A Textbook of Organic Chemistry for JEE Main and Advanced [3, 1 ed.] 9312147013, 9789312147016

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E

arihant

A TEXTBOOK OF w

ORGANIC CHEMISTRY for JEE Main & Advanced

>*

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A TEXTBOOK OF

©RQADJOG CHERflflSTRV for JEE Main & Advanced

Dr. RK Gupta

;{carihant ARIHANT PRAKASHAN (Series), MEERUT

arihant ARIHANT PRAKASHAN (Series), MEERUT All Rights Reserved

« ©AUTHOR !

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PREFACE Success demands a lot from you, more so in the present world of cut-throat competition and information revolution. Every year, a large number of candidates compete for a handful of seats in the IITs, and some other prestigious engineering institutions of India. But to add to the worry, frequently, the formats of these entrance tests keep on changing. As a result, the competition becomes really tough. To come out of these tests in flying colours, one needs to have not only indepth knowledge, but also requisite skills to use the acquired knowledge efficiently, and effectively. Keeping in mind the recent changes in the pattern of examinations, such as JEE Main & Advanced here we are with a complete revised and enlarged New Edition of this popular book of 'Organic Chemistry’. The overwhelming responses to the previous editions of this book has been more than gratifying and have enticed me with the motivation to go for a thorough revision, thereby to make it more useful to the student community. The new edition is built on the proven strength of the previous editions, and is definite to serve its readers in the most effective way.

Special Points about the book • Topical Tests have been provided with individual topics/concepts to facilitate on spot test /practice on the topic. • The numbers of questions in Objective Exercises have been given in 2 levels; Level I & Level II. •

Questions on latest pattern of JEE Main & Advanced- a) Multiple options correct, b) Passage based questions, c) Assertion-reason, d) Matrix matching, and e) Single integer answer type questions etc., all are given in each chapter.

•

To give the exact insight into the level of toughness of the JEE Main & Advanced, previous years' questions of JEE have been provided chapter-wise.

•

Solutions to objective exercises have been given to ensure the sound understanding of the concepts.

•

More weightage has been given to Reaction Mechanism, Stereochemistry, and the Latest Nomenclature.

•

Functional Group Analysis (Chapter 20), has been added to completely cover the JEE Main & Advanced syllabus.

•

Appendices have been added at the end of the book to ensure a solid knowledge base.

Acknowledgement I would like to thank Shri YC Jain, Chairman Arihant Prakashan, for getting the book published in such a nice form. I also acknowledge the support provided by the esteemed teaching fraternity in preparation of the text, through their invaluable suggestions, and critical comments.

I must thank Almighty God for His inspiration, and benevolence, and my family members for thenpatience, and encouragement, without which this work would have been not possible. I welcome constructive comments and suggestions from our readers for the further improvement

of this book.

Dr. RK Gupta [email protected]

CONTENTS 1. Bonding in Organic Compounds-A Basic Concept Covalent Bonding Valence Bond Theory Formal Charge Hydrogen Bonding Acids and Bases Chapter Proficiency Test

• • • • • •

° • • • •

Sigma and Pi Bonds Hybridisation Dipole Moment Resonance Illustrative Solved Examples

31-60

2. Purification and Elemental Analysis Filtration Sublimation Extraction Qualitative Analysis of Elements Molecular Formula Determination Functional Group Identification

• • • • • •

• • • « • •

Crystallization Distillation Chromatography Qualitative A nalys is of Elements Molecular Formula Determination Chapter Proficiency Test

3. Nomenclature of Organic Compounds • Nomenclature of Organic Compounds • IUPAC Names • Chapter Proficiency Test

Structural Isomerism Relative and Absolute Configurations R and S Configurations Illustrative Solved Examples

91-145 • • • •

Stereoisomerism E and Z Configurations Conformations Chapter Proficiency Test

5. Mechanism of Organic Reactions • • • • • •

Bond Cleavage Inductive Effect Mesomeric Effect Types of Reactions (Kharash Peroxide Effect) Substitution Reactions Chapter Proficiency Test

61-89

• Rules of Writing IUPAC Names °' Illustrative Solved Examples

4. Isomerism • • • •

1-30

147-275 Reagents Electromeric Effect Hyperconjugation Addition Reactions Elimination Reactions Illustrative Solved Examples

277-308

6. Alkanes 0 ° •

Isomerism in Alkanes General Properties Chapter Proficiency Test

• •

Methods of Preparation of Alkanes Illustrative Solved Examples

7. Alkenes Preparation of Alkenes Allylic Substitution Ozonolysis-Determination of Structure by Degradation • Degree of Unsaturation (DU)

0 ° •

309-391 • • • • •

Chapter Proficiency Test Properties/Reactions of Alkenes Addition Reactions Stabilities of Alkenes Illustrative Solved Examples

8. Alkynes • • o •

Preparations Acidic Nature Other Reactions Chapter Proficiency Test

393-431 0 ♦ •

Properties and Reactions Oxidative-Hydroboration Illustrative Solved Examples

433-493

9. Alkyl Halides ° • •

Source/Preparation Reactions of Polyhalide Compounds Chapter Proficiency Test

0 •

Properties/Reactions Illustrative Solved Examples

495-553

10. Monohydric Alcohols • 8 •

i

Preparation and Industrial Sources Distinction Chapter Proficiency Test

• »

Properties and Reactions Illustrative Solved Examples

11. Ethers, Oxiranes and Polyhydric Alcohols 8 Methods of Preparation ® ’ Cleavage of Epoxides 8 Trihydric Alcohols or Triols • Chapter Proficiency Test

• n •

Properties Glycols Illustrative Solved Examples

607-691

12. Aldehydes and Ketones o Preparation/Source • Chemical Reactions • Aldol Condensation • Tischenko Reaction • Polymers of Carbonyl Compounds • Illustrative Solved Examples

• • • • • •

Physical Properties Cannizzaro Reaction Claisen Schmidt Condensation Reformatsky Reaction Distinction Chapter Proficiency Test

13. Carboxylic Acids • « • •

Preparation/Source Acidity of Carboxylic Acids Conversions Chapter Proficiency Test

555-606

693-740 • • •

Properties Distinction Illustrative Solved Examples

741-801

14. Functional Derivatives of Carboxylic Acids Preparations Claisen Condensation Illustrative Solved Examples

Properties Urea Chapter Proficiency Test

♦ * •

803-849

15. Amines Preparations Distinction among 1°, 2° and 3° Amines Distinction among 1°, 2° and 3° Amines • Chapter Proficiency Test

Properties Separation of 1°, 2° and 3° Amines Illustrative Solved Examples

• • •

851-880

16. Amino Acids •

Classification Properties of Amino Acids » Isoelectric Point Illustrative Solved Examples

Synthesis of Amino Acids Acid-Base Properties of Amino Acids Peptides Chapter Proficiency Test

