Nomenclature & Isomerism [1-77]

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1. Introduction to organic chemistry

2. Hybridisation of carbon in organic compounds

3. Representation of organic compounds

4. Nomenclature of organic compounds

5. Isomerism

6. Effects operating in organic systems.

7. Fission of a covalent bond: homolysis and heterolysis

8. Reaction intermediates: carbocations, carbanions and free radicals

9. Different reagents: electrophiles, nucleophiles and carbene

10. Types of organic reactions

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Introduction to Organic Chemistry

Organic chemistry is a new branch of chemistry. It mainly deals with compounds made up of carbon. Otherthan carbon some elements like H, O, N, and halogen etc. are also involved. Starting from living beings tocooking gas, all belong to this new branch of chemistry. Let us see how it is important for comingexaminations.

Why are there so many carbon compoundsThis is because a single carbon atom is capable of combining with up to four other atoms. The uniqueproperty of carbon atom is that it can combine with other carbon atoms. The ability of carbon to form a longchain bonding with other carbon atom is called catenation. Therefore carbon atoms can form chains- andrings-type compounds. The simplest organic compounds are made up of only carbon and hydrogenatoms, e.g.

CH4 , CH3 — CH3 , CH3 — CH2 — CH3 and

Methane Ethane Propane

etc.

These properties of carbon can be explained by hybridization.

2p

2p

2s 2sE lectron

Prom otion

sp3

Hybridization sp hybridorbitals

3

1s

Energy

C(groundstate)

Tetravalency of carbon

Since the 2s and 2p orbitals are very close in energy, one electron from the 2s orbital jumps to the 2pz

orbital. The one 2s and three 2p orbitals mix together and give rise to four new different types of orbitals.This is called hybridization and is seen only in carbon atom.

sp3 hybridization in carbon: The one 2s and three 2p orbitals mix together and give rise to four sp3 hybridorbitals, e.g. CH4.Carbon in methane molecule forms four sigma bonds. The carbon here is sp3 hybrid carbon. The methanemolecule has a tetrahedral shape. The C atom is at the centre of the tetragon (three-dimensional equilateraltriangle) and the four H in the four corners of the tetragon. Each C — H bond in methane makes an angleof 109° 28’ with the other.

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2p

2p

2s 2sE lectron

Prom otion

sp3

Hybridization sp hybridorbita ls

3

1s

Ene

rgy

C(ground state)

sp3

sp3

sp3sp3

H

H

HH

1s

1s

1s1s

σ

σ

σ

σ

H

H

H

H

C

109 °28 ´

Orientation of sp3 hybridised orbitals and overlapping with 1s orbitals of H.

sp2 hybridization in carbon: Carbon also exhibits sp2 and sp hybridization.In sp2 hybridization one s and two p orbitals participate. The sp2 hybrid orbital is planar triangular in shapeand the bond angle is 120°.Here the the unhybrid pz orbitals or adjacent carbons overlap to form pi bond, e.g. ethylene molecule.

2p 2p

2s 2sE lectron

Prom otion

sp2

Hybridization sp hybrid orb itals2

1s

Ene

rgy

C(ground state)

Unhybridized p-orbita l

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There are five sigma bonds and one pi bond in ethylene as shown below.

H-atom H-atom

1s 1s

1s1s

sp2

sp2 sp2

sp2sp2

sp2

p z

p z

p z

p z

H-atomH-atom

H

H

H

H

C C 120 °

120 °H

H

H

H

C C

120 °

σ

σσ

σ

σ

120 °OR

120 °π

sp hybridization in carbon:In sp hybridization, only 1 s and 1 p orbital take part. The rest two unhybrid p orbitals form 2 pi bonds. Theshape is linear and the bond angle is 180°.

2p 2p

2p

2s 2sE lectron

Prom otion

sp

Hybridization sp hybridorbita ls

1s 1s

Energy

C(groundstate)

C(excitedstate)

unhybridizedorbita ls

sp hybridization in acetyleneFor sp hybridization the molecule of acetylene is an example. Here three of the bonds lie parallel to eachother CH CH . The angle between the direction of the bonds is 1800.In acetylene in all there are three sigma bonds and two pi bonds.

π-bond

σ σσH H

π-bond

2pz

2py

2py2pz

sp sp sp sp1s1s

HH

H — C — — C — H

π— —σ— —π

σσ

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Sigma and pi bonds

When a carbon atom forms a compound, it always forms covalent bonds. There are two types of covalentbonds: sigma and pi. When the covalent bonds are linear or aligned along the plane containing the atoms,the bond is known as sigma (σ) bond. Sigma bond is stronger than pi bond.

s-s overlap s-p overlap p-p overlap

Whereas sidewise overlap leads to pi bond as shown below

+

+ or

Inte rnuclearaxis

p-orbitals S idew ayoverlapping

Final electroncloud picture

+ + + +

+ + + +

+

+ +

Structural representations of organic compoundsOrganic compounds and their formulae are represented using many structures. Structural representationshould clearly portray the structure, bonding and orientation of different atoms in the molecule. There arevarious representations.

Lewis structure: Here the bond C — H or C — C is represented by .. or xxe.g.

CH HH

HFour electrons of carbons (x)One electron of four hydrogen atoms each ( )(x ) represents a C — H bond.Complete structural formula:Here the two atoms are represented by a dash (—).

C — C (—) represents the bond.

H — C — C — C — C — C — H

H

H

H

H

H

H

H

H

H

H

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Condensed structural formulae:In this formula the bonds are also not shown. This is a simplified version of complete formula. Hereonly the atoms are shown. This is further simplified to

CH3 — CH2 — CH2 — CH2 — CH3 or CH3(CH2)3CH3

Bond line structure:This is further simplification of the condensed formula. Here only the skeleton is shown. Only thoseatoms other than carbon and hydrogen are represented. In these structures it is understood thatthere is a carbon atom at each ‘bend’ and that each carbon atom is attached to as many hydrogenatoms as are needed to complete its valence of four. A zig-zag structure is used for linear compounds.

P en tane

O H

E thy lm e thy l e the rButan–1–01

O

Representing three-dimensional organic structures on a two-dimensional surface (such as a piece ofpaper) presents some challenges.

Wedge and Dash Representation

This reprentation actually represents the bond projections from the plane. The three-dimensional structureof a molecule can be shown through the use of wedged and dashed bonds. Wedged bonds are understoodto be coming out of the plane of the paper towards the reader, whereas dashed bonds are understood to begoing away from the reader. Bonds that drawn as plain lines in one of these structures are in the plane ofthe paper. This is shown for methane.

H

HH

H

C

Fischer projections

These are also used to show three-dimensions. These structures are drawn with the longest carbon chainvertically and it is understood that all horizontal bonds are coming towards the reader, all vertical bonds aregoing away from the reader and that there is a carbon atom at the intersection of four bonds. This type ofdrawing is commonly used for carbohydrates.

C l

H OH

Br

–––––––––––––

H OH

Br

C

––Cl

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Newman and Sawhorse Projections

Rotation around a bond in a linear structure is shown through the use of sawhorse structures or Newmanprojections. The structures below show only two of the many possible conformations. In each pair ofstructures below, the conformation on the left is referred to as staggered and the conformation on the rightis referred to as eclipsed. The front and back groups are drawn slightly offset in the Newman projection ofthe eclipsed structure so that they don’t overlap.

H H

H H

CH3

CH3 H H

H H

CH3

CH3

CH3

CH3

H H

HHCH3

H

H H

H

H3 C

Sawhorse structures are drawn as follows:• Draw three lines that intersect at their ends and are at 120° angles to each other. The intersection

represents the carbon atom at one end of the bond around which the rotation is being shown.• Draw a slanted line from the intersection of the three original lines to a second set of three lines.• Place the groups or atoms that are attached to each of the carbon atoms that make up the bond

whose rotation is being shown on the appropriate carbon.

Newman projection (think of them as a head-on views of sawhorse structures) is drawn as follows:

• Draw a circle.• Draw three lines that intersect at their ends and are at 120° angles to each other so that the

intersection is in the middle of the circle.• Draw a second set of intersecting three lines “behind the circle.”• Place the groups or atoms that are attached to each of the carbon atoms that make up the bond

whose rotation is being shown on the appropriate carbon.

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Nomenclature

Why we need nomenclature specially for organic compounds?

As we have already discussed, carbon atom has unique ability to form stable molecules consisting ofchains of carbon atoms of any length. Coupled with the observation that each carbon atom forms fourbonds to other atoms this leads to incomprehensibly large numbers of possible molecules.

To name such vast numbers of possible molecules requires a systematic approach consisting of a set ofarbitrary rules which are readily learned and applied to each molecule individually to generate the name.

The IUPAC nomenclature system is such a set of rules used widely by organic chemists and this introductionto nomenclature will outline the rules in this system for the simpler organic compounds.

Note that in addition to names invented for compounds because of their origin, there were other attemptsto produce a systematic nomenclature system before the IUPAC system and sometimes compounds haveseveral different names, of which the IUPAC name may not be the one most commonly used.

Earlier Methods of Naming Organic Compounds

Organic compounds were named prior to the end of the 19th century by those that synthesized them.Usually they were named reflecting the source of the compound. These are now called common or trivialnames and some are still in use today. For example, the simplest alcohol has a common or trivial name of‘wood alcohol’. This compound was given this name because the chief source of this alcohol was thedestructive distillation of wood chips. Wood chips can be distilled in this manner and will yield a verycomplex mixture. Analysis of this mixture will result in a large part of this alcohol. Another alcohol, thesecond member of the alcohol family is called ‘grain alcohol’. This trivial name came from the fact that oneof the principle sources of this alcohol is from the fermentation of grains and fruits.

Later in the century, a second nomenclature system was developed called as derived system. In thederived system, the major hydrocarbon portion of the molecule was named as an alkyl group followed bythe family name. The same simplest member of the alcohol family would be called ‘Methyl Alcohol’ becausethe formula CH3OH shows a methyl group attached to the functional group for the alcohol family. Thesecond member’s name in the derived system would be ‘Ethyl Alcohol’ because the formula of this alcoholCH3-CH2-OH shows the ethyl group attached to an OH group.

Nomenclature System for Organic Compounds — IUPAC Nomenclature

The use of common names and derived names did not work because large number of molecules werediscovered and synthesized. The International Union of Pure and Applied Chemists (IUPAC) developed asystematic approach to organic nomenclature. IUPAC came up with a set of rules to name organiccompounds.

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Table 1.2 : Names of Representative Groups

Group Name Group Name

3CH — Methyl CHC ≡− Ethynyl

2 3CH CH — Ethyl

3 2 2CH CH CH — Propyl 22 CHCHCH =− Allyl

CH -CH3

CH3Isopropyl Phenyl

3 2 2 2CH CH CH CH — n-Butyl O Phenoxy

C H 3

CH 3

C H CH 2 Isobutyl C H 2 Benzyl

C H 3

C H

CH 2 C H 3

sec butyl C

O

Benzoyl

– CN Cyano.

– NH2 Amino

tert butyl – NH2OH Hydroxyamino

– OH Hydroxy

32222 CHCHCHCHCH− Pentyl – OCH3 Methoxy

– CHO Formyl

Neopentyl Keto or Oxo

Acetoxy

C H 2Methylene –COOH Carboxy

–CH2OH Hydroxymethyl

CH2 = CH — Vinyl –SO3H Sulphonyl

CH 3 C

CH 3

CH 3

CH 2CH 3 C

CH 3

CH 3

O

C

O

C C H 3

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Naming organic compounds using IUPAC rules

At its simplest, the IUPAC name for an organic compound contains these two parts:

1. A root indicating how many carbon atoms are in the longest continuous chain of carbon atoms.2. A prefix and/or suffix to indicate the family to which the compound belongs.

For example, the name pentanol indicates a carbon chain of length five (pent-), an alkane derivative(-an-) and an OH functional group (-ol).

This system of nomenclature is designed to closely associate the name the compound with thestructure since structure is so important in establishing the physical and chemical behaviour oforganic molecules. General rules established for all of the families:

1. Longest chain rule: The first step in naming compounds is identification of the longest chain ofcarbons. If the compound has functional group, the longest chain with functional group has to beconsidered.This number will be the prefix to the molecule name.

1

2

3

4

5

51

23

4 6

7(i)

(ii)

(iv)

IV

III

II

I

(v)

(vi)

VII

VIII

(v)

VI

Longest chain here is The carbon chains are of 3 different lengths5-carbon chain 6-carbon chain, 7-carbon chain, 8-carbon chainThe only possibility So the largest is 8 carbon chain.

