possibility of the delocalization of its negative charge through interaction with the π orbitals of the aromatic nucleus while alkoxide cannot be stabilized by such effect. Thus pentoxide is a weaker base than alkoxide. Consequently, the phenol would be stronger acid than alcohol. PhO– + H+ phenoxide

PhOH

O

O

O

O

O

phenoxide ion

Delocalization also occurs in the undissociated phenol but as it involves charge separation, it is less effective than in the phenoxide. Thus, phenoxide shows some reluctance to take up proton and the equilibrium lies to the right, indicating greater stabilization of phenoxide and more acidity of phenols with respect to alcohols. Phenols are less acidic than carboxylic acids due to the fact that delocalozation of the negative charge in the carboxylate anion involves structures of identical energy content (as the negative charge resides on more electronegative atom, oxygen) while in pentoxide, the negative charge also resides on the less electronegative atom (carbon) making these structures of high energy content with respect to those structures

O-45 in which the negative charge is on oxygen. Thus, the relative stabilization of the carboxylate with respect to the undissociated carboxylic acid is more effective than the relative stabilization of phenoxide w.r.t. to undissociated phenol. Thus, carboxylate is a weaker base than phenoxide, making the corresponding carboxylic acid, a stronger acid than phenol. RCO2–+H+ carboxylate ion

R–CO2H

The presence of electron– with drawing groups in phenol increases its acidity while the presence of electron releasing groups decreases the acidity of phenol. Now, let us compare the acidic strength of o– , m– and p– nitrophenols. The corresponding phenoxide ions obtained from the three nitrophenols would be stabilized by delocalization as O–

O NO2

O

O NO2

O

O N

N

O

O

o–nitrophenoxide O–

O

O–

N

O N

O

O–

O

O

NO2

O–

O

NO2

NO2

NO2

O

NO2

m–nitrophenoxide O–

O

NO2 NO2 p–nitrophenoxide

O

O

N O

N O

O

O–

O

N O

O

N O

O

O

O-46 From the structures of phenoxide ions, it is evident that o– and p– nitrophenoxides are more extensively stabilized with respect to m–nitro phenoxide (as additional contributing structure is obtained by the dispersal of negative charge over oxygen atom of NO2 group also in o– and p–nitrophenoxide but not in m–nitrophenoxide). Thus corresponding o– and p–nitrophenols are much stronger acids than m–nitrophenol. Out of o– and p–nitrophenols, p–nitrophenol is slightly stronger than o–isomer as o–isomer is bit stabilized by intramolecular hydrogen bonding, thus having a decreased tendency to release proton. O

O N

H O

The order of acidic strength of the nitrophenols is p–nitrophenol > o–nitrophenol > m–nitrophenol (iv) Aliphatic carboxylic acids: The aliphatic carboxylic acids are much stronger acids than alcohols and even phenols. This is due to the fact that the carboxylate ion which is obtained by the loss of proton from carboxylic acid is relatively more stabilized by delocalization than the penoxide and alkoxide ions with respect to the their undissociated molecules. R – C – OH

R – C– O– +H+

O

O

The presence of electron with drawing substituents in simple aliphatic acids increases their acidity while the electron releasing substituents have reverse effect. For instance, let us compare the acidic strength of fluoroacetic acid and acetic acid. CH3 – C – OH

CH3 – C – O– +H+

O

O

F – CH2 – C – OH O

F

CH2

C – O– +H+ O

The fluoro acetate ion is stabilized more due to strong –I effect of fluorine, with respect to acetate ion. Hence, fluro acetate ion is less basic than acetate ion, thereby making fluoroacetic acid stronger than acetic acid. If there is a doubly bonded carbon atom adjacent to the carboxyl group, the acid strength is increased. This will be evident if we compare the acid strength of propanoic acid and propenoic (acrylic) acid.

O-47

α CH2 = CH – C

CH2 = CH – CO 2H

α CH3 – CH2 – C

CH3 = CH2 – CO2H

O O– O

+H+

O–

+H+

As we know that unsaturated α −carbon atom in propionate ion is sp2 hybridized, which means that electrons are drawn closer to the carbon nucleus than in a saturated, sp3 hybridized α −carbon atom, due to greater s contribution in the sp2 hybrid orbital. The result is that sp2 hybridized carbon atoms are less electron donating than saturated sp3 hybridized ones. Thus, propionate ion will be comparatively less stabilized than prepe noate ion, thereby making propanoic acid stronger than propenoic acid (but propenoic will still be weaker than mechanic acid). If there is a triple bonded carbon atom adjacent to carboxyl group, the acid strength is more with respect to the presence of a doubly bonded carbon atom adjacent to –CO2H group. This is attributed to the fact that sp hybridized carbon atoms are less electron– releasing than sp2 hybridized carbon atoms. We can explain very strong acidic character of maleic acid than fumaric acid. If one proton is removed from each of the acids, the corresponding ions are formed. The maleate ion can be stabilized by chelation because hydrogen and oxygen are very near to each other, on the other hand fumarate ion can not be stabilized by chelation because hydrogen and oxygen are on the opposite sides, hence the formation of fumarate ion does not take place. This explains why maleic acid is a stronger acid than its isomer fumaric acid. O COOH

H C

C

C

C

H

H

(Meleic acid) COOH

H

(Maleate ion) (Chelation possible) COO–

H

C

C C

C COOH

O

C O

COOH

H

O–

C

H

H

(Fumaric acid)

HOOC

H

Fumarate ion (Chelation not possible)

Example 9: Compare the acidic strength of the following (i)

CH3COOH, CH3CH2COOH

(ii) CH3CH2COOH, CH2 = CHCOOH, HC ≡ C–COOH. (iii) CH3 COOH, I–CH2 COOH, Br–CH2–COOH, F–CH2–COOH. (iv) Cl2CHCOOH, CCl3COOH (v)

CH3(CH2)2CO2H,ClCH2(CH2)2CO2H,CH3CH(Cl)CH2CO2H,CH3CH2CH(Cl)CO2H

(vi) CH3COOH, C6H5COOH

O-48 Solution : (i) In simple aliphatic acids, more the number of CH3 added, less is the acidic strength. This is due to the +I (inductive) effect of CH3. This effect of adding CH3 decreases with increase in distance between CH3 added to the substituent, which releases H + . Therefore the order of acidic strength is CH3–COOH > CH3–CH2–COOH (ii)

Increase in 's' character increases acidic strength of carboxylic acids. Therefore the order of acidic strength is CH3CH2COOH < CH2 = CHCOOH < HC ≡ C–COOH.

(iii) As the –I effect of the group increases, acidic strength increases Therefore the order of acidic strength is CH3 COOH < I–CH2 COOH < Br–CH2–COOH < F–CH2–COOH. (iv) Increase in number of electron withdrawing substituents into simple aliphatic acids increases acidic strength. Therefore the order of acidic strength is Cl2CHCOOH < CCl3COOH (v)

If Cl is placed nearer to COOH group, more is acidic strength. Therefore the order of acidic strength is CH3(CH2)2CO2H < ClCH2(CH2)2CO2H < CH3CH(Cl)CH2CO2H < CH3CH2CH(Cl)CO2H

(vi) Benzoic acid is more acidic than acetic acid because benzoate anion shows/has more resonance. Therefore the order of acidic strength is CH3COOH < C6H5COOH. (v)

Aromatic carboxylic acids: Benzoic acid is a stronger acid than its saturated analogue cyclohexane carboxylic acid. This is due to the fact that phenyl group is less electron–donating (like a double bond) as compared to saturated carbon atom towards the carboxyl group, as the carbon atom to which –CO2H group is attached is sp2 hybridized in benzoic acid. O

O

H–C

H–C O–H

O–H ⊕

⊕

O–

O C

C O–H

O– ⊕

OH

C OH

As it is evident from the above structures that the lone pair on O atom of the OH group in formic acid undergoes delocalization with the CO group, thereby generating a positive charge center on O atom which destabilize the O–H bond by withdrawing electron density from it. This consequently result in easy release of proton. While in benzoic acid, due to +R effect of phenyl group, the lone pair in O atom of the O–H group is prevented from entering into resonance with CO group to a large extent. This result in a smaller (partial) positive charge on the O atom of the O–H group in benzoic acid with respect to formic acid (the O atom of O–H group has full fledge positive charge), so the release of proton becomes difficult in benzoic acid than formic acid.

O-49 Now let us consider substituted benzoic acids. At this point, we will consider the effect of substituents from m – and p – position only. First let us take methyl substituted benzoic acids. As the methyl group has +I effect, so the net result will be to increase the +R effect in the m– or p– tolyl group. Since, the inductive effect decreases with distance, so the +I effect of CH3 from para position would be less than the +I effect of CH3 group from meta position. Thus, the electron density in the O–H bond of m–methyl benzoic acid would be greater than that in p–methyl benzoic acid. Accordingly, the p–isomer should be more acidic than the m–isomer. But in real practice, the order is reverse. Actually, we have ignored the hyperconjugation effect of CH3 group. Due to hyper conjugation effect of CH3 in p isomer, the electron density in the O–H bond would be increased greatly with respect to that in m isomer. O–H

O C

C

O–H

O

C

C

H

H+

H+

C–H

C–H

C–H

H

H

H

O–H

O

O–H

O

O–H

O C

etc.

O–H

O C

etc.

H–C–H

H–C–H

H – C – H+

H

H

H

Thus, the release of proton facilitated easily in m–isomer than in p–isomer. Secondly, let us take intro substituted benzoic acids. As the –NO2 group exerts –I and –R effect (both reinforming each other), the electron–density from the O–H bond would be withdrawn, thus destabilizing the O–H bond and making m– and p–nitro benzoic acids stronger than benzoic acid.

As it is clear that –NO2 group exerts its electron withdrawing resonance effect from the p–position and not from m–position, thus withdrawal of electron density from the O–H bond is greater in p–intro benzoic acid than m–intro benzoic acid. Thus, p–isomer is stronger acid than m intro benzoic acid.

O-50 Thirdly, we compare the acid strength of m– and p–methoxy benzoic acid. OMe group exerts –I and +R effects. The electron–releasing resonance effect (+R) is operative in p– methoxy benzoic acid and not in m–methoxy benzoic acid while –I effect is operative in both the isomers. Thus, the electron density is greatly increased in the O–H bond in p– isomer than in m– isomer, thereby making m–isomer stronger acid than p–isomer. O–H

O C

OMe

O–H

O

O–H

O

C

O–H

O

C

C

etc.

OMe

⊕

OMe

⊕

OMe

There is a special effect shown by o–substituents called ortho–effect. It has been observed that irrespective of the nature of the group present at ortho position in benzoic acid, nearly all o–substituted benzoic acids are stronger than benzoic acid. As we have seen benzoic acid is a resonance hybrid and so the carboxyl group is coplanar with the ring. Introduction of any o–substituent tends to prevent this coplanarity. Thus, resonance is diminished (or prevented) and so the oxygen atom of the O–H group has a greater positive charge, resulting in increased acid strength. Thus, it follows that greater the steric inhibition of resonance, the stronger is the acid. o–hydroxy benzoic acid (salicylic acid) is far stronger than the corresponding m– and p– isomers. Steric inhibition of resonance can not explain this very large increase, since the corresponding methoxy benzoic acids all have similar strengths. The explanation here is the presence of intra molecular hydrogen bonding. Hydrogen of the –OH group can form a hydrogen bond with the carboxyl group whereas the Me of OMe cannot. Thus the carboxylate ion of o–hydroxy benzoic acid is stabilized by intramolecular hydrogen bonding while that of m– and p–isomer is not, making o–hydroxy benzoic acid stronger than the corresponding m–and p–isomers.

O-51

8.3 Basic Strength of Organic Bases Strength of bases is related to the ease of accepting a proton which inturn depends on the availability of electron pair on the nitrogen atom (or some other basic atom). More is the availability of electron pair, more easily the proton will be accepted and more will be the basic strength.

8.4 Aliphatic Bases As increasing strength in nitrogenous bases is related to the readiness with which they take up protons and which inturn depends on the availability of the unshared electron pair on nitrogen, we expect an increase in basic strength on going from NH3→RNH2 →R2NH→R3N, due to the increasing inductive effect of successive alkyl groups making the nitrogen atom more electron rich. But in actual practice the order of basic strength of amines is NH3 pKa's

9.25 NH3

and

9.25

< Me3N < MeNH2 < Me2NH 9.80

10.64

< EtNH2 < Et3N 10.67

10.77 < Et2NH

10.88

10.93

It can be seen that the introduction of an alkyl group into ammonia increases the basic strength markedly as expected. The introduction of a second alkyl group further increases the basic strength, but the net effect of introducing the second alkyl group is very much less marked than with the first. However, the introduction of a third alkyl group to yield a tertiary amine, actually decreases the basic strength. This is due to the fact that the basic strength of an amine in water is determined not only by electron– availability on the nitrogen atom, but also by the extent to which the cation, (formed by the uptake of a proton), can undergo solvation and so become stabilized. The more hydrogen atoms attached to nitrogen in the cation, the greater the possibilities of extensive solvation via hydrogen bonding between protonated amines and water. H2O......H ⊕

H2O......H

R – N – H......OH2 > H2O......H

⊕

R–N –R> H2O......H

H2O......H ⊕

R–N –R H2O......R

Decreasing stabilization by solving

Thus on going along the series, NH3 →RNH2→R2NH→R3N, the inductive effect will tend to increase the basicity but progressively less stabilization of the cation by hydration will tend to decrease the basicity. The net effect of introducing successive alkyl groups thus becomes progressively smaller, and an actual changeover takes place on going from a secondary to a tertiary amine. If this is assumed to be the right explanation, no such changeover should be observed if measurements of basicity are made in a solvent in which hydrogen bonding cannot take place. This is indeed found to be true. In chlorobenzene, the order of basicity of the butylamines is BuNH2 < Bu2NH < Bu3N while their basic strength in water follows the order Bu3N < BuNH2 < Bu2NH.

O-52 Tetraalkylammonium salts, e.g. R4N⊕I − on treatment with moist silver oxide (AgOH) to yield basic solutions comparable in strength with the mineral alkalis. This is readily understandable as R4N⊕ –OH formed is completely ionized to give R4N+ and free OH − . The effect of introducing electron withdrawing groups, e.g. Cl, NO2, close to a basic center decreases the basicity, due to their electron withdrawing inductive effect. Thus the amine tris (triflora methyl) amine is found to be virtually non basic due to the presence of three powerfully F3C electron withdrawing CF3 groups. N

F3C F3C

The amides are also found to be only very weakly basic in water [pKa for ethanamide(acetamide) is ≈0.5], because of the –I and –R effect of RCO group which makes the electron pair very slightly available on O O– nitrogen atom. ⊕

R – C = NH2

R – C –– NH2

If two C = O groups are present the resultant imides often become sufficiently acidic to form alkali metal salts. For example, benzene –1,2– dicarboximide is not basic but is acidic in nature because of the presence of two electron-withdrawing CO groups. Oδ–

O

N

H

OH–

δ–

N

δ–

O

O

8.5 Aromatic Bases Aniline is a weaker base than ammonia or cyclohexylamine. It is because of the fact that the electron pair on nitrogen is involved in delocalization, making it less available for donation. NH2 NH3

cycloheyxylamine NH2

ammonia ⊕

⊕

⊕

NH2

NH2

NH2

NH2

O-53 If aniline accepts a proton, the anilinium cation formed cannot be stabilized by delocalization as the electron pair on nitrogen is no longer available. Thus, it is 'energitically unprofitable' for aniline to take up a proton. Thus it functions as a base with utmost reluctance. ⊕

NH3

Diphenylamine is even a weaker base than aniline due to the presence of another phenyl group NH

and triphenylamine (Ph3N) is not basic at all by ordinary standards. Introduction of alkyl group (like Me) on the nitrogen atom of aniline results in small increase in the basic strength. C6H5NH2 < C6H5NHMe < C6H5NMe2 Unlike such introduction in aliphatic amines, this small increase in basic strength is progressive, indicating that cation stabilization through hydrogen–bonded solvation, here has less influence on the overall effect. Let us compare the basic strengths of o–, m– and p–nitroanilines. The electron withdrawing NO2 group exerts its –R effect from o– and p–positions while its –I effect from all p–positions while its –I effect from all positions. O

O

NH2

O=N

⊕

⊕

O

NH2

NH2

–O = N

O=N

so on

NH2

⊕

⊕

⊕

NH2

NH2

NH2

so on

N O

N O

O

N O

O

N O

O–

O

Since, the electron pair on nitrogen in o– and p–nitroanilines is delocalized extensively by –I and –R effect combindly, so they are less basic than m–nitroaniline in which only –I effect operates. Out of o– and p– isomers, ortho is less basic than para because –I effect of NO2 from ortho position is stronger than from para position. Thus the basic strength of nitro anilines follow the order m–nitroaniline > p–nitroaniline > o–nitroaniline

O-54 Now if we take o–, m– and p–methoxy anilines. The OMe group exerts –I and +R effects (only from ortho and para position). As OMe group exerts +R from ortho and para position, so o– and p– methoxy anilines are stronger base than m methoxy aniline, in which OMe exerts only –I effect. NH2

NH2

NH2

⊕

OMe

OMe

NH2

⊕

OMe

⊕

OMe etc.

NH2

OMe NH2

NH2

NH2

NH2 etc.

OMe

⊕ OMe

⊕ OMe

⊕ OMe

–I effect of OMe from ortho position is stronger than from para position, thus decreasing the electron density on nitrogen in o–methoxy aniline than in p–methoxy aniline. Thus, the order of basic strength would be p–methoxyaniline > o–methoxyaniline > m–methoxyaniline Example 10 : Arrange the following compounds in the inceasing order of basicity?

N

N

N

H (I)

(II)

H (III)

N H (IV)

Solution : In II, III and IV the lone pair is present in sp3 orbital but in II, the electron pair on nitrogen is

involved in delocalization making it least basic of all. In I, the unshared electron pair is present in sp2 orbital which makes it less basic than III and IV and IV is less basic than III due to –I effect of oxygen atom. Thus, the increasing order of basic strength would be II < I < IV < III.

O-55

Exercise 4.

Long Answer Type Questions 1.

2.

3.

4.

What do you understand by the term "hybrid orbital"? Give geometry of the following: (i)

sp3

(iii)

sp

(ii) sp2

Cl3C.COOH > Cl2CHCOOH > Cl CH2COOH > CH3COOH 5.

(i)

Ethane

(iii)

Ethyne (Acetylene)

OR What do you mean by hydrogen bonding? Discuss intermolecular and intramolecular hydrogen bonding. [Agra 2008]

(ii) Ethylene [Bund. 2009]

6.

An organic compound contains C, H and O its vapour density is 30.0.30 gm of this compounds on combustion yielded 0.44 gm of CO2 and 0.18 gm of H2O. Find out the molecular formula of the compound. [Kanpur 2009]

7.

An organic compound A contains C = 37.2%, H = 7.7% and Cl = 55.03%. This on treatment with KCN gives B. A on treatment with alcoholic ammonia under pressure forms C, D, E and F. What are A, B, C, D, E and F? Give structure.

8.

Write short note on hyperconjugation.

9.

How resonance reduces the besicity and increases the acidity of the molecule? Explain with suitable example. [Kashi 2010]

Write short notes on the following : (i)

Hydrogen bonding

(ii)

Hyperconjugation

(iii)

Resonance energy

(iv)

Inductive effect

[Hazaribagh 2010]

In terms of hybridization of carbon atom discuss briefly the shapes of methane, ethane and ethyne.

5.

(ii)

Write short notes on any two of the following :

(a)

Inclusion compounds

(b)

Clathrates

(c)

Picrates.

[Kanpur 2010] [Avadh 2008]

[Hazaribagh 2011]

Discuss the rules for deciding the relative stabilities of various canonical forms. 10.

What is hyperconjugation? How is it used to determine stability of carbocations and alkenes? [Kashi 2011; Hazaribagh 2008]

Short Answer Type Questions 1.

Explain hyperconjugation and on its basis stability 11. of alkenes. [Bund. 2010; Meerut 2010,11,12; Kanpur 2011] 12.

2.

Write short note on the following

3.

(i)

Hybridisation

(ii)

δ bonds π Inductive effect

(ii)

Resonance

Define aromaticity.

[Lko. 2011]

Explain that inductive effect leads to the development of partial charges while the electromeric effect causes full positive and negative charges. [Rohd. 2009,10]

14.

Explain inter - and intra- molecular hydrogen bonds. [Rohd. 2011]

15.

Describe bond length and bond energy.

[Meerut 2008]

[Meerut 2008,09,10; Agra 2008; Lko. 2008]

What is a charge transfer complex? Give an example. [Lko. 2011]

13.

Write the short note of the following : (i)

[Meerut 2008]

What do you understand by the terms resonance and resonance energy?

[Bund. 2008; Kanpur 2011]

On the basis of orbital theory, explain the structure of the following:

(i)

Explain the following order of acid strength -

[Rohd. 2012]

O-56 Butanol-1 has higher boiling point than diethyl 16. ether, although both have same molecular formula. Explain why? [D.D.U. 2008]

Explain the following : (i)

Chloroacetic acid is stronger acid than acetic acid

17.

Why -OH group of carboxylic acid is more acidic than the -OH group of an alcohol? [D.D.U. 2010]

(ii)

Dimethyl amine is stronger base than trimethyl amine. [Kanpur 2008]

18.

The boiling point of butanol-1 is higher than that of 17. ethoxyethane, although both have same molecular weight. Why? [D.D.U. 2010]

Write structural formulae of the following :

16.

19. 20.

Why chloroacetic acid is stronger acid than acetic acid? [D.D.U, 2011] Explain the acidic nature of hydrogen in acetylene. [D.D.U. 2011; Hazaribagh 2008,11] 18.

Very Short Answer Type Questions 1. 2.

Explain why formic acid is more acidic than acetic 19. acid. [Bund. 2008; Kashi 2010] 20. Write short notes on inductive effect. [Bund 2008; Meerut 2009, 11, Kanpur 2010,11]

3.

Explain why chloroacetic acid is more acidic than acetic acid. [Bund. 2009]

21.

4.

Explain why methyl amine is more basic than ammonia? [Bund. 2010] 22.

5.

Explain why ethyl alcohol is readily soluble in water? [Bund. 2010] 23.

6.

Explain charge transfer complexes.

7.

Explain H-bond and its types.

8.

Define bond energy.

[Meerut 2009]

9.

What is steric effect ?

[Meerut 2009]

10.

Why acetic acid is a stronger acid than propionic 26. acid? [Meerut 2010]

11.

Why ethyl alcohol (mol wt.46) has higher boiling 27. point than diethyl ether (mol.wt.74). [Meerut 2010]

12.

What is resonance?

[Bund. 2010]

[Bund. 2010; Meerut 2010,11]

24.

14. 15.

Iso–octane

(ii)

Dimethyl acetylene

(iii)

Trimethyl carbinol

(iv)

Ethylmethanonate

[Kanpur 2008]

Write short notes on the following : (i)

Hyperconjugation

(ii)

Resonance

[Kanpur 2008]

Explain mesomeric effect.

[Kanpur 2009]

Which is more basic and why? CH3NH2 and (CH3)2NH

[Kanpur 2010]

Write short notes on the following : (i)

sp3 hybridization

(ii)

sp hybridization

[Kanpur 2010]

Why monochloro acetic acid is stronger than acetic acid? [Avadh 2010] Boiling point of 1-Butanol is greater than diethyl ether although both have same molecular formula.Why? [Avadh 2010] Write short note on hyperconjugation. [Kashi 2008]

25.

Draw all the resonating structures of nitro benzene. [Kashi 2010]

Why o-nitrophenol is less soluble in water than in p-nitrophenol? [Kashi 2010] Write hybridization state of each carbon atom of following molecule, CH3– CH = C = CH2

[Meerut 2011; Kanpur 2009,10,11]

13.

(i)

[Kashi 2011]

Define quinol clathrates.

[Lko. 2008]

29. Give an example of intermolecular and intramolecular hydrogen bonding. [Kanpur 2008] 30.

What are charge transfer complexes ?

[Lko. 2008]

Draw the orbital picture of ethane.

[Lko. 2008]

HF and HI have higher boiling point than HCl, 31. explain why? [Kanpur 2008]

sp hybridized carbon is highly electronegative than

What is sp hybridization?

[Agra 2008]

28.

32.

sp 2 and sp 3 . Why?

[Lko. 2009]

Explain inclusion compounds.

[Lko. 2010]

O-57 33.

Why boiling point of p-nitrophenol is more than 5. ortho-nitrophenol? [Lko. 2010]

34.

Chloroacetic acid is stronger as compared to acetic acid. Give reason. [Rohd. 2009,10]

When the hybridization state of carbon atom changes from sp3 to sp2 and finally to sp, the angle between the hybridized orbitals (a)

Decreases gradually

Delocalisation leads stability. Explain with suitable example. [Rohd. 2013]

(b)

Increases progressively

(c)

Is not affected

36.

Write short notes on the bond dissociation energy and bond energy. [Rohd. 2013]

(d)

None of these

37.

Formic acid is stronger than acetic acid. Explain why? [D.D.U. 2008]

38.

Giving reason, arrange

35.

6.

NH3 , CH3 − NH2 (CH3 )2 NH and (CH3 )2 N into their decreasing order of basicties. [D.D.U. 2008] 39.

Identify the compounds from the following which 7. has sp hybrid orbital. HC = CH, C2H6, C6H6

40.

[Hazaribagh 2008]

Identify the compounds from the following which has sp 3 hybrid orbital :

8.

The Cl–C–Cl angle in 1,1,2,2–tetrachloromethane and tetrachloromethane respectively will be about: (a)

120° and 109.5°

(b)

90° and 109.5°

(c)

109.5° and 90°

(d)

109.5° and 120°

Number of π −electrons in naphthalene is: (a)

10

(b) 5

(c)

3

(d) 6

Number of tertiary carbon atoms in:

CH3 H

CH3 − CH3 ≡ CH,CH2 = CH2 [Hazaribagh 2009]

H – C — C — C — C – CH3

Objective Type Questions Multiple Choice Questions 1.

2.

3.

Molecule in which the distance between two 9. adjacent carbon atoms is longest is: (a)

Ethane

(b) Ethene

(c)

Ethyne

(d) Benzene

Number of π − bonds in the following compound: 10. CH3C ≡ C–CN

CH3 CH3 CH3 CH3 (a)

4

(b) 3

(c)

2

(d) 1

In which of the following organic compounds all the carbon atoms do not lie in one plane: (a)

Toluene

(b) 2-butene

(c)

Neopentane

(d) Propene

The ratio of σ and π bonds in benzene is:

(a)

1

(b) 2

(a)

2

(c)

3

(d) 4

(c)

4

The bond between carbon atoms 1 and 2 in 11. 1

2

4.

sp2 and sp2 (b) sp3 and sp

(c)

sp and sp2 (d) sp and sp3

The compound in which all carbons use only sp3 hybrid orbitals for bond formation:

(b) 3 (d) 1

Consider the following compound:

CH2=CH–CH=CH2 4 3 1 2

N ≡ C − CH = CH2 involves the hybridized carbons: (a)

CH3 H

carbon-carbon bond length between C2 and C3 will be: (a)

1.54Å

(b)

1.3 Å

(a)

HCOOH

(b) (H2N)2CO

(c)

Less than 1.54Å and greater than 1.33Å

(c)

(CH3)3COH

(d) (CH3)3C–CHO

(d)

1.21 Å

O-58 12.

Consider the following three halides: (1)

CH3-CH2-Cl

(2)

CH2=CH-Cl

(3)

C6H5-Cl

15.

NH2

Arrange C-Cl bond length of these compounds in decreasing order:

13.

(a)

1>2>3

(b) 1 > 3 > 2

(c)

3>2>1

(d) 2 > 3 > 1

NH2

NH2

NH2

(a)

CH3

(b)

NH2

NH2 CH3

Consider the following compounds:

NH2

Which one of the following is most basic:

NH2

CH3 (d)

(c)

CH3 16.

NO2 (2)

(1)

CN (3)

Arrange these compounds in decreasing order of their basicity:

14.

(a)

1>2>3>4

(b)

2>3>1>4

(c)

4>1>3>2

(d)

4>1>2>3

Among the following the compound having the most acidic alpha - hydrogen is : (a)

CH3 (4)

(b)

CH3COCH3

(c)

CH3 − C − CH2CHO

(d)

Consider the following compounds:

O

N

N

N

H (1)

(2)

H (3)

H (4)

CH3CHO

|| O

17.

N

18.

CH3–CO–CH2–CO2CH3

Allyl isocyanide has: (a)

9 σ bonds and 4 π-bonds

(b)

9 σ bonds, 3 π-bonds and 2-non-bonding electrons

(c)

8 σ bonds and 5 π-bonds

(d)

8 σ bonds, 3 π-bonds and 4– non-bonding electrons

What is the correct increasing order of bond lengths of the bonds indicated as I, II, III and IV in following compound:

II

Order of basicity of these compounds in decreasing order is:

III

I

(a)

4>1>2>3

(b)

1>3>4>2

(c)

2>3>4>1

(a)

I < II < III < IV

1>3>2>4

(b)

I < III < IV < I

(c)

IV < II < III < I

(d)

IV < I < II < III

(d)

CH3

IV

O-59 19.

In which of the following structures the pair of starred carbon atoms do not lie in the same plane (a)

* CH3

H3C C

C

24.

CH3—H2*C

H

25.

*

(c)

CH

* C

H H

* (d)

26.

* CH2–CH 3

H3C C

C H

H3C *

20.

21.

22.

27.

1– butene

(b)

trans 2– butene

(c)

cis –2– butene

(d)

1, 3– butadiene

28.

Which is the strongest carboxylic acid among the 29. following ? (a)

Cl3CCO2H

(b)

Br3CCO2H

(c)

F3CCO2H

(d)

Cl2CHCO2H

The hybridization of carbon atoms in C–C single 30. bond of vinyl acetylene: (CH2 = CH – C ≡ CH) is : (a) (c)

23.

sp3-sp3

(b) sp2-sp

sp-sp2

(d) sp3-sp

31.

Polarization of electrons in acrolein may be written as: (a)

δ–

(b)

–

δ+

CH2 = CH − CH = O δ

+

δ

CH2 = CH − CH = O

(d)

δ–

32.

δ+

CH2 =CH − CH = O δ+

CH2 =CH − CH = O

The compound which has one isopropyl group is : 2,2,3,3 -tetramethyl pentane

(b)

2,2 - dimethyl pentane

(c)

2,2,3 - trimethyl pentane

(d)

2-methyl pentane

No bond resonance is: (a)

Inductive effect

(b)

Mesomeric effect

(c)

Steric effect

(d)

Hyperconjugation

[Bund. 2008]

Which do you expect has higher boiling point? (a)

HF(B)

(b) HCl

(c)

HBr

(d) HI

[Bund. 2008]

Which of the following is more acidic? (a)

Which one of the following has the smallest heat of hydrogenation per mole ? (a)

δ–

(a)

* CH 3

(b)

(c)

CH3COOH

(b)

ClCH2COOH

(c)

CH2CHCOOH

(d)

Cl3CCOOH

[Bund. 2008]

In benzene, C atom exhibits the hybridization: (a)

sp

(c)

sp

(b) sp2 (d) sp3d

3

[Bund. 2008]

In which of the following compounds carbon-carbon bond is shortest in length? (a)

HC = CH

(b)

H2C = CH2

(c)

H3C - C ≡ CH

(d)

H3C - C = CH2

[Bund. 2008]

Which of the following is more basic (a)

NH3

(c)

(CH3)2NH (d) (CH3)3N

(b) CH3NH2 [Kanpur 2009]

The + I effect is shown by (a)

–CH3

(c)

–F

(b) –OH (d) –C6H5 [Rohd. 2009,10]

In a free radical the carbon atom carrying unpaired electron is (a)

sp hybridized

O-60

33.

34.

35.

(b)

sp2 hybridized

(c)

sp hybridized

(d)

dsp2 hybridized

11.

3

[Rohd. 2010]

True/False

In which of the following compounds delocalised 1. bonding is not possible ? 2. (a) 1, 4 – pentadiene (b)

1,3 – butadiene

(c)

1,3,5 – hexatriene

(d)

Benzene

3–cyclopentadiene is highly acidic in character. Melting point of symmetrical molecule is more than the unsymmetrical molecule.

3.

Chloroacetic acid is more acidic than bromoacetic acid.

4.

Tricyanomethane is as strong acid as a mineral acid.

5.

Delocalisation of lone pair of electrons decreases basicity of the compound.

Indicate the type of hybridization of the asterisked 6. carbon : 7. H2C = C* = CH2 8. (a) sp (b) sp2

n-hexanol does not dissolve in water whereas ethanol dissolves.

[Rohd. 2010]

The +I effect is shown by (a)

–C2H5

(b)

(c)

F

(d) –C6H5

(c)

sp3

–OH

(d) no hybridisation [Rohd. 2012]

36.

In allenes the central carbon atom is ……… hybridized.

The no. of σ and π bonds in 1-Pentene-3-yne are (a)

10σ & 2π

(b)

10σ & 3π

(c)

8σ & 3π

(d)

12σ & 3π [Rohd. 2012]

Fill in the Blank 1.

Aliphatic secondary amine is ………basic than tertiary amine in solution.

2.

Melting point of fumaric acid is ……… than the maleic acid.

3.

Tropylium cation is ……… stable than benzyl carbocation.

4.

Cyclopentadiene ……… is aromatic.

5.

Amides are ……… basic than primary amines.

6.

–I group present in a carboxylic acid ……… acidity of the compound.

7.

Pyridine is……… basic than pyrrole.

8.

The boiling point of alcohols is ……… than those of hydrocarbons of comparable molecular masses.

9.

All the carbon atoms in 1,3-butadiene are……… hybridized.

10.

Dipole moment of vinyl chloride is ……… than the ethyl chloride.

All the carbon atoms in 2-butene lie in one plane. Tropylium cation maintains its aromatic character in all of its resonating structures where as benzyl carbocation loses its aromatic character in some of the canonical forms.

9.

Propene is more acidic than propyne.

10.

The chlorine atom in vinyl chloride is less reactive due to +R effect.

11.

No bond resonance is termed as inductive effect.

O-61

Answers Objective Type Questions Multiple Choice Questions 1.

(a)

2.

(d)

3.

(c)

4.

(c)

5.

(b)

6.

(a)

7.

(a)

8.

(b)

9.

(c)

10.

(c)

11.

(c)

12.

(a)

13.

(c)

14.

(d)

15.

(c)

16.

(c)

17.

(b)

18.

(d)

19

(d)

20.

(d)

21.

(c)

22.

(b)

23.

(d)

24.

(d)

25.

(d)

26.

(a)

27.

(d)

28.

(b)

29.

(a)

30.

(d)

31.

(a)

32.

(c)

33.

(a)

34.

(a)

35.

(a)

36.

(b)

Fill in the Blank 1.

more

2.

more

3.

more

4.

anion

5.

less

6.

decreases

7.

more

8.

higher

9.

sp2

10.

less

11.

sp

1.

True

2.

True

3.

True

4.

True

5.

True

6.

True

7.

True

8.

True

9.

False

10.

True

11.

False

True/False

O-62

Hints and Solutions Long Answer Type Questions 5.

Hints: In drawing the canonical forms and deciding about their relative stabilities, following rules are given for your guidance. (i)

All the canonical forms must be bonafide Lewis structures. For example, none of them may have a carbon with five bonds.

(ii)

All atoms taking part in the resonance must lie in a plane or nearly so. The reason for planarity is to have maximum overlap of the p-orbitals.

(iii) All canonical forms must have the same number of unpaired electrons. Thus, CH2 − CH ≡ CH − CH2 is not a valid canonical form for 1,3 butadiene. (iv)

The energy of the hybrid (actual) molecule is lower than that of any canonical form. Obviously then, delocalization is a stabilizing phenomenon. The difference in energy between the hybrid and the most stable canonical structure is called resonance energy.

(v)

All canonical forms do not contribute equally to the actual molecule. Each form contributes in proportion to its stability, the most stable form contributing the most.

(vi)

Structures with more covalent bonds are generally more stable than those with fewer covalent bonds.

(vii) Structures with formal charges are less stable than uncharged structures. For charged structures, the stability is decreased by an increase in charge separation and the structure with two like charges on adjacent atoms are highly unfavourable. (viii) Structures that carry a negative charge on a more electronegative atom are more stable than those in which the charge is on a less electronegative atom. For example,

Structure II is more stable than I. Similarly positive charges are best occupied on atoms of low electronegativity. (ix)

Those structures in which octet of every atom (except for hydrogen which have duplet) is complete are more stable than the others with non-complete octets. For example, ⊕

⊕

R—C

O

R—C

O

IV III Structure IV is more stable than III.

❍❍❍

UnitO-63 -I

C HAPTER

2

Mechanism of Organic Reactions

1. Fundamental Concepts of Organic Reaction The stoichiometry of a reaction consists of the chemical formulas and relative molecular proportions of starting materials and products. Obviously these have a bearing on the mechanism of the reaction, for the overall reaction course must proceed from starting materials to the products. The stoichiometry of the reaction may be misleading, however, because the participants in the overall reaction may not be involved directly in the rate-limiting step. The discovery of intermediates in the course f a reaction is important because these point to the existence of distinct stages, the mechanism of each of which must then be determined. The identification of intermediates that persist only briefly or that are present in only small amounts depends on the availability of powerful, sensitive, and rapid experimental techniques. For this purpose, a number of specialized instrumental procedures (including ultraviolet, infrared,magnetic resonance, and mass spectrometry) are widely used to supplement the more usual chemical and physical methods. The identification of a new chemical substance formed transiently in a reaction mixture, however, does not unambiguously imply that substance is an intermediate in the reaction. In many cases a newly found material is only a temporary repository of the proportion of the reactants and ultimately produces the products by first reverting to the starting material. The identification of the products of a reaction also helps to define the reaction course, because the mechanism in question clearly must account for their formation. Mechanistic theory has been greatly facilitated by the development of powerful methods of separation and purification based on chromatography (separation of compounds on the basis of their relative degrees of adsorption to certain solid substances, such as starch or silica) and also by modern methods available for the analysis of small quantities of materials. These spectroscopic procedures are often used, as is another instrumental method called polarimetry. An important consideration with regard to the products of the reaction is whether the reaction is under kinetic or thermodynamic control. A reaction is said to be kinetically, rather than thermodynamically, controlled when the products are formed in proportions different from those that would prevail at equilibrium between the same products under the same conditions. Thermodynamic control leads to the equilibrium ratio of the products. Often, though not invariably, reactions under kinetic control give a greater amount of the

O-64 thermodynamically less stable of two possible products; if thermodynamic control is then established, the products shift to their equilibrium proportions, which might give a misleading picture of the reaction course. Hence, inferences concerning the nature of the transition state can be drawn from the nature of the products only with a good deal of circumspection.

1.1Reaction Mechanism and Curved Arrow Notation Chemical reactions are a result of bonding changes. These bonding changes are most easily described by considering the changes in electron sharing between atoms. For example, if one considers the collision between two water molecules resulting into the ionization of water and the formation of hydroxide and hydronium ion. H

H—O

+ O H H

H—O +H3O+ Hydroxide Hydronium ion ion

In this reaction, as observed,the oxygen atom of one water molecule collides with a hydrogen atom of the second water molecule. A lone pair of the oxygen atom becomes the new O-H bond in the hydronium ion. Because a hydrogen can only be fully bond to one other atom at a time, the old O-H bond is lost. The electron pair of the old O-H bond becomes a lone pair, sole property of the oxygen atom of hydroxide ion.This very simplistic step-by-step bookkeeping description of all the bonding and electron changes in a reaction is called the reaction mechanism. Study, understanding and prediction of reaction mechanisms is at the very heart of reactions in organic chemistry. Mastering the fundamentals of reaction mechanisms is a fundamental survival skill for students of organic chemistry. Above, it is observed that several lines of text has been used to describe the bonding changes in a single step of a reaction mechanism. A reaction mechanism might have 10-12 steps or more, making such descriptions very cumbersome. A shorthand notation that summarizes these changes has to be developed. This notation, called curved arrow formalism, provides a rapid way of drawing bond and electron changes in a given mechanism step. They are also useful to indicate electron changes between a set of contributing resonance structures.Arrows in chemical drawing have specific meanings. Just as it is important to learn the structural representation and names of molecules, it is important to learn the language of arrow formalism in organic chemistry. Different prominent types are discussed below, 1.

Curved arrows are used to show how electrons are moved in resonance structures and in reactions. Therefore, curved arrows always start at the initial position of electrons and end at their final position. It is important to note that when a pair of electrons in a polar covalent bond is moved to one of the bonded atoms, it is moved to the more electronegative atoms.

2.

A curve arrow with half a head is called a fishhook. This kind of arrow is used to indicate the movement of a single electrons.

3.

Straight arrows point from reactants to products in chemical reactant equations. Straight arrows with half-heads are commonly used in pairs to indicate that a reaction is reversible.

4.

A double-headed straight arrow (↔) between two structures indicates that they are resonance structures. Such an arrow does not indicates the occurrence of a chemical reaction.

O-65 So, as stated earlier,each curved arrow with two barbs on the head

represents the shift of one electron

pair. The curved arrows shows of Direction of electron flow. The tail shows the electron origin, and always come from an electron source, usually a lone pair or bonding pair from a σ or π bond. In this representational mode, the head of the arrow indicates the electron pair destination, either as a new lone pair or a new bond. If the arrowhead points to another atom, that atom must either have an open octet or have an electron pair that can be displaced by the incoming electron pair. It is important to remember that electrons never flow from atoms which are electron-poor to atoms which are electron-rich, so a curved arrow will never point from an atom with a positive charge to an atom with a negative charge.Some of the most common ways are, (i)

New bond formed to X:

(ii)

Bonding pair becomes lone pair; bond broken:

X X

The use of curved arrows for the ionization of water is shown below. This bonding pair becomes a lone pair on oxygen of hydroxide ion.

H H—O H

H

H—O

O H

This lone pair becomes the new O–H bond.

This lone pair was on O–H bond.

+ H—O H

The bonding electron pair was a lone pair on oxygen.

Fig.1: Ionisation of water

The tail of the curved arrow on the right hand side starts at the oxygen lone pair, meaning this curved arrow shows a bonding change for this lone pair. The head of the curved arrow points to the space between the oxygen and hydrogen atoms, meaning the electron pair ends in that space as a bond between the oxygen and the hydrogen. The hydrogen that accepts a new bond to oxygen must give up a pair of electrons because it cannot have more than two valence electrons at a time. Thus, the old O-H bond is being displaced by the new electron pair from the other oxygen atom. The curved arrow on the left indicates the electron pair that was the O-H bond becomes a lone pair on the oxygen of the hydroxide ion. If the arrow starts at a bond between two atoms, then that bond is broken. If the arrow ends between two atoms, then a new bond is formed between those atoms. If the atoms are already attached through a single chemical bond, then a double or triple bond results.The process of drawing a curved arrow mechanism is also commonly called "electron pushing."When drawing curved arrows, the start and stopping points of the arrows are critical. Things that make no difference are where the arrows curve up or down, or whether they start or stop at the top or bottom of an atom. Note also the changes in formal charge that result from the electron changes. If an atom shares a lone pair that it used to have all to itself then its charge decreases by one, so one can conclude that neutral atom become +1 charged. If an atom gains a bonding electron pair all to itself as a lone pair then its charge increases by one unit so that a neutral atom becomes -1charged. The charge is conserved in this mechanism step. The total charge on the left, which is zero, is the same as the total charge on the right (-1 +1 = 0). Charge is conserved in all mechanism steps. You should make a habit of checking your work against this point to minimize errors.

O-66 Lone pairs are often involved in reaction mechanisms, so one must be in the habit of drawing all lone pairs. It is also important that curved arrows be drawn neatly and precisely, clearly showing the atomic origin of the electron pair at the tail of the curved arrow and the electron pair destination at the head of the arrow.

Solved Examples Example 1: Provide the curved arrows for the reaction of hydroxide and hydronium ions to form two molecules of water. H

+ H—O ⊕

H—O

H—O—H

H

Solution : A reasonable approach to an exercise like this is to analyze the bond changes are then draw the corresponding curved arrows. The oxygen of hydroxide ion has shared a lone electron pair with the hydrogen of hydronium ion, forming a new O-H bond. Thus one can draw a curved arrow with the tail at the hydroxide ion lone pair and ending at the hydronium ion hydrogen. This hydrogen atom can only have one covalent bond at a time, so it must sacrifice the bond to the hydronium ion. One can draw a curved arrow to show the bonding electron pair between the hydrogen and oxygen of the hydronium ion moving to become the sole property of the oxygen atom. The oxygen of the hydroxide ion is sharing a pair of electrons that it had all to itself before, so its formal charge drops by one unit (-1 to neutral). The oxygen of the hydronium ion gains a lone pair of electrons that is used to share with the hydrogen, so its formal charge decreases by one unit (neutral to +1). H

+ H—O ⊕

H—O

H—O—H

H

Example 2: Draw the product(s) of the following mechanism step based upon the curved arrows. H2C

CH2 Br

Br

product(s)??

Solution : The curved arrow that starts at the carbon-carbon bond and ends at the bromine atom indicates the bond electron pair has shifted to become a carbon-bromine bond. The left-hand carbon of the double bond has lost an electron pair, so its formal charge becomes one unit more positive (neutral to +1). The curved arrow on the right indicates that the electron pair of the bromine-bromine bond is shifting to reside solely on the bromine on the right, resulting in rupture of the bond and formation of bromine with four lone pairs and a negative charge. ⊕

H2C

CH2 Br

Br

H2C— CH2—Br + Br

Example 3: Provide curved arrows that show how the following mechanism steps might occur. H ⊕ O—H

(i) H

O—H

2H2O

O-67 NO2

⊕

(ii)

+ NO2

⊕

H (iii)

+ HOCH3 +

+ OCH3

Cl

Cl O (iv)

O

C H3C

O

O (v)

H3C

S

O

⊕

C

H + NH3

H3C

O

+ NH4

H3C—Cl + SO2 + Cl

+ Cl

Cl

Solution : (i) The O-H bond of the hydronium ion (H3O+) has been broken. The pair of electrons from this bond have become a lone pair on oxygen of a water molecule, so one may need a curved arrow starting at the O-H bond and ending at the oxygen. One lone pair of hydroxide ion (HO − ) has attacked the hydrogen of the hydronium ion, resulting in a new O-H bond in the second water molecule. One can indicate this with a curved arrow starting at the hydroxide ion lone pair and ending at the hydrogen of the hydronium ion. This is an example of a proton transfer, a mechanism step that is quite common in organic reactions. H ⊕ O—H

+ O—H

2H2O

H

(ii)

A new bond is formed between the nitronium ion (NO2+) and the carbon of the carbon-carbon double bond. The bonding electrons come from the carbon-carbon double bond, so one can draw a curved arrow starting at the π bond portion of the double bond and ending at the nitronium ion. The bottom carbon of the double bond has lost a pair of electrons that it used to share, so its charge becomes one unit more positive. This is a key step in the electrophilic aromatic substitution mechanism. NO2

⊕

+

NO2

⊕ −

(iii) A lone pair on the oxygen atom of the methoxide ion (CH3O ) is now shared with a hydrogen that was taken from the cyclohexane ring. We indicate this with a curved arrow starting at the oxygen lone pair and ending at the hydrogen. The pair of electrons that used to be the C-H bond become the π component of the C=C bond. One can indicate this with a curved arrow starting at the C-H bond and

O-68 ending at the position of the new π bond, in the space between two carbons of the cycloehxane ring. The electron pair that was the C-Cl bond becomes a lone pair on the departing chloride ion. We indicate this by drawing a curved arrow starting at the C-Cl bond and ending at the chlorine atom. This is an example of an E2 reaction. H +

+ HOCH3 Cl

OCH3

Cl

(iv) The lone pair on the nitrogen atom of ammonia (NH3) attacks the hydrogen of acetic acid (CH3CO2H) to form a new N-H bond. We indicate this with a curved arrow starting at the nitrogen lone pair and ending at this hydrogen atom. The hydrogen of acetic acid gives up the pair of electrons that it shares with the oxygen atom. This pair of electrons becomes the π bond component of the new C=O bond. We indicate this with a curved arrow starting at the H-O bond and ending between the carbon and oxygen atoms, the position of the new π bond. The old C=O π bond electron pair is displaced by the incoming π bond, so the old C=O π electron pair shifts to the oxygen at the top of the structure. One can indicate this with a curved arrow starting at the π bond and ending at the oxygen atom. This is another example of a proton transfer reaction. O

O

C H3C

(v)

O

H + NH3

⊕

C H3C

O

+ NH4

The chloride ion forms a new bond with the carbon of the methyl group (CH3) by sharing a lone pair. We indicate this with a curved arrow starting at the chloride ion and ending at the carbon of the methyl group. The carbon of the methyl group has a full octet to begin with, so it must lose a pair of electrons to avoid having ten valence electrons. The electron pair of the C-O bond is displaced, becoming the π component of the new S=O bond. We indicate this with a curved arrow that starts at the C-O bond and ends at the position of the new S-O π bond between the sulfur and oxygen atoms. The sulfur atom has gained a pair of electrons that used to be the C-O bond. This new bonding pair displaces the S-Cl bonding pair and the S-Cl bond is lost. We indicate this with a curved arrow that starts on the S-Cl bond and ends at the chlorine atom. In this case, it was necessary to reposition of the reactants to make the curved arrows easier to draw and understand. As such this is frequently necessary. This reaction is a key stet in the reaction between an alcohol and thianylchloride (SOCl2) to produce a chloroalkane. O Cl

H3C

S

O

H3C—Cl + SO2 + Cl

Cl

1.2 Formal Charge Generally, in chemistry, a formal charge (FC) is the charge assigned to an atomin a molecule, assuming that electrons in a chemical bond are shared equally between atoms, regardless of relative electronegativity. The formal charge of any atom in a molecule can be calculated by the following equation:

O-69

Formal charge=V − N –

B 2

here V is the number of valence electrons of the atom in ground state; N is the number of non-bonding valence electrons on this atom in the molecule; and B is the total number of electrons shared in covalent bonds with other atoms in the molecule. There are two electrons shared per single covalent bond. When determining the correct Lewis structure or predominant resonance structure for a molecule, the structure is chosen such that the formal charge (without sign) on each of the atoms is minimized. Formal charge is a test to determine the efficiency of electron distribution of a molecule. This is significant when drawing structures. Some of the examples are discussed below: 1.

Carbon in methane: FC = 4 - 0 - (8÷2) = 0

2.

Nitrogen in NO2 − : FC = 5 - 2 - (6÷2) = 0

3.

Single bonded oxygen in NO2 − : FC = 6 - 6 - (2÷2) = -1

4.

Double bonded oxygen in NO2 − : FC = 6 - 4 - (4÷2) = 0 An alternative method for assigning charge to an atom taking into account electronegativity is by oxidation number. Other related concepts are valence, which counts the number of electrons that an atom uses in bonding, and coordination number, the number of atoms bonded to the atom of interest.

5.

Ammonium NH4 + is a cationic species. By using the vertical groups of the atoms on the periodic table it is possible to determine that each hydrogen contributes 1 electron, the nitrogen contributes 5 valence electrons, and the charge of +1 means that 1 of the contributed electrons is absent. The final total is 8 total electrons (1 × 4 + 5 –1). Drawing the Lewis structure gives an sp3 (4 bonds) hybridized nitrogen atom surrounded by hydrogen. There are no lone pairs of electrons left. Thus, using the definition of formal charge, each hydrogen has a formal charge of zero (1- (0 + ½ × 2)) and the nitrogen has a formal charge of +1 (5–(0 + ½ × 8)). After adding up all the formal charges throughout the molecule the result is a total formal charge of +1, consistent with the charge of the molecule given in the first place. It is important to note that the total formal charge in a molecule must be as close to zero as possible, with as few charges on the molecule as possible. To explain the concept, one can consider CO2 as an example. CO2 is a neutral molecule with 16 total valence electrons. There are three different ways to draw the Lewis structure

Mode 1:

Carbon single bonded to both oxygen atoms (carbon = +2, oxygens = -1 each, total formal charge = 0)

Mode 2:

Carbon single bonded to one oxygen and double bonded to another (carbon = +1, oxygen double = 0, oxygen single =–1, total formal charge = 0)

Mode 3:

Carbon double bonded to both oxygen atoms (carbon = 0, oxygens = 0, total formal charge =0)

Even though all three structures gave us a total charge of zero, the final structure is the superior one because there are no charges in the molecule at all.

1.3 Comparison of Formal Charge and Oxidation State The concept of oxidation states constitutes a competing but slightly different method to assess the distribution of electrons in molecules. If the formal charges and oxidation states of the atoms in carbon dioxide are compared, the following values are arrived at:

O-70 O

C

O

Formal charge

0

0

0

Oxidation state

2–

4+

2–

The reason for the difference between these values is that formal charges and oxidation states represent fundamentally different ways of looking at the distribution of electrons amongst the atoms in the molecule. With formal charge, the electrons in each covalent bond are assumed to be split exactly evenly between the two atoms in the bond. It is because of this reason the division by two is performed in the formula used, B Formal charge=V − N – 2 The formal charge view of the CO2 molecule is essentially shown below: ''Split'' electrons in covalent bonds evenly O= =C= =O

C

O

O

The covalent hence sharing aspect of the bonding is overemphasized in the use of formal charges, since in reality there is a higher electron density around the oxygen atoms due to their higher electronegativity compared to the carbon atom. This can be most effectively visualized in an electrostatic potential map.With the oxidation state formalism, the electrons in the bonds are "awarded" to the atom with the greater electronegativity. The oxidation state view of the CO2 molecule is shown below: ''Award'' electrons in bonds to oxygen O

2–

O

C

C

O

4–

2–

O

Oxidation states overemphasize the ionic nature of the bonding; most chemists agree that the difference in electronegativity between carbon and oxygen is insufficient to regard the bonds as being ionic in nature.In reality, the distribution of electrons in the molecule lies somewhere between these two extremes modes of charge distribution.

1.4 Types of Bond Cleavage Bond cleavage or scission is the splitting of chemical bonds.A covalent bond between two atoms can be broken in two ways 1. Homolytic cleavage R – X R+X 2.

Heterolytic cleavage R – X

⊕

R + X⊕ or R +X

O-71 1.

Homolytic cleavage: In this cleavage, each atom separates with one electron leading to the formation of highly reactive intermediates called free radicals.

A B

A +B

A B

A +B

Fig. 2: Homolytic cleavage

Homolytic cleavage occurs generally in the gas phase or in solution in non polar solvents and is catalysed by ultra violet light or high temperature or by radical initiators like peroxides. Cleavage of benzoyl peroxide or hydrogen peroxideis an example of homolysis. Mechanism of the reaction in which homolysis takes place is known as homolytic mechanism or free radical mechanism. 2.

Heterolytic cleavage: In this type of fission, one atom may hold on both the electrons leaving none for the other, resulting in a negative and a positive ion (or ion pair).

A B

⊕

A +B

Fig. 3: Heter olytic cleavage

Heterolytic cleavage occurs in polar solvents because of the ease of separation of charge and stabilization of the resultant ion pairs through solvation. In such heterolytic fission, the charged species formed are either carbon bearing positive charge called carbocation or carbon bearing negative charge called carbanion. Mechanism of the reaction in which heterolysis takes place is known as heterolytic mechanism or ionic mechanism. The energy required for heterolysis is always greater than that for homolysis due to electrostatic forces of attraction operational between ions in case of hetrolytic cleavage.

1.5 Types of Attacking Reagents There are basically two types of attacking reagents used in organic chemistry, the electrophiles and nucleophiles. Electrophiles and nucleophiles can be looked upon essentially as acceptors and donors of electron pairs respectively, from and to other atoms which is mostly carbon. 1.

Electrophiles: Electrophilic reagents or electrophiles are the electron–deficient species which tend to attack the substrate at position (or positions) of high electron density.It is observed that electrophiles are positively charged species or must have an atom that carries a partial positive charge that are attracted to an electron rich centre. However, there are cases where electrophiles are found to be neutral as well.In neutral form, the electron loving nature is due to the incomplete octet configuration. As electrophiles accept electrons, they are Lewis acids. The electrophiles are attacked by the most electron-populated part of one nucleophile. The electrophiles frequently seen in the organic syntheses are cations such as H+ and NO+, polarized neutral molecules such as HCl, alkyl halides, acyl halides, and carbonyl compounds, polarizable neutral molecules such as Cl2 and Br2, oxidizing agents such as organic peracids, chemical species that do not satisfy the octet rule such as carbenes and radicals, and some Lewis acids such as BH3 and DIBAL.Some of the common examples are ⊕

⊕

⊕

⊕

*

*

*

H ⊕ ,H3O ⊕ ,N O 2 ,N O,PhN2⊕ ,R ⊕ ,R 2 C H,R 3C ⊕ ,Cl ⊕ ,R − C O,SO 3 ,C O 2 ,B F3 , AlCl3 , *

ICl,Br2 ,O 3 etc. It participates in a chemical reaction by accepting an electron pair in order to bond to a nucleophile.

O-72 2.

Nucleophiles: Nucleophilic reagents or nucleophiles are nucleus loving reagents which are electron–rich species and which tend to attack the substrate at a position of low electron density.All molecules or ions with a free pair of electrons or at least one π bond can act as nucleophiles. For *

••

example, H − , BH −4 , H S O 3− , OH − , RO − , RS − , CN − , RCO −2 , R–C ≡ C − , –CH(CO 2 Et) 2 , R − , H2 O , ••

••

••

*

*

*

R O H, R O R , R MgBr ,R Li, R 2 Cd etc.The star indicates the atom that donates electrons to the ••

••

substrate.Another way of describing is that a nucleophile is a chemical species that donates an electron pair to an electrophile to form a chemical bond in a reaction. As nucleophiles donate electrons, they are, by definition, Lewis bases. 3.

Nucleophilicity: Nucleophilicity, sometimes referred to as nucleophile strength, refers to a substance's nucleophilic character and is often used to compare the affinity of atoms. Neutral nucleophilic reactions with solvents such as alcohols and water are named solvolysis. Nucleophiles may take part in nucleophilic substitution, whereby a nucleophile becomes attracted to a full or partial positive charge.Examples of nucleophiles are anions such as Cl − , or a compound with a lone pair of electrons such as NH3 (ammonia).In the example below, the oxygen of the hydroxide ion donates an electron pair to bond with the carbon at the end of the bromopropane molecule. The bond between the carbon and the bromine then undergoes heterolytic fission, with the bromine atom taking the donated electron and becoming the bromide ion (Br − ), because a SN2 reaction occurs by backside attack. This means that the hydroxide ion attacks the carbon atom from the other side, exactly opposite the bromine ion. Because of this backside attack, SN2 reactions result in a reversal of the configuration of the electrophile. If the electrophile is chiral, it typically maintains its chirality, though the SN2 product's configuration is flipped as compared to that of the original electrophile. H C H

4.

H Br O– H

C

O

H

O–

H + Br –

H

Ambident nucleophile: An ambident nucleophile is one that can attack from two or more places, resulting in two or more products. For example, the thiocyanate ion (SCN–) may attack from either the S or the N. For this reason, the SN2 reaction of an alkyl halide with SCN–often leads to a mixture of RSCN, an alkyl thiocyanate and RNCS which is an alkyl isothiocyanate. Some other examples of ambident nucleophiles are cyanide ion or nitrite ion.

1.5.1 Different Types of Nucleophiles Carbon nucleophiles are alkyl metal halides found in the Grignard reaction, Reformatsky reaction, organolithium reagents, and anions of a terminal alkyne. Enols are also carbon nucleophiles. The formation of an enol is catalyzed by acid or base. Enols are ambident nucleophiles, but, in general, nucleophilic at the alpha carbon atom. Enols are commonly used in condensation reactions, including the Claisen condensation and the aldol condensation reactions. Oxygen nucleophiles are water (H2O), hydroxide anion, alcohols, alkoxide anions, hydrogen peroxide, and carboxylate anions Nuclophilic attack does not take place during intermolecular hydrogen bonding of sulfur nucleophiles, hydrogen sulfide and its salts, thiols (RSH), thiolate O ||

S ||

anions (RS ), anions of thiolcarboxylic acids (R − C − S ), and anions of dithiocarbonates (RO − C − S− ) −

−

O-73 S ||

and dithiocarbamates (R 2N − C − S− ) are used most often. In general, sulfur is highlynucleophilicin nature because of its large size, which makes it readily polarizable, and its lone pairs of electrons are readily accessible. Similarly, nitrogen nucleophiles include ammonia, azide, amines, and nitrites.

1.6 Types of Organic Reactions There is no limitation to the number of possible organic reactions and mechanisms. However, certain general patterns are observed that can be used to describe many common or useful reactions. Each reaction has a stepwise reaction mechanism that explains how it happens. Although this detailed description of steps is not always clear from a list of reactants alone. Organic reactions can be organized into several basic types. Some reactions fit into more than one category. For example, some substitution reactions follow an addition-elimination pathway. This study does not intend to include every single organic reaction. Rather, it is intended to cover all the basic types of organic reactions. The important basic types of reactions are, 1.

Addition reaction

2.

Elimination reaction

3.

Substitution reaction

4.

Rearrangement reaction

There is one more prominent type which refers to organic redox reactions. Organic redox reactions are redox reactions specific to organic compounds and are very common. For example oxidation of aldehydes by Tollen's reagent. 1.

Addition reaction: An addition reaction, in organic chemistry, is in its simplest terms an organic reaction where two or more molecules combine to form a larger one. H

Cl Cl

H

C=C H

+ Cl–Cl H

H

C

C

H

H H

Addition reactions are limited to chemical compounds that have multiple bonds, such as molecules with carbon-carbon double bonds(alkenes), carbon oxygen double bonds, or with triple bonds (alkynes). Molecules containing carbon-hetero multiple bonds like nitrile groups, or imine (C=N) groups, can undergo addition as they too have multiple bond character. An addition reaction expectedly is the opposite of an elimination reaction. For instance the hydration reaction of an alkene and the dehydration of an alcohol are addition-elimination pairs. There are two main types of addition reactions. (i)

Polar addition reactions

(ii)

Non polar addition reactions

Subsequently, there are two main types of polar addition reactions: (a) Electrophilic addition and (b) Nucleophilic addition.

O-74 Two non-polar addition reaction exists as well called ● Free radical addition and 2.

● Cycloaddition reaction. Elimination reaction: An elimination reaction is a type of organic reaction in which two atoms or groups of atoms or substituents are removed from a molecule. This removal of substituents can take place either in one or two-step mechanism. The one-step mechanism is known as the E2 reaction, and the two-step mechanism is known as the E1 reaction. The numbers 1 and 2 refers to the kinetics of the reaction. These numbers refer whether the reaction is bimolecular or unimolecular respectively. These numbers have nothing to do with the number of steps in the mechanism. An example of elimination reaction is shown below,

OH H2SO4 130–140°C

3.

In most organic elimination reactions, at least one hydrogen is lost to form the double bond. It is also possible that a molecule undergoes reductive elimination, by which the valence of an atom in the molecule decreases by two. An important class of elimination reactions are those involving alkyl halides, with good leaving groups, reacting with a Lewis base to form an alkene. Elimination may be considered the reverse of an addition reaction. When the substrate is asymmetric, regioselectivity is determined by Saytzeff's rule or through Hoffmann elimination rule. Substitution reaction: Substitution reaction also known as displacement reaction is a chemical reaction during which one functional group in a chemical compound is replaced by another functional group. Substitution reactions are of prime importance in organic chemistry. Substitution reactions in organic chemistry are classified either as electrophilic or nucleophilic depending upon the reagent involved. There are other classifications as well. Organic substitution reactions are classified in several main organic reaction types depending on whether the reagent that brings about the substitution is considered an electrophile or a nucleophile, whether a reactive intermediate involved in the reaction is a carbocation, a carbanion or a free radical or whether the substrate isaliphatic or aromatic. A good example of a substitution reaction is halogenation. When chlorine gas (Cl-Cl) is irradiated, some of the molecules are split into two chlorine radicals (Cl) whose free electrons are strongly nucleophilic. One of them breaks a weak C-H covalent bond and grabs the liberated proton to form the electrically neutral H-Cl. The other radical reforms a covalent bond with the CH 3 . to form CH 3 Cl (methyl chloride) as shown below, H H H

4.