• ® • •

881-900

17. Carbohydrates Classification Configuration Chapter Proficiency Test

• • •

o •

Preparation of Glucose and Fructose Properties

901-1091

18. Aromatic Compounds • ■ • • • •

Hydrocarbons Nitro Compounds Sulphonic Acid and its Derivatives Phenolic Compounds Carbonyl Compounds Chapter Proficiency Test

• Halogen Compounds ■ Carboxylic Acid and Derivatives • Amino Compounds • Diazonium Salts • Illustrative Solved Examples

1093-1115

19. Cycloalkanes ■ General and Nomenclature • Properties • Illustrative Solved Examples • Preparation/Synthesis

• •

Stability of Cycloalkanes (Baeyer Strain Theory) Chapter Proficiency Test

20. Functional Group Analysis • Study of Physical Properties • Functional Groups • Illustrative Solved Examples

1117-1145 * Elemental Analysis • Identification of the Compound ■ Chapter Proficiency Test

Miscellaneous Problems Appendices

Previous Years' Questions JEE Main & Advanced (2017-2010)

1146-1156 1157-1184 1-32

! — 5 1

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, 1■i

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Ik*.J ffli i JI

Bonding in Organic Compounds -A Basic Concept

Introduction A molecule will only be formed if it is more stable, and has a lower energy than the individual atoms. Normally only electrons in the outermost shell of an atom are involved forming bonds and in this process each atom attains a stab electronic configuration of inert gas. We divide elements in three classes.

Covalent Bonding

Sigma and Pi Bonds Valence Bond Theory Hybridisation Formal Charge

Dipole Moment Hydrogen Bonding Resonance Acidsand Bases Illustrative Solved Examples Chapter Proficiency Test

2 | Organic Chemistry

(A) Electropositive elements, whose atoms give up one or more electrons easily. They have low ionisation potentials. (B) Electronegative elements, which can gain electrons. They have higher value of electronegativity. (C) Elements which have little tendency to lose or gain electrons. Types of Bonds

repulsive interaction between the nuclei becomes so strong that it pushes the atoms apart. Thus, there is an optimum point where net attractive forces are maximized and the H—H molecule is most stable. This optimum distance between nuclei is called the bond length. In theH2 molecule, the bond length is 74 pm. On a graph of energy versus internuclear distance, the bond length corresponds to the minimum energy, most stable arrangement (Fig. 1.2). HH (too close)

(i) (ii) (iii) (iv) (v)

A+B B+ B A+ A Electron deficient molecule or ion (Lewis acid) and electron-rich molecule or ion (Lewis base) H and electronegative element (N, O, F)

Ionic Covalent Metallic Coordiijate

H

H (too far)

Hydrogen H-l

Covalent Bonding If duplet (2) or octet (8) is completed by sharing of electrons between two electronegative elements, the bond formed is called covalent bond. Attractive

Electron cloud

Repulsive

Nucleus

Fig. 1.1 A covalent H—H bond is the net result of attractive and repulsive electrostatic forces. The nucleus electron attractions are greater than the nucleus—nucleus and electron—electron repulsions, resulting in a net attractive force that holds the atoms together to form an H2 molecule.

To see how the formation of a covalent, shared-electron bond between atoms can be described, let’s look at the H—H bond in theH2 molecule as an example. As two hydrogen atoms come close, electrostatic interactions begin to develop between them. The two positively charged nuclei repel each other and the two negatively charged electrons repel each other, but each nucleus attracts both electrons (Fig. 1.1). If the attractive forces are stronger than the repulsive forces, then a covalent bond is formed, with the two shared electrons occupying the region between the nuclei. As you might imagine, the magnitudes of the various attractive and repulsive forces between nuclei and electrons in a covalent bond depend on how close the atoms are. If the hydrogen atoms are too far apart, the attractive forces are small and no bond exists. If the hydrogen atoms press too closely together, the

I

I I Bond length (74 pm)

I

| | Internuclear distance

__L >

Fig. 1.2 A graph of potential energy versus internuclear distance for the H2 molecule. If the hydrogen atoms are too far apart, attractions are weak and no bonding occurs. If the atoms are too close, strong repulsions occur. When the atoms are optimally separated, the energy is at a minimum.

Every covalent bond has its own characteristic length that leads to maximum stability. As you might expect, bond lengths are roughly predictable from a knowledge of atomic radii. For example, since the atomic radius of hydrogen is. 37 pm and the atomic radius of chlorine is 99 pm, the H—Cl bond length in a hydrogen chloride molecule is about 37 pm+ 99 pm = 136 pm.(The actual value is 127 pm)

Strengths of Covalent Bonds Look again at (Fig. 1.2), the graph of energy versus internuclear distance for the H2 molecule, and note how the H2 molecule is lower in energy than two separate hydrogen aoms. When pairs of hydrogen atoms bond together, they form lower-energy H2 molecules and release 436 kJ/mol in the process. Looked at from the other direction, 436 kJ must be added per mole of H2 molecules to split them apart into hydrogen atoms.

/fiigheK

^nerg^ H- + -H

H—H

ZlSveA

V " 4, . ... The amount of energy that must be supplied to break a chemical bond in an isolated molecule in the gaseous state—and thus, the amount of energy that is released when the

Bonding in Organic Compounds—A Basic Concept | bond forms—is called the bond dissociation energy D. Bond dissociation energies are always considered to have positive values because energy must be supplied to break a bond. Conversely, the amount of energy released on forming a bond is always a negative value. Every bond in every molecule has its own specific bond dissociation energy. Not surprisingly, though, bonds between the same pairs of atoms usually have similar dissocation energies. For example, carbon-carbon bonds usually have D values of approximately 350-380 kJ/mol regardless of the exact . structure of the molecule. H H H H H

.. Ill

1 Hl

H H Ethane D = 376 kJ/mol

H

H— C— C—H H— C—C—C— H H H Propane D = 356 kJ/mol

• Main-group elements tend to undergo reactions that leave them with eight outer shell electrons. That is, main-group elements react so that they attain a noble gas electron configuration with filled 5 and p sub-levels in their valence electron shell. . electrons shared • Number of bonds between two atoms =------------------2 -- ,—--- —

H

H

H

I I 111 II I I H H H H

H—C—C—C— C— II H Butane D = 352 kJ/mol As a result, we can construct a useful table of average bond dissociation energies (Table 1.1) to compare different kinds of bonds. Keep in mind, though, that the actual value in a specific molecule might vary by ±10% from the average.

The Lewis Theory The Octet Rule For many light atoms a stable arrangement is attained when the atom is surrounded by eight electrons— the octet rule. (In case of H2, duplet is completed) 1. Group 1 A(l) elements tend to lose their ns} valence-shell electron, thereby adopting the electron configuration of the noble gas element in the previous row of the periodic table. 2. Group 2A(2) elements tend to lose both of their ns2 valence-shell electrons and thus, adopt a noble gas electron configuration. 3. Group 3A(13) elements tend to lose all three of their ns2np] valence-shell electrons and adopt a noble gas electron configuration. 4. Group 7A(17) elements tend to gain one electron, changing from ns2np5 to ns2np6, thereby adopting the configuration of the neighbouring noble gas element in the same row. 5. Group 8A( 18) (noble gas) elements are essentially inert; they rarely gain or lose electrons. All these observations can be gathered into a single statement called the octet rule.