So prefix will be oct

Greek prefixeso di for 2o tri for threeo tetra for fouro penta for fiveo hexa for sixo hepta for seveno octa for eight

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2. Identify the various branches and groups attached to the following compound.5

1

2

3

4

6

7

8

Only one branch - 2 carbon - length branch

3. Lowest number rule: Fix the position of branches. To do this make sure you follow the lowestnumber rule. First number the longest carbon chain you have identified from one end to the other.Remember to number them from both the ends. Whichever has the lowest number that is the finalpostion.The branch always gets the lowest possible position.

5

13

6

7

8

2

5

4

3

6

7

8

1

2

Branch is on C-4 carbon Branch is on C-5 carbon

4. Identify the suffix — the family name. To do this check out the functional group in your carbon chain.

Do you know names of carbon chainso Meth for one carbono Eth for a two carbon continuous chaino Prop for a three carbon chaino But for a four carbon chaino Pent for a five carbon chaino Hex for a six carbon chaino Hept for a seven carbon chaino Oct for an eight carbon chaino Non for a nine carbon chaino Dec for a ten carbon chain

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5

13

6

7

8

2

4

Saturated compound so suffix is (-ane)Name - 4-ethyl-octane

These rules are adequate to name simple carbon compounds. But when it comes to multiplefunctional group compounds, multiple branch compounds some special rules are required.

Special rules:There are many compounds with complicated and branched structures. Let us take each of thespecial structures and learn the specific rule to name them.

1. Naming a compound with more than one substituent: When there are more than one substituentor branch they can be of various types.

a. Substituents are same: If the substituents are same, then the name need not be repeated.Just name the branch once and add prefix di, tri etc., depending on how many times the substituentis getting repeated. But the position of the substituent needs to be repeated as many times, with acomma in-between.

5

4

36

7

1

2

I

IIIII

IVVVI

VII

The position of branches are same. So either way we can assign the number.Name is 3, 5-diethyl-heptane

b. Substituents are different: If more than one kind of substituents are present in the parentchain, then name them in alphabetical order. Indicate the carbon number of each group is attached,followed by the substitute name as a series separated by commas. Remember to place a hyphenbetween a position number and the name. The substituents should get the lowest possible numbers.

5

4

3

6

7

1

2

5

4

3

6

7 1

2

Name is 3-ethyl-5-methyl heptane Order of substituents is written in alphabetical order.

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Similar and different sets of substituents are present.

5

3

87

1

2

2

6

4

2-ethyl-4- 5-dimethyl octane

Naming of Compounds Containing One Functional Group

Additional rules are desired for naming compounds containing functional groups.

Rules

1. The name of a functional group containing compound is obtained by adding a suffix to the ‘root’derived from the name of the hydrocarbon of longest carbon chain. The suffix replaces the endingane in most cases. The suffixes for various groups are listed in Table 1. 3.

2. The chain is numbered so that the functional group gets the lowest number.

3. A functional group containing a C - atom is always assigned number 1 and its position is notmentioned while writing the name of the compound. Such groups are:

– COOH, – CHO, – COO R, – CN, – CONH2 and – COCl

Exceptions are: C = C, C ≡C and

O

— C —

Table 1.3 : Suffixes for groups.

Group Suffix Prefix Group Suffix Prefix

C = C – ene Alkenyl – COOH – oic acid Carboxyalkyl

C≡C – yne Alkynyl – COOR – oate Alkoxycarbonyl

– OH – ol Hydroxy – NH2 – amine Amino

– CHO – al Formyl- C - Cl

O

–oyl chloride Chloroformyl

– C –

O

– one Oxo – CN – nitrile Cyano

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Naming of Compounds Containing Two Functional Groups :

Similar rules as discussed in naming of compounds containing one functional group are applied for namingcompounds with two functional groups with slight modifications.

Rules

1. Priority order of functional groups

Carboxylic acid > Sulphonic acid > Acid anhydride > Esters > Acid chlorides > Amides >Nitriles > Aldehydes > Ketones > Alcohols > Amines > Ethers

All the remaining groups such as halo (fluoro, chloro, bromo, iodo), nitro (–NO2), nitroso (–NO) andalkoxy (–OR) are always treated as substituents group.

For compunds with both double and triple bonds, numbers as low as possible are given to doubleand triple bonds even though this may at times give “-yne” a lower number than “-ene”. When thereis a choice in numbering, the double bonds are given the lowest numbers.

2. The name of the compound ends with the suffix of the principal functional group.

3. All other groups including the functional groups are used as prefixes (Table 1.2) . Double and triplebonds are not used as prefixes.

The following examples will illustrate these rules.

CH2 = CH - CH2 - CH2 - OH

4 123

3 - Buten -1-ol

(-OH is the principal functional group)

Note if the ending does not occupy the terminal position, then the letter ‘e’ is omitted in writing thename of the compound.

HC C CH2 C CH

541 2 3

1,4 - Pentadiyne

Note here letter ‘e’ is retained because it occupies a terminal position.

HOOC - CH2 - CH2 - COOH

4 123OHHCHCHO 2

1

2

2−

1,4-Butanedioic acid 1, 2 - Ethanediol

CH3

CH2 - CH - CH2 - CH2 - CH2 - CONH2

Cl

4 3 1256

OH|

OOHCHCHCHC1

2

23

3

4−−−

6- Chloro-5- methylhexanamide 3- Hydroxybutanoic acid

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CH3 - CH2 - CH - CH2 - CH2CN

NO2

56 4 3 12

CHOCHCHCCH||O

223 −−−−15 4 3 2

4- Nitrohexanenitrile 4- Oxopentanal

OH

1

2

7

3

86

4

59

CH3 - CH2 - CH = CH - CH - NCH3

CH3

3 1245

4, 5-diethyl-6-methylnon-5-en-1-ol N, N - Dimethyl-2-pentenamine

H

H

COOH

COOH

1

2

3

4

H

HOOC

COOH

H

12

3

4

cis-But-2-ene-1, 4-dioic acid trans-but-2-ene-1, 4-dioic acid

Same rules apply to compounds containing three functional groups.

CH2 CH CH2

OH OH OH

3 12 CH2 — CH — CH2

COOH COOH COOH

4 3 2

5 1

1, 2, 3 - Propanetriol propane - 1, 2, 3-tricarboxylic acids

These rules can be illustrated by the following examples.

CH3

CH2 = CH - CH2 - CH - CH2 - CH3

5 61 2 3 4

3

5

2

432

3

1HCHCCCHC −−≡−

4 - Methylhex -1 - ene 2 - Pentyne

CH 3 - CH 2 - CH 2 - CH - CH

CH 3

CH 3OH

56

1

234

CH3 - CH - CH2 - COOH

CH3

1234

2 - Methylhexan - 3 - ol 3 - Methylbutanoic acid

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CH3

CH3 - C -CH2 - COOC2H5

CH31234

CH3 - CH2 - CH2 - CH - CH3

CHO

5

1

234

Ethyl 3, 3 - dimethylbutanoate 2 - Methylpentanal

CH3

CH3

CH3 - CH2 - CH2 - N

123

N, N - Dimethyl -1- propanamine

Naming of Monocyclic Compounds

Rules

1. For naming monocyclic compounds same rules apply as for naming straight chain hydrocarbons.The difference being the word cyclo is prefixed before the name of the parent straight chainhydrocarbons.

Cyclopropane Cyclobutane Cyclopentane Cyclohexane

2. If a single substituent is present on the ring, no numbering is needed.

CH 3C H 2C H 3

Methylcyclopentane Ethylcyclohexane

3. Numbering is needed if two or more groups are present on the ring.

CH 3

C 2H 514

23C H 3

C H 31

23

4

5

6

1-Ethyl-2 methylcyclobutane 1, 2-dimethylcyclohexane4. If a single double bond present in the ring, then no numbering is needed.

Cyclopentene Cyclohexene

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5. If one or more double bonds and a group or groups are present on the ring, numbering is neededand the lowest number is assigned to the carbon atoms linked with the double bond. The numbersto the groups are assigned by going around the ring.

C H 3

6

5

43

2

1

CH 3

CH 34

32

1

3 - Methylcyclohexa -1, 4 -diene 1, 3-Dimethylcyclobut-1-ene

6. If two functional groups are present on the ring, then the principal functional group gets the lowestnumber

CH3CH3

O

6

5

43

21

C HC

O H

6

54

3

2

1

4, 4- Dimethylcyclohex 2-Ethynylcyclohex-1-ol2-ene-1-one

The following cyclic compounds have special nomenclature.

OC O O H

Cyclopentanone Cyclohexanecarboxylic acid

C O O C H 3

CO NH 2

C O C l

Carbomethoxycyclopentane Cyclobutanecarbonitrile Cyclohexane carbonyl chloride

Naming of Bicyclic Compounds

Compounds containing more than one ring and close rings have two or more of the same carbon atomscalled polycyclic. These are of three main types and named differently. A compound containing rings withone common carbon atom is called spiro cyclic.A compound containing rings with an adjacent carbon atom in common is called a condensed ringsystem.

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A compound containing more than two atoms is called a bridgehead ring system.

Spirocyclic Condensed Bridgehead

In order to name a bridgehead hydrocarbon, we must identify the carbons to which the rings are joined,these carbon are called bridgehead carbons.

Bridgehead carbon1-Carbon

2-Carbon 2-Carbon

Bridgehead carbon

The total number of carbon atoms are counted. In the above case there are seven. The compound’s nameis derived from heptane and is called bicycloheptane(2 - cyclic rings). There are two carbons on each sideand one above, therefore, the compound is named bicyclo [ 2.2.1] heptane. The numbers are written inparenthesis and in decreasing order. For substituted bicyclic ring systems, the number 1 position isassigned to the bridgehead carbon. Numbering of the remaining carbons proceeds around each ring in turnbeginning with the largest ring. In doing this the lowest number is assigned to groups and double bonds.

1

2

345

6

7

1

2

345

6

7

Bicyclo [2. 2. 1] heptane Bicyclo [2. 2. 1] hept -2- ene

O1

2

345

6

7

C H 3

1

2

3

456

7

8

1, 7, 7-Trimethylbicyclo [2. 2. 1] hept-2-one 3-Methylbicyclo [3. 2. 1] octane

Note that the right hand ring is larger.

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1

2

3

4

5

6

7

8

Bicyclo [2. 2. 2 ] octane Bicyclo [1. 1. 0] butane Bicyclo [4. 3. 0] nonane

End the name by attaching the characteristic ending that has been established foreach family. These characteristic endings are as follows:o ‘ane’ for the alkane familyo ‘ene’ for the alkene familyo ‘yne’ for the alkyne familyo ‘ol’ for the alcohol familyo ‘al’ for the aldehyde familyo ‘one’ for the ketone familyo ‘oic acid’ for the carboxylic acid family

IsomerismThe existence of different compounds with the same molecular formula but different structural formula iscalled isomerism. Thus 1-chloropropane and 2-chloro propane depict the relationship of isomerism.Compounds which have the same molecular formula but differ in physical and chemical properties arecalled isomers. The two chloropropanes are isomers. Isomerism has been classified into two types,structural and stereoisomerism which are further subdivided as shown in the diagram.

Configurationalisom erism

Conform ationalIsom erism

Structrual Isom erism Stereoisom erism

Isom erism

Geom etricalisom erism

Opticalisom erism

Chain Isom erism

Functional Isom erism

Positional Isom erism

Ring Chain Isom erism

M etam erism

Tautom erism

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Chain Isomerism

In this type of isomerism the compounds possess same molecular formula but differ in the arrangement ofcarbon chains.

3223 CHCHCHCH −−−

Butane

m .p. - 0.5°C

CH 3

Isobutane

CH — CH 3

CH 3

m .p . -12°C

Pentane has the following three chain isomers.C H 3 C H 2 C H 2 C H 2 C H 3

C H 3

CH C H 2 CH 3

CH 3

C H 3 C C H 3

CH 3

CH 3

C 5H 12m .p. 27°

m .p. 36°

Pentane

n-pentane

Iso-pentane

Neo-pentane

CH3 CH2 CH2 CH2 OH CH3 CH CH2 OH

CH3

1-Butanol

2-Methyl propanol

Functional Isomerism

It is the type of isomerism in which the compounds possess the same molecular formulae but differ in thefunctional groups.

CH3 CH2 OH CH3OCH3

Note: Alcohols and ethers are always functional isomers, so are carboxylic acids and their esters.

CH3 CH2 CH2COOH 23CHCH 3OCHC||O

What other compounds would show functional isomerism?