H + Cl–Cl

C H

hν Cl

H

+ H–Cl

C H

H

Rearrangement reaction: A rearrangement reaction is a broad class of organic reactions where the carbon skeleton of a molecule is rearranged to give a structural isomer of the original molecule. Often a substituent moves from one atom to another atom in the same molecule. In the example below the substituent R moves from carbon atom 1 to carbon atom 2: —C–C–C— R

—C–C–C— R

O-75 Intermolecular rearrangements also take place. A rearrangement is not well represented by simple and discrete electron transfers. The actual mechanism of alkyl groups moving, as in Wagner-Meerwein rearrangement, probably involves transfer of the moving alkyl group fluidly along a bond, not ionic bond-breaking and forming. In pericyclic reactions, explanation by orbital interactions gives a better picture than simple discrete electron transfers. It is, nevertheless, possible to draw the curved arrows for a sequence of discrete electron transfers that give the same result as a rearrangement reaction, although these are not necessarily realistic. In allylic rearrangement, the reaction is indeed ionic. Three key rearrangement reactions are 1,2-rearrangements, pericyclic reactions and Olefin metathesis. Table 1: Different types of organic reactions and their characteristics Reaction Type

Subtype

Comment

1. Addition reaction

a.Electrophilic addition(polar addition)

Include such reactions as halogenation, hydrohalogenation and hydration.

b.Nucleophilic addition (polar addition) c.Radical d. Addition(non polar addition) 2.Elimination reaction

3. Substitution reaction

Include processes such as dehydration and are found to follow an E1, E2 or E1cB reaction mechanism a. Nucleophilic aliphatic substitution b. Nucleophilic aromatic substitution c. Nucleophilic acyl substitution d. Electrophilic substitution

With SN1, SN2 and SNi reaction mechanisms

e. Electrophilic aromatic substitution f. Radical substitution 4.Rearrangement reactions

a.1,2-rearrangements b. Pericyclic reactions c. Metathesis

In addition to all these types of reactions, there are condensation reactions in which a small molecule, usually water, is split off when two reactants combine in a chemical reaction. The opposite reaction, when water is consumed in a reaction, is called hydrolysis. Many Polymerization reactions are derived from organic reactions. They are divided into addition polymerizations and step-growth polymerizations.

O-76

2. Various Reaction Intermediates Whenever a reaction takes place in organic chemistry, the reactants are not directly transformed to the products. The reaction proceeds via some intermediates which are formed during the course of reaction and are not very stable species.The most common transient intermediates formed are radicals, carbocations and carbanions but other reactive intermediates possible are carbenes, nitrenes and benzynes. The formations of these intermediates and their characteristic reactions have been dealt here in brief.

H H

C H

Structure of CH3(sp2 hybridised)

2.1 Free Radicals Their formation is initiated mostly by ultra violet light. For example, •

hν

Cl − Cl   → 2 Cl •

•

R − H + Cl → R + HCl This alkyl group bearing an odd unpaired electron is called alkyl radical. The radicals are electron deficient and paramagnetic in nature. A free radical has carbon in the sp2 hybridised state, thus exhibit planar (flat) structure. This type of hybridization is possible in most of the cases especially in case of conjugated free radicals. However, in case of un conjugated free radicals, the hybridization state may be sp3. sp2

sp3

Pyramidal

Planar

Fig. 4: Hybridization states and shape of free radical

If a free radical with three different atoms or group of atoms is attacked by another radical, it leads to the generation of a stereocentre, existing as enantiomeric pair. R C R' R''

R Cl

R'

R Cl + Cl

R'

R'' R'' Enantiomeric pair

The free radicals can act as electrophiles which attack at the site of high electron density on the substrate.Rearrangement among radicals is found to be much less common than otherwise similar rearrangements that involve carbocations. In this they resemble carbanions and the reason for their

O-77 resemblance becomes apparent when we compare the transition states (T.S.) of the three for a 1,2 alkyl shift (for detail see carbocation section). ≠

⊕

≠

R C

≠

R C

C

Carbocation T. S.

R C

C

Radical T. S.

C

Carbanion T. S.

These transition states involve two, three and four electrons, respectively and electrons in excess of two can be accommodated only in an anti–bonding molecular orbital of much higher energy. However as with carbanions, 1,2–aryl shifts are also known in radicals, which involve stabilized, bridged transition state. The driving force for the rearrangement is to attain stability.

Stabilty of Free Radicals 1.

The stability of free radicals increases with increase in the number of alkyl groups. Thus the stability of different free radicals is: H

CH3

CH3

CH3




>

2°

>

3°

Hybridization of the charge-bearing atom: The greater the s-character of the charge-bearing atom, the more stable the anion. The extent of conjugation of the anion: Resonance effects can stabilize the anion so that its stability is increased manifold as shown below in the case of benzyl carbanion,

–CH2

CH2

CH2 δO δP δO

CH2

δB H C δB H

One more example of the same concept and hence subsequent order of stability is, < < no resonence (less stable)

< resonance (more stable)

resonance (more stable)

Similarly, allyl carbanion is more stable than any non conjugated carbanion in the following order,

O-90 H

H

C H

allylic

methyl

>

>

1°

>

>

2°

3°

Some more examples of greater stability are, O

O R1

RO

N≡ R1

R3 R2

O

N

N= R1

R3

R2

O C

R1

N O–

R2

R1

⊕

R2

O R1

R1 O

R2

RO

O C

R2

R2

R1

⊕

R2

Aromaticity: The stabilization is particularly enhanced when the negative charge is delocalised through aromaticity. An example is shown below, – or –

–

cyclopentadienyl anion

Example 5 : Compare the stabilities of phenyl (C6H5) and cyclohexyl (a) cations and (b) anions. + Solution: (a) C 6 H5+ is a vinyl type carbocation and is less stable than C 6 H11 (Cycloheroyl cation), a 2° (b)

carbocation. In C 6 H5− :, the electron pair is in an sp2 hybrid orbital. The carbanion has more s–character and is more – stable than C 6 H11 : whose unshared electron pair is in an sp3 hybrid orbital.

+

H

– +

C6H+ 5

+

C6H11

C6H5– :

sp2

sp3

H +

–

C6H11

2.9 Carbene Carbenes are electrically neutral carbon containing reactive intermediates with two bond pairs and one unshared pair of electrons. They are electron deficient species and acts as electrophiles. The common

O-91 reactions in which they are formed as intermediates are Reimer Tiemann reaction, carbylamine reaction etc.Carbenes exist transiently due to their less stabilty.

Methods of preparation Some familiar processes by which carbenes are generated are discussed below. 1.

These intermediates are generated by the α–elimination of trihalomethane. For example, Cl

Cl H– C

Cl

OH– or OR–

Cl

2.

C

–H2O or–ROH

Cl

∆

CCl2+CO2+Cl–

Photolysis of ketene and diazomethane yield carbenes as well.

⊕

CH2 = N = N

⊕

hν

CH2 = C = O

4.

Cl–C– Cl dichloro carbene

Thermolysis of trichloro acetate ion also gives rise to carbenes. CCl3–COO–

3.

–Cl–

Cl

CH2 + C ≡ O

hν

CH2 + N ≡ N

Scheme of one more method of preparation is given below, H R

N–Ts

Bronsted base

R

C=N

C + N2 + HTs

R

R

2.10 Types of Carbenes Carbenes are very reactive and it is often difficult to prove that they are actually present in a given reaction.The two non bonded electrons of a carbene can be either paired or unpaired as shown below, R

R'

R

R

Singlet carbene

Triplet carbene

Fig. 6: Singlet and triplet carbenes

If they are paired, the species is a singlet while if they are unpaired, the species is referred as triplet.The carbon of singlet and triplet carbenes are sp 2 and sp hybridized respectively.The structures of the two alongwith the bond angles are represented as 180°

103°

H

C

singlet carbene

H

C

H

triplet carbene

O-92 The singlet CCl2 and CBr2 are also bent with angles of 100° and 114° respectively. Singlet carbene is the less stable form due to the interelectronic repulsion between the electrons kept in the same orbital. However, it is often (not always) the singlet form which is first generated in the initial photolysis.

Chemical reactions involving carbenes 1.

Ring closure reactions with alkenes are one of the most common reactions of carbenes as shown below, R' C

R C

+

C

R

C

C

C

R' An alkene

2.

A carbene

A cyclopropane

There are some reactions reported in which carbene is not isolated as such but involves carbene formation. In these reactions, it is not the free carbene intermediate, the carbenes are generated in situ called as carbenoids. A familiar example of this is Simmon–Smith reaction. C C

ZnI

C

+CH2

C

I

ZnI CH2 I

C C

ZnI CH2 I

2.11 Nitrene Generally it is assumed that a nitrene (R-N:) is the nitrogen analogue of a carbene. Nitrenes are electrically neutral nitrogen containing species with two unshared pairs and one bond pair of electrons. Here, nitrogen is electron deficient atom therefore it is considered as an electrophile. A nitrene is a reactive intermediate and is involved in many chemical reactions.

Methods of preparation As nitrenes are highly reactive, they are not isolated. Instead, they are formed as reactive intermediates during a reaction. There are some common ways to generate nitrenes. Such intermediates result from the elimination of a proton and a leaving group (generally halogen) from the same atom of a substrate. For example, R–CH2–NH–Br

OH– –H2O

R–CH2–N–Br

–Br–

R–CH2–N Nitrene

Other ways of generating nitrenes are α–elimination of N alkyl sulphonate esters and from azides by thermolysis or photolysis, with expulsion of nitrogen gas. This method is analogous to the formation of carbenes from diazo compounds.These can also be prepared from isocyanates, with expulsion of carbon monoxide. This method is analogous to the formation of carbenes from ketenes.

2.12 Types of Nitrenes In the most simple nitrene, the linear imidogen (:N-H), two of the 6 available electrons form a covalent bond with hydrogen, two other create a free electron pair and the two remaining electrons occupy two degenerate p orbitals.

O-93 triplet

singlet

RN

R–N

R

R2C

C

R

R–N

R

C

R

Fig. 7.Singlet and triplet nitrene compared with carbenes

Consistent with Hund's rule the low energy form of imidogen is a triplet with one electron in each of the p orbitals and the high energy form is the singlet state with an electron pair filling one p orbital and the other one vacant.

2.13 Benzyne Benzyne is an example of an aryne. The word -yne definitely means a triple bond, so that benzyne is a benzene ring containing a triple bond as shown in the structure below, H H

H H

This is no ordinary triple bond as the second interaction results from a weak interaction of sp2 hybrid orbitals lying in the plane of the ring. The triple bond is non-linear due to the constraints of the 6-membered ring. Hence, benzyne is strained and highly reactive. Normally, arynes in general, are best described as having a strained triple bond; however, they possess some biradical character as well. Benzyne was first postulated by Georg Wittig in 1940.The discovery of benzyne led to rapid developments in synthetic methodologies to make this highly reactive intermediate useful for organic synthesis.

Methods of preparation of benzyne Due to extreme reactivity of arynes they must be generated in situ. In the early days of benzyne chemistry, harsh conditions were needed to generate arynes–many of the methods require strong base or high temperatures. Aryl halide can be treated with strong base to remove an aromatic proton and generate benzyne via elimination of two ortho substituents.The reaction scheme is shown below,

O-94 Cl

NH2 NaNH2 NH3(1)

Benzyne

Mechanism: As observed, these intermediates are generated by the elimination of a proton and a leaving group (generally halogens) from the adjacent carbons from a benzene ring system. Cl

Cl –Cl–

NH2

⊕

–NH3

H

1.

Benzyne

Arenediazonium-2-carboxylates can serve as precursors to benzynes. The main drawback of this method is the explosive nature of diazonium compounds. ⊕

N2 heat CO2

2.

Milder methods for benzyne generation have been developed. Aryl triflates have been widely used in synthesis.Fluoride displacement of the trimethylsilyl group, as shown below, allows for generation of benzyne under mild conditions. TMS Bu4NF OTf

In a reaction where benzyne is formed, if the solution does not have sufficient nucleophile, they undergo dimerization to give biphenylene. Dimerization

Biphenylene

2.14 Bonding in Benzyne Benzynes are very reactive. Neither benzyne nor any other aryne (substituted benzyne) has yet been isolated under ordinary conditions. The triple bond in benzynes is not identical with the formal triple bond of alkynes because here the two π bonds are formed by the overlapping of p z − p z and sp2–sp2 orbitals whereas in alkynes, they are formed by the orbital overlap of p z − p z and p y − p y .

O-95 °

1.39 A

sp2

°

134 A

R–C≡C–R

180° angle

1.41 A° 138° A

120° angle

Poor overlap –looks much like a diradical

R

R –Bond formed is quite weak –Hybridization sp 2 –Formed only as an internediale Fig. 8. Comparison of bonding in benzyne and alkyne

It should be noted that the π bond (resulted from sp2–sp2 overlap) is a very weak bond which can be easily ruptured by the attack of a nucleophile, thus making the benzynes very reactive.

Another feature to be noted is that the extra pair of electrons does not affect the aromaticity. The original sextet still functions as a closed ring and the two additional electrons are merely located in a α orbital that covers only two carbons.

Chemical reaction of benzynes Even at low temperatures arynes are extremely reactive. Their reactivity can be classified into certain major classes like nucleophilic additions, ring closure or pericyclic reactions, bond-insertion etc. 1. Nucleophilic additions Upon treatment with basic nucleophiles, aryl halides are deprotonated α to the leaving group, resulting in dehydrohalogenation. The resulting benzyne forms addition products, usually by initial protonation. Generation of the benzyne intermediate is the slow step in the reaction. Br

NH2 slow

NH2

NH2 NH2

H NH2

H–NH2

In case of disubstitution, the chemical behavior of benzynes is impacted by the group previously present in the benzene ring as shown below,

O-96 OMe

OMe Cl

H

2.

OMe

OMe

Cl

OMe H–NH2

NH2

NH2

NH2

NH2

Coupling or aryne couplingre actions allow for generation of biphenyl compounds which are valuable in pharmaceutical industry, agriculture and as ligands in many metal-catalyzed transformations. Br

Ar

Ar

Li

Br

Li–Ar Br

3.

Pericyclic reaction:Benzynes can undergo [4+2] cyclization reactions. The concerted mechanism of the Diels-Alder reaction between benzyne and furan is shown below. However, many benzyne [4+2] cycloadditions are thought to proceed via a stepwise mechanism.

O

O

3. Methods of Determination of Reaction Mechanism In determining the mechanism of a reaction, one of the major problems is to deduce the spatial or three-dimensional changes that occur to the molecules involved as they proceed from their initial state through the intermediate stages and transition states to the final products. Knowledge about such changes can generally be deduced from knowledge of the stereochemistry of the starting materials, intermediates, and final products provided these are obtained under kinetic control. Information of this kind is obtained by determinations of optical activity and analysis of the structures of the compounds by standard means. In certain instances, information about the movement of atoms between molecules during the course of a reaction can be gained by using compounds containing isotopes of certain of the atoms. These isotopes behave much like the ordinary atoms they replace, but they can be identified by their behaviour. For example, in the hydrolysis of ethyl acetate, it is crucial to a determination of the mechanism to be able to establish which of the two reactants - ethyl acetate or water provides the oxygen atom that ends up in the product - ethyl alcohol. In this case, the use of water labelled with oxygen-18 reveals that the oxygen atom in the alcohol comes from the ethyl acetate molecule. Some of the prominent terms used in the methods to determine the reaction mechanism are discussed below,

3.1 Isotope Effect Isotope effect or kinetic isotope effect (KIE) refers to a difference in reaction rate due to a difference in the isotope present in the substrate. There is a definite change in the rate of a chemical reaction when one of the atoms in the reactants is substituted with one of its isotopes. So, if hydrogen atom is replaced by deuterium, then there is a difference in the value of rate of reaction. Definitionwise, it is the ratio of rate constants for the reactions involving the light (kL) and the heavy (kH) isotopically substituted reactants:

O-97 KIE =

kL kH

It is generally of two types, 1.

Primary kinetic isotope effect

2.

Secondary kinetic effect

1.

Primary kinetic isotope effects: A primary kinetic isotope effect may be found when a bond to the isotopicallylabeled atom is being formed or broken. For the previously mentioned nucleophilic substitution reactions, primary kinetic isotope effects have been investigated for both the leaving groups, the nucleophiles, and the α-carbon at which the substitution occurs. Kinetic isotope effects at the α-carbon can be used to develop some understanding into the symmetry of the transition state in S N 2 reactions, although this kinetic isotope effect is less sensitive than what would be ideal, also due to contribution from non-vibrational factors.

2.

Secondary kinetic isotope effects: A secondary kinetic isotope effect is observed when no bond to the isotopically substituted atom in the reactant is broken or formed in the rate-determining step of a reaction. By its definition, secondary kinetic isotope effects tend to be much smaller than primary kinetic isotope effects. However, since kinetic isotope effects can be calculated and measured to very high precision, secondary kinetic isotope effects are still very useful for elucidating reaction mechanisms. For the aforesaid nucleophilic substitution reactions, secondary hydrogen kinetic isotope effects at the α-carbon provide a direct means to distinguish between SN1 and SN2 reactions.

Value of Kinetic Isotope Effect It has been found that SN1 reactions typically lead to large secondary kinetic isotope effects, approaching to their theoretical maximum at about 1.22, while SN2 reactions typically yield kinetic isotope effects that are very close to or less than unity. Kinetic isotope effects that are greater than 1 are referred to as normal kinetic isotope effects, while kinetic isotope effects that are less than one are referred to as inverse kinetic isotope effects. In general, smaller force constants in the transition state are expected to yield a normal kinetic isotope effect, and larger force constants in the transition state are expected to yield an inverse kinetic isotope effect when stretching vibrational contributions dominate the kinetic isotope effect. ⊕

C +Nu: H(D)

SN1 –LG:

H(D) Nu H(D)

C LG

SN2 +Nu:

Nu

C

H(D) C

LG –LG:

The magnitudes of such secondary isotope effects at the α-carbon are largely determined by the C α − H(D) vibrations. For an SN1 reaction, since the carbon is converted into an sp2 hybridized carbonium ion during the transition state for the rate-redermining step with an increase in C α − H(D) bond order, an inverse kinetic isotope effect would be expected if only the stretching vibrations were important. The observed large normal

O-98 kinetic isotope effects are found to be caused by significant out-of-plane bending vibrational contributions when going from the reactants to the transition state of carbenium formation. For SN2 reactions, bending vibrations still play an important role for the kinetic isotope effect, but stretching vibrational contributions are of more comparable magnitude, and the resulting kinetic isotope effect may be normal or inverse depending on the specific contributions of the respective vibrations.

3.2 Kinetic and Thermodynamic Study of Organic Reactions Thermodynamic reaction control or kinetic reaction control in a chemical reaction can decide the composition in a reaction product mixture when competing pathways lead to different products and the reaction conditions influence the selectivity. The distinction is relevant when product A forms faster than product B because the activation energy for product A is lower than that for product B, yet product B is more stable. In such a case A is the kinetic product and is favoured under kinetic control and B is the thermodynamic product and is favoured under thermodynamic control. The conditions of the reaction, such as temperature, pressure, or solvent, affect which reaction pathway may be favored: either the kinetically controlled or the thermodynamically controlled one. Note this is only true if the activation energy of the two pathways differ, with one pathway having a lower Ea (energy of activation) than the other. Prevalence of thermodynamic or kinetic control determines the final composition of the product when these competing reaction pathways lead to different products. The reaction conditions as mentioned above influence the selectivity of the reaction - which pathway is taken. The concept of kinetic and thermodynamic control can be exemplified with the help of Diels-Alder reaction. The Diels-Alder reaction of cyclopentadiene with furan can produce two isomeric products. At room temperature, kinetic reaction control prevails and the less stable endo isomer 2 is the main reaction product. At 81°C and after long reaction times, the chemical equilibrium can assert itself and the thermodynamically more stable exo isomer1 is formed. The exo product (1) is more stable by virtue of a lower degree of steric congestion, while the endoproduct is favoured by orbital overlap in the transition state.

H O

O H H

+ 1

H

O 2

O-99

Exercise 5.

Write short note on stability of tropyliumcation over triphenylmethylcation. [Avadh 2008]

Write short notes on :

6.

Discuss the stability of carbocations. [Avadh 2009]

(i)

Inclusion compounds

7.

Write notes on the following :

(ii)

Nitrene

(i)

Carbene

(iii)

Free Radical

(iii)

Benzyne

(iv)

Carbene

(v)

Charge transfer complexes.

Long Answer Type Questions 1.

8. [Bund. 2009]

What are carbenes? What do you understand by 9. singlet and triplet carbenes? [Agra. 2008]

3.

Write notes on :

5.

Cross over experiment.

(ii)

Isotopic labeling.

(iii)

Trapping of reactive intermediates.

What are free radical? Discuss their characteristics and order of stability? [Rohd. 2011]

11.

Discuss the role of hyperconjugation to explain the relative stabilities of ethyl, isopropyl and tert.butylcarbocations. [D.D.U 2008]

What is carbocation? Describe the geometry of electron deficient carbon of carbocation.Explain the 12. mechanism of any such reaction in which the rate determining step is the slow ionization of substrate molecule which generates carbocation as an 13. intermediate. [Lko. 2009]

(ii)

CH3C +H2 is planar whereas CH3 CH2 and •

CH3 CH2 are pyramidal. Why?

CH(CH3 )2 ,CH3 ,C(CH3 )3 ,CH2CH3

What are carbocations? Arrange the 14. following carbocations in order of increasing stability:

Write short notes on

+

[D.D.U 2011]

(i)

Homolytic and heterolytiec fission

(ii)

Electrophiles and Nucleophiles

+

+

(CH3 )2 CH

[Hazaribagh 2008]

Very Short Answer Type Questions 1.

Write short notes on nitrene. [Bund. 2008; Meerut 2009,10,11; Agra 2008; Kanpur 2011]

Write a short note on isotopic effects. [Hazaribagh 2011]

Short Answer Type Questions 1.

2. 3.

4.

[D.D.U 2009]

Arrange with reason the following carbanions in order of their decreasing stability.

Explain heterolytic and hemolytic bond fission with suitable examples.

C6H5CH+ ,(C6H5 )2 CH,(CH3 )3 C, 2

(iii)

(ii) Carbene [Kashi 2011]

[Lko. 2010,11]

(i)

(i)

Benzyne

What is carbene? How is it formed? Give the examples of the reaction with structure and stability.

10.

[Avadh 2009]

4.

[Kashi 2010]

Write notes on the following : (i)

2.

(ii) Nitrenes

2.

Discuss the stability of carbonium ion. [Bund. 2009]

3.

Explain the following :

Write notes on the following : (i)

Carbonium ion.

(ii)

Free radical.

(i)

Electrophile

(ii) Nucleophile [Bund. 2010]

[Meerut 2008,10,11]

4.

Write a short note on homolytic fission of a covalent bond. [Meerut 2011] 5. What are cabocations? How do you account for the 6. relative stability of 1°, 2° and 3° carbocations. [Agra 2008] 7. Write short note on structure and stability of carbenes. [Avadh 2008]

Calculate the formal charge at the carbon of +CH3 and at the oxygen of H2O. [Bund. 2010] Explain homolytic and heterolytic bond fission. [Bund. 2010]

What are arynes?

[Bund. 2010]

Explain structure and stability of singlet and triplet carbene. [Bund. 2010]

O-100 8.

Why propyl carbonium ion less stable than 23. allylcarbonium ion? [Meerut 2008]

Types of hybridization of the following : (i)

Carbonium ion

9.

Write a note on carbene.

[Meerut 2009,11,12]

(ii)

Free radical

10.

Write a short note on carbanion. [Meerut 2009,10]

(iii)

Nitrene

11.

The order of stability amongst the carbanions is :

(iv)

Ethyene

Primary > Secondary >Tertiary.Explain.

24.

[Meerut 2011; Agra 2009]

12.

Which is more stable and why? ⊕

⊕

CH2 − CH = CH2 and CH2 = CH

25.

[Kanpur 2011]

13. 14. 15.

Why a tertiary carbocation is more stable than a primary carbocation? [Avadh 2010] 26. Write short note on homolytic and heterolytice fission. [Kashi 2008] Write short note on electrophiles and nucleophiles.

[Lko. 2010]

Pick up electrophile and nucleophile from following : (i)

:CCl2

(iii)

N2+

(ii) SO3 ••

(iv) R N H2

[Lko. 2010]

Why is a tertiary free radical more stable as compared to primary and secondary free radicals ? [Rohd. 2010]

27.

Explain free radicals with suitable example. [Hazaribagh 2009]

[Kashi 2008]

16.

17.

[Lko. 2010]

Why tertinary carbonium ion is more stable than secondary and primary carbonium ion?

Select electrophiles and nucleophiles from the 28. following:

Arrange the following according to their increasing stability :

NH3 , AlCl3 , BF3 , C −N,N +O2 and H2O

(i)

CH3 − CH2 − CH2 − CH2⊕

[Kashi 2011]

(ii)

CH3 − CH2 − CH ⊕ − CH3

Arrange the following carbanions according to their decreasing order of statbility :

(iii)

(CH3 )3 − C ⊕

29.

⊕

(i)

[Hazaribagh 2009]

What is benzyne? Give its structure.

CH3

[Hazaribagh 2011] ⊕

(ii)

CH3 − CH2

Objective Type Questions

⊕

(iii) 18.

CH3 − CH – CH3

[Kashi 2011]

What is A in the given reaction?

[Lko. 2008]

Multiple Choice Questions 1.

2 Dimerization A

19. 20. 21.

(a)

:CH3+

(c)

:CH2CH3

–

–

CH

(b) :CH R CH2 > R 2 CH > R 3 C + + + + R 3 C >R 2 CH > R CH2 > CH3 + + + + R 3 C >R 2 CH < R CH2 > CH3

(d)

None of these

+

(a) (b)

6.

+

+

13.

7.

8.

(c)

NH3

[Rohd. 2011]

(c)

CH+ 3 NO+ 2

[Rohd. 2013]

⊕

(a) p − NO2C6H4 − CH2 ⊕

(b) C6H5 CH2

⊕

(c) p − Cl2 – C6H4 − CH2 ••

⊕

(d) H3CO − C6H4 − CH2 Which among the following carbocations is most stable: ⊕

(a)

(b) C 6H5 − C H2 ⊕

(c) C6H5 − CH − C6H5 ⊕

(d) CH3 − C − CH3 | CH3

12.

(b) 2 and 4

1, 2 and 3

(d) 2, 3 and 4

Tropylium carbocation is …… stable than benzyl carbocation. Benzyne contains …… π electrons in the plane perpendicular to the molecular plane. A singlet carbene is …… stable than a triplet carbene. Out of carbocations and carbanions, it is …… which is more willing to rearrangement reactions. Propyl carbonium ion is …… stable than allylcarbonium ion.

7.

Conversion of alcohol to alkene in the presence of conc. sulfuric acid proceeds via…… .

8.

Chlorobenzene when treated with sodamide yields aniline via …… mechanism.

9.

Dehydration of alcohol is a/an …… reaction.

10.

Oxidation of aldehydes by Tollen's reagent is an example of ……..

True/False 1.

Addition reaction is just reverse of substitution reaction.

2.

Benzyne is a non aromatic compound.

3.

Ammonia is a nucleophile.

4.

All triplet carbenes are sp hybridized.

C

⊕

1, 2 and 4

(c)

4.

••

11.

Singlet carbene has planar geometry

(a)

Methyl carbocation is …… in shape.

Which of the following reagents may be described as electrophilic reagent? 5. (a) NH3 (b) AlCl3 (c) (d) CH3CH2OH CH 3 CH −2 6. Which of the following species is most stable ?

••

3.

[Rohd. 2013]

10.

••

Singlet C H2 is more stable than triplet C H2

Fill in the Blank

(b) Cl+ (d) NH3

••

C Cl2 is more stable than C Br2

(4)

Which of the following is not a nucleophile? 1. (a) NH3 (b) CO (c) CO2 (d) H2O [Rohd. 2012] 2. Which of the following is not an electrophile? (a)

9.

(b) N H4 (d) AlCl3

••

(3)

[Rohd. 2011]

••

C F2 is more stable than C Cl2

(2)

+

BF3

••

(1)

A nucleophile is (a)

Which among the following statement is correct?

5.

•

CH3 CH2 are pyramidal.

6.

Nucleophiles are Lewis acids.

7.

SO3 is an electrophile.

8.

Phenyl carbanion is less stable than benzyl carbanion.

9. Which one of the nitrogen containing compounds is an electrophile? (b) NH2–OH (a) NH 2 − NH 2 10. (c) NF3 (d) NH3

+

+

+

+

R 3 C > R 2 CH2 > R CH2 > CH3 is the order of stability of carbonium ions. Properly substituted carbanion can show optical activity.

O-102

Answers Objective Type Questions Multiple Choice Questions 1.

(d)

2.

(b)

3.

(b)

4.

(a)

5.

(b)

6.

(c)

7.

(b)

8.

(d)

9.

(b)

10.

(d)

11.

(a)

12.

(c)

13.

(a)

Fill in the Blank 1.

more

2.

6

3.

trigonal planar

4.

less

5.

carbocation

6.

less

7.

carbocation

8.

benzyne

9.

elimination

10.

redox reaction.

True/False 1.

False

2.

False

3.

True

4.

True

5.

True

6.

False

7.

True

8.

True

9.

True

10.

True

Hints and Solutions Long Answer Type Questions 5.

+

+

+

(iii) (C 6H5 )2 C H > C 6H5C H2 > (CH3 )3 C + > (CH3 )2 C H

Short Answer Type Questions

13. CH −3 > CH3CH2 − > (CH3 )2 CH − > (CH3 )3 C −

Very Short Answer Type Questions 12.

+

+

CH2 = CH − C H2 > CH2 = C H +

17.

+

+

CH3C H CH3 > CH3 − C H2 > C H3 19. CH3 22. CH2 ↑ ↓ → sp2 CH2 ↑ ↑ → sp ⊕

C H3 → sp3

Objective Type Questions Multiple Choice Questions 5. Electron donating group attached to carbonium ion increases stability. 11. + C is highly stable due to bent orbital overlap. 3

❍❍❍

Unit -I O-103

C HAPTER

3

Alkanes and Cycloalkanes

1. Alkanes The alkanes or paraffins (a still-used historical name that also has other meanings as well) are the saturated hydrocarbons comprising of homologous series with the general formula, CnH2n+2 where n is an integer. Many of them occur naturally and the chief source of the alkanes is mineral oil or petroleum. Alkanes consist of hydrogen and carbon atoms and all bonds are single bonds. Alkanes belong to a homologous series of organic compounds in which the members differ by a molecular mass of 14.00u (mass of a methanediyl group, –CH2–, one carbon atom of mass 12.01u, and two hydrogen atoms of mass ≈1.01u each). Each carbon atom has 4 bonds which may be either C-H or C-C bonds, and each hydrogen atom is joined to a carbon atom. A series of linked carbon atoms is known as the carbon skeleton or carbon backbone. The number of carbon atoms is used to define the size of the alkane. An alkyl group, generally abbreviated with the symbol R, is a side-chain that, like an alkane, consists solely of single-bonded carbon and hydrogen atoms, for example methyl or ethyl group. The simplest possible alkane is methane, CH4. There is no limit to the number of carbon atoms that can be linked together, the only limitation being that the molecule is acyclic, is saturated, and is a hydrocarbon. Saturated oils and waxes are examples of larger alkanes where the number of carbons in the carbon backbone is greater than 10. Alkanes are not very reactive and have little biological activity. All alkanes are colourless and odourless. Alkanes normally are viewed as a molecular tree upon which can be hung the more biologically active portions (functional groups) of the molecule.

1.1 Isomerism and Nomenclature The IUPAC nomenclature for alkanes is based on identifying hydrocarbon chains. Unbranched, saturated hydrocarbon chains are named systematically with a Greek numerical prefix denoting the number of carbons and the suffix "-ane".

1

2 3

4

3 4

2

1

3–methylbutane 2–methylbutane (correct numbering) (incorrect numbering)

O-104 Some examples are,

H3C

H3C

CH3

CH3

H3C

CH3

H3C

CH3

H3C

Propane

C–3

CH3

H3C C–4

CH3 2–methyl propane

CH3

H3C

2–methyl butane

2–dimethyl butane

Butane CH3

CH3 CH3 CH3 CH3 CH3

H3C

CH3

Pentane

C–5

H3C C–7

H3C

CH3 CH3

H3C C–10

CH3 CH3 2,2,3,3–tetramethylbutane

CH3 CH3 CH3 CH3

Nonane

C–9

CH3

H3C

CH3 CH3 CH3 2,2,3,3,4,4–hexamethylpentane

Octane

C–8

CH3 CH3 CH3

H3C

CH3 CH3

H3C

CH3

CH3 CH3 CH3

Heptane

CH3

2,2,3,3,4,4,5,5,6,6–decamethylheptane 2,2–dimethylpropane neopentane

Hexane

C–6

H3C

CH3 CH3 CH3 CH3 CH3

CH3 H3C

CH3

H3C

CH3

H3C CH3 CH3 CH3 CH3

2,2,3,3,4,4,5,5–Octamethylhexane

Decane

Numbers in the name, referring to which carbon a group is attached to, should be as low as possible, so that 1- is implied and usually omitted from names of organic compounds with only one side-group. Some more examples are, 7 6 5 4 3 2 1 CH3–CH2–CH–CH–CH–CH–CH3 5' CH3 CH2 CH3CH3 6' CH3 7'CH 3

CH3 CH CH2 CH2 methylcycloproane

2,3,5–trimethyl–4–propylheptane (NOT: 2,3–dimethyl–4–sec–butylheptane)

CH3

CH3

CH3–CH–CH2–CH2–CH–CH2–CH–CH2–CH3 CH–CH3 CH2–CH3 5–sec–butyl–2, 7–dimethylnonane

CH3–CH2–CH–CH–CH2–CH3 CH3 CH2 CH3

3–ethyl–4–methylhexane

O-105 Alkanes with more than three carbon atoms can be arranged in various different ways, forming structural isomers. The simplest isomer of an alkane is the one in which the carbon atoms are arranged in a single chain with no branches. This isomer is sometimes called the n-isomer, here n is used for "normal", although it is not necessarily the most common. However the chain of carbon atoms may also be branched at one or more points. The number of possible isomers increases rapidly with the number of carbon atoms. For example C1 is methane only,C2 is ethane only, C3 is propane only, C4 has 2 isomers : n-butane and isobutane, C5 has 3 isomers: pentane, isopentane, and neopentane, C6 has 5 isomers: hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane and C12 has a staggering 355 isomers. Branched alkanes can give rise to chirality. For example 3-methylhexane and its higher homologues are chiral due to their stereogenic center at carbon atom number 3. In addition to these isomers, the chain of carbon atoms may form one or more loops. Such compounds are called cycloalkanes.

1.2 Methods of Preparation of Alkanes A number of methods are available for the synthesis of alkanes. Some of these are discussed below,

1.2.1 Methods Involving No Change in the Carbon Skeleton Substitution of halogen by hydrogen can be done in as discussed below, 1.

Reduction by dissolving metals: Reduction can be achieved by dissolving metals like zinc and acetic or hydrochloric acid, zinc and sodium hydroxide, zinc-copper couple and ethanol, etc. Initially, it was presumed that nascent hydrogen is the actual reducing agent in this reaction. Now it is believed that there is an electron–transfer from the metal to the substrate leading to the formation of carbanion which is followed by the abstraction of a proton from the solvent. Thus, reduction with a zinc–ethanol couple may be formulated as Zn → Zn2+ + 2e– e−

RX + e–→ X– + R.   → R :– R:– + C2H5OH → R—H + –OC2H5 2.

Reduction by reducing agents like LiAlH4, NaBH4 etc.: Primary and secondary alkyl halides are readily reduced to alkanes by lithium aluminium hydride (LiAlH4) whereas reduction of tertiary halides with LiAlH4 gives mainly alkenes. On the other hand, sodium borohydride (NaBH4) reduces secondary and tertiary halides, but not primary, whereas TPTH, triphenyltin hydride (Ph3SnH) reduces all three types of alkyl halides. So each reducing agent is specific in its action.

(i)

4R – X + LiAlH4 → 4R–H + LiX + AlX3 (X ≠ F) Or,R–X + H:(–) →R–H + X–

Hydride ion comes from LiAlH4. (ii)

R–X + (n - C4H9)3SnH →R – H + (n - C4H9)3SnX

3.

Using organometallic compounds: Alkyl halides react with either Mg or Li in dry ether to give organometallic compounds having a carbanionic site. These organometallic compounds upon hydrolysis yield corresponding alkanes. dry ether

R – X + 2Li     → R– Li+ + LiX

O-106 then,

R– Li+ + H2O →R – H + LiOH dry ether

R – X + Mg     → R– Mg+X then,

RMgX + H2O → R – H + MgX(OH) strong acid weak acid

The net effect is replacement of X by H. This reaction can be done with any compound that is more acidic than alkane. Thus the net effect of the reaction is the displacement of a weak acid from its salt by a strong acid. Using this reaction, we determine the number of acidic or active hydrogens present in a given compound. This quantitative estimation of number of active hydrogens is called Zerewittnoff's method. For example, O 3RMgX+HC≡C–CH(OH)–CO2H→3R–H+XMgC≡C–CH–C OMgX

OMgX

Three moles of alkane formed shows that the compound contains three active or acidic hydrogens.

1.2.2 Hydrogenation of Alkenes in the Presence of Pd or Pt This addition is an example of heterogeneous catalysis involving syn addition. CH3

CH3 CH3–C=CH2+CH2+H2

Pt

CH3–CH–CH3

Dihydrogen in the presence of Raney Ni is also used for the purpose of reduction. Raney Ni is use of nickel aluminium alloy in the presence of caustic soda. The stereospecificity of the reaction is that the addition of hydrogen to the double bond occurs in syn fashion without disturbing the configuration at the chiral carbon. For example, CH3 H H

CH3

CH3

OH

OH

D2 /Ni

D

H CH=CH2

H

OH

D

H

+

CH2D

CH2D Diastereomers

The addition of both the deuterium atom occurs from the same side. In some molecules, the attack is from the bottom side and in other molecules, D2 attacks from the top side leading to the formation of two isomers called diastereomers.

1.3 Methods Involving Change in the Carbon Skeleton 1.3.1 Methods in Which number of Carbons Atoms Increases w.r.t. Starting Compound 1.

Wurtz reaction: An ethereal solution of an alkyl halide, preferably the bromide or iodide, is treated with sodium, when alkane is obtained. For example,

O-107 dry ether

R1 − X + R 2 − X + 2Na     → R1 − R 2 + 2NaX In this reaction , two R groups are coupled by reacting RBr, RCl or RI with Na or K. The yields of the product are best for 1° alkyl halides (60%) and least for 3° alkyl halides (10%). Looking further in the above reaction, it was found that in addition to the desired alkane R 1 − R 2 , there will also be present the alkanes R 1 − R 2 and R 2 − R 2 . Unsaturated hydrocarbons are also obtained. Obviously, then, the best yield of an alkane will be obtained when R 1 & R 2 are same, i.e., when the alkane contains an even number of carbon atoms and is symmetrical. It has been found that the Wurtz reaction gives good yields only for 'even carbon' alkanes of high molecular weight, and that the reaction generally fails with tertiary alkyl halides. The reaction involves the intermediate formation of free radicals. The suggested mechanism for the Wurtz reaction is shown as Na → Na + + e− R − Br + e − → R • + Br – 2R • → R − R Na + + Br − → NaBr (Alkane with even carbon atoms) It is important that note metal other than sodium, which can be employed in Wurtz reaction are Ag and Cu in finally divided state. 2.

Corey House synthesis: A superior method for coupling is the Corey House synthesis which could be employed for obtaining alkanes containing odd number of carbon atoms also known as unsymmetrical alkanes. For example, R – X + 2Li → R–Li++ LiX 2R–Li+ + CuX → (R)2CuLi + LiX R2CuLi + 2R' – X → 2R – R' + LiX + CuX

Some examples of the reaction are shown below, CH3I

1. Li 2. CuI

(CH3)2CuLi

(CH3)2CuLi+CH3(CH2)8CH2I

0°

1. Li CH3–CH2–Cl 2. CuI (CH3–CH2)2LiCu

CH3(CH2)9CH3 CH3–CH2–Cl

CH3–CH2–CH2–CH3

3.

Kolbe's electrolytic method: A concentrated solution of the sodium or potassium salt of a carboxylic acid or a mixture of carboxylic acids is electrolysed. For example, R 1CO2K + R 2 CO2K + 2H2O → R 1 – R 2 + 2CO2 + H2 + 2KOH

If R1 and R2 are different, then hydrocarbons R 1– R 1 and R 2 – R 2 are also obtained along with R 1 – R 2 . Earlier several mechanisms have been proposed for the Kolbe reaction. The free– radical theory is the one

O-108 now favoured, having strong evidences in support of it. For example, when sodium propionate is electrolysed, n–butane, ethane, ethylene and ethyl propionate are obtained. Mechanism C2H5CO2Na → C2H5CO −2 +Na + At anode, the propionate ion discharges to form a free radical. C2H5CO −2 → C2H5CO •2 +e– This propionate free radical then breaks up into the ethyl free radical and carbon dioxide. C2H5CO •2 → C2H •5 + CO2 Then, the following type of dimerisation can take place, (i)

2C2H •5 → C4H10

(ii)

C2H •5 +C2H •5 → C2H6 + C2H4

(iii)

C2H •5 +C2H5CO •2 → C2H5CO2C2H5

Reaction (i) gives n butane; (ii) gives ethane and ethylene by disproportionation and (iii) gives ethyl propionate. Thus at anode, gases evolved are CO2, ethane, ethene and butane. At cathode, hydrogen ions accept an electron and are converted to H2 gas. 2H+ + 2e– → H2

1.3.2 Methods in Which Number of Carbon Atoms Decreases w.r.t. Starting Compound: 1.

Decarboxylation of carboxylate salts: By heating a mixture of the sodium salt of a carboxylic acid and soda-lime, alkanes can be obtained. ∆

R CO2Na + NaOH (CaO) → RH + Na2CO3 This process of eliminating CO2 from a carboxylic acid is known as decarboxylation. This reaction can be employed for decreasing the length of carbon chain i.e. to descend a homologous series. This decarboxylation reaction probably involves following mechanistic steps. Mechanism O

O R–C

OH OH

2.

R–C O

CO2 + R

H–OH R–H+OH

Homolytic bond dissociation energies and the relative stabilities of radicals: Homolytic bond dissociation energies provide a convenient way to estimate the relative stabilities of radicals. The energy required to break covalent bonds homolytically is called homolytic bond dissociation energy and abbreviated by the symbol ∆H°. In the reaction given below, the designated C – H bonds are broken homolytically and the values of ∆H° for these bonds are also given. CH3CH2CH2–H → CH3CH2CH •2 + H • ∆H° = + 98 k calmol–1 (n–propyl radical–a 1° radical)

O-109 ∆H°=+94.5 k cal mol–1

CH3CHCH3+H

CH2–C–CH2–H

(Isopropyl radical –a 2° radical)

H

These reactions differ in the amount of energy required and in the type of carbon radical being produced. More energy must be supplied to produce a primary alkyl radical from propane than is required to produce a secondary carbon radical from the same compound. This mean that the primary radical has absorbed more energy and thus has greater potential energy. As the relative stability of a chemical species is inversely related to its potential energy, the secondary radical must be more stable than the primary radical by 3.5 k cal mol–1 CH3 CH3 (98–94.5) CH3–C–CH2–H

∆H°=+91 k cal mol–1

CH3–C–CH3+H tert–butyl radical (3°)

H

CH3

CH3

CH3–CH–CH2+H

CH3–C–CH2–H

∆H°=+98 k cal mol–1

Isobutyl radical (1°)

H

The tertiary radical is more stable than the primary radical by 7 k cal mol–1. The kind of pattern that we find in these examples is found with alkyl radicals generally. Overall their relative stabilities are as follows Tertiary > Secondary > Primary > Methyl

Physical Properties of the Alkanes 1.

The first four alkanes: Methane to butane are colourless gases, the next thirteen - pentane to heptadecane are colourless liquids and those containing 18 carbon atoms and more are solids at ordinary temperatures. Table 1:Physical properties of alkanes Alkane

Formula

Boiling point [°C]

Melting point [°C]

Density [g·cm–3] (at 20 °C)

Methane

CH4

–162

–182

gas

Ethane

C 2 H6

–89

–183

gas

Propane

C 3 H8

–42

–188

gas

Butane

C4H10

0

–138

gas

Pentane

C5H12

36

–130

0.626 (liquid)

Hexane

C6H14

69

–95

0.659 (liquid)

Heptane

C7H16

98

–91

0.684 (liquid)

Octane

C8H18

126

–57

0.703 (liquid)

Nonane

C9H20

151

–54

0.718 (liquid)

O-110 Decane

C10H22

174

–30

0.730 (liquid)

Undecane

C11H24

196

–26

0.740 (liquid)

Dodecane

C12H26

216

–10

0.749 (liquid)

Hexadecane

C16H34

287

18

Icosane

C20H42

343

37

solid

Triacontane

C30H62

450

66

solid

Tetracontane

C40H82

525

82

solid

Pentacontane

C50H102

575

91

solid

Hexacontane

C60H122

625

100

solid

2.

0.773 (liquid)

Boiling and melting point: The boiling point of alkanes show gradual rise as the carbon content increases. In general, the boiling point difference between two successive members of the homologous series, except for the first few members, is about 20–30°C. Among the isomeric alkanes, the straight chain isomer has a higher boiling point than the branched chain isomer. The greater the branching of the chain, the lower the boiling point.

In fact, the lowering of boiling point with the branching of the carbon chain is a feature characteristic of all the families of organic compounds. The van der Waal's forces which hold non–polar molecules are weak and have a very short range. Therefore, within a family of compounds the strength of intermolecular forces would be directly proportional to the size or the surface area of the molecule. In other words, larger the molecule, stronger the intermolecular forces would be. The process of boiling requires overcoming these intermolecular forces of attraction. As the molecules become larger, the intermolecular forces increase and the boiling points must rise with increase in the number of carbons atoms. As the branching increases in a molecule, its shape approaches that of a sphere and there is a reduction in surface area. This renders the intermolecular forces weaker and they are overcome at relatively lower temperature. Therefore, a branched–chain isomer must boil at a temperature lower than that of a straight chain isomer. Their melting points also show a rise with the increasing number of carbon atoms, but the rise is not as regular as in the case of boiling points. It is, however, significant that as we move from an alkane having an odd number of carbon atoms to a higher alkane, the rise in melting point is much higher than that when we move

O-111 up from an alkane with an even number of carbon atoms. The intermolecular forces in a crystal depend not only on the size of the molecules but also on how they are packed into a crystal. During melting, these intermolecular forces have to be overcome. Since breaking of crystal structure is a more complicated process, it is understandable that the rise in melting point with increasing molecular weight is not as regular as in the case of boiling points. 3.

Solubility: Alkanes are made up of carbon and hydrogen atoms only. Since these two elements have almost similar electronegativities, alkanes are non–polar. Therefore, non–polar alkanes are soluble in non–polar solvents like carbon tetrachloride, benzene, etc. but insoluble in polar solvents like water, alcohol, etc.

4.

Density: The densities of alkanes show a definite rise with increasing molecular weight, but they reach a limiting constant value of about 0.8 with n hexadecane (C16H34). Thus, alkanes are always lighter than water.

5.

Molecular geometry: The molecular structure of the alkanes directly affects their physical and chemical characteristics. It is derived from the electron configuration of carbon, which has four valence electrons. The carbon atoms in alkanes are always sp3 hybridized, that is to say that the valence electrons are said to be in four equivalent orbitals derived from the combination of the 2s orbital and the three 2p orbitals. These orbitals, which have identical energies, are arranged spatially in the form of a tetrahedron, the angle of 109º.28' between them. H H

H

H Fig. 2: The tetrahedral structure of methane.

An alkane molecule has only C - H and C - C single bonds. The former result from the overlap of a sp3-orbital of carbon with the 1s-orbital of a hydrogen; the latter by the overlap of two sp3-orbitals on different carbon atoms. The bond lengths amount to 1.09×10–10 m for a C - H bond and 1.54×10–10 m for a C - C bond. The spatial arrangement of the bonds is similar to that of the four sp3-orbitals-they are tetrahedrally arranged. Structural formulae that represent the bonds as being at right angles to one another, while both common and useful, do not correspond with the reality. Chemical Properties of the Alkanes 1.

Nitration: Under certain conditions alkanes react with nitric acid, when a hydrogen atom will be replaced by a nitro–group, NO2. This process is known as nitration. Nitration of the alkanes may be carried out in the vapour phase between 150° and 475°C, when a complex mixture of mononitroalkanes is obtained. The mixture consists of all the possible mononitroderivatives and the nitro compounds formed by every possibility of chain fission of the alkane. For example, propane gives a mixture of 1–nitropropane, 2–nitropropane, nitroethane and nitromethane. HNO3

NO2 |

CH3CH2CH3   → CH3CH2CH2NO 2 + CH3 − CHCH3 + C 2H5NO 2 + CH3NO 2 400 ° C

O-112 2.

Sulphonation: It is the process of replacing hydrogen atom by a sulphonic acid group, SO3H. Sulphonation of a normal alkane from hexane onwards may be carried out by treating the alkane with oleum (fuming sulphuric acid). The ease of replacement of H– atoms in tertiary compounds is more than secondary, and in secondary compounds replacement of H–atoms by sulfonic acid group is easier than primary. Replacement of a primary hydrogen atom in sulphonation is very slow indeed. Isobutane, which contains a tertiary hydrogen atom, is readily sulphonated to give t–butylsulphonic acid. (CH3)3CH + H2SO4/SO3 → (CH3)3CSO3H + H2SO4

3.

Oxidation: All alkanes readily burn in excess of air or oxygen to form carbon dioxide and water. CnH2n+2+

(3n + 1) (2n + 2) O2(g) → nCO2(g) + H2O(l) 2 2

On the other hand, controlled oxidation under various conditions, leads to different products. Extensive oxidation gives a mixture of acids consisting of the complete range of C1 to Cn carbon atoms. Less extensive oxidation gives a mixture of products in which no chain fission has occurred. Under moderate conditions mixed ketones are the major products, and oxidation in the presence of boric acid produces a mixture of secondary alcohols. The oxidation of alkanes in the vapour state occurs via free radicals, Oxidising reagents such as potassium permanganate readily oxidises a tertiary hydrogen atom to a hydroxyl group. For example, isobutane is oxidised to t–butanol. KMNO

4 → (CH ) COH (CH3)3CH + [O]      3 3

4.

Halogenation: Chlorination is brought about by photo irradiation, heat or catalysts, and the extent of chlorination depends largely on the amount of chlorine used. A mixture of all possible isomeric monochlorides is obtained, but the isomers are formed in unequal amounts, due to difference in reactivity of primary, secondary and tertiary hydrogen atoms. The order of ease of substitution of different types of H is Tertiary Hydrogen > Secondary Hydrogen > Primary Hydrogen Chlorination of isobutane at 25°C gives a mixture of two isomeric monochlorides. CH3

CH3 CH3–CH–CH2–Cl and (64%)

CH3–C–CH3 Cl (36%)

The tertiary hydrogen is replaced about 5 times as fast as primary hydrogen. Bromination is similar to chlorination, but not so vigorous. Iodination is reversible, but it may be carried out in the presence of an oxidising agent such as HIO3, HNO3 etc., which destroys the hydrogen iodide as it is formed and so drives the reaction to the right. CH4+I2 5HI+HIO3

CH3I+HI 3I2+3H2O

O-113 Iodides are more conveniently prepared by treating the chloro or bromo derivative with sodium iodide in methanol or acetone solution. For example, acetone

RCl + NaI   → RI + NaCl This reaction is possible because sodium iodide is soluble in methanol or acetone, whereas sodium chloride and sodium bromide are not. This reaction of halide exchange is known as Conant–Finkelstein reaction. Direct fluorination is usually explosive. So, special conditions are necessary for the preparation of the fluorine derivatives of the alkanes. UV light

RH + X2     → RX + HX or ∆

Reactivity of X2: F2 > Cl2 > Br2 > I2 The mechanism of chlorination of methane is as follows Chain initiation step: Cl – Cl     → 2Cl • ; ∆H = + 243 kJ mol–1 UV light or ∆

The required enthalpy comes from Ultraviolet (UV) light or heat supplied. Chain propagation step: (i)

H3C – H + Cl • → H3C • + H – Cl ; ∆H = – 4 kJmol–1 (rate determining step)

(ii)

H3C • + Cl – Cl → H3C – Cl + Cl • ; ∆H = – 96 kJmol–1

The sum of the two chain propagation steps in the overall reaction is CH4 + Cl2 → CH3Cl + HCl

; ∆H =–100 kJmol–1

In propagation steps, the same free radical intermediates, here Cl • and H3C • , being formed and consumed. Chain termination step: Chains terminate on those rare occasions when any two free-radical intermediates collide to form a covalent bond. Cl • + Cl • → Cl2 H3Cl • + Cl • → CH3 – Cl H3C • + • CH3 → H3C CH3 Radical inhibitors stop chain propagation by reacting with free radical intermediates. For example, ••

••

••

••

••

••

••

••

H3C • + • O – O • → CH3 – O – O •

O-114 The potential energy curve for the halogenation (chlorination) of alkane is shown as Eact

Potential energy

R–H+Cl HCl+R R+Cl2

R–Cl+Cl Progress of reaction Fig. 3: Potential energy curve for chlorination of alkane

In more complex alkanes, the abstraction of each different kind of hydrogen atom gives a different isomeric product. Three factors determine the relative yields of isomeric product. (i)

Probability factor: This factor is based on the number of each kind of hydrogen atoms in the alkane molecule. For example, in CH3CH2CH2CH3 there are six equivalent 1° H and four equivalent 2° H. The probability of abstracting a 1°H to 2° H is 6 to 4, or 3 to 2.

(ii)

Reactivity of H • : The order of reactivity of hydrogen atoms is 3° > 2° > 1°.

(iii) Reactivity of X • : The more reactive Cl • is less selective and more influenced by the probability factor. The less reactive Br • is more selective and less influenced by the probability factor, as summarized by the Reactivity-Selectivity Principle. If the attacking species is more reactive, it will be less selective, and the yields will be determined by the probability factor as well as reactivity of hydrogen atoms while if the species attacking is less reactive and more selective, the yield of the product is governed exclusively by reactivity of hydrogen atoms. Cl / hν

CH3CH2CH2CH3  2→ CH3CH2CH2CH2Cl + CH3CH2CH(Cl)CH3 n butane

25 ° C

(28%)

(72%)

Cl / hν

(CH3)2CHCH3  2→ (CH3)2CHCH2Cl + (CH3)3C – Cl isobutane

25 ° C

(64%)

(36%)

Br2 / hν

CH3CH2CH2CH3   → CH3CH2CH2CH2Br + CH3CH2CH(Br)CH3 n butane

127° C

(2%)

(98%)

Br2 / hν

(CH3)2CHCH3   → (CH3)2CHCH2Br + (CH3)3C Br isobutane

127° C

traces

(over 99%)

In the chlorination of isobutane abstraction of one of the nine primary hydrogens leads to the formation of isobutyl chlorides, whereas abstraction of a single tertiary hydrogen leads to the formation of tert-butyl chloride. The probability favours formation of isobutyl chloride by the ratio of 9:1. But the experimental results show the ratio roughly to be 2:1 or 9:4.5. Evidently, about 4.5 times as many collisions with the tertiary

O-115 hydrogen are successful as collisions with the primary hydrogens. The Eact is less for abstraction of a tertiary hydrogen than for the abstraction of a primary hydrogen.The rate of abstraction of hydrogen atoms is always found to follow the sequence 3° > 2° > 1°. At room temperature (25°C), the relative rates in chlorination are 5.0 : 3.8 : 1.0 respectively for 3°, 2° and 1° hydrogen atoms. Using these values, we can predict the ratio of isomeric chlorination products from a given alkane. For example, Cl

2 CH3CH2CH2CH3   → CH3CH2CH2CH2Cl + CH3CH2CHClCH3

light, 25 ° C

n – butylchloride number of 1° H reactivity of 1° H = × sec − butyl chloride number of 2 ° H reactivity of 3 ° H =

28% 6 1.0 3 equivalent to × = 72% 4 3.8 7.6

Inspite of these differences in reactivity, chlorination rarely yields a great excess of any single isomer. In most cases, both the products are formed in considerable amounts. The same sequence of reactivity, 3° > 2° > 1°, is found in bromination, but with enormously larger reactivity ratios. At 127° C the relative rates per hydrogen atom in bromination are 1600 : 82 :1 respectively for 3°, 2° and 1° hydrogen atoms. Here, differences in reactivity are so marked that it outweighs probability factor. Hence bromination almost exclusively gives selective product. In bromination of isobutane at 127°C, isobutyl bromide number of 1° H reactivity of 1° H = × tert − butyl bromide number of 3 ° H reactivity of 3 ° H =

9 1 9 0.5% × = equivalent to 1 1600 1600 99.5%

Hence, tert butyl bromide happens to be the exclusive product (over 99%) with traces of isobutyl bromide. The reason for the higher selectively in bromination as compared to chlorination is due to the following explanation. According to the general principle, for comparable reactions, the more endothermic (or less exothermic) reaction has a transition state (TS) which more closely resembles the intermediate. The less endothermic (or more exothermic) reaction has a TS which less closely resembles the intermediate and may more closely resemble the ground state (reactants). Since attack by Br • on an alkane is more endothermic than attack by Cl • , its TS shows more C–H bond breaking and more H– Br bond formation. Any stabilization in the intermediate radical also occurs in the corresponding TS. Therefore, a TS leading to a 3° R • , has a lower enthalpy than one leading to a 2°R • , which in turn has a lower enthalpy than one leading to a 1°R • , the relative rates of H–abstraction by Br • are 3° > 2° > 1°. The TS for H abstraction by Cl • has less C–H bond breaking and less H–Cl bond formation. The nature of the incipient radical has less effect on the enthalpy of the TS and on the rate of its formation. Hence there is less difference in the rate of formation of the three kinds of R • 's. In the attack by the comparatively unreactive bromine atom, the transition state is reached late in the reaction process, after the alkyl group has developed considerable radical characater. In the attack by the highly reactive chlorine atom, the transition state is reached early, when the alkyl group has gained very little radical character. Thus bromination is more selective than chlorination.

O-116

δ R

R–H+Br low reactivity, high selectivity

R–H+Cl

δ Br

H

R+H+Br

Transition state reached late, much radical character δ R

H

δ Br

R+H–Cl

Transition state reached early, little radical character

2. Cycloalkanes Cycloalkanes also called naphthenes(but different from naphthalene)are saturated hydrocarbons in which the carbon atoms are joined by single covalent bonds to form a ring. Cycloalkanes consist of only carbon (C) and hydrogen (H) atoms and there are no multiple C-C bonds to hydrogenate. They are also called alicyclic compounds. The prefix ali-is added because of their similarity to aliphatic compounds. The first member of the series is cyclopropane,C3H6 . Cycloalkanes might have one or more rings of carbon atoms in the chemical structure of their molecules. A general chemical formula for cycloalkanes would be CnH2(n+1-g) where n = number of C atoms and g = number of rings in the molecule. Cycloalkanes with a single ring are named analogously to their normal alkane counterpart of the same carbon count. So, these are named as cyclopropane, cyclobutane, cyclopentane, cyclohexane, etc. The larger cycloalkanes, with greater than 20 carbon atoms are typically called cycloparaffins. Cycloalkanes are classified into small, common, medium, and large cycloalkanes. So, cyclopropane and cyclobutane are the small ones, cyclopentane, cyclohexane, cycloheptane are the common ones. On the other hand, cyclooctane through cyclotridecane are the medium ones, and the rest are the large ones.

2.1 Nomenclature The IUPAC rules for naming cycloalkanes are as follows :

2.1.1 Monocyclic Alkanes Rule 1: The name of an unsubstituted cycloalkanes is obtained by attaching the prefix cyclo-to the name of the corresponding normal alkane having the same number of carbon atoms as in the ring. For convenience and simplicity, cycloalkanes are often represented by simple geometric figures :

cyclopropane

cyclobutane

cyclopentane

cyclohexane

cycloheptane

A triangle is used for for cyclopropane, a square for cyclobutane, a pentagon for cuyclopentane a hexagon for cyclohexane and so on. It is understood that each corner represents one carbon and two hydrogens.

O-117 Rule 2 : Substituents on the ring are named and their positions are indicated by numbers. The ring is numbered so that the carbon bearing the substituents will have the lowest numbers. If possible stereochemistry is also indicated. Me

Br

Me

Et

OH 3–Bromo cyclopentene

3–Methylcyclo hexanol

Cl

H

H

Cl Trans-1,2dichloro cyclopropane

1– Ethyl– 3– methyl cyclohexane.

Rule 3: This rule primarily deals with naming of alkyl-subsituted cycloalkanes. One has to count the number of carbon atoms in the ring and also in the largest alkyl substituent. If the number of carbon atoms in the ring is equal to or greater than the number in the largest substituent, the compound is named as alkyl-substituted cycloalkane (case 1 shown below). If the number of carbon atoms in the ring is less than the number of carbon atoms in the substituent, it is named as a cycloalkyl-substituted alkane (case 2).

Ethyl cyclo propane (case 1)

1–Cyclopropyl butane (case 2)

1–P ropyl Cyclopropane (case 3)

If there is a tie between the number of carbons on the cycloakanes part of the molecule and the acyclic alkane part, choose the cycloalkane as the parent or main chain(case 3).Some other examples are, CH3–CH–CH2–CHO

Br 2–Cyclopropyl 1–Bromo–2–butyl butane cyclobutane.

3–Cyclobutyl butanal

2.1.2 Polycyclic Alkanes The naming of polycyclic alkanes such as bicyclic alkanes and spiro alkanes is more complex, with the base name indicating the number of carbons in the ring system, a prefix indicating the number of rings (e.g., "bicyclo"), and a numeric prefix before that indicating the number of carbons in of each ring, excluding of junctions. For instance, a bicyclooctane that consists of a six-member ring and a four-member ring, which share two adjacent carbon atoms that form a shared edge, is Fig. 4: [4.2.0]–bicyclooctane [4.2.0]-bicyclooctane.

O-118 That part of the six-member ring, excluding of the shared edge has 4 carbons. That part of the four-member ring, exclusive of the shared edge, has 2 carbons. The edge itself, excluding of the two vertices that define it, has 0 carbons. As IUPAC nomenclature system is constantly being revised, so in the above example [4.2.0]-bicyclooctane would be written as bicyclo[4.2.0]octane. Another example of the IUPAC method is discussed below. In this example the base name is listed first, which indicates the total number of carbons in both rings including the carbons making up the shared edge is heptane, which means hept or 7 carbons, and ane, which indicates only single bonding between carbons. Then in front of the base name is the numerical prefix, which lists the number of carbons in each ring, excluding the carbons that are shared by each ring, plus the number of carbons on the bridge between the rings. In this case there are two rings with two carbons each and a single bridge with one carbon, excluding the carbons shared by it and the other two rings. There is a total of three numbers and they are listed in descending order separated by dots, thus: [2.2.1]. Before the numerical prefix is another prefix indicating the number of rings (here the prefix is bicyclo). Thus, the name is bicyclo [2.2.1]heptane.