'"1

i' h:h

2

1 (single)

H—H

:o: :o:

4

2 (double)

0=0

••

6

3 (triple)

N=N

;o-xh

2 (between H and 0)

1 (two single bonds)

H*x NXH • X H

2 (between H andN)

1 (three single bonds)

H H*x C-XH X• H

2 (between H and C)

1 (four single bonds)

• X

H

3

H

0—H

I

H

H— N— H H

1

H— C— H

H

• All the valence (outer shell) electrons of the atoms in a Lewis structure must be shown in the structure. • Usually, all the electrons in a Lewis structure are paired.

:ci:ci:

or

XI—C1J^Z> lone Pairs

bond pair

*o’*o*

or

:o=o ’

This is actually incorrect since, it does not have unpaired electrons which account for the paramagnetic character of 02 (by MO theory, 02 is said to have two unpaired electrons.) A resonance hybrid of the following two contributing structures has unpaired electrons and also confinns to the fact that the experimentally measured bond length in 02 shows to have some multiple bond character.

:o—o:

:6=o:

• If there are odd number of electrons (15 as in NO), then there is at least one unpaired electron somewhere in the structure, • At least one atom lacks a completed octet of electrons and species is paramagnetic.

4 | Organic Chemistry

complete octet 1

unpaired electron —»• N=O Z

..

•n: .o. • •

1

7 electrons in N complete octet of N and O electrons shared between N and 0 = 5 number of bonds = 2.5

• A mathematical relationship can be used to determine total number of electrons shared : S = N-A where, 5 is the total number of electrons shared in the molecule or polyatomic ion; N is the number of valence shell electrons

needed by all the atoms in the molecule or ion to achieve noble gas configurations (TV = 8 x number of atoms (H excluded) + 2 x number of H atoms); A is the number of electrons available in the valence shell of all of the representative atoms. This is equal to the sum of their periodic group numbers. In C02, A for each O atom is 6 and for C atom is 4.

Thus,

,4=4 + 2x6 = 16 7V=8x3=24

Thus,

5=24-16 = 8

Thus, there are, eight electrons shared to form bonds

:o: :c: :o:

Topical Test 1 Step-Through Questions

1. In the following table, write (a) covalency of each atom (b) structure of the compounds with H I..,!/. I 3 i ' ■

Lewis symbol

i

-

___ 4.

H* »Be*

•B*

i___

• c*

.N.

-7 .0. ••

HI

a.

•FJ

Covalency Compounds ... with H

:x = x: 3. Show bond structure of CH3CH2OH

2. Consider following figure representing potential energy change in the formation of a covalent compound say X2-

+

Answer the following questions (a) Which point identifies the bond length between the two atoms of the diatomic molecule? (b) Which point identifies the maximum repulsion between the two-atoms of the diatomic molecule? (c) Which element is most likely to be X in the diatomic molecule shown

A

Zk

4. Name the following processes (a) M(s) ---- > M(g) (b) X2(g) ---- > 2X(g) (c) M(g) ---- > M+(g)+e~ (d) X(g)+e" ---- > X’(g) 5. Answer as directed Element Atomic Number

B

A

i

D

C I

I

Illi Intemuclear distance -

I

H (a) He (b) F (c) 0 (d) N (e) (f) . / C

1 2 9 8 7 4

Valence Electrons (P)

i

Element short of electrons to complete duplet/octet (Q)

Deep Understanding 1. Bond length can be calculated by merely adding covalent bond radii which are H=0.28A, C=0.77A, N=0.70A, 0= 0.66A, Cl= 0.99A, (C=)=0.67A, (C=) = 0.61A, (N =) = 0.55 A. Calculate bond lengths in (a) NH3 (b) CH2C12 (c) HOC1 (d) HCN (e)CH2=CH2 (f) CH^CH

2. In which molecule is the van der Waals’ force likely to be the most important in determining the melting point and boiling point IC1, Br2, HC1, H2S, CO ! 3. Write Lewis structure of •i (a) Cf (b) PF6(c) ici;

Bonding in Organic Compounds—A Basic Concept | 5 4. Atomic number of the four elements A, B, C and D are respectively Z -1, Z, Z + 1 and Z + 2. If Z = 9, what is type of bonding between (a) A and D (b) B and D (c) A and A (d) B and B 5. Valence electrons in the elements X and Y are 3 and 6. What is the formula of the compound formed from X and Y ?

6. Classify Lewis acid and Lewis base out of (b) NH3 (a) H20 (c) H+ (d) BF3 (e) A1H3 (f) ROH (g) CH3NH2 (h) CN" (i)

Sigma and Pi Bond g

• There can be following types of overlapping along the axes (end to end):

H— H

N^N 2x

0=0

a

It

H

C=N Sigma overlap (lobes point along nuclei)

I

N=C—C—C=N

i

• (s-s) overlapping when ^orbital overlaps with another s-orbital • (s - p) overlapping, (p can be px or py or pz) • (Px-Px) overlapping • any of the hybrid orbitals overlaps with another hybrid orbital or 5 or p orbital. Bond formed in this manner is called sigma (o) bond in which electron density is concentrated in between the two atoms, and on a line joining the two atoms. • Double or triple bonds occur by sideways overlap of orbitals (like (pv - py)and(pz - pz) orbitals) giving pi (tc) bonds in which electron density also concentrates between the atoms, but on either side of the line joining the atoms. • The shape of the molecule is determined by the o-bonds (and lone pairs) but not by the k bonds. Pi bonds merely shorten bond length. Thus,

(C=C)< (C=C) H—C—C—H =c K H H H

IoI II

H H

Thus, rc-bond is more reactive than a bond. • While determining type of hybridisation on the atom, tc bonds are never taken into account, but lone pairs are always considered. H

1°

H—C —H

H— C=C—H G I X I

H-^-C=C— H 2k

H four 0 bonds (sp3)

H H three —r

52

N-atom

Px

Hybridisation

/T\

b(H)

i—r

\T/‘

s - p overlap

Pz

Py

2s|iT

We consider formation of CH4 molecule based on valence bond approach. To account for tetravalent nature of carbon, we .take it in excited state in which there are four unpaired electrons and four C—H o bonds are supposed to be by overlap ofs-s,s-px ,s~Py ands-pz types.