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Positional Isomerism

It is the type of isomerism in which the compounds possess the same molecular formula but differ in theposition of the same functional group.

CH3 — CH2 — CH2 — CH2 — OH CH3 — CH2 — CH — CH3

CH3 — CH2 — CH — CH2 — CH3

ClCl

CH3 — CH — CH2 — CH2 — CH3

OH1 - Butanol2 - Butanol

2 - Chloropentane 3 - Chloropentane

CH3 — CH CH — CH3

2-butene CH3 — CH2 — CH CH2

1-butene

A disubstituted benzene has three positional isomers. e.g.

OH

NO2

OH

NO2

OH

NO2

p-nitrophenol o-nitrophenol m-nitrophenol

Metamerism

It is the type of isomerism in which the compounds possess the same molecular formula but the distributionof alkyl groups on either side of the functional group is dissimilar. Metamerism occurs in amines, ketones,ethers and esters.

CH 3 CH 2 O CH 2 CH 3 CH 3O CH 2 CH 2 CH 3

Ethoxyethane 1 - M ethoxypropane

CH 3 — C — O CH 2 — CH 2 — CH 3

O

CH 3 — CH 2 — C — OCH 2 — CH 3

O

CH 3 — CH 2 — NH — CH 2 — CH 3 CH 3 — NH — CH 2 — CH 2 — CH3

Ethyl propanoate Propyl e thanoate

N-Ethyl ethanam ine N-M ethyl propylam ine

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How many isomers are possible by compound having formula C4H10O?

Ring chain isomerism

This type of isomerism is due to the difference in linkage of carbon atoms in the form of ring or open chainstructure, i.e. C3H6.

CH3 — CH CH2

propeneCH2

CH2 CH2

CyclopropaneTautomerism

It is the phenomenon of existing of single compound in two readily interconvertible structures called astautomers which appears in acid catalysed or base catalysed conditions.This is dynamic isomerism because the structures are in equilibrium with each other. It is due to thepresence of labile hydrogen atom. Some common type of tautomerism are(i) keto-enol tautomerism (ii) nitro-aci tautomerism (iii) nitrite-nitro tautomerism.

Keto-enol Tautomerism

In the particular case of keto-enol tautomerism, the isomers or the tautomers contain a keto and anenol form. For example, in the presence of an acidic or basic catalyst a rapid equilibrium is establishedbetween an aldehyde or ketone and its isomeric (tautomeric) forms.

CCH

C

C

OH

Enol-formKeto-form

O

Presence of at least one hydrogen atom is essential adjacent to the carbonyl group for a compoundto exhibit keto-enol tautomerism.

CCH 3 C H 3 C C H 2 CH 3

O O H

O

Shows keto-enol tautomerism Does not show keto enol tautomerism

��

... 23��

These tautomeric forms are two different molecules. One form can be more stable than the other. Generally,the keto form is of lower energy than the enol form and thus more stable. These exist in solutionsimultaneously unlike the resonance structures.

An important example of this type of tautomerism is acetoacetic ester.

CH3 C

O

CH2COO C2H5 CH3 C

OH

CHCOOC2H5

Percentage of enol form decreases in the following order.Phenols > β-diketones with phenyl group > β-diketones > β-ketoesters > normal aldehyde and ketones.In the following compound, H-bond formation results in appreciable stability of the enol-form and sometime,resonance also stabilises enol form.

C H 3 C C H 2 C C H 3

OO

C H 3C C H C C H 3

OH

O

Keto enol tautomerism is also exhibited by cyclic compounds.

O

O H

OH

O H

There are certain cases when enol forms are more stable.

(i) If the enolic double bond is in conjugation with other double bond: For example,

H 3C — C — C H 2 — C — O E t

O O

H 3C — C C H — C — O E t

O — H O

H 3C — C — C H C — O E t

O — H O+ –

36% 64%

Enol form is stabilized by resonance and intramolecular H-bonding.

... 24��

(ii) Enols with strongly electron-withdrawing groups are stable at room temperature as the tautomerizationbecomes slow.

F3C — C — CF2H

O

F3C — C

OH

CF2

(iii) Aromaticity can stabilize the enol forms in ring compounds.

H

HH H

H

H

O

O O

OH

O HOH

(iv) Presence of bulky groups can destabilise keto compounds, e.g.

CH C R

Ph

PhO

C C R

Ph

PhOH

sp3 sp2

In keto form, sp3 hybridised carbon atom cannot arrange two bulky groups (bond angle 109 28'°� ) while in

enol form sp2 carbon atom (bond angle 120°) can arrange the bulky groups at greater distance and hencestabilises it.

Difference between resonance and keto-enol tautomerism

Resonance Keto-enol tautomerism

1. Resonance involves a shift in the Tautomerism involves a change in the positionposition of electrons only. of an atom, generally a H-atom.

2. Resonating structures are arbitrary and Tautomers exist in solution as they aredo not exist. different compounds.

3. These structures do not exist in equilibrium. They exist in dynamic equilibrium. 4. In resonance structures, the functional The tautomeric forms possess different

group does not change. functional groups. 5. Resonating structures lower the potential The tautomeric structures have no stabilization

energy and thus stabilize the molecule. effect on the molecule.

... 25��

Illustrative Examples

Example 1: Which of the following compounds does not show tautomerism?

(a) O

O

(b) OO

(c) O

O(d) C H C H O H

Solution: Among the given options, benzoquinone is very stable through conjugation. Hence, it willnot tautomerize.

O O O O+ –

S o on Answer: (b)

Example 2: Keto-enol tautomerism is observed in

(a) Ph C H

O

(b) Ph C CH 3

O

(c) P h C P h

O

(d) P h C C P h

O C H 3

C H 3

Solution: Ph C CH 3

O

Ph C

O H

C H 2

Only this compound has α-H atom which can involve in tautomerism. Answer: (b)

Nitro-aci tautomerism

CH 3 N O

O

CH 2 N O H

O

(Nitroform ) (Aci form )

... 26��

Nitrite-nitro tautomerism

H O N H N

O

O O

N itrite fo rm N itro fo rm

Introduction to Stereoisomerism

There are some reactions, which gives few products having same molecular formula, but their physical andchemical properties are different. Their products can only be distinguished on further analysis. Theircompounds differ in arrangement of atoms or groups in 2-D or 3-D space. Such compounds are known asisomers and the chemistry which deals with these compounds is called stereochemistry.

Definition: It is the study of spatial arrangement (arrangement of atoms or groups in two-dimensional orthree-dimensional space) of atoms in any given compound.

Stereoisomerism and Its ClassificationThis has the same structural formulae but different spatial arrangements.

Stereoisom ers

Configurational isom erism Conform ational isom erism

Geom etrical isom erism Optical isom erism

Conformations

In alkanes, the carbon-carbon bond is formed by the axial or head on overlapping sp3 hybrid orbitals of theadjacent carbon atoms. The electron distribution in the bond is symmetrical around the internuclear axis(molecular axis). The carbon-carbon single bond, therefore, allows freedom of the free rotation about the C— C bond because of axial symmetry. As a result of rotation about the C — C single bond, the moleculecan have different spatial arrangements of atoms or groups attached to the carbon atoms. The differentarrangements of a molecule can be obtained by rotation around C — C single bonds are called conformersor conformational isomers. The molecular geometry corresponding to a conformer is known as conformation.

Conformations of Ethane

In ethane, CH3 — CH3, the methyl groups can rotate about the C — C bond. As a result, an infinite numberof spatial arrangements of ethane are possible. There are two extreme forms. The staggered and eclipsedconformations. They can be represented by Sawhorse or Newman projections.

... 27��

HH

HH

HH

H

eclipsed fo rm

H

H

H

H

HH

staggered form

Conformations of ethane (Sawhorse model)

HH

HH

H

H

repulsion

H

H

H

H

H

H

eclipsed form staggered formConformations of ethane (Newman projection)

The staggered conformation is associated with minimum energy because the repulsive interactions betweenthe hydrogen atoms attached to the two carbon atoms are minimum due to the maximum distance betweenthem. The reverse is true with the eclipsed conformations due to the minimum distance between hydrogenatoms. Thus, staggered conformation is more stable than the eclipsed form. However, the energy differencebetween the staggered and eclipsed conformations is only 12.6 kJ mol–1. At any given time, we wouldexpect a greater percentage of ethane molecules to be in staggered conformation because of its lowerenergy.

12.6 kJ mol–1

eclipsed

staggered

Rotation about C – C bond of ethane

Pot

entia

l ene

rgy

However, at room temperature, it is usually not possible to separate them from one another.

... 28��

Conformations of Cycloalkanes

In cyclopropane and cyclobutane, there are large differences between the tetrahedral angle (109° 28´) andthe actual bond angles. In cyclopropane, the bond angle is 60° while in cyclobutane it is 90°. So thesemolecules have considerable strain and hence they are less stable and highly reactive. Cyclopropane andcyclobutane are planar molecules.

Higher cycloalkanes are not planar they are puckered. Thus, the normal bond angle (109° 28’) is retainedand they are free from any strain. Consequently they are called strainless rings.

In the case of cyclohexane, two strainless ring conformations are possible. They are called chair andboat form.

H

H

H H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

Chair form Boat form

The energy difference between the boat and chair conformations is 44 kJ mol–1. The chair form is morestable than the boat form and has the lower energy. In the boat form, any hydrogen atoms on the adjacentcarbon atoms correspond to the unfavourable eclipsed conformation. The chair form does not have theseunfavourable interactions and all the hydrogen atoms are fully staggered. Cyclohexane exists in chairconformation at room temperature.

44 kJ m ol– 1

22 kJ m ol– 1

28.4 kJ m ol– 1

half chair Tw ist boat half chair

chair formchair form

Pot

entia

l ene

rgy

Relative energies of the conformations of cyclohexane molecule

... 29��

Geometrical isomerism

Geometrical isomerism or the cis-trans isomerism is a consequence of restriction to rotation about aC = C bond.

H

H

C

H

C

H

H

H H

HC C

H

H

Rotation about C – C bond possible Rotation about C = C bond not possible

Let us consider two different compounds which contain the same atomic pattern: one is maleic acid andthe other fumaric acid. The two groups around the C == C bond can be arranged in two different ways to givetwo types of isomers. If similar groups are placed on the same side, then it is called cis and if on oppositeside then it is trans isomer.

H

C

H O O CC

C O O H

H H

CH O O C

CC O O H

Hcis – butenedioic acid trans – butenedioicacidMaleic acid Fumaric acidm.p. 130° C m.p. 287°µD = 1.69 µD = 0.0

These two acids thus show geometric or cis-trans isomerism. The geometric isomers differ in their physicaland chemical properties.The trans-isomer being symmetrical has zero dipole moment while the cis-form has appreciable value ofdipole moment

H

C C

C l

H

C l

H

C C

C l H

C l

µD = 1.89 µD = 0

Chemically, these isomers also behave differently. Maleic acid on heating to 150° C is converted to maleicanhydride while fumaric acid is recovered unchanged. Another example is

H

C C

H

H 3C CH 3

H

C C

HH 3C

CH 3

cis-2-butene trans-2-butene

... 30��

It is stated that the geometric isomerism is due to the presence of a C = C bond, but presence of a doublebond is not enough. In addition, the two groups attached to a carbon atom should be dissimilar.

H

C C

H

C 2H 5 C 2H 5

C C

H

C H 3C 2H 5

C H 3

C C

HC l

C l H

(Shows geometrical isomerism) (No geometrical isomerism) (No geometrical isomerism)

Other compounds containing a C = N, N = N also show geometrical isomerism.

Compounds containing a C – C bond also exhibit geometrical isomerism as the cyclic compounds. In suchcompounds the rotation about a C – C bond is not possible due to rigidity of the ring.

H

C O OH

H

C O OH HC O O H

H C O O H

cis-isomer trans-isomer

O H

HH

O H

O H H

H O H

cis-isomer trans-isomer

To find out the number of geometrical isomers, the following idea will help.

(i) If an unsaturated compound has two or more double bonds and at least two of the end groups aredifferent, then number of geometrical isomers are 2x, where x = Number of double bonds. e.g.2, 4-heptadiene has four geometrical isomers.

C

C

H C

H

H C 2H 5

H 3C — CH CH — CH CH — C 2H 5

H CH 3

C

C

C

H C

H

H 5C 2 H

H CH 3

C

cis-trans trans-trans

C

C

H C

H

H5C 2H

H 3C H

C

C

C

H C

H

H C 2H 5

H 3C H

C

trans-cis cis-cis

... 31��

(ii) If the compound consists of two or more double bonds and when two end groups are same, the

number of geometrical isomers is x 1 y 12 2− −+ ,

wherex

y2

= when even number of double bonds are present,

x 1y

2+= when odd number of double bonds are present,

x = Number of double bonds.e.g. 3, 5-octadiene has three isomers.

