Nomenclature & Isomerism [1-77]

Nomenclature and Isomerism Chapter Objectives 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

Chemistry / Nomenclature and Isomerism

... 1

Concept Notes Introduction to Organic Chemistry Organic chemistry is a new branch of chemistry. It mainly deals with compounds made up of carbon. Other than carbon some elements like H, O, N, and halogen etc. are also involved. Starting from living beings to cooking gas, all belong to this new branch of chemistry. Let us see how it is important for coming examinations. Why are there so many carbon compounds This is because a single carbon atom is capable of combining with up to four other atoms. The unique property of carbon atom is that it can combine with other carbon atoms. The ability of carbon to form a long chain bonding with other carbon atom is called catenation. Therefore carbon atoms can form chains- and rings-type compounds. The simplest organic compounds are made up of only carbon and hydrogen atoms, e.g.

etc.

CH4 , CH3 — CH3 , CH3 — CH2 — CH3 and Methane

Ethane

Propane

These properties of carbon can be explained by hybridization.

2p 2p E lectron

2s

P rom otion

2s

sp

3

H ybridization

sp3 hybrid orbitals

1s

E nergy

C (ground state)

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 hybrid orbitals, e.g. CH4. Carbon in methane molecule forms four sigma bonds. The carbon here is sp3 hybrid carbon. The methane molecule has a tetrahedral shape. The C atom is at the centre of the tetragon (three-dimensional equilateral triangle) and the four H in the four corners of the tetragon. Each C — H bond in methane makes an angle of 109° 28’ with the other. Chemistry / Nomenclature and Isomerism

... 2

2p 2p E lectron

E nergy

2s

sp

2s

P rom otion

3

sp3 hybrid orbita ls

H ybridization

1s C (ground state) H 1s H σ

sp 3

109 °28 ´ sp 3

sp 3

H 1s

σ

σ

C

sp 3

H

H

H

1s

H

σ

1s H

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 shape and the bond angle is 120°. Here the the unhybrid pz orbitals or adjacent carbons overlap to form pi bond, e.g. ethylene molecule.

En ergy

2p

2s

U nhybridized p-orbita l

2p E lectron P rom otion

2s

sp 2 H ybridization

sp2 hybrid orb itals

1s C (ground state)


... 3

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

H -atom

H -atom 1s

pz

sp 2 sp 2

sp 2

sp 2 sp 2

pz

1s H -atom

pz

1s H -atom

H

H

H

120 ° C

1s

pz

sp 2

σ 120 °

120 ° O R 120 °

C

C

σ C π

σ

H

σ H H H H 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. The shape is linear and the bond angle is 180°. 120 °

σ

2p

2p 2p E lectron

2s

P rom otion

1s

E nergy

unh yb ridized orbita ls

sp

2s

H ybridization

sp hybrid orbita ls

1s C (ground state)

C (excited state)

sp hybridization in acetylene For sp hybridization the molecule of acetylene is an example. Here three of the bonds lie parallel to each other 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 2p y H sp 1s σ

2p y 2p z

2p z sp

σ

sp

sp H σ 1s

H

H

π σ —σ— σ H — C —— C — H —π—

π-bond Chemistry / Nomenclature and Isomerism

... 4

Sigma and pi bonds When a carbon atom forms a compound, it always forms covalent bonds. There are two types of covalent bonds: 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

+

+

+ +

Inte rn uclear axis

+

+

p-orbitals

+

S idew ay overla pping +

+ +

Final ele ctron cloud picture

+

+

or

+

Structural representations of organic compounds Organic compounds and their formulae are represented using many structures. Structural representation should clearly portray the structure, bonding and orientation of different atoms in the molecule. There are various representations. Lewis structure: Here the bond C — H or C — C is represented by .. or xx e.g.