Is U Py

Py

' s

? i ph rn rn kofH

Px

Px

Fig. 1.3 Structure of N2 by Valence Bond Method

• Only the orbitals of an isolated single atom can undergo hybridisation. • The hybrid orbital generated are equal in number to that of the pure atomic orbitals which mix up. • A hybrid orbitals, like the atomic orbitals, cannot have more than two electrons of opposite spins. Hybrid orbitals do not make tc bonds. If there are n bonds, equal number of atomic orbitals must be left unhybridised for n bonding.

T|T

w njz,2 15

2p2 ground state

excited state

What do you think about these (C—H) bonds? • are they equivalent? • are they having different bond lengths and bond energies (due to different types of overlap)? our answer is : Number of orbitals and type of hybridisation have been summarised in Table 1.2.

Table 1.2. Hybrid Orbitals and their Geometric Orientation

Stag!

ill*

sp (two) sp2 (three) sp3 (four)

linear trigonal planar

BeCl?, CO„ C?H> BF3,C2H4, SO2,SO3

180° 120°

tetrahedral

ch4 nh3 h2o

109° 28' 107° 48' 104° 27'

sp3d (five)

trigonal pyramidal

PC15 sf4

120° and 90° 101° 36'and 86° 33'

s+ p + p + p + d + d (six)

sp3d2

d+ d+s+ p+ p+ p (six) •

d2sp3

octahedral

[Fe(CN)6]4~ [Co(NH3)6]3+

90° 90°

linear trigonal planar tetrahedral trigonal pyramidal (due to Ip - bp repulsion) angular (V-shaped) due to Ip - Ip and Ip - bp repulsion trigonal pyramidal irregular tetrahedral linear octahedral octahedral square planar (with two lone pairs) octahedral octahedral

d + s+ p + p (four)

dsp2 (four)

square*planar

[Ni(CN)4]2-

90°

square planar '

[Pt(NH3)4]2+

90°

square planar

5 + p (tWO)

5 + p + p (three) 5 + p + p + p (four)

(five)

5 octahedral

SF*

[Fe(H2O)6]3 XeF4

180° 90° 90° 90°

8 | Organic Chemistry 9

How to Determine Type of Hybridisation

3 => sp hybridisation

For this, one is need to know first of all number of lone pairs (by use of VSEPR theory). We adopt two methods for study of type of hybridisation : Method I : Count number of atoms directly attached to central atom + lone pairs + single electrons Method II : Count o bonds +----- > (coordinate) bonds + lone pairs + single electrons This number comes out to be 2 => sp hybridisation

4 => sp3 hybridisation

dsp2 hybridisation (decided by way of orbital used) 5 => sp3 d hybridisation 6 => sp3d2 hybridisation (based on outer J

or inner rf-orbitals are used). d sp hybridisation.

We illustrate methods in Table 1.3. —--------------------—

Table 1.3.

Method of Determination of Type of Hybridisation ■

•

.

.

/

•

■.-'•■

...

■■■•TV-:-; Th -

.

H—Bi

-H

Be

2

2

B

3

3

C

4

0=C=0

C

H— C=C—H

H H

-f

■

TrV-.-

2

sp

3

3

.Jsp

4

4

4

SP

2

2

2

2

sp

C(any)

3

3

3

3

sp2

C(any)

2

2

2

2

sp

N

3

3

1

4

4

sp3

0

2

2

2

4

4

sp

P

5

5

5

5

sp3d

S

6

6

6

6

sp3d2

H

H C

H-

H

H

H

H

H—C=C— H

H 0

H

■H

Cl Cl

Cl

Cl

F

F\l F—S—F lXF

F

Bonding in Organic Compounds—A Basic Concept | 9 r

0

N

2

1

N(any)

1

I

1

s

2

1

I

C *

3

ch2=ch—ch=ch2

C(any)

0

1

1

3

3

2

2

3

3

sp2

3

3

3

sp2

3

3

3

3

sp2

S

4

2

4

4

sp3

S

4

4

4

4

sp3

N *

2

2

2

2

0

1

1

2

3

3

sp1

1

1

1

2

2

sp

sp2

0

:n=n:

0

1

0

H3C— C— 0— H

T H—O—S—0—H I

2

0

H—0—S—0—H 0

:n=n=o: a

•

Topical Test 2 'fy Step-Through Questions 1. (a) C2H2 acetylene react with Na liberating H2 H—C=C—H+2Na ---- > Na—C=C—Na + Hl2; (1) Thus, compare electronegativity of C atom in C2H2, C2H4, C2H6 based on hybridisation. (2) Select most acidic H in propyne H

I

H—C—C=C-~H H (c) Relate electronegativity of C-atom based on s-character in C-atom.

2. Following structure has all types of bonds H H H H H—C=C— C— C = C — C — C = CN

T T I T a

bhc

f

d

TI t T eHJ

&

(A) Which bond has maximum bond length? (B) Which bond has maximum bond energy? (C) Generalise a rule to relate bond energy with bond length

10 | Organic Chemistry What are hybridisation of C+ and C" atoms?

3. Benzene has following structure:

C® C®.

4. Determine percentage of s-character, p-character and d-character in the following:

s (a) (b) (c) (d) (e)

H

Benzene

(a) What is hybridisation of each C-atom? (b) There is delocalisation of n-electrons as shown below

d

' P

sp sp~ sp sp3d dsp2 ■>

©

Deep Understanding 2. When ethene reacts with H2, ethane is formed. Which bond is affected? 3. Although both carbon and silicon are in group IVA (14), very few Si =Si bonds are known. Explain.

1. What are the total number of o and it bonds in (a) Methyl isocyanate CH3NCO? (b) Tetra cyanomethane C(CN)4? (c) Naphthalene C10H8?

Formal Charge The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms in the bond. Formal charge (F) is determined using „------------- ... F = v------- M k ___ pNH3 (0 = 107°) > pS02 (0 = 120°)

Cl

Cl

/9/2

B

P- resultant

^AB

\e/2

where, p AB is the vector moment of the side AB.

~

Cl

Cl

p = 8xt?

but in polyatomic molecule with angle 0, resultant dipole moment is the vector summation of the vector moments.

0 2

0 = 60°

0=120°

ortho

meta

Cl

Cl

0=180°

para

p=6.30D p=0.00D p=3.80D Theoretical p=0.00D p=6.00D p=3.79D Experimental In this case p (o-isomer) > p (w-isomer) > p (p-isomer) but experimental value of o-isomer < theoretical value.

12 | Organic Chemistry

• This is due to dipole-dipole repulsion in oisomer that increases bond angle greater than 60° and p decreases.

ij. ...