C

C

H C

H

H C 2H 5

H5C 2 — CH CH — CH CH — C 2H 5

H 5C 2 H

C

C

C

H C

H

H 5C 2 H

H C 2H 5

C

cis-cis trans-trans

C

C

H C

H

H C2H 5

H C 2H 5

C

C

C

H C

H

H 5C 2 H

H 5C 2 H

C

cis-trans trans-cis

Here cis-trans and trans-cis are same isomers.Again for 2, 4, 6-octatriene, let us find out the number of geometrical isomers.

H3C — CH CH — CH CH — CH CH — CH3

Number of geometrical isomers = x 1 y 12 2− −+ = ( ) ( )3 1 2 1 2 12 2 2 2 6− −+ = + =Since the number of double bonds are 3,

∴ x = 3; y = 3 1

22+ =

Optical IsomerismThe optical isomers have the same chemical reactions and will be alike in all physical properties. Theycan only be distinguished by their ‘action on plane-polarized light’. This activity is referred asoptical activity.Now a question arises: What is optical activity? So let us seek the answer to this question.

Light is propagated by a vibratory motion of the particles present in the atmosphere. Thus, in ordinary light,vibration occur in all planes at right angles to the line of propagation. In plane polarised light, the vibrationtakes place only in one plane, vibration in other plane being cut off by passing it through nicol prism.

... 32��

Ordinary lightwaves vibratingin all directions

N icolprismNicolprism

Plane po larized lightwaves vibratingin one direction

A nicol prism is constructed of iceland spar, a substance which is doubly refracting.It is so made that light moves vibrating in all directions except one cut off by refraction to oneside of the prism.

Certain organic compounds, when their solutions are placed in the path of a plane polarised light, have theproperty of rotating its plane through certain angle which may be either to the left or to the right. If thepolarised light has its vibration in the plane XY before entering such a solution, the direction on leaving it willbe changed to say X’Y’, the plane having been rotated through the angle ‘α‘.

Solution

X

Y

X'

Y '

X

Y

X'

Y '

α

This property of a substance of rotating the plane of polarised light is called optical activity and the substancepossessing it is said to be optically active.

Specific rotationIt is used for the measurement of optical rotations. It is defined as the number of degrees of rotationobserved when light is passed through one decimetre of its solution having concentration 1 g per millilitre.

[ ]t C

obsD C

° αα =

×� ,

where α → Specific rotation,

obsα → Observed angle of rotation,

→� Length of the solution in decimetre,

C → Concentration of the active compound in grams per millilitre.The observed rotation of the plane of polarised light produced by a solution depends on the following.1. The amount of substance in tube.2. Length of solution examined.3. Temperature of the experiment.4. Wavelength of light used.

��

... 33��

It is customary to denote the temperature and wavelength of light used as a subscript. [ ]t C

D

°α → stands for

specific rotation at temperature t degrees celsius and using D-line of sodium light.The specific rotation may be different in different solvents and different concentration, e.g.

[ ]25

D24.7

°α =

The positive sign indicates the direction of rotation towards right, and the negative sign indicates thedirection of rotation towards left.

P lane rotatedto the left

(Laevo rotato ry)

P lane po larisedlight

P lane rotatedto the right

(Dextro rotatory)

The organic compound showing optical activity may exist in following three forms.1. Laevorotatory — rotating the plane of polarised light to the left [(–) form]2. Dextrorotatory — rotating the plane of polarised light to the right [(+) form]3. Racemic mixture — an inactive form which does not rotate the plane of polarised light at all. It is a

mixture of equal amounts of (+) and (–) forms.For example: Lactic acid may exist as:(i) (+) — Lactic acid(ii) (–) — Lactic acid(iii) ( ± ) — Lactic acid

Optical isomers have identical physical properties like boiling point, melting point, solubility, etc. Chemicalproperties of optical isomers are same towards achiral reagents, solvents and catalysts, etc., but towardschiral reagents they reacts at different rates.Such forms of the same compounds which differ only in theiroptical properties are called optical isomers and the phenomenon is termed as optical isomerism.

Physical properties of lactic acids

Name Melting point (°C) Density [ ]25

D

°α

(+) - Lactic acid 26 1.248 +2.24°(–) - Latic acid 26 1.248 –2.24°( ± ) Lactic acid 26 1.248 0.00°

... 34��

ChiralityA compound of carbon is said to be asymmetric or chiral if four different atoms or groups are bonded to it.The necessary condition for a molecule to exhibit optical isomerism is dissymmetry or chirality.Any molecule or compound is said to be chiral or dissymmetric if:(i) It is non superimposable on its mirror image,

For example:

(a) H CH C23

CH 3

C Cl

H

M irror p lane

Cl

CH 3

C CH CH2 3

H

(b)

M irror p lane

Br

F

C Cl

I

C l

F

C Br

I

(ii) It does not contain any element of symmetry.

A molecule as a whole is asymmetric if it does not possess any element of symmetry such as:1. Plane of symmetry: The plane of symmetry of a molecule represents a plane bisecting the molecule

such that each half of the molecule is the mirror images of the other half. It is represented by ‘σ’.For example:

HO OCF

C

C

F

F F

Plane of sym m etry

HO

H H

OH

CO OH

Plane of sym m etry

2. Centre of symmetry: The centre of symmetry is defined as that point (atom) in a molecule fromwhich if a straight line is drawn from any part of the molecule through that point (atom) and extendedto an equal distance by a straight line on the other side, a like point (atom) is encountered. It issymbolised as C, and is also called centre of inversion.For example benzene

C

C C

C

C CCentre of symm etry

... 35��

These stereoisomerism which are related to each other as non superimposable mirror images and do notcontain any element of symmetry are called as enantiomers.In general, the presence of a chiral centre is considered as the cause of optical activity, but there arecertain compounds such as:Allenes (R — CH == C == CH — R), which are chiral or dissymmetric but do not contain any asymmetriccarbon and show optical activity.

CH 3

HC C C

CH 3

H

CH 3

HC C C

CH 3

H

1, 3 - dim ethylallene

C CC

H

HH

H

The shaded and unshaded orbitals are at right angles to each other, and thus lie in different planes.

Fischer Projections

When we attempt to depict configuration, we face the problem of representing three-dimensional structureson a two-dimensional surface. To overcome this difficulty we use Fischer projections.The following points are followed in writing the Fischer formula.(1) The carbon chain of the compound is arranged vertically, with the most oxidized carbon at the top.(2) The asymmetric carbon atom is in the paper plane and is represented at the intersection of crossed

lines.(3) Vertical lines are used to represent bonds going away from the observer, i.e. groups attached to the

vertical lines are understood to be present behind the plane of the paper.(4) Horizontal lines represent bonds coming toward the observer, i.e. groups attached to the horizontal

lines are understood to be present above the planes of the paper.For example:

(1) C

CO OH

H C3 NH 2

H NH 2 C

CO OH

CH 3

H

C H

CH 3

CO OH

NH 2

... 36��

(2) C

CHO

CH 2OH

H C

CHO

CH 2OH

OH

H OH

CH 2OH

CHO

H

OH

Visua lise the structuresuch tha t the m ains

Carbon cha in is vertical

In order to convert one Fischer projection to another identical form, any two pairs of substituents (eveninterchanges) linked to the chiral carbon need to be interchanged. One interchange leads to an enantiomericstructure.

CH 2OH

OHH

CHO

erchangein t

Firs t

CH 2OH

OH

H

OHCerchangein t

Second C H 2O H

O H

H

O H C

(I) (II) (III)

Structres (I) and (III) are identical. Structures (I) and (II) are enantiomers.

A Fischer projection may be rotated by 180° in the plane of the paper (around a point) but not 90° or 270°to obtain an identical structure.

CH 2O H

O HH

CHO

180°

C H 2O H

H O H

C H O

(I) (II)

Structures (I) and (II) are identical. A 90° rotation results in an enantiomeric form.

... 37��

Absolute Configuration

While discussing optical isomerism, we must distinguish between relative and absolute configurations,about the asymmetric carbon atoms. Let us consider a pair of enantiomers, say (+) and (–) lactic acid.

H C OH

CH 3

CO OH

HCOH

CH 3

CO OH

(+) – Lactic acid (–) – Lactic acid

We know that they differ from one another in the direction in which they rotate the plane of polarised light.But we have no knowledge of the absolute configuration of the either isomer, i.e. we cannot tell as to whichof the two possible configurations corresponds to (+) acid and which is the (–) acid.Before we discuss about any type of absolute configuration, it should be remembered that there is norelation between configuration and sign of rotation.

D and L systemThis system has been used to specify the configuration at the asymmetric carbon atom. In this system,the configuration of an enantiomer is related to the two forms of glyceraldehyde were arbitrarily assignedthe absolute configuration.

H C OH

CH 2OH

CHO

HCOH

CH 2OH

CHO

(+) - glyceralodehydeD Configuration

(–) - glyceralodehydeL Configuration

If the configuration at the asymmetric carbon atom of a compound is related to D(+) glyceraldehyde, itbelongs to D-series and if it can be related to L(–) glyceraldehyde, the compounds belong toL-series.

For example:

H OH

CH 2OH

CHO

OH

OH

HHO

H

H

HOH

CH 2OH

CHO

OH

OH

H HO

H

H

D - (+) - g lucose L - (–) - glucose

... 38��

Compare the dotted brackets of glyceraldehyde and glucose, the positions of H, OH and CH2OH aresimilar in both cases and hence we can assign D and L configuration to glucose.For example:

H N2 H

CH 2OH

CO OH

L - (–) - SerineIn this example, if we replace - NH2 by OH, the configuration in the dotted bracket is similar to that ofL-glyceraldehyde. Thus, it belongs to L-serine

Optical Isomerism in Compounds with More Than One Asymmetric Carbon Atoms

An organic compound which contains two dissimilar asymmetric carbons, can give four possiblestereoisomeric forms. Thus, 2-bromo-3-chlorobutane may be written

CH 3 CH CH* *

Br C l1

CH 3 2 3 4

The two asymmetric carbons in its molecule are dissimilar in the sense that the groups attached to eachof these are different.

C2 has 3 3CH ,H,Br,CHClCH

C3 has 3 3CH ,H,Cl,CHBr CH

Such a substance can be represented in four configurational forms.CH 3

H C Br

H C Cl

CH 3

(I)

CH 3

Br C H

HCCl

CH 3

(II)

CH 3

CH 3

(III)

CH 3

H C Br

CH 3

(IV)

Br C H

H C Cl HCCl

The forms I and II are optical enantiomers and so are forms III and IV. These two pairs of enantiomers willgive rise to two possible racemic modifications.The I and III are not mirror images or enantiomers and yet they are optically active isomers. Similarly, theother two forms, i.e. II and IV are also not enantiomers but optically active isomers.Such stereoisomers which are optically active isomers, but not mirror images, are called diastereoisomersor diastereomers.Diastereomers have different physical properties and differ in specific rotation. They may have same oropposite signs of rotations.

... 39��

Isomerism in Tartaric Acid

The two asymmetric carbon atoms in tartaric acid are

CH(OH)CO OH*

CH(OH)CO OH*The possible arrangements are

CO OH

H C OH

HO C H

CO OH

(I)

CO OH

HCHO

H C OH

CO OH

(II)

CO OH

HO C H

CO OH

(III)

CO OH

H C OH

H C

CO OH

(IV)

HCHO

OH

Of these, formula IV when rotated through 180° in the plane of the paper becomes identical with formula III.Therefore, for tartaric acid we use only I, II and III arrangements.

Now, if the force which rotates the plane of polarized light is directed from H to OH.

(I) Structure I will rotate the plane of polarized light to the right and will represent (+) – tartaric acid.(II) Structure II will rotate the plane of polarized light to the left and will represent (–) – tartaric acid.(III) Structure III will represent optically inactive tartaric acid, since the rotatory power of the upper left

half of the molecules is balanced by that of the lower half.COO H

H C OH

COO H

H C OH

Plane of sym metry

Such arrangement is also termed as meso forms.

The physical properties of the four tartaric acids are tabulated below.