H H C H H Four 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 H H H H H—C—C—C—C—C—H H

H

H

H

H


... 5

Condensed structural formulae: In this formula the bonds are also not shown. This is a simplified version of complete formula. Here only 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 those atoms other than carbon and hydrogen are represented. In these structures it is understood that there is a carbon atom at each ‘bend’ and that each carbon atom is attached to as many hydrogen atoms as are needed to complete its valence of four. A zig-zag structure is used for linear compounds. O OH P e n ta n e

E th y lm e th y l e th e r

B utan– 1–01

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

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

C H

H H

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

Cl OH

H Br


H

Cl –––– –– – C –– ––– –– – Br

OH

... 6

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

C H3

H

H

CH3

H CH3

H

C H3 H

H

H

H

C H3 H

H

H

C H3 H3 C H

H H C H3

H H

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.


... 7

Nomenclature Why we need nomenclature specially for organic compounds? As we have already discussed, carbon atom has unique ability to form stable molecules consisting of chains of carbon atoms of any length. Coupled with the observation that each carbon atom forms four bonds 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 of arbitrary 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 introduction to 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 attempts to produce a systematic nomenclature system before the IUPAC system and sometimes compounds have several 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 trivial names 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 the destructive distillation of wood chips. Wood chips can be distilled in this manner and will yield a very complex mixture. Analysis of this mixture will result in a large part of this alcohol. Another alcohol, the second member of the alcohol family is called ‘grain alcohol’. This trivial name came from the fact that one of 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 the derived system, the major hydrocarbon portion of the molecule was named as an alkyl group followed by the family name. The same simplest member of the alcohol family would be called ‘Methyl Alcohol’ because the formula CH3OH shows a methyl group attached to the functional group for the alcohol family. The second member’s name in the derived system would be ‘Ethyl Alcohol’ because the formula of this alcohol CH3-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 were discovered and synthesized. The International Union of Pure and Applied Chemists (IUPAC) developed a systematic approach to organic nomenclature. IUPAC came up with a set of rules to name organic compounds.


... 8

Table 1.2 : Names of Representative Groups Group

Name

Group

Name

CH3 —

Methyl

−C ≡ CH

Ethynyl

CH2CH3 —

Ethyl

CH3 CH2 CH2 —

Propyl

−CH2 CH = CH2

Allyl

CH3 CH -

Isopropyl

CH3

CH3 CH2CH2CH2 —

Phenyl

n-Butyl

O

Phenoxy

Isobutyl

CH 2

Benzyl

CH 3 CH

CH 2

CH3 CH3

O

CH 2 CH

C

sec butyl

Benzoyl

CH3

CH3 CH 3

C

tert butyl

CH 3 −CH2CH2CH2CH2CH3

Pentyl

CH3 CH3

C

CH2

Neopentyl

– CN

Cyano.

– NH2

Amino

– NH2OH

Hydroxyamino

– OH

Hydroxy

– OCH3

Methoxy

– CHO O

Formyl Keto or Oxo

C O

CH3

Acetoxy C

CH3

CH2

Methylene

–COOH –CH2OH

Carboxy Hydroxymethyl

CH2 = CH —

Vinyl

–SO3H

Sulphonyl


... 9

Naming organic compounds using IUPAC rules At its simplest, the IUPAC name for an organic compound contains these two parts: 1. 2.

A root indicating how many carbon atoms are in the longest continuous chain of carbon atoms. 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 the structure since structure is so important in establishing the physical and chemical behaviour of organic 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 of carbons. If the compound has functional group, the longest chain with functional group has to be considered.This number will be the prefix to the molecule name. 4

2 1

4

2

(ii)

3 IV

(i)

(v)

VI

1

5

III

VII

7 VIII

(vi)

(iv)

3

6 5

(v)

I

II

Longest chain here is 5-carbon chain The only possibility

The carbon chains are of 3 different lengths 6-carbon chain, 7-carbon chain, 8-carbon chain So the largest is 8 carbon chain. So prefix will be oct

Do You Know? Greek prefixes o di for 2 o tri for three o tetra for four o penta for five o hexa for six o hepta for seven o octa for eight


... 10

2.