Fig. 5: Bicyclo [2.2.1] heptane

Methods of Preparation There are a number of methods available for the preparation of cycloalkanes. Some of the important methods of preparation are discussed below, 1.

From calcium salts of dicarboxylic acids : When the calcium or barium salts of dicarboxylic acids are heated, cyclic ketones are formed. The cyclic ketones can be readily converted into the corresponding cycloalkanes by Clemmensen reduction. O CH2–C–O– CH2–C–O–

Ca2+

∆

–CaCO3

O

Clemmensen reduction

Cyclopropane

O Calcium succienate

Another example is O CH2–CH2–C–O– CH2–CH2–C–O– O Calcium adipate

Ca2+

∆

O

Clemmensen reduction

Cyclopentane

O-119 2.

From dihalides : Terminal dihalides when allowed to react with sodium or zinc from cycloalkanes. This reaction is an extension of Wurtz reaction. However, there is a limitation that the reaction is useful for the preparation of 3-to 6 membered rings only. CH2–CH2–Cl

+2Na

∆

+2NaCl

CH2–CH2–Cl 1,4– Dichlorobutane –Cl

Another example is:

∆

+2Na

+2NaCl

–Cl 1,3–Dichloropropane

3.

From esters of dicarboxylic acids: Esters of dicarboxylic acids when treated with sodium metal undergo intramolecular acetoacetic ester condensation and a β − kentoester is formed. This is an important name reaction termed as Dieckmann Reaction. The β −ketoesters on hydrolysis yield cyclic ketones. These on reduction yield the corresponding cycloalkanes. O CH2–CH2–C–OE CH2–CH–H

Na

O

–C2H5OH

COOEt Diethyl adipate

H3O+

O

COOH

COOEt

heat, –CO2 Zn(Hg) conc HCl

O

Cyclopentane

4.

From aromatic hydrocarbons: Six-membered cycloalkanes can be prepared by the catalytic reduction of benzene and its derivatives. The reaction scheme is shown below, H2, Ni ∆, High pressure

Benzene

Cyclohexane

CH3

CH3 H2, Ni ∆, High pressure

Toluene

Methylcyclohexane

O-120 5.

From alkenes: When alkenes are allowed to react with methylene iodide (CH2I2) in the presence of zinc-copper couple, cyclopropane derivatives are formed. Some of the examples are shown here, CH3–CH2CH=CH2+CH2I2

Zn+Cu Ether

CH3–CH2–CH–CH2+ZnI2

But–l–ene

CH2 Ethyl cyclopropane

CH3–CH=CH–CH3+CH2I2

Zn+Cu Ether

CH3–CH–CH–CH3+ZnI2

But–2–ene

CH2 1,2–Dimethyl cyclopropane

It is again a commonly used name reaction used in synthetic organic chemistry known as Simmons-Smith Reaction. This reaction is stereospescific with respect to an alkane. Substituents which are trans in the alkenes are trans in the cyclopropane as well. The reaction scheme is shown below: Me

Me C=C

H

H

Me Me CH2I2 Zn–Cu,Ether

CH2 cis–1,2– Dimethyleydo propane

cis–2–Butene

Me

Me C=C

H

H

HC — CH

Me CH2I2 Zn–Cu,Ether

HC — CH CH2 Me trans–1,2– Dimethyl cyclopropane

Mechanism: It is a one step or concerted reaction. Both new carbon-carbon bonds are formed simultaneously. The nucleophilic carbon-carbon double bond causes loss of the iodide leaving group and the electrons from the nucleophilic C – Zn bond is used to form the other carbon-carbon bond. Cu

I − CH 2 − I + Zn   → I − CH 2 − ZnI ether

Me

Me C=C

H

Me

Me C—C

+ Zn I2

H

I–CH2–ZnI

H CH2 H cis–1,2– Dimethylcyclopropane

Physical Properties 1.

Physical state: Cyclopropane and cyclobutane are expectedly gases at room temperature due to subdued intermolecular forces. The remaining cycloalkanes are liquids.

O-121 Solubility: Cycloalkanes are insoluble in water but dissolve in ethanol and ether. It is solely due to their non-polar nature. Melting and boiling points: Melting and boiling points of cycloalkanes show a gradual increase with the increase in molecular weight just like their acyclic counterparts. Table 2: Melting and boiling point of some cycloalkanes

2. 3.

Compound

bp°C

mp°Cc

Cyclopropane

–32.8

–127.4

Cyclobutane

12.5

–90.7

Cyclopentane

49.3

–13.9

Cyclohexane

80.7

6.6

IR spectrum: Like alkanes, they show characteristic C - H stretching absorption at 2850-3050 cm −1.

4.

Wavelength (µm) 2.5

3

4

5

6

7

8

9

10

12

14

16

20

24

Transmittance(%)

100 80 60 40 20 0 4000 3500 3000

2600 2200 2000 1800 1600 1400 1200 1000

800

600

400

Wavenumber (cm–1) Fig. 6: IR spectrum of cyclohexane

Chemical Properties Cycloalkanes resemble alkanes in their chemical inertness towards various reagents. However, cyclopropane and cyclobutane are the exceptions due to a clear cut reason. It is the high bond angle strain which makes these two compounds highly reactive towards a host of reagents. With certain reagents they undergo ring-opening and give addition products. Ring–Opening Reactions: 1. Addition of Cl2 and Br2 : Cyclopropane can add to halogens like chlorine and bromine in the dark to form addition products. CCl4 is used as a solvent. CH2 +Br2

CH2 CH2

CCl4 dark

Br Br 1,3–Dibromo propane

O-122 Cyclobutane and higher homlogues do not give this reaction due to the relaxed bond angle strain. 2.

Addition of HBr and HI: Cyclopropane reacts with concentrated HBr and HI to yield 1-bromopropane and 1-iodopropane respectively. Cyclobutane and higher members do not give this reaction. I +HI 1–Iodopropane

3.

Addition of H2SO4 : Cyclopropane reacts with concentrated H2SO4 to yield acyclic sulfonic acid formed in the following manner. It is important to remind that cyclobutane and higher members again like the previous cases do not give this reaction.

4.

Addition of hydrogen: Cyclopropane and cyclobutane reacts with hydrogen in the presence of Ni catalyst to give propane and n-butane respectively. It is important to note that higher temperature is required for cyclobutane to take part in the reaction.

5.

Oxidation: Cycloalkanes undergo oxidation with hot alkaline potassium permanganate to form dicarboxylic acids. One such example is shown below, alk. KMnO4

∆ Cyclohexane

6.

CH2–CH2–COOH CH2–CH2–COOH Adipic acid

Addition to an unsymmetrical cycloalkane: The addition of a unsymmetrical reagent such as HX like concentrated H2SO4 or HBr etc to an unsymmetrical substituted cyclopropane ring such as methylcyclopropane causes the breaking of a C-C bond in the ring along the dotted line to yield a 2-substituted butane as the major products. This product is derived from the most stable carbocation intermediate according to a reaction scheme which is being shown here,

O-123 CH3

CH3 H+

CH H2C

CH+

H2C

CH2

CH3 x–

CH — X CH2–CH3

CH2 H

Substitution Reactions: Substitution with Cl2 and Br2: Cycloalkanes react with halogens like chlorine and bromine even in the presence of UV light to given substitution products. +Cl2

UV radiation

Cyclopropane

+Cl2

Cl

+HCl

Chlorocyclopropane UV radiation

Cyclopentane

Cl

+HCl

Chlorocyclopentane

Mechanism : Following steps are involved: Step 1. Initiation: Chlorine undergoes hemolytic fission to form chlorine free radicals.

Step 2. Chain-Propagation Steps: (a) Chlorine free radical attacks the cyclohexane molecule to give HCl and cyclohexyl free radical. Here, it is important to note that generally fish-hook arrows are used to indicate the migration of single electrons. Steps (a) and (b) are continuously repeated. Cl• + H

•

H–Cl +

Cyclopentane

Cyclopentyl free radical

Step 3. Chain-Termination Steps: The above chain reaction comes to halt when any two free radicals combine. For example, Cl• +Cl•

Cl–Cl •

+

•

•

+ Cl•

—

—Cl

O-124

3. Stability of Rings and Ring Strain The carbon atoms in cycloalkanes are sp3 hybridized and therefore a deviation is observed from the ideal tetrahedral bond angles of 109°28'. This causes an increase in potential energy and an overall destabilizing effect. Eclipsing of hydrogen atoms is an important destabilizing effect, as well. The strain energy of a cycloalkane is the theoretical increase in energy caused by the compound's geometry, and is calculated by comparing the experimental standard enthalpy change of combustion of the cycloalkane with the value calculated using average bond energies. Cycloalkanes tend to give off a very high and non-favorable energy, and the spatial orientation of the atoms causes the ring strain. When atoms are close together, their proximity is highly unfavorable and causes steric hindrance. The reason why one does not want ring strain and steric hindrance is because heat will be released due to an increase in energy; therefore, a lot of that energy is stored in the bonds and molecules, causing the ring to be unstable and reactive. Another reason we try to avoid ring strain is because it will affect the structures and the conformational function of the smaller cycloalkanes. One way to determine the presence of ring strain is by its heat of combustion. By comparing the heat of combustion with the value measured for the straight chain molecule, one can determine the stability of the ring. There are different types of strain, but the types which are the most important are eclipsing or torsional strain and bond angle strain. Bond angle strain causes a ring to have a poor overlap between the atoms, resulting in weak and reactive C-C bonds. An eclipsed spatial arrangement of the atoms on the cycloalkanes results in high energy.

3.1 Types of Strain There are different types of ring strain, as discussed earlier. Some of these are, 1.

Transannular strain is defined as the crowding of the two groups in a ring.

2.

There is also eclipsing strain, also known as torsional strain, which means that the intramolecular strain present is due to the bonding interaction between two eclipsed atoms or groups.

3.

Another type of strain is called the bond angle strain, and it is present when there is a poor overlap between the atoms. There must be an ideal bond angle to achieve the maximum bond strength and that will allow the overlapping of the atomic orbitals.

Cyclopropane is necessarily planar or flat, with the carbon atoms at the corners of an equilateral triangle. The 60º internal bond angles are much smaller than the optimum 109° 28' angles of a normal tetrahedral carbon atom. Consequently there is huge angle strain as stated by Baeyer strain theory and the resulting angle strain dramatically influences the high chemical reactivity of cyclopropane.

O-125 Cyclopropane also suffers substantial eclipsing strain or torsional strain, since all the carbon-carbon bonds are fully eclipsed. Cyclobutane reduces some bond-eclipsing strain by folding so that the out-of-plane dihedral angle is about 25º, but the total eclipsing and angle strain remains high.

Cyclopentane has very little angle strain as the internal angles of a regular pentagon are 108º, but its eclipsing strain would be large (about 10 kcal/mol) if it remained planar.

Consequently, the five-membered ring adopts non-planar puckered conformations whenever possible, as shown above. Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations, such as those shown below. Cycloheptane and cyclooctane have greater strain than cyclohexane. The major contributory of this strain is due to steric hindrance by groups on opposite sides of the ring also termed as transannular crowding as stated earlier. cyclohexane

cycloheptane

cyclooctane

O-126

Solved Examples Example 1:

Trans-1,2-dimethylcyclopropane

is

more

stable

than

cis-1,2-dimethyl–

cyclopropane. Why? Drawing a picture of the two will help your explanation. Solution : The cis isomer suffers from steric hindrance and has a larger heat of combustion. Example 2: Out of all the cycloalkanes, which one has the most ring strain and which one is strain free? Explain. Solution : Cyclopropane- ring strain. Cyclohexane chair conformation- ring strain free. Example 3: What does it mean when people say "increase in heat leads to increase in energy" and how does that statement relate to ring strains? Solution : When there is an increase in heat there will be an increase of energy released therefore there will be a lot of energy stored in the bond and molecule making it unstable. Example 4: Why is that the bigger rings have lesser strains compared to smaller rings? Solution : Smaller rings are more compacts, which leads to steric hindrance and the angles for these smaller rings are harder to get ends to meet. Bigger rings tend to have more space and that the atoms attached to the ring won't be touching each other as much as atoms attached to the smaller ring.

3.2 Ring Strain Energy Ring strain is highest for cyclopropane, in which the carbon atoms form a triangle and therefore have 60° C-C-C bond angles. There are also three pairs of eclipsed hydrogens. The ring strain is calculated to be around 120 kJ/mol. Cyclobutane has the carbon atoms in a puckered square with approximately 90°bond angles; "puckering" reduces the eclipsing interactions between hydrogen atoms. Its ring strain is therefore slightly less, at around 110 kJ/mol. For a theoretical planar cyclopentane the C-C-C bond angles would be 108°, very close to the measure of the tetrahedral angle. Actual cyclopentane molecules are puckered, but this changes only the bond angles slightly so that angle strain is relatively small. The eclipsing interactions are also reduced, leaving a ring strain of about 25 kJ/mol. In cyclohexane the ring strain and eclipsing interactions are negligible because the puckering of the ring allows ideal tetrahedral bond angles to be achieved. As well, in the most stable chair form of cyclohexane, axial hydrogens on adjacent carbon atoms are pointed in opposite directions, virtually eliminating eclipsing strain. After cyclohexane, the molecules are unable to take a structure with no ring strain, resulting in an increase in strain energy, which peaks at 9 carbons (around 50 kJ/mol). After that, strain energy slowly decreases until 12 carbon atoms, where it drops significantly; at 14, another significant drop occurs and the strain is on a level comparable with 10 kJ/mol. After 14 carbon atoms, sources disagree on what happens to ring strain, some indicating that it increases steadily, others saying that it disappears entirely. However, bond angle strain and eclipsing strain are an issue only for smaller rings.

3.3 Baeyer Strain Theory In 1885, Baeyer enunciated a theory to explain the relative stability of the first few cycloalkanes. He based his theory on the fact that the normal angle between any pair of bonds of a carbon atom is 109°28' . Baeyer postulated that any deviation of bond angles from the normal tetrahedral value would impose a condition of

O-127 internal strain on the ring. He also assumed that all cycloalkanes were planar and thus calculated the angles through which each of the valence bond was deflected from the normal direction in the formation of the various rings. This he called bond angle strain, which determined the stability of the ring.

In cyclopropane, the three carbon atoms occupy the corners of an equilateral triangle. Thus cyclopropane has C-C-C bond angles of 60°. This implies that the normal tetrahedral angle 109°28' between any two bonds 24.75° 1 is compressed to 60° , and that each of the two bonds involved is pulled in by 5° (109° 5°−60° ) = 24 ° 44 2 (Fig. 7) The value 24°44' then represents the bond angle strain or the deviation through which each bond bends from the normal tetrahedral direction. The angle strain for other cycloalkanes can be calculated in the same way. The value are given in Table 3. Whether the angle strain is positive or negative, its magnitude determines the extent of strain in the ring. Table 3: Strain of some common cycloalkane ring–sizes Ring size

Strain energy (Kcal/mol)

Ring size

Strain energy (Kcal/mol)

3

27.5

10

12.4

4

26.3

11

11.3

5

6.2

12

4.1

6

0.1

13

5.2

7

6.2

14

1.9

8

9.7

15

1.9

9

12.6

16

2.0

The angle strain is maximum in the case of cyclopropane. Thus according to the Baeyer strain theory, cyclopropane must be highly strained molecule and consequently most unstable. The cyclopropane ring should therefore be expected to open-up on slightest provocation and thus releasing the strain within it. This is actually so. Cyclopropane undergoes ring-opening reactions with Br2,HBr and H2(Ni) to give open-chain addition products. The angle strain in the case of cyclobutane is less than that in case of cyclopropane. This results into enhanced stability. Thus as expected, cyclobutane undergoes ring opening reactions but only under more drastic conditions. The angle strain is minimum in the case of cyclopentane. This implies that cyclopentane is under least strain and should be most stable. Thus it is not surprising that cyclopentane does not undergo ring-opening reactions. The angle strain in the case of cyclohexane is higher than that in case of cyclopentane. This strain increases continuously with increases in the number of carbon atoms in the ring.

O-128 According to the Baeyer Strain theory, cyclohexane and the higher cycloalkanes should becomes increasingly unstable and hence more reactive. Contrary to this prediction, cyclohexane and the higher members are found to be quite stable. They do not undergo ring-opening reactions. Instead they resembles open-chain alkanes in being highly unreactive. Like acyclic alkanes, these compounds react by substitution. Thus the Baeyer Strain theory satisfactorily accounts for the exceptional reactivity of cyclopropane, cyclobutane.

3.4.Sache-Mohr Theory In order to account for the stability of cyclohexane and other higher members, Sache and Mohr (1918) proposed that such rings can becomes free form strain if all the ring carbons are not forced into one plane, as was assumed by Baeyer. If the ring assumed a 'folded' or 'puckered' condition and the normal tetrahedral angles of 109°28' are retained then as a result, the strain within the ring is relieved. For example, cyclohexane can exist in two non-planar puckered conformations both of which are completely free from strain. These are called the chair form and the boat form because of their shape. Such non-planar strain-free rings in which the ring carbons can have normal tetrahedral angles are also possible for higher cycloalkanes.

The chair form of cyclohexane reveals that the hydrogen atoms can be divided into two categories. Six of the bonds to hydrogen atoms point straight up or down almost perpendicular to the plane of the moelcule. These are called axial hydrogens. The other six hydrogens lie slightly above or slightly below the plane of the cyclohexane ring, and are called equatorial hydrogens.

3.5 Molecular Orbital Theory and Banana Bonding One can explain the relative stability of cycloalkanes in terms of MO theory. It is known that a covalent bond between two atoms is formed by the overlap of orbitals of the atoms involved. The greater the extent of overlap the stronger is the bond formed. The atomic orbitals overlap to the maximum extent if they overlap along their axes. As the axes of sp3 orbitals are at angles of 109°28' to each other, the C – C bonds will have their maximum strength if the C - C - C bond angles have a value of 109°28′ .

O-129 Cyclopropane has C - C - C bond angles of 60°. Cyclobutane has C - C - C bond angles have a value of 90°. The higher cycloalkanes and alkanes have C - C - C bond angles of 109°28′ . As shown in figure no 9. the small bond angles of cyclopropane indicate that the overlap of sp3 orbitals of carbon in alkanes for example in propane. The bond angles of cyclopropane are less than the bond angles of cyclobutane, which in turn are less than the bond angles of higher cycloalkanes or n-alkanes. Therefore, the overlap of orbitals in cyclopropane is less than that in cyclobutane, which in turn is less than that in higher cycloalkanes or n-alkanes. The overlap of sp3 orbitals of carbon in cyclopentane, higher cycloalkanes or n-alkanes is maximum because in these cases it is possible for the sp3 orbitals to overlap along their axes, the bond angle being approximately equal to 109°28' .

This implies that the C - C bonds in cyclopropane are weaker than the C - C bonds of cyclobutane, which in turn are weaker than the C - C bonds in higher cycloalkanes and n-alkanes. Thus, it is not surprising that cyclopropane undergoes ring-opening reactions very readily and more drastic conditions are required to bring about the cleavage of the cyclobutane ring. Cyclopentane and higher members, as expected, do not undergo ring opening reactions and behave very much like the alkanes.

O-130

Exercise (iv)

Long Answer Type Questions 1.

2.

3.

Describe Baeyer's strain theory. What are the defects in it and what modification have been made by Sachse in it. [Bund. 2008; Meerut 2010] 6.

Cl (i)

Explain the mechanism of Kolbe reaction.

(iii)

How cyclo hexanol is converted to cyclo hexane. [Kashi 2010]

(ii)

(ii)

Write the mechanism of free radical halogneation of alkane. Write Baeyer's strain theory for cycloalkanes. What are the limitations of this theory? 7.

Account for the ring strain of cyclopropane in terms of orbital overlap.

and

D

give

a

conc H2SO4

Write short notes on the following : Acidity of alkynes

(b)

Kinetic and thermodynamic control. [Lko. 2011]

Explain the mechanism of free radical halogenations of alkanes taking chlorination of methane as an example. [Hazaribagh 2008, 2010]

Why Corey House reaction is better than Short Answer Type Questions 1. Discuss stability of cycloalkanes on the basis of Wurtz reaction ? Identify A, B, C and D. Li / ether Baeyer's strain theory. 2 Cholorobutane  →

or

(iii)

Which of the isomeric pentane can form only one monochloro substitution product? justify. [Lko. 2009] 2. Identify the products A, B, C and D formed in each 3. of the following reactions and also name the type of reaction involved in each step : [Lko. 2011] 4.

CH3

(i)

CH3–CH2–C=CH–CH2–CH3 aq KOH

CH3–CH2–C H–CH2–CH3

B

HBr

A

H2SO4

aq KOH

B

C6H5COOOH4

C

alc KOH

A C

H2SO4

alc KOH

[Bund. 2009; Agra 2008]

What is Corey House synthesis ?

[Meerut 2012]

Give the mechanism of the halogenation reaction. [Agra 2008,09]

Discuss the stability of cycloalkanes on the basis of Baeyer's strain theory. Explain the following : (i)

Sachse-Mohr theory.

(ii)

Alkanes are relatively unreactive. [Kanpur 2011]

6.

Write short note on Baeyer strain theory.

7.

Give two methods for the preparation of large ring cycloalkanes. [Avadh 2010]

8.

Explain the mechanism of free redical halogenation of alkanes taking chlorination of methane as example. [Kashi 2008]

[Avadh 2008, 10]

CH3OH

NBS

5.

A

OH H2SO4

Discuss banana bond in cyclopropane.

[Kanpur 2010; Hazaribagh 2009]

C

170°c

Cl H2SO4

(iii)

alc KOH

B

(a)

C H 1Br

(ii)

C

Br2

OH

cut 5 1 → D + C A  → B  

5.

A

∆

[Kashi 2011]

(i)

alc KOH

CH3

(ii)

(i)

H2SO4

Identify the product plausible mechanism :

(i) Explain the mechanism of chlorination of methane.

(ii)

4.

CH3–CH2–C H–CH3

B D

O-131 9.

What is Saytzeff rule? Explain it taking example of 9. the mechanism of dehydration of alcohols.

Why boat form of cyclo hexane is less stable than chair form? [Kashi 2010]

10.

Write a note on peroxide effect.

Write IUPAC name of the given compound:

11.

Draw structure of all possible isomers of alkane having molecular formula C5H12. [Kashi 2011]

12.

What is the difference between conformation and configuration? Give boat and chair conformation of cyclohexane. Explain the potential energy 11. relationship among conformation of cyclohexane.

[Kashi 2008]

[Lko. 2010]

13.

Give the mechanism of the following : (i) (ii)

10.

H3C–CH–CH2–CH3

12.

CH3–CH–CH–CH2–CH3 H3C–CH–CH3

[Lko. 2008]

Chlorination of methane in presence of light is an exothermic reaction.Why? [Lko. 2009] Why butene-2, is more stable than butene-1? [Lko. 2010]

Kolbe reaction

13. Chlorination of methane in the presence of 14. sunlight. [Lko. 2011]

Explain banana bonds.

[Lko. 2010]

The melting point of n-alkanes with even number of carbon atoms is higher than those with odd number of carbon atoms.Why? [Lko. 2011]

14.

Write a short note on banana bond.

15.

Prove that the mechanism of chlorination of 15. methane involves the formation of free radicals as intermediates. [Rohd. 2009, 10]

Write down all possible isomers of n-pentane and arrange than is decreasing order of their boiling point. [Lko. 2011]

16.

Give the mechanism of Wurtz reaction. It this 16. reaction suitable for the preparation of unsymmetrical alkane? [Rohd. 2009,10]

Give the name and structure of alkanes that yield the following products on ozonolysis. (i)

Only acetone

17.

Write a note on cracking.

(ii)

Acetaldehyde and formaldehyde

18.

n-Alkanes generally have higher boiling and melting points than corresponding branched chain alkanes, 17. why? [Rohd 2012,13] 19.

[Lko. 2011]

[Rohd. 2011]

[Rohd. 2009]

Write short note on Baeyer and Perkin's synthesis. [Rohd. 2010]

What is Clemmensen reduction? [Hazaribag 2009,11]

Very Short Answer Type Questions 1.

Why propane is more stable than ethane?

20.

Calculate the angle strain in cyclobutane ring. [Hazaribag 2010]

[Meerut 2010]

2.

Objective Type Questions

Give the mechanism of the Wurtz reaction. [Meerut 2011]

3.

Give the reaction mechanism of Kolbe reaction. [Kanpur 2008]

4.

Write notes on Clemmensen's reduction reaction.

Multiple Choice Questions 1.

[Kanpur 2008]

5.

°C  →? + ? + ? + ? CH3CH2CH3 + HNO3 400

[Kanpur 2010]

6.

Write notes on stability of cycloalkanes? [Avadh 2009]

7.

Define Wurtz reaction. [Kashi 2008; Hazaribag 2008]

8.

What is Wolff-Kishner reduction? [Kashi 2008, 10; Hazaribag 2008]

2.

Which of the following shows no ring strain : (a)

Cyclo propane

(b)

Cyclo butane

(c)

Cyclo heptanes

(d)

Cyclo hexane

[Bund. 2008]

Sodium salt of an acid on distillation with sodalime gives ethane. The acid is (a)

CH3COOH

(b)

HCOOH

(c)

CH3CH2COOH

(d)

C6H5COOH

[Rohd. 2011]

O-132 3.

4.

5.

If an acyclic alkane hydrocarbon contains n carbon atoms, how many hydrogen atoms it must contain?

Fill in the Blank

(a)

n

(b) 2n

1.

(c)

n+2

(d) 2n + 2

The reducing agent used in case of Clemmenson reduction is……….

During debromination of meso–dibromobutane, 2. the major compound formed is

The chlorination of ethane proceeds via……… mechanism.

(a)

n–butane

(b) 1– butene

3.

n–hexane can be aromatized to ………. .

(c)

cis–2–butene

(d) trans–2–butene

4.

The chlorination of butane yields……… structural isomers.

The reaction conditions leading to the best yield of C2H5Cl are 5. UV light (a) C2H6(excess) + Cl2  → (b)

dark

C2H6 + Cl2   →

6.

7.

8.

9.

10.

11.

UV light

C2H6 + Cl2(excess)  →

(d)

C2H6 + Cl2  →

UV light

The highest boiling point is expected for (a)

Isooctane

(b)

n–octane

(c)

2,2,3,3–tetramethylbutane

(d)

n–butane

magnesium

6.

Branched chain alkanes have………boiling point than isomeric straight chain isomer.

7.

Red P and HI converts propanoic acid into ………. .

8.

The bond angle strain is the highest in………. .

9.

The value of bond angle strain is calculated with the help of………. .

10.

Decarboxylation of cyclopropane carboxylic acid with soda lime yields………. .

room temperature

(c)

In Zerevitinoff reaction, methyl bromide is converted into ………. .

Bromination of an alkane as compared to True/False chlorination proceeds 1. Kolbe's electrolytic process can be used to prepare methane. (a) At a slower rate (b

At a faster rate

2.

(c)

With equal rate

Amalgmated zinc and conc HCl can be used to convert propanoic acid into propane.

(d)

With equal or different rate depending upon 3. the temperature

To prepare odd carbon alkane, Wurtz reaction is preferred over Corey House synthesis.

Which of the following radicals has maximum stability ? (a)

3°

(b)

2°

(c)

Vinyl

(d)

Benzyl

4.

Bigger rings have lesser strains compared to smaller rings.

5.

Soda lime is used for decarboxylation of aldehydes to alkanes.

Which of the following will have least hindered rotation about carbon carbon bond? 6. (a)

Ethane

(b) Ethylene

(c)

Acetylene

(d) Hexachloroethane 7.

Electrolysis of concentrated solution of sodium propanoate produces the hydrocarbon 8. (a)

Methane

(b)

Ethane

(c)

Propane

(d)

Butane

9.

The reactivity of hydrogen atom in an alkane 10. towards substitution by bromine atom is (a)

1° H > 2° H > 3° H

(b)

1° H < 2° H < 3° H

(c)

1° H > 2° H < 3° H

(d)

1° H < 2° H > 3° H

Hydrogenation of alkene to alkane is a syn hydrogenation. Wurtz reaction mechanism.

proceeds

via

free

radical

A side product of Kolbe's electrolytic process is formation of ester. The bond angle strain value for cyclopropane is 30°. Bromination of alkanes is more selective than chlorination.

O-133

Answers Objective Type Questions Multiple Choice Questions 1.

(d)

2.

(c)

3.

(b)

4.

(d)

5.

(a)

6.

(b)

7.

(a)

8.

(d)

9.

(a)

10.

(d)

11.

(b)

Fill in the Blank 1.

amalgmated zinc and conc. HCl

2.

free radical

3.

benzene

4.

4

5.

methane

6.

lower

7.

propane

8.

cyclopane

9.

Baeyer strain theory

10.

cyclopropane

True/False 1.

False

2.

False

3.

False

4.

True

5.

False

6.

True

7.

True

8.

True

9.

False

10.

True

O-134

Hints and Solutions Long Answer Type Questions CH 3 |

5.

HBr

(i) CH 3 − CH 2 − C = CH − CH 2 − CH 3  → CH 3 |

CH 3 − CH 2 − C − CH 2 − CH 2 − CH 3 | Br

↓ aq KOH CH 3 |

CH 3 − CH 2 − C − CH 2 − CH 2 − CH 3 | OH

Conc.

↓ H 2 SO4 , 170° C CH 3 |

CH 3 − CH = C − CH 2 − CH 2 − CH 3 6.

CH3

(i)

OH

conc. H2SO4, ∆

Short Answer Type Questions 18.

Due to more surface area.

Very Short Answer Type Question 5. 11. 19.

It is an example of vapor phase nitration C–Cl bond is very strong. O ||

R − C− R ′

     →

NH 2 −NH 2, KOH, CH 2 − OH | CH

2

R − CH 2 − R ′

− OH

Objective Type Questions Multiple Choice Questions 2. 8. 9.

(c) It is a decarboxylation technique and the carboxylate carbon is last in the form of carbon dioxide. (d) Stability order of C free radical benzylic 3°>2°>1°> vinyl (a) C–C (Single bond) has least hindrance ❍❍❍

Unit -II O-135

C HAPTER

4

Stereochemistry of Organic Compounds

1. Introduction In organic chemistry, a number of organic compounds having different physical and chemical properties can be represented by the same molecular formula. The property by virtue of which organic compounds possess different physical and chemical properties and have the same molecular formula is known as isomerism and the different compounds are known as isomers or isomerides.Since the isomers are constituted of the same number of atoms, so it is clear that the difference in their properties must be due to the difference in the relative arrangement of the atoms within the molecules.

1.1 Types of Isomerism The difference in the relative arrangement refers to either different structural arrangements or different spatial arrangements. On this basis, two types of isomerism are observed mainly. These are 1.

Structural isomerism

2.

Stereoisomerism

The complete flow sheet of classification of isomerism is, Isomerism Structural isomerism Chain Position isomerism isomerism

Stereoisomerism

Functional Metamerism Geometrical group isomerism isomerism

Optical isomerism

Some more types of isomerism are also possible like tautomerism. It is also considered as a type of structural isomerism. Some text treats conformational isomerism as a type of stereoisomerism although it is not a true type because conformational isomers are rapidly interconvertible.

O-136

2. Structural Isomerism If the isomers have the same molecular formula but they differ in the relative arrangement of atoms, it is called structural isomerism. In structural isomers, the structural formulae of the isomers differ whereas the molecular formula remains same. This type of isomerism is further divided into various types.

2.1 Chain Isomerism Chain isomerism arises due to the difference in the structure of carbon chain. It is sometimes also referred as nuclear isomerism. The difference is in the length of the carbon chain or in the size of the carbon ring. For example, 1. n–butane and isobutane are chain isomers. CH3 – CH2– CH2 – CH3 CH3 – CH – CH3 | n–butane CH3 isobutane 2. Cyclohexane and methylcyclopentane are nuclear isomers. CH3 CH2

CH

CH2 CH2

CH2 CH2

CH2

cyclohexane

3.

CH2

CH2 CH2

CH2

methycyclopentane

C5H12 has three chain isomers. CH3 CH3CH2CH2CH3

CH3CH2–CH–CH3

n–pentane

4.

CH3

CH3 Isopentane

C

CH3

CH3 neopentane

C4H9NH2 also shows chain isomerism. Some of the examples are, CH3CH2CH2NH2 n–butylamine

CH3 CH3 C–NH2 CH3 tert.butylamine

2.2 Positional Isomerism This isomerism arises due to the difference in position of either substituent like halogen atom or nitro group or functional group in the same carbon chain.For example, 1. CH3–CH2 –CH = CH2 and CH3–CH = CH–CH3 → → 1-Butene are positional isomers.

2-Butene

O-137 2.

CH3– CH2– CH2– I and CH3 –CH–CH3 | I 1–iodopropane 2–iodopropane are also positional isomers.

3.

CH3–CH2–CH–CH2–CH3 | CH3 3–methylpentane are positional isomers.

4.

C6H4(NO2)2 exhibits following three positional isomers.

and CH3–CH–CH2–CH2–CH3 | CH3 2–methylpentane

NO2

NO2

NO2

NO2

NO2 NO2 m–dinitrobenzene

o–dinitrobenzene

p–dinitrobenzene

2.3 Functional Isomerism Functional isomers have the same molecular formula but they differ due to the presence of different functional groups. As an example, carboxylic acids and esters are functional isomers. Similarly,alcohols and ethers are also functional isomers. Even aldehydes, ketones, unsaturated alcohols, unsaturated ethers, cyclic ethers and cyclic alcohols are functional isomers to each other. Cyanides and Isocyanides are also functional isomers. Some more examples are shown below, 1.

C3H6O exhibits the following functional isomers. O

O

CH3–CH2–C–H , CH3–C–CH3 , CH3–OCH=CH2 , CH2=CHCH2OH Propanal

2.

Propanone

Methoxyethene

2–propen–1–o1 (allyl alcohol)

C3H6O2 shows following functional isomers. O

O

O

CH3–CH2–C–OH , CH 3–C–OCH3 , H–C–OC 2H5 Propanoic acid Methyl ethanoate Ethyl methanoate (I) (II) (III)

3.

C2H6O exhibits two functional isomers. CH3–CH2 –OH ;

CH3–O–CH3

Ethanol

Dimethyl ether

O-138

2.4 Metamerism It is caused by the attachment of different radicals/groups to a polyvalent atom (i.e. an atom having more than one valency). A metamer can be obtained by shifting one or more CH2 groups from one side of the polyvalent functional group to the other side. These isomers are of same homologous series. Metamerism is found to occur in amines, ethers, ketones, esters etc. For example, (i)

C4H10O exhibits two metameric ethers as shown below, CH3 – CH2 – O– CH2 – CH3

and

Diethyl ether (ii)

CH3 – CH2 – CH2 – O – CH3 Methyl n–propyl ether

C5H10O2 exhibits following metamers as shown, O

O

CH3CH2CH2–C–OCH3 , CH3CH2–C–OCH2CH3 Ethyl propanoate Methyl butanoate O

O

CH3–C–OCH2CH2CH3 , H–C–OCH2CH2CH3 n–propyl ethanoate n–butyl methanoate

(iii)

Similarly, C4H10NH exhibits following metamers. C2H5–NH–C2H5

CH3–NH–CH2CH2CH3

Diethyl amine

N–methylpropyl amine

2.5 Tautomerism This is a special type of functional isomerism where, functional isomers exist in equilibrium with each other. Such isomers are called tautomers. The necessary condition for this type of isomerism is the presence of hydrogen, α − (a hydrogen on a carbon adjacent to carbon of functional group) to the carbonyl group.A very common, form of tautomerism is that between a carbonyl compound containing an α hydrogen and its enol form. This type of isomerism is also known as keto enol isomerism. R'

R – C – C – R'' O

H Keto form

R – C = C – R'' O

R'

H

Enol form

In simple compounds (when R" = H, alkyl, OR, etc) the equilibrium lies more towards the left. The reaction can be seen by examining the bond energies of various bonds. The keto form differs from the enol form in possessing a C–H, a C–C, and a C = O bond whereas the enol form has a C = C, a C–O and an O–H bond. The keto form is thermodynamically more stable than enol form and enol forms are normally not isolated. However in certain cases, a larger amount of the enol form is present, and it can even be the predominant form.The percentage of enol form increases in the order simple aldehydes and ketones anti. Explain. [D.D.U. 2009]

Choose optically inactive compound from 19. the following :

Discuss the optical isomerism of a compound having two asymmetric carbons. [D.D.U. 2009]

H

(iii)

(i) (ii)

H

enantiomers

Arrange the following groups according to Cahn-Ingold-Prelog sequence rule (from highest to lowest) :

CH3 HO

between

[Lko. 2011]

CH3 (b)

[Lko. 2008]

-COOH, -CH2OH, OH, -CHO, NH2, CN, C3H7.

OH

H

Me

diastereomers.

CH3

(a)

H (i)

(iii)

COOH

H

Explain the following :

H

H3C

Me

HO CH3

(a)

20.

Me

Me

H

H

H

Me

Me

H

Write the configuration of following compounds : (i)

Z-1-Iodo-2-hydroxypropene.

(ii)

E-2-Nitro-3-bromobut-2-enoic acid. [D.D.U. 2009]

(b) 21.

Meso-tartaric acid is optically inactive though it has asymmetric carbon atom. [D.D.U. 2010]

22.

Discuss the methods used for determination of configuration in geometrical isomers. [D.D.U. 2010]

23.

Write the structure of isomeric amines having molecular formula C3H9N. [D.D.U. 2011]

O-189 24.

Explain the geometrical isomerism in maleic acid 6.

Write notes of the following :

and fumaric acid. Assign their configurations.

(i)

Tauatomerism

(ii)

Geometrical isomerism

[Kanpur 2008]

7.

Explain asymmetric synthesis.

[Kanpur 2010]

8.

Explain the following :

[D.D.U. 2011]

25.

Draw the potential energy diagram for the various conformations of C2H6 and explain their relative stability.

26.

[D.D.U. 2011]

Draw conformations of ethane. [Hazaribagh 2008]

27.

Meso compound

(ii)

Racemic mixture

What is E and Z system of nomenclature? Explain

isomerism in alicyclic compounds.

with examples.

[Kanpur 2010]

10.

Draw conformations of ethane.

[Kanpur 2008]

11.

Define resolution.

12.

Define racemic mixture.

Write short notes on the following : (i)

Resolution on enaniomorphs.

(ii)

E and Z system of nomenclature.

[Kashi 2008; Hazaribagh 2008]

[Kashi 2008; Hazaribagh 2008]

[Hazaribagh 2009]

29.

[Kanpur 2010]

What is stereoisomerism? Discuss geometrical 9. [Hazaribagh 2009]

28.

(i)

Distinguish between enantiomer and diastereomer. 13.

Draw the structure of R(-)-2 butanol. [Kashi 2010]

[Hazaribagh 2010]

30.

Write short note on chair and boat confirmation of 14. cyclohexane.

31.

Draw all the stereoisomers of 1,2–

and 1,3–

15.

Explain stereoisomerism in tartaric acid.

33.

Explain erythro and threo system of nomenclature

stereoisomers, their R, S configuration and optical

CH3

Very Short Answer Type Questions Represent the configuration compounds by R and S system.

I HO

F

Cl 2.

Write

of

the

C

and

H

of

Write a detail note on geometrical isomerism. Write a detail note on resolution. [Meerut 2009,10,12]

5.

16.

What are enantiomers of lactic acid?

17.

Depict axial and equatorial bonds in cyclohexane

18.

Give the concept of conformation of ethane. [Agra 2008]

[Lko. 2010]

[Rohd. 2013]

What are enantiomorphs? Explain. [Hazaribagh 2009]

19.

What is an asymmetric carbon atom?

[Bund. 2010] [Meerut 2009]

4.

CH3

different

conformation of cyclobutane (only by Newman

3.

CH3

chair conformation. [Bund. 2010]

structure

projection).

Na

[Lko. 2009]

CH2OH name

2ClCH2–CHCH2CH3

H3C–CH2–CH–CH–CH2–CH–CH2–CH3

following

CHO

C

In the given reaction, given the number of isolated activity.

32.

Br

[Lko. 2009]

[Hazaribagh 2010]

dimethylcyclopentanes.

1.

Homologues can never be isomer.Why?

[Hazaribagh 2011]

20.

What is the condition to be satisfied for a compound to be chiral?

[Hazaribagh 2008]

O-190 7.

Objective Type Questions Multiple Choice Questions 1.

The isomers related to each other as object and 8. mirror image are (a) Enantiomer (b) Diastereomer (c) Rotamer (d) None of the above

The molecules represented by the above two structures are (a)

Identical

(b)

Enantiomers

(c)

Diastereomers

(d)

Epimers

Which of the following compound is incapable of exhibiting tautomerism. (a)

O

[Bund. 2008]

The compound which exhibits optical isomerism is : (a) (CH3 )2 CH − CH2Br

(b)

CH3 |

(b)

O

(CH3 )2 − CH − CH2Br C2H5

O

|

(c)

(CH3 )2 − CH − CH2Br

(d) 3.

O

(CH3 )3 − C − Br

(c)

—CH=CH—OH

[Agra 2008]

How much chiral carbon atom in following structure CH3

(d)

|

H − C − OH

O

2.

O

|

(a) (c)

COOH One Three

(b) Two (d) No chiral atom

9.

[Agra 2009]

4.

5.

R and S system used in which isomerism? (a) Conformational (b) Configuration 10. (c) Geometrical (d) Optical [Agra 2009] What is IUPAC name of compound below, CH2OH

11.

CH3CH2CH2CHCH2CH3

(a) (b) (c) (d) 6.

3-Methaneol hexane 2-Ethyl pentanol 5-Ethyl pentanol None of these

12. [Agra 2009]

Benzil on reduction with NaBH4 gives (A). The number of optically active stereoisomers possible for (A) are (a)

1

(b)

2

(c)

3

(d)

4

Which of the following compounds will exhibit cis-trans isomerism? (a)

2-butene

(b)

2-butyne

(c)

2-butanol

(d)

butanal

The number of isomers of C6H14 is: (a)

4

(b)

5

(c)

6

(d)

7

The isomers which can be inter converted through rotation around a single bond are:

Which of the following compounds is optically active ?

(a)

Conformers

(b)

Diastereomers

(a)

Bromochloromethene

(c)

Enantiomers

(b)

2-Bromo-2-Chloropropane

(d)

Positional isomers

(c)

1–Bromo–2–Chloropropane

(d)

All of the above.

13. [Rohd. 2013]

An isomer of ethanol is: (a)

Methanol

(b)

Diethyl ether

(c)

Acetone

(d)

Dimethyl ether

O-191 14.

H

possible for butane-2,3-diol?

15.

(a)

1

(b)

(c)

3

(d)

2

16.

17.

4

(c)

3

(d)

2

3

(b) 4

(c)

5

(d) 6

20.

H H

OH H

C6H5CHO + HCN → C6H5CH(OH)CN

OH OH CH3 CH3

CH3

(d)

H OH

and

H H

HO HO CH3

CH3

(a)

Optically active

(b)

A meso compound

Which of the following Fischer's projection formula is identical to D-glyceraldehyde?

(c)

A racemate

(a)

(d)

A mixture of diastereomers

22.

CH2OH HO

The number of geometrical isomers of

(a)

4

(b) 6

(c)

8

(d) 10

C

H

CHO

(c)

(a)

R3CNO2

(b) RCH2NO2

(c)

(CH3)2NH

(d) (CH3)3CNO

HO

(a)

If in a compound, one is present then so is the other.

(b)

Each forms equal number of isomers for a given compound.

(c)

Both are included in stereo isomerism.

(d)

They have no similarity.

H

H

CH3

OH CH3

and

HO H

H OH CH3

C

CH2OH

HO 23.

enantiomers?

OH H

CHO

(d)

Which of the following pairs of compounds are

H

CH2OH

C

A similarity between optical and geometrical

CH3

OH

C

CHO

Tautomerism will be exhibited by

(a)

CHO

H CH2OH

(b)

isomerism is that

21.

and

CH3

CH3CH = CH - CH = CH - CH = CHCl is

19.

CH3

OH H

The product would be

18.

H H

CH3

H OH

The total number of isomeric trimethylbenzene is (a)

HO OH

CH3

(c)

molecular formula C4H9OH is (b)

and

CH3

The total number of isomeric alcohols with the 5

OH H

OH

4

(a)

CH3

CH3

(b)

How many optically active stereoisomers are

24.

The name of the following compound is

(a)

(2Z, 4Z)-2, 4-hexadiene

(b)

(2Z, 4E)-2, 4-hexadiene

(b)

(2E, 4Z)-2, 4-hexadiene

(d)

(2E, 4E)-2, 4-hexadiene

Number of structural isomers of C6H14 are (a)

4

(b) 5

(c)

6

(d) 7

O-192 25.

26.

27.

(a)

9 σ bonds, 1 π bond and 2 lone pairs

The number of enantiomers of the compound CH3CHBrCHBrCOOH is

(b)

8 σ bonds, 2 π bonds and 2 lone pairs

(a)

0

(b) 1

(c)

9 σ bonds, 2 π bond and1 lone pair

(c)

3

(d) 4

(d)

10 σ bonds, 1 π bond and1 lone pair

The enolic form of acetone contains

30.

31.

The compound which is not isomeric with diethyl ether is (a)

Methylpropyl ether

(b)

Butanol

(c)

2 methyl 2 propanol

(d)

Butanone

32.

The compound

H3C

33.

H C=C

H3C

Isomerism between CH3OC2H5 is (a)

Chain

(b) Positional

(c)

Functional

(d) Structural

(a)

Enantiomers

(b) Diastereoisomers

(c)

Tautomers

(d) Metamers

Which has a symmetric carbon atom

Cl Br

C

(a)

H–C–C–H

COOH

will exhibit (a)

Geometrical isomerism

(b)

Optical isomerism

(c)

Geometrical and optical isomerism

(d)

Tautomerism

(c)

In following

28.

COOH OH

H H

OH COOR (A) H HO

H H

COOR OH OH COOH

35.

(B)

(b)

H H

H Cl

H H

H–C–C–D

(d)

Br OH

(a)

1–butene

(b) trans–2–butene

(c)

cis–2–butene

(d) 1, 3–butadiene

Which of the following compounds will show geometrical isomerism ?

COOH OH

(b)

propene

(c)

1–phenylpropene

H COOR

(d)

2– methyl–2–butene

(a)

A and B are identical

(b)

A and B are enantiomers

(c)

A and B are diastereoisomers

(d)

A and C are enantiomers

How many optically active stereoisomers are possible for 2,3– butanediol ? (a)

1

(b) 2

(c)

3

(d) 4

37.

H–C–C–H

Which one of the following has the smallest heat of hydrogenation per mole ?

2–butene

36.

H–C–C–Cl

H H

(a)

(C)

29.

H Cl

H H 34.

and

The optical isomers which are not mirror images of each other

H

H3C

CH3CH2CH2OH

The number of isomers for the compound with molecular formula C2BrClFI is (a)

3

(b) 4

(c)

5

(d) 6

Isomers which can be interconverted through rotation around a single bond are (a)

Conformers

(b) Diastereoisomers

(c)

Enantiomers

(d) Positional isomers

O-193 38.

39.

Of the following compounds, which will have a zero dipole moment ?

True/False

(a)

1,1 - dichloroethylene

1.

(b)

cis - 1,2 dichloroethylene

Geometric isomers are diastereomeric to each other.

(c)

trans - 1,2 - dichloroethylene

2.

Maleic acid is more stable than fumaric acid.

(d)

none of these compounds 3.

Oximes are optically active around the double bond.

4.

All allenes are optically active.

5.

A racemic mixture is optically inactive.

6.

Chair form of cyclohexane is equally stable to its boat conformation.

The compounds :

CH3

H3C C=C H

H3C and

H

H C=C CH3

H

are examples of : (a)

Enantiomers

(b)

Geometrical isomers

7.

Butane has infinite rotamors.

(c)

Metamers

8.

Methanol can show conformational isomerism.

(d)

Optical isomers 9.

All the rotamors are rapidly interconvertible.

10.

1,3-dimethyl cyclobutane can show geometric isomerism.

Fill in the Blank 1.

Lactic acid is an optically ……… compound.

2.

Meso form of tartaric acid contains a/an ……… plane of symmetry.

3.

The compound which rotates the plane of plane polarized light in clockwise direction is ……….

4.

Glucose molecule has ……… optical isomers.

5.

Tartaric acid has……… chiral carbon atoms.

6.

The absolute method of configuring optically active compound is referred as……….

7.

The reference molecule for D,L configuration is ……….

8.

The angle of rotation caused by an optically active substance is……… proportional to the length of polarimetric tube.

9.

Threo and erythro isomers are……… to each other.

10.

1,2-Dimethylcyclopentane stereoisomers.

consists

of

………

O-194

Answers Objective Type Questions Multiple Choice Questions 1.

(a)

2.

(c)

3.

(a)

4.

(d)

5.

(b)

6.

(c)

7.

(c)

8.

(c)

9.

(b)

10.

(a)

11.

(b)

12.

(a)

13.

(d)

14.

(b)

15.

(b)

16.

(a)

17.

(c)

18.

(c)

19.

(b)

20.

(c)

21.

(a)

22.

(c)

23.

(d)

24.

(c)

25.

(a)

26.

(d)

27.

(b)

28.

(b)

29.

(b)

30.

(d)

31.

(c)

32.

(b)

33.

(c)

34.

(d)

35.

(a)

36.

(d)

37.

(a)

38.

(c)

39.

(b)

Fill in the Blank 1.

active

2.

internal

3.

dextrorotatory

4.

16

5.

2

6.

R,S nomenclature

7.

glyceraldehyde

8.

directly

9.

diastereomeric

10.

3

True/False 1.

True

2.

False

3.

False

4.

False

5.

True

6.

False

7.

True

8.

True

9.

True

10.

True

O-195

Hints and Solutions Long Answer Type Questions 15.

There are two ways to draw cyclohexane because it can be in a hexagon shape or in a different conformational form called the chair conformation and the boat conformation. The chair conformation drawing is more favored than the boat because of H H H the energy, the steric hindrance, and a new strain called the H transannular strain. Here is what the boat conformation looks like. The boat conformation isn't the favored conformation because it is less stable and has a steric repulsion between the two H's, shown with the pink curve. This is known as the transannular

H

H H

strain, which means that the strain results from steric H crowding of two groups across a ring. The boat is less stable than the chair by 6.9 kcal/mol. The boat conformation, however, is flexible, and when we twist one of the C-C bonds, it reduces the transannular strain. When we twist the H C-C bond in a boat, it becomes a twisted boat as drawn H below.

H

H

H

H H H

H

H

H H

Although there are multiple ways to draw cyclohexane, the most stable and major conformer is the chair because it has a lower activation barrier from the energy diagram. Half–Chair Half

E Boat

Twist–boat Chair

Twist–boat Chair

Reaction coordinate to conformations of cyclohexane

The transition state structure is called a half chair. This energy diagram shows that the chair conformation is lower in energy; therefore, it is more stable. The chair conformation is more stable because it does not have any steric hindrance or steric repulsion between the hydrogen bonds. By drawing cyclohexane in a chair conformation, we can see how the H's are positioned. There are two positions for the H's in the chair conformation, which are in an axial or an equitorial formation.

O-196 H

H

H

H

H

H

H

H

H

H

H

H

This is how a chair conformation looks, but you're probably wondering which H's are in the equitorial and axial form. Here are more pictures to help. H

H H H

H

H

These are hydrogens in the axial form. H H

H

H

H

H

These hydrogens are in an equitorial form. Of these two positions of the H's, the equitorial form will be the most stable because the hydrogen atoms, or perhaps the other substituents, will not be touching each other. This is the best time to build a chair conformation in an equitorial and an axial form to demonstrate the stability of the equitorial form. 16.

The study of the relative energies of conformational isomers can be done nicely taking butane.Let us have a Newmann projection by looking down the C(2)–C(3) bond of butane. 1

2

3

4

CH3–CH2–CH2–CH3

H

H C CH3

CH3 C H

H CH3

Let us assume that relatively large methyl groups will best be kept as far from each other as possible then we have the conformation called the anti-form H shown in fig. (A). 120° clockwise rotation passes over a transition state to a new minimum energy conformation of butane shown in fig. (B).

H

H

H CH3 Fig (A)

O-197 CH3

Conformation (B) will be less stable than (A) because in (B) the two methyl groups are closer than they are in (A). Conformation (2) is called H3C 'gauche' form. Gauche form of butane (B) is about 0.6 kcal mol–1 higher in energy than the anti form (A). There was an eclipsed transition state for the interconversion of (A) to (B) which lies about 3.4 kcal mol–1 higher than (A). H

H

H H Fig (B) CH3

Another 120° clockwise rotation brings us to a second gauche form of butane (B). The transition state for the conversion of (B) to (B') involves an H eclipsed state of two

CH3

H

H H Fig (B') H3C

C–CH3 bonds shown in fig. (C) and lies 3.8 kcal mol–1 above (B) and (B'), a total of about 4.4 kcal mol–1 above (A). Final 120° clockwise rotation return us to (A). H

CH3

H

H

H Fig (C)

The angle versus energy plot for butane can be shown as. H3C

Potential energy

H3C H

CH3 H H

H3C H H H

CH3 A

H H

H H

3.4 kcal mol–1

CH3

C

H H

H H3C

3.4 kcal

–1 B' mol CH3

B CH3

H3C

H

H

H

H

H

60°

H3C H

3.8 kcal mol–1

120°

CH3

A

H H

H 0°

H H

180°

240°

Rotation about the central C2–C3bond

300°

360°

O-198

Short Answer Type Questions 31.

Cyclopentane is best considered as a flat ring. All trans–isomers exist as a pair of enantiomers. All cis isomers are meso. Me

Me Me

Me Me racemate trans–1,2–Dimethylcyclopentane

Me meso trans–1,2–Dimethylcyclopentane

Me

Me Me

Me

Me

racemate trans–1,3–Dimethylcyclopentane

32.

Me

meso trans–1,3–Dimethylcyclopentane

Tartaric acid contains two asymmetric centers, but two of the "isomers" are equivalent and together are called a meso compound. This configuration is not optically active, while the remaining two isomers are D- and L- mirror images, i.e., enantiomers. The meso form is a diastereomer of the other forms. COOH H HO

OH

COOH HO

H

OH

H

COOH HO

H

COOH

COOH H

OH

H

OH

COOH HO

COOH

COOH

Meso tartaric acid HO

COOH HO

(Natural) tartaric acid L–(+)–tartaric acid dextrotartaric acid

COOH

D–(–)–tartaric acid levotartaric acid

(1:1) DL–Tartaric acid "racemic acid"

33.

This nomenclature is used only in those compounds which have (a)

Only two chiral carbons and

(b)

The following structure, R' - Cab - Cbc - R"

O-199 i.e., out of six substituents on two asymmetric carbons, at least two should be same. When two like groups in Fischer projection formula are drawn on the same side of the vertical line, the isomer is called erythro form; if these are placed on the opposite sides, the isomer is said to be threo form. R'

CH3

CH3

a

C

b

H

C

Cl

H

C

Cl

c

C

b

H

C

Br

H

C

H

C6H5

R"

Brythroform Brythroform

C6H5 Thero form

Objective Type Questions Multiple Choice Questions CH3

6.

CH3 OH

H Br

H C2H5

HO

H

Br

H C2H5

7.

Since the two stereoisomers are not mirror image, they are diastereomers.

8.

Benzoquinone is very stable as it is highly conjugated system so it does not get enolized.

9.

Ph–C–C–Ph O O

NaBH4

* * Ph–CH–CH–Ph OH OH

The reduction product will have 3 stereoisomers of which only 2 would be optically active. The third isomer (meso) will be optically inactive. 10.

Logic: The compounds with each doubly bonded carbon attached to two different groups (like Cab=Cab, Cab=Ccd) exhibit geometrical isomerism i.e., cis and trans forms. The geometrical isomerism arises due to restricted rotation of double bond. However, even though there is restricted rotation for triple bond, alkynes do not exhibit geometrical isomerism, since the triply bonded carbons are attached to one group each only.

11.

C6H14 indicates saturated hydrocarbon i.e., an alkane. There are 5 chain isomers possible with this formula i.e., one linear, two monosubstituted, two distubstituted.

O-200 Linear H3C–CH2–CH2–CH2–CH3 hexane

Disubstituted

Mono substituted CH3

CH3 H3C–C–CH2–CH3

H3C–CH2–CH2–CH2–CH3 2–methylpentane

CH3 2,2–dimethylbutane

H3C–CH2–CH2–CH2–CH3 CH3 3–methylpentane

12.

H3C–CH–CH–CH3 CH3 CH3 2,3–dimethylbutane

Logic: Isomers have same molecular formula but differ in their structures or in spatial arrangement.

Solution: Compound

Structural formula

Molecular formula

Ethanol

C2H5OH

C2 H6 O

Methanol

CH3OH

CH4O

Diethyl ether

C2H5OC2H5

C4H10O

Dimethyl ether

CH3OCH3

C2 H6 O

Only dimethyl ether has same molecular formula as that of ethanol. 14.

Logic: Theoretically, the total number of enantiomers possible for a molecule is equal to 2n, where n = number of chiral centers. However, to show optical activity, the molecule must be asymmetric. Hence some of the configurations may be optically inactive due to symmetry.

Solution: Butane-2,3-diol, CH3-CH(OH)-CH(OH)-CH3 has two chiral centers. Hence, theoretically, there must be 2n = 22 = 4 enantiomers possible. However, one of the configuration has plane of symmetry and hence is optically inactive. It is called meso isomer. It has no enantiomer. Therefore only three stereoisomers are possible among which only are two optically active. ❍❍❍

UnitO-201 -III

C HAPTER

5

Alkenes, Cycloalkenes, Dienes and Alkynes

1. Alkenes Alkenes are hydrocarbons whose molecules contain the carbon–carbon double bond. They are also called olefins. Alkenes or olefins contain the structural unit C = C and have the general formula CnH2n. These unsaturated hydrocarbons are isomeric with the saturated cycloalkanes, so it can be said that one ring is equivalent to a double bond. CH3CH=CH2 Propylene

H2C—CH2 CH2 Cyclopropane

This can be further explained by the example C4H8. It has a number of isomers as shown below, 3

1

2

3

4

H2C=CH–CH2–CH3

–CH3 Cyclobutane Methylcyclopropane

1-butene

1

2

H2C=C

CH3 CH3

2–methylopropene

1.1 Nomenclature and Isomerism There are certain rules which have to be followed to write the IUPAC name of a compound containing a double bond or more than one double bonds. Rule number 1-Name the longest chain with the double bond. However, it is not always the longest chain of carbons. End the name of longest chain carbon with "-ene". C 2 H4

H H

C3 H6

H H

H C=C

ethene

H HH C = C–C–H H

propene

O-202 Rule number 2- When necessary, use the lowest number to give the location of double bond in the longest chain. Some of the demonstrative examples are shown below, C 4 H8

HHH

H

C = C–C–C–H

H

HH HH

HH

C 4 H8

H

H HH

HH

C 4 H8

2-butene

C–C = C–C–H

H

H

H HHH H

H

1-pentene

C=C–C–C–C–H

H C5H10

2-butene

C–C = C–C–H

H

C5H10

1-butene

HH H HH HH H

2-pentene

H–C–C=C–C–C–H HH

H

H

C5H10

3-methyl-1-butene

H C H H H

H

C=C–C–C–H H

H H H

C5H10

H

C

H

2-methyl-1-butene

H HH

C=C–C–C–H H

H H H

C5H10

2-methyl-2-butene

H C H HH H H–C–C=C–C–H H

H

O-203 H

C5H11

H

C

H

2,2-dimethyl propene

H H

C=C–C–H H H

C

H H

H

Alkenes normally show cis - trans or in general geometric isomerism, apart from other types of isomerism, which has been discussed in detail in the chapter 4. Some of these geometric isomers in alkenes are shown below, CH2CH3 H3C

H3C C H

C

C H

Cis–2–pentene

H

H

H

H

C CH2CH3

Trans–2–pentene

H Cis–cyclooctene

H Trans–cyclooctene

1.2 Bonding in Alkenes Like a single covalent bond, double bonds can be described in terms of overlapping atomic orbitals, except that, unlike a single bond which comprises of a single sigma bond, a carbon-carbon double bond consists of one σ and one π bond. This double bond is stronger than a single covalent bond(611 kJ/mol for C=C vs. 347 kJ/mol for C-C) and also shorter with an average bond length of 1.33 Angstroms (133 pm). Each carbon of the double bond uses its three sp² hybrid orbitals to form sigma (σ) bonds to three atoms. As predicted by the VSEPR model of electron pair repulsion, the molecular geometry of alkenes includes bond angles about each carbon in a double bond of about 120°. The angle may vary because of steric strain introduced by non bonded interactions created by functional groups attached to the Fig. 1: π bond in the perpendicular carbons of the double bond. For example, the C-C-C bond angle in plane of ethene propylene is 123.9°. The unhybridized 2p atomic orbitals, which lie perpendicular to the plane created by the axes of the three sp² hybrid orbitals, combine to form the pi (π) bond. This bond lies outside the main C-C axis, with half of the bond on one side and half on the other. Rotation about the carbon-carbon double bond is restricted because it involves breaking the π bond, which requires a large amount of energy (264 kJ/mol in ethylene). As a consequence, substituted alkenes may exist as one of two isomers, called cis or trans isomers. More complex alkenes may be named using the E-Z notation, used to describe molecules having three or four different substituents or the side groups. If one considers but-2-ene, it has two geometric isomers. These two isomers of butene are slightly different in their chemical and physical properties.

O-204

1.3 Methods of Preparation of Alkenes A good number of methods are available for the preparation of alkenes. Some of the more important methods are being discussed below,

1.3.1 Partial Hydrogenation of Alkynes Alkynes can be partially reduced to alkenes by the use of reagents like Lindlar's catalyst, Na in liquid NH3 etc. An alternate to Lindlar's catalyst is P–2 catalyst with H2. 1.

Lindlar's catalyst: Specially designed catalysts can be used to prepare cis–alkenes from disubstituted alkynes. Metallic palladium deposited on calcium carbonate can be used in this way after it has been conditioned with lead acetate and quinoline. This special catalyst is known as Lindlar's catalyst.

2.

P-2 catalyst: Another catalyst that permits the hydrogenation of an alkyne to an alkene is nickel boride called P–2 catalyst. This catalyst can be prepared by the reduction of nickel acetate with sodium borohydride. NaBh

4 → Ni B (P–2) (CH3COO)2Ni    2

EtOH

Hydrogenation of alkynes in the presence of P–2 catalyst causes syn addition of hydrogen to take place, and the alkene that is formed from an alkyne with an internal triple bond has the (Z) or cis configuration. Reduction with Lindlar's catalyst or P-2 catalyst R–C ≡ C – R

H2,Pd/CaCO3 quinoline (syn addition)

R

R C=C

H

H

The alkene obtained in case of catalytic reduction is essentially cis. 3.

Reduction with lithium or sodium metal in ammonia: When alkynes are reduced with lithium or sodium metal in ammonia or ethylamine at low temperatures, the addition of hydrogen atom occurs in anti–fashion. This reaction is called as dissolving metal reduction which produces trans–alkene. R–C ≡ C – R

(i) Li,C2H5NH2 (ii) NH4Cl (anti addition)

R

H C=C

H

R

Mechanism The mechanism for this reaction involves successive electron transfer from Li or Na atom and proton transfers from NH3 or RNH2. In the first step, lithium atom transfers an electron to the alkyne to produce an intermediate that bears a negative charge and has an unpaired electron, called as radical anion. In the second step, the amine transfers a proton to produce a vinylic radical. Then, transfer of another electron gives a vinylic anion. It is this step that determines the stereochemistry of the reaction. The trans–vinylic anion is formed preferentially because it is more stable as the bulky alkyl groups are farther apart. Protonation of the trans–vinylic anion leads to the formation of trans–alkene.