Name Melting point (°C) Density [ ]2∝

(+) – tartaric acid 179 1.760 +12°(–) – tartaric acid 170 1.760 –12°(±) – tartaric acid 206 1.697 0°meso – tartaric acid 140 1.666 0°

... 40��

Number of Possible Stereoisomers in Compounds ContainingDifferent Number of Asymmetric Atoms

1. When the molecule cannot be divided into two equal halves, i.e. the molecule has no symmetry and(n) is the number of asymmetric carbons atoms, thenthe number of d- and l- forms, a = 2n,The number of meso- forms, m = 0,Total number of optical isomers = a + m = 2n.

2. When the molecule can be divided into equal halves, i.e. the molecule has symmetry and thenumber of asymmetric carbon atoms is even, thenThe number of d- and l- forms, a = 2n–1,

The number of meso- forms, m

n1

22

− = .

Total number of optical isomers = a + m n

1n 1 22 2−−= + .

3. When the molecule can be divided into two equal halves and the number (n) of asymmetric carbonatom is odd, then

the number of d- and l- forms, a n 1

n 1 2 22 2−−= − ,

The number of meso- forms, m

n 12 2

2−

= .

Total number of optical isomers n 1a m 2 −= + = .In all the above cases,The number of racemic forms will be

a

2.

Resolution of Racemic Mixture

The separation of a racemic mixture into its (+) or (–) components is termed resolution. Since the opticalisomers have identical physical properties, they cannot be separated by ordinary methods such as fractionalcrystallization or fractional distillation.

The following methods can be used for resolution.1. Chemical methods2. Mechanical methods3. Biochemical methods4. Kinetic methods5. Selective adsorbtion

In general, we use chemical method of resolution in which ± mixture is mixed with another suitable opticallyactive isomer when the products are no longer mirror image isomers and may be separated by crystallization.For example, a solution of lactic acid may be treated with an optically active base such as the alkaloid(–) – Brucine. The resulting product will consist of two salts:

1. (+) – Acid (–) – Base2. (–) – Acid (–) – Base

... 41��

Inspection of the configuration of the two salts shows that they are not enantiomorphic. Therefore, theyhave different solubility in water and can be separated by fractional crystallization.

Chemical Reaction and Stereochemistry

Stereoselective reaction: A stereoselective reaction is one in which the reactant is not necessarilychiral (as in the case of alkyne), but in which the reaction produces predominantly or exclusively onestereoisomeric form of the product.Stereospecific reaction: A stereospecific reaction is one that produces predominantly or exclusively one

stereoisomers of the product when a specific stereoisomeric form of the reactant is used.It should be noted that all stereoisomeric reaction are stereoelective, but the reverse is not necessarilytrue.

For example synthesis of trans-2-butene from 2-butyne.

CH 3 C C CH 3

2-Butyne

(1) L i, E tN H 2

(2) N H Cl4

CCH 3

HC

CH 3

H

Trans - 2 - bu tene

Here the possible products can be trans-2-butene as well as cis-2-butene. But trans-2-butene is formedpredominantly thus showing that it is a stereo selective reaction.

Example of stereospecific reactionReaction of cis- and trans-2-butene with Br2. One stereoisomeric form of the reactant, say trans-2-butene,gives one product, i.e. meso compound. Whereas the other stereoisomeric form of cis-2-butene givesracemic mixture.

C

HH C3

C

H C H 3

Br2

CCl4

C

HBr

C

CH 3

Br H

trans - 2 - butene 2, 3 - D ibrom obutane(m eso com pound)

CH 3

... 42��

C

HH C3

C

Br2

CCl4

C

HBr

C

CH 3

Br Br

cis - 2 - bu tene(Racem ic mixture)

H C3 H CH 3

+C

BrH

C

CH 3

Br HC H 3

Regioselective Reactions: When a certain reaction that can potentially yield two or more constitutional

isomers actually produces only one (or a predominance of one), the reaction is said to be regioselective,

e.g. alkynes react with HCl and HBr to form haloalkenes or geminal dihalides following Markownikoff’s rule.

R1

C C R2

C

R1

HC

R2

X

HX HX R1

C C R2

H X

H XAsymmetric synthesisWhen a compound containing an asymmetric carbon atom is synthesized by ordinary laboratory methodsfrom a symmetric compound, the product is a racemic mixture. If, however, such a synthesis is carriedunder the ‘asymmetric influence’ of a suitable optically active reagent, one of the optically active isomersis produced in preference.This kind of process in which an asymmetric compound is synthesized from a symmetric compound toyield the (+) – or (–) – isomer directly, is termed as asymmetric synthesis.

For example, when pyruvic acid is reduced as such, it yields (±) – lactic acid. However, when pyruvic acidis first combined with an optically active alcohol, e.g. methanol (R OH) to form an ester, which is thenreduced the product upon hydrolysis yields (–) – lactic acid in exess.

CH — CO — COOH +3

Pyruvic acidCH — CO — COOR3

M enthyl pyruva te

+H O2

CH — C — CO OH + ROH3

H

OH(–) – lactic acid (excess)

CH

CH 3

H C2 CH 2

H C2 CHO H

CH

CH(CH )23

(M enthol)

... 43��

Where

R =

CH

CH 3

CH 2 CH 2

CH 2 CH — O —

CH

CH(CH )23

Importance of Stereochemistry1. Biological importance

Chirality is a phenomenon that pervades the universe.(a) The human body is structurally chiral with the heart lying the left of the centre and the lever to

the right.(b) Helical seashells are chiral.(c) All but one of the 20 amino acids that make up naturally occurring protein are chiral and have

L-configuration. The synthesized D-protein is resistant to break down by enzymes.

2. Chemical importanceStereochemistry plays an important role in deciding the physiological properties of compounds, e.g.(a) (–) – nicotine is much more toxic than (+) – Nicotine.(b) (+) – adrenaline is very active in construction of blood vessels than (–) – adrenaline.

3. Chiral drugsChirality is crucial for the effect of drugs. In a majority of cases only one enantiomers is found tohave the desired effect, e.g.(a) (+) – thalimide has the intented effect of curing morning sickyness.

(b) (s) – enantiomers of ibuprofen has the pain-relieving action.

There are several factors that influence the behaviour of organic molecules. Many of the properties such asacidic and basic strengths, boiling point, solubility in water, rates of chemical reactions and stability ofintermediates can be predicted on the basis of the following effects.(i) Resonance(ii) Inductive effect(iii) Electromeric effect(iv) Hydrogen bond(v) Steric efftect(vi) Hyperconjugation

... 44��

Resonance

Ethylene can be represented by a single Lewis structure. However, many molecules cannot be representedadequately by a single Lewis structure. Two or more structures must then be combined to provide a gooddescription of the molecule.

Examples are (C6H6), −2

3CO ion and SO2, etc.

Different structures of a molecule which differ in the position of electrons are called resonance contributingstructures or canonical structures. The actual structure of the molecule is the resonance hybrid of all thepossible resonance structures. The following structure is for benzene.

In case of an ion ( )−23CO the charge is equally distributed on all the atoms. This distribution is called

dispersal of charge and it leads to greater stability. Therefore, this mode of stabilizing substances is calledas resonance.

O C O– –

O C–

O

O O

O –

C O–

O

–

O C –O

O –

–Resonance hybrid structure

Resonance is also called mesomerism. It is represented by a double headed arrow ( )↔ . The resonance

hybrid is more stable than the contributing structures. The resonance energy of a system is the differencein energy between the actual energy of the hybrid and the energy of the most stable contributing structure.The resonance energy is measured by taking a model molecule. The resonance structures are only arbitraryor imaginary. Dispersal (or delocalization) of electrons decreases the potential energy of a molecule andenhances its stability.The resonance energy is a measure of the stability of the molecule. The larger is this energy, the morestable is the molecule. Benzene has resonance energy of 36 K cal per mole.One can draw different resonance structures of a molecule by using the following rules.

(i) The molecule should be planar.(ii) It contains an alternating system of single and double bonds (a conjugated system).(iii) The relative positions of nuclei should remain unchanged (e.g. tautomerism).(iv) The negative charge must preferably lie on the most electronegative atom.(v) The charge needs to be preserved in all the resonating structures.

... 45��

(vi) The electrons always move away from a negative charge.(vii) Arrows should be drawn to indicate the direction of the movement of electrons.

Resonance (mesomeric) effect is of two types:

(A) If an atom or group of atoms donates electrons through resonance, it is considered to exhibit+R or +M effect, e.g.

C C OH — C — C–

O H

(+M or +R effect of –OH group)

Groups showing +R effect are –NHR, –NR2, –NH2, –OR, –NHCOR, –X (halogens) etc.

(B) If atoms or groups withdraw electrons through resonance, then it is known as –R or –M effect.

e.g. C C CH O — C — C CH — O–

(–R effect of –CHO group)

Other groups showing –R effect are –NO2, –COOH, –CN, –COR, –CO2R, –CONH2, –COCl,–SO3H, etc.

For substituted benzene, if the substituent bears at least one lone pair of electrons, it shows +Meffect, while the substituent with partial or complete positive charge exhibits –M effect, e.g.

NH2 NH2 NH2

+ +

etc.

+R effect of –NH2 group.

N

O O

N

O O

N

O O

etc.+

+–R effect of –NO2 group.

... 46��

Inductive EffectThe electron shift takes place by resonance in a conjugated system. There is an additional way for similartransmission of electrons and this is done through the inductive effect (I). This effect takes place when agroup attached to the carbon chain has the tendency to release or withdraw electrons through the chain. Ittakes place in a saturated carbon chain unlike resonance. It is of two types: + I (i.e. the group attached tothe chain is electron-donating) and – I (i.e. the group attached to the chain is electron-withdrawing).

Group showing +I effect: CH3, C2H5, CH(CH3)2, C(CH3)3, O–,

Group showing –I effect: –NO2, –CN, –N+(CH3)3, –F, –Cl, –Br, –I, –OCH3, –OH, –C6H5

Chloroacetic acid is a stronger acid than acetic acid because

Cl - CH2 C

OH

OCl-CH2 - C

O

O

H++

-I

Ka = 1.4 × 10-3

CH3-C

OH

OCH3-C

O

O

H++

+I

Ka = 1.75 × 10-5

The chloro-acetate ion is more stable than the acetate ion because of – I effect of the Cl group. In short,inductively an electron-withdrawing (– I) group has an acid strengthening effect and an electron-donatinggroup (+ I) has an acid weakening effect. The reverse is true for bases.

Features of inductive effect

The extent of inductive effect can be predicted by noting the following points.

(i) The larger is the electron-withdrawing effect of a group, the greater is the –I (inductive) effect.F CH2 COOH Br CH2 COOH

Ka3105.2 −× 3103.1 −×

(ii) Inductive effect is additive, i.e. more the electron-withdrawing groups attached to the chain,larger is the –I effect.

Cl3CCOOH Cl2CHCOOH

Ka1103.2 −× 2104.5 −×

(iii) Since this effect is transmitted through a chain it becomes less effective with distanceClCH2CH2COOH ClCH2CH2CH2COOH

Ka41032.8 −× 51002.3 −×

... 47��

Electromeric Effect

It is a temporary or time-variable effect which is observed in presence of reagents involving transfering ofelectrons in an unsaturated system.

X Y X Y

in presence of reagent +

in absence of reagent

The common example of electromeric effect is the partial charge development in an alkene during additionreaction.

R

C

H

CH

HC — C

R

H

H

H

R — CH — CH3

Br

C — CH3

R

H

HBr1 2

+ –

Br – + H+

C1-atom is having positive charge as it has been neutralized by +I effect of R group and has been attackedby Br – to give the addition product.

The Hydrogen Bond

A hydrogen bond is formed when a hydrogen atom lies between two electronegative atoms. The force ofattraction between a hydrogen atom attached by a covalent bond in one molecule and an electronegativeatom attached in the same or different molecule is called the hydrogen bond.

F

H H

F

H

F

The bond is formed due to electrostatic attraction and is represented by a dotted line. It is a weak bond andthe bond energy is of the order of 2-10 Kcal per mole.

There are two types of H-bonds.(i) Intermolecular, i.e. among different molecules.(ii) Intramolecular, i.e. within the same molecule.

Hydrogen bond effects the properties of a molecule such as boiling points, solubility in water, acidicstrength, basic strength, etc.

... 48��

The boiling point of H2S is 40° C while that of water is much 100° C. This large difference is explained on thebasis of intermolecular H-bonds in water leading to a large size cluster.

OH

H

O

O

O

H

H

H H

H

H

H-bond formation in water

Such an association does not exist in H2S as the sulphur atom has a poor electronegativity and large size.Similarly, boiling points of alcohols are higher since they are hydrogen bonded. Alcohols are soluble inwater because they form hydrogen bond with water.