Linear Linear Bent Trigonal planar Trigonal pyramidal T-shaped Tetrahedral Square planar Seesaw Trigonal bipyramidal Square pyramidal Octahedral

AX AX2 ax3

repulsion attraction due to increases 0, H-bonding 0 decreases, p increases p decreases

f>

A

CH3

Cl

H

'•rTtv1*!*'#' .■

‘

CH

Cl

OH

Table 1.4 Relation between Molecular Geometry and Dipole Moment

no repulsion or attraction no change in actual value

• % ionic character in a molecule _ observed value of p * theoretical value of p

AX4 ax5

AXb

can be non-zero zero can be non-zero zero can be non-zero can be non-zero zero zero can be non-zero zero can be non-zero zero

-J.,

HF HC1 HBr HI

1-91

101

103 0.78 0.38

136

151 170

39.4% 16.0%

All X atoms are assumed to be identical. • Dipole moment of czs-geometrical isomer is greater than Zrans-geometrical isomer

HF > HC1 > HBr > HI

• Ionic character can also be decided based on electronegativity difference. Molecule H—X is 50% ionic if electronegativity difference of X and H is 1.7 eV. • Based on Pauling’s electronegativity scale (EN)x ~ (EN)h = 0.208 -Jah_a-

where, AH_% is the extra bond energy in kcal mol (0.208 is a conversion factor when electronegativities (EN) are given in eV). ah-X = (actual bond energy (BE) of H-% bond) ”7(®E)h_h (BE)x_z

I

I

CH3—C—H p = + ve

4.7%

Thus, HF is having only approximately 40% ionic character, the order being

CH,—C—H

CH,—C—H

10.8%

H—C—CH3 p=0

Because the polar C—Cl bond in chloromethane has two charged ends—(a) positive end and a negative end—we describe it as being a bond dipole, and we often represent the dipole using an arrow with a cross at one end (----->) to indicate the direction of electron displacement. The point of the arrow represents the negative end of the dipole (S-), and the crossed end (which looks like a plus sign) represents the positive end (8+). Cl5"

I.

H"SC ^H H

or

iV

H" C^H H

Chlorine is at negative end of bond dipole

Carbon is at positive end of bond dipole

Chloromethane, CH3CI

• Geometry of the molecule can be decided based on the value of the dipole moment as given in Table 1.4.

Hydrogen Bonding • J he concept of the hydrogen bonding was introduced by Latimer and Rvdebush (1920). » Hydrogen bonding is said to be formed when slightly acidrc hyorogen attached to a strongly electronegative a«i?H >wCi as i ■ K aNi

CH3—C=NZ

,N=C—CH 3

I 'N=C—CH 3

0...H—b

I i

i

H5+

H—0 8+

H

6+h

nickel dimethyl glyoximate (a chelate) The necessary conditions for the formation of intramolecular hydrogen bonding are : (a) the ring formed as a result of hydrogen bonding should be planar. (b) a 5 or 6- membered ring should be formed. (c) interacting atoms should be placed in such a way that there is minimum strain during the ring closure.

8+

8+

8-

...H

H5+

8-

H...0 H8*

H

I

0,

0— H-0 (C)

8-

H

o-nitro phenol

CH3—C=N

8+

If we actually observe, one water molecule is joined to four water molecules — two with H-atoms and other two with O-atoms. Thus, coordination number of water molecule in water is four.

I I

H

8-

Illi I HS+ H6+ H5+

°XH n^° I o

(b)

I I

8+

H8+

This type of H-bonding occurs when polar H and electronegative atom are present in the same molecule. c%

8-

H—0...H—0...H—0...H—0

Intramolecular H-Bonding

(a)

13

• When ice is formed from liquid water, some air gap is formed (in tetrahedral packing of water molecules). Due to this volume of ice is greater than liquid water and thus, ice is lighter than water. We can say that density decreases when ice is formed. Reversely when ice melts, density increases but only upto 4°C, after this intermolecular H-bonding between water molecules breaks hence, volume increases and hence, density decreases. Thus, water has maximum density at 4°C. (b) In hydrofluoric acid (HF), there is again association by H-bonding. H\

8+

8+

8+ \F'' 8-

8+

,-H'

,-H

SF''' 8-

8-

F

8-

14 | Organic Chemistry

However, in the gaseous state, several polymeric forms of the HF molecules exist in which the monomers are held together through H-bonding. A pentagonal arrangement of H-F molecules is shown below : ,,-F---- H^ 101 pm .X^’nso0 15jDpm F 'Xrj 108° F H

(d) Due to hydrogen bonding, viscosity (r|) of the liquid increases as given below (in centipoise (CP).

CH3CH2OH

I I

I

I

I

3-

8—

6+

8+

8-

S**"

However, isomeric ether is less soluble in water due to its (ether) non polar nature. 8-

8+

CH3—CH2—O—H polar

8+

8-

ch3—o—ch3

less polar

(I) Though the hydrogen atoms in a methyl groups are not polarised, if an electronegative group like chloro, carbonyl, nitro or cyano is attached to it, the C—H bond gets polarised due to the inductive effect and the hydrogen atom becomes slightly acidic resulting in the formation of weak hydrogen bonds. H H H

400-

300-

HF

200-

T| = 1070CP

H2Te SbH3 H2sjx HI ^SnH4 H2Se

PH3

.

GeH4

/SiH4 CH4

100-

co

Fig. 1.4

Resonance We can write two electronic structures of ozone, O 3-

o.

:o:

...0—H...O—H...O—H...

ch2oh

T] = 17CP h2o\

NH3

I I

CH2OH

ch2oh

i] = 1.2CP

(c) There is also similar H-bonding in alcohol (R—OH), ammonia (NH3) and phenol (C6H5OH) molecules. (d) Carboxylic acid dimerises in gaseous state due to H-bonding 100 pm 160 pm H H O/.H—O H— C—C ,C— C —H 0— FL.O H H (e) Alcohol is said to be highly soluble in water due to crossed intermolecular H-bonding (between H2O and R—OH • molecules). R H R

CH2OH

ch2oh

o:: '9.

::oP/

P.:

A B In A, the oxygen—oxygen bond on the left is a double bond and the oxygen—oxygen bond on the right is a single bond. In B, the situation is just the opposite. Experiment shows, however, that the two bonds are identical. Therefore, neither structure A nor B can be correct. • One of the bonding pairs in ozone is spread over the region of all three atoms rather than associated with a particular oxygen—oxygen bond. This is called delocalised-bonding — a type of bonding in which a bonding pair of electrons is spread over a number of atoms rather than localised between two.

H—C—C=N ... H—C—C=N...H—C—C=N

H

H

?r

H ch3

CH3—C—C—H...0=C

H

ch3

Consequences of the Hydrogen-Bonding (a) Due to H-bonding boiling points of water, ammonia, hydrofluoric acid are abnormally high (Fig. 1.4.) (b) B.P.: H0O> HF>NH3 >CH4 M.P.: H2O>NH3 > HF>CH4 (c) Solubility of the organic compounds in water is due to H-bond formation.