O-205

••

R Li•

–

C=C

+R–C≡C–R

••

H–NHEt

R

R

•

•

••

R

R

Li•

H

– ••

C=C R

Trans–vinylic anion

Vinylic radical

Radical anion

H–NHEt

R

H C=C

H C=C R

H

Trans–alkene

1.3.2 Cracking of Petroleum Cracking of petroleum hydrocarbons is the source of commercial alkenes. Rather, it is an industrial method of preparation of alkenes. The reaction scheme is.

1.3.3

Dehydrohalogenation of Alkyl Halides (1, 2-Elimination)

When an alkyl halide is subjected to elimination in the presence of a strong base, it loses a molecule of hydrogen halide and yields an alkene. As an example, when isopropyl bromide is treated with hot concentrated alcoholic solution of a strong base like potassium hydroxide, propylene is obtained. C H OH

CH3–CHBrCH3 + KOH  2 5 → CH3CH = CH2 + KBr + H2O heat

This is an example of dehydrohalogenation also known as 1,2-elimination. Dehydrohalogenation involves elimination of the halogen atom and the hydrogen atom from adjacent carbons in this case. This is called 1,2-elimination. For the double bond to form, the hydrogen must come from a carbon that is adjacent to the carbon holding the halogen. Now the carbon holding the halogen is commonly the α-carbon. Any carbon adjacent to the α-carbon is a β-carbon, and its hydrogens are β-hydrogens. Elimination, thus involves loss of a β-hydrogen, so it is very frequently termed as β–elimination. So, the general scheme of the reaction is, –C–C–

KOH alcohol

C=C

+KX+H2O

H X

In some cases, dehydrohalogenation, yields a single alkene and in other cases a mixture as shown below, alc.

CH3CH2Cl + KOH  → CH2 = CH2 CH3CH2CH–CH3+KOH Cl

alc.

CH3CH2CH=CH2+CH3CH=CH–CH3 1-butene (minor)

2–butene (major)

In case of competing products, the formation of an alkene by dehydrohalogenation of alkyl halide is governed by Saytzeff's rule which states that "The more substituted alkene is formed in greater amount during an elimination process." All 1,2–elimination reactions are characterized by the following common traits:

O-206 (i)

The substrate contains a leaving group (an atom or group that leaves the molecule) takes its electron pair along with it.

(ii)

In a position β–to the leaving group, the substrate contains an atom or group nearly always hydrogen that can be abstracted by a base, leaving its electron pair behind.

(iii)

Reaction is brought about by the action of a base. Typically, the base is a strongly basic anion like, an alkoxide, derived from an alcohol and KOH. For example, ethoxide, C2H5O − , tert-butoxide, (CH3)3CO − etc. Heterolytic bond dissociation energies show that strength of carbon-halogen bonds follow the sequence, Bulky bases like text–butoxide salts give to anti Saytzeff or Hoffmann elimination product.

Heterolytic bond (Dissociation energy) R–F > R – Cl > R – Br > R – I In these elimination reactions the reactivity of alkyl halides follows the sequence, Reactivity towards (E2 elimination) R–I > R – Br > R – Cl > R – F Some of the examples are, (a)

CH3

(b)

CH3

CH3

CH2=C–CH2CH2CH3+CH3–C=CHCH2CH3

CH3–C–CH2CH2CH3

(major)

(minor)

Br

CH3CH3

(c)

CH3

CH3–C – CHCH3

(CH3)2C=C(CH3)2+CH2=C–CH(CH3)2 (minor)

(major)

Br

E2 Mechanism This reaction involves a single step in which base pulls a proton away from the carbon and simultaneously a halide ion departs, thus a double bond is formed. Halogen takes an electron pair with it while hydrogen leaves its electron pair behind to form the double bond. What characterizes this particular mechanism is that all the steps are taking place simultaneously, in a single step, via a single transition state. So, it is a concerted mechanism. #

Xδ–

X –C–C– H

–C

+

• – •B

• – •X +C=C+B–H

C– Hδ

Bδ

–

In this transition state two bonds are being broken that is C – H and C – X. The energy required for bond breaking comes from the bond formation of the bond between the proton and the base, and the formation of π bond. As the base begins to pull proton away from the molecule, the β-carbon, armed with the electron pair begin to form a π bond with another carbon atom. As the π bond starts to form, the carbon-halogen bond starts to break. Thus the π bond making helps to supply energy for the carbon-halogen bond breaking.

O-207 The rate-determining step involves reaction between a molecule of alkyl halide and a molecule of base, and its rate is proportional to the concentration of both the reactants. This mechanism is named E2, as molecularity is 2. Rate = k[RX][:B–]

E2 reaction (Bimolecular elimination)

Moreover, the order of reaction is also two because the rate of reaction depends upon the concentration of alkyl halide and base. The best stereospecific conformation for E2 elimination is an anti-coplanar conformation with H and X being 180° apart which permits the approaching electron-rich base to be at a maximum distance from the electron-rich leaving group. B•• – H

anti –coplanar

L L=Electron rich leaving group

In certain cases, exception to Saytzeff's rule have also been observed. If one carries out dehydrohalogenation with a strong and bulky base such as potassium tert butoxide in tert butyl alcohol, it leads to the formation of the less substituted alkene in higher amount. Such kind of product formation is governed by Hoffmann's elimination rule, so it is termed as Hoffmann's product. The dehydrohalogenation follows Hoffmann's rule in two cases, either 1.

When the base employed is very bulky like potassium tertiary butoxide and the leaving group is any halide (Cl, Br, I) or

2.

When the base employed is ethoxide and the leaving group is a poor one like fluoride ion.

1.3.4 Acid Catalysed Dehydration of Alcohols Heating an alcohol with a strong acid like H2SO4 causes elimination of water and form an alkene. Elimination of water is called dehydration. +

–C–C–

H

heat

C=C

+ H2O

H OH

The reaction is an elimination process and is favoured at higher temperatures. The most commonly used acids in the laboratory are Bronsted acids which act as proton donors such as sulphuric acid and phosphoric acid. Mechanism The mechanism of acid catalyzed dehydration is E1 (unimolecular elimination). In this case, the substrate is a protonated alcohol also termed as an alkyl oxonium ion. It gives rise to the formation of carbocation intermediate by the loss of weaker base, H2O. The carbocation formed may undergo rearrangement (if

O-208 possible) and it finally loses a proton to give an alkene. A catalytic role is assigned to the acid H2SO4 as it is regenerated at the end of the reaction. Step 1: Protonation of alcohol

Step 2: Formation of a carbocation slow

CH3CH–CH3

+

CH3CHCH3+H2O [rate determining step] Isopropyl carbocation

+

H2O

Step 3: Deprotonation + fast CH CH=CH +H SO CH3CH–CH2+HSO4– 3 2 2 4

H

It is important to note that the formation of the more stable alkene is the general rule as suggested by Saytzeff's or Zaitsev rule in the acid-catalyzed dehydration of alcohols. However, Saytezeff's rule could predict the formation of product only when the generated carbocation does not undergo rearrangement. In case of carbocation undergoing rearrangement, dehydration does not follow Saytzeff's rule. Dehydration reactions of alcohols show several important characteristics which are as follows, 1.

The experimental conditions like temperature and acid concentration that are required to bring about dehydration are closely related to the structure of the individual alcohol. Alcohols in which the hydroxyl group is attached to a primary carbon (in case of primary alcohols) are the most difficult to dehydrate. For example, dehydration of ethanol requires concentrated sulphuric acid and a high temperature of 180°C. Conc. H SO

4 → CH = CH + H O CH3 – CH2 – OH    2 2 2 2

180 ° C

Secondary alcohols usually dehydrate under milder conditions. For example, cyclohexanol, which is a secondary alcohol, dehydrates in 85% phosphoric acid at 165-170°C. OH 85% H3PO4 165–170°C

+H2O

Tertiary alcohols are usually so easily dehydrated that extremely mild conditions is be used. For example, tert butyl alcohol dehydrates in 25% H2SO4 at a temperature of 85°C only. CH3 CH3–C–OH

CH3 25%H2SO4 85°C

CH3–C=CH2+H2O

CH3

Thus, overall, the relative ease with which alcohols undergo dehydration is in the following order 3° alcohol > 2° alcohol > 1° alcohol

O-209 2.

Some primary and secondary alcohols also undergo rearrangements of their carbon skeleton during dehydration since the intermediate involved is a carbocation which readily undergoes rearrangement. Reactions involving carbocations undergo rearrangements wherever possible. They occur almost invariably when the migration of an alkanide ion or hydride ion can lead to a more stable carbocation. For example,

CH3

CH3 +

CH3–C–CH2

methanide migration

+

CH3–C–CH2 CH3

CH3 H +

hydride

CH3–C–CH–CH3 migration

+

CH3–C–CH2–CH3 CH3

CH3

Rearrangements of carbocations can also lead to the change in ring size, as the following example shows ring expansion. CH3 OH

CH3 +

CH–CH3 CH3 H+,heat (–H2O) 2°Carbocation

In order to predict that a given ring size would undergo ring expansion or contraction, the following points can be used as guiding principle. (i)

(ii)

Formation of cyclic compounds by intramolecular cyclizations depends on two factors. (a)

Thermal stability which is determined by ring strain.

(b)

The entropy factor (probability of bringing together the terminal carbons) that decreases with chain length, making it successively more difficult to synthesize rings of increasing size.

Using the above two factors, we will summarize the ease of ring closure for cyclic compounds. (a)

Thermal stability:

6 > 7, 5 > 8, 9 >> 4 > 3

(b)

Entropy (probability) of ring closure:

3>4>5>6>7>8>9

Ease of synthesis (net effect of 1 and 2) :

5 > 3,6 > 4, 7, 8, 9

O-210 (iii)

In 3–membered ring synthesis, the entropy factor is more important than the thermal stability, so their formation is readily attained despite greatest ring strain. For rings larger than six carbons, ring stability is outweighed by the highly unfavorable entropy factor, so larger rings with more than six carbons are stable but difficult to prepare. Table1: Comparative details of E1 And E2 Reactions E1 reaction

(a) Steps

(i)

H–C–C++

H–C–C–X (ii)

(b) Transition state

(i)

(ii)

(c) Kinetics

(d) Driving force (e) Stereo specificity (f) Effect of R

+ H–C–C+ –H

H–C–Cδ

Hsδ

E2 reaction

+

C=C δ–

B••

–

Xδ

H

B•• H+C=C+×

B••–+H–C–C–X

δ+

C

H

C

Xδ

C

–

C

First order, unimolecular.

Second order, bimolecular.

Rate = k1[RX]

Rate = k2[RX][:B − ]

Ionization of R–X Non–stereospecific Stability of R

+

Attack by B:

−

on H

Anti elimination Alkene stability (Saytzeff Rule)

3° > 2° > 1° RX (g) Rearrangement (h) Regio-selectivity

Common Saytzeff

None Usually, Saytzeff but Hoffmann elimination taken place with bulky bases like (Me3CO − )

(i) Alkyl group (j) Base strength (k) Leaving group

3° > 2° > 1°

3° > 2° > 1°

Weak

Strong

Weak base favours reaction

Weak base favoures reaction

−

(l) Catalyst

−

I − > Br >Cl − > F −

I − > Br >Cl − > F −

Ag+, AlCl3

No specific catalyst

O-211

Solved Examples Example1: Compare the mechanisms of E1 and E2 reactions. Solution: Concerted (One-step) mechanism–bimolecular elimination –E2 HO H

H

CH3

CH3

CH3 CH3CH2

Na+

I

H +H2O+NaI

C=C CH3

C2H5

Unconcerted (Two–step) mechanism–unimolecular elimination–E1 O H H

C

H

CH3 CH3CH2

CH3 I

CH3

O

–

H

CH3 CH3 CH3CH2 or

Na+

H +

CH3 (CH3–CH2)

C=C

C2H5 CH3

CH3 H CH3CH2

H

+

CH3 C=C C2H5

CH3 CH3 H

1.3.5. Debromination of Vicinal Dibromides Vicinal or vic dibromides are dibromo compounds in which the bromine are placed on adjacent carbon atoms. The name geminal or gem dibromide is used for those dibromides when both bromine atoms are attached to the same carbon atom. Vic-dibromides undergo debromination when they are treated with a solution of sodium iodide in acetone or a mixture of Zn dust in acetic acid or ethanol. –C–C– +2NaI

acetone

C=C

+ I2+2NaBr

Br Br

–C–C–

Zn in CH3CO2H or CH3CH2OH

C=C

+ ZnBr2

Br Br

Neither method is used for preparing alkenes since the necessary dibromo compounds are not readily accessible, in fact they are prepared by the reaction of alkenes with Br2.

O-212 Mechanism A possible mechanism of debromination with NaI in acetone is Br C—C– fast Br –+ C—C

slow

C=C + IBr

+

I–

Br

Br I

Probable mechanism of debromination with Zn in acetic acid is Zn → Zn2+ + 2e– Br

Br

2e–+ —C – C–

Br –+ –C—C–

C=C + Br–

Br

1.3.6 Wittig Reaction This reaction involves a special class of compounds called alkylidene phosphoranes also known as phosphorus ylides which have a carbanionic site. This carbanion attacks the carbonyl carbon giving an intermediate called betaine which undergoes ring closure and then undergoes fission to give alkenes. δ+

+

PPh3

O

O–

PPh3

O

PPh3

C

C

C

C

–

C R1

C H

H

δ

R2

R1

H H Betaine

R2

R1

H

H

Ph3P=O R2 H

+

H

C=C R2

R1 Alkene

If one sees the reaction closely in terms of reactants, it can be concluded that the alkene can be generated by putting the two reactants in such a manner that >C = O and Ph3P = C< part faces each other and then removing O and PPh3 and joining the two carbon atoms. C=O+Ph3P=C

C=C+Ph3P=O

For example, the product formation of the reaction below is predicted on the basis of Wittig reaction.

O-213 O

+Ph3P=CH–CH3

X; now we have to identify x

Going with the mechanism of Wittig reaction, the product 'X' can be easily written as Some more examples of Wittig reaction are,

CH–CH3

•• –

••

O ••

+

– ••

••

+ CH2—Ph3P O ••

+

PPh3

CH2

••

+

O ••

•• –

••

CH2+ Ph3P— O ••

PPh3

CH2

CH3

H3C

H3C Br Ph3P+-CH3 KOBu 100°C,2h

H3C

O Camphor

CH3

91% yield H3C

CH2

1.3.7 Heating Quaternary or Tetra Alkylated Ammonium Hydroxides Alkenes can also be formed by heating tetra alkyl ammonium hydroxides with the simultaneous elimination of trialkyl amine. This reaction involves Hoffmann's elimination and product will be governed by Hoffmann's rule. H

OH– ⊕

C2H5

CH2–CH2–N–C2H5 α β C2H5

∆

CH2=CH2+(C2H5)3N+H2O

OH–

CH3 β α ⊕ CH3–CH2–CH–N–CH3 CH3 CH2–H

∆

CH3CH2CH=CH2+(CH3)3N+H2O (major) + CH3CH=CH–CH3 (minor)

In those cases, when the two alkyl groups bear β −hydrogen atoms, the elimination occurs from that alkyl part which has more number of β −hydrogens.

O-214 β OH– α CH2–CH2CH3 β ⊕ α ∆ CH3CH2CH=CH2+N(Me)2n–Pr+H2O CH3–CH2–CH–N–CH3 (major) CH2–H CH3 CH3CH=CH–CH3 β (minor)

Physical properties of alkenes The first three members are gases, the next fourteen members are liquids and the higher ones are solids. They are colourless and odourless except ethylene which has a faint sweet smell. These are practically insoluble in water but fairly soluble in non–polar solvents like benzene, petroleum ether, etc. They show a regular gradation in physical properties, such as boiling points, with increasing carbon content. The boiling points of two successive members of the homologous alkene series differ by about 20–30°C, except for very small homologues. The branched chain alkenes have lower boiling points than the corresponding straight chain alkenes. Like alkanes, alkenes are generally non–polar, but certain alkenes are weakly polar. For example, dipole moment of propene and 1–butene is about 0.35D due to their unsymmetrical geometry. Cis–alkenes, in contrast to trans–alkenes, also have a small dipole moment. Therefore cis–alkenes, boil at somewhat higher temperature than the trans–alkene. Cis–alkenes have poor symmetry and as such do not fit into the crystalline lattice, with respect to the trans–isomers. Consequently, cis–alkenes have generally lower melting points. Chemical properties of alkenes Alkenes commonly participate in electrophilic addition reaction due to presence of π electron cloud, which makes them an electron abundant system inviting attack from electrophile. Out of alkenes and alkynes, alkenes are more reactive than alkynes towards electrophilic addition reaction. It is definitely important to remember that hydrogenation of alkene is not an electrophilic addition reaction. Moreover, towards hydrogenation, out of alkenes and alkynes, alkynes are more reactive than alkenes.

1.5 Hydrogenation Generally, the addition of hydrogen to unsaturated compounds like alkenes or alkynes is among the most common and almost certainly the most useful of all their addition reactions. Direct addition of hydrogen always involves heterogeneous catalysis by finely divided metals such as Ni, Pt, Pd, Ru, Rh etc. It is significant in this context that both alkenes and hydrogen react exothermically and reversibly with the catalytic metals. With the alkene this presumably involves its π electrons as alkanes are not similarly adsorbed. No π electrons are available in the hydrogen molecule either and its adsorption must involve considerable weakening of its σ bond, though not necessarily complete fission to yield H • atoms. The actual spacings of the metal atoms in the surface will clearly be of importance in making one face of a metal crystal catalytically effective and another not, depending on how closely the actual atom spacings approximate to the bond distances in alkene and hydrogen molecules. In actual practice, only a relatively small proportion of the total metal surface is found to be catalytically effective, so called 'active sites'. These adsorb alkene strongly, and then desorb the resultant alkane molecule immediately, thus becoming free for further alkene absorption. In agreement with this 'lining up' of alkene molecules on the catalyst surface and the probable approach of activated hydrogen molecules from the body of the metal, it might be expected that hydrogenation would proceed stereoselectively syn. For example,

O-215

Me

H Ni/H2

H Me

Me Cis isomer Me3CC≡CCMe3

Me H2 Lindlar's catalyst

H

H C=C

Me3C

(1)

CMe3

not

H C=C Me3C

CMe3 H

(2)

Alkynes can often be reduced selectively to the alkene by use of the Lindlar's catalyst (Pd supported on CaCO3 partialy poisoned with Pb(OAc)2). Here again syn stereoselectivity is observed despite the fact that this will lead to the more crowded, thermodynamically less stable cis–alkene (1) rather than trans alkene(2). It has also been demonstrated that cis alkenes are hydrogenated much more rapidly than trans alkenes. In any case, the rate of hydrogenation falls with increasing substitution in the alkene. More recently homogeneous hydrogenation catalysts, such as RhCl(Ph3P)3, (Wilkinson's catalyst) have been developed which are soluble in the reaction medium. These are believed to transfer hydrogen to an alkene via a metal hydride intermediate. Hydrogenation with Wilkinson's catalyst too leads to a considerable degree of syn stereoselectivity in hydrogen addition. Example 2: (i) Complete the following reactions Pd,Pt, or Ni (a) CH3CH = CH2 + H2     → ? [RhCl(Ph 3 P)3 ] (b) CH3CH = CH2 + H2      → ?

The catalyst in (ii) is Wilkinson's catalyst. (ii)

In terms of the catalysts used, classify these two reactions.

(iii)

Show the steps in the mechanism of reaction (ii) involving the Wilkinson's catalyst.

Solution : (i) (a) and (b) CH3CH2CH3. (ii)

Both reactions are catalytic hydrogenations (addition of H2). (i) Heterogeneous catalytic hydrogenation in which the solid catalyst is in a different phase from the reactants. (ii) Homogeneous catalytic hydrogenation because reactants and catalyst are in the same phase. (c) In step 1, an H2 adds to the rhodium complex and one Ph3P ligand (L) is lost, resulting in a five–coordinate rhodium complex, (A) (L = Ph3P). In this oxidative addition, the Rh changes oxidation state from +1 to +3. In step 2, the alkene reacts with (A) to form a π complex, (B) which undergoes rearrangement (step 3) of an hydrogen atom to one of the carbons of the double bond, the other carbon forming a σ bond to the Rh (C).

O-216 H Cl

L Rh L

L

R2 –L

H Cl

L Rh

H Cl

L

C=C

Rh

Rh

L

L

L

L

Cl

L

C–C–H

L

C=C (A)

(B)

(C)

In the last step, a second hydrogen atom is transferred to the other carbon, and the alkane is lost with simultaneous regeneration of the metal catalyst. (C)

+L

[RhCl(Ph3P)3]+HCH2CH2H.

The catalysis depends on the ability of rhodium (a transition metal), to form a π complex with alkene.

1.5.1 Addition of Hydrogen Halides Hydrogen halides (HCl, HBr and HI) add to the double bond of alkenes. R R –C–C–

C = C+HX

X H

Mechanism for addition of hydrogen halide to an alkene is a normal electrophilic addition mechanism. It involves the following two steps, Step 1: Formation of carbocation R H R

slow C = C+H–X (RDS) –C–C–+X– ⊕

RDS is rate determining step. Step 2 : Attack on carbocation R X

R –C–C–+X

–

fast

–

–C–C–

⊕

H

X H

The addition of HBr to some alkenes gives a mixture of the expected alkyl bromide. If there is a possibility of rearrangement, then some more isomers are also formed.

O-217

CH3

CH3 CH2=CH CH–CH3

H+

CH3–CH–C–CH3 ⊕

H

CH3 Br–

CH3–CHBr–CH–CH3 ∼H– 2–Bromo–2–methylbutane CH3 CH3CH2–C–CH3

–

Br

⊕

CH3

CH3CH2–C–CH3 Br

3°

2–Bromo–2–methylbutane

In the ionic addition of an unsymmetrical reagent to a double bond, the positive portion of the reagent attaches itself to a carbon atom of the double bond so as to yield the more stable carbocation as intermediate. Because this is the step that occurs first, it is the step that determines the overall orientation of the reaction. In those cases where rearrangement does not occur, the addition of HX to alkenes follow Markovnikov's rule. The rule states that during the addition of an unsymmetrical reagent to an unsymmetrical alkene "the negative part of the unsymmetrical reagent goes to that carbon atom which bears lesser number of hydrogen atoms". But in those cases, where rearrangement occurs, the overall addition of HX to alkenes does not follow Markovnikov's rule. For example, when HI is added to 1-butene the reaction leads to the formation 2-iodobutane. 2-iodobutane contains a stereocenter as shown here with asterisked carbon, *

CH3CH2–CH–CH3

CH3CH2CH=CH2+HI

I

The product, therefore, can exist as a pair of enantiomers. The carbocation that is formed in this first step of the addition is trigonal planar (sp2 hybridized) and is achiral. When the iodide ion reacts with this flat carbocation, reaction is equally likely at either face. Thus reaction leads to the formation of two enantiomers and both the enantiomers are produced in equal amounts. Thus, product of the reaction is a racemic mixture. Example 3: Explain the regioselectivity of the products when HBr reacts with (a) CF3CH = CH2, (b) BrCH = CH2 and (c) CH3–O–CH = CHCH3. Solution: An electron–withdrawing group attached to the carbon that has lesser number of hydrogens will tend to destabilize a positive charge on this carbon and cause the addition to be anti–Markovnikov in nature. Such is the case in (a), where the major product is CF3CH2CH2Br. However, in (b) the electron–withdrawing inductive effect of Br is superseded by extended π delocalization of an unshared electron pair from Br to C+, ••

+

+

thereby stabilizing the R+,[•• Br − C H – CH3 ↔[•• B R = CH − CH3 ] .Thus, the Markovnikov product, ••

••

Br2CHCH3, is formed. (c) Protonation of either C of the C = C gives a 2° R+. However, O atom behaves just + + like Br in delocalizing a pair of electrons to the adjacent C , stabilizing the positive charge and H bonds to the carbon β to the O as follows ••

+

+

[CH3– O − C H–CH2CH3 ↔ CH3 – O = CH – CH2CH3] ••

••

O-218 The product is CH3 – O – CHBrCH2CH3. In general when the electron withdrawing group attached to the doubly–bonded carbon has an unshared pair of electrons, Markovnikov addition is observed.

1.5.2 Addition of Hydrogen Bromide in Presence of Peroxides The addition of HBr to propene, MeCH = CH2(1), under polar conditions ( in absence of peroxide) yields 2–bromopropane. However, in the presence of peroxides like benzoyl peroxide or under other conditions that promote radical formation the addition yields 1–bromopropane. This addition of HBr in presence of peroxide is generally referred to as the peroxide effect leading to Anti Markvnikov addition. It is also referred as Kharasch effect. This difference in orientation of HBr addition is due to the fact that in the first case, the reaction is initiated by H+ and proceeds via the more stable secondary carbocation while in the second case, it is initiated by Br • and proceeds via the more stable secondary carbon free radical. Mechanism Anti– Markvnikov addition is a free radical mechanism and it is normally comprised of 3 steps. Step 1:Chain initiation step R–O–O–R → 2RO • (–O–O–bond is weak) RO • + H–Br → RO – H + Br • Step 2:Chain propagation step •

•

MeCH=CH2+Br• (1)

MeCH–CH2Br+MeCH–CH2 H–Br (2)

(4)

Br

•

Br +MeCH2–CH2Br (3)

The initiation is by Br • , as hydrogen abstraction by RO • from HBr is energetically much more favourable than the alternative of bromine abstraction to form ROBr + H • . The alternative addition of Br • to (1) to form •

MeCH(Br)CH2 • (4) does not occur, as secondary radical MeCHCH2Br (2) is more stable than primary radical (4). It is very important to note that anti –Markovnikov addition is applicable only for HBr and that too in the presence of a peroxide.The reason for this is reflected in the ∆H values (kJ mol–1) for the two steps of the chain reaction for addition of HX to CH2 = CH2, as shown by data below, (1) X • + CH2 = CH2

(2) XCH2CH2 • + HX

H–F

–188

+155

H – Cl

–109

+ 21

H – Br

–21

–46

+29

–113

H–I

O-219 Only for HBr, both the chain steps are exothermic while for HF the second step is highly endothermic, reflecting the strength of the H–F bond and the difficulty of breaking it. For HCl, it is again the second step that is endothermic (though not to such a great extent) while for HI it is the first step that is endothermic, reflecting the fact that the energy gained in forming the weak I–C bond is not compensatory to the energy being lost in breaking the C = C double bond. Thus only a few radical additions of HCl are known, but the reactions are not very rapid and the reaction chains are short at ordinary temperatures. Even with HBr addition, the reaction chains tend to be rather short, much shorter than those in halogen addition and more than a trace of peroxide is thus needed to provide sufficient initiator radicals. For preparative purposes up to 0.01 mol peroxide per mol of alkene is required. Once inititated, reaction by this pathway is very much faster than any competing addition via the polar pathway and the Anti–Markovnikov product like (3) will thus predominate. If the Markovnikov product, e.g. MeCH(Br)CH3 from propene, is required it is necessary either to purify the alkene rigorously before use or to add inhibitors (good radical acceptors such as phenols, quinones, etc) to mop up any radicals formed in the reaction. Essentially complete control of orientation of HBr addition, in either direction, can thus be achieved, under preparative conditions, by incorporating either peroxides (radical initiators) or radical inhibitors in the reaction mixture. This is particularly useful as such control is not confined purely to alkenes themselves. Example 4 : (i) Rule out the following mechanism for the peroxide induced addition of HBr to a symmetrically R–substituted alkene such as RCH = CHR. Initiation step: RO • + HBr → ROBr + H • •

Propagation step 1: RCH = CHR + H • → R C HCH 2R • Propagation step 2: R C HCH 2R + HBr → RCHBrCH 2R + H • Bond energies of H – Br and C – Br bond are 88 and 69 kcal mol–1 respectively. Example: (ii) Why is this mechanism inappropriate for an unsymmetrically R–substituted alkene such as R2C = CH2 ? Solution : (i) The actual initiation step, RO • + HBr → ROH + Br • , is thermodynamically more favorable because O–H has a much greater bond energy than the very weak O–Br. Also, step 2 is endothermic, ∆H = (bond energy of H–Br)–(bond energy of C–Br) = 88–69 = +19 kcal/mol. Solution: (ii) This mechanism would lead to the unobserved Markovnikov addition.

1.5.3 Addition of Bromine and Chlorine Alkenes react rapidly with bromine at room temperature and in the absence of light. If bromine in CCl4 is added to an alkene, the reddish-brown colour of bromine disappears almost instantly as long as the alkene is present in excess. This serves as a classical test for detection of unsaturation in alkenes or alkynes.

O-220 (a)

+ Br2

C=C

CCl4 r.t.

–C–C–

rapid decolorization of Br 2/CCl4 is a test for alkenes and alkynes

Br Br vicinal–dibromide

(b)

CH3–CH–CH–CH3

CH3CH=CH3+Cl2 –9°C

Cl

Cl

Mechanism The mechanism proposed for halogen addition is an ionic mechanism proceeding via cyclic halonium ion in general. In the first step the exposed electrons of the π bond of the alkene attack the halogen in the following way. R C 1

C

2

Br2 in CCl4

R

+

C

1

–

Br

–

C

–Br–

2

R C

Br

C

C

Br

+

Cyclic bromonium ion

Br

Br

R

C

Br Vicinal dibromide

In this reaction, when the attacking reagent (bromine) approaches alkene, the temporary polarization develops on the alkene with C2 atom gaining a negative charge and C1 atom acquiring positive charge as it can be compensated by the +I effect of R group. The alkenes, being electron rich compounds due to the presence of π–electron cloud, are attacked by the electrophile (Br+) to give a cyclic bromonium ion. Here, the formation of cyclic bromonium ion as intermediate is possible because bromine is of considerably large size having lone pairs to be bonded to both the carbons simultaneously. The cyclic bromonium ion is then attacked by Br– from the top as lower side is already blocked. Now, the three membered ring is cleaved by trans opening giving vicinal dibromide as the product. Thus, the overall addition of Br2 to alkene follows trans stereoselectivity. For example, when cyclopentene reacts with bromine in CCl4, anti-addition occurs and a mixture of two enantiomeric molecules of trans-1,2-dibromocyclopentane are obtained giving rise to a racemic modification as shown below,

Br2 CCl4

Br H H

Br

+

H

Br

Br

H

Similarly, when cis-2-butene adds bromine, the product is a racemic form of 2-3-dibromobutane. When trans-2-butene adds bromine, the product is the meso compound. Thus it is observed that a particular stereoisomeric form of the starting material react in such a way that it gives a specific stereoisomeric form of the product. Thus the reaction is stereospecific.

1.5.4 Addition of Water Conversion of an alkene into alcohol can be achieved by hydration of alkene. Couple of more methods like hydroboration - oxidation and oxymercuration -demercuration also convert an alkene into alcohal.

O-221 1. Acid catalyzed hydration: The acid catalyzed addition of water to the double bond of an alkene is a method of preparation of low molecular mass alcohols. The addition of water to the double bond follows Markovnikov's rule in those case where rearrangement is not involved. CH3

CH3 CH3–C=CH2+HOH

H+

CH3–C–CH3

25°C

OH

Mechanism As the reactions follow Markovnikov's rule, acid catalyzed hydration of alkenes do not yield primary alcohols except in the special case of the hydration of ethene. The occurance of carbocation rearrangements limits the utility of alkene hydrations as a laboratory method for preparing alcohols. +

OH

OH2 MeCH=CH2

H⊕

⊕

MeCH –CH2

H2O

MeCH–CH2

H (1)

–H⊕

MeCH–CH2 H

H

The formation of the carbocationic intermediate (1), either directly or via an initial π complex, appears to be rate limiting and the overall orientation of addition is Markovnikov. Acids that have weakly nucleophilic anions (like HSO −4 from dilute aqueous H2SO4) are chosen as catalysts, so that their anions will offer little competition to the actual nucleophile H2O. In case, if any ROSO3H is formed, it will be hydrolysed to ROH under the conditions of the reaction. 2. Hydroboration-oxidation (HBO): Whenever an alkene is allowed to react with diborane followed by hydrogen peroxide an alcohol is obtained as shown below, BH3. ET2O diglyme

B

91% H2O2 / NaOH / H2O

OH

So, as observed, in the overall hydroboration-oxidation reaction, H2O is added to the double bond. This is reiterated with the examples below, (i)

(ii)

CH3

CH3 CH3–C=CH2 CH3

CH3–CH=C–CH3

(BH3)2

(BH3)2

H2O2 / OH–

H2O2 / OH–

CH3–CH–CH2OH CH3

CH3–CH–CH–CH3 OH

O-222 The reaction is free from rearrangment because the reaction mechanism does not involve carbocation intermediate. Moreover, the reaction involves syn addition of the two groups - H and OH. The reaction scheme can be presented as below, R'

R'

R' C=C+H–B

H Hydroboration

H2O2/OH–

–C–C–

H

H

H B

Oxidation

–C–C– H OH

H

Mechanism Step1:Hydroboration With the reagent diborane, (BH3)2 alkenes undergo hydroboration to yield trialkylboranes, R3B. CH3–CH=CH2

(BH3)2

CH3–CH–CH2

CH3-CH2-CH2 BH2

BH2

H

(CH3–CH2–CH2+3B)

Step2: Oxidation Trialkylboranes, R3B on oxidation gives alcohols. The oxidation step involves conversion of the C – B bond into C – OH without inversion. The probable mechanism involves the following steps, H

R B

O–O

R

R–B–O–O

R R H

R

H

R–B–O–R + OH HOO–

R R

R–B–O–R

B–O–R

O–O–B–O–R O

R

R

HOO– O

O

O

R + H

R

O H

R–O–B–O–R O OH + R

H–O H2O

O–H B

H

+ R–O

O H

Interestingly, it can be said that hydroboration–oxidation process gives products corresponding to anti-Markovnikov addition of water to the carbon–carbon double bond. Some more examples are,

O-223 OH 1. B2H6

+

2. H2O2

OH

85%

15%

H BH3

H3C C

CH2

BH3 THF

H OH H2O2

H3C—C—CH2

H3C

OH–

H3C—C—CH2

CH3

CH3

3. Oxymercuration-Demercuration: In the overall oxymercuration-demercuration reaction, H2O is added to the double bond. The reaction is again free from rearrangment like hydroboration-oxidation. It is because the reaction does not involve carbocation intermediate.Moreover the reaction involves syn addition using Markovnikov's rule. Alkenes react with mercuric acetate in the presence of water to give hydroxymercurial compounds which on reduction yield alcohols. H HgOAc

R

H Oxymercuration C = C+H2O+Hg(OAc)2 H R

R–C–C–H

HH NaBH4 Demercuration

OH H

R–C–C–H OH H

OAc is CH3–COO– group

The first stage, oxymercuration involves addition to the carbon–carbon double bond of -OH and Hg(OAc)2. The electrophile of Hg(OAc)2 is AcOHg+ that adds C = C to form a mercurinium ion similar to a bromonium ion. The mercurinium ion then reacts with H2O (not OAc–) at the more substituted carbon. +HgOAc ••

••

H2O

[RCH–CH2]

HgOAc OAc–

Mercurinium ion

RCH–CH2+HOAc OH

Then, in demercuration, HgOAc is replaced by H. The reaction sequence amounts to hydration of the alkene, but is much more widely applicable than direct acid catalysed hydration due to the distinct advantage of yielding non rearranged products. Oxymercuration-demercuration gives alcohols corresponding to Markovnikov addition of water to the carbon-carbon double bond. For example,

O-224 (i)

CH3(CH2)3CH=CH2

Hg(OAc)2,H2O

CH3(CH2)3CH–CH2 OH HgOAc NaBH4

CH3(CH2)3CH–CH3 OH

CH3

(ii)

CH3

CH3–C–CH=CH2

NaBH4

Hg(OAc)2,H2O

CH3–C–CH–CH3 CH3 OH

CH3

1.5.5 Halohydrin Formation If the halogenation of an alkene is carried out in aqueous solution rather than in CCl4, the major product of the reaction is a halohydrin molecule. In this case, the molecules of the solvent become reactants. –C–C–+–C–C–+HX

C=C+X2+H2O

X OH X X X2=Cl2 or Br2

Mechanism Halohydrin formation can be explained by the following mechanism C=C+X–X

–C–C–+X– X+ +OH

–C–C–+H2O X+

–C–C–

OH

2

–H+

X

–C–C– X

The addition of X and OH occurs in the trans manner, as the reaction proceeds by the formation of halonium ion intermediate. If the alkene is unsymmetrical, the halogen adds up on the carbon atom with greater number of hydrogen atoms so that the overall addition follows Markovnikov's rule. H3C H3C

C = CH2

Br2,H2O or HOBr

CH3 CH3

C–CH2Br OH

1.5.6 Hydroxylation There are a number of reagents that overall add two OH groups to alkenes. The two OH groups can be either added from the same side known as syn hydroxylation or from the opposite side which is generally termed as anti hydroxylation.

O-225 1.

Syn hydroxylation: Syn hydroxylation can be achieved by the use of osmium tetroxide (OsO4) or by the use of Baeyer's reagent Baeyer's reagent is cold, alkaline KMn O 4 .

Addition of OsO4 OsO 4 adds to yield cyclic osmic esters (2) which can be made to undergo ready hydrolytic cleavage of their Os–O bonds to yield the1,2–diol (3). Me

O

H

Me

OsO4

O

Os HO

O O

H

Me

H

HO Me

H +(HO)2OsO2

Me H

Me H (1)

H2O

(meso compound)(3)

(2)

Cis 2–butene (1) thus yields the meso 1,2 diol–(3), i.e., the overall hydroxylation is stereoselectively syn, as would be expected from Os–O cleavage in a necessarily cis cyclic ester (2). The disadvantage of this reaction as a preparative method is the expense and toxicity of OsO4. However, this disadvantage can be nullified by using it only in catalytic quantities in association with H2O2 which re–oxidises the osmic acid, (HO)2OsO2, formed to OsO4. Addition of Baeyer's reagent (cold, alkaline permanganate, Mn O −4 ) Alkaline permanganate, MnO −4 (a reagent used classically to test for unsaturation), will also effect stereoselective syn addition and this by analogy with the above, is thought to proceed via cyclic (cis) permanganic esters. It has not proved possible actually to isolate such species but use of Mn18O −4 , was found to lead to a 1,2–diol in which both oxygen atoms were O18 labeled. Thus both were derived from MnO −4 , and neither from the solvent H2O, which provides support for a permanganic analogue of (2) as an intermediate, provided that Mn18O −4 undergoes no O18 exchange with the solvent H2O under these conditions. The disadvantage of MnO −4 for hydroxylation is that the resultant 1,2 –diol is very much susceptible to further oxidation by it. 2.

Anti hydroxylation and epoxidation: Peroxyacids, RCOOOH will also oxidize alkenes, e.g. trans 2–butene (4), by adding an oxygen atom across the double bond to form an epoxide (5). O

O C R

C R

– Oδ –

H–Oδ Me

O

H

H

H

Me Me

H (4)

O

O

H

Me

H

Me

Me H (5)

+RCO2H

O-226 Epoxides though uncharged have a formal resemblance to cyclic bromonium ion intermediates, but unlike them are stable and may readily be isolated. However, they undergo nucleophilic attack under either acid or base catalysed conditions to yield the 1,2–diol. In either case attack by the nucleophile on carbon atom will be from the opposite side of the oxygen bridge in (5). Such attack on the epoxide will involve inversion of configuration. O

O–

H

Me

Me

Me

H

OH H2O

OH–

Me

HO

OH Me

H

H

Me

H

H

(7) (5) H⊕

H OH

O

⊕

H2O

H

Me H

Me

Me

H

OH Me H

Me –H+

⊕

H

Me

OH

OH2

(6)

H

(7)

Attack has been shown on only one of the two possible carbon atoms in (5) and (6), though on different ones in the two cases. In each case, attack on the other carbon will lead to the same product, the meso 1,2–diol (7). By comparing the configuration of (7) with that of the original alkene (4), it can be seen that in overall terms setereoselective anti hydroxylation has been effected.Thus by suitable choice of reagent, the hydroxylation of alkenes can be made stereoselectively syn or anti. For example, 3CH2–CH2+2MnO2+2KOH (Syn addition)

3CH2=CH2+2KMnO4+4H2O

OH OH (Ethylene glycol)

CH3CH=CH2

(1) OsO4 (2) NaHSO3/H2O

CH3–CH–CH2 (Syn addition) OH OH (Propylene glycol) (Anti addition)

OH H CH3 H

CH3 C=C H

Cis–2–butene

RCO3H H+, H2O

CH3–C–C–CH3+enantimer H OH (Trans–,2–diol)

O-227 Example 5: Write the products formed when cis and trans - 2 -butene are treated with peroxy acetic acid. Solution :

CH3

CH3 C=C

C–C

CH3

CH3

H

H

O

CH3CO3H H

H

Cis–2-butene CH3

H C=C CH3

C–C

H

H

O

O

CH3CO3H

CH3

Trans–2–butene

CH3

+

C–C CH3

H CH3

H

H

Example 6 : Write the products formed when 1-butene is treated with mcpba. O

Solution :

O O–OH CHCl3

H2C–CH2–CH=CH2 +

H3C–CH2–CH–CH2 + 1,2–epoxybutane

O Cl OH

mcpba

Cl

1.5.7 Oxidation by Hot Alkaline KMNO4 Alkenes are oxidatively cleaved by hot alkaline permanganate solution. The terminal CH2 group of 1-alkene is completely oxidized to CO2 and water. A disubstituted carbon atom of a double bond becomes \/ C = O group of a ketone. A monosubstituted carbon atom of a double bond becomes aldehyde group which is further oxidised to salts of carboxylic acids. For example, (i)

CH3CH=CHCH3

KMnO4,OH– heat

(cis or trans)

(ii)

CH3 CH3–CH2–C=CH2

(c)

O 2CH3–C

H+

2CH3CO2H

O– Acetate ion

KMnO4, OH– H+

CH3 CH3CH2–C=O +CO2+H2O

O-228 CH3

(d)

CH3–CH2–CH2HC=CH–CH

KMnO4, OH– H+

CH3

CH3 CH3CH2CH2CO2H + CH–CO2H CH3

1.5.8 Ozonolysis A more widely used method for locating the position of double bond in an alkene involves the use of ozone, O3. Ozone reacts vigorously with alkenes to form unstable compounds called molozonides, which rearranges spontaneously to form compounds known as ozonides. Ozonides, themselves are highly unstable so these are reduced directly with Zn and water. The reductive ozonolysis produces carbonyl compounds - aldehydes and ketones, that can be isolated and identified. O –C=C–

O3

–C–C– O

O O

C

C O

2 C = O + Zn(OH)2

O

Ozonide

Aldehydes and /or Ketones

Ozonolysis can be of either of reductive type or of oxidative type. The difference lies in the fact that products of reductive ozonolysis are aldehydes and/or ketones while in oxidative ozonolysis, the products are carboxylic acids and/or ketones. This is because H2O2 formed would oxidize aldehdyes to carboxylic acids, however ketones are not oxidized. In reductive ozonolysis, we add zinc which reduces H2O2 to H2O and thus H2O2 is not present at all to oxidize any aldehyde formed. Zn + H2O2 → ZnO + H2O For example, (i)O ,CH Cl ,

3 2 (CH3)2CH – CH = CH2     2  → (CH3)2CHCHO + HCHO

(ii)Zn / H2 O

(i)O ,CH Cl ,

3 2 (CH3)2C = CH – CH3     2  → (CH3)2C = O + CH3CHO

(ii)Zn / H2 O

Example 7: Write structural formula for the compounds which yield the following products on reductive ozonolysis. (i)

Two moles of O = C(CH3)CH2CH3

(ii)

(C2H5)2C = O + O = CHCH = O + O = CHCH2CH3

(iii)

H2C = O + O = CHCH(CH3)CH(CH3)2

(iv)

Two moles of O = CHCH2CH = O

(v)

O = CHCH2CH2CH2CH = O

Solution : (i) The formation of a single carbonyl compound signals a symmetrical alkene. Write twice the structural formula of the ketone with the C = O groups facing each other. Omit the O's and join the C's with a double bond.

O-229 CH3

CH3

H3C H3C

CH3CH2C=O O=CCH 2CH3 (ii)

CH3CH2C=CCH2CH3(cis or trans)

A total of four C = O's in the products indicates a diene, a compound with two C = C's. (C2H5)2C = O + O = CHCH = O +O =CHCH2CH3 ← (C2H5)2C = CHCH = CHCH2CH3

(iii)

Two different carbonyl compounds means the alkene is unsymmetrical. H2C = O+O = CHCH(CH3)CH(CH3)2 ← H2C = CHCH(CH3)CH(CH3)2

(iv)

Two moles of a dicarbonyl compound signals a symmetrical cycloalkadiene. HC=O O=CH CH2

H2C

HC=O O=CH (v)

The presence of two C = O's in the same product indicates a cycloalkene, in this case cyclopentene. HC=O O=CH CH2

H2C CH2

1.5.9 Substitution Reactions at Allylic Position high

Cl2 + H2C = CHCH3  → H2C = CHCH2Cl + HCl temperature

low concentration

Br2 + H2C = CH – CH3       → CH2 = CH – CH2Br + HBr of Br2

These halogenations are like free radical substitution of alkanes. The order of reactivity of H-abstraction is allyl>3 >2 >1 >vinyl. Allylic substitution by chlorine is carried out using Cl2 at high temperature and alkene with α −carbon in gaseous phase. Allylic bromination can be carried out using N–bromosuccinimide. Propene undergoes allylic bromination when it is treated with N–bromosuccinimide (NBS) in CCl4 in the presence of peroxides or light. O

CH2=CH–CH3 +

O

N–Br

light or ROOR CCl4

O N–Bromosuccinimide (NBS)

N–H

CH2=CH–CH2–Br+ O 3–Bromopropene (allyl bromide)

Succinimide

O-230 Mechanism The reaction is initiated by the formation of a small amount of Br • (possibly formed by dissociation of Br2 molecule). The chain propagation steps for this reaction are the same as for chlorination. CH2=CH–CH2–H+Br

CH2=CH–CH2+HBr

CH2=CH–CH2–Br–Br

CH2=CH–CH2Br+Br

N–Bromosuccinimide is nearly insoluble in CCl4 and provides a constant but very low concentration of bromine in the reaction mixture. It does this by reacting very rapidly with the HBr formed by the reaction of NBS with traces of H2O present in it. Each molecule of HBr is replaced by one molecule of Br2. O

O

N–H+Br2

N–Br+HBr

O

O

Under these conditions, that is, in a non–polar solvent and with a very low concentration of bromine, very little bromine adds to the double bond, instead it undergoes substitution and replaces an allylic hydrogen atom. To understand why high temperature favours allylic substitution over addition requires a consideration of the effect of entropy changes on equilibria. The addition reactions has a substantial negative entropy change because it combines two molecules into one. At low temperatures, the T∆S term in ∆G = ∆H –T∆S, is not large enough to offset the favorable ∆H term. But as the temperature is increased, the T∆S term becomes more significant, ∆G becomes more positive, and the equilibrium becomes more unfavorable for addition and subsequently favours allylic substitution.

1.6 Polymerisation of Alkenes Polymerisation of alkenes to produce polymers like polythene, polypropene, PVC, PTFE etc. the general method of preparation is shown below,

O-231 Some of the common polymers of alkenes are (a) Notice that the principle is the same for each addition polymerisation H

H C

C

n

*

H

H

C

*

H n Poly(ethene)

H C

C

C

Each of the polymer structures shows the repeat unit in brackets. This is repeated thousands of times in each polymer molecule.

H n

H

H

ethene (b)

H

*

H

Cl

H

H

C

C

*

H

Cl n Poly(choloroethene)

chloroethene

(c) It is important to show the repeat unit with 'side–links' to indicate that both sides are attached also to other repeat units. H

H C

n H

C

* C6H5

phenylethene

H

H

C

C

H

*

C6H5 n

Poly(phenylethene)

Most frequently encountered polymers of alkenes are generally addition polymers. Some of these types are described below.

1.6.1 Poly(ethene) or Polythene or Polyethylene It is generally of two types; 1.

LDPE (low density polythene)

2.

HDPE (high density polythene)

1. Low density poly(Ethene), LDPE: Formation of LDPE is an example of addition polymerisation. An addition polymerisation reaction is one in which two or more molecules join together to give a single product. During the polymerisation of ethene, thousands of ethene molecules join together to make poly(ethene) commonly called polythene.

O-232 nCH2=CH2

[–CH2–CH2]n

The number of molecules joining up is variable, but it is in the region of 2000 to 20000. Reaction conditions are drastic and the temperature required is 200°C, pressure is high and a small amount of oxygen as an impurity is required to initiate. Properties and uses Low density poly(ethene) has quite a lot of branching along the hydrocarbon chains, and this prevents the chains from lying tidily close to each other. Those regions of the poly(ethene) where the chains lie close to each other and are regularly packed are said to be crystalline. Where the chains are a random jumble, it is said to be amorphous. Low density poly(ethene) has a significant proportion of amorphous regions. One chain is held to its neighbours in the structure by Van der Waals dispersion forces. Those attractions will be greater if the chains are close to each other. The amorphous regions where the chains are inefficiently packed lower the effectiveness of the Van der Waals attractions and so lower the melting point and strength of the polymer. They also lower the density of the polymer hence, "low density poly(ethene)". Low density poly(ethene) is used for familiar things like plastic carrier bags and other similar low strength and flexible sheet materials. 2. High density poly (Ethene), HDPE: This is made under quite different conditions from low density poly(ethene). Conditions required are temperature is kept at about 60°C, pressure is low - a few atmospheres and the reaction is catalysed by Ziegler-Natta catalysts or other metal compounds. Ziegler-Natta catalyst Ziegler-Natta catalysts are mixtures of titanium compounds like titanium(III) chloride, TiCl3, or titanium(IV) chloride, TiCl4, and compounds of aluminium like aluminium triethyl, Al(C2H5)3. There are all sorts of other catalysts constantly being developed. These catalysts work by totally different mechanisms from the high pressure process used to make low density poly(ethene). The chains grow in a much more controlled - much less random - way. Properties and uses of HDPE High density poly(ethene) has very little branching along the hydrocarbon chains - the crystallinity is 95% or better. This better packing implies that van der Waals attractions between the chains are greater and so the plastic is stronger and has a higher melting point. Its density is also higher because of the better packing and smaller amount of wasted space in the structure. High density poly(ethene) is used to make things like plastic milk bottles and similar containers, washing up bowls, plastic pipes and so on. Look for the letters HDPE near the recycling symbol.

1.6.2 Polyvinyl Chloride,PVC Poly(chloroethene) or poly vinyl chloride is commonly known by the initials of its old name, PVC. Poly(chloroethene) is made by polymerising chloroethene, CH2=CHCl.

O-233

The equation is usually written: Cl

H

Cl H

nC

C

—C–C—

H

H

HH n

It doesn't matter which carbon is attached to chlorine in the original molecule. The polymerisation process produces mainly atactic polymer molecules - with the chlorines orientated randomly along the chain.

2. Cycloalkene (or Cycloolefin) A cycloalkene is a type of alkene hydrocarbon which contains a closed ring of carbon atoms, but has no aromatic character. Some cycloalkenes, such as cyclobutene and cyclopentene, can be used as monomers to produce polymer chains. In IUPAC nomenclature cyclic alkenes named as cycloalkenes. The general formula of cyclic alkene is CnH2n-2. Due to geometrical considerations, smaller cycloalkenes are almost always the cis isomers, and the term cis tends to be omitted from the names. Some of the cyclic compounds containing double bonds is shown below, 6

Cyclobutene

Cyclopentene

Cyclohexene

5

1

4

2

3 1, 4–Cyclohexadiene

Some frequently observed examples of cyclic alkenes are cyclobutene and cyclopentene. Cycloalkenes have wide applications in synthesis of different polymers.

2.1 Conformation in Cycloalkene Conformation is a spatial arrangement of a molecule of a given constitution and configuration. In the case of a four atom molecule linked in a chain manner, rotation of atom A or D about the inner B-C bond by an angle ω leads to a different mutual relation of atom A and D and results in population of a set of different rotational isomers or "conformations".

O-234 D

rotation through ω

C

C A

A

D

B

B ω

The single parameter differentiating such conformers is an angle between two planes that contain atoms ABC and BCD in themselves. This dihedral angle ω is called a "torsion" angle and is most frequently used for specification of the type of conformations.

ω

Fig. 3. Torsional angle

Just like aliphatic alkenes, cycloalkenes can show geometrical isomerism. Generally cis-form of cyclic alkene is more common compare to trans-form. In larger rings containing around 8 atoms, cis-trans isomerism around the double bond is observed as shown below,

H

H

H

H

cis–form

H

from–form

Moreover, the non planar or trans form shows chirality as shown below,

H

O-235 mirror plane H

H

H

H

H

H

Nonplanar trans cyclooctene (Chiral forms)

H Planar cis–cyclooctene (Achiral) (a)

(b)

2.2 Methods of Preparation of Cycloalkenes Just like aliphatic alkenes, cycloalkenes can be synthesized by following methods. 1.

By dehydration of cyclic alcohols: Cyclohexanol undergoes dehydration in the presence of concentrated acids like phosphoric acid forms to form cyclohexene. The reaction scheme is shown below, OH

Heat conc. H3PO4

+ H2O Cyclohexene

Cyclohexanol

Mechanism Dehydration of cyclic alcohol involves the formation of cyclohexyl carbocation which further loses a proton to form cyclic alkene. H+

⊕

OH

OH2

H2O

⊕

H ⊕

2.

–H+

By Diels-Alder cycloaddition: Diels-Alder cycloaddition is a-cycloaddition of a conjugated diene and an alkene or alkyne also termed as dienophile to form cyclic alkenes. It is a type of pericyclic or an electrocyclic reaction which involves the 4 π-electrons of the diene and 2 π–electrons of the dienophile to form cycloalkenes. Reaction involves the cleavage of less stable π bond to more stable σ bonds.

O-236

diene 4 electron

dienophile 2 electrons

The most common example of Diels-Alder cycloaddition is additional reaction of 1,3-butadiene and ethene to form cyclohexene.

+

700°C

Chemical properties of cycloalkenes: These compounds go through the electrophilic addition reactions that are characteristic of alkenes, when the ring remains intact. Cycloalkenes decolourise the purple colour of dilute cold KMnO 4 or red colour of bromine in carbon tetrachloride. 1.

Addition of proton acids to cycloalkenes: It has been observed that electrophiles like protic acid can react with alkenes to form carbon-halogen bonds by donating positive halogen, Br ⊕ , Cl ⊕ , or I ⊕ .In a similar manner these reagents can add to cycloalkenes as well to yield similar products. Likewise, carbon-hydrogen bonds can be formed by the use of typically strong proton acids. These acids are more effective in the absence of large amounts of water because water can compete with the alkene as a proton acceptor. Hydrogen chloride addition to a cycloalkene occurs by way of a proton-transfer step to give the carbocation and a chloride ion followed by a step in which the nucleophilic chloride ion combines with the carbocation: H CH2=CH2+H—Cl ⊕

CH3–CH2+Cl

slow fast

⊕

CH2–CH2+Cl

CH3–CH2 –Cl

All of the hydrogen halides (HF, HCl, HBr, and HI) will add to alkenes. Addition of hydrogen fluoride, while facile, is easily reversible. However, a solution of 70% anhydrous hydrogen fluoride and 30% of the weak organic base, pyridine, which is about 1/10,000 times as strong as ammonia, works better, and with cyclohexene gives fluorocyclohexane. With hydrogen iodide, care must be taken to prevent I2 addition products resulting from iodine formed by oxidation reactions such as 4HI + O2 -> 2I2 + 2H2O. With hydrogen bromide, radical-chain addition may intervene unless the reaction conditions are controlled carefully later. The stereochemistry of addition depends largely on the structure of the alkene, but for simple

O-237 alkenes and cycloalkenes, addition occurs predominantly in an antarafacial manner. For example, hydrogen bromide reacts with 1,2-dimethylcyclohexene to give the trans addition product as shown below, + HBr CH3

.

CH3

1,2–dimethylcyclohexene

CH3

Br

trans–1–bromo–1,–dimethyl–cyclohexane

Hydroxylation and oxidation of alkenes hydroxylation reaction: Several oxidizing reagents react with cycloalkenes just like alkenes under mild conditions to give, as the overall result, addition of hydrogen peroxide as HO—OH. Of particular importance are alkaline permanganate (MnO4-) and osmium tetroxide (OsO4), both of which react in an initial step by a suprafacial cycloaddition mechanism like that postulated for ozone. Both the addition mechanisms are shown below, with acyclic isomers. C

C

O Mn

C

O

O

O

Mn

O

O

O

C

C

H2O

C +MnO3

HO OH

O

2.

H H3C

Unstable C

O

C

O

O

Os

O

O

O

C

Os

C mild reduction O (Na2SO3)

C

+Os(0)

HO OH

O

C

Stable osmate ester

Each of these reagents produces cis-1,2-dihydroxy compounds (diols) with cycloalkenes: 1. OsO4, 25° 2. Na2SO3 (Or KMnO4, OH, H2O) Cyclopentene

H

H

HO

OH

Cis–1,2–cyclopentanediol

3.

Hydrogenation of cycloalkenes: Cycloalkenes upon reduction with dihydrogen in the presence of a metallic catalyst produce cycloalkanes. As an example, cyclohexene is subjected to reduction giving rise to cyclohexane as shown below, + H2 Cyclohexene

Pd/C

Cyclohexane

O-238

3. Dienes Systems with more than one C=C is referred to as polyenes. The simplest types of polyenes are those in which there are two double bonds, known as dienes. A diene is a molecule that has two double bonds. If the molecule is also a hydrocarbon, it is called an alkadiene. The relative arrangement of the double bonds defines the characteristic reactions of the systems.

3.1 Types of Dienes There are three possible arrangements for double bonds in the order of from most to least stable; 1. Conjugated diene 2. Isolated diene 3. Cumulated diene

3.1.1 Conjugated Diene The double bond units occur consecutively giving a continuous π system. Here, all the adjacent "p" orbitals can overlap with each other. The result is that conjugated diene reactivity differs to that of simple alkenes. This extra bonding interaction between the adjacent π systems makes the conjugated dienes the most stable type of diene. Conjugated dienes are about 15kJ/mol or 3.6 kcal/mol more stable than simple alkenes. The simplest conjugated diene is 1,3–butadiene as shown here, H2C=CH–CH=CH2 The extended conjugation in 1,3–butadiene can be shown below,

3.1.2 Isolated Diene The double bond units occur separately. The π systems are isolated from each other by sp3 hybridised centers. The result is that isolated dienes have reactivity that is characteristic of simple alkenes. One of the simpler isolated diene is , H2C=CH–CH2–CH=CH2 1,4–pentadiene

3.1.3 Cumulated Diene The double bond units share a common sp hybridised C atom. The result is that cumulated dienes have reactivity more like simple alkynes. It is important to note the relative spatial positions of the two CH2 units at the ends of the cumulated system. The simplest of such compounds are allenes. The structural arrangement of allene is shown below,

O-239 H

H C=C=C H

H

Bonding in allene The bonding arrangement of allenes is interesting and shown as, H 1s

H

py

py

1s

sp2 C

sp2

pz

sp2 1s

sp

C

sp

sp2 pz

H

C sp2 H

1s

Fig. 5. Bonding arrangement in allene

The two carbon atoms on the left and right are supposed to have sp2 hybridization. This leaves the two extreme carbon atoms, left and right, with one un-hybridised p-orbital each. The carbon on the left hand side is rotated such that the unhybridized p orbital is py. The carbon on the right is rotated such that the unhybridized p orbital is pz. The central carbon atom is supposed to be sp hybridised. The p orbital used for sp hybridization is px. Expectedly, the central carbon is left with two un-hybrid p-orbital. One is pz and another is py. The middle carbon atom forms one π bond due to pz - pz overlap and another π bond due to py - py overlap as shown in the above diagram. Due to this reason the two extreme carbon atoms belong to different planes causing asymmetry. This asymmetry further results into chirality in properly substituted allenes. The relative instability of allenes probably reflects extra strain as a result of one carbon atom forming two double bonds. 1,2-Propadiene is slightly more strained than propyne, its heat of hydrogenation being about 2 kcal mol-1 greater than that of propyne.

3.2 Methods of Preparation of Dienes Dienes are prepared from the same reactions that form ordinary alkenes. The two most common methods are the dehydration of diols also known as dihydroxy alkanes and the dehydrohalogenation of dihalides. The generation of either an isolated or conjugated system depends on the structure of the original reactants. Vicinal diols, which have two hydroxyl groups on adjacent carbon atoms, and vicinal dihalides, which have halogen substituents on adjacent carbons, always become conjugated systems in elimination reactions. Other reactant configurations can lead to products that include both conjugated and isolated systems.

O-240 Dehydration of diols: HOCH2CH2CH2CH2OH

H–

H2C=CH–CH=CH2 1,3–butadiene

butane–1,4–diol

+ HOCH2CH2CH2CH2CH2OH H H2C=CH–CH2–CH=CH2

1,4–pentadiene

pentane–1,5–diol

Dehydrohalogenation of dihalides: KOH XCH2CH2CH2CH2X alcohol H2C=CH–CH=CH3 1,3–butadiene

1,4–dihalobutane

KOH

XCH2CH2CH2CH2CH2X alcohol 1,5–dihalobutane

H2C=CH–CH2–CH=CH3 1,4–pentadiene

Chemical properties of dienes 1. Diels alder reaction: In the Diels Alder reaction a double bond adds to a 1,3–conjugated diene to give a 6-membered ring. This reaction is known as 4+2 cycloaddition. The diene is labelled as dienophile. The simplest example is reaction between a 1,3-butadiene and ethene to yield a cyclohexene as shown below, The dienes show some interesting chemical reactivities. Some important ones are discussed below,

+

The analogous reaction of 1,3-butadiene with ethyne to form 1,4-cyclohexadiene is also known and the reaction scheme is shown below,

heat

Since the reaction forms a cyclic product, via a cyclic transition state, it can also be described as a "cycloaddition". Mechanism The reaction normally is a concerted or one step process:

The reaction is widely used in synthetic organic chemistry due to high degree of regio– and stereoselectivity which is attributed to the concerted mechanism. The reaction usually is thermodynamically favourable due to the conversion of 2 π–bonds into 2 new stronger σ-bonds. The two reactions shown above require drastic reaction conditions, but the normal Diels-Alder reaction is favoured by electron withdrawing groups on the electrophilic dienophile and by electron donating groups on the nucleophilic diene. e.g..

O-241 O

O

O

+

O

O maleic anhydride

O

Some common examples of dienes and dienophiles are shown below:

Dienes

O CN

CO2Me

CO2Me

Dienophiles

O

CO2Me

O

O MeO2C

CO2Me

O

CO2Me

O

Benzene does not undergo a Diels Alder reaction, despite the fact that one can 'locate' a diene fragment in its structure. It is definitely because of the aromaticity of benzene ring. If it participates in cycloaddition reaction, aromaticity is lost. O

O

+

No reaction

O

But interestingly anthracene responds to the reaction as shown below, O O O O +

O

O RE=350kj/mol Therefore loss of RE=50kj/mol

RE=2×150= 300kj/mol

O-242 Why anthracene exhibits diene behaviour is definitely because the product still has 2 benzenoid rings left. In the heteroaromatic systems like furan or thiophene the Diels Alder reaction is given by furan but not by thiophene. It is being shown below, O O easy

O

O

O

O O

O

O

S

No Reaction

O O

Why thiophene does not respond is due to the fact that thiophene has more aromatic character than furan.