Any molecule which can form hydrogen bond with water will be soluble in it.

o-nitrophenol has a lower boiling point than the p-isomer because the ortho isomer can form hydrogen bondintramolecularly and intramolecular H-bonding reduces the boiling point.

H

O

O

�N

O

O H

N

OO H

H

O

–

+

+–

p-N itrophenol(intermolecular H -bonding)

o-N itrophenol

– +

δ

δ

δ δ

Hydrogen bonds exist in primary amines, carboxylic acids, proteins, etc.

��

... 49��

Steric Effects

An energetically unfavourable effect on any physical or chemical property that results from van der Waal’srepulsion is generally termed as steric effect. The energy barrier to rotation in butane is more than that ofethane, as in the former, there are two –CH3 groups. Steric effects result from interactions between atomsor groups that are non-bonded to each other. It is the result of the presence of bulky groups. Thus, betweenquinuclidine and triethylamine the former is more basic. In triethylamine the three ethyl groups offerinterference for the donation of the electron pair but in quinuclidine these groups are pinned back and theelectron-pair is available for donation.

..N

..N

Quinuclidine Triethylamine

The presence of alkyl groups on the benzene ring also affects adversely the acidity of phenols.The ionization constant of phenols is of the following order.

O H

N O 2

O H

N O 2

C H 3CH 3

O H

N O 2

C H 3C H 3

2,6-dimethyl-4-nitrophenol has a value of ionization constant comparable to p-nitrophenol. But the acidityof 3,5-dimethyl-4-nitrophenol is almost 10 times lower. This reduced acidity is explained in terms of stericeffects. In this compound the two methyl groups twist the nitro group out of the plane of the benzene ring.As a result, the phenoxide ion cannot be stabilized by resonance with the nitro group. This effect is alsotermed as steric inhibition of resonance.

Illustrative Example

Example 1: N, N-dimethylaniline and its 2,6-dimethyl derivative, which one is readily undergoingazo-coupling reaction? Explain.

Solution:

N

Me Me

N

N

M e M e

H N N Ph

N

Me Me

N N Ph

–H+ Ph N+

+

+

... 50��

The lone pair on N-atom is available for electrophilic attack due to the presence of twoelectrons donating —CH3 groups.

N

Me Me

H3C CH3

N

Me Me

H3C CH3

–

+

Due to the presence of four bulky —CH3 groups —N(CH3)2 cannot be in the same planewith benzene ring. Hence, it is not capable of azocoupling reaction due to steric hindrance.

Hyperconjugation

The relative magnitude of +I effect of alkyl groups follow the following order.

3 2 2 3Me C– Me CH– MeCH – CH –> > >However, in certain cases, particularly in case of conjugated double bond or benzene ring, the reverse ordercan be found. It can be explained by the extension of resonance, where delocalisation occurs in thefollwing manner:

H — C — CH CH2

H

H

H — C CH — CH2

H

H

H — C — CH CH2

H

CH3

H — C CH — CH2

H

CH3

(3 similar structures)

(2 structures)

H3C — C — CH CH2

H

CH3

H3C — C CH — CH2

H

CH3

(only one structure)

... 51��

H3C — C — CH CH2

CH3

CH3

No such extended structurepossible which can explainthe stability

This type of delocalization involves σ− and π−bond orbitals and is given the name hyperconjugation or nobond resonance. Like resonance more is the hyperconjugative structures, more stable is the ion ormolecule. The number of hyperconjugative structures in an alkene is obtained by the number of C — Hbonds attached to the carbon bonded directly to the double bonded carbon atoms.

Significance of this effect

a. Alkyl groups containing C — H bonds if attached to benzene ring, increases electron density bothat the ortho and para positions, e.g.

H — C — H

H

H — C — H

H

H — C H

H

H — C H

H

+ ++

b. Let us compare the stability of 1-butene and 2-butene.H3C — CH — CH CH2

H3C — CH CH — CH2

H+

(2 hyperconjugative structures)

H3C — CH CH — CH3 H3C — CH — CH

H

CH2

+

(6 hyperconjugative structures)

Hence, non-terminal alkenes or more substituted alkenes have better stability compared to terminal alkenes.

Relative strength of organic acids

To determine whether a molecule is acidic or not, first ionize the molecule. Then determine if the resultantion can be stabilized by any means (resonance, inductive effect, H-bonding, etc). If the ion can be stabilizedthen the parent molecule is always acidic.

Phenols are stronger acids than alcohols (the ion cannot be stabilized) but weaker than the carboxylicacids. This is so because after the carboxyl group loses a proton, the carboxylate anion is stabilized byresonance in which the negative charge remains on the oxygen atom unlike in phenol in which it isdelocalized over the C-atoms as well.

... 52��

R - C

OH

O

R - CO

OH

+R - C

O

O

Resonance structures

+

Electron-withdrawing groups ( )CN,NO2 −− present on the ring in phenols help to delocalize the negativecharge on the oxygen atom and stabilize the anion and hence increases the acidic strength. Electron-donating groups ( )33 CH,OCH −− do just the opposite.

N+

O H

O

N+

OO

N+

O O O

etc

O

O

N+

O

O

O

Resonance has opposite effect on the basicity of molecules. It rather decreases the basicity of aromaticamines. Basicity implies the capacity of a molecule to donate its electron pair. The electron-pair is resonatedwith the ring and is thus not available for donation. Aniline for this reason is less basic than cyclohexylamine.

2HN2

..HN 2HN

No resonancepossib le

2HN2HN

–

–

–

Now if we compare the acidic strength of o-, m- and p-nitrophenols, then

O

N

O

O

O

N

O

O

O

N

O

O

O

N

O

O

O

N

O

O

–

–

–o-nitrophenoxide ion

– –

... 53��

O O

NO2NO2

O

NO2

O

NO2

m-nitrophenoxide

O

NO2

O O

NO2 NO

O

–

O

NO

O

O

NO2

O

NO2

p-nitrophenoxide

Hence, it is clearly observed that o- and p-nitrophenoxides are stabilised through –R effect of –NO2 group.While such kind of stabilisation is absent in m-nitrophenoxide. Among ortho and para isomers,p-isomer is more acidic as ortho compound is stabilised through intramolecular H-bonding. Hence, it hasless tendency to donate proton (H+).

OH

NO

O

δ+

δ−

Aliphatic and aromatic carboxylicacids

The strength of an organic acid RCOOH depends on the stability of RCOO– (carboxylic ion) as comparedto RCOOH in aqueous solution.For an aliphatic acid, the stability of carboxylate ion determines the strength of the acid, e.g.

CH3 — CH2 — C — OH

O

H3C CH2 C — O + H

O

– +

... 54��

Since ethyl (CH3CH2) group is electron releasing, it will reduce the stability of carboxylate anion and henceacidic strength decreases.

CH3 — CF2 — C — OH

O

H3C CF2 C — O + H

O

– +

Here CH3CF2 group is much less electron donating than previous one and hence stabilises carboxylateanion. So the acid strength increases.

What is the decreasing order of acid strength of following?

H3C — CH2 — COOH H2C CH — COOH HC C — COOH

I II III

Acidic strength in aromatic acids depends on the stability of the benzoate anion. Electron-withdrawinggroups substituted in benzene ring disperse the negative charge of –COO– group and increases the stabilityof the anion while electron-donating groups have the adverse effect on the acid strength, e.g.if we consider the m- and p-isomers of methyl benzoic acid, m-isomer is a stronger acid than p-isomer.

C

OH O

H — C — H

H

C

HO O

H — C H

H

C

HO O

H

C

OH O

C — H

H

C

HO O

C

HO O

H

C — H

H

HC — H

H

H

H — C H+

etc.

+ + etc.

+

In p-isomer, the carbon atom attached to –COOH group gets additional negative charge due tohyperconjugation, which will increase the charge density on oxygen atom of –OH group and hence decreasesacidic strength. But such kind of destabilising factor is absent in m-isomer.

��

... 55��

On the other hand, m- and p-nitro benzoic acids have different characteristics. Since –NO2 group is anelectron-withdrawing group exerting –I as well as –R effect on the benzene ring, we can explain the acidicstrength as follows.

C

HO

O

N

HO

NOO

C

HO

O

etc.

N

O

O

C

OH

O

N — O

O

C

OH

O

etc.

In m-isomer, the carbon of –COOH group has not carried positive charge in any resonating structure.Here only –I effect is predominating. Hence, p-isomer is more acidic.

All the ortho-substituted benzoic acids show a special effect called ortho effect. Irrespective of the natureof the group, most of the o-substituted benzoic acid is stronger than benzoic acid. It can be explained bythe fact that due to the presence of o- substituent coplanarity of –COOH group with phenyl ring is beingdisturbed and resonance effect of phenyl ring does not increase the electron density on–COOH group. So the acid strength increases.

Again, o-hydroxy benzoic acid (salicylic acid) is more acidic than its two isomers because of intramolecularH-bonding as follows.

OH

CO

O

(Carboxylate ion is stabilised by H-bonding)

... 56��

Same effect is observed with phthalic acid.

C — O

C O

O

H

O

(Anion is stabilised by intramolecular H-bonding)


Example 1: Arrange the following compounds according to the decreasing order of acidic strength.

(a)

COOH COOH COOH

OCH3

I

OCH3

II

OCH3

III

(b) HCOOH and

COOH

Solution (a): –OCH3 group has –I effect as well as +R effect. Due to +R effect its acidic strength

decreases in o- and p-isomers as compared to m-isomer where only –I effect predominates.

C

HO

O C

HO O

OCH3OCH3+

etc.

... 57��

C

HO

O C

HO

O

OCH3

OCH3+

etc.

Between o- and p-isomers, o-methoxy benzoic acid is more acidic due to ortho effect.Therefore, the decreasing order is II > I > III.

(b) In formic acid, the carboxylate ion is stabilised by equivalent resonating structure. Hence,its energy distribution is very regular.

HCOOH H H — C

O

O

H — C

O

O

+ +–

–

Whereas in benzoic acid, due to the +R effect of phenyl ring more negative charge will beaccumulated on oxygen atom of –COOH group and hence decreases the acid strength.

C O

HO

C O

HO–

etc.

∴ HCOOH >

COOH

Relative Strength of Organic Bases

Strength of organic bases increases as the ability to donate lone pairs of electrons increases for aliphaticand aromtic amines.

For aliphatic amines in aqueous solution, the electron-donating groups increase the basic strength.The decreasing order of basic strength of aliphatic amines should be

R3N R2NH RNH2 NH3> > >

But the stability of the protonated base in aqueous solution through H-bonding is also important. Tertiaryamine has minimum H-bonding stability.

... 58��

R — NH

H

OH2

H OH2

OH2

N

R

R

H

H

OH2

OH2

R — N — H OH2

R

R+

I

+

II

+

III

Now the combined factor is highest for secondary-amine, then primary amine and then tertiary-amine.So the decreasing order of basic strength isII > I > III.In gaseous medium, R3N > R2NH > RNH2 > NH3.

Electron-donating groups increase the basic strength and electron-withdrawing groups have adverse effecton basic strength. Consider the relative strength of aniline, nitroaniline and methoxyaniline.

NH2 NH2

+

etc.

Lone pair on N-atom involves in resonance. Hence, it is less available for protonation.

NH2 NH2

NO

O

NO

O

+

etc.

–

–R effect of –NO2 group decreases the basic strength to a greater extent.

NH 2 NH 2

O Me OM e

–etc.

+

–OMe group (methoxy) has –I effect but stronger +R effect which increases the charge density on –NH2

group and increases the basic strength.

So in case of aromatic bases to find out the strength of bases it should be considered if the lone pair onN-atom is involved in resonance or not. Involvement in resonance decreases the basic strengthand vice versa.

... 59��


Example 1: Arrange the following according to the increasing order of basic strength.

NH2

OMe

NH2

OMe

NH2

OMe

IIIIII

Solution: –OMe group has –I effect and also +R effect when it is present to o- and p-positions withrespect to other substituent.Again –I effect is least in p-isomer.Therefore, the increasing order of basic strength is II < I < III.

A reaction is said to have taken place when one compound has been converted to another compound sothat it no longer shows its initial properties. The organic compounds are bonded by covalent bonds andthey can undergo reactions only by breaking some bonds and forming new bonds. There is more than oneway in which this can happen but all reactions involve electrons. Electrons are unit negative charges andthe magnitude and disruption of charge are important consideration in the study of organic reactions, andeventually their mechanisms. A reaction mechanism is simply a detailed description of the sequence ofsteps involved in going from reactants to products.