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

7

4 6 3

8

1 2

Only one branch - 2 carbon - length branch 3.

Lowest number rule: Fix the position of branches. To do this make sure you follow the lowest number 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 final postion.The branch always gets the lowest possible position. 5

7

4

2

5 6 3

8

1 2

Branch is on C-4 carbon 4.

1

3 8

6 7

Branch is on C-5 carbon

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

Do You Know? Do you know names of carbon chains o Meth for one carbon o Eth for a two carbon continuous chain o Prop for a three carbon chain o But for a four carbon chain o Pent for a five carbon chain o Hex for a six carbon chain o Hept for a seven carbon chain o Oct for an eight carbon chain o Non for a nine carbon chain o Dec for a ten carbon chain


... 11

5

7

4 6 3

8

1 2

Saturated compound so suffix is (-ane) Name - 4-ethyl-octane These rules are adequate to name simple carbon compounds. But when it comes to multiple functional 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 the special structures and learn the specific rule to name them. 1.

Naming a compound with more than one substituent: When there are more than one substituent or 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 substituent is getting repeated. But the position of the substituent needs to be repeated as many times, with a comma in-between.

1

VII

VI

II

4

2

V

III

IV

3

I

6

5

7

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 parent chain, 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 hyphen between a position number and the name. The substituents should get the lowest possible numbers. 4

2

4

6

6

2

1 3

5

7

Name is 3-ethyl-5-methyl heptane Chemistry / Nomenclature and Isomerism

7

5

3

1

Order of substituents is written in alphabetical order. ... 12

Similar and different sets of substituents are present.

2 3 5

7 8

4

2

6

1

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 ending ane 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 not mentioned while writing the name of the compound. Such groups are: – COOH, – CHO, – COO R, – CN, – CONH2 and – COCl O

Exceptions are: C = C, C ≡ C and — 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

–oyl chloride

Chloroformyl

– nitrile

Cyano

O – CHO

– al

Formyl

– one

Oxo

- C - Cl

O –C–


– CN

... 13

Naming of Compounds Containing Two Functional Groups : Similar rules as discussed in naming of compounds containing one functional group are applied for naming compounds 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) and alkoxy (–OR) are always treated as substituents group. For compunds with both double and triple bonds, numbers as low as possible are given to double and triple bonds even though this may at times give “-yne” a lower number than “-ene”. When there is 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 triple bonds are not used as prefixes. The following examples will illustrate these rules. 4

2

3

1

3 - Buten -1-ol

CH2 = CH - CH2 - CH2 - OH (-OH is the principal functional group)

Note if the ending does not occupy the terminal position, then the letter ‘e’ is omitted in writing the name of the compound. 1