Stereoselectivity in Diels Alder reaction The Diels Alder reaction is stereospecific with respect to both the diene and the dienophile. Addition is syn on both components as bonds formed from same species at the same time. 2.

Electrophilic addition to a diene: The addition of an electrophile to a diene yields various products depending upon the nature of diene -whether it is conjugated or isolated, temperature etc. It is discussed below one by one.

Kinetic and thermodynamic control and effect of temperature

Energy

A very interesting and significant discussion arises from the fact that while the pathway to the 1,4 product occurs with greatest free energy decrease, the pathway to the 1,2 product is achieved with less activation energy as shown below in the energy profile,

R

Kinetic control

Thermodynamic control

A B Reaction Fig. 6: Energy profile of different product formation paths

O-243 This means that formation of 1,4 product is favoured thermodynamically, but 1,2 product is favoured kinetically. Higher temperatures, 'thermodynamic conditions', promote the formation of 1,4 product, ecause if a larger fraction of the molecules possess activation energy for either pathway, as would occur at higher temperature, a large portion of diene concentration continuously moves down the 1,4 pathway and forms the more stable 1,4 product. Lower temperatures, however, are 'kinetic conditions'. At lower temperatures, fewer reagent diene particles have enough energy to get over the activated complex energy hump to form the 1,4 product, so 1,2 addition predominates. Case1: Addition to Conjugated Diene The addition of halogen to 1,3–butadiene at temperatures below zero leads mainly to the 1,2–addition product, while addition reactions run at temperatures above 50°C with these chemicals produces mainly the 1,4–addition product. Br

Br

H2C=CH–CH–CH2 (low temp.) 1,2 product

H2C=CH–CH=CH2

Br2

Br

Conjugated diene

+

Br

H2C–CH=CH–CH2 (high temp.) 1,4 product

If the reaction is initially run at 0°C and then warmed to 50°C or higher and held there for a period of time, the major product will be a 1,4 addition. These results indicate that the reaction proceeds along two distinct pathways. At high temperatures, the reaction is thermodynamically controlled, while at low temperatures, the reaction is kinetically controlled. Mechanism Br

Br H2C=CH–CH=CH2 + Br–Br Conjugated diene

Br +

+

+

H2C–CH=CH–CH2+Br–

H2C=CH–CH–CH2 Allylic corbocation intermediate

Br

Br +

Br

H2C=CH–CH–CH2 or H2C–CH=CH–CH2

+

Br

Br

Br

Br

H2C=CH–CH–CH2 + H2C–CH=CH–CH2 1,2 product

1,4 product

Instead of forming the triangular halonium ion which is typical of electrophilic addition of halogen to alkene, this reaction follows the standard carbocation mechanism for addition across a double bond. So, halogen adds to a conjugated diene to form a resonance stabilized allylic carbocation intermediate. The two resonance forms of the resonance stabilized allylic carbocation lead respectively to two different possible products, the 1,2 and 1,4 products of addition. The energy diagram of the reaction of 1,3–butadiene with hydrogen bromide shows the pathways of the two products generated from the intermediate.

O-244

Addition of other electrophiles When other electrophiles are added to conjugated dienes, 1,4 addition also occurs. Many reactants, such as halogens, halogen acids, and water, can form 1,4–addition products with conjugated dienes. Like the addition of hydrogen halides to conjugated dienes, halogens add to dienes via direct and conjugate addition pathways: X–X

X

1 2

X

3

1

+ X

2

X 4

Direct or 1,2–addition Conjugate or 1,4–addition

Addition in isolated dienes Both isolated and conjugated dienes undergo electrophilic addition reactions. In the case of isolated dienes, the reaction proceeds in a manner identical to alkene electrophilic addition. The addition of hydrogen bromide to 1,4–pentadiene leads to two products. Br H2C=CHCH2CH=CH2+HBr

H2CCHCH2CH=CH2+BrH2CCH2CH2CH=CH2 4–bromo–1–pentene (Markovnikov product)

3.

5–bromo–1–pentene (anti–Markovnikov product)

Isomerisation: As propadiene is less stable than propyne, so, 1,2-propadiene naturally isomerizes to propyne. This isomerization occurs under the influence of strongly basic substances such as sodium amide in liquid ammonia or potassium hydroxide in ethyl alcohol:

O-245

CH2=C=CH2

NaNH2, NH3

CH3C≡CH

∆G°(g)=–2 kcal KOH,C2H5OH (CH3)2C=C=CH2 (CH3)2CHC≡CH 3–methyl–1,2–butadiene ∆G°(g)=–2.5 kcal

4. Alkynes The alkynes are unsaturated hydrocarbons that contain one triple bond. They have the general formula CnH2n–2 and the triple bond is known as the 'acetylenic bond'. Many alkynes have been found in nature. Some of the commonest alkynes are shown below, H |

H−C ≡C −H

H – C ≡ C – C– H | H

ethyne

H |

propyne

H |

H |

H − C− C ≡ C – C− H H − C– C ≡ C – C – C – H | H

| H

| H

2 butyne

2,4 pentayne

Alkynes are isomeric to alkadienes, cycloalkenes and bicycloalkanes. An example of C5H8 is shown below, 2 1 2 3 4

5

1

2

3

4

5

H3C–C≡C–CH2CH3 H2C=CH–CH2–CH=CH2 2–pentyne

1,4–pentadiene Cyclopentene

1 5

4 3 Bicyclo[2,1,0]pentane

4.1 IUPAC Rules for Nomenclature of Alkyne 1.

The yne suffix (ending) indicates an alkyne or cycloalkyne.

2.

The longest chain chosen for the root name must include both carbon atoms of the triple bond.

3.

The root chain must be numbered from the end nearest a triple bond carbon atom. If the triple bond is in the center of the chain, the nearest substituent rule is used to determine the end where numbering starts.

4.

The smaller of the two numbers designating the carbon atoms of the triple bond is used as the triple bond locator.

5.

If several multiple bonds are present, each must be assigned a locator number. Double bonds precede triple bonds in the IUPAC name, but the chain is numbered from the end nearest a multiple bond, regardless of its nature.

O-246 6.

Because the triple bond is linear, it can only be accommodated in rings larger than ten carbons. In simple cycloalkynes the triple bond carbons are assigned ring locations 1 and 2. Which of the two is 1 may be determined by the nearest substituent rule.

7.

Substituent groups containing triple bonds are: HC ≡ C (Ethynyl group)

HC ≡ CH-CH2 (Propargyl group)

4.2 Bonding in Alkynes In acetylene, the H-C ≡ C bond angles are 180°. By virtue of this bond angle, alkynes tend to be rod-like. Correspondingly, cyclic alkynes are rare. Benzyne is highly unstable. The C ≡ C bond distance of 121 picometers is much shorter than the C=C distance in alkenes (134 pm) or the C-C bond in alkanes (153 pm). The triple bond is very strong with a bond strength of 839 kJ/mol. The sigma bond contributes 369 kJ/mol, the first pi bond contributes 268 kJ/mol and the second pi-bond of 202 kJ/mol bond strength. Bonding usually discussed in the context of molecular orbital theory, which recognizes the triple bond as arising from overlap of s and p orbitals. In the language of valence bond theory, the carbon atoms in an alkyne bond are sp hybridized: they each have two unhybridized p orbitals and two sp hybrid orbitals. Overlap of an sp orbital from each atom forms one sp-sp sigma bond. Each p orbital on one atom overlaps one on the other atom, forming two pi bonds, giving a total of three bonds. The remaining sp orbital on each atom can form a sigma bond to another atom, for example to hydrogen atoms in the parent acetylene. The two sp orbitals project on opposite sides of the carbon atom. pz–Orbital py–Orbital σ–bond

H

C

C

H

π–bond π–bond Fig. 8: Molecular orbital diagram of ethyne

Terminal and non terminal or internal alkynes Internal alkynes feature carbon substituents on each acetylenic carbon. Symmetrical examples include diphenylacetylene and 3-hexyne. Terminal alkynes have the formula RC2H. An example is propyne. Terminal alkynes, like acetylene itself, are mildly acidic, with pKa values of around 25. They are far more acidic than alkenes and alkanes, which have pKa values of around 40 and 50, respectively. The acidic hydrogen on terminal alkynes can be replaced by a variety of groups resulting in halo-, silyl-, and alkoxoalkynes. The carbanions generated by deprotonation of terminal alkynes are called acetylides.

O-247

4.3 Methods of Preparation of Alkynes There are a number of methods available for the synthesis of alkynes. Some of these are discussed below in detail. 1.

From calcium carbide: Calcium carbide is the compound CaC2, which consists of calcium ions (Ca2+) and acetylide ions, C 2− 2 . It is synthesized from lime and coke in the following reaction: CaO + 3C → CaC2 + CO This reaction is very endothermic and requires a temperature of 2000°C. For this reason it is produced in an electrical arc furnace. As a matter of fact, calcium carbide may formally be considered a derivative of acetylene which is an extremely weak acid though not as weak as ammonia. The double deprotonation means that the carbide ion has very high energy. Instant hydrolysis occurs when water is added to calcium carbide, yielding acetylene gas, the hydrolysis reaction taking place is shown below, CaC2 + 2H2O → Ca(OH)2 + C2H2

2.

Dehydrohalogenation of vic-dihalides or gem-dihalides: A geminal (1,1) or vicinal (1,2) dihalide upon dehydrohalogenation with alc. KOH and eventually with NaNH2 yields an alkyne as shown below, H X

H H

H

–C–C– or –C–C– alc. KOH H X A gem-dihalide

KX+H2O+–C=C

X X

NaNH2

X

A vic-dihalide

NaX+NH3+–C≡C– Alkyne

A vinyl halide

The vinyl halide requires the stronger base, sodamide (NaNH2) to carryout elimination process of dehydrohalogenation. 3.

Dehalogenation of vic-tetrahalogen compounds: Even tetrahalogenated alkanes can also give rise to alkynes using strong base in high concentration as shown below, EtOH

CH3-CBr2-CBr2-CH3 + 2Zn    → CH3-C ≡ C-CH3 + 2ZnBr2 2,2,3,3-tetrabromobutane

heat

2-butyne

This reaction has the drawback that the halogen compound is itself prepared by halogen addition to alkynes. The mechanism for dehalogenation has been taken up in the topic "alkenes". 4.

From Kolbe's electrolytic process: The starting compound is a salt already containing a carbon-carbon double bond. One such compound is maleic acid. Mechanism is similar to that of formation of ethane using Kolbe's electrolysis. CH–COONa || CH − COONa sodium maleate is a sodium salt of maleic acid. It is subjected to electrolysis using inert electrodes and the products at respective electrodes are analysed.

O-248 At Anode: CH-COO-

-2e →

|| CH-COO* CH-COO* ||

CH*

-2CO2

||

→

CH-COO*

CH →

CH*

||| CH

(free radical formation) At cathode: 2Na+ + 2e– → 2Na 2Na + H20 → 2NaOH + H2 In this manner, alkyne is obtained at anode, while NaOH is formed at cathode and hydrogen gas is liberated. 5.

Alkyl substitution in acetylene or 1-alkynes: In this reaction lower alkynes are used to synthesise higher alkynes or alkynes of higher molecular mass. NaNH2 R— C ≡ C—H +

NH3 or →R—C ≡ C: Na + or 1 Na 2 H2 +

–

R —C: + R'— CH2 – X → R—C C— CH2— R' + X (SN2 mechanism) 1° alkyl holide

There is wide variety possible using this method. Acetylene itself may be alkylated either once to make a terminal alkyne or twice to make an internal alkyne. The basis of the reaction is that acetylene or 1–alkyne has the terminal hydrogen which is weakly acidic as discussed earlier, so it is easily removed either by an amide base or an electropositive metal. liq.NH

n-C H Br

3 4  9 HC ≡ CH + NaNH2     → HC ≡ C– Na+    → CH3(CH2)3C ≡ CH

−33 ° C

1-hexyne 2NaNH

+

2 n − C H Br

2 → Na–C≡C–Na +    3 HC≡CH       7→ CH3CH2CH2C≡CCH2CH2CH3 liq.NH3

4-octyne

−33 ° C

Since acetylide ions are highly basic, competing elimination is a common side reaction. The products of such an elimination reaction are an alkene from the alkyl halide alongwith an alkyne by substitution reaction pathway.

Br R–C≡C+H–CH2–CHR

CH2=CHR+Br–+RC≡C–H

O-249

Limitation of the reaction So, in order to get good yield of alkynes, the reaction should take the course of substitution, for which primary alkyl halides would be the right reagent. In practice, the alkylation of acetylene or another terminal alkyne is only a good method for the synthesis of higher alkynes when reacted to primary halides that do not have branches close to the reaction centre. With secondary halides, and even with primary halides that have branches close to the reaction centre, elimination is usually competing or major reaction. 6.From chloroform: When two moles of chloroform is condensed with silver metal, acetylene is produced as shown below, HCCl3

Ag, Heat

HC≡CH Ethyene

Chloroform

Example 8: Outline a synthesis of propyne from isopropyl or n–propyl bromide. The needed vic-dihalide is formed from propene, which is prepared from the alkyl halides. Solution : CH3CH2CH2Br n–Propylbromide

or CH3CHBrCH3

Br alc. alc. CH3–CH=CH2 2 CH3CHBrCH2Br KOH KOH

CH3CH=CHBr NaNH2

Isopropylbromide

CH3C=CH

Example 9: Synthesize the following compounds from HC ≡ CH and any other organic and inorganic reagents (do not repeat steps): (a)

1-pentyne,

(b) 2-hexyne.

Solution : NaNH

CH CH CH I

(a)

2 →H–C ≡ C• − Na+    3 2  2→ H– C ≡ C– CH CH CH H–C ≡ C–H     • 2 2 3

(b)

3→ CH –C ≡ C–H     2 → CH –C ≡ C − Na+ Na+ :C– ≡ C–H    3 3

CH I

NaNH

CH CH CH I

3 2  2→ CH –C ≡ C–CH CH CH  3 2 2 3

Physical properties of alkynes The physical properties of alkynes are similar to those of the corresponding alkenes. The lower members are expectedly gases with boiling points somewhat higher than those of the corresponding alkenes. Terminal alkynes have lower boiling points than isomeric internal alkynes and can be separated by careful fractional distillation. The CH3-C bond in propyne is formed by overlap of a sp3 hybrid orbital from the methyl carbon with a sp hybrid orbital from the acetylenic carbon. The bond is sp3–sp type. Since one orbital has more s-character than the other and is thereby more electronegative, the electron density in the resulting bond is not symmetrical. The unsymmetrical electron distribution results in a dipole moment larger than that observed for an alkene, but still relatively small.

O-250 CH3CH2C≡ CH µ= 0.80 D

CH3CH2CH=CH2 µ= 0.30 D

CH3C≡CCH3 µ= 0.0 D

Symmetrically disubstituted acetylenes, of course, have no net dipole moment.

4.4 Acidity of Terminal Alkynes The acidic nature of hydrogen in acetylene is characteristic of hydrogen in the group ≡ CH, and it has been suggested that this is because the C ≡ H bond has considerable ionic character due to resonance. There is, however, evidence to show that the electronegativity of a carbon atom depends on the number of bonds by which it is joined to its neighbouring carbon atom. Since π − electrons are more weakly bound than σ − electrons, the electron density round a carbon atom with π bonds is less than that when only σ– bonds are present. Thus a carbon atom having one π −bond has a slight positive charge compared with a carbon atom which has only σ – bonds. Hence, the electronegativity of an sp2 hybridised carbon atom is greater than that of an sp3 hybridised carbon atom. Similarly, a carbon atom which has two π −bonds carries a small positive charge which is greater than that carried by a carbon atom with only one π −bond. Thus the electronegativity of an sp hybridized carbon atom is greater than that of an sp2 hybridised carbon atom. Thus more the's' character a bond has, the more electronegative is that carbon atom. Therefore the attraction for electrons by hybridized carbon will be sp > sp2 > sp3. Therefore the hydrogens in terminal alkynes are relatively acidic. Acetylene itself has a p K a of about 25. It is a far weaker acid that water ( p K a 15.7) or the alcohols ( p K a 16-19), but it is much more acidic than ammonia ( p K a 34). A solution of sodium amide in liquid ammonia readily converts acetylene and other terminal alkynes into the corresponding carbanions. RC ≡ CH + NH −2 → RC ≡ C– + NH3 This reaction does not occur with alkenes or alkanes. Ethylene has a p K a of about 44 and methane has a p K a of about 50, which means that they are weaker acid than NH3. From the foregoing p K a 's one can see that there is a vast difference in the basic character of the carbanions RC ≡ C − , CH2 = CH − , and CH −3 . This difference may readily be explained in terms of the character of the orbital occupied by the lone-pair electrons in the three anions. Methyl anion has a pyramidal structure with the lone-pair electrons in an orbital that is approximately sp3. In vinyl anion, the lone-pair electrons are in an sp2 orbital. In acetylide ion, the lone pair is in an sp orbital. sp3

H

H

••

Methyl anion

H

sp

••

H–C≡C

C=C H

Vinyl anion

••

H

C

sp2

H Acetylide ion

Electrons in s-orbitals are held closer to the nucleus than they are in p-orbitals. This increased electrostatic attraction means that s-electrons have lower energy and greater stability than p-electrons. In general, the greater the amount of s-character in a hybrid orbital containing a pair of electrons, the less basic is that pair of electrons, and the more acidic is the corresponding conjugate acid. Alternatively, since the greater the electronegativity of an atom, the more readily it can accommodate a negative charge and hence less basic the

O-251 −

species would be. Basicities of the following carbanions follow the order: CH≡ C − < CH2=CH CH2 = CH2 > CH4. Of course, the foregoing argument applies to hydrogen cyanide as well. In this case, the conjugate base, N ≡ C − , is further stabilized by the presence of the electronegative nitrogen. Consequently, HCN is sufficiently acidic ( p K a 9.2) that it is converted to its salt with hydroxide ion in water. HCN + OH − → CN − + H2O

Chemical properties of alkynes Alkynes frequently undergo electrophilic addition reactions like alkenes. Owing to the presence of a triple bond, acetylene is more unsaturated than ethylene, and forms addition products with two or four univalent atoms or groups. A triple bond consists of one σ bond and two π bonds which produces a linear molecule. When two univalent atoms add on to a triple bond, the diagonal arrangement sp hybridized lobes of atomic orbitals changes into the trigonal lobes due to sp2 hybridization, and the further addition of two univalent atoms changes the trigonal arrangement into the tetrahedral arrangement due to sp3 hybridization. Under suitable conditions, it is possible to isolate the intermediate alkene. Arguments based on bond strengths would lead to the conclusion that acetylene should be more reactive than ethylene: C ≡ C (808.3) → C=C (606.7), ∆H = +196.6 kJ mol–1 ; C=C (606.7) → C–C (347.3), ∆H = +259.4 kJ mol–1. This order of reactivity is observed for catalytic hydrogenation. The mechanism of this reaction, however, does not involve electrophiles. However, the reactivity of alkynes towards general electrophilic addition reaction is lesser than that of alkenes. This is unexpected in view of the fact that the π electron density in a triple bond is higher than that in a double bond. A possible contributing factor for decreased reactivity of a triple bond to electrophiles may be the fact that the bridged carbonium ion from acetylene (assuming its formation) is more strained than the bridged carbonium ion from ethylene. Br+ –C≡C–

Br2

–C

C–+Br–, Br+

C

C

Br2

C

C+Br–,

Justification for the assumption of these intermediate bridged ions being involved in addition to acetylene by the polar mechanism lies in the fact that the additions are stereoselectively trans. 1.

Hydrogenation: Acetylene adds on hydrogen in the presence of a catalyst, the reaction proceeding in two stages. H / Ni

H / Ni

C2H2  2 → C2H4  2 → C2H6

O-252 Most catalysts like Ni, Pt, Pd, Rh, etc. cause hydrogenation of a triple bond. As such the reaction is so fast that alkene is not isolated, rather the reaction stops after the formation of alkane. However, it is possible to isolate the intermediate alkene by using suitable conditions. One such method uses Lindlar's catalyst, which consists of a Pd supported on CaCO3 catalyst partially poisoned with lead acetate. A better catalyst is Pd supported on BaSO4 partially poisoned with quinoline. Dialkylacetylenes can be catalytically reduced by Lindlar's catalyst to a mixture of cis– and trans–alkenes, the former predominating while reduction with sodium in liquid ammonia produces predominantly the trans–alkene, which is also produced by reduction with lithium aluminium hydride. These mechanism have been dealt in the topic on alkenes. cis–reduction can also be carried out with diborane or with di–isobutylaluminium hydride. 2.

Addition of halogen: Halogens are added to alkynes just like alkenes with a slight difference that the product formed is a tetrahalogenated alkane as shown in the scheme below, X X R–C≡C–H

X2 CCl4

X X X

R–C=C–H CCl2 4

R–C–C–H X X

The electrophilic addition mechanism for this reaction has been already taken up in the topic "alkenes". This reaction is used for detecting unsaturation of the type C = C or C ≡ C as the reddish–brown colour of Br2 disappears due to the formation of colourless vicinal tetrabromo alkane. 3.

Addition of HX: Alkynes can also add on the halogen acids, their order of reactivity being HI > HBr > HCl > HF. HF addition requires high pressure. The addition of the halogen acids can take place in the dark but is catalysed by light or metallic halides. The addition is in accordance with Markovnikov's rule. For example, acetylene combines with hydrogen bromide to form vinyl bromide first and then ethylidene dibromide in the following manner, HBr

CH ≡ CH + HBr → CH2=CHBr  → CH3CHBr2 Peroxides have the same effect on the addition of hydrogen bromide to acetylene as they have on alkenes. The mechanism of the addition of halogen acids is the same as that for the alkenes. CH≡CH+H–Br

+ Br–+CH=CH2

CH2=CHBr

Addition of another molecule of hydrogen bromide could give either the intermediate carbonium ion, +

CH3C HBr (a stable carbocation due to the stabilization imparted through pπ–dπ bonding with +

bromine), or C H2CH2Br (a primary carbocation). Since the former is more stable than the later, the reaction proceeds via the former to form ethylidene dibromide. Thus, Br–H+CH2=CHBr

+ CH3–CHBr+Br–

CH3CHBr2

O-253

Example 10: Give the equation for the reaction of one equivalent of HBr with 1– pentene–4–yne. Solution : The reaction proceeds through a carbocation. Since the alkyl cabocation from the alkene group is more stable than the vinyl carbocation from the alkyne group, the ∆H# for its formation is less and alkenes react at a faster rate than alkynes toward electrophilic addition. H2C=CHCH2C≡CH

4.

H+

+

H3C–C HCH2C≡CH

Br–

H3CCHBrCH2C≡CH

Addition of H2O (Nucleophilic addition reaction): When passed into dilute sulphuric acid at 60°C in the presence of mercuric sulphate as catalyst, acetylene adds on a molecule of water to form acetaldehyde. The mechanism of this hydration probably takes place via the formation of vinyl alcohol as an intermediate. CH ≡ CH + H2O

H2SO4 Hg2+

[CH2=CHOH]

tautomerize

CH3CHO

Mechanism The function of the mercuric ion imparhing nucleophilic addition character. A possible explanation is that the mercuric ion, because of its large size, can readily form a bridged ion or a π complex. Moreover, mercuric ion provides electro–philicity to alkyne thereby inviting attack from water as a nucleophile. So, this becomes an example of nucleophilic addition reaction. HC≡CH+Hg

2+

HC

CH Hg2+

–H

+

HC=CH–OH

H3O+

H2O

+

HC=CHOH2 Hg+

Hg2++[CH2=CHOH]

tautomerize

CH3CHO

Hg+

The homologues of acetylene form ketone when hydrated using this method. For example, propyne gives acetone in the following manner, OH H2SO4 CH3–C≡C–H+H2O HgSO 4

CH3–C=C–H H A vinyl alcohol (unstable)

O CH3–C–CH3 (Markovnikov addition) Acetone

O-254

Phenyl acetylene yields acetophenone as shown below, C≡C–H

O=C–CH3 H SO

2 4 +H2O HgSO 4

Acetophenone

5.

Addition of Boron Hydride: Monoalkylacetylene on treatment with R2BH gives vinyl adduct which on hydrolysis gives alkene but on alkaline oxidation yields aldehyde. R'

H C=C

R'—C≡C–H+R2BH

BR2

H

CH3COOH hydrolysis

R'CH=CH2

Oxidation H2O2NaOH R'CH2CHO

With dialkylacetylenes, the products of hydrolysis and oxidation are cis-alkenes and ketones, depending upon the choice of reagent as shown below,

3CH3–C≡C–CH3

BH3

CH3

CH3 C=C

B

H a vinylborane

3

H2O2 CH3–CH2–CO–CH3 NaOH 2–butanone

CH3COOH CH3

CH3 C=C

H H Cis–2–butene

6.

Dimerization: Acetylene upon dimerization with Cu+ salt yields vinylacetylene as shown, Cu(NH3 )+ Cl −

2  2 H – C ≡ C – H  → H2C=CH–C ≡ C–H

H2 O

7.

Vinylacetylene

Oxidation to carboxylic acids: Oxidation with hot potassium permanganate yields smaller carboxylic acids. The site of triple bond is the site of cleavage, 3 CH3C ≡ CH + 8KMnO4 + KOH → 3CH3COOK + 3K2CO3 + 8MnO2 + 2H2O CH3CH2C ≡ C–CH3 + 2KMnO4 + KOH → CH3CH2COOK + CH3COOK + 2MnO2 + 2H2O

O-255 CH3

CH3 CH3–CH–C≡CCH2CH2CH3

oxidation

CH3CH–CO2H + CH3–CH2CH2COOH (Two isomeric acids each having M.F. C4H8O2)

Ozonolysis: Even alkynes undergo ozonolysis like alkenes to yield a variety of products,

8.

1.O

3 → CH COOH + HOOCCH CH CH3C ≡ CCH2CH3   3 2 3

2.hydrolysis

2–pentyne

Acetic acid

Propanoic acid

Acetylene and its homologues form ozonides with ozone, and these compounds are decomposed by water to form diketones, which are then oxidised to acids by the hydrogen peroxide formed in the reaction. C 1

2

1

2

R C≡CR +O3→R C

C–R

O

O

H2O

1

2

R –C–C–R +H2O2

1

2

1

2

R –C–C–R +H2O2→R CO2H+R CO2H

O O

Acetylene is exceptional in that it gives glyoxal as well as formic acid. (i)O

3 → OCHCHO + HCO H CH ≡ CH    2

(ii)H2 O

Example 11: List the alkynes needed to synthesize the following ketones in the best possible yields (i)

(CH3 )2 CH C CH3 || O

(ii) CH3CH2 C CH2CH2CH3 || O

(iii) CCH2 O

Solution : (i)

(ii)

Methyl ketones, CH3COR are best made from terminal alkynes. Use (CH3)2CHC ≡ CH to prepare given ketone.

Use the symmetrical internal alkyne CH3CH2C ≡ CCH2CH3 rather than the unsymmetrical CH3C ≡ CCH2CH2CH3, which also gives CH3COCH2CH2CH2CH3.

(iii) C≡C

O-256 9.

Polymerization reactions: When acetylene is passed into dilute hydrochloric acid at 65°C in the presence of mercuric ions as catalyst, vinyl chloride is formed. Hg 2+

CH ≡ CH + HCl    → CH2 = CHCl Acetylene adds on hydrogen cyanide in the presence of cuprous chloride in hydrochloric acid as catalyst to form vinyl cyanide. CH ≡ CH + HCN →CH2=CHCN Vinyl cyanide is used in the manufacture of Buna-N-synthetic rubber, which is a copolymer of vinyl cyanide and butadiene. When acetylene is passed into warm acetic acid in the presence of mercuric ions as catalyst, vinyl acetate and ethylidene diacetate are formed. Hg 2+

CH ≡ CH + CH3CO2H    → CH2=CHOOCCH3 Hg 2+

CH2=CHOOCCH3 + CH3CO2H    → CH3CH(OOCCH3)2 Vinyl acetate (liquid) is used in the plastics industry. Ethylidene diacetate (liquid), when heated rapidly to 300-400°C, gives acetic anhydride and acetaldehyde. When passed through a heated Cu or Fe tube, acetylene polymerises to a small extent to benzene as shown here,

3C2H2→

Homologues of acetylene behave in a similar manner. For example, propyne polymerises to 1,3,5-trimethylbenzene, and but-2-yne to hexamethylbenzene. CH3

CH3 3CH3C≡CH→ H3C

; CH3

H3C

CH3

3CH3C≡CCH3→ H3C

CH3 CH3

Under suitable conditions acetylene polymerises to cyclooctatetraene (COT) as well in the following manner,

4C2H2→ COT

10.

Reaction with hypochlorous acid: When acetylene is passed into hypochlorous acid solution, dichloroacetaldehyde is formed. HOCl

CH ≡ CH + HOCl → CHCl=CHOH    → Cl2CHCH(OH)2 → Cl2CHCHO +H2O Dichloroacetic acid, Cl 2CHCO2H, is also formed by the oxidation of dichloroacetaldehyde in the presence of hypochlorous acid.

O-257 11.

Reaction due to acidic nature: Due to their acidic nature, alkynes are quantitatively deprotonated by alkyllithium compounds, which may be viewed as the conjugate bases of alkanes. CH3(CH2)3C ≡ CH + n–C4H −9 Li+ → CH3(CH2)3C ≡ C–Li++n-C4H10 The foregoing transformation is simply an acid-base reaction, with l-hexyne being the acid and n-butyllithium being the base. Since the alkyne is a much stronger acid than the alkane (by over 20 pK units!), equilibrium lies essentially completely to the right. Terminal alkynes give insoluble salts with a number of heavy metal cations such as Ag+ and Cu+. The alkyne can be regenerated from the salt, and the overall process serves as a method for purifying terminal alkynes. However, many of these salts are explosively sensitive when dry and should always be kept moist. –C ≡ C – H + M+ → –C ≡ C – M + H + For example, when a terminal alkyne like acetylene is treated with [Ag(NH3)2]+ white ppt. of silver acetylide is formed. So, this reaction can be used for identification of acetylene or any terminal alkyne through formation of silver acetylide (white ppt.) or silver alkynide. H – C ≡ C – H+2[Ag(NH3)2]+ → Ag – C ≡ C – Ag +2NH +4 Another characteristic reaction is shown below when a terminal alkyne is treated with [Cu(NH3)2]+ to yield a red precipitate as shown below in the case of propyne, CH3–C ≡ C – H + [Cu(NH3)2]+ → CH3–C ≡ C–Cu + NH +4 + NH3 Cuprous methylacetylide (red ppt.) Some more reactions pertaining to acidic nature of terminal alkynes are shown below like reaction with sodium metal or with sodamide, liq.NH

3 → H–C ≡ C: Na++ ½H H–C ≡ C–H + Na     2 Sodium acetylide

CH3–CH–C≡C–H+NaNH2 CH3

ether

CH3–CH–C≡C:–Na++NH3 CH3

Sodium isoprophylacetylide

Example 12 : Dehydrohalogenation of 3-bromohexane gives a mixture of cis-2-hexene and trans-2- hexene. How can this mixture be converted to pure (i)

cis-2-hexene?

(ii)

trans-2-hexene?

Solution : Relatively pure alkene geometric isomers are prepared by stereoselective reduction of alkynes. (i)

Hydrogenation of 2-hexyne with Lindlar's catalyst gives 98% cis-2-hexene.

O-258

CH3CH=CHCH2CH2CH3

Br2

CH2CH–CHCH2CH2CH3

2NaNH2

Br Br

CH3C≡CCH2CH2CH3 2–hexyne

H3C

H2

CH2CH2CH3

C=C

Lindlar's catalyst

H

H

Cis–2–hexene

Reduction with Na in liquid NH3 gives the trans product.

(ii)

H3C CH3C≡CCH2CH2CH3

Na NH3

H

C=C CH2CH2CH3

H

Trans–2–hexene

Example 13 : (i)

Give the equations for the reactions of propyne with (a) Na (in hexane), −

(b) Na+NH −2 (in liquid NH3), (c) Ag(NH3) +2OH − and (d) Cu( NH 3 )2OH − . (ii)

On which chemical property of propyne do these reactions depend?

Solution : (i)

(a) 2CH3 ≡ CH + 2Na → 2[CH3C ≡ C: Na+ ] (sodium propynide) + H2 an alkynide anion liq.NH3 (b) CH3C ≡ CH + NaNH2    → CH3C ≡ C: − Na+ + NH3

(c) CH3C ≡ CH + Ag(NH3) +2 OH − → CH3C ≡ CAg(s) + 2NH3 + H2O white ppt. (d) (ii)

CH3C ≡ CH + Cu(NH3) + OH − → CH3C ≡ CCu(s) + 2NH3 + H2O 2

red ppt.

The acidity of the terminal hydrogen of a terminal alkyne.

Example 14: Write down the structures of (A) and (B). NaNH , MeI

Na / NH (I)

2 PhC ≡ CH     → (A)    3 → (B)

Solution : (A): Ph – C ≡ C – CH3

(B):

Ph

H

H

CH3

(Pure trans)

O-259 Example 15: What would be the major product in each of the following reactions? CH3

(i) CH3

CH2Br

C

C2H5OH ∆

CH3

(ii)

CH3

H2 Lindlar catalyst

Solution : CH3

(i) CH3

C

CH2CH3

OC2H5

(ii)

CH3 H

H

Example 16: (i) Identify (A), (B) and (C) in the following schemes and write their structures: Br2 / CCl4

(ii)

(A)

NaNH2

(B)

HgSO4 / H2SO4

(C)

Identify (X), (Y) and (Z) in the following synthetic scheme and write their structures. Explain the formation of labelled formaldehyde (H2C*O) as one of the products when compound (Z) is treated with HBr and subsequently ozonolysed. Mark the C* carbon in the entire scheme. BaC*O3 + H2SO4 → (X) gas [C* denotes C14] (i)Mg/ ether

LiAlH

4 → (Z) CH2 = CH – Br      → (Y)     

(ii)X (iii)H 3 O +

Solution: (i)

Br NaNH2

Br2/CCl4

CH–CH2 (dehydrohalogenation)

C≡CH

Br (A)

(B)

C–CH3 (C)

O

HgSO4 / H2SO4

O-260 *

(ii)

*

Ba C O 3 + H2SO4 → C O 2 (g) + BaSO4 + H2O (x) CH2=CH–Br

(Z)

HBr

(i) Mg/ether *

(ii) CO2 (iii) H3O+

LiAlH4 * CH2=CH–COOH (Y)

* + CH2=CH–CH 2

* CH+2–CH=CH2

(Allyl carbocation)

Br –

* =O Br–CH2CHO+CH 2

O3 /H2O/Zn

* OH CH2=CH–CH 2 (Z)

* CH2–CH=CH 2 Br

Example 17: The reactivity of hydrogen atoms in an alkane towards substitution by bromine atom is: (i)

1°H > 2°H > 3°H

(ii) 1°H < 2°H < 3°H

(iii) 1°H > 2°H < 3°H

(iv) 1°H < 2°H > 3°H

Solution: Homolytic bond dissociation energy is inversely related to the stability of free radicals. Thus the order of reactivity is 3°H > 2°H > 1°H. ∴ (ii) Example 18: The treatment of CH3OH with CH3MgI releases 1.04 ml of a gas at S.T.P. The mass of CH3OH added is: (i)

1.485 mg

(ii) 2.98 mg

(iii) 3.71 mg Solution:

(iv) 4.47 mg

CH3OH + CH3MgI → CH4 + MgI(OH)

Moles of CH4 =

1.04 = 4.642 10 −5 22400

∴ No. of moles of CH3OH = 4.642 ×10–5 Mass of CH3OH = 4.642 10–5 ×32 = 1.485 mg ∴ (i) Example 19: Anti Markovnikov addition of HBr is not observed in: (i)

Propene

(iii) 2–butene

(ii) Butene (iv) 2–pentene

Solution: If double bond is attached to two similar groups, then anti–Markovnikov addition is not possible. In 2–butene, double bond is attached to two similar alkyl (CH3) groups. ∴

(iii)

O-261 Example 20: Dipole moment is shown by: (i)

1,4–dichlorobenzene

(ii) cis –1,–2– dichloroethene

(iii) Trans –1,2– dichloroethene Solution:

(iv) Trans –1,–2 dichloro–2–pentene

Symmetrical compounds will have zero dipole moment. Cl H C=C

&

Cl H

Cl Cl

Unsymmetrical compounds will show dipole moment. H

H C=C

Cl

Cl

Cis–1,2–dichloroethene

∴

H3C

Cl C=C

Cl

CH2CH3

Trans–1,2–dichloro–2–pentene

(ii), (iv)

Example 21: The dehydrohalogenation of 2–bromobutane with alcoholic KOH gives: (i)

Only 2–butene

(ii) Only 1–butene

(iii) 2–butene as the major product

(iv) 1–butene as the major product

Solution: H

H Br H3C–C–C–CH3

alcoholic KOH

H3C–C=C–CH3 + H3C–C–C=CH2

H H

H H Major (more substituted alkene)

H H Minor

(iii)

∴

Example 22: The addition of HCl in the presence of peroxide does not follow anti Markovnikov rule because (i)

HCl bond is too strong to be broken homolytically

(ii)

Cl atom is not reactive enough to add on to a double bond

(iii) Cl combines with H to give back HCl (iv)

HCl is a reducing agent

Solution: Peroxide effect involves free radical mechanism. HCl bond is too strong to be broken homolytically, so addition of HCl in presence of peroxide does not follow anti Markovnikov's rule. ∴

(i)

O-262 Example 23: The ozonolysis of an olefin gives propanone. The olefin is: (i)

propene

(ii) but –1– ene

(iii) but –2– ene

(iv) 2,3–dimethyl but–2–ene

Solution: Terminal double bond on ozonolysis will produce only aldehyde. To form ketone it should not be a terminal alkene. Olefin forms only one ketone so it is a symmetrical alkene. O /H O

2 → 2CH COCH (CH3)2C = C(CH3)2  3 2 3 3

(iv)

∴

Example 24: The ozonolysis of a triple bond produces (i)

A mixture of aldehydes/ketones and carboxylic acids

(ii)

A mixture of aldehydes/ketones

(iii) A mixture of carboxylic acids (iv)

CO2 and H2O

Solution : Ozonolysis of a triple bond produces carboxylic acids. O CH3CH2C≡CCH3

O3

CH3CH2C–CCH3 O

O H2O

CH3CH2COOH+CH3COOH Mixture of carboxylic acid

∴

(iii)

Example 25: The addition of water to propyne in the presence of HgSO4–H2SO4 produces: CH3COCH3

(i)

(ii) CH3 – C(OH) = CH2

(iii) CH3CH2CH2OH Solution:

(iv) CH3CH2CHO

The homologues of acetylene form ketone when hydrated in presence of HgSO4 and H2SO4. OH H2SO4 CH3–C≡C–H+H2O HgSO 4

CH3–C=C–H H

Vinyl alcohol (Markovnikov addition) (Unstable) tautomerize

CH3–C–CH3 O

∴

(i)

O-263 Example 26: The compounds 1–butyne and 2–butyne can be distinguished by using: (i)

Bromine water

(ii) KMnO4 solution

(iii) Tollen's reagent

(iv) Chlorine gas

Solution: Terminal and non terminal alkynes can be distinguished using ammonical Ag+ solution (Tollen's reagent). The metal ion Ag+ reacts with acetylinic hydrogen to form insoluble acetylides. This reaction is possible only for terminal alkynes. (iii)

∴

Example 27: Propyne is produced during the hydrolysis of: (i)

CaC2

(ii) Al4C3

(iii) Mg2C3

(iv) SiC

Solution: On hydrolysis CaC2 gives acetylene where Mg2C3 gives propyne. Mg2C3 + 4H2O → CH3C ≡ CH + 2Mg(OH)2 (iii)

∴

Example 28: Which of the following will produce C2H6 on reaction with C2H5MgBr in significant amount? (i)

Toluene

(ii) Benzyne

(iii) Phthalimide

(iv) Cyclopropene

Any compound having active H–atom on reaction with RMgX evolves alkane.

Solution: (iii)

∴

Example 29: When isobutane is monochlorinated in the presence of ultra-violet light, the product obtained in higher yield is: (i)

n-butyl chloride

(ii) Isobutyl chloride

(iii) sec-butyl chloride

(iv) Tert-butyl chloride Cl

Solution: CH3–CH–CH3 CH3

∴

∴

Cl2/hν

CH3–CH–CH2Cl+CH3–C–CH3 CH3

CH3

isobutyl chloride no. of 1°H× reactivity of 1°H = = tert butyl chloride no. of 3°H× reactivity of 3°H

9 × 1 1 5

(ii)

Example 30: An organic compound (A) contains only C, H and O atoms. On reaction with excess of CH3MgI, (A) gives a gas (X) whose volume under STP conditions was found to be 67.2 litre. The number of hydroxyl groups in the compound (A) are:

O-264 (i)

1

(ii) 2

(iii) 3

(iv) 4

Solution: 67.2 = mole of gas(X) evolved = 3 mole of gas (X) 22.4 The number of active H–atom present in a molecule is equal to the number of moles of gas obtained i.e., no. of –OH groups in compound (A)=3 (iii)

∴

Example 31: Ph3SnH can be used to reduce: (i)

Primary alkyl halide

(ii) Secondary alkyl halide

(iii) Tertiary alkyl halide

(iv) All of these

Solution: Ph3SnH is used to reduce all types of alkyl halides i.e., primary, secondary and tertiary alkyl halides. (iv)

∴

Example 32: Identify the correct order of boiling points of the following compounds: CH3CH2CH2CH2OH,

CH3CH2CH2CHO,

CH3CH2CH2COOH

2

3

1 (i)

1>2>3

(ii) 3 > 1 > 2

(iii) 1 > 3 > 2

(iv) 3 > 2 > 1

Solution: Carboxylic acids are intermolecularly hydrogen bonded forming stable cyclic dimer, while alcohols are associated with intermolecular hydrogen bonding to form associated molecules. Aldehydes are unable to form H–bonds. Thus, boiling point should be highest for butanoic acid, followed by butanol and then butanal. (ii)

∴

Example 33: Identify the set of reagents/reaction conditions 'X' and 'Y' in the following set of transformations: X

Y

CH3 – CH2 – CH2Br  → Product  → CH3 –C H – CH3 | Br

(i)

X = dilute aqueous NaOH, 20°C; Y = HBr/acetic acid, 20°C

(ii)

X = concentrated alcoholic NaOH, 80°C; Y = HBr/acetic acid, 20°C

(iii) X = dilute aqueous NaOH, 20°C; Y = Br2/CHCl3, 0°C (iv)

X = concentrated alcoholic NaOH, 80°C; Y = Br2/CHCl3, 0°C

O-265 Solution: X can be either dilute aqueous NaOH or concentrated alcoholic NaOH giving 1 propanol or propene respectively. 1–propanol on reaction with Y (HBr/acetic acid) would have given 2–bromopropane as the major product through substitution while propene on reaction with Y would have given the same product via addition. Since addition reactions are faster than substitution, so the most appropriate option is. (ii)

∴

Example 34: Consider the following reaction: •

H3C − CH − CH − CH3 + B R → ' X' +HBR | D

| CH3

Identify the structure of the major product 'X' •

(i)

•

(ii) H3C − C H − C − CH3

H3C − C H − CH − C H2 | D

(iii)

| D

| CH3

•

| CH3

•

(iv) H3C − C H − C H − CH3

H3 C − C − CH − CH3 | D

| D

| CH3

| CH3

Solution: Attack of Br • free radical on tertiary carbon is thermodynamically more favoured, leading to the formation of tertiary free radical, which is stabilized by hyperconjugation. (ii)

∴

Example 35: Identify a reagent from the following list which can easily distinguish between 1–butyne and 2–butyne: (i)

Bromine, CCl4

(ii) H2, Lindlar catalyst

(iii) Dilute H2SO4, HgSO4

(iv) Ammoniacal Cu2Cl2 solution

Solution: 1–butyne has acetylenic hydrogen; so it will react with ammoniacal Cu2Cl2 solution to give red precipitate of copper butynide. 2–butyne does not respond to this test. (iv)

∴

Example 36: Which one of the following alkenes will react fastest with H2 under catalytic hydrogenation condition? R

(i)

(iii)

R

(ii)

R

H H

H

H

R

R

R

(iv) R

R

R

H

R

R

Solution: Rate of Hydrogenation of an alkene decreases with increasing alkyl substitution on C = C. Choices (C) and (D) are ruled out due to steric hindrance (as the number of R groups are more). Among (A) and (B) two alkyl groups on the same carbon atom offer more steric hindrance than alkyl groups on two separate carbon atoms. ∴

(i)

O-266

Exercise Long Answer Type Questions 1.

2.

7.

(i)

Classify different types of dienes. Discuss 1, 2 and 1, 4 addition reactions

Write short notes on the following :

of conjugated dienes.

(i)

Markovnikov's rule

(ii)

Diel's Alder reaction [Bund. 2008; Agra 2009]

(ii)

Write a note on acidity of 1-alkynes. [Kashi 2011]

How is buta-1, 3-diene is prepared? Why is it more 8. stable than non-conjugated dienes? Explain the formation of 1, 2- and 1,4-addition product when buta 1, 3-diene reacts with hydrobromic acid.

(i)

Separately ethylene and bromine both are non-polar. How will you explain that addition of bromine to ethylene takes place by ionic

[Meerut 2009]

3.

mechanism? Outline the mechanism.

What is ozonolysis? Explain with examples. Write the structure and IUPAC name of alkene which give the following produced on ozonolysis:

(ii)

An olefine compound 'A' on ozonolysis gives one

5.

Acetone only

(ii)

Acetone and formaldehyde

(iii)

Methyl ethyl ketone and acetone

How will you convert acetylene into the following compounds?

(a)

1-Butyne

(c)

Vinyl acetone

(iii)

10.

(b) Neoprene

following compounds? Allene (b) Ethene (c) n-propyl alcohol [Kanpur 2011]

6.

formaldehyde.

Compete

11.

reaction.

Outline

the

Peroxide H3C – C =CH =CH3 + HBr   → ?

[Lko. 2009]

C(–)

is a waker base than H3C - CH2(–).

(i)

H–C ≡ Explain.

(ii)

How will you synthesise the following: (a)

Methyl acetylene from isopropyl alcohol

(b)

1,4 dibromo –butene from acetylene

(c)

Chloroprene from acetylene. [Lko. 2009]

How will you conversions?

bring

about

the

following

(i)

Ethanol → Chloroform

(ii)

Acetylene → Ethylidene dibromide

(iii)

Chloral →DDT

(iv)

Acetylene → Cyclooctatetraene [Rohd. 2012]

(i) Explain Markovnikov's rule and its mechanism with example.

Complete the following reactions & write down the mechanism of any one of them

(ii)

Discuss the stability and 1,4 addition reactions in conjugated butadiene.

(i) BH / THF

CH3 – CH =CH2  →

[Hazaribagh 2010]

(ii) H2 O/ NaOH

HBr CH2 – CH – CF3   →

the

mechanism of the given reaction:

(ii) How will you convert propene into the (a)

of

structure for A.

A compound (A) C5H12, on chlorination at 300°C gives a mixture of four different monochorinated derivatives (B), (C), (D) and (E). Two of these 9. derivatives give the same stable alkane (F) of dehydrohalogneation. (F) on oxidation with hot alkaline KMnO4 followed by acidification, gives two products (G) and (H). Identify (A) to (H) and write all reactions. [Kanpur 2011] (i)

each

Acetaldehyde and methylglyoxal. Assign the

(i)

[Kanpur 2010]

4.

molecule

12. [Avadh 2008]

Discuss the mechanism of 1,2 and 1,4 addition of HBr to buta-1,3 diene with the relative yield of the two products.

O-267 13.

On ozonolysis followed by the hydrolysis of an organic compound AC7H14 gave and aldehyde BC3H6O and a kentone CC4H6O Identify the compounds A, B and C and give all chemical reactions . [Kanpur 2008]

14.

Give name and structure of such alkenes that yield on ozonolysis followed by the hydrolysis with chemical reaction :

Short Answer Type Questions 1.

(i)

(ii) 2.

Acetylene is passed through red hot iron tube in presence of catalyst dicarbonyle di (triphenyl phosphino) nickel. Calcium carbide reacts with water. Write reactions only. [Bund. 2008]

Arrange the following in decresing order of their acidity and explain. [Bund. 2008] CH3 – CH3, – CH ≡ CH, CH2 = CH2

3.

Explain ozonolysis.

4.

Describe the types of polymerisation in acetylene.

15. 16.

Explain Markovnikov's rule with suitable example.

7.

Explain the following : (i)

Saytzeff rule

(ii)

Markovnikov's rule.

17. [Bund. 2010]

What are dienes ? Classified the dienes. Give methods of preparation, properties and structure 18. and stabillity of Butadiene - 1,3. [Meerut 2008, 10, 11]

8.

(ii)

CH3COCH3 and HCHO

Explain the mechanism of any two of the following : (i)

Elimination reactions

(ii)

Addition reaction to C=C

(iii)

SN reactions

[Meerut 2008]

9.

Discuss Kharasch effect.

10.

Explain ozonolysis with mechanism.

11.

Complete the following reactions.

[Kanpur 2008]

What is Diel's Alder reaction? Give its mechanism. [Kanpur 2009]

[Bund. 2009]

6.

CH3CHO and HCHO

[Bund. 2008] [Bund. 2009]

5.

(i)

Give structure of alkenes from which on ozonolysis followed by hydrolysis the following are obtained : (i)

CH3COCH3 and HCHO

(ii)

Two molecules of CH3CHO [Kanpur 2009]

Write short notes on the following: (i)

Hydroboration

(ii)

Hydroxylation

[Kanpur 2009]

Write the reaction of HBr on propene. Give the mechanism in the presence of peroxide and in absence of peroxide. [Kanpur 2009]

19.

What is Markovnikov's rule? Explain with example and give mechanism. [Kanpur 2010]

20.

Complete the following reactions : Zn (a) CH3–CH=CH2+O3→?   → ? + ?

[Meerut 2012]

H2 O

[Agra 2009]

(b) CH ≡ CH +2CuCl +2NH4OH → ? +? +? [Kanpur 2011]

(i)

CH2 =CH2 +CS2Cl2 → ?

(ii)

CH3 – CH =CH2 +HBr → ?

(a)

(iii)

1 Ag CH2 =CH2 + O2  → ? 2 200 –400 ° C

CH2 =CH2 +O3  4→

(b)

CH2 =CH2 +HOCl → ?

21.

[Kanpur 2008]

12.

22.

Complete the following reactions : CCl

[Kanpur 2011]

Discuss the mechanism of E1 reaction. [Avadh 2008]

Complete the following reactions :

(i)

CH ≡ CH +2CuCl +2NH4OH → ?

Complete the following reaction and give its mechanism. [Avadh 2010]

(ii)

Ni → ? R – CH =CH2 +2H  

CH2 =CH2 – CH =CH2 +Br2 →

23.

200 ° C

[Kanpur 2008]

O-268 24.

Give the mechanism of the reaction:

32.

Conc. CH2SO4

→ CH3 – CH2 – OH  

(i)

Dehydration of Alcohols.

CH3 – CH =CH2

(ii)

Saytzeff Rule

(iii)

Dehydrohalogenation

∆

25.

Explain the following with suitable examples :

[Avadh 2010]

Complete the following reactions : (i)BH / THF

3 (i) CH3 – CH2 – C ≡ C – H   →

33.

(ii)H2 O2 / NaOH

Explain : (i)

Unimolecular and biomolecular elimination reactions.

(ii)

Dehydration of alcohol with mechanism.

H O/ Hg++

2 (ii) CH3 – CH2 – C ≡ CH   →

[Avadh 2009]

26.

[Lko. 2010]

Explain: (i)

Alkene is more reactive than alkyne for electrophilic addition.

(ii)

1-Butyne is more acilic than 1-Butene.

34.

[Avadh 2009]

27.

28.

Write note on:

Complete the following reactions and indicate the 36. major product,

Explain the addition of halogen acid on an unsymmetrical alkene. Give the mechanism of the reaction. [Rohd. 2011]

(i)

Hydration of alkyne.

(ii)

[Avadh 2009]

(i) BH / THF

3 CH3 – CH – CH2   →

37.

Explain the mechanism of elimination (E1 and E2) reaction of alkyl halides. [Rohd. 2011]

38.

Explain ozonolysis of benzene.

Cl2 /500 ° C

CH3 – CH =CH2  → [Avadh 2009]

Discuss the structure of ethylene and acetylene 39. based on molecular orbital theory with the help of 40. suitable diagram. [Kashi 2008] (i) Explain why hydrogen atom of acetylene is more acidic than that of ethane. (ii)

31.

[Lko. 2010]

In the presence of diethyl peroxide the addition of HBr to propene is against Markownikoff's rule. Give mechanism also. [Rohd. 2009, 10]

Hofmann elimination reaction.

(ii) H2 O2 / NaOH

30.

Explain: (i) Oxymercuration Demercuration reaction. (ii) Antimarkovnikov's rule. (iii) Hydroboration-oxidation of propene.

35.

(i)

(i)

29.

[Lko. 2008]

41.

Delocalisation leads to stability.Why? [Rohd. 2012] How is 1,3-butadiene industrially prepared? [Rohd. 2012]

Complete the following reactions : (i)

Write short notes on : (a)

Epoxidation (b) Ozonolysis

(c)

Hydroxylation

[Kashi 2010]

CH3 H3C–C–Br EtOH, ∆

(ii)

(iii)

KMnO

H3C–CH=CH–CH3  4→ [D.D.U. 2009]

42.

In butadiene, C– C bond length is 1.46 A° and C = C bond length is 1.39 A°, Why? [D.D.U. 2011]

43.

(i)

?

Give the product and mechanism of their formation of the following : Conc, → CH3 – CH2 – Cl  

CH3

(ii)

O .Ag

2 H2C=CH2   →?

250 ° C

Complete the following reaction :

(i)

[Rohd. 2012]

NaOH

CH2 = C ≡ CH2 + Br2 → ? (i) O3 / CH2 Cl2 ∆

CH3 – C ≡ CH  →

(ii) ZnIH2 O [Lko. 2008]

(ii)

Write the structure of alkene which on ozonolysis gives acetone molecule only. [D.D.U. 2011]

O-269 44.

45.

Discuss the structure of ethylene and acetylene 9. based on molecular orbital theory with the help of suitable diagram. [Hazaribagh 2008]

Indicate the type of hybridization on each carbon atom in the following molecules. (i)

CH3 − CH2 − C ≡ CH

What is Saytzaff rule? Explain it taking example of the mechanism of dehydration of alcohols? 10.

(ii)

HC ≡ C − C ≡ CH

[Hazaribagh 2008,10]

46.

Write a note on peroxide effect. [Hazaribagh 2008,10]

47.

CH3 − CH2 = CH – CH ≡ CH 11.

What is Saytzeff rule? Explain it taking example of the mechanism of dehydration of Butanol-2.

[Agra 2008]

Complete the following reactions: CH2 = CH − CH =CH2 +CH2 = (i) Heat CH2   → ?

(ii)

[Hazaribagh 2009,11]

48.

[Agra 2008]

Give the IUPAC names of the following compound :

CH2 = CH − CH =CH2 + low temp

HCl  → ?

Explain polymerization of alkenes. [Hazaribagh 2009]

[Kanpur 2009]

Explain the action of H2O on propyne in presence 12. of HgSO4/H2SO4. [Hazaribagh 2009]

Why acetylene is acidic while ethane is neutral ?

50.

What are the conjugated diens? Discuss the relative 13. stabilities of different dienes. [Hazaribagh 2011]

51.

Alkynes undergo nucleophilic additions but alkenes do not. Explain. [Hazaribagh 2011] 14.

Show hybridization on each carbon atom in CH3 − C ≡ H . What is the difference between bonds ? [Kanpur 2009]

49.

Very Short Answer Type Questions 1.

Give reactions only: (i)

Cyclohexane reacts with hydrogen presence of Ni at 80°C to 100°C.

Convert: HC = CH → CHOCHO

16.

(i)

in

What is Diels-Alder reaction?

5.

(a) CH2 = CH − CH =CH2 +CH2 =CH2 → ? (b)

CH3 − CH2 − CH(–OH) – CH3 60%H SO

2 4 → ? +?  

100 ° C

18.

[Meerut 2008]

[Kanpur 2010]

19.

What happens when (a)

dil H SO

2 4 →.............. CH ≡ CH  

HgSO4

(b)

oxidation

 → C3COOH

7.

Discuss Saytzeff's rule.

8.

Complete the following reactions :

20.

(i)

CH ≡ CH to acetaldehyde

(ii)

CH ≡ CH to chloroprene.

[Kanpur 2011]

Complete the following reactions : HOH → CH3 – CH =CH – CH2Cl  

[Avadh 2008]

[Meerut 2012]

21.

[Agra 2008]

How will you convert the following ?

(i)

[Meerut 2008]

Peroxide → ? CH3 − CH = CH2 + HBr  

[Kanpur 2010]

Convert: Chloroform into acetylene.

Write the I.U.P.A.C names of the following : CH2 = CH − C ≡ CH

6.

Complete the following reactions :

Explain why acetylene is more acidic than ethylene. [Bund. 2010]

Methane into Carbon tetrachloride [Kanpur 2009]

Cyclobutane reacts with HBr. [Bund. 2008] 17. Explain why hydrogen of alkyne is acidic?

[Bund. 2010; Avadh 2009]

4.

[Kanpur 2009]

Ethyne into 1-Butyne

(ii)

[Bund. 2009; Avadh 2009]

3.

How will you distinguish between butyne-1 and butyne-2? [Kanpur 2009]

15.

(ii) 2.

[Kanpur 2009]

Which one of the following alkenes is most stable? (i) CH3 – CH2 – CH2 – CH =CH2 (ii) CH3 – CH =CH – CH2 – CH3 [Avadh 2010]

O-270 22.

Write IUPAC name of CH2= CH–CH.=CH2. [Kashi 2008]

23.

Explain why hydrogen atom of acetylene is acidic. [Kashi 2008]

24.

Complete the following reaction : CH3 – CH =:CH2 – HO – Cl →

[Kashi 2010]

25.

Why alkynes are less reactive than alkenes towards electrophilic reagent? [Kashi 2010]

26.

Complete the following reaction CH3 – CH =CH2 +HBr →

27.

35.

CH3

36.

CH3

2. [Kashi 2008]

CH3–CH=CH–CH–CH3 29.

Ethylene is more reactive than acetylene towards electrophilic reagents. Explain. [Lko. 2009]

30.

What happens when 2-methyl -1- butanol is 3. dehydrated with conc. H2SO4? [Lko. 2009] Give the peroxide effect on the given reaction: CH3CH =CH2 +HBr → Product.

32.

[Lko. 2009]

Write the product of ozonolysis of Butene-1. [Lko. 2009]

33.

34.

Name the product of reductive ozonolysis of the 4. following : (i)

CH3CH = CHCH3

(ii)

R2C = CHR

Explain, why ethylene is more reactive than acetylene towards electrophilic addition reaction. Indicate the product and explain its formation in

the

equation

given,

H2 O

[Lko. 2011]

Write mechanism of epoxidation. Write IUPAC name of CH2 = CH – C = CH.

Objective Type Questions

CH3

31.

[Lko. 2011]

[Hazaribagh 2008]

1.

CH3–C–CH2–CH3

CH3

CH2 = C(CH3) – CH = CH2.

Multiple Choice Questions

CH3

(ii)

CH3 – CH2 – CH2 – CH = CH2

(iv)

?

Write IUPAC name of the following :

(i)

(iii)

CH2 =C =CH2  →

[Kashi 2011]

28.

CH2 = CH – CH = CH2

H+

38.

(I) Zn/H2O

H

(ii)

mechanistically [Kashi 2011]

(I) O3

C=C

CH2 = C = CH2

[Lko. 2011]

Write product of ozonolysis in the following reaction: 37.

CH3

(i)

Alkenes have general formula: (a)

CnH2n

(c)

CnH2n+2

(b) C2nH2n (d) CnH2n-2

[Bund. 2008]

Which one of the following alkenes will give acetone on ozonolysis? (a)

CH3–CH2–CH=CH2

(b)

CH3–CH=CH–CH3

(c)

CH3–CH=CH–CH3

(d)

none

[Avadh 2010]

Reaction of cold alkaline KMnO4 solution with alkene is called : (a)

Hydration

(b)

Hydroxylation

(c)

Oxidative Cleavage

(d)

Hydrogenation

[Avadh 2010]

Which of the following compounds forms a white ppt. with ammonical silver nitrate? (a)

CH3CH2C ≡ CCH3

[Lko. 2010]

(b)

CH2 =CHCH2CH3

Which amongst the following has highest heat of hydrogenation and why?

(c)

CH3CH2C ≡ CH

(d)

CH3CH ≡ CHCH3

[Rohd. 2012,13]

O-271 5.

The reaction of propene with HOCl proceeds via

10.

the addition of:

6.

(a)

H + in the first step

(b)

Cl + in the first step

(c)

OH

(d)

+

−

in the first step

Cl and OH

−

in a single step

11.

Which of the following compound will be least susceptible to elimination of hydrogen bromide ?

propanol

(b)

propyl hydrogen sulphate

(c)

acetone

(d)

propanol

Water can be added across a triple bond in the presence of: (a)

Acidic medium

(b)

Alkaline medium

Br–CH2–CH2–NO2

(c)

Neutral medium

(b)

Br–CH2–CH2–CH3

(d)

Acid and HgSO4

(c)

Br–CH2–CH2–CN

(d)

Br–CH2–CH2–CO2Et

12.

Me H Me H Me

13.

H

Hydrogenation of the above compound in the presence of poisoned palladium catalyst gives: (a)

An optically active compound

(b)

An optically inactive compound

(c)

A racemic mixture

(d)

A diastereomeric mixture

14.

The treatment of R'MgBr with RC≡CH produces an alkane: (a)

RH

(b) R'H

(c)

R–R

(d) R'–R

The ozonolysis of an olefin gives only propanone. The olefin is: (a)

Propene

(b)

but–1–ene

(c)

but–2–ene

(d)

2,3–dimethyl –but–2–ene

Anti–Markovnikov's addition of HBr is not observed in: (a)

propene

(b) butene

(c)

2–butene

(d) 2–pentene

The treatment of ethene with cold alkaline

n–Propyl bromide on treating with ethanolic potassium hydroxide produces:

potassium permanganate produces:

(a)

propane

(b) propene

(c)

propyne

(d) propanol

(a)

9.

(a)

(a)

7.

8.

When propyne is treated with aqueous H2SO4 in the presence of HgSO4, the major product is:

15.

Ethylene glycol 16.

1–Chlorobutane on reaction with alcoholic potash gives

(b)

Formaldehyde

(c)

Formic acid

(a)

1–butene

(b) 1–butanol

(d)

Carbon dioxide and water

(c)

2–butene

(d) 2–butanol

Propyne and propene can be distinguished by: (a)

conc. H2SO4

(b)

Br2 in CCl4

(c)

dil. KMnO4

(d)

AgNO3 in ammonia

17.

Which of the following compounds gives the same carbocation on ionization ?

Br Br

(1)

(2)

O-272 Br Br

(3)

18.

(4)

(a)

1 and 3

(b) 2 and 4

(c)

1 and 2

(d) 1 and 4

(a)

Electrophilic addition reaction

(b)

Nucleophilic addition reaction

(c)

Free radical addition reaction

(d)

The formation of carbocation as the intermediate

20.

Isolated diene is ……… stable than cumulated diene.