• Structures A and B are called resonating or canonical structures and C is the resonance hybrid of these two structures. This phenomenon is called Resonance : a situation in which more than one plausible structure can be written for a species, and in which the true structure cannot be written at all. • 0=0 bond length is 121 pm and O—O bond length is: 145 pm but in ozone experimental measured O—O bond, length is 128 pm suggesting that resonance hybrid is a true; representation. • In benzene, D and E are extreme resonating structure,, while F is a resonance hybrid of these two structures,

Bonding in Organic Compounds—A Basic Concept

•• e

:o:

=

D E F and again carbon—carbon bond length is 140 pm which is intermediate of C—C (154 pm) and C=C (134 pm) bond lengths. • Difference in the energies of the canonical forms and resonance hybrid form is called resonance stabilisation energy (AEJ, and provides stability to the molecule.

E

:o

e

.• Thus, phenol (which is also an oxo acid) is more acidic than an alcohol because of resonance charge delocalisation of the phenoxide ion. • Carbon-oxygen bond lengths in carboxylate ion are equal due to resonance, and thus, stability of the carboxylate ion is increased. ^0 oe R—C R— C 0

Conditions for Resonance

• In nitrate ion, nitrogen-oxygen bond lengths are equal due to resonance,

o:

:o:

:o—n— o:

:o—n=o:

.. I ..

;o:

.. I ..

The canonical forms should not differ in the atomic arrangements. Only positions of the electrons can change. Different canonical forms are separated by ' (double headed arrow) • The number of unpaired electrons should be the same in all the canonical forms. Thus, following structure of benzene is incorrect.

• The positive charge should reside as far as possible, on the less electronegative element and vice-versa. Thus, H—F can be represented as

;O=N— o: • 1,3-Butadiene can have following structures: e e ch2=ch—ch=ch2 CH2=CH—CH—ch2 ®

e

ch2—ch=ch—ch2

H—F

..

ch2=ch—ci:

HF i

Negative p£o

Weak conjugate base

Weak acid

ao

Fi

Filtration

a Crystallisation a Sublimation a

Distillation

a

Extraction

a

Chromatography

a Qualitative Analysis of Elements a Quantitative Analysis of Elements a

Molecular Formula Determination

a

Molecular Weight Determination

a

Functional Group Identification

a

Illustrative Solved Examples

a

Chapter Proficiency Test

32 I Organic Chemistry

Properties of organic compounds can be correlated to their structures. Organic compounds under quantitative and structural investigations should be in its purest form. Following methods can be adopted for the purification of the organic compounds. Selection of the method depends on the nature of the organic compound and the types of impurities present in it. • Filtration • Recrystallisation • Sublimation • Distillation • Fractional Distillation • Differential Extraction • Chromatography. Melting points and boiling points are the best criteria of assessing the purity of the compounds since, pure compounds have sharp melting point and boiling point.

Filtration Filtration is a simplest process of separating solid material from the solution using some form of filter medium (solvent). The simplest filtration apparatus is a filter funnel fitted with a filter paper. Filtration process is very slow and takes long time. Large quantities of material are usually filtered through a Buchner funnel under reduced pressure using a filter pump. Examples of some mixtures and solvents required to separate their components are given below :

__ aESS.

th

____ . ii •

■■

.

.

'

urea (recovered by evaporation)

1. Naphthalene and urea

water

naphthalene

2. Anthracene and benzoic acid 3. Oxalic acid and para-dichlorobenzene

hot water hot water

anthracene benzoic acid (recovered on cooling) para-dichlorobenzene oxalic acid (recovered by evaporation)

i

In examples (2) and (3) above, filtration of hot solution is required. In such case, a specially designed (as shown in Fig. 2.1) funnel is used. Flat filter paper

Neoprene adapter Clamp

Thick walled tubing

“

— Vacuum release tap

t Trap

To vacuum

Buchner flask

Fig. 2.1 Suction filtration using a Buchner funnel

Crystallisation Crystallisation has long been one of the prime methods used by organic chemiststo clean up impure compounds during reaction or synthetic sequences. It takes advantage of the fact that solubility of a solid in a solvent at higher temperature is more than at lower temperature. Before carrying out crystallisation, it is an advantage to have an idea about the degree of purity of the substance and the nature of impurities. • Crystals obtained are in purest form if the impurities present in the impure solid dissolves in the solvent and remains dissolved when solution is cooled. • Impurities may remain undissolved in hot solution and can be removed by filtration. The filtrate is subjected to crystallisation! after concentration.

Purification and Elemental Analysis | 33

Fig. 2.2 outlines the flow-sheet of the process of crystallisation. Sometimes, repeated crystallisation is carried out to make it more yield-effective. Impure substance

Solution of impure substance

1

Filtrate

3

Insoluble impurities Pure crystals

6W

Crystals in mother liquor

Damp crystals Mother liquor

Second yield of crystals

Fig. 2.2 Schematic flow-sheet for crystallisation

Step 1 : Dissolution Impure substance is dissolved in the selected solvent based on the nature of solid and impurities present, and saturated solution is made. The purer the substance the larger the crystals, and more slowly it will dissolve. If the solution is strongly coloured due to impurities, activated charcoal may be added to decolourise the solution.

Step 2: Filtration Dust and insoluble impurities can be removed by filtration using a fluted filter paper which has been preheated by pouring through ' it a small volume of hot solvent. This is called simple or gravity filtration. To prevent premature crystallisation, a slight excess of the solvent may be used.

Step 3: Crystallisation Crystallisation from a super-saturated solution can be induced by : • slowly cooling hot saturated solution to room temperature or below by using ice-cold water. • by addition of a seed crystal (compound in the pure form).

Step 4: Separation of Crystals Crystals are separated from the mother liquor by filtration using a suction pump. In smaller quantity of the compound, separation . can be affected by centrifugation.

Step 5 : Crystallisation and Separation Step 3 and step 4 are repeated to get second yield.

Step 6: Drying Solvent used in crystallisation is removed by evaporation (like ether) or using drying agents (desiccants) like silica gel, phosphorus pentoxide or fused (anhydrous) calcium chloride. Hydrocarbon solvents are removed by using paraffin.

Choice of Solvent It is a common saying that like dissolves like. Polar compounds are soluble in polar solvents and nonpolar compounds are soluble in . nonpolar solvents. It is always better to try out on a small scale first. One can consider following guidelines while carrying out purification of organic compounds by crystallisation. • Substances tend to be more soluble in chemically similar solvents. ; • Good crystallisation medium means that the compound is very soluble in hot and insoluble in cold. ; • Polar solvents (as water) tend to produce better crystals than hydrocarbon solvents. • Table 2.1 summarises common solvents (in increasing polarity) and the class of compounds for which they can be used.

34 | Organic Chemistry Table 2.1 Solvents to be used for Crystallisation of Different Substances. --------- - -----. . . ,A . . EL ■ ■

--------- ---------1. Hydrocarbons

. ___

7 .■ .. '

-j

hydrophobic (nonpolar)

pentane, hexane, benzene, petroleum ether

2. Ethers

diethyl ether; methylene chloride (CH2C12)

3. Halohydrocarbons

chloroform

4. Tertiary amines

acetone

5.