2

HC

C

4

3

CH2

C

5

1,4 - Pentadiyne

CH

Note here letter ‘e’ is retained because it occupies a terminal position. 4

2

1

2

3

HOOC - CH2 - CH2 - COOH 1,4-Butanedioic acid 6

5

4

3

1, 2 - Ethanediol 2

1

CH2 - CH - CH2 - CH2 - CH2 - CONH2 Cl

1

HO C H2 − C H2OH

CH3

6- Chloro-5- methylhexanamide


4

3

2

1

C H3 − C H − C H2 − C OOH | OH 3- Hydroxybutanoic acid

... 14

6

5

3

4

1

2

CH3 - CH2 - CH - CH2 - CH2CN

4- Nitrohexanenitrile 6

O

4- Oxopentanal 8

7

9

5

1

3

1

4, 5-diethyl-6-methylnon-5-en-1-ol

N, N - Dimethyl-2-pentenamine

1 2

CH3 CH3

OH

4

H

COOH

1 2

COOH

3

H

4

4

H

2

3

4

CH3 - CH2 - CH = CH - CH - N

2

H

1

2

3

4 ||

CH3 − C − CH2 − CH2 − CHO

NO2

5

5

HOOC

COOH cis-But-2-ene-1, 4-dioic acid 3

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

Same rules apply to compounds containing three functional groups. 4

2

3

CH2 — CH — CH2

3

2

1

CH2

CH

CH2

OH

OH

OH

COOH COOH COOH 1

5

1, 2, 3 - Propanetriol

propane - 1, 2, 3-tricarboxylic acids

These rules can be illustrated by the following examples. 2

1

3

5

4

6

CH2 = CH - CH2 - CH - CH2 - CH3

1

2

3

4

5

C H3 − C ≡ C− C H2 − C H3

CH3 4 - Methylhex -1 - ene 6

5

4

3

2 - Pentyne 2

CH3

3

2

1

CH3 - CH - CH2 - COOH

CH3 - CH2 - CH2 - CH - CH 1

OH 2 - Methylhexan - 3 - ol

4

CH3


CH3

3 - Methylbutanoic acid

... 15

CH3 4

3

4

5

1

2

CH3 - C -CH2 - COOC2H5

3

2

CH3 - CH2 - CH2 - CH - CH3 1

CH3

Ethyl 3, 3 - dimethylbutanoate 3

2

1

CHO

2 - Methylpentanal

CH3

CH3 - CH2 - CH2 - N CH3

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 chain hydrocarbons.

Cyclopropane 2.

Cyclobutane

Cyclopentane

Cyclohexane

If a single substituent is present on the ring, no numbering is needed. C H 2C H 3 CH 3

Methylcyclopentane 3.

Ethylcyclohexane

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

4.

3

2

CH3

4

1

C 2H 5

4 5 6

2

CH3

1

CH3

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

Cyclopentene Chemistry / Nomenclature and Isomerism

Cyclohexene ... 16

5.

If one or more double bonds and a group or groups are present on the ring, numbering is needed and the lowest number is assigned to the carbon atoms linked with the double bond. The numbers to the groups are assigned by going around the ring. CH3

CH3

3 4

3

5

1

1

4

6

3 - Methylcyclohexa -1, 4 -diene 6.

2

2

CH3

1, 3-Dimethylcyclobut-1-ene

If two functional groups are present on the ring, then the principal functional group gets the lowest number O OH 1

6

1

2

C

6 5

CH

2

3 4

3

5

CH 3

4

CH 3 4, 4- Dimethylcyclohex 2-ene-1-one

2-Ethynylcyclohex-1-ol

The following cyclic compounds have special nomenclature. COOH O

Cyclopentanone

Cyclohexanecarboxylic acid COCl

CO OCH 3

CONH2

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 atoms called polycyclic. These are of three main types and named differently. A compound containing rings with one common carbon atom is called spiro cyclic. A compound containing rings with an adjacent carbon atom in common is called a condensed ring system. Chemistry / Nomenclature and Isomerism

... 17

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.

B ridge head carbon 1-C arbo n

2-C arbo n

2-C arbo n

Bridgehead carbon The total number of carbon atoms are counted. In the above case there are seven. The compound’s name is derived from heptane and is called bicycloheptane(2 - cyclic rings). There are two carbons on each side and one above, therefore, the compound is named bicyclo [ 2.2.1] heptane. The numbers are written in parenthesis and in decreasing order. For substituted bicyclic ring systems, the number 1 position is assigned to the bridgehead carbon. Numbering of the remaining carbons proceeds around each ring in turn beginning with the largest ring. In doing this the lowest number is assigned to groups and double bonds. 7

7 1

1 6

5

6

2 4

3

2 4

5

Bicyclo [2. 2. 1] heptane

3

Bicyclo [2. 2. 1] hept -2- ene 8

7 1

7 1

6

2

O 2

5

4

6 3

1, 7, 7-Trimethylbicyclo [2. 2. 1] hept-2-one

5

CH3

4 3

3-Methylbicyclo [3. 2. 1] octane Note that the right hand ring is larger.