8.

Ethyne is ……… acidic than ethane.

9.

Properly substituted allenes are ………. .

10.

Cyclooctene can show ……… isomerism around double bond.

The addition of HBr to an alkene in the presence of peroxide is an example of:

19.

7.

The peroxide effect involves: (a)

Ionic mechanism

(b)

Free–radical mechanism

(c)

Heterolytic fission of double bond

(d)

Homolytic fission of double bond

The addition of HCl to 3,3,3–trichloropropene

True/False 1.

diene. 2.

Diels Alder reaction is a pericyclic reaction.

3.

Bromine water test is confirmatory test for alkenes and alkynes.

4.

Cl3CCH2CH2Cl

(b)

Cl3CCH(Cl)CH3

(c)

Cl2CHCH(Cl)CH2Cl

(d)

Cl2CHCH2CHCl2

Fill in the Blank 1.

Alkenes are ……… hydrocarbons.

2.

Alkenes are ……… reactive than alkynes towards catalytic hydrogenation.

3.

Buta-1,3-diene undergo ……… addition at 273K.

4.

Conversion of propyne into acetone is an example of ……… .

5.

Baeyer's reagent is ………. .

6.

Cyclohexene and hexyne are ……… isomers.

Reductive ozonolysis of but-2-ene yields 2 moles of acetone.

5.

Alkynes are more reactive than alkenes towards electrophilic addition reaction.

6.

Cyclohexa-triene is a conjugated triene.

7.

Ethene upon treatment with peroxyformic acid yields an epoxide.

gives: (a)

Cumulated diene is less stable than conjugated

8.

Out of cis 2-butene and trans 2-butene, trans isomer is less stable.

9.

Dehydration of alcohols is an E1 mechanism.

10.

Dehydrohalogenation of 2-Bromobutane yields 2-butene as major product.

O-273

Answers Objective Type Questions Multiple Choice Questions 1.

(b)

2.

(d)

3.

(b)

4.

(c)

5.

(b)

6.

(b)

7.

(b)

8.

(a)

9.

(d)

10.

(c)

11.

(d)

12.

(b)

13.

(d)

14.

(c)

15.

(b)

16.

(a)

17.

(c)

18.

(c)

19.

(b)

20.

(a)

Fill in the Blank 1.

unsaturated

2.

less

3.

1,2

4.

nucleophilic addition reaction.

5.

cold alkaline potassium permanganate solution

6.

ring–chain

7.

more

8.

more

9.

chiral

10.

cis–trans

True/False 1.

True

2.

True

3.

True

4.

False

5.

False

6.

True

7.

True

8.

False

9.

True

10.

True

Hints and Solutions Long Answer Type Questions 12.

Conjugated dienes undergo addition reactions in a similar manner to simple alkenes, but two modes of addition are possible. These differ based on the relative positions of H and X in the products: H–X

1

H

2

+

1

H

3 2

X 4

X

Direct or 1,2–addition

Conjugate or 1,4–addition

O-274 Direct

H-X adds "directly" across the ends of a C=C

Conjugate

H-X adds across the ends of the conjugated system

The numbers 1,2- and 1,4- denote the relative positions of H and X in the products. The distribution of the products depends on the reaction conditions as shown by the example below. The corresponding mechanism is also given here with. HBr –30°C

19%

81%

Br

Br

3–bromo –1–butene

HBr 20°C

1–bromo –2–butene

44%

56% Br–

Br–

mechanism

H+ H

H +

+ 2° carbocation

1°carbocation

At low temperature, the reaction is under kinetic control and the major product is the one which is obtained from fastest reaction. The fastest reaction is definitely the one in which the bromide ion reacts with the secondary carbocation yielding 3-bromo-1-butene. At higher temperature, the reaction is under thermodynamic control and the major product is the more stable system. The more stable system refers to the more highly substituted alkene which is 1-bromo-2-butene.

Very Short Answer Type Questions 38. C–C

C=C O

O O

H C

C R

R

H O

O

mmm

UnitO-275 -IV

C HAPTER

6

Arenes and Aromaticity

1. Arenes There are hydrocarbons which consist of both aliphatic and aromatic groups and these compounds are known as arenes. Toluene, ethyl benzene and isopropyl benzene are alkyl benzenes while vinyl benzene also known as styrene is an alkenyl benzene. The aliphatic portion of these compounds is commonly called the side chain. CH3

C2H5

Methyl benzene Ethyl benzene

CH(CH3)2

Isoprophyl benzene

CH=CH2

Phenyl ethene

1.1 Methods of Preparation of Arenes Preparation of Alkyl Benzenes There are a number of methods available for the formation of arenes. Some of these are discussed below, 1.

Friedel−Crafts alkylation: Substitution of hydrogen atom by alkyl group in aromatic compound is known as Friedel–Crafts alkylation. Friedel–Crafts alkylation is initiated by the attack of alkyl carbocation on aromatic ring. The carbocation can be generated either by R–X and AlX3, alcohol and acid or by alkene and acid. AlX

3 → R++Alx − R – X    4

H+

R – OH  → R + + H2O The Friedel–Crafts acylation involves following steps, H+

Alkene  → Alkyl carbocation

O-276 H K1

+

+R

R +

R –H

+

K–1

K2

This reaction has the limitation that linear alkyl benzenes (LAB) cannot be prepared by this method as the carbocations undergoes extensive rearrangements. 2.

Friedel−Crafts acylation: Substitution of hydrogen by acyl group in aromatic compound is known as Friedel Crafts acylation. The most commonly used catalyst is anhydrous AlCl3, although other Lewis acids can also be used. The acylating reagents are normally an acid derivative like acid halides or acid anhydrides. The Friedel–Crafts acylation involves following steps, O +

–

R–C≡O +AlCl4 or RCO+AlCl4– H

(ii) Slow RDS

H

COR

⊕

+R–C+=O

(iii)

••

R–C=O

••

+

R–C–Cl + AlCl3

••

(i)

COR AlCl4–

⊕

COR

+HCl+AlCl3

fast

This reaction can be employed for preparing lab. In the first step, Friedel–Crafts acylation is done which is followed by Clemmenson's reduction in the second step.

1.2 Preparation of Alkenyl Benzenes 1.

Styrene (vinyl benzene) can be prepared as CH(Br)CH3 CH2=CH2 H+

CH2CH3 NBS in CCl4 alc. KOH

CH=CH2

(Styrene)

O-277 Alkenyl benzenes that have their side-chain double bond conjugated with the benzene ring are more stable than those that do not have. —C=C–C– is more stable than

—C–C=C– Non conjugated system

Conjugated system

Evidence is support of this comes from acid-catalyzed dehydration of alcohols, which are known to yield the most stable alkene. For example, dehydration of the following alcohol yields exclusively the conjugated system. H

H

—C=C–C–

H H+,heat

—C=C–C–

(–H2O)

H

H OH

As conjugation always lowers the energy of an unsaturated system by allowing the p electrons to be delocalized, this behavior is just what we would expect. 2.

Alkenyl benzenes can also be obtained by simple elimination process like the dehydrohalogenation of haloalkyl groups attached to benzene ring. The elimination is generally governed by Saytzeff rule i.e., more substituted alkene is formed in greater amount. Couple of examples are given below, CH=CH–CH3 (major) +

CH2–CH–CH3 Br

alc. KOH

CH2CH2–CH–CH3 Br

CH2CH=CHCH3 (major)

alc. KOH

CH2–CH=CH2 (minor)

CH2CH2CH=CH2 (minor)

+

Solved Examples Example 1: Give the products and account for the major product of dehydrohalogenation of (i)

ArCH2CHBrCH2CH2CH3

(iii) CH3CHArCHBrCH2CH3

(ii)

CH3CBrArCH2CH2CH3

(iv)

BrCH2CHArCH2CH2CH3

Solution : (i) Conjugated ArCH = CHCH2CH2CH3 (major), unconjugated ArCH2CH = CHCH2CH3 (minor). (ii)

Conjugated, trisubstituted CH3CAr = CHCH2CH3 (major), conjugated disubstituted H2C = CArCH2CH2CH3 (minor)

O-278 (iii)

Conjugated CH3CAr = CHCH2CH3(major) unconjugated CH3CHArCH = CHCH3 (minor).

(iv)

H2C = CArCH2CH2CH3. A 1° Br can give only a single product.

In general the most substituted, least sterically–hindered alkene with its C = C conjugated with the ring is the major product because of its greatest stability. This stability is a contributing factor towards faster rate of formation.

2. Structure of Benzene Benzene is formed from a six-membered ring constituted by carbon atoms with alternating single and double bonds. Interestingly, this description fits well for a cyclohexatriene ring. This bonding arrangement was developed by Kekule. According to Kekule model, benzene has two resonating structures as shown below, H

H

C

H

C

C C

H

C C

H

H

H

H

C C

C C

C C

H

H

H

H

Fig. 1: Two resonating Kekule structures of benzene

These two resonating forms correspond to the double and single bonds superimposing to give rise to six one-and-a-half bonds. As is standard for resonance diagrams, a double-headed arrow is used to indicate that the two structures are not distinct entities, but merely hypothetical possibilities. Neither of the two structures is an accurate representation of the actual compound. Rather the actual compound is the best represented by a resonance hybrid or average of these two Kekule structures, and the whole phenomenon can be shown below,

≡

nb:

not

Or, it can be compared to a real life situation below, Unicorn

Dragon ≡ Rhinoceros Real structure Resonance structures (hybrid)

O-279 H H

H

H

H H

is a far better representation of benzene molecule. Here, the circular π–bond also known as Armstrong's inner cycle represents a delocalised π-bond above and below the ring. The electron density is evenly distributed throughout the molecule. This model more correctly represents the location of electron density within the aromatic ring.

2.1 σ and a π-bonding Arrangement The single bonds are formed with electrons in line between the carbon nuclei–these are called σ-bonds. Double bonds consist of a σ-bond and a π-bond. Following diagram shows the σ bonding arrangement among only carbon atoms in the ring. σ

σ

σ

σ

σ

σ

Fig. 2: σ bonding arrangement in the hexagonal ring of benzene

However, the benzene ring contains two types of σ–bonds–type I involve C-C σ bonds which is being shown in fig above and the type II involve C-H σ bonds. These are shown together in the diagram below,

O-280 The π-bonds are formed from overlap of atomic p-orbitals above and below the plane of the ring. The following diagram shows the positions of these p-orbitals:

Fig. 4: Position of p–orbitals in benzene ring

These p–orbitals are out of the plane or more precisely in the plane perpendicular to the molecular plane of the carbon atoms. These p–orbitals can interact with each other freely, and give rise to delocalized π bonding. This means that instead of being tied to one atom of carbon, each electron is shared among all six carbon atoms in the ring. Thus, the "extra" electrons strengthen all of the bonds on the ring equally.

H

H H

H H

H 6 p–orbitals

H

H H

H H

H Delocalized

Fig. 5: Delocalized π – ring in the perpendicular plane

2.2 Bond length Comparing the two contributing Kekule structures of benzene, it can be observed that all the single and double C-C bonds are interchanged in the two canonical forms as observed in fig1. Bond lengths can be measured using X-ray diffraction technique. The average length of a C-C single bond is 154 pm; that of a C=C double bond is 133 pm. In localized cyclohexatriene, the carbon-carbon bonds should be alternating 154 and 133 pm. Instead, all carbon-carbon bonds in benzene are found to be about 139 pm, a bond length intermediate between single and double bond. This mixed single and double bond character is typical for all molecules in which bonds have a different bond order in different contributing structures. Bond lengths can be compared using bond orders. For example in cyclohexane the bond order is 1 while that in benzene is 1+(3/6)= 1.5 . Consequently, benzene has more double bond character and hence has a shorter bond length than cyclohexane.

O-281

2.3 Resonance Energy of Benzene Every structure is associated with a certain quantity of energy, which determines the stability of the molecule or ion. So, lower is the energy, the greater is its stability. A resonance hybrid has a structure that is intermediate between the contributing structures; the total quantity of potential energy, however, is lower than the intermediate. Hybrids are therefore always more stable than any of the contributing structures would be and the same is true for benzene also. Delocalization of the π-electrons lowers the orbital energies, imparting this stability. The difference between the potential energy of the actual structure (the resonance hybrid) and that of the contributing structure with the lowest potential energy is called the "resonance energy".

2.4 Measurement of Resonance Energy of Benzene It is known that resonance or delocalization energy is the amount of energy needed to convert the true delocalized structure into that of the most stable contributing structure. The empirical resonance energy can be estimated by comparing the enthalpy change of hydrogenation of the real substance with that estimated for the contributing structure. The complete hydrogenation of benzene to cyclohexane via 1,3-cyclohexadiene and cyclohexene is exothermic; 1 mole benzene delivers 208.4 kJ (49.8 kcal).

1 mol

H2

H2

H2

∆H=+23.4 kJ ∆H=–112.1 kJ ∆H=–119.7 kJ (+5.6 kcal) (–26.8 kcal) (–28.6 kcal) Benzene 1,3–Cyclohexadiene Cyclohexene Cyclohexane

Hydrogenation of one mole of double bonds delivers 119.7 kJ (28.6 kcal), as can be deduced from the last step, the hydrogenation of cyclohexene. In benzene, however, 23.4 kJ (5.6 kcal) are needed to hydrogenate one mole of double bonds. The difference, being 143.1 kJ (34.2 kcal), is the empirical resonance energy of benzene. Because 1,3-cyclohexadiene also has a small delocalization energy (7.6 kJ or 1.8 kcal/mol) the net resonance energy, relative to the localized cyclohexatriene, is a bit higher: 151 kJ or 36 kcal/mol. This measured resonance energy is also the difference between the hydrogenation energy of three 'non-resonance' double bonds and the measured hydrogenation energy: (3 × 119.7) – 208.4 = 150.7 kJ/mol (36 kcal).

2.5 Molecular Orbital Representation of Benzene The bond angle of 120° in benzene suggests that C atoms are sp2 hybridised. An alternative representation therefore starts with a planar framework and considers overlap of the p orbitals (π electrons).

O-282

Unused p orbital ×6

sp2 hybridised C

Delcocalised π system

According to simple MO rules, mixing n atomic orbitals yields the same number of molecular orbitals or stated mathematically, Mix n × π atomic orbitals → nπ molecular orbitals If applied in the case of ethene, a predicted result is obtained. So, two 2p atomic orbitals are mixed together to yield a pair of molecular orbitals labelled as bonding molecular orbitals and antibonding molecular orbitals,

+

2p orbitals

Bonding π–orbital

Antibonding π–orbital (π*)

π∗

Energy

overlap

π

Similarly, in benzene ring, there are 6 overlapping pure atomic p orbitals, so there will be 6 molecular orbitals formed as well. Out of these 6 molecular orbitals, 3 are bonding and 3 are antibonding. So, it can be summarised as, 3 Higher energy π* In benzene there are 6 x overlapping p–orbitals → 6 π MO's 3 Lower energy π

The exact calculation of their position as shown below is beyond the scope of this text book.

O-283

π∗ overlap

Energy 6 isolated p–orbitals

π

Each MO can accommodate 2 electrons, so for benzene, one can observe that all electrons are paired and occupy low energy MO's or bonding MO's. All bonding MO's are filled. Benzene is therefore said to have a closed bonding shell of delocalised π electrons and this accounts in part for the stability of benzene. There is a simple "trick" for working out the orbital energies which is termed as Frost-Musulin diagrams. In this diagram, a polygon is drawn in a circle. One has to draw the molecular framework of a cyclic system of overlapping p-orbitals, making sure that an atom must be kept at the bottom. Atomic positions, in the present case the positions of p-orbitals then map on to the energy level diagram as shown below,

3. Aromaticity and Huckel's Rule 3.1 Aromaticity Historically, the earliest use of the term "aromatic" was in an article by August Wilhelm Hoffmann in 1855. Hoffmann used the term as a class term for compounds of benzene, many of which do have odors unlike pure saturated hydrocarbons. Today, there is no general relationship between aromaticity as a chemical property and the odour or smell of such compounds. In terms of the electronic nature of the molecule, aromaticity describes the way conjugated ring of unsaturated bonds, lone pairs of electrons, or empty molecular orbitals exhibits a stabilization stronger than would be expected by the stabilization of conjugation alone. Huckel's rule helps us to decide that a given compound would be aromatic or not by observing its structure.

O-284

3.2 Huckel's Rule Huckel's rule deals with the aromaticity of a compound. According to Huckel's rule, an aromatic compound contains a set of covalently bonded atoms with specific characteristics. These characteristics are outlined below, 1.

A compound must contain a delocalized conjugated π system so that it can become aromatic. A delocalized conjugated π system is most commonly an arrangement of alternating single and double bonds in a compound. However, conjugation can be achieved in a number of other ways like a negative charge is conjugated with a double bond or a lone pair is conjugated with a double bond etc.

2.

The compound must have a coplanar structure, which means that all the contributing atoms must be in the same plane.

3.

Constituting atoms arranged in one or more than one rings must constitute a closed loop of electrons.

4.

Most importantly, the species must have (4n + 2) π electrons, where n=0, 1, 2, 3, and so on.

Whereas benzene is aromatic due to the presence of 6 π electrons, from 3 double bonds, cyclobutadiene is not. It is because the number of π delocalized electrons is 4 and does not confirm to the prerequisite of (4n + 2) π electrons. The cyclobutadienide (2–) ion, however, is aromatic because it has 6π electrons. Aromatic molecules typically display enhanced chemical stability, compared to similar non-aromatic molecules. A molecule that can be aromatic will tend to alter its electronic or conformational structure to be in this situation. This extra stability changes the chemistry of the molecule. Aromatic compounds undergo electrophilic aromatic substitution and nucleophilic aromatic substitution reactions very frequently. However, these compounds do not show electrophilic addition reactions as happens with carbon-carbon double bonds. It is definitely attributed to the fact that taking part in electrophilic addition reactions causes loss of aromatic stability of the compound.

3.3 Aromaticity in Some Other Compounds 1.

The cases of naphthalene (I) & anthracene (II) are considered to check for their aromaticity.

(II)

(I)

These compounds are definitely planar due to the sp² hybridized carbon atoms. Both these compounds have conjugation in carbocyclic ring leading to cyclic delocalization and the closed loop formed would have 10 π and 14 π electrons respectively. So, both the compounds are aromatic as both the conditions are fulfilled. 2.

If four species are provided, which are shown as

•

H Cyclopropene (III)

Cyclopropenyl radical (IV)

⊕

H Cyclopropenyl cation (V)

• •

H Cyclopropenyl anion (VI)

O-285 Of these four species, (III), (IV) and (VI) are non-aromatic, due to the failure of first condition, second condition and second condition respectively. While the species (V) is aromatic as it will have permitted cyclic delocalization and closed π–orbital is having 2 π electrons corresponding to n = 0. 3.

If following species are provided then the aromatic species has to be picked out. H •

H ••

⊕

Cycloheptatriene Cycloheptatrienyl Cycloheptatrienyl cation (VII) radical (VIII) (Tropyliumion) (IX)

H

Cycloheptatrienyl anion (X)

Of the given species, only species (IX) is aromatic since it satisfies both the conditions required by a species to be aromatic, while (VII), (VIII) and (X) are non-aromatic. (VII) is non–aromatic since it does not undergo cyclic delocalization as one of the carbon atoms of the ring is sp3 hybridized inhibiting extended conjugation. (VIII) is non–aromatic since in the closed loop, it has 7 π electrons which is not a Huckel number so it is not satisfying Huckel's rule. (X) is non–aromatic since in the closed loop; it has 8 π electrons which is also not a Huckel number. 4.

Now, it has to be analysed whether pyrrole and pyridine are aromatic or not. It will be analysed one by one.

N Pyridine (XI)

5.

N H

Pyrrole (XII)

In pyridine, the nitrogen atom is sp2 hybridized with one of the sp2 hybrid orbital having a lone pair of electrons. Thus, the sp2 hybrid lobe with lone pair is out of the plane with respect to the delocalized π bond present in the ring. It is because the 6π electron ring is present in the Z plane as the ring is formed by the overlap of pz orbital of adjacent carbon and the molecular plane is XY, so the lone pair of nitrogen is not at all involved in delocalization. Thus pyridine is aromatic with no involvement of lone pair in cyclic delocalization. While in pyrrole, the nitrogen atom is sp3 hybridized with one of the sp3 hybridized orbital having a lone pair of electrons. This sp3 lobe with lone pair is nearly planar with respect to pz–orbital of the adjacent carbon, so the lone pair of nitrogen is involved in cyclic delocalization and the closed loop will have 6 π–electrons. Thus, pyrrole is aromatic in nature. Furan (IX) and thiophene(X) are aromatic as well due to the same line of explanation as applicable to pyrrole.

O Furan (XIII)

S Thiophene (XIV)

O-286 Both furan and thiophene are typical in one aspect. These two compounds represent cases when an atom in any aromatic system can have more than one lone pair of electrons and out of those lone pairs only one is the part of aromatic sextet and the rest are not therefore ignored for the 4n + 2 rule. So, in furan, the oxygen atom is sp² hybridized. One lone pair is in the π system and the other in the plane of the ring lying in one of the sp² hybridized lobes. So, only one lone pair of electrons is considered as part of the aromatic ring of furan and the other one which is lying in the molecular plane is excluded from the aromatic sextet. Since there are 6 π electrons, so furan is aromatic with one lone pair of electrons available on oxygen atom which is not a part of aromatic ring. The same explanation is applicable for thiophene.

NMR signal The circulating π electrons in an aromatic molecule produce ring currents that oppose the applied magnetic field in NMR. The NMR signal of protons in the plane of an aromatic ring are shifted substantially further down-field than those on normal non-aromatic sp² carbon atoms. This is an important way of detecting aromaticity. By the same mechanism, the signals of protons located near the ring axis are shifted up-field.

3.4 Antiaromaticity Planar monocyclic molecules containing 4n π electrons are called antiaromatic and are, in general, destabilized. Molecules that could be antiaromatic will tend to alter their electronic or conformational structure to avoid this situation, thereby becoming non-aromatic. For example, cyclooctatetraene (COT) distorts itself out of planarity, breaking π overlap between adjacent double bonds. Relatively recently, cyclobutadiene was discovered to adopt an asymmetric, rectangular configuration in which single and double bonds indeed alternate but there is no resonance and the single bonds are markedly longer than the double bonds, reducing unfavorable p-orbital overlap. This reduction of symmetry lifts the degeneracy of the two formerly non-bonding molecular orbitals, which by Hund's rule forces the two unpaired electrons into a new, weakly bonding orbital and also creates a weakly antibonding orbital. Hence, cyclobutadiene is non-aromatic; the strain of the asymmetric configuration outweighs the anti-aromatic destabilization that would afflict the symmetric, square configuration. Example 2: Which species is the smallest aromatic substance? Solution : The cyclopropenyl cation is an aromatic substance because it has two π electrons (n = 0). It forms a stable salt that can be stored in a bottle. H–C=C–H C+ H

Example 3: Which of the following species is/are aromatic (i)

(ii) ⊕

SbCl6

O-287 (iii)

Ph

(iv)

⊕

⊕

H

Ph

•

Solution : For a compound to be aromatic, there must be cyclic delocalization with the closed loop formed must have (4n + 2) π electrons. In the given case, (i), (ii) and (iii) qualify to be called as aromatic. ∴

(i), (ii) & (iii)

4. Electrophilic Aromatic Substitution Reactions Most substitutions at an aliphatic carbon are nucleophilic. However, in aromatic systems the situation is reversed, because the high electron density at the aromatic ring attracts positive species and negative ones not so easily. In electrophilic substitutions, the attacking species is an electrophile-normally a positively charged ion or sometimes a neutral species as well. The leaving group must necessarily depart without its electron pair; therefore in electrophilic substitutions the most important leaving groups are those that can best exist without the pair of electrons necessary to fill the outer shell, which is thereby the weakest Lewis acids. The most common leaving group in electrophilic aromatic substitutions is a proton. These reactions in the general form are shown below:

4.1 General Mechanism Normally an electrophilic aromatic substitution reaction has some common steps like in the first step, generation of electrophile takes place. In the second step, the electrophile attacks the benzene ring to give a σ −complex and in the last step, retention of aromaticity takes place which leads to the eventual product formation.These steps are discussed below one by one. Step -1: Generation of electrophile In this step, normally electrophiles like Cl + , nitronium ion , alkyl carbocation, acylium ion etc are produced as per the reaction. Step-2: Formation of σ – complex The electrophile (E + ) attacks the ring, removing a pair of electrons from the sextet to give a carbocation which is a resonance hybrid as shown in step–1. Ions of this type are called Wheland intermediate, σ –complexes or arenium ions. Normally σ –complexes are highly resonance stabilized as shown below, H

H E

E

+ Slow

+ H

H ≡

+ E

+

+ Arenium ion ( σ complex)

E

E +

O-288

Step-3: Retention of aromaticity The arenium ion loses a proton from the carbon atom that bears the electrophile. This step results into retention of aromaticity and yield the eventual product. +

E E

H fast

+

+H

The energy profile for general mechanism is shown as,

Some of the important electrophilic aromatic substitution reactions are halogenation, sulphonation, nitration, Friedel Craft reactions-alkylation and acylation both etc.These are discussed one by one below,

4.2 Halogenation Benzene does not react with bromine or chlorine unless a Lewis acid is present in the mixture. It is because of this, benzene does not decolorize a solution of bromine in carbon tetrachloride despite having three double bonds. When Lewis acids are present, however, benzene reacts readily with bromine or chlorine, and the reaction yields bromobenzene and chlorobenzene in good percentage as shown,

O-289

The Lewis acids most commonly used to effect chlorination and bromination reactions are FeCl3, FeBr3, and AlCl3 in the anhydrous form. Ferric chloride and ferric bromide are usually generated in the reaction mixture by adding iron to it. The iron then reacts with halogen to produce the ferric halide. 2Fe + 3X2 → 2FeX3

Mechanism The mechanism for aromatic bromination proceeds in the following manner. Step–1: Generation of electrophile Bromine combines with FeBr3 to form a complex that dissociates to form a positively charged bromine ion and FeBr4– .

Step–2: Formation of σ − complex The positive bromine ion attacks benzene to form an arenium ion also known as σ − Br

Br

+

+ Br

+ H

H

H Slow

+

+ Arenium ion

Step–3: Retention of aromaticity The arenium ion loses a proton to become bromobenzene. – Br –FeBr3

H Br

Br +H–Br +FeBr3

+

complex. Br

O-290 The general energy profile is, Transition state

⊕

∆G#

Br H

Energy

Br

∆G°

Reaction progress Fig. 8: Energy profile for halogenation of benzene

Iodination Iodine, on the other hand, is so unreactive that a special technique has to be used to effect direct iodination. The reaction has to be carried out in the presence of an oxidizing agent such as nitric acid or iodic acid. The oxidizing agent has an important role to play- it removes whatsoever HI formed in the reaction. As such HI is strongly reducing in nature and it can reduce iodobenzene back to benzene.

+ I2

HNO3

I

4.3 Nitration Benzene reacts slowly with hot concentrated nitric acid to yield nitrobenzene. The reaction is much faster if it is carried out by heating benzene with a mixture of concentrated nitric acid and concentrated sulfuric acid.

Concentrated sulphuric acid increases the rate of the reaction by increasing the concentration of the electrophile–the nitronium ion. Step–1: Generation of electrophile Nitric acid acts as a base and accepts a proton from the stronger acid, sulfuric acid as shown below,

O-291

The protonated nitric acid dissociates and produces a nitronium ion. Step 2: Formation of σ − complex The nitronium ion reacts with benzene by attacking the π cloud and forming an arenium ion.

Step 3: Retention of aromaticity The arenium ion then transfers a proton to some base in the mixture such as HSO −4 and becomes nitrobenzene. +

NO2 H+HSO4–

NO2 fast

+H2SO4

The general energy profile is, ≠ ≠

Energy

NO2 H +

NO2 +HNO3 +H2SO4 Reaction coordinate Fig. 9: Energy profile for nitration of benzene

O-292

4.4 Sulfonation Benzene reacts with fuming sulfuric acid at room temperature to produce benzenesulfonic acid. Fuming sulfuric acid is an equimolar mixture of sulfuric acid and sulfur trioxide (SO3). Sulfonation also takes place in concentrated sulfuric acid alone, but the reaction rate is very slow.

Step–1: Generation of electrophile In concentrated sulphuric acid, sulfur trioxide is produced in the following manner in which H2SO4 acts both as an acid and a base. SO3+H3O++HSO4–

2H2SO4

Step–2: Formation of σ −complex The neutral SO3 molecule which is an electrophile reacts with benzene by attacking the π cloud and forming an arenium ion. H

O +

S

H

SO3–

SO3–

+ H

SO3–

Slow

+

O O Sulfur trioxide

+ Arenium ion

Step–3: Retention of aromaticity The arenium ion thus formed then transfers a proton to the conjugate base of sulphuric acid, HSO −4 and yields unprotonated benzene sulphonic acid. – + SO3

H+HSO4– fast

SO3– +H2SO4

Step– 4 Eventually the protonation of the compound takes place to yield benzene sulphonic acid. –

SO3H

SO3 +H3O+

fast

+H2O

O-293 All the steps in sulphonation are reversible, including step 1 in which sulfur trioxide is formed from sulfuric acid. This means that the overall reaction is an equilibrium process as shown below, SO3H +H2O

+H2SO4

4.5 Friedel–Crafts Alkylation A general schematic representation for a Friedel - Crafts alkylation reaction is expressed as: R +R—X

AlCl3

+HX

Mechanism Step–1: Generation of electrophile Alkyl halide molecule in the presence of a Lewis acid catalyst gives rise to a carbocation as shown below. The carbocation is the effective electrophile in the reaction. (CH3)2CHCl+AlCl3 Isopropyl cholride

(CH3)2CH+ +AlCl4– Carbocation

Step–2: Formation of σ −complex Benzene ring attacks on the carbocation and yield the σ −complex also known as arenium ion as shown here, H CH3 +

+CH

CH3 CH

+

CH3

CH3

Arenium ion

Step–3: Retention of aromaticity The arenium ion thus formed then transfers a proton to the AlCl4 and yields alkylated benzene. CH3

H

CH3 CH

+

CH3

CH +AlCl4–

CH3+HCl+AlCl3 Isopropylbenzene

O-294 Some more examples are, CH(CH3)2 +CH3CH=CH2 0°C HF Propene

Isopropylbenzene

A mixture of an alcohol and an acid may also be used to yield a carbocation. OH 60°C

+

BF3

Cyclohexanol

Cyclohexylbenzene

4.6 Friedel Crafts Acylation The RCO–group is called an acyl group, and a reaction where by an acyl group is introduced into a compound is called acylation reaction. The Friedel - Crafts acylation reaction is an effective means of introducing an acyl group into an aromatic ring. The reaction is often carried out by treating the aromatic compound with an acyl halide. Unless the aromatic compound is one that is highly reactive, the reaction requires the addition of at least one equivalent of a Lewis acid, such as AlCl3,as well. The product of the reaction is an aryl ketone.

+CH3COCl Acetyl Cholride

AlCl3 excess benzene 80°C

COCH3 +HCl

Acetophenone

Acyl chlorides, also called acid chlorides, are easily prepared by treating carboxylic acids with thionyl chloride (SOCl2) or phosphorus pentachloride (PCl5). 80 ° C

CH3CO2H

+ SOCl2   →

CH3COCl + SO2 + HCl

PhCO2H

+ PCl5

PhCOCl + POCl3 + HCl

Benzoic

Phosphorus

Benzoyl

acid

pentachloride

chloride

→

Friedel - Crafts acylations can also be carried out using carboxylic acid anhydrides. For example : AlCl3 +(CH3CO)2O excess benzene 80°C

Acetic anhydride (a carboxylic acid anhydride)

COCH3 +CH3COOH Acetophenone

O-295

Mechanism In most Friedel - Crafts acylation the electrophile appears to be an acylium ion formed from an acyl halide in the following way: Step 1: Generation of electrophile Acyl halide molecule in the presence of a Lewis acid catalyst gives rise to an acylium ion which is a highly stable carbocation as shown below. As such, the carbocation-acylium ion is the effective electrophile in the Friedel–Crafts acylation reaction. O

O

+

–

R – C – Cl – AlCl3

R – C – Cl + AlCl3 O +

+

–

+

R – C ≡ O + AlCl4–

R–C=O

R – C – Cl – AlCl3

An acylium ion (a resonance hybrid)

Step–2: Formation of σ −complex Benzene ring attacks the carbocation and yield the σ −complex also known as arenium ion as shown here, H

R + C O+

+

C–R O

Arenium ion

Step–3: Retention of aromaticity The arenium ion thus formed then transfers a proton to the AlCl4 and yields acylated benzene as shown below H +

COR

COR +AlCl – 4

+HCl+AlCl3

Step–4 In the last step aluminum chloride which is a Lewis acid as well forms a complex with the ketone (a Lewis base). After the reaction is over, treating the complex with water liberates the ketone. C–R O

R C6H5

C–R +AlCl3

C=O AlCl3+3H2O

O

–

AlCl3

+

R C6H5

C=O +Al(OH)3+3HCl

O-296

4.7 Limitations of Friedel-Crafts Reaction 1.

The carbocation formed from an alkyl halide, alkene, or alcohol can rearrange to a more stable carbocation and the major product obtained from the reaction is usually the one from the more stable carbocation. When benzene is alkylated with butyl bromide, some of the developing 1° butyl carbocation rearrange by a hydride shift -- to become more stable 2° carbocation. Then benzene reacts with both kinds of carbocations to form both n–butylbenzene and sec-butyl-benzene. δ+

CH3CH2CH2CH2Br AlCl3 CH3CH2CHCH2

δ

BrAlCl3

+

(–BrAlCl–3)

CH3CH2CHCH3

H (–AlCl3) (–HBr)

CH3CH2CHCH3

CH3CH2CH2CH2

n–Butylbenzene (32–36% of mixture)

2.

(–H+)

sec–Butylbenzene (64–68% of mixture)

Friedel - Crafts reactions do not occur when powerful electron-withdrawing groups are present on the aromatic ring or when the ring bears an -NH2, -NHR, or -NR2 group. This applies to both alkylation and acylation.

NO2

+

N(CH3)3

O

O

C–OH

C–R

CF3

SO3H

+

NH3

It is always important to remember that groups present on an aromatic ring can have large effect on the reactivity of the ring towards electrophilic aromatic substitution. Electron - withdrawing groups make the ring less reactive by making it electron deficient. Any substituent more electron withdrawing or deactivating than a halogen, that is, any meta-directing group, makes an aromatic ring too electron deficient to undergo a Friedel - Crafts reaction. 3.

Friedel–Crafts reactions do not take place even in the case of the amino groups like -NH2, -NHR, and -NR2 attached directly to benzene ring. It is because these groups are changed into powerful electron withdrawing groups by the Lewis acids used to catalyze Friedel - Crafts reaction. For example:

O-297 H

H

H–N

H–+N––AlCl3 + AlCl3 Does not undergo a Friedel–Crafts reaction

4.

Aryl and vinylic halides cannot be used as the halide component because they do not form carbocations readily. Cl

no Friedel – Crafts reaction AlCl3 no Friedel – Crafts reaction –C=C–Cl

4.8 Birch Reduction The Birch reduction is an organic reaction which is particularly useful in synthetic organic chemistry. The reaction was reported in 1944 by the Arthur Birch. It converts aromatic compounds having a benzenoid ring into a product, 1,4-cyclohexadienes, in which two hydrogen atoms have been attached on opposite ends of the molecule. It is the organic reduction of aromatic rings like benzene or naphthalene or their derivatives in liquid ammonia with sodium, lithium or potassium and an alcohol, such as ethanol and tert-butanol. Though the original reaction reported by Arthur Birch in 1944 utilized sodium and ethanol. H H

NH3

Na ROH

H H

This reaction is quite dissimilar to the catalytic hydrogenation, which usually reduces the aromatic ring all the way to a cyclohexane. Another example is the reduction of naphthalene, Na, NH3,–78°C EtOH, Et2O

The Birch reduction also offers access to substituted 1,4-cyclohexadienes.

78–80%

O-298

Mechanism

e–

ROH –RO–

H

H e– H H

H ROH –RO–

H H

H H

H H

The question of why the 1,3-diene is not formed, even though it would be more stable through conjugation, can be rationalized in valence bond terms. It is very important to observe that the electron-electron repulsions in the radical anion will prefer to have the nonbonding electrons separated as much as possible, in a 1,4-relationship rather than 1,3-relationship. Reactions of arenes with +I- and +M-substituents lead to the products with the most highly substituted double bonds: OMe

OMe Li, NH3 EtOH

Na,NH3 EtOH

On the other side for arenes with –I and –M substituents, the effect of electron-withdrawing substituents on the Birch Reduction varies. For example, the reaction of benzoic acid leads to 2,5-cyclohexadienecarboxylic acid. It can be rationalized on the basis of the carboxylic acid stabilizing an adjacent anion which is placed alpha to it: COOH

COOH Na,NH3 EtOH

H3O+

Alkene double bonds are only reduced if they are conjugated with the arene.

Na,NH3 EtOH

O-299

5. Disubstitution in Benzene Ring and Theory of Substituent Effect It has been observed that certain groups activate the benzene ring toward electrophilic substitution, while other groups deactivate the ring. When it is said that a group activates the ring, it means that the group increases the relative rate of the reaction. So, an aromatic compound with an activating group reacts faster in electrophilic substitutions than benzene. When it is said that a group deactivates the ring, it means that an aromatic compound with a deactivating group reacts slower than benzene or the rate of reaction is retarded. It is generally found that the relative rates of the reactions depend on whether the group S shown below withdraws or releases electrons towards the benzene ring.

If S is an electron-releasing group (relative to hydrogen), the reaction occurs faster than the corresponding reaction of benzene. If S is an electron-withdrawing group, the reaction occurs slower than that of benzene. It appears, then, that the substituent (S) must affect the stability of the transition state relative to that of the reactants. Electron-releasing groups apparently make the transition state more stable, while electron-withdrawing groups make it less stable. This is fairly reasonable, because the transition state resembles the arenium ion, and the arenium ion is a delocalized carbocation. On this basis of electron release or withdrawal, groups attached to the ring can be put into two categories, 1.

Ortho-para directing

2.

Meta directing

O-300

The general energy profile for disubstitution of a benzene ring is, Y

Y

H

Ortho attack

E

⊕

+E

Y ⊕ SO3H Sulfonation

Br⊕ Bromination

R⊕ F–C alkylation

⊕ NO2 Nitration

⊕ RC=O F–C acylation

Meta attack

⊕

H E

Energy

Cl Chlorination

Y Para attack

∆E1

⊕

⊕ +E

E

H

∆E2#

H

Y

Y

⊕

E

∆E Y

⊕ E+H

Reaction progress

Fig. 10: Energy profile for disubstitution of a benzene ring

5.1 Ortho - Para - Directing Groups Except for the alkyl and phenyl substituents, all of the ortho-para directing groups are of the following general type:

All of these ortho-para directors have at least one pair of nonbonding also known as lone pair of electrons on the atom adjacent to the benzene ring. This structural feature - a lone pair of electrons on the atom adjacent to the ring - determines the orientation and influences reactivity in electrophilic substitution reactions.

NH2 E

E

⊕

⊕

H

NH2

E

E NH2

NH2

H E

⊕

⊕

⊕

NH2

H

⊕

H E Para attack

E

Relatively stable contributor

NH2

NH2

NH2 Meta attack

E

⊕

⊕

NH2

H

H

H ⊕

Ortho attack

NH2

NH2

NH2

H

⊕

O-301

H E

H E

H E Relatively stable contributor

It is observed that four reasonable resonating structures can be written for the arenium ions resulting from ortho and para attack, whereas only three can be written for the arenium ion that results from meta attack. This, in itself, suggests that the ortho- and para-substituted arenium ions must be more stable. Of greater importance, however, are the relatively stable structures that contribute to the hybrid for the ortho- and para-substituted arenium ions. In these structures, nonbonding pairs of electron from nitrogen form an extra bond to the carbon of the ring. This extra bond and the fact that every atom in each of these structures has a complete outer octet of electrons - makes these structures the most stable of all of the contributors. Because these structures are unusually stable, they make a large - and stabilizing- contribution to the hybrid. This means, of course, that the ortho- and para-substituted arenium ions themselves are considerably more stable than the arenium ion, that results from the meta attack. In case of halogens inductive effect predominates over the mesomeric effect, due to which they act as ring deactivators instead of ring activators.

5.2 Meta-Directing Groups All meta-directing groups have either a partial positive charge or a full positive charge on the atom directly attached to the ring. As a typical example let us consider the trifluoromethyl group.

O-302 CF3

CF3 H E

E

Highly unstable contributor CF3

CF3

CF3 ⊕

⊕

Meta attack

E

⊕

⊕

CF3

H ⊕

Ortho attack

CF3 H

E CF3

E

CF3

H E

⊕

Para attack

⊕

⊕

CF3

H

⊕

H

H E

H E Highly unstable contributor

H E

From above it is clear that in the resonance structures for the arenium ion arising from ortho and para attack that one contributing structure is highly unstable relative to all the others because the positive charge is located on the ring carbon that bears the electron-withdrawing group. No such highly unstable resonating structure can be drawn in the arenium ion arising from meta attack. This means that the arenium ion formed by meta attack should be the most stable of the three.

5.3 Ortho - Para Direction and Reactivity of Alkylbenzenes Alkyl groups are ortho - para directors. One can also account for this property of alkyl groups on the basis of their ability to release electrons - an effect that is particularly important when the alkyl group is attached directly to a carbon that bears a positive charge.

O-303 CH3

CH3 H E

E

Relatively stable contributor CH3

CH3

CH3 ⊕

⊕

Meta attack

H

⊕

H

H E

E

E

CH3

CH3

H E

⊕

Para attack

⊕

⊕

CH3

E

⊕

⊕

CH3

H ⊕

Ortho attack

CH3 H

H E Relatively stable contributor

H E

In ortho attack and para attack it is observed that one can write resonating structures in which the methyl group is directly attached to a positively charged carbon of the ring. These structures are more stable relative to any of the others because in them the stabilizing influence of the methyl group (by electron release) is the most effective. These structures, therefore, make a large stabilizing contribution to the overall hybrid for ortho - and para - substituted arenium ions.

5.4 Summary of Substituent Effects on Orientation and Reactivity One can summarize the effects which the groups have on orientation and reactivity in the following way. Classification of substituents: Table 1:Effect of substituents on electrophilic aromatic substitution Ortho - Para Directors Strongly activating -NH2, -NHR, -NR2 ••

-OH, −O •• −

Meta Directors Strongly deactivating -NO2 − NR +3

••

-CF3, -CCl3 Moderately activating -NHCOCH3, -NHCOR

Moderately deactivating

O-304 -OCH3, -OR

-C ≡ N

Weakly activating

-SO3H

-CH3, C2H5, -R

-CO2H, -CO2R

C 6 H5

-CHO, -COR

Weakly deactivating -F, -Cl, -Br, -I

5.5 Synthetic Applications 5.5.1 Nuclear Halogenation or Halogenation of the Benzene Ring When halogenation of the alkyl or alkenyl benzenes is to be carried out in ring, a lewis acid catalyst is required as discussed in the topic of halogenations of benzene ring. Since alkyl and alkenyl groups are ortho and para directing, so the incoming electrophile - Cl + or Br + respectively will be oriented at either ortho or para position. C2H5

C2H5

C2H5 +

Cl2/FeCl3

Cl

Cl

(minor)

CH=CH2

(major)

CH=CH2

CH=CH2 +

Cl2/FeCl3

Br (minor)

Br (major)

For the addition of Br2 to double bond of the side chain, the reaction should be carried out in the presence of Br2 in CCl4.

5.5.2 Halogenation of the Side Chain It has already been observed that bromine and chlorine replace hydrogen atom on the ring of toluene when the reaction takes place in the presence of a Lewis acid. In ring halogenation, the electrophiles are positively charged chlorine or bromine ions or they are possibly Lewis acid complexes that have positive halogens. These positive electrophiles attack the π electrons of the benzene ring and aromatic substitution take place. Chlorine and bromine can also be made to replace hydrogen atoms of the methyl group of toluene or in general the alkyl group of the side chain. Such a chemical activity is referred as side chain halogenations. Side chain halogenation takes place when the reaction is carried out in the absence of Lewis acids and under conditions that favour the formation of radicals. When toluene reacts with N-bromosuccinimide (NBS) in the presence of light, the major product is benzyl bromide. N- Bromosuccinimide furnishes a low concentration

O-305 of Br2 and the reaction is analogous to that for allylic bromination. Here, it is referred as benzylic bromination. Mechanisms of such substitutions have already been discussed in the topic "alkenes". O

O

CH3 +

C

H2C

NBr

H2C

light

CH2Br +

CCl4

H2C

C O

H2C

C NH C O

Benzyl bromide

Side chain chlorination of toluene can also takes place in the gas phase at 400 - 600°C or in the presence of UV light. When an excess of chlorine is used, multiple chlorinations of the side chain occur. CH3

CH2Cl

CHCl2

Cl2 heat / light

Cl2 heat / light

Benzyl chloride

CCl3

Cl2 heat / light

(Dichloromethyl)– benzene

(Trichloromethyl)– benzene

These halogenations take place through the same radical mechanism as for halogenation of alkanes. The halogens dissociate to produce halogen atoms and then the halogen atoms initiate chains by abstracting hydrogen atoms of the methyl group.

Mechanism Expectedly the reaction mechanism has following steps. Chain initiation step: Step 1

Peroxides,heat

X2   → 2X • or light

Chain propagation step: Step 2 C6H5CH3 + X • → C6H5CH2 • + HX Benzyl radical Step 3 C6H5CH2 • + X2 → C6H5CH2X + X • Benzyl halide Abstraction of hydrogen from the methyl group of toluene produces a benzyl radical. The benzyl radical then reacts with a halogen molecule to produce a benzyl halide and a halogen atom. The halogen atom then brings about a repetition of step 2, then step 3 occurs again and so on. Benzylic halogenations are similar to allylic halogenations in that they involve the formation of unusually stable radicals. Benzylic and allylic radicals are more stable than tertiary radicals.

O-306 The greater stability of benzylic radicals accounts for the fact that when ethyl benzene is halogenated the major product is the 1-halo-1-phenylethane as benzylic radical is formed faster than the 1° radical. X fast

CH2CH3

X2

CHCH3+X

Benzylic radical 1–halo–1–phenyletane (more stable) (major product)

–CH2CH3 X

(–HX)

CH2CH2

X2

CH2CH2X+X

slow 1° radical (less stable)

1–halo–1–phenyletane (major product)

Example 4: (i)

Give the product of the reaction of p–xylene with NBS or with 2Br2 at 125°C in presence of light.

(ii)

Why does not dibromination at the one Me occur ?

(iii) Which reacts faster, p–xylene or toluene? Explain. Solution: (i)

CH2Br

CH2Br α–α'–dibromo–p–xylene

(ii)

Electron withdrawal by one Br makes removal of an additional H • more difficult and thus the second substitution occurs at other Me group.

(iii)

p–xylene reacts faster than toluene as the electron–releasing Me provides additional stability to the radical.

5.5.3. Addition to the Double Bond of Alkenyl Benzenes In the presence of peroxides, hydrogen bromide adds to the double bond of 1-phenyl propene to give 2-bromo-1-phenyl propane as the major product. HBr

CH=CHCH3 peroxides

CH2CHCH3 Br

In the absence of peroxides, HBr adds in just the opposite way.

O-307

HBr

CH=CHCH3 (no peroxides)

.

CHCH2CH3 Br

The addition of hydrogen bromide to 1–phenylpropene proceeds through a benzylic radical in the presence of peroxides, and through a benzylic cation in their absence.

5.5.4 Oxidation of the Side Chain Strong oxidizing agents oxidize toluene to benzoic acid. The oxidation can be carried out by the action of hot alkaline potassium permanganate. This method gives benzoic acid in almost quantitative yield. (1) KMnO ,OH–

4 → PhCO H PhCH3      2

heat

Benzoic acid

An important characteristic of side chain oxidations is that oxidation takes place initially at the benzylic carbon. Alkyl benzenes with alkyl groups longer than methyl are ultimately degraded to benzoic acids. (i) KMnO ,OH– / heat

4  PhCH2CH2CH2R      → PhCO2H

(ii) H3 O+

An alkylbenzene

Benzoic acid

Side chain oxidations are similar to benzylic halogenations, because in the first step the oxidizing agent abstracts a benzylic hydrogen. Once oxidation has started at the benzylic carbon, it continues at that site. Ultimately, the oxidising agent oxidizes the benzylic carbon to a carboxyl group and in the process; it cleaves off the remaining carbon atoms of the side chain. In light of the proposed mechanism, tert butyl benzene must be resistant to such kind of oxidation activity as it does not contain a benzylic hydrogen and this has been found to be practically true also. Side chain oxidation is not only restricted to alkyl groups but other groups in the side chain can also be degraded to benzoic acid. Alkenyl, alkynyl and acyl groups are oxidized by hot alkaline potassium permanganate in the same way. C6H5CH=CHCH3 O

or C6H5C≡CCH3 O

or

KMnO4, OH–, heat H3O+

C6H5COH

C6H5CCH2CH3

Some more applications are discussed below. 1.

The substitution reaction of aromatic rings and the reactions of the side chains of alkyl and alkenyl benzenes, when taken together, offer us a powerful set of reactions for organic synthesis. By using these reactions skillfully, one will be able to synthesize a large number of benzene derivatives. Part of the skill in planning a synthesis is in deciding the order in which reactions should be carried out. For example, let us suppose that one wishes to synthesize o-bromonitrobenzene. One can see very quickly that the bromine has to inserted into the ring first as it is an ortho-para director.

O-308 Br

Br

Br NO2

Br2

HNO3 H2SO4

FeBr3

+

NO2

The ortho and para compounds obtained as products can be separated by various methods. However, if one had introduced the nitro group first, one would have obtained m–bromonitrobenzene as the major product. 2.

Other examples in which choosing the proper order for the reactions are important are the syntheses of the ortho, meta and para-nitrobenzoic acids. One can synthesize the ortho and para-nitrobenzoic acids from toluene by nitrating it, then separating the ortho and para-nitrotoluenes, and then oxidizing the methyl groups to carboxyl groups. CH3 CO2H KMnO4, OH–,heat H3O+

CH3 HNO3 H2SO4

NO2 p–nitrotoluene

NO2 p–nitrobenzoic acid

CH3

CO2H

NO2

NO2 KMnO4, OH–,heat H3O+

o–nitrobenzoic acid

o–nitrotoluene

One can synthesize m-nitrobenzoic acid by reversing the order of the reactions. CH3

CO2H (i) KMnO4,OH–,heat (ii)

H3O+

CO2H

HNO3 H2SO4,heat NO2 Benzoic acid

m–nitrobenzoic acid

Suppose one needs to synthesize m-chloroethylbenzene from benzene. Cl

CH2CH3

O-309 One must begin by chlorinating benzene and then follow with a Friedel–Crafts alkylation using CH3CH2Cl and AlCl3, or one could begin with a Freidel Crafts alkylation followed by chlorination. However, neither method will give the desired product. The product obtained in both the cases is a mixture of ortho and para chloro ethyl benzene. Actually, there is a three-step method that will work if the steps are done in the right order. First of all, one will carry out Friedel Crafts acylation using CH3COCl + AlCl3 on benzene. This is followed by chlorination reaction with Cl2 in presence of FeCl3, in which Cl + will enter at meta position. Finally, the product is reduced by Zn–Hg and conc. HCl (Clemmenson reduction) to give the desired product. COCH3 CH3COCl AlCl3

COCH3

Cl2/ FeCl3

CH2CH3 Zn–Hg/HCl

Cl

Cl

Very powerful activating groups such as amino groups and hydroxyl groups cause the benzene ring to be so reactive that undesirable reaction may take place. Some reagents used for electrophilic substitution reactions, such as nitric acid, are also strong oxidizing agents. Both electrophiles and oxidizing agents seek electrons. Thus, amino groups and hydroxyl groups not only activate the ring towards electrophilic substitution, they also activate it towards oxidation. For example, nitration of aniline results in considerable destruction of the benzene ring because it is oxidized by the nitric acid. Consequently, direct nitration of aniline is not a satisfactory method for the preparation of o- and p-nitroaniline. Treating aniline with acetyl chloride, CH3COCl, or acetic anhydride, (CH3CO)2O, converts aniline to acetanilide. The amino group is converted to an acetamido group (NHCOCH3), a group that is only moderately activating and one that does not make the ring highly susceptible to oxidation. With acetanilide, direct nitration becomes possible. NH2

NHCOCH3

NHCOCH3

NHCOCH3

NO2 CH3COCl Base

Aniline

HNO3 H2SO4

Acetanilide

NO2 p–nitroacetanilide

+

o–nitroacetanilide (trace) (1) H+,H2O (2) OH–

NH2 O This step removes the CH3CO group and replaces it with an –H

+CH3COH

NO2 p–nitroaniline

O-310 Nitration of acetanilide gives p-nitroacetanilide in excellent yield with only a trace of the ortho isomer. Acidic hydrolysis gives p-nitroaniline, also in good yield. However, if one needs ortho-nitroaniline, then the synthesis pathway discussed right now would not be a satisfactory method as only a trace of o-nitroacetanilide is obtained. It is because the acetamido group is purely a para director due to steric hindrance factor. However, one can synthesize o-nitroacetanilide through the reaction sequence below, NHCOCH3

NHCOCH3

NO2 HNO3

conc. H2SO4

Acetanilide

SO3H

NH2

NHCOCH3

NO2

H2O

H2SO4, distill,OH–

SO3H

o–nitroaniline

Here it can be observed that how a sulfonic acid group can be used as a "blocking group". One can remove the sulfonic acid group by desulfonation at a later stage. In this example, the reagent used for desulfonation (dilute H2SO4) also conveniently removes the acetyl group that is employed to "protect" the benzene ring from oxidation by nitric acid.

5.6 Ortho - Para Ratio An aromatic substitution reaction of a benzene derivative bearing an ortho, para-directing group would give twice as much ortho as para product if substitution were completely random, because there are two ortho positions and only one para position available for substitution. However, this situation is rarely observed in practice. It is often found that the para substitution product is the major one in the reaction mixture. In some cases this result can be explained by the spatial demands of the electrophile. For example, Friedel-Crafts acylation of toluene gives essentially all para substitution product and almost no ortho product. The electrophile cannot react at the ortho position without developing van der Waals repulsions with the methyl group that is already on the ring. Consequently, reaction occurs at the para position, where such repulsions cannot occur. Typically, para substitution predominates over ortho substitution, but not always. For example, nitration of toluene gives twice as much o-nitrotoluene as p-nitrotoluene. This result occurs because the nitration of toluene at either the ortho or para position is so fast that it occurs on every encounter of the reagents; that is, the energy barrier for the reaction is insignificant. Hence, the product distribution corresponds simply to the relative probability of the reactions. Because the ratio of ortho and para positions is 2 :1, the product distribution is 2 :1. In fact, the ready availability of o-nitrotoluene makes it is a good starting material for certain other ortho-substituted benzene derivatives. In summary, the reasons for the ortho, para ratio vary from case to case, and in some cases these reasons are not well understood. Whatever the reasons for the ortho - para ratio, if an electrophilic aromatic substitution reaction yields a mixture of ortho and para isomers, a problem of isomer separation arises that must be solved if the reaction is to be useful. Usually, syntheses that give mixtures of isomers are avoided because, in many cases, isomers are difficult to separate. However, the ortho and para isomers obtained in many electrophilic aromatic substitution reactions have sufficiently different physical properties that they are readily separated. For example, the boiling points of o- and p-nitrotoluene, 493 K and 510 K, respectively, are sufficiently different that these isomers can be separated by careful fractional distillation. Thus, either isomer can be obtained relatively pure° from the nitration of toluene. The melting points of o- and p-chloronitrobenzene, 34°C and

O-311 84°C, respectively, are so different that the para isomer can be selectively crystallized. Generally,the para isomer of an ortho, para pair typically has the higher melting point, often considerably higher. Most aromatic substitution reactions are so simple and inexpensive to run that when the separation of isomeric products is not difficult, these reactions are useful for organic synthesis despite the product mixtures obtained. Example 5: Give structure of the principal organic products expected from mononitration of (i) o-nitrotoluene (ii) m-dibromobenzene (iii) p-nitroacetanilide (iv) m-dinitrobenzene (v) m-cresol (vi) o-cresol (vii) p-cresol (viii)m-nitrotoluene (xi) p-xylene (x) terephthalic acid (p-C6H4(COOH)2) (xi) anilinium hydrogen sulphate (C6H5NH +3 HSO4–) Solution : (i)

CH3

(ii)

Br

NO2

Br NO2

(iii)

NO2

NHCOCH3

NO2

(iv)

NO2

O2N

NO2

NO2

(v)

OH

(vi)

OH CH3

CH3 NO2

(vii)

NO2

OH

OH

(viii) NO2

O2N

NO2 CH3

O-312 CH3

(ix)

CO2H

(x) NO2

NO2 CH3

CO2H

– NH+ 3 HSO4

(xi)

NO2

6. Fused or Condensed Aromatic Hydrocarbons Fused aromatic hydrocarbons contain more than one ring and have two carbons shared by two or more aromatic rings. The most important members of this class are naphthalene, anthracene,phenanthrene, durene etc. Some of these are shown below,

• •

•

• • •

• • •

•

•

•

•

•

•

•

•

Naphthalene

•

•

•

•

• • •

•

•

•

•

• •

• • •

•

•

•

•

•

•

•

•

Anthracene

• •

• •

• • •

• •

•

• •

•

•

•

•

•

• •

•

•

•

•

•

•

Phenanthrene

Naphthalene has two carbon atoms shared by two rings. Anthracene and phenanthrene have two pairs of carbon atoms shared by two rings, each pair being shared by a separate pair of rings.

6.1 Naphthalene Naphthalene is the largest constituent of coal-tar. It is obtained industrially by chilling the middle oil fraction (bp 170-230°C), when crude naphthalene crystallizes out. It is purified by washing it successively with dilute sulfuric acid so that the basic impurities can be removed, sodium hydroxide to remove acidic impurities, and water. Finally the solid is sublimed or steam-distilled to yield pure naphthalene. Naphthalene as moth balls has been used to protect woolen goods from moths for many years. Recently, p-dichlorobenzene has replaced naphthalene in the manufacture of moth balls, as it has a less obnoxious odor. It is used for increasing the illuminating power of coal gas. Naphthalene is used in the manufacture of phthalic anhydride, carbaryl for insecticides, 2-naphtol for dyes and several medicinal products.

O-313

6.2 Nomenclature of Naphthalene Derivatives In naming derivatives of naphthalene, the numbering system shown below is used. The numbers selected to denote the position of a substituent on the naphthalene rings should be as small as possible. CH3

Br 8

1

7

2 3

6 5

H3C

4

1–Bromonaphthalene

Naphthalene

1,6–Dimethylnaphthalene

An alternative nomenclature system is occasionally used when there is only one substituent on the naphthalene skeleton. This system uses the Greek letters α and β to designate the two possible orientations of the single substituent. NO2 α

α

β

β

β

β α

Br

α

β–Bromonapthalene

α–Nitronapthalene

Naphthalene

6.3 Structure of Naphthalene All the ten carbon atoms in naphthalene are sp2 hybridized. The sp2 hybrid orbitals overlap with each other and with s orbitals of the eight hydrogen atoms forming C - C and C - H σ bonds (figure 11). Since the σ bonds result from the overlap of trigonal sp2 orbitals, all carbon and hydrogen atoms in naphthalene lie in one plane. This has been confirmed by X-ray diffraction studies. sp2–sp2 C–C σ bond H

C C

H

H sp2–sp2 C–C σ bond C

H

H

C

C

H

H 120°

≡ H C

C

C H

H

C H

C

H

sp2–sp2 C–C σ bond

Fig. 11: Molecular orbital diagram of naphthalene

H H

H

H

O-314 Also each carbon atom in naphthalene possesses an unhybridized p orbital containing one electron. These p orbitals are perpendicular to the plane containing the σ bonds. The lateral overlap of the p orbitals produces a π molecular orbital containing ten electrons as shown in figure 12. One half of this π molecular orbitals lies above and the other half lies below the plane of the σ bonds. Naphthalene shows aromatic properties because the resulting π molecular orbitals satisfies the Huckel's rule and for naphthalene n = 2 in (4n + 2) π electron rule. π–molecular orbital

p–orbitals

p–p overlap

Fig. 12: π overlap in naphthalene

A common shorthand representation of naphthalene is simply two fused hexagons with a circle inside each hexagon. The circle represent the π molecular orbital. π–molecular orbital

According to the resonance theory, naphthalene is considered to be a hybrid of the following three forms. 8

1

7

2

6

3 5

4

(A)

(B)

(C)

X-Ray diffraction studies show that, unlike benzene, all carbon-carbon bonds in naphthalene are not of the same length. In particular, the C1 − C2 bond is considerably shorter (1.36 A) than the C1 − C2 bond (1.40 A°). This difference can be understood if one examines the three resonance forms given above. It is important to note that the C1 − C2 bond is double in two structures (A and B) and single in only one (C); whereas the C1 − C2 bond is single in two structures (A and B) and double in only one (C). One would, therefore, expect the C1 − C2 bond to have more double-bond character (shorter bond length), and the C2 − C3 bond to have more single-bond character (longer bond length). The resonance energy of naphthalene is about 61 kcal/mole. This value is less twice the amount of a single benzene ring (36 kcal/mole). As a result, naphthalene is somewhat less aromatic (more reactive) than benzene.

O-315

6.4 Synthesis of Naphthalene Naphthalene may be obtained: 1.

From petroleum: When petroleum fractions are passed over copper catalyst at 680°C, naphthalene and methylnaphthalenes are formed. Methylnaphthalenes are separated and converted into naphthalene by heating with hydrogen under pressure (dealkylation). CH3 Cu Petroleum Fractions 680°C

+ Naphthalene

Methylnaphthalenes

H2, ∆, pressure (–CH4)

Prior to 1960 all naphthalene came from coal-tar, but now almost half of it is produced from petroleum by the above method. 2.

Haworth synthesis: This involves the following steps :

Step 1: Benzene and succinic anhydride are heated in the presence of aluminium chloride to form β-benzoylpropionic acid. (Friedel-Crafts Acylation). O

O H2C

+

H2C

C

C O C

HO–C

O Succinic anhydride

Benzene

CH2

AlCl3

CH2

O β–Benzoylpropionic acid

Step 2: β– Benzoylpropionic acid is treated with amalgamated zinc in the presence of hydrochloric acid to give γ–phenylbutyric acid. (Clemmensen Reduction). O

H2 C

C CH2 HO–C

CH2

O β–benzoylpropionic acid

CH2

Zn(Hg) HCl

HO–C

CH2

O γ–phenylbutyric acid

Step 3: γ-Phenylbutyric acid is heated with concentrated sulfuric acid or polyphosphoric acid form α-tetralone. (Ring closure Reaction).

O-316 H2 C CH2 H HO–C

CH2

H2SO4

+ H2O

∆

O

O γ–phenylbutyric acid

Tetralone

Step 4: α-tetralone is heated with amalgamated zinc and hydrochloric acid to give tetralin. (Clemmensen Reduction). Zn(Hg) HCl

O α–tetralone

Tetralin

Step 5: tetralin is heated with palladium to yield naphthalene. (Aromatization reaction).