Bond Cleavage

A covalent bond is a pair of electrons shared by two atoms. We are concerned with the manner a bond isbroken and the nature of the resulting fragments. There are fundamentally two different modes of bondcleavage or bond fission.

1. Heterolytic Cleavage

In this type of bond fission, the shared pair of electrons is retained by one of the separating fragments.

−+→ B:AB:A

A : B A– : B+

The species obtained after heterolytic cleavage or heterolysis are charged ions. The carbon containingions are of two types: (i) carbocations and (ii) carbanions.

2. Homolytic Cleavage

In this type of bond fission the bond is broken in such a manner that the shared pair of electrons is dividedequally between the two fragments, known as free radicals.

⋅⋅ +→ BAB:A

... 60��

Free radicals are uncharged. The species so produced above are called reaction intermediates. Reactionsinvolving heterolytic fission are known as ionic or polar reactions and those involving homolytic fissionare called as non-ionic or non-polar reactions.

The Carbocation or Carbonium ion

An ion with a positive charge on the carbon atom is called a carbocation. In a carbocation carbon atomhas six electrons in the valence shell, it is electron deficient, the bond angle is 120° and it is sp2 hybridized— it is planar, i.e. all the bonds lie in one plane. A carbocation can be stabilized by resonance or inductiveeffect or hyperconjugation, For example benzyl cation is stabilised through resonance as

Benzyl cation

CH2 CH2 CH2

etc

+

+ +

CH 2 CH CH 2 CH 2 CH CH 2 + +

Stabilised through resonance

Allyl cation

H2C CH

Vinyl cation

no resonance hence unstable.

Inductive effect:

Since a carbocation has a positive charge, an electron-releasing group with +I effect will stabilize it.

C H 3 C

C H3

C H 3

C H 3 C

C H 3

H

> > H C

C H 3

H

> >

3° 2°

+

1°

+ +

A tertiary (3°) carbocation is more stable than secondary (2°), which in turn is more stable than primary(1°). The resonance effect is always more predominant than the inductive effect in stabilizing an ion.

... 61��

Carbocations are stabilised by hyperconjugation also.

H 3C — C

CH 3

CH 2 — H

H 3C — C

CH 3

CH 2H

H 3C — C

CH 2H

C

CH 3

CH 3CH 3

CH 2H

+

+

etc. +

+

Thus, tertiary carbocation is more stable than secondary and so on.

Rearrangement of carbocations

H 3C — CH — CH 2 OH

CH 3

Conc. H 2SO 4

H+

H 3C — CH — CH 2 — OH 2

H 3C — CH — CH 2CH 3

H 3C — C — CH 2

CH 3

H

~H –

hydrideshift

H 3C — C — CH 3

CH 3

CH 3

+

Adition

+

2°(less stab le than 3°)

–H2O

+

1° cationvery less stable and w ill go for

rearrangem ent for better stability

~CH 3 sh ift–

+

(3° m ost stable cation)

Hence, 3° carbocation is the stable intermediate.

Is tertiary cation most stable cation always?

��

��

... 62��

The Carbanion

An ion with a negative charge on the carbon atom is called a carbanion. In a carbanion the carbon atomhas eight electrons. Hence, it is electron-rich. It is trigonal pyramidal like NH3. Carbon atom is sp3 hybridized.A carbanion can be stabilized by resonance or inductively by electron-withdrawing groups.

Resonance :

H

-

H

-

H

Cyclopentadienyl carbanionInductive Effect :

CH3C H

CH3

CH3

CH3 C

CH3

H

C

CH3

H

3° 2° 1°Electron-donating groups destabilize a carbanion while electron-withdrawing groups stabilize it.

NO 2 3OC H

>

−2CH −

2CH

Illustrative Examples

Example 1: Which one is more stable, 6 5 6 11C H or C H+ + ? Explain.

Solution:

+ sp

+ sp2

The positive charge is present on electronegative sp carbon atom. Hence, phenylcarbocation is less stable.While cyclohexyl cation is a secondary cation with sp2 hybridised carbon atom which ismore stable than a phenyl cation.

... 63��

The Free Radical

A species which contains an unpaired electron is called a free radical. A free radical has a total of sevenelectrons. It is considered electron deficient and neutral in nature. Geometrically a free radical may beplanar or pyramidal. All the factors that stabilize a carbocation also stabilize a free radical. A tertiary (3°)radical is more stable than the other two. The stability of tertiary radical can be explained by hyperconjugation.

Classification of Reagents

Nucleophilic Reagents (Nucleophiles)

A reagent which attacks the positive end of a polar bond or nucleus-loving is known as nucleophile.Generally, negatively charged or electron rich species are nucleophilic.

e.g. −Θ

−−−−− −− 3233 CH,HN,COOCH,I,CN,OCH,OH , 2 3 3 2H O, NH , NH — NH

+

N..

..N H 3,CH 3 — O — CH 3,

..

. ...

C 2H 5 — O H,. ...H 2O, etc.

All nucleophiles are in general Lewis bases.

Electrophilic Reagents (Electrophiles)

A reagent which attacks a region of high electron density or electron-loving is known as electrophile. Allpositively charged or electron deficient species are electrophilic.

3 2H , CH , NO , Cl , Br , Ag+ + + + + +

Neutral reagents which contain an electron-deficient atom are also electrophiles.AlCl3, SO3, BF3, SOCl2, POCl3, FeCl3, ZnCl2

All electrophiles are in general Lewis acids.

Carbenes

A carbene may be described as a divalent carbon compound. Here the carbon atom is linked to twoadjacent groups by covalent bonding. A carbene is neutral and possesses two free electrons, i.e. a total ofsix electrons. It is also considered as electron deficient.Carbene is of two types

(i) Singlet carbene: CH2 hybridisation sp2

it is v-shaped

(ii) Triplet carbene: CH2 hybridisation spit is linear shaped

Triplet carbene is more stable than single carbene.

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Carbene is an electrophile.

Types of organic reactions

Organic compounds undergo a large number of interesting reactions. These reactions are broadly classifiedinto following:

1. Substitution 2. Addition 3. Elimination

4. Rearrangement 5. Condensation 6. IsomerisationSubstitution reaction: During substitution one atom or group of atoms is replaced by other group or atom.It can be done by nucleophile, electrophile, or free radical.

Nucleophilic substitution: When a substitution reaction is brought about by a nucleophile, the reactionis called nucleophilic substitution reaction.

−− +→+− XOHROHXRSN1 Reaction: Unimolecular nucleophilic substitution reaction.It is a two-step process. First step involved the formation of carbocation which is slow and rate determiningstep.The rate of substitution depends on the concentration of the substrate.

CH3 — C — CH2Cl

CH3

CH3

OH–

slow

CH3 — C — CH2

CH3

CH3SN1 +

The carbonium ion formed can undergo rearrangement to give more stable carbonium ion before attack ofthe nucleophile.

CH3 — C — CH2

CH3

CH3

CH3 — C — CH2 – CH3

CH3

Fast OH–CH3 — C — CH2CH3

OH

CH3

+

+

1, 2-Methyl anion shift

In SN1 reaction, there can be racemisation. Order of reactivity of RX as SN1 3° > 2° > 1° > CH3XSN2 Reaction: This is called bimolecular nucleophilic substitution and it is one-step process. SN2 issecond order reaction, because substrate and nucleophile both are involved in the rate determining step.

H — C — B r + OH

CH 2CH 3

CH 3

–O H C Br

H CH 3

.

CH 2CH 3

FastH O — C — H

CH 3

CH 2CH 3

d– d–

Transition stateunstable

slow

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There is complete stereochemical inversions during SN2.For SN2 reaction, the order of reactivity is CH3X > 1° > 2° > 3° (Alkyl halide)

High concentration of the nucleophile favours SN2 reaction while low concentration favours SN1 reaction.The higher the polarity of the solvent, the greater is the tendency for SN1 reaction.

Addition

Two molecules react to form one. The reagent often adds to NCorOC,CC,CC ≡−=>−≡−<=>bond and the π bond is converted into σ bond. It can be electrophilic addition and nucleophilic addition.

ClCHCHClCHCH 22CCl

Cl22

4

2 − →=

OH2+H

OH

+

(Hydration)

Elimination Reactions

In most elimination reactions, two groups on adjacent atoms are lost as a double bond is formed.

CH3 – CH – CH – CH3

OH H

Conc. H2SO4

– H2OCH3 — CH CH – CH3

We divide elimination reactions into three classes.

(i) E1 reaction. It involves two steps. In first step, the C – L bond is broken heterolytically to form acarbocation (as in SN1 reaction).In second step, carbocation loses a proton from an adjacent carbon atom to form a π bond inpresence of base.

First step

CH3

CH3

C — CCH3

CH3

Slow

CH3

CH3

C — CCH3

CH3– L

H

L

H

–+

CarbocationSecond step

Base +CH3

CH3

C — CCH3

CH3

Fast

CH3

CH3

CCH3

CH3– H

H

C

++

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The first step is slow and rate determining step E1 reaction is favoured in compounds in which one leavinggroup is at a secondary or tertiary position.Evidence for E1 mechanism:(a) Follow the first order kinetics.(b) Where the structure permits are accompanied by rearrangement.

(ii) E1 CB

This reaction is called unimolecular conjugate base elimination reaction. First step consists of theremoval of a proton, H+ by a base generating a carbanion (II).In the second step, leaving group releases from carbanion (II) to form alkene.

– H

Fast

+Ph

HC — C

CH3

CH3–

(II)

Slow

– L–Ph

HC = C

CH3

CH3Ph

HC — C

CH3

CH3

H

L L

+ Base

Because step I (deprotonation) is fast and reversible, the reaction rate is controlled by how fast theleaving group is lost from the carbanion (II) (conjugate base). The loss of L– from (II) is rate determiningstep and is unimolecular. Hence, we call it E1 CB reaction.

(iii) E2 ReactionThis is one step process, which includes breaking of 2 σ bonds and formation of one π bondsimultaneously.

CH 3

CH 3

C —

H

LCH 3

CH 3

CH 3

CH 3

C = CCH 3

CH 3 –+ H -Base + L

C C

HB

L

Baseslow

+δ–

δ–

Transition state

–C

It is a bimolecular since substrate and base are involved in the rate determining step. E2 reaction does notproceed through an intermediate carbocation but through a transition state.Evidence for the E2 mechanism:(a) Follow the second order kinetics (b) These are not accompanied by rearrangement

The order of reactivity of alkyl halides in E1 and E2 are 3 ° > 2 ° > 1 °

E , E1 2

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1. The correct order of electronegativity of carbon atom in different hybrid orbitals is

(a) 2 3sp sp sp> > (b) 3 2sp sp sp> >

(c) 3 2sp sp sp> > (d) 3sp sp sp> >

2. The number of sigma and pi-bonds in 1-buten-3-yne are: (1989)(a) 5 sigma and 5 pi (b) 7sigma and 3 pi (c) 8 sigma and 2 pi (d) 6 sigma and 4 pi

3. Which of the following represent the given mode of hybridization sp2–sp2–sp–sp from left to right? (2003)

(a) CHH 2C C N (b) CH C C CHC(c) C C CH C2 CH 2 (d) H C2 CH 2

4. The compound in which C uses its sp3-hybrid orbitals for bond formation is: (1989)(a) HCOOH (b) (H2N)2CO (c) (CH3)3COH (d) CH3CHO

5. The enolic form of acetone contains: (1990)(a) 9 simga bonds, 1 pi bond and 2 lone pairs(b) 8 sigma bonds, 2 pi-bonds and 2 lone pairs(c) 10 sigma bonds, 1 pi bond and 1 lone pair(d) 9 sigma bonds, 2 pi bonds and 1 lone pair

6. Discuss the hybridization of carbon atoms in allene (C3H4) and show the orbitalπ − overlaps. (1999)

7. If two compounds have the same empirical formula but different molecular formulae, they must have: (1987)

(a) different percentage composition (b) different molecular weight(c) same velocity (d) same vapour density

8. The IUPAC name of the compound having the formula is: (1984)

C

CH 3

CH CH 2H C3

CH 3

(a) 3,3,3-trimethyl-1-propene (b) 1,1,1-trimethyl-2-propene(c) 3,3-dimethyl-1-butene (d) 2,2-dimethyl-3-butene

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9. The UPAC name of the compound (1987)

( )2 3 2CH CH – CH CH= is :

(a) 1,1-dimethyl-2-propene (b) 3-methyl-1-butene(c) 2-vinyl propane (d) none of the above

10. The IUPAC name of C6H5COCl is :(a) benzoyl chloride (b) benzene chloro ketone(c) benzene carbonyl chloride (d) c hloro phenyl ketone