... 18

8 7

1

6

2 4

5

3

Bicyclo [2. 2. 2 ] octane

Bicyclo [1. 1. 0] butane

Bicyclo [4. 3. 0] nonane

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

Isomerism The existence of different compounds with the same molecular formula but different structural formula is called 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 are called 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. Isom erism

Structrual Isom erism

Stereoisom erism C hain Isom erism Functional Isom erism P ositional Isom erism

C onform ational Isom erism

C onfigurational isom erism

R ing C hain Isom erism M etam erism

G eom etrical isom erism

O ptical isom erism

Tautom erism


... 19

Chain Isomerism In this type of isomerism the compounds possess same molecular formula but differ in the arrangement of carbon chains. CH3

CH — CH3

CH3 − CH2 −CH 2 − CH3

CH3

B utane m .p. - 0.5°C

Isobu tane m .p . -12°C

Pentane has the following three chain isomers. CH3 CH2 CH2 CH2 CH3

m .p. 3 6°

n-penta ne

CH3 CH C H 2 CH 3

C 5 H 12 P entane

m .p. 2 7°

CH3

Iso -p entane

CH 3 CH3

C

CH3

N eo-pen tane

CH 3

CH3 CH2 CH2 CH2 OH

CH3 CH CH2 OH

1-Butanol

CH3 2-Methyl propanol

Functional Isomerism It is the type of isomerism in which the compounds possess the same molecular formulae but differ in the functional groups. CH3OCH3 CH3 CH2 OH Note: Alcohols and ethers are always functional isomers, so are carboxylic acids and their esters.

CH3 CH2 CH2COOH

O || CH3 CH2 C OCH3

Ask Yourself What other compounds would show functional isomerism?


... 20

Positional Isomerism It is the type of isomerism in which the compounds possess the same molecular formula but differ in the position of the same functional group. CH3 — CH2 — CH — CH3

CH3 — CH2 — CH2 — CH2 — OH

OH 2 - Butanol

1 - Butanol CH3 — CH — CH2 — CH2 — CH3

CH3 — CH2 — CH — CH2 — CH3

Cl

Cl 3 - Chloropentane

2 - Chloropentane CH3 — CH

CH — CH3

CH3 — CH2 — CH

2-butene

CH2

1-butene

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

OH

OH

NO2 NO2 NO2 o-nitrophenol

p-nitrophenol

m-nitrophenol

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

CH 3 C H 2 O C H 2 C H 3

C H 3O C H 2 C H 2 C H 3

E thox ye thane

O CH3 — CH2 — C — OCH2 — CH3 E thyl propanoate CH 3 — CH 2 — N H — C H 2 — C H 3 N -E thyl ethanam ine Chemistry / Nomenclature and Isomerism

1 - M eth oxypropane

O CH3 — C — OCH2 — CH2 — CH3 P ropyl e than oate C H 3 — N H — CH 2 — CH 2 — C H 3 N -M ethyl propylam ine ... 21

Ask Yourself 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 chain structure, i.e. C3H6. CH3 — CH CH2

propene CH2 CH2

CH2

Cyclopropane Tautomerism It is the phenomenon of existing of single compound in two readily interconvertible structures called as tautomers 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 the presence 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 an enol form. For example, in the presence of an acidic or basic catalyst a rapid equilibrium is established between an aldehyde or ketone and its isomeric (tautomeric) forms. O OH C C CH C Keto-form Enol-form

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

CH3 C

CH 3

CH3 C

CH2

Shows keto-enol tautomerism


Does not show keto enol tautomerism

... 22

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 solution simultaneously unlike the resonance structures. An important example of this type of tautomerism is acetoacetic ester.

O CH3

OH CH2COO C2H5

C

CH3

C

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. H O

O

O

O

CH3 C CH2 C CH3

C

CH3

CH

C

CH3

Keto enol tautomerism is also exhibited by cyclic compounds.

O

OH OH

OH

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, O O O — H O H 3C — C — C H 2 — C — O E t

H 3C — C

CH — C — OEt

36%

64% +O — H

O

–

H 3C — C — C H C — OEt Enol form is stabilized by resonance and intramolecular H-bonding.