Ketones, aldehydes, esters

ethyl acetate/methyl acetate

6.

Phenols, alcohols.

ethanol

7.

Carboxylic acids

methanol

8. Sulphonic acid

water

9. Organic salts

water

___ J 1

|

Ii hydrophilic (polar)

Sublimation If you keep naphthalene balls in an open test tube you may observe that size of naphthalene balls decreases day by day. It is due to its slow vaporisation without converting into liquid. Rate of vaporisation is increased on increasing temperature. Such substances like naphthalene, ammonium chloride, etc., are known as sublimable. Sublimation is an alternative to crystallisation for purifying some solids. The criteria for effective purification by sublimation require that: (i) the compound to be purified must have a relatively high vapour pressure. (ii) the impurities must have vapour pressure substantially lower than the compound to be purified. The technique involves placing the impure solid in a dish and heating it slowly to a temperature higher than that of the cold surface (generally inverted funnel) on which it is to be collected, but lower than its melting point. Thus, solid is vaporised and at the same time vapours condense to solid on the cold surface. Crystals thus, obtained are usually pure, since, impurities are non volatile in nature.

sublimate , Solid ---------- > Vapours

cool

* Solid

Sublimation may be carried out in a China-dish on which inverted glass-funnel is placed. In between the two (China-dish and glass; funnel) a filter paper is placed. A cotton plug is inserted on the mouth of the stem. When the compound is heated, vapours are; formed and sublimate on the cold surface. Fig. 2.3 (a) Rate of sublimation may be increased if the process is carried out under reduced pressure as shown in Fig. 2.3 (b).

I

Cotton plug

Water in

Sublimate

Water out

U, T

Impure compound

1

" 7D —►Vacuum Cold finger

Sublimed material

Sample l

Heat source

t

Water condenser

Clamp

*

Clamp

Distillation flask

Water out

Receiver adapter

*

t

Water in


1, thus, repeated extraction will extract organic substance from water to organic solvent. The above principle based method is called differential method. It is carried out in a separating funnel as shown in Fig. 2.8.

Solvent layer Aqueous layer

Solute

Solvent rich in solute

Aqueous _ 5* layer C_

•I

Fig. 2.8 Three-stages of differential extraction

4—W

If the organic compound is less soluble in the organic solvent, then extraction is made feasible only when large amount of organic solvent is used. The technique called continuous extraction is used and the apparatus used in this process is called Soxhlet extractor. List consists of a glass cylinder (C), having a side tube (T) and syphon (S). A water condenser (W) at the top and a boiling flask (B) has been fitted with the cylinder. The solid mixture is taken in the cylinder (C) fitted with a thick filter paper. The solvent is taken in the flask B. When heated, solvent vapours rise and condense (with the help of condenser). When condensed vapours of the solvent pass into the cylinder (C), c-ompound is extracted. When the level of the liquid in cylinder (C) reaches upto the top in syphon S, it gets siphoned and is collected in the flask which is being heated continuously. Solvent is again vaporised and extraction is repeated. Soxhlet extraction is useful in the extraction of oils and fats from the flowers and seeds and alkaloids from the plants. Solvents used in extraction technique have been summarised in Table 2.3.

T

S

□ o

® c

CL

S.2

> O

ICl

>

O CO

> O Q

i

w

Solvent + Extracted substance

Heat

Fig. 2.9 Soxhlet extraction

38 ■ Organic Chemistry

Table 2.3.

I-----

Solvents used in Liquid-Liquid Extraction

___________

.

Pure water Acidic solution (pH 0-6) Basic solution (pH 8-14)

Chloroform Dichloromethane Hydrocarbons • Aliphatics : pentane and above • Aromatic : toluene and xylenes Alcohols : C6 and above are immiscible with water Esters • Ketones : C6 and above • Ethers : diethyl ether and above.

Chromatography The term chromatography owes its origin to the fact that the Russian scientist Tswett (1906) reported separating different coloured constituents of leaves by passing an extract of the leaves through a column of calcium carbonate, alumina and sucrose. Thus, the word chromatography is from the Greek words meaning ‘colour’ and ‘to write’. Chromatographic separation depends on the differences in the partition coefficients of the components of a mixture between the two immiscible phases. One of these is the mobile phase which moves relative to the other i.e., the stationary phase. The substances being separated are transported with the mobile phase. The partition coefficient K of a substance, in such a two-phase system is given by

C„ Where, Cs is the concentration of the substance in the stationary phase and C,n is the concentration of the substance in the mobile phase. Thus, greater the partition coefficient of a substance, the greater would be its concentration in the stationary phase, as a result retention in the stationary phase would be higher and its movement with the mobile phase would be slower.

Types of Chromatography Based on the physical states of the mobile and stationary phases, various chromatographic methods are classified as:

z

Solid Liquid Solid Liquid

zzzzzzxzz. i

Vapour Vapour Liquid Liquid

Gas chromatography (gas-solid) Gas chromatography (gas-liquid) Adsorption chromatography Liquid-Liquid partition chromatography

3

3

The removal of the adsorbed substance from the adsorbent by washing is known as elution. The solution, coming out of the chromatographic column is called the eluate. Fig. 2.10 illustrates the distribution of the two species A and B along a column, as they move down the column. If we measure the concentration of eluted molecules as they emerge from the column and plot this as a function of time or of the volume of the mobile phase passed through the column, a chromatogram results. 0)

ro o. &

A+B

CD C

o '"5 w

e c o

f 0)

o

c o Q

I

BA

A

MMJM Distance along column

Fig. 2.10 Distribution of two substances, A and B, along a chromatographic column in a chromatographic separation

Purification and Elemental Analysis | 39

Adsorption Chromatography In adsorption chromatography ,the stationary phase (adsorbent) is a solid (for example silica and alumina) on which the sample components are adsorbed. The mobile phase may be liquid (liquid-solid chromatography) or a gas (gas-solid chromatography). The components distribute between the two phases through a combination of sorption and desorption processes. Based on differential adsorption, this technique is further classified as : * fc • Column chromatography • Thin-layer chromatography

Column Chromatography In this technique, column (having a stop-cock) used for the separation is packed with a stationary phase (adsorbent) as shown. The mixture under study in a minimum amount of solvent is applied at the top of the adsorbent. A suitable solvent or a mixture of solvents (depending on the nature of components) is allowed to flow down the column slowly. Depending upon the degree to which the compounds are adsorbed, partial or complete separation takes place. The most readily adsorbed substances are retained near the top and unadsorbed come down to various distances in the column as shown in Fig. 2.11.