Pd

+ 2H2

Naphthalene

Tetralin

Substituted naphthalenes can also be obtained by using toluene, bromobenzene, or anisole instead of benzene in Step 1. 3.From 4-phenyl-3-butenoic acid: When 4-phenyl-3-butenoic acid is heated with concentrated sulfuric acid, 1-naphthol is produced. 1-naphthol on distillation with zinc gives naphthalene. H C CH H HO–C

CH2

H2SO4 (–H2O)

O 4–Phenyl–3–butenoic acid

+ H2O

Rearrangement

H O

OH

(Unstable)

1–Naphthol

Zn ∆

+

Naphthalene

ZnO

O-317 4.From 4-Phenyl-1-butene: When 4-phenyl-1-butene is passed over red hot calcium oxide, naphthalene is produced. H2 C CH2

CaO ∆

+ H2

CH H2C

Naphthalene

4–Phenyl–1–Butene

Physical properties of naphthalene Naphthalene is a colorless crystalline solid. It melts at 82°C and boils at 218°C. Naphthalene is insoluble in water, but dissolves in ether, benzene, and hot ethanol due to its non polar nature. It sublimes readily when warmed and is volatile with steam. Naphthalene has a characteristics moth ball odor. Figure show UV and 1H NMR spectra. 0.9 0.8

5000

Absorbance

0.7 0.6

4000

0.5 0.4

3000

0.3

2000

0.2 1000

0.1

230 240 250 260 270 280 290 300 310 320 330 340 350 360 nm Fig.13: Ultraviolet spectrum of naphthalene shows a pattern typical of an extended conjugated system with peaks at wavelengths as lon g as 320 nm

Intensity Intensity

H

10

7.6 7.6

7.9 ppm 7.9 7.87.8 PPM

4H

9

8

7.5 7.4 7.4ppm PPM

H

H

H

H

H H H

4H

H TMS TMS

7

6

4 3 5 Chemical shift (δ)

2

1

0

ppm

Fig. 14: HNMR spectrum of naphthalene, two symmetric multiplets can be observed at δ=7.49 and 7.86 ppm, charcteristic of aromatic hydrogens deshielded by the ring cur rent effect of the delocalized π electrons

O-318

Chemical properties of naphthalene Some of the important chemical reactions of naphthalene are described below : 1.

Electrophilic substitution reactions: Naphthalene, like benzene, undergoes electrophilic substitution reactions. Substitution occurs primarily at C-1 (α-position). This can be understood if we examine the intermediate carbonium ion. Two resonance forms can be written for the intermediate carbonium ion obtained from the attack at C-1 (without involving the other ring), whereas only one such form is possible for substitution at C-2. E+ in the following equations represents an electrophile.

Attack at C-1 H

1

H

E +

2

–H+

+E+ +

Naphthalene

E

E

More stable

1–Substitution product

Attack at C-2 1

+

2

H E

+E+

E –H+

Less stable

Naphthalene

2–Substitution product

Consequently the former intermediate is more stable and the product with a substituent at C–1 predominates. Substitution at C-2 (β-position) occurs only when the reactions are carried at higher temperatures or when bulkier solvents are used. (i) Sulfonation: Naphthalene undergoes sulfonation with concentrated sulfuric acid. If the reaction is carried at 165°C, 2-naphthalenesulfonic acid is obtained. SO3H SO3H H2SO4 60°C

H2SO4 60°C

1–naphthalene– sulfonic acid

Naphthalene

2–naphthalene– sulfonic acid

(ii) Nitration: Naphthalene undergoes nitration with concentrated nitric acid in the presence of sulfuric acid at 60°C to produce 1-nitronaphthalene.

O-319 • •

NO2

• •

• • •

• •

H2SO4 60°C

+HNO3

• •

•

•

• •

• •

1–nitronaphthalene

Naphthalene

• •

•

Mechanism The mechanism of this reaction involves the following three steps: Step 1: The electrophile NO 2+ is generated. 2H2SO 4 + HONO 2 → NO 2+ + 2HSO 4− + H3O + Step 2: The electrophile attacks naphthalene at C-1 to form the resonance-stabilized carbonium ion.Here,only those resonance forms are shown below which do not involve the other ring. H

1 2

H

NO2

NO2

+

+

+ NO2 +

Step 3: A hydrogen ion is eliminated to yield 1-nitronaphthalene. –

–

HSO4 H

HSO4 H

NO2

NO2

NO2

+

+ +

H2SO4

1–nitronapthalene

Similar steps may be written for other electrophilic substitution reactions of naphthalene. (iii) Halogenation: Naphthalene undergoes chlorination or bormination in boiling carbon tetrachloride to give 1-chloronaphthalene. Unlike benzene, no Lewis acid catalyst is required. Cl + Cl2 Naphthalene

CCl4 ∆

+ HCl 1–chloronaphthalene

When this reaction is carried at room temperature, naphthalene dichloride (1,4-addition product) is obtained. On heating, naphthalene dichloride loses HCl to yield 1-chloronaphthalene.

O-320 Cl

H + Cl2

Cl

CCl4 Room temperature

–HCl Cl

H Naphthalene

Naphthalene dichloride 1–chloronaphthalene

(iv) Friedel-Crafts acylation: Naphthalene undergoes acylation with acetyl chloride and AlCl3 in carbon disulfide to give 1-acetylnaphthalene. When nitrobenzene is used as a solvent, 2-acetylnaphthalene is obtained. CH3

O C

O

O

O

2–acetylnaphthalene

naphthalene

1–acetylnaphthalene

C–CH3

CH3–C–Cl, AlCl3 in C6H5NO2

CH3–C–Cl, AlCl3 in CS2

(v) Friedel-Crafts alkylation: Naphtahelene undergoes alkylation with alkyl halides in the presence of AlCl3 to give 2-alkylnaphthalene. + CH3CH2Cl

CH2CH3 + HCl

AlCl3

Naphthalene

2–ethylnaphthalene

(vi) Chloromethylation: Naphthalene reacts with formaldehyde and HCl in the presence of zinc chloride to form 1-chloromethylnaphthalene. CH2Cl

O + H—C—H + HCl Naphthalene

Formaldehyde

ZnCl2

1–chloromethylnaphthalene

6.5 Other Reactions 1.Oxidation: Naphthalene is much more easily oxidized than benzene. With chromium trioxide in acetic acid at room temperature, it gives 1,4-naphthaquinone.

O-321 O CrO3 CH3COOH

O 1,4–naphthaquinone

Naphthalene

Oxidation of naphthalene with oxygen and vanadium pentoxide at 475°C yields phthalic anhydride. This method is used industrially. O C Air, V2O5 475°C

O

COOH

H2O/H+

COOH

C O Phthalic anhydride

Naphthalene

Phthalic acid

Phthalic acid is an important industrial material. It is used in the manufacture of resins paints, dyes and plastics. 2.Reduction: Naphthalene undergoes reduction more readily than benzene. With sodium and ethyl alcohol (bp 78°C) it gives 1,4-dialin or 1,4-dihydronaphthalene. With sodium and isopentanol (bp 130°C) it gives tetralin or 1,2,3,4-tetrahydronaphthalene. Na/C5H11OH Boil

Na/C2H5OH Boil

Naphthalene

Tetralin

1,4–dialin

Catalytic reduction completely hydrogenates both rings and produces decalin or decahydronaphthalene. 8

+ 5H2 Ni or Pt ∆, Pressure

7 6 5

Naphthalene

9

10

1 2 3 4

Decalin

Decalin consists of two cyclohexane molecules (in chair forms) used together along a common side. It exists in cis and trans isomeric forms which differ in the space relationship of hydrogen atoms at C-9 and C-10 positions. When nickel is used as a catalyst, the main product is trans-decalin. When platinum is used as a catalyst, the main product is cis-decalin. Decalin sold commercially is the mixture of both these forms, and is used as a solvent in the varnish and lacquer trade.

O-322 H

H

H H trans–decalin (bp 185°C)

cis–decalin (bp 195°C)

7. Anthracene Anthracene occurs in coal-tar and is obtained from the green oil fraction (boiling point 270-360°C). On cooling this fraction crude anthracene crystallizes out. Crude anthracene contains phenanthrene and carbazole as impurities. It is purified by washing it successively with solvent naphtha in order to remove phenanthrene and pyridine to remove phenanthrene and pyridine to remove carbazole. Finally the solid is sublimed to give pure anthracene. In naming derivatives of anthracene, the numbering system shown below is used. The numbers selected to denote the position of a substitutent on the anthracene rings should be as small as possible. Anthracene is used in the manufacture of anthraquinone. SO3H

Br 8

9

1

7

2 3

6 5

10

Anthracene

4

9–bromoanthracene

1–anthracenesulfonic acid

7.1 Structure of Anthracene All 14 carbon atoms in anthracene are sp2 hybridized, so it is a planar molecule. The sp2 hybrid orbitals overlap with each other and with s orbitals of the ten hydrogen atoms forming C - H C and C - H σ bonds. Since the bonds result from the overlap of trigonal sp2 orbitals, all carbon and hydrogen atoms in H anthracene lie in the same plane (figure 13). This has been confirmed by X-ray diffraction studies.

H

H

H H H

H

H

H

σ bonding framework in anthracene

Also each carbon atom in anthracene possesses an unhybridized p orbitals containing one electron. These p orbitals are perpendicular to the plane of the σ bonds. The lateral overlap of these p orbitals produces a π molecular orbital containing fourteen electrons (figure 15). One half of this π molecular orbital lies above and the other half lies below the plane of the σ bonds. Anthracene shows aromatic properties because the resulting molecular orbital satisfies the Huckel's rule as n = 3 in (4n + 2) π electron.

O-323 π molecular or bital

Fig. 15: π molecular orbital diagram of anthracene

A common shorthand representation of anthracene is simply three linearly fused hexagons with a circle inside each hexagon. The circles represent the molecular orbital. π molecular orbital

According to the resonance theory, anthracene is considered to be a hybrid of the following canonical forms,

X-ray diffraction studies show that, like naphthalene, all carbon-carbon bonds in anthracene are not of the same length. In particular, the C1 − C2 bond is considerably shorter (1.37 A°) than the C2 − C3 bond. (1.42 A°). This difference in bond lengths can be understood if we examine the four resonance forms given above. Notice that the C1 − C2 bond is double in three structures (A, B and C) and single in only one (D); whereas the C2 − C3 bond is single in three structures (A, B and C), and double in only one (D). We would, therefore, expect the C1 − C2 bond to have more double-bond character (shorter bond length), and the C2 − C3 bond to have more single-bond character characterized by longer bond length.The resonance energy of anthracene is 84 kcal/mole. This averages to 28 kcal/mole per ring, which is substantially lower than that of benzene (36 kcal/mole). As a result, anthracene is much less aromatic than benzene and behaves more like an unsaturated aliphatic hydrocarbon.

7.2 Synthesis of Anthracene Anthracene is obtained with the help of a number of reaction. Some of these reactions are discussed below, 1.

By Haworth synthesis: This involves the treatment of benzene with phthalic anhydride in the presence of aluminium chloride to form o-benzoylbenzoic acid. This is then heated with concentrated sulfuric acid to give 9, 10-anthraquinone. Distillation of the anthraquinone with zinc dust yields anthracene.

O-324 O

O

C

C AlCl3

O+

H C

C O Phthalic anhydride

H2SO4 (–H2O)

OH O o–benzoylpropionic acid

Benzene

O Zn Distil

O 9,10–anthraquinone

2.

Anthracene

By Friedel-Crafts reaction: Benzyl chloride reacts with itself to form 9,10-dihydroantheracene, which readily loses two hydrogen atoms to yield anthracene. H H CH2–Cl H + H Cl–CH2

AlCl3 (–2HCl)

–2H

H Benzeyl chloride (2 molecules)

H

9,10–dihydroanthracene

Anthracene

Anthracene may also be prepared by the Friedel-Crafts reaction between benzene and 1,1,2,2-tetrabromoethane (acetylene tetrabromide), or between benzene and dibromomethane. H AlCl3 Br–C–Br + + (–4HBr) Br–C–Br H Benzene Benzene Acetylene tetrabromide H Br Br

C +

H H

H Br

Br C H H Dibromomethane

Anthracene

–2H

AlCl3 (–4HBr)

H

H

9,10–dihydroanthracene

Anthracene

O-325 3.

By Elbs reaction: The conversion of a diaryl ketone containing a methyl or methylene group ortho to the carbonyl function is known as the Elbs Reaction. For example, when o-methylbenzophenone is heated at 450°C, anthracene is formed. O C 450°C

+ H2O

CH3 o–methylbenzophenone

4.

Anthracene

By Diels-Alder reaction: This involves the reaction of 1,4-naphthaquinone with 1,3-butadiene. The product of this reaction is oxidized with chromium trioxide in glacial acetic acied to form 9,10-anthraquinone. Distillation of anthraquinone with zinc dust yields anthracene. O

H2 C CH + CH C H2

O H CrO3 CH3COOH

O H

O

1,4–naphthaquinone 1,3–butadiene

Diels–Alder adduct

O Zn Distil

O

9,10–Anthraquinone

Anthracene

Physical properties of anthracene Anthracene is a colorless solid. It melts at 218°C and boils at 340°C. Anthracene is insoluble in water, but dissolves in benzene. It shows a strong blue fluorsence when exposed to ultraviolet light. The fluorescent property of anthracene on clothing, skin, money, etc., is not detected under ordinary light but easily noticed when exposed to ultraviolet light.

Chemical properties of anthracene Anthracene undergoes addition and electrophilic substitution reactions. These reactions preferentially occur at the C-9 and C-10 positions. This can be understood if one examines the intermediate carbonium ions obtained from attack at C-1, C-2 and C-9 (all other positions at equivalent to either 1 or 2 or 9 by symmetry). E+ in the following equations represent an electrophile.

O-326 Attact at C-1 8

9

H

1

7

2

6

3 5

10

E

H

E +

+ E+

+

4

Anthracene

Attact at C-2 +

+ E+

H E

Anthracene

Attact at C-9 H

E

+ E+ +

Anthracene

Attack at C-9 yields a carbonium ion intermediate in which two benzene rings are retained; whereas attack at C-1 or C-2 yields an intermediate in which a naphthalene system is retained. The former intermediate is more stable and its formation favored because the resonance energy of two benzene rings (2 × 36 = 72 kcal) exceeds that of naphthalene (61 kcal). This carbonium ion intermediate can lose a proton to give the corresponding substitution product, or it can react with a nucleophile to form the 9,10-addition product. Nu : in the following equations represents a nucleophile. E

E

H

+

–H H + H

Substitution product H

E

E

+ Nu:– H

+

Nu H Addition product

Substitution at C-1 or C-2 occurs only when the reaction is reversible as in the case of sulfonation. The main chemical reactions of anthracene are described below: 1.

Diels-Alder reaction: Anthracene undergoes a Diels-Alder reaction with maleic anhydride to yield the corresponding adduct.

O-327 O

+

HC HC

Anthracene

C O C

O

O Maleic anhydride

HC HC

C O C

O Diels–Alder adduct

2.

Reaction with halogens: Anthracene reacts with chlorine in carbon tetrachloride at room temperature to give 9,10-dichloro-9,10-dihydroanthracene (an addition product). On heating, this addition product loses a molecule of hydrogen chloride by 1,4-elimiantion to form 9-chloroanthracene (a substitution product). H + Cl2

Cl

Cl

In CCl4

∆ (–HCl)

H Anthracene

Cl 9–chloroanthracene

9, 10–chloro–9,10– dihydroanthracene

9-Chloroanthracene may also be obtained by chlorinating anthracene at 100°C, or by treating anthracene with cupric chloride in CCl4. Bromine reacts similarly. 3.

Friedel-Crafts acylation: Anthracene undergoes acylation with acetyl chloride and aluminium chloride to form 9-acetylanthracene. H3C

O C

O + CH3–C–Cl Anthracene

4.

AlCl3

9–acetylanthracene

Nitration: Anthracene undergoes nitration with concentrated nitric acid in acetic anhydride at room temperature to yield a mixture of 9-nitroanthracene and 9,10-dinitroanthracene. The usual nitrating mixture (HNO3 + H2SO4) is not used because it leads to the formation of 9,10-anthraquinone by oxidation.

O-328 NO2

NO2 (CH3CO)2O ∆

+ HNO3

NO2 9,10–dinitroanthracene

9–nitroanthracene

Anthracene

5.

+

Sulfonation: Anthracene undergoes sulfonation with concentrated sulfuric acid to yield a mixture of 1-anthracenesulfonic acid and 2-anthracenesulfonic acid. At lower temperatures 1-anthracenesulfonic acid is the major product; whereas at higher tempertures 2-anthracenesulfonic acid is the major product. SO3H H2SO4

6.

+

1–anthracenesulfonic acid

Anthracene

SO3H

2–anthracenesulforic acid

Oxidation: Anthracene undergoes oxidation with sodium dichromate and sulfuric acid to form 9,10-anthraquinone. Other oxidizing agents like nitric acid and air in the presence of V2O5 also lead to the formation of 9,10-anthraquinone. O •

Na2Cr2O7 H2SO4

Anthracene

7.

•

• •

• •

• •

O Anthraquinone (9, 10–anthraquinone)

• •

• • • •

• •

• • • •

• •

• •

Reduction: Anthracene undergoes reduction with sodium and ethyl alcohol to form 9,10-dihydroanthracene. H

H

Na/C2H5OH ∆

Anthracene

H H 9,10–dihydroanthracene

Catalytic reduction using nickel at 225°C first gives 9,10-hydroanthracene, and on continued hydrogenation this is converted into 1,2,3,4-tetrahydroanthracene and 1,2,3,4,5,6,7,8-octahydroanthracene. Notice that the 9,10-hydrogen atoms migrate to the neighboring ring.

9,10–dihydroanthracene

1,2,3,4–tetrahydroanthracene

1,2,3,4,5,6,7,8–Octa– hydroanthracene

O-329 Reaction with Sodium: Anthracene reacts with metallic sodium in liquid ammonia to form a deep blue 9,10-disodioanthracene. H + 2Na

– ••

Liq. NH3

+

2Na H

– ••

8.

9,10–dihydroanthracene (Deep blue)

Anthracene

When the disodio-derivative is heated with an alkyl halide, it gives the corresponding 9,10-dialkylanthracene. H

CH2CH3 – •• +

2Na + CH3CH2Cl – ••

H

∆

CH2CH3 9,10–disodioanthracene

9,10–diethylanthracene

8. Biphenyls Biphenyl or phenylbenzene or 1,1'-biphenyl is an organic compound that forms colorless crystals. It has a distinctively pleasant smell. Biphenyl is an aromatic hydrocarbon with a molecular formula (C6H5)2. It is notable as a starting material for the production of polychlorinated biphenyls (PCBs), which were once widely used as dielectric fluids and heat transfer agents. Biphenyl is also an intermediate for the production of a host of other organic compounds such as emulsifiers, optical brighteners and plastics. Biphenyl is insoluble in water, but soluble in typical organic solvents. The biphenyl molecule consists of two connected phenyl rings. Biphenyl occurs naturally in coal tar, crude oil, and natural gas and can be isolated from these sources via distillation.

8.1 Methods of Preparation of Biphenyl Some important methods of preparation of biphenyls are discussed below, Biphenyl is produced industrially as a byproduct of the dealkylation of toluene to produce benzene: 1. C6H5CH3 + C6H6 → C6H5-C6H5 + CH4 The other principal route is by the oxidative dehydrogenation of benzene: 2. 2 C6H6 + 1/2 O2 → C6H5-C6H5 + H2O 3.

Fittig reaction: It is an extension of Wurtz reaction and consists of heating an ethereal solution of bromobenzene with metallic sodium.

O-330

–Br+2Na+Br–

4.

+2NaBr

Ulmann biaryl synthesis: Iodobenzene on heating with copper in a sealed tube forms biphenyl. –I+2Cu+I–

+2Cul

Chloro–or bromobenzene fails to undergo this coupling reaction unless some strong electron withdrawing group (e.g. NO2) is present in the ortho or para position. In the laboratory, biphenyl can also be synthesized by the treatment of PhMgBr with copper salts. Chemical properties Lacking functional groups, biphenyl is fairly non-reactive, which is the basis of its main application. Biphenyl is mainly used as a heat transfer agent and as a eutectic mixture with H3C diphenylether. This mixture is stable to 400 °C. Biphenyl does undergo sulfonation followed by base hydrolysis produces CH3 p-hydroxybiphenyl and p,p'-dihydroxybiphenyl, which are useful fungicides. In another substitution reactions, it undergoes halogenation. As a general rule of thumb, chiral molecules must have at least one chiral center-that is, a carbon that has four unique substituents coming off of it. However, like most rules of thumb, there OH are exceptions, and there are indeed examples of chiral molecules that HO have no chiral centers, biphenyl's belong to that category. As discussed earlier, biphenyls or in general bi-aryls are the compounds that have Fig. 16: A biphenyl that can be separated into enantiomers two aromatic rings joined by a single bond, can be chiral if they have bulky groups in their ortho positions that provide a barrier to the single bond free rotation. These compounds are chiral because the steric bulk of the groups in the ortho positions provide a large energy barrier to the free rotation around the carbon-carbon single bond, and the conformations are not easily interconverted. Such stereoisomers that can be interconverted through a single bond rotation are called atropisomers. Butane, for example, has conformations that are atropisomers; however, unlike the biaryls, the barrier to rotation is so small that they are interconverted rapidly at room temperature, and they are, for practical purposes, achiral. HO

H3C CH3

CH3

OH

OH HO

H3C

Fig. 17: Interconversion between the enantiomers of this biphenyl is extremely slow at room temperature because of the high barrier to rotation.

O-331

Exercise (iii)

Long Answer Type Questions 1.

Cl

SbCl5

O

(iv) N2O5

?

CH2CH2CH3 Cl2/FeCl3

(A)

CH3

(v)

Cl

2.

?

Write structures of the products A, B, C, D and E in the following scheme.

Na–Hg/HCl

(B)

HNO3/H2SO4

+CH3-CH–CH2–Cl

Given structures and names of the principal organic products expected from mononitration of (i)

4.

m–dinitrobenzene

CH3

(ii)

Anhydrous AlCl3

?

Give the major product obtained from the reaction of each of the following compound with conc. HNO3/conc. H2SO4 (i)

NO2

?

OH (iii)

m–dibromobenzene

COOH

NHCOCH3

(iv)

(ii)

CH3

?

NO2 3.

NO2

Complete the following sequence of reactions with appropriate structures.

(iii)

Cl

O

(i)

CH2–CH2–C–Cl

? AlCl3

?

Cl

(iv)

(ii)

Ph

H C=C

H

OCH3

CH3

+ Br2(Fe)

?

?

Cl

O-332 (v)

Cl

8.

The ionization potential of cyclopropene is 9.95 volt while that of cyclopentadiene is 9.27 V. Explain

9.

Fill in the blanks with appropriate structures of reaction products in the following transformation.

?

SOCl

2→ O– HOOC – C6H4 – CH2 – C6H5  

NO2

5.

Give the structures of the principal products 10. expected from the ring monobromination of each of the following compounds. In each case tell whether bromination will occur slower or faster than with benzene itself. 11. (i) Acetanilide (ii) Iodobenzene

HCl

Toluene reacts with bromine in the presence of light to give benzyl bromide while in the presence of FeBr3 it gives p–bromotoluene. Give explanation for the above observations. What do you understand by aromaticity? State and explain Huckel's rule.

or

Sec-butylbenzene

(v)

Ethylbenzoate (vi) acetophenone

Explain the following mechanism of benzene :

(vii)

Phenetole (C6H5-O-CH2CH3)

(i)

(iv) N-methylaniline

(viii)

Diphenylmethane (C6H5CH2C6H5)

(ix)

Benzonitrile

(x)

Benzotrifluoride (C6H5CF3)

O ••

Explain the mechanism of Friedel Crafts alkylation and acylation both in benzene ring.

13.

Discuss the mechanism of chlorination in benzene. Explain why chlorination takes place at ortho and para-positions in toluene and not at meta-position.

14.

What is meant by term orientation ? Discuss the di and tri-substitution products of benzene.

[Meerut 2008]

NO2

N2O5 CH3CN

[Meerut 2008, 09, 11]

15.

What do you understand by electrophilic substiution reactions ? Discuss the mechanism of nitration, sulphonation and Fredal-Crafts alkylation of benzene. [Meerut 2009,11,12]

16.

Discuss the methods of preparation, properties and uses of benzene. [Meerut 2011]

17.

Describe the naphthalene.

NH2 Cons. HNO3 +H2SO4

NO2 Identify the products in given reaction sequence:

Zn–Hg/HCl

(B)

O

SOCl2

and

properties

of

[Meerut 2011]

Give preparation, properties and uses of toluene.

19.

Explain why nitration of phenol occurs at o- and p-posit ions whereas nitrobenzene gives m-distributed product only. [Agra 2008]

20.

How will you convert benzene into naphthalene ? Explain its structure in detail explaining that in naphthalene two benzene rings are fused at ortho position. [Kanpur 2008]

[Meerut 2012] anhydrous AlCl3

O (A)

preparation

18.

O +

Nitration

[Bund. 2010; Agra 2008; Kanpur 2008,09,11]

O ••

NH2

(ii)

12.

(C6H5CN)

Propose the probable mechanism of following reactions.

(ii)

Sulphonation

[Bund. 2009; Agra 2009]

Biphenyl (C6H5-C6H5)

(i)

7.

Zn − Hg

AlCl3

(iii)

(xi) 6.

anhydrous

(G)   → (H)   → (I)

(C)

AlCl3/CS2

(D)

O-333 21.

22.

23.

24.

25.

Explain the following

26.

(i)

Write four methods for the prepartation of chlorobenzene.

(ii)

How will you prepare the following from chlorobenzene?

(i)

Phenol is brominated easily than benzene.

(ii)

Nitration of toluene is easier than benzene.

(iii)

Ortho and para products are formed in substitution reaction of chlorine

(a)

Phenol

(iv)

Formic acid is stronger than acetic acid.

(b)

Aniline

(v)

Mehtylamine is stronger base than ammonia

(c)

Phenyl cyanide

(vi)

Chlorine of methyl chloride is more reactive than chlorine of chlorobenezne. [Kanpur 2008] 27.

(d)

D.D.T

How can ethyl benzene and styrene be distinguished ? Give the products on the expected reaction of each of the following reagents with ethyl benzene. Give equation : (i)

Hot KMnO4 (aq)

(ii)

Bromine / Iron

(iii)

Br2 (hv)

(iv)

Conc. H NO3/H2SO4

(i)

[Kashi 2008]

Discuss molecular orbital structure of benzene.

(ii)

Give the reaction.

mechanism

of

Friedel-Crafts

(iii)

Write short notes on Brich reduction. [Kashi 2010]

28.

(i) What is Huckel rule of aromaticity? (ii)

[Kanpur 2009]

Write mechanism of the following (a)

Friedel–Crafts reaction,

(b)

Birch reduction.

[Kashi 2011]

What is Haworth synthesis? How will you prepare the following from naphthalene? 29.

How will you convert the following :

(i)

Teralin

(i)

Benzene to biphenyl,

(ii)

1,4 Naphtha-quinone

(ii)

Phenylacetyene to acetophenone

(iii)

α–naphthol

(ii)

Nitrobenzene to chlorobezene

(vi)

β −naphthol

[Kanpur 2010]

30.

How will you obtain naphthalene from 4-phenyl butylenes? Obtain the following compounds from the naphthalene :

(i)

(ii)

Why nitrobenzene is used as solvent instead of benzene for Friedel - Crafts alkylation of bromobenzene ? Explain. How will you synthesise the following from benzene or toluene ?

(i)

β −naphthol

(ii)

1,4-napthaquinone

(a)

p – Br. C6H4 – CH2 – CI

(iii)

β −naphthylamine

(b)

p – Br. C6H4 CH = CH2

(iv)

Benzene

(c)

(v)

Phthalic acid

(i)

Discuss the mechanism of halogenation and sulphonation of benzene.

(ii)

Classify the following groups as activating or deactivating when attached to benzene. [Kashi 2008]

CH3 C6H5

[Kanpur 2011]

C

OH

CH3 31.

[Lko. 2008]

[Lko. 2009]

(i)

Discuss the mechanism of halogenation and sulphonation of benzene.

(ii)

Classify the following groups as activating or deactivating when attached to benzene. –NO2, –NH2, –Cl, –CH3,–COCH3 [Hazaribagh 2008]

O-334 32.

33.

(i)

What are aromatic compounds? Discuss the 6. mechanism of the following reactions : (a)

Friedel Crafts reaction

(b)

Nitration

7.

(ii)

How is toluene prepared? Describe its important reactions. [Hazaribagh 2008]

(i)

What is deactivating groups? Explain the deactivating effect of NO2 group when it is present in benzene ring.

(ii)

8.

35.

Explain orbital structure of benzene molecule. Discuss Huckel's rule.

(iii)

Starting from benzene how will you prepare :

directing?

C6H5COOH

Phenol

(b)

Benzoic acid?

[Hazaribagh 2010]

Tetralin

11.

What happens when : (i)

Naphthalene is heated with conc.H2SO4.

(i)

Relative methods of orientation.

(ii)

(ii)

Action of chlorine in toluene under different conditions.

Naphthalene is heated with O2 in the presence of V2O5. [Kanpur 2011]

(iii)

σ and π complexes.

(iv)

–OH group in phenol is o- and p-directing.

12.

Explain the structure of benzene on the basis of molecular orbital theory and resonance theory. [Kanpur 2011]

On the basis of σ − complex stability , explain why −NH3 ⊕ group is meta – directing group whereas OCH3 group is ortho and para – directing group.

Explain in detail why cyclopentadienyl anion is aromatic while cyclo octatetraene is not aromatic.

[Avadh 2008]

14.

Discuss the mechanism of Friedel – Crafts reaction. [Avadh 2008,09]

Discuss the mechanism of nitration of benzene. [Bund. 2008; Agra 2009] 15.

5.

(ii) β–Napthol

Write short notes on the following :

[Bund. 2008,09]

4.

[Kanpur 2008]

[Kanpur 2009]

(a)

Short Answer type Questions

3.

[Kanpur 2008]

How you will get the following from napthalene (i)

13.

2.

group is ortho and para

How can you prepare the following from toluene (i) T.N.T (ii) Xylene

[Hazaribagh 2011]

1.

[Kanpur 2008]

How can you explain that –NO2 group is meta

(iii) 10.

(ii)

How will you prepare the following compounds

directing while –NH2

What are the different isomers represented by the molecular formula C7H7Cl? Write them. 9.

(i)

[Agra 2008]

from benzene ? (i) Benzaldehyde (ii) Acetophenone

[Hazaribagh 2008]

34.

What is aromaticity? Explain with examples.

What happens when toluene reacts with chlorine : (Give reactions only) (i) At low temperature, in absence of light and 16. in presence of Lewis acid. (ii) At high temperature, in presence of light and absence of catalyst. [Bund. 2009] 17. Discuss the difference in reaction of benzyl chloride and chlorobenzene. [Bund. 2009] Give the preparation and uses of B.H.C. B.H.C is obtained by heating benzene with chlorine in 18. presence of sunlight. Write the reaction. [Meerut 2009; Agra 2009]

Why phenol is more reactive than nitrobenzene for electrophilic aromatic substitution reactions? [Avadh 2009]

Give the mechanism of nitration of benzene. [Avadh 2010]

Explain why -OH group is ortho, para director and activator whereas -NO2 group is meta director and deactivator for aromatic electrophilic substitution reaction.

[Avadh 2010]

Chlorine is ortho/para directing but still deactivating in nature. Why?

[Lko. 2011]

O-335 19.

Identify the product A, B and C and name the type

,

of reaction involved in each step :

⊕

CH3COCl AlCl3

(i) Zn/Hg HCl

A

B

NBS

[Bund. 2010]

4.

C

Cl2/hν Alcl3

A

B

5.

(a)

Alcl3

C

Write note on electrophilic reagent. [Kanpur 2009]

7.

How can you convert toluene into benzoic acid? [Kanpur 2009]

8.

[Kanpur 2009]

9.

Explain the directive influence of –NH2 and –NO2

benzene ring ? Explain. –NO2 ,−C2H5 ,−SO3H 10.

[Rohd. 2012]

Describe mechanism of sulphonation of benzene. [Rohd. 2012]

24.

Explain ortho/para ratio.

25.

Explain giving reason why NO2 and group is

[Rohd. 2012]

meta-directing and NH2 group is ortho and para-directing. 26.

[Hazaribagh 2010]

Bromobenzene into benzoic acid

(iii)

Benzene into Benzyl chloride

(iv)

Chlorobenzene into D.D.T

[Kanpur 2009] [Kanpur 2010]

deactivates the benzene ring? -NO2 and -CH3. 13.

14.

[Kanpur 2010]

Explain that in naphtahlene two benzene rings are fused at ortho positions.

15.

[Kanpur 2010]

Explain why phenol is more easily brominated than benzene.

[Kanpur 2010]

How can you get the following? (i)

Benzoic acid from benzene.

(ii)

Saccharine from toluene.

Define and explain Huckel's rule. [Bund. 2008; Meerut 2012; Kanpur 2011]

Benzene has three double bonds but it does not 16. decolourise bromine water. Why?

3.

(ii)

Which group of the following two activates and

[Hazaribagh 2011]

2.

Ethyl bromide into diethyl ether

12.

Give the mechanism of sulphonation of benzene. In

Very Short Answer Type Questions

(i)

Convert:Benzene into T.N.T.

[Hazaribagh 2010]

what respect does it differ from nitration?

1.

Convert

11.

Give the mechanism of electrophilic substitution in benzene.

27.

[Kanpur 2009]

Prove that two benzene rings are fused in a naphthalene molecule in ortho positions.

23.

Which of the following groups activate or deactivate

groups on the basis of electron displacement. [Rohd. 2011]

22.

Convert: Benzene → Acetophenone.

Why do o- and p- directing groups usually [Rohd. 2011]

[Kanpur 2009]

6.

State Huckel rule of aromaticity and

activate the ring? 21.

Complete the following reaction : 3→ C6H + CH3Cl  

explain it with two examples. (b)

[Meerut 2012]

AlCl

NH2SO4/HNO3

[Lko. 2011]

20.

What happens when benzene is oxidized in presence of V2O5 ?

CH3Cl AlCl3

(ii)

,

[Bund. 2010]

Which among the followings are aromatic and why? 17. Explain it.

[Kanpur 2011]

Convert: Chlorobenzene into picric acid. [Kanpur 2011]

Give one example of Friedel - Crafts reaction. [Kashi 2008; Hazaribagh 2008,11]

O-336 18.

What is Huckel' s rule of aromaticity ? [Kashi 2008] 2.

19.

Define aromaticity and state Huckel's rule.

The major product obtained when Br2/Fe is treated with

O

[Kashi 2010]

20.

Benzene is planar molecule and burns with a sooty flame. Why?

21.

HN H3C

[Lko. 2009]

CH3

Alkoxy group in benzene is ring activating though oxygen is more electronegative. Why? [Lko. 2009]

22.

is

Choose the correct answer then justify: The compound which will give only meta derivative on

(a)

nitration

23.

H3C

(i)

Anisole

(ii)

Chlorobenzene

(iii)

Nitrobenzene

25.

(b)

Why nitrobenzene gives m-dinitro benzene on [Lko. 2010]

H3C

CH3

[Lko. 2011]

Write a note on activating nature of NH2 group and

Br

deactivating nature of COOH group. [Lko. 2011]

27.

O

HN

Explain, why aniline does no undergo Friedelcraft's reaction.

26.

Br

[Lko. 2010]

nitration?

CH3

[Lko. 2009]

Explain acetylation of benzene in presence of Lewis acid.

24.

O

HN

(c)

State Huckel's rule of aromticity and explain it with

HN

O

H3C

two examples.

CH3

[Rohd. 2009; Hazaribagh 2008]

28.

Why –NO2 group is meta directing group?

Br

[Rohd. 2009]

29.

(d)

Benzene does not give deal Alder reaction. Explain. [Rohd. 2013]

HN H3C

O CH3

Objective Type Questions Br

Multiple Choice Questions 1.

3.

The compound that is most reactive towards electrophilic nitration is (a) Toluene (b) Benzene (c) Benzoic acid (d) Nitrobenzene

4.

The reaction of toluene with chlorine in presence of ferric chloride gives predominantly (a) Benzoyl chloride (b) M–chlorotoluene

Among the following, the compound that can be most readily sulphonated is (a)

Benzene

(b)

Nitrobenzene

(c)

Toluene

(d)

Chlorobenzene

O-337 (c) (d) 5.

Benzyl chloride O– and p–chlorotoluene

8.

The correct product of mono nitration of

Which of the following carbocations is expected to be most stable?

CH3

(a)

CH2–COO +

NO2 H Y

is (a)

CH3

(b)

CH2–COO

+

(b) CH2–COO

H Y

NO2

(c)

NO2

CH3

(c)

+

CH2–COO

H Y

(d) 6.

When nitrobenzene is treated with Br2 in presence of FeBr3, the major product formed is m–bromonitrobenzene. Statements which are related to obtain the m–isomer are 9. (a) The electron density on meta carbon is more than that on ortho and para positions. (b)

(c)

(d)

7.

CH3

(d)

NO2

CH2–COO

The intermediate carbonium ion formed after initial attack of Br+ at the meta position is least destabilized Br+

Loss of aromaticity when attacks at the 10. ortho and para positions and not at meta position Easier loss of H + to regain aromaticity from the meta position than from ortho and para positions

An activating group (a)

Activates only ortho and para positions

(b)

Deactivates meta position

(c)

Activates ortho and para more than meta

(d)

Deactivates meta more than o– & p–

H Y +

A deactivating group (a)

Deactivates only ortho and para positions

(b)

Deactivates meta position

(c)

Deactivates ortho and para more than meta

(d)

Deactivates meta more than o– & p–

In aniline, the –NH2 group (a)

Activates the benzene ring via both inductive and resonance effects

(b)

Deactivates the benzene ring via both inductive and resonance effects

(c)

Activates the benzene ring via resonance effect and deactivates it via inductive effect

(d)

Activates the benzene ring via inductive effect and deactivates it via resonance effect

O-338 11.

12.

13.

14.

In the Friedel –Crafts acylation, the electrophile is (a)

C6H5+

(b)

AlCl3−

(c)

CH3CO+

(d)

C6H5CH2+

The main reaction product when toluene is treated with concentrated HNO3 in the presence of concentrated H2SO4 is (a)

2,4,6 –trinitrotoluene

(b)

2,4 –dinitrotoluene

(c)

p–nitrotoluene

(d)

m–nitrotoluene

NO2

(d)

H +

15.

Y

Which of the following species is expected to have maximum enthalpy in an electrophilic aromatic substitution reaction? +

Eδ

H

δ+

In chlorobenzene, the –Cl group (a)

Activates the benzene ring more via resonance effect than deactivating it via inductive effect

(b)

Deactivates the benzene ring more via inductive effect than activating it via resonance effect

(c)

Activates the benzene ring via resonance effect and deactivates it via inductive effect. Both these effects are evenly matched.

(d)

Is a net deactivating group with meta director characteristics.

+E+

(IV)

17.

+

H Y

+

18.

H Y

(V)

(a)

Species (II)

(b) Species (III)

(c)

Species

(d) Species (V)

(IV)

Which of the following sequence regarding the rate of halogenation is correct ? (a)

Fluorination > chlorination > bromination

(b)

Fluorination < chlorination < bromination

(c)

Fluorination < chlorination > bromination

(d)

Fluorination > chlorination < bromination

For the electrophilic substitution reaction involving nitration, which of the following sequence regarding the rate of reaction is true? (a)

KC H > KC D > KC T 6 6 6 6 6 6

(b)

KC H < KC D < KC T 6 6 6 6 6 6 KC H = KC D = KC T

(c)

NO2

E δ+

H Y

(c)

(III)

(II) δ+H

+

NO2

+

(I)

Which of the following carbocations is expected to be most stable? NO2 (a) 16.

(b)

H

6

6

6

6

6 6

(d)K C H > K C D < K C T 6 6 6 6 6 6 Which of the following statements made on nitration is/are false? (a)

Nitration of ethylbenzene

benzene

is

faster

than

(b)

Nitration of benzene C6H5NHCOCH3

is

slower

than

O-339

19.

(c)

Nitration of benzene chlorobenzene.

is

faster

(d)

Nitration of chlorobenzene m–chloronitrobenzene

gives

(e)

Nitration of benzene hexadeuterobenzene

than

is

faster

than

22.

23.

Friedel Crafts alkylation and acylation leads to new carbon-carbon bond formation.

(b)

Arylhalide/anhydrous AlCl3 combination can be used as reagent for Friedel-Crafts alkylation.

(c)

When benzene and nitrobenzene, both liquids, are mixed and treated with CH3COCl/AlCl3 under proper reaction conditions the products obtained after acidification are meta acetyl nitrobezene and acetophenone.

(d)

24.

n-propylbenzene can be made in good yield 25. from benzene using n-propyl chloride and anhydrous AlCl3 combination as reagent.

20.

To transform

(a)

1.54 A°

(b)

1.34 A°

(c)

1.39 A°

(d)

1.21 A° [Bund. 2009]

Which of the following statements made on Friedel-Crafts reaction is/are false? (a)

What is the bond length between C-C in benzene

into

26.

Nitration of Benzene is (a)

Nucleophilic substitution

(b)

Electrophilic substitution

(c)

Nucleophilic addition

(d)

Free radical substitution

[Rohd. 2009,10]

The –COOH group in benzoic acid is (a)

Ortho directing

(b)

Meta directing

(c)

Para directing

(d)

Ortho and Para directing

[Rohd. 2009,10]

The –CHO group in benzaldehyde is : (a)

Ortho directing

(b)

Meta directing

(c)

Para directing

(d)

Ortho and para directing

[Rohd. 2010]

When chlorobenzene reacts with methyl chloride and sodium metal in ethereal solution, gives

, two initial steps could be

toluene. The reaction is called as :

O2N (a) (b) (c) (d)

21.

Nitration followed by Friedel - Crafts alkylation. Friedel - Crafts alkylation followed by nitration. 28. Nitration followed by Friedel - Crafts acylation. Friedel - Crafts acylation followed by Clemmensen's reduction followed by nitration

-NH2 group is

29.

(a)

o-directing

(b) p-directing

(c)

o and p-directing (d)m-directing [Bund. 2008]

(a)

Wurtz reaction

(b)

Friedel Craft reaction

(c)

Wurtz-Fitting reaction

(d)

Perkin's reaction

[Rohd. 2011]

The electrophile in aromatic nitration is (a)

Nitronium ion

(b)

Nitrite ion

(c)

Nitrinium ion

(d)

Nitrate ion

[Rohd. 2012]

In Friedel-Craft acylation the electrophile is (a)

CH3CO +

(b)

C6H5+

(c)

AlCl3

(d)

CH3+ [Rohd. 2013]

O-340

Fill in the Blank 1.

Benzene is ……… due to 6 π electrons.

2.

Benzene normally does not participate in ………... .

3.

True /False 1.

If a ring is planar then it will be aromatic.

2.

Aromatic hydrocarbons are immiscible in water.

3.

Mixture

of

concentrated

nitric

acid

and

The effective electrophile in sulphonation of

concentrated hydrochloric acid is termed as

benzene is ………. .

nitrating mixture. 4.

4.

Cyclobutadiene is ………. .

5.

Cyclooctatetraene is ………. .

6.

In the Friedel–Crafts acylation, the electrophile is

Acoording to Huckel's rule, for a ring to be aromatic it should have (4n+2) π electrons.

5.

In halogenation of benzene using Cl2 along with anhydrous AlCl3 as a catalyst, Cl2 is electrophile.

………. . 7.

Out of aniline and toluene, ……… is more reactive towards Friedel–Craft reaction.

8.

6.

presence of anhyd. AlCl3 give n-propyl benzene. 7.

In Friedal craft reactions, anhyd. AlCl3 is used as a catalyst.

Friedel–Crafts acylation consumes……… amount of 8. Lewis acid catalyst than Friedel Crafts alkylation.

9.

Out of p–xylene and toluene, p-xylene is ……… 9. reactive towards nitronium ion.

10.

Out of naphthalene and anthracene ……… does not participate in Diels Alder reaction.

Reaction of benzene with 1- chloropropane in the

10.

At high temperature, benzene on reaction with hydrogen with nickel as catalyst undergo addition reaction to give cyclohexane. Activating groups direct the upcoming electrophile to meta position in benzene ring during electrophillic substitution. Benzene is the only aromatic ring in organic chemistry.

O-341

Answers Objective Type Questions Multiple Choice Questions 1.

(c)

2.

(a)

3.

(a)

4.

(d)

5.

(d)

6.

(a), (d)

7.

(c)

8.

(b)

9.

(c)

10.

(c)

11.

(c)

12.

(c)

13.

(b)

14.

(d)

15.

(a)

16.

(a)

17.

(c)

18.

(a),(d),(e)

19.

(b,c,d)

20.

(d)

21.

(c)

22.

(c)

23.

(b)

24.

(b)

25.

(b)

26.

(c)

27.

(a)

28.

(a)

29.

(a)

Fill in the Blank 1.

aromatic

2.

electrophilic

3.

SO3

4.

antoaromatic

5.

non aromatic

6.

CH3CO +

7.

toluene

8.

more

9.

more

10.

naphthalene

True/False 1.

False

2.

True

3.

False

4.

True

5.

False

6.

False

7.

True

8.

True

9.

False

10.

False

Hints and Solutions Long Answer Type Questions 1.

O-342 Na Hg and HCl reduces the \/ C = O to \/ CH2 . 2.

NO2

(i)

OH

(ii)

Br

(iii)

NHCOCH3

(iv)

NO2

NO2

NO2

CH3

Br

NO2

3.

NO2

(i)

NO2

CH=CH–CH3

(ii)

O Br

(iii) +

4.

NO2

(iv)

(v)

H3C C

SbSl6–

CH3

NO2

(i)

CH3

(ii)

CH2

NO2

COOH

O2N

NO2

(iii)

Cl

(iv)

OCH3

(v)

Cl

NO2

NO2

Cl

5.

(i)

NO2

Cl

NO2 NHCOCH3

(ii)

I

, faster

Br

, slower

Br

O-343 (iii)

CH3

NH–Me

(iv)

CH2CH3

CH

, faster , faster Br Br

(v)

O C

(vi)

O–C2H5

O

CH3

C

, slower

, slower

Br

(vii)

OC2H5

(viii)

Br CH2–C6H5 , faster

, faster

(ix)

Br CN

(x)

Br CF3

, slower

, slower

Br

Br

(xi) Br, faster

6.

CH2

(i) CH2

O

CH3

CH2 CH2

••

⊕O

NO2–ONO2

NO2

–NO3–

CH2 CH2

O

H

CH3

CH2 CH2

NO2

⊕ –H+

CH3

NO2 H

O ••

CH3

O-344 ⊕

(ii)

⊕

NH3

NH2

NH3 NO+ 2

H+

NO2

7.

O

(A):

O

(B):

NH2 OH – NO2 NO2 CH2CH2CH2COOH

C–CH2–CH2–C–OH

O

(C):

(D):

CH2CH2CH2–C–Cl

O

8.

Cyclopropene on ionization forms cyclopropene anion, which is an antiaromatic compound. While ionization of cyclopentadiene leads to the formation of an aromatic species cyclopentadiene anion, . So it is formed more easily.

It is why the energy required for the ionization i.e., ionization potential of cyclopropene is higher than that of cyclopentadiene. O

O

9.

C

C Cl

(G)=

CH2

(H)= CH2

CH2 (I)= CH2

10.

Side chain bromination is favoured under photochemical irradiation and involves a free radical mechanism. In the presence of FeBr3, electrophilic substitution in the benzene ring occurs. ❍❍❍

UnitO-345 -IV

C HAPTER

7

Alkyl and Aryl Halide

1. Introduction Organic halogen compounds are a large class of natural and synthetic chemicals that contain one or more halogens (fluorine, chlorine, bromine, or iodine) combined with carbon and other elements. The simplest organochlorine compound is chloromethane, also called methyl chloride (CH3Cl). Other simple organohalogens include bromomethane (CH3Br), chloroform (CHCl3), and carbon tetrachloride (CCl 4 ). There is an interesting aspect-if the halogen atom is directly attached to aliphatic hydrocarbon, the resultant compound is alkyl halide. Similarly, if the halogen atom is directly attached to aromatic hydrocarbon, the resultant compound is termed as aryl halide.

2. Alkyl Halides Alkyl halides are the derivatives of alkanes which can be obtained by replacing one hydrogen atom by halogen atom. The general formula of monohalogen derivative of alkanes is CnH2n+1X, here X is any halogen atom. The monohalogen derivatives of alkanes are commonly called as alkyl halides.

2.1 Nomenclature and Classification of Alkyl Halides Common names of alkyl halides are derived from the name of parent hydrocarbon. Br

CH3Br

CH3CH2Br

(methyl bromide)

(ethyl bromide)

Br

(isopropyl bromide)

Br

(n–butyl bromide)

(n–propyl bromide)

Br

(sec–butyl bromide)

O-346

Br

Br

Br (isobutyl bromide)

(n–pentyl bromide)

(tert–butyl bromide)

Br Br Br (isopentyl bromide)

(neopentyl bromide)

(tert–pentyl bromide)

2.2 Classification Alkyl halides are formally derived from alkanes by exchanging hydrogen for halogen atoms like fluorine, chlorine, bromine, and iodine. Depending on the degree of substitution at the carbon atom carrying the halogen, alkyl halides are classified into primary, secondary and tertiary alkyl halides. R R–CH2–X

R–CH–R

R–C–R

Primary alkyl halide

X Secondary alkyl halide

X Tertiary alkyl halide

Fig. 1: Primary, secondary and tertiary alkyl halides

Vinyl halides are often classified as the fourth type of alkyl halides.

2.3 IUPAC Nomenclature In the generally accepted nomenclature of alkyl halides, the name of the alkyl residue is followed by the name of the halide, for example methyl iodide and ethyl chloride. However, the IUPAC nomenclature considers an alkyl halide a substituted alkane, halogens are treated the same way as alkyl groups. An example is shown here in which the IUPAC name is written in brackets in the illustration below. CH3–Cl Methyl chloride (chloromethane)

CH3–CH2–F Othyl fluorido (fluoro ethane)

CH3–CH–CH3

CH3–CH2–CH–CH3

I isopropyl iodide (2–iodopropane)

Br sec–butyl bromide (2–bromobutane)

CH3 H3C–C–CH3

Br tert–butyl bromide (2–bromo–2–methylpropane)

O-347 The name of the halogen is followed by the name of the alkane, for example, iodomethane and chloromethane. If an alkyl halide contains more than one halogen, the halogen names are noted in alphabetical order, such as 1-chloro-2-iodobutane. Some more names are, CH3

CH3

C—I

Cl—CH2—C—CH3 H 1–chloro–2–methylpropane

H (1–odoethyl) cyclooctane

In the nomenclature of unsaturated alkyl halides, unsaturated bonds have a higher priority than halogen substituents. This requires special attention when numbering the carbon chain. CH3 H3C—C=CH—CH2—CH2—Cl 5–chloro–2–methylpent–2–ene

Some commonly used chlorinated organic solvents, such as carbon tetrachloride (tetrachloromethane), chloroform (trichloromethane) and methylene chloride (dichloromethane), are known more by their trivial name than by their IUPAC name. Hydrocarbons in which all hydrogens are replaced by halogens are commonly named as perhaloalkanes or perhaloalkenes e.g., perfluoroethylene or 1,1,2,2 -tetrafluoroethene.

2.4 Isomerism in Haloalkanes Isomerism in haloalkanes is primarily of two types one is chain isomerism and another is position isomerism. In addition to these two types, haloalkanes always show optical isomerism when properly substituted. 1.

Chain isomerism: Haloalkanes containing four or more carbon atoms exhibit chain isomerism in which the isomers differ in the chain of carbon atoms. For example, C4H9Br has three chain isomers, such as: CH3 CH3 CH3CH2CH2CHBr 1–bromobutane

2.

CH3–C–Br

CH3–CH–CH2Br CH3 1–bromo–2–methylpropane 2–bromo–2–methylpropane

Position isomerism: Haloalkanes with three or more carbon atoms exhibit position isomerism in which the isomers differ in the position of halogen atom. For example, C3H7I has two position isomers: CH3CH2CH2I 1–iodopropane

CH3CHCH3 I 2–iodopropane

O-348

Solved Examples Example1: How many structural isomers are possible when C5H12 is subjected to monobromination? Solution: 8.

2.5 Methods of Preparation of Alkyl Halides There are a number of methods available for the preparation of alkyl halides. This is an important category of organic compounds and these can be prepared from alkanes, alkenes, alkynes, alcohols, aldehydes, ketones, carboxylic acids etc. Some of the prominent methods are discussed below, 1.

From alcohols: A variety of reagents can be applied to synthesize alkyl halides from alcohols. These are discussed below,

(i) By using hydrogen halide HX

R — OH  → RX + H2O It must be noted that the HX used should be dry, which is produced, in situ, as follows heat

2NaCl + H2SO4  → 2HCl↑ + Na2SO4 heat

2NaBr + H2SO4  → 2HBr↑ + Na2SO4 heat

6NaI + 2H3PO4  → 6HI↑ + 2Na3PO4 It may be noted that H3PO4 instead of H2SO4 is used to prepare HI. This is because HI is a reducing agent and H2SO4 is an oxidising agent. The above substitution reactions proceed via SN1 and SN2 mechanism. Both the above mechanisms are operative during reaction having competition among them.

Mechanism R–OH

H+

⊕

–H O

R–O–H S 21 N

R

+X–

R–X+(Some rearranged product if possible)

H X–

SN2

R–X+H2O

Details about SN1 and SN2 will be discussed later in this capter (ii) By using phosphorous halides: Both phosphorus pentahalides and trihalides when allowed to react with alcohol yield alkyl halide as shown below,

O-349 ROH + PCl5 → RCl + POCl3 + HCl 3ROH + PCl3 → 3RCl + H3PO3 3ROH + PBr3 → 3RBr + H3PO3 3ROH + PI3 → 3R–I + H3PO3 Phosphorous halides are prepared by treating red phosphorous and halogen. The merit of using phosphorous halides is that free carbocation is not generated and so rearrangement does not take place.

(iii) By using SOCl2 (thionyl chloride) Pyridine

R — OH + SOCl2   → R– Cl + SO2↑ + HCl↑ The reaction follows SN2 pathway. The usefulness of this method is that there is no side product which has to be separated. It makes this reaction of converting an alcohol into alkyl halide as one of the better synthetic pathways. The side products are gaseous SO2 which escape from the reaction mixture and HCl which forms a salt with the base (pyridine) named pyridinium chloride (C 5H5N +Cl − ). In the absence of pyridine reaction follows SN1 mechanism in which configuration of the alcohol is retained. 2.

X

2 → R—X + HX Reactivity of above reaction By direct halogenation of hydrocarbons: R—H   

hv

with respect to type of hydrogen to be replaced follows the order, Tertiary hydrogen > Secondary hydrogen > Primary hydrogen As far as the reactivity of halogen is concerned, F 2 is most reactive while I 2 is least reactive. In fact reaction with I 2 is reversible and is carried out in the presence of some oxidising agents like HIO 3 , HNO 3 etc. to oxidise HI.HI is being scavenged by oxidising agents like HIO 3 , HNO 3 thereby inhibiting the reversibility of the reaction and making it useful. Mechanisms of these reactions has been discussed in the topic "alkanes". 3.

By halide exchange: Alkyl iodides can be prepared from alkyl chlorides and alkyl bromides by nucleophilic substitution. This can be achieved by treating them with NaI, using acetone as a solvent. It is known as Finkelstein Reaction. Feasibility of this reaction is due to the solubility of NaI in acetone and more nucleophilic character of I − ion. It follows SN2 mechanism. NaI

R — Cl  → R — I + NaCl Acetone

R − Br  → R − I + NaBr NaI

Acetone

4.

By addition of HX to alkenes: Generally alkyl halides like alkyl chlorides, bromides and iodides can be prepared by treating an alkene with corresponding hydrogen halides (HCl, HBr and HI). The addition of these compounds to alkene takes place according to Markovnikov's addition rule as discussed earlier in the topic "alkenes". The reaction proceeds by electrophilic addition of H + to give

O-350 more stable carbocation followed by attack of X − . Anti–Markovnikov addition of HBr can be achieved if reaction is carried out in presence of peroxide like H2O 2 or benzoyl peroxide. This reaction proceeds by the attack of bromine free radicals ( • Br). These mechanisms are schematically shown below, X−

HX

CH 3 — CH = CH2  → CH 3 — CH — CH 3  → CH3 − C H − CH3 | X

⊕

HBr

HBr

CH 3 — CH = CH 2  → CH 3 — CH — CH 2 — Br  → CH 3 — CH 2 — CH 2 Br + Br• Peroxide

5.

•

From silver salts of carboxylic acid: Even carboxylic acids can be directly converted to alkyl halides by an appropriate choice of reagents as shown. Although the the reaction is limited to chlorine and bromine only. CCl

4 → RX + AgX + CO RCOOAg + X 2    2

(X 2 = Cl 2 or Br 2 ) This reaction is popularly known as Hunsdieker reaction.

Mechanism The mechanism is uncertain but probably in the first step acyl hypohalite is formed which subsequently decomposes into free radicals. RCOO 2 Ag + X2 → RCOOX + AgX RCOOX → RCOO• + X• RCOO• → R• + CO 2 R• + X 2 → RX + X• R• + RCOOX → RX + RCOO• etc. In Hunsdieker reaction, Br 2 in CCl4 gives better yield than Cl 2 in CCl4. — X is

The yield and ease of formation of R

1°RX > 2°RX > 3°RX If I 2 is used in Hunsdieker reaction in CCl4, then an ester is formed. This is known as Birnbaum—Simonini reaction. O ||

2RCOOAg + I2 → R − C − Or + 2AgI + CO 2 6.

Preparation of allylic or benzylic halides (i)

Direct halogenation of any aromatic hydrocarbon preferably gives benzylic halide. This is because benzyl radical is resonance stabilized.

O-351 X–CH–CH3

CH2–CH3 X2 hν

(ii)

If an alkene is treated with halogen at high temperature or in presence of radiations or any reagent which is able to provide halogen radicals in low concentrations, then allyl halides are formed. X•

CH3 — CH = CH2   → X — CH2 — CH = CH2 high temperature

CH3 — CH = CH2 + Cl2  → C H2 − CH = CH2 | Cl

400 to 500 ° C

NBS

CH3 — CH = CH2  → CH2 — CH = CH2 | Br

SOCl

2 → CH − CH = CH CH3 — CH = CH2   2 2

| Cl

7.

Preparation of polyhalides (i)

Addition of halogen to alkenes produces vicinal dihalides. (X = Cl or Br )

2 2 2 CH2 = CH2 + X2   → CH2 − CH2

| Cl

(ii)

| Cl

Addition of halogen (Cl2 or Br2) to alkynes produces tetra halo derivatives. X2

X2

X |

X |

HC ≡ CH   → C H = C H   → H − C − C − H | X

| X

| X

| X

Mechanism of addition of halogen to alkenes has been discussed in the topic "alkenes". 8.

Preparation of vinyl halides HCl

CH ≡ CH  → CH2 = CH CuCl2

| Cl

The above reaction proceeds via electrophilic addition. The mechanism of the reaction is discussed below.

Mechanism

O-352

2.6 Properties of the Alkyl Halides The properties of alkyl halides can be of two types - one is physical properties like boiling point, density, water solubilty etc. The second type is definitely the chemical reactivities shown by these compounds. These are discussed below one by one.

Physical properties of the alkyl halides 1.

2.

3.

4.

Boiling point: Due to the presence of slight polarity in alkyl–halides as carbon halogen bond is polar due to difference in electronegativities of the two atoms as well as greater molecular weights, alkyl halides have considerably higher boiling points than alkanes containing same number of carbon atoms. Branching leads to decrease in surface area due to which Van der Waals force decreases. So boiling point of isomeric alkyl halides decreases with branching. For similar alkyl halides with different halogen atoms, boiling point increases with molecular mass. Consequently, alkyl iodides have maximum boiling point whereas alkyl fluorides have minimum boiling point. Boiling points of alkyl fluorides are lower than those of hydrocarbons of comparable molecular weight. The difference is due to the small size of fluorine, the tightness with which its electrons are held, and their particularly low polarizability, so, 2-methylpropane has a bp =–1° whereas the bp of 2-fluoropropane is –11°C. Water solubility: Alkyl halides are negligibly less soluble in water because of their inability to form hydrogen bonds. However alkyl fluoride are comparatively more soluble in water. Their solubility further decreases with increase in the size of alkyl group. They are soluble in typical organic solvents. Density: The density of alkyl halides is more than that of alkane and decreases from alkyl iodides to alkyl fluorides. Alkyl iodides and alkyl bromides are denser than water whereas monochloro alkanes and monofluoroalkanes are lighter than water. More the number of halogen atoms more will be the density if carbon skeleton remains same. Dipole moment: Dipole moment of RX depends on: (i)

The sizes of the partial charges,

(ii)

The distance between them, and

(iii)

The polarizability of the unshared electrons on halogen.

As such, dipole moments of all the haloalkanes are comparable. Example 2: Write a short note on the dipole moment of haloalkanes. Solution : l

Electronegativities of the halides: F > Cl > Br > I

l

H

Bond lengths increase as the size of the halogen increases:

+ C

C—F < C—Cl < C—Br < C—I l

Bond dipoles: C—Cl > C—F > C—Br > C—I 1.56D

l

1.51D 1.48D

δ

H

1.29D

Molecular dipoles depend on the geometry of the molecule.

H

+

µ

X δ–

O-353

Chemical properties of the alkyl halides Carbon halogen bond in alkyl halide is polar owing to partial positive charge on carbon and partial negative charge on halogen. Due to the positive charge on carbon, any nucleophile can attack the molecule and if it is a stronger nucleophile than halide, it can substitute halide. So, nucleophilic substitution reactions are the most common reactions of alkyl halides. Some of the prominent chemical reactivity of alkyl halide is discussed below, 1.

Preparation of Organometallic compounds Alkyl halide on treatment with Mg metal in presence of dry ether yields Grignard reagent. Using the same pathway these alkyl halides can be converted to alkyl lithium as shown below, Mg

R — X  → RMgX ether

(Grignard reagent) Li

R — X  → R — Li + LiX ether

(Organo lithium compound)

Mechanism R — X + Li → [R — X] − + Li+ [R — X] − →

R• + X−

R • + Li → R – Li The extra electron in free radical anion occupies an antibonding molecular orbital. It must be noted that the reactivity of alkyl halides to form organometallic compounds increases from alkyl fluorides to alkyl iodides. Infact alkyl fluorides do not respond to the reaction and alkyl chloride and aryl halides give considerable yield only if THF (Tetrahydrofuran) is used as a solvent. Further, more the stability of free radical, more will be reactivity of alkyl halide to form organometallic compound. Alkyl organometallic compounds is a good sources of carbanions (as carbon is more electronegative than the metals) and are very useful in organic synthesis. 2.

Reduction of alkyl halides Reduction

R — X  → R — H The reagents used for reduction of alkyl halide to alkane are Zn and CH3COOH, Zn and HCl, Zn and NaOH, Zn — Cu couple and alcohol, aluminium amalgam and alcohol etc. 3.

Wurtz reaction ether

2R — X + Na   → R — R Details about this reaction had already been discussed during preparation of alkanes. 4.