11. What will be the IUPAC name of the compound given below?

CH 3

C H6 5

CHC C CH 3

CH 3 CHCl

CH 3

O

12. Suggest an IUPAC names for the following compounds.

(i)

3232

32223

3

CHCHCHCH||

CHCHCCHCHCHCH|

CH

(ii)

(iii)

3

3

C H|

C HC HC H

322 C HC HC H

3C H

52HC

C H

(iv)

13. Suggest an IUPAC name for each of the following.

(i) 3CH

3CHCH2=CHCH2CH2CH2N (ii) O CH 2 C H 2 O H

(iii)

O

C N(iv)

C N

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14. Write the IUPAC names of following compounds

(i) Br — CH — CH — (CH ) — CHClBr2 2 5

CH C l2

(ii)

CH — CH — CH 22

OH OH OH

(iii) *(iv)

NO 2

CO OH

HO OC

NO 2

(v)

(CH ) CC OOC H3 2 2 5

NO 2

15. An isomer of ethanol is: (1986)(a) methanol (b) diethyl ether (c) acetone (d) dimethyl ether

16. The number of isomers of C6H14 is: (1987)(a) 4 (b) 5 (c) 6 (d) 7

17. Only two isomeric monchloro derivatives are possible for: (1986)(a) n-butane (b) 2,4 diemthyl pentane(c) benzene (d) 2-methyl propane

18. How many monochloroderivatives are possible for the following?

(i) 3 2 2 2 2 3CH — CH — CH — CH — CH — CH (ii) C H 3 C H C H 2 C H 2 C H 3

C H 3

(iii) C H 3 C H 2 C H C H 2 C H 3

C H 3

(iv) C H 3 C C H 2 C H 2 C H 3

C H 3

C H 3

(v) CH 3 C CH2 CH3

CH3

CH 3

(vi) CH 3 CH CH CH 3

CH 3

CH 3

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19. Write the structure of all the possible isomers of dichloroethene. Which of them will have zero dipolemoment? (1985)

20. The number of geometric isomers in case of the following compoundCH3CH = CHCH = CHCH2CH3 is

(a) 3 (b) 2 (c) 4 (d) 5

21. Which of the following compounds will exhibit cis-trans (geometrical) isomerism? (1983)(a) 2-butene (b) 2-butyne (c) 2-butanol (d) butanal

22. Which of the following compounds will exhibit geometrical isomerism? (2000)(a) 1-phenyl-2-butene (b) 3-phenyl-1-butene (c) 2-phenyl-1-butene (d) 1,1-diphenyl-1-propene

23.

C

H3C

H3CC

H

C

H

H3C COOH

The above compound will exhibit(a) geometrical isomerism (b) optical isomerism(c) geometrical and optical isomerism (d) tautomerism

24. Which one of the following compounds will show geometrical isomerism?

(a) 3 2CH — CH CH= (b)

(c) C N — OH

CH 3

CH 3

(d) HO — N N — OH=

25. H

Br

OH

H

CH3

C2H5

HO

Br

H

H

CH3

C2H5

The molecules represented by the above structures are(a) diastereomers (b) enantiomers(c) meso-compounds (d) None of these

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26. Which of the following have asymmetric carbon atom? (1989)

(a) C

Cl

CH

H

H

H

Br

(b) C

H

CH

H H

Cl

C l (c) C

H

CH

H H

Cl

D (d) C

H

CH

Br OH

CH 3

H

27. How many optically active stereoisomers are possible for butan-2,3-diol ? (1997)(a) 1 (b) 2 (c) 3 (d) 4

*28. Which of the following compound will be optically active?

(a)

CH 3

H H

CH 3

(b)

F

F

(c) CH 3

CH 3

(d) H H

COO HCH 3

29. Write down the structures of the stereoisomers formed when cis-2-butene is reacted with bromine. (1995)

30. Which one of the following has maximum enol content?(a) Acetaldehyde (b) Butanone(c) Propanone (d) Acetyl acetaldehyde

31. Keto-enol tautomerism is observed in; (1988)

(a) C

O

HC H6 5

(b) C

O

C H6 5 CH 3

(c) C

O

C H6 5 C H6 5

(d) C

O

C H6 5 CH 2 C

O

CH 3

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32. The compound exhibits

(a) geometrical Isomerism (b) tautomerism(c) optical isomerism (d) conformational isomerism

33. Isomers which can be interconverted through rotation around a single bond are: (1992)(a) conformers (b) diastereomers (c) enantiomers (d) positional isomers

34.

CH 3

H

HH

H 3 2

CH 3

C2 is rotated anticlockwise 120° C about C2—C3 bond. The resulting conformer is (2004)(a) partially eclipsed (b) eclipsed (c) gauche (d) staggered

35. Boat structure of cyclohexane is unstable due to.(a) Non-bonding interaction of flagpole hydrogen.(b) Non planar structure.(c) Eclipsed conformation(d) Maximum van der Walls interaction.

*36. 2-Bromopropan-1-ol has almost zero dipole moment whereas 2-fluoropropan-1-ol has a measurabledipole moment. Explain.

37. The kind of delocalization involving σ -bond orbitals is called(a) inductive effect (b) resonance(c) hyperconjugation (d) None of these

38. Phenol is more acidic than

(a) C2H5OH (b)

OH

NO 2

(c)

O H

N O 2

(d) CH3COOH

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39. Arrange the following in the order of their increasing basicity. (1986)p-toluidine, N, N-dimethyl-p-toluidine, p-nitroaniline, aniline.

40. Which of the following species has the highest hyperconjugative stability?

(a) H3C — CH CH2(b) H3C — C CH2

CH3

(c) H3C — C CH

CH3CH3

(d) H3C — C C — CH3

H3C CH3

41. Which of the following is the most stable carbocation?

(a) +

3 3(CH ) C (b) +

3 2(CH ) CH

(c) CH 2

+(d)

+2 2 2CH CH NO

*42. Which one of the following free radicals has the least bond angle?

(a) • 3CF (b) • 3CH

(c) • 3 2CH(CH ) (d) • 3 3C(CH )

43. Inductive effect involves(a) σ-electron (b) π-electron(c) Both (a) and (b) (d) None of these

44. In which of the following, the resonance is possible?

(a) 2 2CH CH — CH — CHO= (b) 3 3CH COCH

(c) 2CH CH — CHO= (d) 2 2 2CH CH — CH — CH CH= =

45. The most stable free radical, will be

(a) (b)

(c) (d)

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*46. Account for the basicity order of the following compounds.

H

N

NH2

N

47. Arrange the following compounds in the increasing order of acidity.

O H

N O 2

O H

N O 2

O H

N O 2

, ,

(i) ( ii) ( iii)

48. Give the enol form of CH3 COCH2 COCH3 with intramolecular hydrogen bonding.

49. A nucleophile must possess(a) two lone pairs of electrons (b) an unpaired electron(c) an overall positive charge (d) tendency to donate electron pair

50. H3C — CH CH2

Br2/H2O H3C — CH — CH2

OH Br

The above reaction is an example of(a) addition reaction (b) condensation reaction(c) elimination reaction (d) substitution reaction

51. Which of the following is an electrophile?(a) RNH2 (b) : CH2

(c) CN– (d) H2O

52. The molecule which behaves as electrophile as well as nucleophile is(a) CH3Cl (b) CH3CN (c) CH3OH (d) CH3NH2

53. Which of the following is not a nucleophile?(a) CN– (b) OH– (c) NH3 (d) BF3

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54. Which of the following bond will undergo heterolytic cleavage most readily?(a) C — C (b) C — H (c) C — O (d) O — H

55. Carbanion is iso-structural with(a) free radical (b) ammonia (c) carbene (d) carbonium ion

56. Which of the following has the highest nucleophilicity?

(a) F− (b) OH− (c) 3CH − (d) 2NH −

57. Separate the following into electrophiles and nucleophiles.

3 2 3 2 3CH OH, NO , AlCl , H S, RSH, Br , OH , NH+ + −

58. Vinyl carbocation is unstable while vinyl carbanion is more stable. Explain.

*59. Triplet carbene is more stable than singlet carbene. Why?

*60. is more stable carbocation than . Why?

61. Give the correct order of stability of the following anions.

CH – C C , CH – CH – CH and CH – CH = CH3 3 2 2 3

– – –

62. Which of the following compounds exhibit geometrical isomerism?(a) C2H5Br (b) (CH)2(COOH)2(c) CH3CHO (d) (CH2)2(COOH)2

63. The lowest molecular weight alkane which is capable of showing enantiomorphism must have(a) four carbon atoms (b) five carbon atoms(c) six carbon atoms (d) seven carbon atoms

64. Geometrical isomerism is possible in(a) Acetone-oxime (b) Isobutane(c) Acetophenone-oxime (d) Benzophenone-oxime

*65. The specific rotation of a pure enantiomer is +10°. If it is isolated from a reaction with 30% racemisationand 70% retention, the observed rotation is(a) +10° (b) +14°(c) +7° (d) –7°

66. The number of possible racemic form of glucose is(a) 4 (b) 8 (c) 12 (d) 16

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67. Which of the following statements is not correct?(a) Enantiomers are similar chemically, but their rates of reaction with other optically active substancesare usually different.(b) A sample of sec-butyl chloride is optically inactive due to the absence of chiral carbon atom.(c) Diastereomers differ in their physical properties.(d) None of these

68. Predict the number and kind of stereoisomers possible for each of the following compounds.(a) 2-Butenoic acid(b) 1, 2-Dimethyl cyclopentane(c) 2, 3-Butane diol(d) Bromocyclohexane

69. The geometrical isomerism shown by

is

(a) cis, cis (b) trans, cis(c) trans, trans (d) None of these

70. Statement – I : p-hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid.Statement – II : o-hydorxybenzoic acid has intramolecular hydrogen bonding. (2007)

(a) Statement–I is True, Statement–II is True, Statement–II is a correct explanation for Statement–I(b) Statement–I is True, Statement–II is True, Statement–II is not a correct explanation forStatement–I(c) Statement–I is True, Statement–II is False(d) Statement–I is False, Statement–II is True

71. Statement – I : Molecules that are not superimposable on their mirror images are chiral.Statements – II : All chiral molecules have chiral centres. (2007)(a) Statement–I is True, Statement–II is True, Statement–II is a correct for Statement–I(b) Statement–I is True, Statement–II is True, Statement–II is not a correct explanation forStatement–I(c) Statement–I is True, Statement–II is False(d) Statement–I is False, Statement–II is True

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Nomenclature and Isomerism

1. a 2. b 3. a 4. c,d 5. a 7. b 8. b 9. b 10. c

11. 4-methyl-4-phenyl-3-(1’-chloroethyl)-2-pentanone

12. (i) 3 - Ethyl -3, 6 - dimethyloctane (ii) 5, 6 - Diethyl - 3 - methyldecane

(iii) 3 - Methyl - 4 - (1 - methylethyl) heptane (iv) 5 - (2 - Methyl propyl) nonane

13. (i) N,N-Dimethyl-5-amino-1-pentene (ii) 2-Phenoxyethan-1-ol (iii) 4, 4-Dimethyl-3-oxo-hexanenitrile (iv) 3 - Hexenenitrile

14. (i)1,8 dibromo-1-chloro-7-(chloromethyl) octane (ii) Propane 1, 2, 3-triol

(iii) Cyclo-dodeca-1,4,7,10-tetraene (iv) 66’-Dinitrodiphenyl-2,2’-dicarboxylic acid

(v) Ethyl 2-methyl-2-(3-nitrophenyl) propanoate

15. d 16. b 17. a,d 18. (i) Three (ii) Five (iii) Four (iv) Four (v) Three (vi) Two

20. c 21. a 22. a 23. b 24. d 25. a 26. c,d 27. b 28. d 31. d 32. b 33. a 34. c 35. a, c

37. c

38. a 39. d

40. a 41. a

42. a 43. c

45. b 46. Pyridine > Aniline > Pyrrole

47. III > I > II 48. C H C CH C C H3 3

OH

O

49. d

50. a 51. b 52. b 53. d54. d 55. b 56. c

57. Electrophiles: NO2+, AlCl3, Br+, Nucleophiles: CH3OH, H2S, RSH, OH–, NH3

61. C H – C 3

–C > C H – C H = CH 3

– > C H – C H – CH 3 2 2–

62. b 63. d 64. c 65. c

66. b 67. b

68. (a) two geometrical isomers (b) three stereosiomers (c) two enantiomers

(d) two conformational isomers 69. b 70. d 71. c

Nomenclature & Isomerism [1-77]

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