... 23

(ii)

Enols with strongly electron-withdrawing groups are stable at room temperature as the tautomerization becomes slow. OH

O F3C — C — CF2H

(iii)

F3C — C

Aromaticity can stabilize the enol forms in ring compounds.

O H H

OH

H H

O

O H

(iv)

CF2

OH

OH

H

Presence of bulky groups can destabilise keto compounds, e.g. Ph Ph

CH

C

C

R Ph

Ph sp 3

O

sp 2

C

R

OH

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 hence stabilises it. Difference between resonance and keto-enol tautomerism

1. 2. 3. 4. 5.

Resonance

Keto-enol tautomerism

Resonance involves a shift in the position of electrons only. Resonating structures are arbitrary and do not exist. These structures do not exist in equilibrium. In resonance structures, the functional group does not change. Resonating structures lower the potential energy and thus stabilize the molecule.

Tautomerism involves a change in the position of an atom, generally a H-atom. Tautomers exist in solution as they are different compounds. They exist in dynamic equilibrium. The tautomeric forms possess different functional groups. The tautomeric structures have no stabilization effect on the molecule.


... 24

Illustrative Examples Example 1:

Which of the following compounds does not show tautomerism? (a)

(b) O

O

O

O O

(c) Solution:

(d)

CH

OH

O Among the given options, benzoquinone is very stable through conjugation. Hence, it will not tautomerize.

O Example 2:

CH

+ O

O

– O

S o on

Answer: (b)

Keto-enol tautomerism is observed in

O

O

(a) P h

C

(b) P h

H

O

(c) P h

C

(d) P h

Ph

C

CH3

O

CH3

C

C

Ph

CH3

OH

O Solution:

Ph

C

CH3

Ph

C

CH2

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

O

O CH3

N

O

(N itroform )


CH2

N

OH

(A ci form )

... 25

Nitrite-nitro tautomerism O H

O

N

H N O N itro fo rm

O

N itrite fo rm

Introduction to Stereoisomerism There are some reactions, which gives few products having same molecular formula, but their physical and chemical properties are different. Their products can only be distinguished on further analysis. Their compounds differ in arrangement of atoms or groups in 2-D or 3-D space. Such compounds are known as isomers 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 or three-dimensional space) of atoms in any given compound.

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

Ste re oisom ers

C onfigu rational isom erism

G e om etrical isom erism

C onform ational isom erism

O p tical isom erism

Conformations In alkanes, the carbon-carbon bond is formed by the axial or head on overlapping sp3 hybrid orbitals of the adjacent 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 molecule can have different spatial arrangements of atoms or groups attached to the carbon atoms. The different arrangements of a molecule can be obtained by rotation around C — C single bonds are called conformers or 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 number of spatial arrangements of ethane are possible. There are two extreme forms. The staggered and eclipsed conformations. They can be represented by Sawhorse or Newman projections.


... 26

H

H

H H

H H

H

H

H

H

H

H eclipsed form

stagge re d form

Conformations of ethane (Sawhorse model)

H

H H repulsion

H H

H H

H

H

H

H

H eclipsed form stagge re d form Conformations of ethane (Newman projection)

P otential energy

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

12.6 kJ m ol – 1

stagge red R otatio n ab out C – C b ond of ethane

However, at room temperature, it is usually not possible to separate them from one another. Chemistry / Nomenclature and Isomerism

... 27

Conformations of Cycloalkanes In cyclopropane and cyclobutane, there are large differences between the tetrahedral angle (109° 28´) and the actual bond angles. In cyclopropane, the bond angle is 60° while in cyclobutane it is 90°. So these molecules have considerable strain and hence they are less stable and highly reactive. Cyclopropane and cyclobutane are planar molecules. Higher cycloalkanes are not planar they are puckered. Thus, the normal bond angle (109° 28’) is retained and 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 and boat form. H