Dehydrohalogenation of alkyl halides alcoholic KOH

R — X   → Alkene

O-354 alcoholic KOH

CH3 − CH − CH2 − CH3      → CH3 − CH = CH − CH3 + CH2 = CH − CH2CH3 | X

(major)

(minor)

Details of this reaction had been discussed during preparation of "alkenes". A very important reaction of alkyl halides i.e., Aliphatic nucleophilic substitution is dealt as a separately heading. Example 3 : 36.4 g of 1,1,2,2– tetrachloropropane was heated with Zn dust and the product was bubbled through ammonical AgNO 3 . What is the weight of the precipitate obtained? Give equations for the reactions involved. Solution : Ag O

Zn dust

2 CHCl2 — CCl2CH3   → CH ≡ C — CH3    →

Ammonical AgNO3

∆

∵

182 g compound gives 147 g precipitate

∴

36.4 g compound gives

+

AgC − ≡ CCH3

147 × 36.4 = 29.4 g 182

3. Aliphatic Nucleophilic Substitution In nucleophilic substitutions, the attacking reagent which is a nucleophile brings an electron pair to the substrate, which uses this pair to form a new bond and the leaving group comes away with an electron pair. When nucleophile is a solvent, the reaction is called solvolysis. R — X + Y → R — Y + X Y is a nucleophile which may be neutral or negatively charged while RX may be neutral or positively charged. R — I + HO − → R – OH + I − +

R — I + NMe3 → R — N Me3 + I− +

R — N Me3 + OH − → R — OH + NMe3 +

+

R — N Me3 + H2S → R — SH2 +NMe3

O-355 aq. NaOH or KOH or moist Ag2O R'O–Na+ KSH or NaSH R'S–Na+ NH3

R–X

R'NH2 R'2NH AgCN/KCN

R–OH+NaX or KX R–O–R'+NaX R–SH+Kx or NaX R–S–R'+NaX RNH2+HX R–NHR'+HX R–NR'2+HX R–C≡N+R–N

C+X–

–

C≡C—R'

I–,acetone NO2– –

CH (CO2Et)2

R–C≡C–R' R–I+X– R–O–N=O+R–NO2+X– R–CH(CO2Et)2

Before going on for discussion of the mechanism of nucleophilic substitution reactions in alkyl halides, one must have an idea about the few basic terms. These are mentioned below.

Basicity and nucleophilicity Basicity is the tendency of a base of abstract H + and nucleophilicity is its tendency to attack as a nucleophile on carbon. All the bases are nucleophiles and all the nucleophiles are bases.

3.1 S N 1Reaction S N 1 stands for substitution nucleophilic unimolecular. The ideal version of S N 1 consists of two steps. The first step is the slow ionization of the substrate and is rate determining step. The second step is the rapid reaction between the carbocation and nucleophile. As the rate determining step involves only one molecule (it is alkyl halide), so it becomes an unimolecular process. RDS

R — X  → R + + X − +

−

Step 1 (slow)

R + Y → R — Y Step 2 (fast) Moreover, the order of reaction is also equal to 1. So, the rate expression of the reaction is, Rate = k1[RX]; k1 is rate constant of first order reaction. The ionization is always assisted by solvent, since the energy of activation required for breaking of the bond is largely recovered from solvation of ions produced. For example, ionization of t–BuCl in gas phase requires about 150 kcal/mol but in water it requires only 20 kcal/mol. So, a polar solvent facilitates S N 1 reaction. Moreover, in pure S N 1 reactions, solvent molecules assist the departure of leaving group from the front side. As the reaction involves formation of carbocation, it is expected that only those alkyl halide par take in S N 1 mechanism which can yield a stable carbocation. The carbocation generated by first step has an sp2 hybridized carbon i.e. the structure is trigonal planar. Thus nucleophile will attack the carbocation from the front side as well as from the rear side with equal ease, leading to the formation of two isomers, if the chiral carbon is present in the substrate.

O-356

3.2 Stereochemical Aspects of S N 1Mechanism Let us understand the mechanistic details of S N 1 reaction in the following manner keeping the spatial arrangement of groups in mind, R' C R' C R

R"

–

R'

Br

slow r.d.s. –Br–

C+ R

R

OH front side attack OH

–

R"

OH

R'

R" front side attack

C HO

R"

R

Fig. 3: Stereochemical view of S N1 mechanism

δ

δ

(CH3)3C---Cl

⊕

⊕

A planar carbocation is formed during the slow rate determining step of the reaction, thus it may be expected − that the nucleophilic attack by OH (or solvent H2O) would occur from either side of the carbocation leading to the formation of 50/50 mixture of substituted products having same and opposite configuration as that of starting material. It means that the reaction is accompanied by racemisation leading to an optically inactive (±) product. However, in practice, the complete racemisation is never observed. It is always accompanied by some degree of inversion. The relative proportion of the two depend on (a) the stability of carbocation (b) the ability of solvent as nucleophile (c) the ionizing ability of the solvent and (d) the bulkiness of the groups on the substrate. Till the formation of free carbocation, the nucleophile attacks from the rear side of the compound leading to inverted product while once the carbocation is formed the probability of attack from both sides are equally feasible. As a result, overall inverted product exceeds the retained product. So, it is a case of racemisation with inversion of configuration. The energy profile of the reaction can be shown as,

δ

δ

(CH3)3C---OH ⊕

Potential energy

E2a E1a (CH ) C 3 3

(CH3)3CCl (CH3)3COH Progress of the reaction Fig. 4: Energy profile of SN1 reaction

O-357

3.3 S N 2 Reaction Mechanism S N 2 stands for substitution nucleophilic bimolecular. In this mechanism, there is a backside attack of the nucleophile. The nucleophile approaches the substrate from a position 180° away from the leaving group. The reaction is a one–step process with no intermediate. As both the molecules (alkyl halide as well as nucleophile participates in the step) the molecularity of the reaction becomes two. Moreover, the order of reaction is also equal to 2 as suggested by its rate law expression, Rate =k2[RX][:Nu—] k2 is rate constant of first order reaction. In the mechanism below, the C — Y bond is formed as C — X bond is broken, Y– +C – X RDS

δ–

[Y

δ–

C

X]#

Y – C + X–

Transition state Fig 5: SN2 mechanism

The energy necessary to break the C — X bond is supplied by the simultaneous formation of C — Y bond. The position of atoms at the top of the curve of activation energy can be represented as transition state (TS). The group X must leave as Y comes in, because at no time the carbon atom can have more than 8 electrons in its outermost shell. In the transition state, the carbon is sp 2 hybridized. One lobe of unhybridised p orbital is available for overlapping with an orbital of the incoming Nu − while the other lobe overlaps with an orbital of the leaving group X − . These overlaps account for the partial bonds (shown by dotted lines) drawn in figure 5. The reaction is initiated by Nu − beginning to overlap with the small lobe (tail) of the sp 3 hybrid orbital bonding with X. In order to provide more bonding volume to give a stronger bond, the tail becomes the larger lobe (head) and the head becomes the tail, inverting the configuration of carbon. The configuration of the original compound is the opposite to that of the compound obtained. As Nu − starts to bond to carbon, it loses some of its full charge and in the transition state has a δ − charge, as does X as it begins to leave as an anion. The corresponding energy profile is shown below,

O-358 In S N 2 mechanism, the front side attack has never been observed. In a hypothetical front side attack, both the nucleophile and nucleofuge or leaving group would have to overlap with the same lobe of p–orbital whereas the back side attack involves the maximum amount of overlap throughout the course of reaction. During the transition state, the three non–reacting groups and the central carbon atom are approximately coplanar. R Nu:

R

δ– X

C

δ–

#

R Nu

C

Nu

H

H

D sp3 hybridized

:X

–

H D

D

sp3 hybridized with p orbital S Configuration

+

C

Transition State

sp3 hybridized R Configuration (Nu and X of same priority)

Fig. 7: Lobal arrangement of carbon in S N2 mechanism

In the above representation it is clear that the carbon atom is linked to five groups so it is highly overcrowded. Due to high overcrowding it is easier to understand that bigger the groups attached to carbon, more unstable will be the transition state. Hence steric factors will have important role in S N 2 substitution. Due to this reason the reactivity towards S N 2 is as follows. H

R

H

H

R–C–X < R–C–X 2° > 1° > Me

Me > 1° > 2° > 3°

(h) Nature of Leaving group

+

Inversion

Weakest base is best leaving group, Weakest base is best leaving group, i.e.. i.e., I − > Br − > Cl − > F − I − > Br − > Cl − > F −

(i) Solvent effect

Rate α H–bonding ability and dielectric constant

(j) Determining factor

Stability of R

(k) Rearrangement

Observed

Not observed, except for allylic

Lewis and Bronsted acids:

No specific catalyst

(l) Catalysis

+

Ag , AlCl3 and strong HA +

Depends on charge type.Polar aprotic solvents leaves "freest" most reactive Nu. Steric hindrance

O-361 The energy profile of the two reactions can be compared as below,

⊕

[SN2] [SN1]

C L

⊕

Free energy

First step in the RDS for [S N1]

C=L + –

Nu

–

C–Nu + L

Progress of the reaction Fig. 9: Comparision of energy profiles of SN1 and SN2 reactions

Example 5:

CH3 CH3 – C – CH2Br

CH3

C2H5O– SN2 C2H5OH SN1

Identify the products of both reactions. Solution : In SN2 reaction, the rate determining step involves the simultaneous bond making and bond breaking through a transition state. Hence there is no scope of rearrangment and the product in neopentyl ethyl ether.

SN1 reaction involves the formation carbonium ion in the rate determining step. It can undergo rearrangement for better stability as follows.

O-362

Hence, the rearranged product will be obtained.Since the carbonium ion is sterically hindered, elimination will occur at a faster rate than the nucleophile attack by C2H5OH. So the alkene will be the major product. Example 6: When CH3 — CH = CH — CH2Cl reacts with alcoholic KCN, a mixture of isomeric products is obtained. Explain. Solution : It can undergo both SN1 and SN2 reaction. By SN2 reaction only one product is formed. But by SN1 reaction, intermediate is carbonium ion, which may lead to rearrangment for better stability. CH3 CH CH CH3–CH=CH–CH2Cl+CN–

SN2 slow

δ–

Cl

C

δ–CN

H H Transition state

slow SN1

fast ⊕

CH3–CH=CH–CH2

CH3–CH–CH=CH2 ⊕

CN– 2°Carbonium ion (more stable)

CN–

CH3–CH=CH–CH2CN (A)

CH3–CH=CH–CH2CN

(minor)

(A)

CH3–CH–CH=CH2 (B)

Cn

(major)

Thus we get two isomeric products using SN1 reaction.

3.5 Ambident Nucleophiles Some nucleophiles have lone pair of electrons on more than one atom and can attack through more than one site. Such nucleophiles are called ambident nucleophiles. In such cases different products due to attack by different sites are possible. Attack by a specific site can be promoted under special conditions. Two well known examples are discussed in detail. Case 1:Attack by CN − nucleophile (: − C ≡ N :) CN −

R — X  → R – CN + R – NC + X − Nitriles Isonitriles In CN − , carbon atom, which is negatively charged, will be a soft base as compared to nitrogen which is a hard base. So if the reaction proceeds via SN1 mechanism, it produces a free carbocation which is a hard acid. It is very natural that attack takes place exclusively through nitrogen, the hard base. But if the reaction proceeds via SN2 mechanism which yields small positively charged carbon which is a soft acid then attack takes place

O-363 through carbon atom which is the soft base. So if one wishes to increase the relative yield of nitriles, use of NaCN or KCN etc in a less polar solvent is prescribed because these conditions facilitates SN2 substitution. Similarly if one wishes to increase the yield of isonitrile, use of AgCN is prescribed. Ag+ has very high affinity +

for X − so it favours the formation of R and the reaction proceeds via SN1 mechanism. This will result in attack by hard base giving R – NC. Further if one compares primary, secondary and tertiary alkyl halides, formation of R – NC should be favoured due to more favourable SN1 substitution in tertiary alkyl halide. Case 2: Attack by NO 2− nucleophile ( − O– N = O) NO−

2 → R – O – N = O + R – X 

Alkane nitrite

R – NO2 + X − Nitro alkane

In NO 2− , oxygen atom, which is negatively charged, will be a hard base as compared to nitrogen. So if the reaction proceeds via SN1 mechanism, then attack through oxygen atom (hard base) will take place yielding alkane nitrite. But if the reaction takes place via SN2 mechanism then attack takes place through nitrogen (soft base) to yield nitro alkane. If we want to increase the yield of nitro alkane the reaction should proceed via SN2 mechanism. i.e. we can use NaNO2, KNO2 etc. Moreover the yield will be best if we use primary alkyl halide and less polar solvent. Formation of nitrite will dominate if we use tertiary alkyl halide, more polar solvent and AgNO2 because Ag+ has strong affinity for X − and can form a carbocation to force the reaction to proceed via SN1 mechanism. Primary alkyl halide with AgNO2 chiefly gives nitroalkane but if secondary and tertiary alkyl halides are used then AgNO2 will yield nitrite as the major product. Table 2: Comparative details of E2 and SN2 reactions Factor

Favors E2

Favour SN2

(1)Alkyl group

3° > 2° > 1°

1° > 2° > 3°

(2)Leaving group

I − > Br − > Cl − > F −

I − > Br − > Cl − > F −

(3)Reagent

Strong, bulky Bronsted base

Strong nucleophile

(4)Solvent

Favored

Slightly favored

(i)

Low polarity

Disfavored

Disfavored

(ii)

Protic Polar

Strongly favored

Favored

(iii)

Aprotic polar

O-364

Table 3: Comparative details of E2 and SN1 reactions Factor

Favors E2

Favors SN1

3° > 2° > 1°

3° > 2° > 1°

I − > Br − > Cl − > F −

I − > Br − > Cl − > F −

Strong and High

Very weak and low

(1) R group (2) X (3)Base (i)

Strength and

(ii)

Concentration

Example 7: Give the major product (with proper explanation) when following halogen compounds are treated with sodium ethoxide? (i)

CH3–CH–CHCH3 Br

(ii)

(iii)

CH3

CH3

CH2Br CH3

Cl

Solution : (i)

Since sodium ethoxide is a strong base, it has a tendency to abstract proton to give alkene rather than a nucleophilic attack.

O-365 (ii)

H

slow

CH3 ⊕

CH3 Cl

2° carbonium ion (A) +

–H

CH3

⊕

1, 2 hydride shift

CH3 3° carbonium ion (A) C2H5O–

CH3 ⊕

OC2H5

CH3

(iii)

CH2Br CH3

⊕

(minor)

slow

CH2

(major)

1,2 methyl shift

CH2CH3 ⊕

⊕

H

CH3 H H 3° carbonium ion (more stable)

1° carbonium ion

–H+

–H+ ⊕

⊕

∼ ring expansion

CH3

CH2CH3

(minor)

CH3

Example 8: (i) Give the product from the reaction of Mg/Et2O with (a) BrCH2CH2Br, (b) BrCH2CH2CH2Br and (c) BrCH2CH2CH2CH2Br. (ii)

Discuss the difference in behavior of (b) and (c) in part (i).

(iii) Give the product of the reaction of CBr4 + MeLi + cyclohexene. Solution : (i) (a) H2C = CH2 (elimination of vic X's), (b) (ii)

(a)

cyclopropane, (c) BrMg(CH2)4MgBr

Unlike compound (b), compound (c) does not react intramolecularly because a 4 –membered ring does not form as readily as does a 3–membered ring.

O-366 (iii)

halogen metal

CBr4 + MeLi     → [Br3CLi] + MeBr, exchange

Br3CLi → Br2C: + LiBr Dibromocarbene The carbene then adds to the alkene in the typical fashion giving Br Br 7, 7–dibromo [4.1.0]–bicycloheptane

Example 9 : Give the organic products of the following reactions •O• • •

(i)

• • • •

– n–PrBr+••N–O •• i–PrBr+[••SC ≡ N••–] (isocyanate)

(iii)

EtBr+[••SSO3]2–(thiosulfate)

• •• •

• • • •

(ii)

(iv)

ClCH2CH2CH2I + CN − (One mole each) →

(v)

H2NCH2CH2CH2CH2Br   →

− H+ ∆

Solution : The nucleophiles in (i), (ii) and (iii) are ambident since they each have more than one reactive site. In each case, the more nucleophilic atom reacts even though the other atom may bear a more negative charge. n–PrNO2

(iv)

ClCH2CH2CH2CN. I − is a better leaving group than Cl − .

(v)

(ii) i–PrSCN

−

(i)

(iii) [EtSSO3] (with its cation it is called a Bunte salt).

When the nucleophilic and leaving groups are part of the same molecule, an intramolecular displacement occurs if a three, a five– or a six–membered ring can N H

form.

O-367

4. Polyhalogen Compounds Polyhalogenated compounds are organic compounds with multiple substitutions of halogens - F, Cl. Br and I. They are of particular interest and importance because halogens are highly reactive and comprise a superset of which has many toxic and carcinogenic industrial chemicals as members. They are generally non-miscible in organic solvents or water, but miscible in some hydrocarbons from which they often derive. PHCs are highly useful in a vast array of manufactured products, from wood treatments, to cookware coatings, to non-stick, waterproof, and fire-resistant coatings, cosmetics, medicine, electronic fluids, food containers and wrappings, in everything from furniture and furnishings. Two representative examples are chloroform and carbon tetra chloride discussed below one by one.

4.1 Chloroform Chloroform is an organic compound with formula CHCl 3 . It is one of the four chloromethanes yielded post chlorination of methane. The colorless, sweet-smelling, dense liquid is a trihalomethane, and is considered hazardous. Now a days, its use for refrigerants is being phased out. Chloroform volatilizes readily from soil and surface water and undergoes degradation in air to produce phosgene, dichloromethane, CO, carbon dioxide, and hydrogen chloride. In chloroform carbon atom is sp3 hybridised and the tetrahedral structure of the compound is shown below, H C Cl

Cl Cl

Methods of preparation In industry, chloroform is produced by heating a mixture of chlorine and either chloromethane or methane. At 400-500°C, a free radical halogenation occurs, converting these precursors to progressively more chlorinated compounds: CH4 + Cl2 → CH3Cl + HCl CH3Cl + Cl2 → CH2Cl2 + HCl CH2Cl2 + Cl2 → CHCl3 + HCl Chloroform undergoes further chlorination to yield CCl4 as shown below, CHCl3 + Cl2 → CCl4 + HCl

Uses 1.

The major use of chloroform today is in the production of the chlorodifluoromethane, a major precursor to tetrafluoroethylene: CHCl3 + 2 HF → CHClF2 + 2 HCl The reaction is conducted in the presence of a catalytic amount of antimony pentafluoride. Chlorodifluoromethane is then converted into tetrafluoroethylene, the main precursor to teflon.

O-368 2.

Chloroform is also used as a solvent for fats, oils, rubber, alkaloids, waxes, gutta-percha, and resins, as a cleansing agent, grain fumigant, in fire extinguishers, and in the rubber industry. CDCl 3 is a common solvent used in NMR spectroscopy.

3.

As a reagent, chloroform serves as a source of the dichlorocarbene CCl 2 group. It reacts with aqueous sodium hydroxide usually in the presence of a phase transfer catalyst to produce dichlorocarbene, CCl 2 ,a potent electrophile. This reagent affects ortho-formylation of activated aromatic rings such as phenols, producing aryl aldehydes in a reaction known as the Reimer-Tiemann reaction. Alternatively the carbene can be trapped by an alkene to form a cyclopropane derivative. In the Kharasch addition chloroform forms the CHCl2 free radical in addition to alkenes.

4.

Chloroform was once a widely used anesthetic. however, it was quickly abandoned in favor of ether upon discovery of its toxicity, especially its tendency to cause fatal cardiac arrhythmia analogous to what is now termed "sudden sniffer's death".

4.2 Haloform Test Haloform reaction is a characteristic test reaction shown by all methyl ketones. The schematic diagram of the reaction is shown here, O + CH3

R

3I2 + 4NaOH

O R

+ 3NaI + 3H2O + CHI3 (s)

O– Na+

Among aldehydes ethanol is a rare aldehyde to respond to haloform reaction. CH3COR are readily oxidsed by NaOX (NaOH + X2) to haloform, CHX3, and RCO2Na. Even alcohols having CH3CH(OH)- functionality respond positively to haloform test. For example, C2H5CO2– Na+ + CHI3+2OH– Iodoform yellow; mp. 119°

C2H5–C–CH3+3NaOI O

KOCl

(CH3)2C=CH–C–CH3 60°C (CH3)2C=CHCO2K + O

Example 10 : (i)

H2SO4

CHCl3

(CH3)2C=CHCO2H 3–Methyl–2–butenoic acid

Which of the following compounds does not give haloform test?

CH3CH2COCH3

(iii) CH3CHO

(ii) CH3OH (iv) PhCOCH3

Solution : CH3OH will not respond to haloform test as it is devoid of CH3CO– or CH3CH(OH)– structural unit. ∴

(ii)

O-369

Mechanis As stated earlier, methyl ketones are characterized through the haloform test. The mechanism of haloform test is shown below, O

O + CH2

Ph

–

+

OH Ph

H

Br

O

Br

O + H2O CHBr –

Ph

H2O

CH2 –

–

+ CHBr

Ph

OH

Br

H Br

Br

O Br

–

O

+ Ph

CBr2

–

OH

+ CBr2 –

Ph

H2O

H Br •• O ••

••

CBr3

Ph

•• OH ••

O + Ph

O

H

Br

O –

–

OH

+ Br – Ph

CBr3 O

CBr3

Ph

– + H O

CBr3

4.3 Carbon Tetrachloride Carbon tetrachloride, also known by many other names like tetrachloromethane (IUPAC name), is the inorganic compound with the chemical formula CCl 4 . It was formerly widely used in fire extinguishers, as a precursor to refrigerants, and as a cleaning agent. It is a colourless liquid with a "sweet" smell that can be detected at low levels.

Method of Preparation 1.

Carbon tetrachloride was originally synthesized by the French chemist Henri Victor Regnault in 1839 by the reaction of chloroform with chlorine, but now it is mainly produced from methane: CH4 + 4 Cl2 → CCl4 + 4 HCl

2.

The production often utilizes by-products of other chlorination reactions, such as from the syntheses of dichloromethane and chloroform. Higher chlorocarbons are also subjected to "chlorinolysis:" C2Cl6 + Cl2 → 2 CCl4

O-370 3.

Prior to the 1950s, carbon tetrachloride was manufactured by the chlorination of carbon disulfide at 105 to 130 °C: CS2 + 3Cl2 → CCl4 + S2Cl2 The production of carbon tetrachloride has steeply declined since the 1980s due to environmental concerns.

Structure In the carbon tetrachloride molecule, four chlorine atoms are positioned symmetrically as corners in a tetrahedral configuration joined to a central carbon atom by single covalent bonds. Cl C Cl

Cl Cl

Due to this symmetrical geometry, CCl4 is non-polar. Methane gas has the same structure, making carbon tetrachloride a halomethane.

Uses 1.

2. 3. 4.

5. 6.

As a solvent, it is well suited to dissolving other non-polar compounds, fats, and oils. It can also dissolve iodine. It is somewhat volatile, giving off vapors with a smell characteristic of other chlorinated solvents. Historically, carbon tetrachloride was used as a pesticide to kill insects in stored grain, but in 1970 it was banned in consumer products in different countries. Carbon tetrachloride has practically no flammability at lower temperatures. Under high temperatures in air, it forms poisonous phosgene. As it has no C-H bonds, carbon tetrachloride does not easily undergo free-radical reactions. Hence, it is a useful solvent for halogenations either by the elemental halogen or by a halogenation reagent such as N-bromosuccinimide (these conditions are known as Wohl-Ziegler bromination). It is sometimes useful as a solvent for infrared spectroscopy, because there are no significant absorption bands > 1600 cm–1. As carbon tetrachloride does not have any hydrogen atoms, it was historically used in proton NMR spectroscopy. However, carbon tetrachloride is toxic, and its dissolving power is low. Its use has been largely superseded by deuterated solvents.

5. Aryl Halides Aryl halides are compounds containing halogen attached directly to an aromatic ring. The general formula of aryl halides is ArX, where Ar is phenyl, substituted phenyl or any aryl groups. Some of the frequently observed aryl halides are shown below, Cl

I

Br

NO2 Chlorobenzene

Iodobenzene

m–bromonitrobenzene

O-371 Cl

CO2H Br

NO2 p–chloronitrobenzene

o–bromobenzoic acid

Example11: Draw the structures of bromobenzene, fluorobenzene and 2,4-dichloromethyl benzene. Br

Solution:

F

CH3 Cl

Cl Bromobenzene

Fluorobenzene

2,4–dichloromethylbenzene (2,4–dichlorotoluene)

An aryl halide is not just any halogen compound containing an aromatic ring. For example, benzyl chloride is not an aryl halide, because halogen is not directly attached to the aromatic ring. Its structure and properties resemble substituted alkyl halide. The aryl halides differ a lot from alkyl halides in their preparation and properties. Aryl halides are comparatively unreactive towards the nucleophilic substitution reactions, which are characteristics of the alkyl halides. However, the presence of electron withdrawing groups on the aromatic ring greatly increases the reactivity of aryl halides towards nucleophilic substitution reaction. However, in the absence of such groups, reaction can still be brought about by very strongly basic reagents or under drastic conditions. Nucleophilic aromatic substitution follow two very distinct mechanistic pathways. 1.

The bimolecular displacement mechanism–involving the carbanion as intermediate and applicable to activated aryl halides due to the presence of electron withdrawing groups like nitro group.

2.

The elimination addition mechanism–involving the bizarre reaction intermediate called benzyne and applicable to unactivated (due to the absence of any electron withdrawing group) and deactivated aryl halides(due to the presence of electron donating group).

Example12: Why a vinyl halide show an interesting parallelism to aryl halides? Solution: Vinyl halides are the compounds in which halogen is attached directly to a doubly bonded carbon as shown here, |

|

− C = C− X A vinyl halide is equally unreactive towards nucleophilic substitution reactions as an aryl halide, for basically the same reason. Moreover, this low reactivity is caused partly by the partial double bond character of the carbon-halogen bond. The partial double bond character is attributed to resonance as shown below,

O-372

5.1 Structure and Reactivity of Aryl and Vinyl Halides The low reactivity of aryl and vinyl halides towards nucleophilic substitution reaction has been attributed to two different factors, 1.

Delocalization of electrons by resonance and

2.

Differences in σ bond energies due to difference in hybridization of carbon. Cl

Cl

+Cl

+Cl

+Cl

H

H

H (I)

(II)

(III)

(IV)

(V)

Chlorobenzene is considered to be hybrid of not only the two Kekule structures I and II, but also structures, III, IV, and V, in which chlorine is joined to carbon by a double bond. In structures III, IV, and V chlorine bears a positive charge and the ortho and para positions bear a negative charge. In a similar way, vinyl chloride is considered to be a hybrid of structure VI and structure VII (in which chlorine is joined to carbon by a double bond). In (VII), chlorine bears a positive charge and C-2 bears a negative charge. Other aryl and vinyl halides have structures exactly analogous to these.

Contribution from III, IV and V, and from VII stabilizes the chlorobenzene and vinyl chloride molecules, and gives a double bond character to the carbon–chlorine bond. Carbon and chlorine (and halogen, in general) are thus held together by something more than a single pair of electrons, and the carbon chlorine bond is stronger than if it were a pure single bond. It becomes difficult to break a double bond and eventually results into low reactivity. So, it can be said that the low reactivity of these halides towards nucleophilic substitution is due to resonance stabilization of the halides. In alkyl halides, the carbon holding halogen atom is sp3 hybridized whereas in aryl and vinyl halides, carbon is sp2 hybridized. It results into the shorter and stronger carbon–halogen bond. Thus the C-X bond becomes more stable towards nucleophilic substitution which reinforces the low reactivity of aryl and vinyl halides. Chlorobenzene and bromobenzene have dipole moments of only 1.7 D, and vinyl chlorides and vinyl bromides have dipole moments of only 1.4 D. This is consistant with the resonance picture of these molecules. In the structures that contain double bonded halogen (III, IV, V and VII) there is a positive charge on halogen and a negative charge on carbon, to the extent that these structures contribute to the hybrids, they tend to oppose the usual displacement of electrons towards halogen. Although there is still a net displacement of electrons towards halogen in aryl halides and in vinyl halides but it is less than other organic alkyl halides.

O-373

5.2 Preparation of Aryl Halides 1.

From diazonium salts: Benzene diazonium salt is an important compound in the synthesis of various aryl halides. Depending upon the nature of reagent selected various products can be obtained as shown below, BF4– ,∆ HNO

redn.

C6H6 H SO3 C6H5NO2 Sn/HCl C6H5NH2 2 4

HONO C6H5N2+ 0°C

CuCl,∆

C6H5Cl

CuBr,∆

C6H5Br

I–,∆

2.

C6H5F

C6H5I

By Halogenation of arenes or substituted arenes: Benzene can be directly halogenated by the application of dihalogen in the presence of a Lewis acid catalyst as shown here, Lawis acid

ArH + X2  → ArX + HX Here X2 = Cl2 or Br2 , Lewis acid = FeCl3, AlCl3, Tl (OAc)3 etc. For example, NO2

NO2 Cl2.FeCl3

Cl m–chloronitrobenzene NHCOCH3

NHCOCH3 Br2

Acetanilide

Br p–bromo acetanilide (Major product)

Chemical properties of the aryl halides 1.

Low reactivity of aryl and vinyl halides towards SN reaction: An alkyl halide is conveniently detected by the precipitation of insoluble silver halides when it is warmed with alcoholic AgNO3. The reaction occurs instantaneously with tertiary alkyl or benzyl bromides, and within five minutes or so with primary and secondary bromides. But bromobenzene or vinyl bromide does not react at all with alcoholic AgNO3. So, these can be heated with alcoholic AgNO3 for days without the slightest trace of AgBr being detected.

O-374 Ag+

X

OH

–

–

OR

–

–

Aryl halide or – C=C – X Vinyl halide

NH3

No reaction under normal conditions.

CN– ArH, AlCl3

The typical reaction of alkyl halides, is nucleophilic substitution as discussed earlier in the chapter, R – X +:Z − → R – Z + :X − Here,

Z = OH − , OR − , NH3, CN − , NH2− , ROH, H2O etc.

On the other side, aryl halides undergo nucleophilic substitution only with extreme difficulty. Except for certain industrial processes where very severe conditions are applied, one does not ordinarily prepare phenols (ArOH), ethers (ArOR), amines (ArNH2), on nitriles (ArCN) by nucleophilic attack on aryl halides. Moreover, the aryl halides cannot be used in the Friedel-Crafts reaction for arylation. However, aryl halides undergo nucleophilic substitution readily if the aromatic ring is activated by the placement of strongly electron withdrawing groups like –NO 2 , –CF 3 at ortho or para position to the halogen atom. Even unactivated aryl halides can undergo nucleophilic substitution reaction if a strong base is applied at high temperature. In such cases, the reaction proceeds via the intermediate benzyne. For example, Dow's process used for the manufacture of phenol involves benzyne intermediate. 2.

Electrophilic aromatic substitution: Aryl halides can always undergo the typical electrophilic aromatic substitution reactions like nitration, sulphonation, halogenation, Friedel–Craft reactions etc. The unusual feature is that the halogen atom which is attached directly to the benzene ring is deactivating the benzene ring towards an incipient electrophile, yet ortho and para-directing towards disubstitution.

5.3 Formation of Grignard Reagent Aryl halides on treatment with magnesium metal in dry ether yields a Grignard reagent. The reaction scheme is shown below, dry ether

ArBr + Mg   → ArMgBr tetrahydrofuran

ArCl + Mg  → ArMgCl 1.

Electrophilic aromatic substitution: Although halogen is a deactivating group but still it directs the incoming electrophile to ortho and para position as discussed earlier.

O-375 For example, Cl

Cl

Cl NO2

+ HNO3

H2SO4

+

NO2 Cl

Cl

Cl CH3

CH3Cl AlCl3

+

CH3 Cl

Cl

Cl Br

Br2/Fe

+

Br

2.

Nucleophilic aromatic substitution: The nucleophilic aromatic substitution can be done in two possible ways, (i)

Bimolecular addition elimination mechanism

(iii)

Benzyne or elimination -addition mechanism

These two are discussed below one by one.

(i) Bimolecular SNAr or Addition-Elimination mechanism ArX + Z − → ArZ + X − For facile reaction, aryl group must contain strongly electron withdrawing groups at ortho and/or para position to the halogen atom. The reaction involves the formation of intermediate as carbanion. More importantly, the presence of electron withdrawing group stabilizes the carbanion and facilitates the reaction.

Cl

OC2H5 NO2

C2H5OH 300°C, 10 atm

NO2

O-376 Cl

OC2H5 NO2

NO2

C2H5OH 150°C, 5 atm

NO2

NO2 Cl

OEt

O2N

NO2

C2H5OH

O2N

NO2

75°C, 2 atm

NO2

NO2

Reaction proceeds by carbanion intermediate. The rate of the reaction increases with the increase in number of electron withdrawing groups, since the carbanion formed would be stabilized more. The mechanism of the reaction is of addition–elimination type. Step: 1 X X NO2 Nu– slow (RDS)

Nu

Nu

X NO2

••

X NO2

O

Nu ••

N O

••

X

O

Nu N

O

Step: 2 X ••

Nu

Nu NO2 X– fast

NO2

It can be seen that the presence of –NO2 group at ortho or para position would facilitate to disperse the negative charge of the carbanion, thus stabilizing it more and making the reaction to occur fast.

O-377 Example 13: Identify (A), (B), (C), (D) (E) and (F) in the following series of reactions. Br2 aq. KOH (A) (B) hν alc. KOH

Na

+(C) (D) NBS (E)

(C)

(F)

Solution: Br Br2 hν

OH

ONa

aq.KOH

Na

(A)

(B)

(C)

alc.KOH

NBS

Br

ONa

Br (D)

(E)

Williamson synthesis

+

O

(E)

(C)

(F)

Example 14: 1–Fluoro–2,4–dinitrobenzene is highly reactive toward nucleophilic substitution through an SNAr mechanism. What product would be formed when 1– fluoro–2,4–dinitrobenzene reacts with each of the following reagents? (i)

CH3CH2ONa

(ii) NH3

(iii) C6H5NH2

(iv) CH3CH2SNa

Solution: (i)

F

OCH2CH3 NO2

NO2 + CH3CH2ONa

NO2

NO2

O-378 (ii)

(iii)

Since lone pair on N–atom of aniline will not be available for nucleophilic attack as it is in resonance with benzene ring. (iv)

F NO2 + C6H5NH2SNa

no reaction

NO2

Due to large size of S–atom, the overall nucleophile becomes a bulky one and attack of it on the given substrate gives a sterically hindered species which is not stable at all. Hence, the substitution will not occur. (ii) Benzyne or elimination: addition mechanism: Ring not activated toward nucleophilic substitution reaction participates through benzyne mechanism. The reaction scheme is shown below, strong base

ArX + :Z −  → ArZ + :X − Some of the examples are shown below,

O-379

Unactivated aryl halide means either no electron withdrawing substitutent is present or it is present at meta position. On the other hand, deactivated aryl halides mean the presence of electron releasing groups at any position. Rate determining step of the reaction is the formation of benzyne intermediate.

Mechanism Br

–

•NH – • 2 fast

NH2

H

–Br–,–NH3 slow step

NH2 ••

NH2

NH3 –NH2– ,fast

Benzyne

Some of the examples are, (a)

(b)

14

Chlorobenzene with chlorine bonded to C gives almost 50% aniline having NH2 bonded to aniline with NH2 bonded to normal 12 C atom.

14

C and 50%

O-380 (c)

CH3

CH3

CH3 NH2

Cl NH2– –Cl,–NH3

(d)

OCH3

NH2–

NH3

OCH3

OCH3 NH2–

NH2– –Cl,–NH3

NH3

NH2

Cl

(e)

CH3

CH3 NH2– –Cl,–NH3

CH3 NH2–

NH3

NH2

Cl

(f)

(g)

CH3

CH3 NH2– –Cl,–NH3

CH3 NH2–

NH3

NH2 Cl

Example15: When 2–bromo–1,3–dimethylbenzene is treated with sodium amide in liquid ammonia, no substitution takes place. Explain this fact as an evidence of elimination-addition reaction. Br

Solution : H3C

CH3

NaNH2 liq. NH3

2–bromo–1,3–dimethyl benzene

no reaction

O-381 In deactivated/unactivated halobenzene, substitution takes place by elimination–addition mechanism, i.e. benzyne mechanism. But for this, an α −H atom is required which is not present in the given compound. Hence, it will not react with NaNH2 to give any substituted product. Example 16: Explain the success of the following reaction where hydride is the leaving group in nucleophilic aromatic substitution (SNAr).

Solution:

In the case SNAr occurs with hydride (H − ) substitution. This is not a normal process. The reaction occurs by following mechanism. Step–I NO2

NO2 OH–

+H

NO2 H

–

NO2 OH

Step–II H − + Fe3+ + H3O+ →Fe2+ + H2 + H2O Oxidizing agent Thus the oxidizing agent ferricyanide makes the H − transfer feasible by oxidizing it to H2 gas in presence of acid. This is an unusual nuclephilic substitution reaction.

O-382 3.

Fittig reaction: It is an extension of Wurtz reaction and consists of heating an ethereal solution of bromobenzene with metallic sodium.

4.

Ulmann biaryl synthesis: Iodobenzene on heating with copper in a sealed tube forms biphenyl. +2CuI

I+2Cu+I

Chloro–or bromobenzene fails to undergo this coupling reaction unless some strong electron withdrawing group (e.g. NO2) is present in the ortho or para position.

6. Synthesis and Uses of DDT DDT is acronym for dichloro diphenyl trichloroethane and its IUPAC name is 2, 2-bis (p-Chlorophenyl) -1,1,1-trichloroethane. D.D.T. is almost insoluble in water but it is moderately soluble in polar solvents. The structural formula of the compound is, Cl

Cl

Cl

Cl

Methods of preparation One important synthetic pathway is being outlined below,

Cl

O-383

Uses of D.D.T: D.D.T. is a powerful insecticide. It is widely used as an insecticide for killing mosquitoes and other insects.

Hazards of D.D.T: D.D.T. is not biodegradable. Its residues accumulate in environment and its long term effects could be highly dangerous. It has been proved to be toxic to living beings. Therefore, its use has been abandoned in many western countries. However, inspite of its dangerous side effects, D.D.T. is still being widely used in India due to non-availability of other cheaper insecticides. Example 17: Full name of DDT is (i)

1, 1, 1-trichloro-2, 2-bis(p-chlorophenyl) ethane

(ii)

1, 1-dichloro-2, 2-diphenyl trimethylethane

(iii) 1, 1-dichloro-2, 2-diphenyl trichloroethane (iv)

None of these.

Solution: (iii)

7. Synthesis and Uses of BHC Benzene hexachloride was first prepared in 1825; the insecticidal properties were identified in 1944. Benzene hexachloride (BHC) is any of several stereoisomers of 1,2,3,4,5,6-hexachlorocyclohexane. The structural formula of the compound is, Cl Cl

Cl

Cl

Cl Cl

The Y-isomer is about 1,000 times more toxic than any of the other diastereomers formed in the reaction. The structural differences between these individuals are in the orientations of the chlorine atoms with respect to the ring of carbon atoms. The chemical addition of chlorine to benzene produces a mixture of several stereoisomers of 1,2,3,4,5,6-hexachlorocyclohexane. The Y-isomer, which makes up 20-25 percent of this mixture, is more soluble than the other isomers in certain solvents and can be separated from them. More volatile than D.D.T., BHC has a faster but less protracted action upon insects.

Hazards of BHC Lindane has been shown to accumulate in the food chain. This occurs because animals, including humans, eat foods grown in lindane-contaminated soils, and fishes and other marine life are exposed to lindane-contaminated waters. In fishes and mammals, exposure to high levels of lindane may cause acute poisoning, which is evidenced by nervous system dysfunction. Chronic exposure may adversely affect liver

O-384 function in humans. Use of lindane as an insecticide has been banned in many countries. Topical use in lotions to combat lice is permitted. Example 18: The halide which undergoes nucleophilic substitution most readily is (i)

p – H3C C6H4Cl

(ii) o – MeOC6H4Cl

(iii) p – ClC6H4Cl

(iv) C6H4CH(Cl)CH3

Solution : Alkyl halides undergoes nucleophilic substitution faster than aryl halides. ∴

(iv)

Example 19: In order to obtain (CH3)3CCl(g), which reactants are most suitable? (i)

(CH3)3CBr + HCl

(ii) (CH3)2C = CH2 + Cl2

(iii) (CH3)2C = CH2 + HCl

(iv) (CH3)2CO + CH3OH

Solution : Alkenes in presence of an acid generates a carbocation which will be attacked by nucleophile Cl − to give the required product. Cl −

H+ Cl −

(CH3)2C = CH2  → (CH3)2C ⊕ – CH3  → (CH3)3C – Cl ∴

(iii)

Example 20: Formation of CH3CH = CHCH2Br on treatment of 1–butene with N–Bromo succinimide is an example of (i)

Hoffmann rearrangement

(ii) 1,3–shift

(iii) Allylic rearrangement

(iv) α −chlorination

Solution : NBS is a reagent used selectively to carryout allylic bromination. NBS

•

•

CH3 – CH2 – CH = CH2  → CH3CH –CH=CH2 ↔CH3CH=CH–C H2

Such substitutions are referred as allylic rearrangements. ∴

(iii)

Example 21: In the reaction of p–chlorotoluene with KNH2 in liq. NH3, the major product is (i)

o – toluidine

(iii) p – toluidine

(ii) m – toluidine (iv) p – chloroaniline

O-385 Solution : Cl

CH3

NH2

–

NH2

KNH2 –NH3, –Cl

CH3

NH2

+NH3

CH3

CH3 m–toluidine

The position of -NH2 group is controlled by +I effect of -CH3 group. ∴

(ii)

Example 22 : Following halide(s) does not give precipitate with alcoholic silver nitrate (i)

Ethyl iodide

(iii) Vinyl chloride

(ii) Tert butyl chloride (iv) n–propyl bromide

Solution : Precipitation of silver halide could take place if halogen is expelled from the halide readily, which will be feasible only when there is a single bond between R and X. In vinyl chlorides, there are double bond characteristics between C — Cl bond due to resonance, which makes it difficult to be expelled, thus precipitation does not occur with it.

∴

(iii)

Example 23 : A carbon compound (A) forms (B) with sodium metal and again (A) forms (C) with PCl5 but (B) and (C) form diethylether. Therefore (A), (B) and (C) are (i)

C2H5OH, C2H5ONa, C2H5Cl

(iii) C2H5OH, C2H6, C2H5Cl

(ii) C2H5Cl, C2H5ONa, C2H5OH (iv) C2H5OH, C2H5Cl, C2H5ONa

Solution : The compound reacting with sodium would be an alcohol (A) which gives alkoxide (B) with it. Alcohol reacting with PCl5 gives an alkyl halide (C). (B) and (C) reacts by nucleophilie substitution to give diethyl ether. Since diethylether is a symmetrical ether, so both the alkyl parts of (B) and (C) should be ethyl.

∴

(i)

O-386 Example 24: Which chloroderivative of benzene among the following would undergo hydrolysis most readily with aq.NaOH to furnish the corresponding hydroxy derivative? NO2

(i) O2N

(ii) O2N

Cl

Cl

NO2

(iii)

(iv) C6H5Cl Me2N

Cl

Solution : The nucleophilic substitution in aryl halides is assisted by the presence of electron withdrawing groups at ortho para–positions. More is the number of electron withdrawing groups, more easier would be the hydrolysis. (i) ∴ Nu |

Example 25 :

CH 3 − C H − CH 2 − X gives the CH 3 − C − CH 3 on treatment with a |

|

CH 3

CH 3

nucleophile. In which condition, the yield of product will be maximum (i) H 2O as a solvent, X = Cl (ii) CHCl 3 as a solvent, X = Br (iii) H 2O as a solvent, X = I

(iv)

CHCl 3 as a solvent, X = Cl

Nu Nu −

|

Solution : CH3 − CH − CH2 − X  → CH3 − C − CH3 + X − |

CH3

|

CH3

As it is evident that reaction involves rearrangement, this means that reaction proceeds via carbocation formation. The carbocation formation is possible by the ionization of given alkyl halide. The ionization is assisted by a higher polar solvent like H2O and a weaker base as the leaving group like I − . (iii) ∴ Example 26: A hydrocarbon (A) having molecular weight 70 gives a single monochloride but three dichlorides on chlorination in the presence of ultra violet light. The hydrocarbon (A) is (i) 2–pentene (ii) cyclopentane (iii) 2–methyl–2–butene (iv) methylcyclobutane Solution: Since the hydrocarbon (A) gives only a single monochloride, it implies that all the C—H bonds must be of the same type. So the given hydrocarbon should be cyclopentane.

∴

(ii)

O-387 Example 27: In the following groups, (I)

–OAc

(II) –OMe

(III) –OSO2Me

(IV) –OSO2CF3

the order of leaving group ability is (i)

(I) > (II) > (III) > (IV)

(ii)

(IV) > (III) > (I) > (II)

(iii) (III) > (II) > (I) > (IV)

(iv)

(II) > (III) > (IV) > (I)

Solution: The anions of strong acids, i.e. weak conjugate base always act as good leaving groups particularly in polar solvent. Since the decreasing order of acidic strength is following. F3CSO3H > CH3SO3H > CH3COOH > CH3OH The order of leaving group ability is F3CSO −3 > CH3SO −3 > CH3COO − >H3CO − ∴

(ii)

Example 28:

Identify (Z) in the following reaction series, NaOH(aq.)

Al O

Cl / H O

2 3 2 2 CH3.CH2CH2Br  → (X)   → (Y)   → (Z)

heat

Mixture of CH 3 − C H − C H 2 and CH 3 − C H − C H 2

(i)

(ii)

|

|

Cl

Cl

|

|

OH

Cl

CH 3 − C H − C H 2 |

|

OH

Cl

(iii) CH 3 − CH − C H 2 |

Cl

(iv)

|

OH

CH 3 − C H − C H 2 |

|

Cl

Cl Al O

aque . NaOH

2 3 Solution: CH3CH2CH2Br  → CH3CH2CH2OH   → CH3CH = CH2

∆

(X) Cl / H O

2 2   → CH3 − CH − CH2

| OH

∴

(ii)

| Cl

(Y)

O-388 Example 29: For the reaction, CH3CH CH2CH3

CH3CH=CHCH3

NaNH2

CH2=CHCH2CH3

(X)

(i)

CH3 – CH = CH – CH3 predominates.

(ii)

CH2 = CH – CH2 – CH3 predominates.

(iii) Both are formed in equal amounts. (iv)

The product ratio is dependent on the halogen (X).

Solution: When alkyl halide undergoes elimination reaction in presence of base, more substituted alkene (Saytzeff's product) will be the major product. But either in presence of bulky base or the poor leaving group may give rise to the less substituted alkene (Hoffmann's product) as the major product. Since base is a non–bulky strong one, the ratio of product is determined by the halogen atom. ∴

(ii)

Example 30: 1–phenylethyl chloride undergoes alkaline hydrolysis to give 1–phenylethanol with (i)

Complete inversion of configuration

(iii) Retention of configuration

(ii)

Racemization plus some inversion

(iv) Complete racemization

Solution:

So there is racemization plus some inversion. ∴

(ii) Me

Example 31: HO

Me

–

H

OH (II)

Et

Me Cl

H Et

OH– H (I)

Me HO + HO

H Et

Et

Steps (I) and (II) are (i)

both SN1

(iii) (I): SN1, (II): SN2

(ii)

both SN2

(iv)

(I): SN2, (II): SN1

Solution: By (I) there is a product with no change in configuration while by (II) there is inversion hence (I) is SN1 and (II) is SN2. ∴

(iii)

O-389 Example 32: Br

CH3

H

(i)

HO

H

OH– SN2

(A). The product (A) is

(ii)

CH3

H

H

(iii) both

CH3

H

H

HO

(iv) none

Solution: Inversion takes place by SN2 reaction. ∴

(ii)

Example 33:

(I) CH3CH2CH2CH2Cl (II) CH3CH2–CH(Cl)–CH3

(III) (CH3)3C – Cl

Increasing tendency for SN1 and SN2 reaction is (A)

SN1: (I) < (III) < (II)

(B)

SN2: (II) < (III) < (I)

(i)

(A) and (B) both are correct

(ii)

Only (A)

(iii) Only (B)

(iv)

Both (A) and (B) are incorrect

Solution: For SN1, order of reactivity is 1° < 2° < 3° For SN2, order of reactivity is 1° > 2° > 3° ∴

(ii)

Example 34: (i)

SN2 reaction at an asymmetric carbon of a compound always gives

An enantiomer of the substrate.

(iii) A mixture of diastereomers.

(ii)

A product with opposite optical rotation.

(iv)

A single stereoisomer.

Solution: SN2 reaction proceeds with inversion of configuration. Since the attacking nucleophile is not same as that of leaving group, the product cannot be enatiomer of the substrate so the product will not necessarily have opposite optical rotation. Moreover only one product is obtained, so we cannot obtain diastereomers. ∴

(iv)

Example 35: (i)

The hydrolysis of 2– bromo 3– methylbutane yields

3–methyl –2– butanol

(iii) 3– methylbutanol

(ii)

2– methyl –2– butanol

(iv) 1– methylbutanol

Solution: The intermediate is a 2° carbocation which rearranges to a more stable 3° carbocation by the shift of hydride.

O-390 CH3

CH3

⊕

CH3 CH3–CH–CH–CH3

CH3CH2–C–CH3 ⊕

1,2hydride CH3–CH–CH–CH3 shift

Br

H2O

CH3 CH3CH2–C–CH3 OH 2–methyl–2–butanol

∴

(ii) NEt 2 |

OH −

Example 36: CH 3 − CH − CH 2   → ? |

Cl

The product of the above reaction is (i)

CH3 CHCH2NEt2

(ii) CH3 CHCH2 OH

(iii)

CH3CHCHNEt2

(iv) CH3 CHCH2 – N– OH

| NEt2

| OH

| OH

Cl

Solution:

Et

Et N OH– CH3–CH–CH2 CH3CH–CH2NET2 ⊕

NEt2 CH3–CH–CH2

| Et

–Cl–

OH

Cl

This is an example of neighbouring group participation. ∴

(i)

Example 37:

C H OK/C H OH

2 5 2 5 → ? CH3CH2CH(F)CH3  

The product of the above reaction is (i)

CH3CH2CH = CH2

(iii) CH3CH2CH(OEt)CH3

(ii) CH3CH = CHCH3 (iv) None of the above

Solution: Since F − is a poor leaving group, attack of base (C2H5O − ) will form less substituted alkene. ∴

(i)

O-391

Long Answer Type Questions 1.

How does S N 1 reaction differ from S N 2 and discuss S N 1 mechanism.

[Bund. 2009; Hazaribagh 2009]

2.

3.

7.

Explain the mechanism of S N 2 and S N 1 reaction and also discuss the stereochemistry of these 8. reactions. [Bund. 2010; Meerut 2009, 12]

(ii)

Starting with cholorbenzene, how will

p- methoxy - benzyl chloride

(c)

p - nitro benzyl chloride.

(i)

Write

four

methods

chlorobenzene? (a)

Phenol

(b)

Aniline

(c)

Phenyl cyanide

(d)

D.D.T.

[Hazaribagh 2008]

Short Answer Type Questions

(a)

Aniline

(ii)

(b)

D.D.T.

(iv) Benzene

1.

Phenol

(i)

Explain why chloroform is stored in dark coloured air tight bottle.

(ii)

Bleaching powder reacts with acetone. [Bund. 2009]

2.

Write short note on the following :

S N 1 reaction,

(ii)

Chlorination of toluene in presence of

BHC

Write chemical reactions only (i)

(i)

Oxidation of chloroform in presence of air and sun light.

(ii)

sunlight

6.

[Lko. 2008]

Reimer-Tiemann reaction.

3.

Write a short note on BHC.

4.

Discuss the preparation and properties of D.D.T.

(i) Which one of the following is an example of S N 2 reaction? Give the product and outline the mechanism of

(a)

H3C - CH = CH - CI + NaNH2 →?

(b)

(CH3)3 C - Br + C2H5 OH → ?

(c)

CH3. CH2 - CH2 - Br + LiAiH4 → ?

[Bund. 2010]

[Meerut 2012; Agra 2008]

5.

Why

benzyl

chloride

undergoes

nucleophilic

displacement easily like allyl chloride? [Agra 2008]

reaction :

(ii)

the

How will you prepare the following from

[Kashi 2010, 11]

(iii)

for

you

obtained the following :

5.

[Lko. 2009]

preparation of Chlorobenzene.

Differentiate between S N 1 and S N 2 reaction.

(i)

(b)

[Lko. 2010]

[Kanpur 2011]

4.

Benzyl chloride

Explain with examples S N 1 and S N 2 reactions.

(ii)

Describe the action of chlorine on toluene under difference experimental conditions. Name different products formed and short also how they behave towards (i) Caustic soda and (ii) Oxidising agents?

(a)

6.

How will you convert the following ? (i)

C6H5CH2Cl into C6H5CH2OC2H5

(ii)

C6H5Br into C6H5CH3

Arrange the following compound in order of 7.

Write short notes on the following :

decreasing reactivity towards S N 1 reaction and justify:

(i)

Westron (CHCl2 - CHCl2)

(ii)

Freon (CCl2F2)

[Kanpur 2010]

[Kanpur 2011]

O-392 8.

Discuss the mechanism of S N 2 reaction.

3.

[Avadh. 2008]

9.

What is Haloform reaction?

10.

Give the mechanism of S N 1 reaction of alkyl

following (i)

[Avadh. 2009]

halides.

[Avadh. 2010]

Give method of preparation and uses of the Chloretone

(ii)

D.D.T [Kanpur 2009]

4.

Write note on iodoform reaction.

5.

How can you convert the following C2H5Cl into T.E.L?

11.

How

will

you

synthesize

chlorobenzene,

bromobenzene, iodobenzene and fluorobenzene

6.

7. 12.

[Kanpur 2010]

How will you distinguish between benzyl chloride and p-chlorotoulene ?

from benzene diazonium salt? Explain haloform reaction and its mechanism

[Kanpur 2010]

Give method of preparation and uses of the following : (i) D.D.T.

[Rohd. 2010]

[Kanpur 2009]

(ii) Westron [Kanpur 2010]

Describe mechanism of S N 2 giving energy profile 8. diagram. [Rohd. 2012]

Chlorine of methyl chloride is more reactive than

14.

Write a note on Walden inversion.

Convert: Ethyl alcohol into chloroform.

15.

Write short notes on the synthesis of DDT and BHC.

13.

[Rohd. 2012]

[Rohd. 2013]

9.

chlorine of chlorobenzene. Why?

[Kanpur 2011]

10.

Why vinyl choride is inert towards nucleophilic substitution reaction?

16.

17.

Discuss

[D.D.U. 2008]

the

mechanism

of

1-bromomethane with aq. NaOH. 19.

Define haloform reaction.

20.

Define Ulmann reaction.

21.

11.

hydrolysis

13.

14.

[Hazaribagh 2009]

Explain that chlorine of vinyl chloride is inert.

Cl

the

preparation

of

[Kashi 2011]

Allyl chloride hydrolyses more rapidly than tertiary butyl halide. Why?

[Lko. 2009]

Allyl halides are more reactive than alkyl halides

Write a note on the synthesis and used of DDT and Lindane.

15.

[Lko. 2011]

CH3Br and (CH3)3CBr hydrolyze involving different mechanisms. What are these mechanisms? Explain. [Lko. 2011]

16.

Explain S N 2 nucelophilic substitution reaction in alkyl halide. [Hazaribagh 2010]

17.

Explain why chlorine of vinyl chloride in inert. [Hazaribagh 2010]

18.

Give the IUPAC name of following compounds : BrCH 2 − CH 2 − CH 2 − CHO and

of

[Lko. 2011]

[Hazaribagh 2009,11]

Discuss the factors which affect S N 1 and S N 2 reactions. [Hazaribagh 2011]

CH 3 CH − CH 2 − CH 2 − Br |

methods

[D.D.U. 2010]

[Meerut 2011]

2.

two

towards nucleophilic substitution reaction. Why?

of

Very Short Answer Type Questions 1.

Write

chlorobenzene.

Discuss the mechanism of S N 2 reaction by taking 12. the example of hydrolysis of ethyl bromide by aq.NaOH.

18.

[Kashi 2011]

Cl atom of ethyl chloride is more reactive than Cl atom of vinyl chloride. Explain why? [D.D.U. 2008]

[Kanpur 2010]

Why are S N 1 and E1 competing reactions? [Hazaribagh 2011]

19.

Distinguish between the transition state and an intermediate.

[Kanpur 2009]

20.

Classify haloalkanes.

[Hazaribagh 2011]

O-393 7.

Objective Type Questions Multiple Choice Type Questions 1.

2.

3.

4.

5.

Which compound has the lowest boiling point (a)

n-pentyl chloride

(b)

n-butyl chloride

(c)

sec-butyl chloride

(d)

t-butyl chloride

8.

CH2 = CH – CH2Cl + OH − → (A). The compound 9. (A) in the reaction is (a)

Ethyl chloride

(b)

Ethyl alcohol

(c)

Ethylene and methyl alcohol

(d)

Allyl alcohol

10.

If 1,3–dibromopropane reacts with zinc, the product is (a)

Propene

(c)

Cyclopropane (d) Hexane

Which of the following is an organometallic compound? (a)

Lithium methoxide

(b)

Lithium acetate

(c)

Lithium dimethylamide

(d)

Methyl lithium

(CH3)3CMgCl on reaction with D2O produces (a)

(CH3)3CD (b) (CH3)3COD

(c)

(CD3)3CD (d) (CD3)3COD

The exchange reaction given below Cl – CH2 – CH = CH2  Nal →I – CH2 – CH = CH2 Acetone

Ethylidene chloride on treatment with aqueous KOH gives (a)

acetaldehyde

(b) ethylene glycol

(c)

formaldehyde

(d) none of them

+ NaCl↓ is known as

Which compound reacts fastest by S N 1 mechanism? (a)

(CH3)2CHI

(b) CH3CH2I

(c)

(CH3)3CI

(d) CH3CH2CH2I

(a)

Wurtz reaction

(b)

Fittig reaction

(c)

Feinklestein reaction

(d)

Koble's reaction

CH2

11.

RCl + NaOH (aq) → ROH + NaCl HBr (A) ROOR

The rate law for this reaction is given by Rate = k' [RCI] The rate of the reaction will be (a)

6.

(b) Propane

Theproduct (A) is

doubled on doubling the concentration of sodium hydroxide

(b)

halved on reducing the concentration of alkyl halide to one-half

(c)

increased on increasing the temperature of the reaction

(d)

unaffected by increasing the temperature of the reaction

(a)

CH2Br

(b)

CH3 Br

In the reaction, Acetylene + HBr (excess) → ? The product is (a)

Vinyl bromide

(c)

Both are correct

(b)

Ethyl bromide

(d)

None is correct

(c)

Ethylene bromide

(d)

Ethylidene bromide

12.

What is the end product of the following sequence of reaction?

O-394 OH H2SO4 ∆

(a)

BrCH2CH = CHCH2Br

(b)

CH3CH =CH – CHBr2

(c)

CH3 – C(Br) = C(Br) –CH3

(d) 13.

(A)

2NBS

Product 18.

None of the above

When 3– ethyl –2– pentene reacts with HBr in the presence of peroxide, the major product is 19.

Br H

(a)

CH3–CH–C–CH2CH3

(b)

14.

Methyl chloride

(c)

Vinyl chloride

(d)

Propyl chloride

[Agra 2008]

What is full name of DDT (a)

Diphenyl diethyl trichloroethane

1.

Both of these

(d)

None of these

S N 2 reaction at an asymmetric carbon of a compound always gives its ………. .

OH

(c)

[Agra 2008]

Fill in the Blank

2.

OH Br

17.

Ethyl chloride

(b)

None of these

C2H5

Which is inert towards nucleophilic substitution reaction? (a)

CH2 = CHCl

(b) CH2 = CH – CH2Cl

(c)

C6H5CH2Cl

(d) C6H5CH2CH2Cl

To prevent oxidation of chloroform one of the following is added : CH3Cl

(a)

(d)

When 3– ethyl –2– pentene reacts with Br2/H2O, the 3. major product is 4. C2H5 (a) 5.

(a)

Formation of PVC takes place from which compound?

Br

(c) CH2Cl2

(b) C2H5OH (d) CH3COCH3 [Bund. 2008] Which will be more reactive towards nucleophilic substitution?

Vinyl chloride is ……… reactive than ethyl chloride towards nucleophilic substitution reaction. Benzyne is ……… in nature.

S N 1 reaction is accompanied with ………. . Butane on dichlorination yields………… structural isomers.

6.

Pentane on monobromination yields ……… chiral carbon centres.

7.

Out of –OAc and –OMe, ……… is a better leaving group.

8.

Chlorine is a/an ……… director to nitroniumion.

9.

Bromobenzene is ……… reactive than allyl bromide towards nucleophilic substitution reaction.

10.

In the reaction of p–chlorotoluene with KNH2 in liq. NH3, the major product is ………. .

CH3–CH–C–CH2CH3

16.

(d) HC≡ C–Cl [Agra 2008]

Diamine dichloro trichlorobenzene

CH3–CH–C–CH2CH3

15.

C6H5–Cl

(c)

None of these

(b)

(c)

Dichloro diphenyl trichloroethane

Both of these

Br

(b) CH=CH–Cl

(b)

C2H5 (d)

CH3CH2Cl

C2H5

CH3CH2–C–CH2CH3 (c)

(a)

O-395

True/False

6.

1.

Order of reactivity of alkyl halides for S N 1 7. mechanism is 1° >2° > 3°.

2.

Allylic and benzylic halides are more reactive towards nucleophillic substitution. 8.

3.

Preperation of alkyl chloride in the presence of SOCl2 from alchohol is preferred.

4.

Reaction of Br2 /CCl4 with alkene is one of the 9. method to test unsaturation in the compound. For the same alkyl group, boiling points of alkyl halide increases in the order RF > RCl > R Br > RI. 10.

5.

Due to the polar nature of haloalkanes, they are highly soluble in water. In S N 1 mechanism, rate of reaction depend upon the concentration of nucleophile. According to Saytzeff rule, " In dehydrohalogena– tion reactions, the preferred product is that alkene which has gretaer number of alkyl groups attched to the doubly bonded carbon atoms." Aqueous and alchoholic KOH both are used to convert alkyl halides to alchohols. Aryl halides are less reactive towards nucleophillic substitution reactions.

Objective Type Questions Multiple Choice Questions 1.

(d)

2.

(d)

3.

(a)

4.

(c)

5.

(b)

6.

(d)

7.

(c)

8.

(d)

9.

(a)

10.

(c)

11.

(a)

12.

(a)

13.

(a)

14.

(a)

15.

(a)

16.

(c)

17.

(a)

18.

(c)

19.

(b)

Fill in the Blank 1.

enantiomer

2.

Less

3.

aromatic

4.

racemisation with inversion

5.

9

6.

3

7.

–Oac

8.

O/P

9.

less

10. m–Toluidine

True/False 1.

False

2.

True

3.

True

4.

True

5.

False

6.

False

7.

False

8.

True

9.

False

10.

True

O-396

Hints and Solutions Short Answer Type Questions 6.

C H O − Na +

(i)

C6 H 5 CH 2 − Cl  2 5  → C6 H 5 CH 2 − O − C2 H 5

(ii)

C6 H 5 Br + CH 3 Br

Na 1 C6 H 5 CH 3 dry ether Cl

11.

CuCl Br CuBr

+ N2

I

KI

HBF F

Very Short Answer Type Questions 6.

AgNO3 yields white curdy ppt with CH2–Cl

9. 20.

CH 3 − CH 2 − OH

I

2+

NaO4

CHI 3 + HCOONa

H

R

R

R–C–X

R–C–X

R–C–X

H 1°

H 2°

R 3°

Multiple Choice Questions 6.

(d) It is an electrophilic addition reaction involving 2 moles of HBR.

15.

(a) Vinyl halides are reluctant to participate in S N reaction. mmm