Solvent

s

(s) Mixture of (a + b + c)

■>

1

s

F—1

L_ 5 — (a) i

t

1

i[ Jj

J

1 --- I

■

Adsorbent

Fig. 2.11

■J

(a)

k------

LZ^

(b + c)

(b)

(c) I

I

■

L

¥

Different stages of separation of components of mixture of (a), (b) and (c) by column chromatography

Thin-Layer Chromatography (TLC) As the name implies, adsorbent is the thin layer of silica gel or alumina spread over a glass plate of suitable size. The plate thus, formed is TLC plate, on which solution of the mixture is placed in the form of a spot about 2 cm above one end of the TLC plate. This treated TLC plate is placed in a closed jar having a solvent (Fig 2.12). As the solvent moves up the plate, individual components move up along the plate to different distances depending on the degree of adsorption and separation takes place. R retention factor, which gives relative adsorption of each component in the mixture can be defined by the following equation : distance moved by the spot centre from the base line distance moved by the solvent from the base line Lid-*-]------------ Q------------ r I I I I I

Solvent front

F"

O O

■

!

■+

-0-0—-► Base line ■ Solvent —►->; ____ V

(a)

Start

Start L_

(b)

—

(C)

Fig. 2.12. TLC plate spotted with the sample (a) TLC before development (b) Development of TLC plate (c) TLC after development

40 | Organic Chemistry

values depend on the conditions under which the chromatogram was run, namely, type of plate, eluent, temperature. Coloured spots are immediately visible. Colourless spots can be made visible by : • UV, if the substance absorbs UV light (as in case of aromatic compounds). • standing the plate in iodine vapour in a chromatographic jar; organic compounds generally give coloured spots with iodine • spraying with 1 : 1 H2SO4—water mixture and then heating strongly to carbonise the compounds. • spraying with suitable reagents which give coloured spots with the substances under observation—for example, ninhydri in case of amino acids.

Partition Chromatography

The stationary phase of partition chromatography is a liquid supported on an inert solid. Again, the mobile phase may be a liquii' (liquid-liquid partition chromatography) or a gas (gas-liquid chromatography, GLC). Paper chromatography is a type of partitio chromatography in which the stationary phase is a layer of water adsorbed on a sheet of paper. In the normal mode of operation of liquid-liquid partition, a polar stationary phase (for example methyl alcohol on silica) i used with a nonpolar mobile phase (for example hexane). This favours retention of polar compounds and elution of nonpol.a compounds and is called normal phase chromatography. If a nonpolar stationary phase is used, with a polar mobile phase the nonpolar solutes are retained more and polar solutes can be more readily eluted. This is called reversed-phas' chromatography. The spots of the coloured compounds in case of paper chromatography are visible at different heights from the position of initiia spot on the chromatogram. The spots of the separated coloured compounds may be observed by the methods as in TLC.

Topical Test 1 Step-Through Questions

1. A given mixture contains sugar and sand. Sugar can be recovered by Step I Step II Step III 2. Select suitable solvent in each case for the crystallisation of (b) benzene sulphonic acid (a) benzoic acid (c) benzophenone 3. Suggest suitable distillation methods for separation of mixture of (a) aniline (b.p. 457 K) and chloroform (b.p. 334 K) (b) Crude oils mixture (petroleum) (c) glycerol in spent-lye (d) aniline and water 4. Mention adsorbent and eluent in the following : r-

i-

(A) (B)

(C)

Adsorption chromatography Liquid-liquid partition chromatography. (Gas-liquid) gas chromatography

Qualitative Analysis of Elements The qualitative analysis of organic compounds means the detection of all major elements present in it with the help of suitaib chemical tests. The detection of carbon and hydrogen is done by the copper oxide test. Carbon is an essential constituent of an organic compound whereas hydrogen is seldom absent. On ignition with cupric oxiic (CuO), carbon and hydrogen are converted into CO2 and H2O respectively and can be tested by their usual tests: 2CuO + C —2Cu+CO2 (turns lime water milky) CO2+Ca(OH)2 ---- > CaCO3+H2O

Purification and Elemental Analysis | 41

CuO + 2H

A

* Cu + H2O (drops appear on the cooler part of the ignition tube also turns anhydrous copper sulphate blue)

4H2O + CuSO4 H2O

CuSO45H2O

blue Organic compounds (supposed to contain phosphorus) are heated with oxidising agents (like sodium peroxide Na2O2). The phosphorus present in the organic compound is oxidised to phosphate. The detection of nitrogen, halogen, phosphorus and sulphur is done by Lassaigne method.

J 1

I i

Lassaigne Method This is used to detect nitrogen, halogen, phosphorus and sulphur. Organic compound is fused with dry sodium in a fusion-tube and fused mass after extraction with H2O is boiled and filtered. Filtrate called sodium extract (S.E.) is used to detect elements (other than C and H) and the tests are given in Table 2.4. Instead of sodium metal, mixture of sodium carbonate and zinc can be used. It is called Middleton's fusion method. • Organic compounds being covalents normally do not have ionisable groups, hence direct test is not possible. • Fusion with Na forms soluble salt (like NaCl, NaCN, etc.) which can be easily detected. Table 2.4. Lassaigne Method (Detection of Elements)

Ii_____________ L Nitrogen

Na + C + N

A. xr^xi NaCN S.E.+ FeSO4 + NaOH, boil and (S.E.) cool + FeCl3 + cone. HC1 Blue or green colour.

_ w.-________ 3

____________________________

FeSO4 + 2NaOH---- > Fe(OH)2 + Na2SO4 Fe(OH)2 + 6NaCN---- > Na4[Fe(CN)6] + 2NaOH

NaFe[Fe(CN)6] + 3NaCl Prussian blue

Na4[Fe(CN)6] + FeCl3

Fe4[Fe(CN)6]3+12NaCl Prussian blue

or 3Na4[Fe(CN)6] + 4FeCl3

Sulphur

Halogen

2Na + S

Na + Cl

Na,S (S.E.)

NaCl (S.E.)

(i) S.E. + sodium nitro prusside (A). A deep violet colour. (ii) S.E. + CH3COOH + (CH3COO)2Pb A black ppt.

S.E.+HNO3 + AgNO3 (i) White ppt soluble in aq NH3 confirms Cl (ii) Yellow ppt partially soluble in aq. NH3 confirms Br. (iii) Yellow ppt insoluble in aq NH3 confirms I As in test for nitrogen; instead of green or blue colour, blood red coloration confirms presence of N and S both.

Nitrogen A and sulphur Na + C+N+S—-» together NaCNS (S.E.) Sodium thiocyanate L Solution is boiled with nitric Na-,0-, boil Phosphorus P > Na3PO4 acid and then treated with ammonium molybdate (NH4 )2MoO4. Formation of yellowppt indicates presence of phosphate hence, phosphorus in organic compound.

Na4[Fe(CN)5NOS] deep violet

(i) Na2S + Na,[Fe(CN)5NO] (4)

CH3COOH . (ii) Na2S + (CH3COO)2Pb —‘3------> PbS>L + 2CH3COONa black PP'-

Na% + AgNO3

HNO3

* AgA" X ppr.

AgCl + 2NH3(a