H

H

H

H H

H

H

H

H

H H

H

H

H

H H

H

H

H

H

H

H C hair form

H

B oat form

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

Tw ist boat

P otential ene rg y

half chair

44 kJ m ol – 1 22 kJ m ol chair form

half chair

28.4 kJ m ol– 1 –1

chair form

Relative energies of the conformations of cyclohexane molecule


... 28

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

H

C

C

H

H

H

H C

C

H

H H Rotation about C – C bond possible

H

Rotation about C = C bond not possible

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

H

H C

COOH C

C

HOOC

C

HOOC

COOH

cis – butenedioic acid Maleic acid m.p. 130° C µD = 1.69

H

trans – butenedioicacid Fumaric acid m.p. 287° µD = 0.0

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

H C

H

C

Cl C

Cl

Cl

C H

Cl

µD = 1.89

µD = 0

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

H C

H

C

H3C

C

CH3 cis-2-butene

CH3

H 3C

C H

trans-2-butene


... 29

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

CH 3

H C

C

C

C2H5

C2H5

H

C

C

C 2H 5

CH3

(Shows geometrical isomerism)

Cl

H

C H

Cl

(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 such compounds the rotation about a C – C bond is not possible due to rigidity of the ring.

H

H

H

COOH

COOH COOH cis-isomer H

COOH

H

trans-isomer H

H

OH OH cis-isomer

OH

OH H 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 are different, then number of geometrical isomers are 2x, where x = Number of double bonds. e.g. 2, 4-heptadiene has four geometrical isomers.

H 3C — C H

CH — CH

C 2H 5

H

C H — C 2H 5

C H C

C

C H

H

cis-trans

H

C H

CH3

H 3C

H C

C

tra ns-trans


H C

C CH3

C

C H

C 2H 5

H

C

C H

H

H 5C 2

C

C H

H

H 5C 2

C H

tran s-cis

H 3C

H

cis-cis ... 30

(ii)

If the compound consists of two or more double bonds and when two end groups are same, the number of geometrical isomers is 2x −1 + 2y −1 , x y = when even number of double bonds are present, where 2 x +1 y= when odd number of double bonds are present, 2 x = Number of double bonds. e.g. 3, 5-octadiene has three isomers. H 5C 2 — C H C H — C 2H 5 CH — CH

C 2H 5

H

H 5C 2

C H C

C H

H 5C 2

H

H C

C H

H

H

C 2H 5 tra ns-trans

H C

C

C H

H C

C

H C

C

H 5C 2

C

C

C H

C 2H 5

H

H

C C 2H 5

cis-cis cis-trans Here cis-trans and trans-cis are same isomers. Again for 2, 4, 6-octatriene, let us find out the number of geometrical isomers. H 3C — CH CH — CH CH — CH CH — CH

H 5C 2

H

tra ns-cis

3

3 −1 2−1 Number of geometrical isomers = 2x −1 + 2y −1 = 2( ) + 2( ) = 22 + 21 = 6 Since the number of double bonds are 3,

∴ x = 3; y =

3 +1 =2 2

Optical Isomerism The optical isomers have the same chemical reactions and will be alike in all physical properties. They can only be distinguished by their ‘action on plane-polarized light’. This activity is referred as optical 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 vibration takes place only in one plane, vibration in other plane being cut off by passing it through nicol prism.


... 31

N icol prism

O rdinary light w aves vibrating in all directio ns

P lane po larized lig ht w aves vibrating in one dire ction

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

S olution

Y

Y'

Y'

Y

This property of a substance of rotating the plane of polarised light is called optical activity and the substance possessing it is said to be optically active. Specific rotation It is used for the measurement of optical rotations. It is defined as the number of degrees of rotation observed when light is passed through one decimetre of its solution having concentration 1 g per millilitre. t °C α [α ]D = obs , l×C where α → Specific rotation,

αobs → Observed angle of rotation, l → 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. Chemistry / Nomenclature and Isomerism

... 32

t °C

It is customary to denote the temperature and wavelength of light used as a subscript. [α ] → stands for D 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°

[α]D

= 24.7

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

P lane rotated to the left (Laevo rotato ry)

P lane po larised light

P lane rotated to the right (D extro 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. Chemical properties of optical isomers are same towards achiral reagents, solvents and catalysts, etc., but towards chiral reagents they reacts at different rates.Such forms of the same compounds which differ only in their optical properties are called optical isomers and the phenomenon is termed as optical isomerism. Physical properties of lactic acids 25°

Name

Melting point (°C)

Density

[α]D

(+) - Lactic acid (–) - Latic acid ( ± ) Lactic acid

26 26 26

1.248 1.248 1.248

+2.24° –2.24° 0.00°


... 33

Chirality A 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:

M irror plane CH3

(a)

H 3C H 2C

CH3

C

Cl

C

Cl

C H 2C H 3

H

H M irror p lane F

F (b)

Cl

C

Br

Br

C I

I (ii)

Cl

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:

F

F

COOH

HOOC HO

C

OH H

H

C F

2.

F

P lane of sym m etry

P lane of sym m etry Centre of symmetry: The centre of symmetry is defined as that point (atom) in a molecule from which if a straight line is drawn from any part of the molecule through that point (atom) and extended to an equal distance by a straight line on the other side, a like point (atom) is encountered. It is symbolised as C, and is also called centre of inversion. For example benzene C

C

C

C

C

C entre of sym m etry C Chemistry / Nomenclature and Isomerism

... 34

These stereoisomerism which are related to each other as non superimposable mirror images and do not contain 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 are certain compounds such as: Allenes (R — CH == C == CH — R), which are chiral or dissymmetric but do not contain any asymmetric carbon and show optical activity.

CH3

CH3 C

C

CH3

CH3 C

C

H

H

C

C

H

H

1, 3 - dim ethylallene

H H C

C

C H

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 structures on 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:

COOH COOH

COOH

NH2 C

(1)

H 3C

H NH2

NH2

C

H

C

H

CH3

CH3


... 35

CHO CHO

CHO (2)

H

Visua lise the structure such tha t th e m ain s

C

C H 2O H

OH

H H

C arbo n cha in is vertical

C

OH

OH C H 2O H

C H 2O H

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

H

CHO H

OH

First in t e rch an ge

OH

OHC

CH2OH (I)

H S e co nd in t e rch an ge

C H 2O H

OHC

CH2OH

OH

(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.

CHO H

OH C H 2O H (I)

C H 2O H

1 8 0° HO

H CH O

(II)

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


... 36

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.

COOH H

C

COOH

OH

OH

CH3

C

H

CH3

(+) – 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 which of 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 no relation between configuration and sign of rotation. D and L system This 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 assigned the absolute configuration.

CHO H

C

CHO OH

OH

C H 2O H

C

H

C H 2O H

(+) - glyceralodehyde D C onfiguration

(–) - glyceralodehyde L C onfiguration

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

CHO

CHO

H HO H

OH H OH

OH H OH

H HO H

H

OH

OH

H

C H 2O H D - (+) - g lucose Chemistry / Nomenclature and Isomerism

C H 2O H L - (–) - glucose ... 37

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

H 2N

H C H 2O H

L - (–) - S erin e In this example, if we replace - NH2 by OH, the configuration in the dotted bracket is similar to that of L-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 possible stereoisomeric forms. Thus, 2-bromo-3-chlorobutane may be written

CH3 1

C* H

*H C

2

3

Br

Cl

CH3 4

The two asymmetric carbons in its molecule are dissimilar in the sense that the groups attached to each of these are different. C2 has CH3 ,H,Br,CHClCH3 C3 has CH3 ,H,Cl,CHBr CH3 Such a substance can be represented in four configurational forms. CH3 CH3 CH3 CH3

H

C

Br

Br

C

H

Br

C

H

H

C

Br

H

C

Cl

Cl

C

H

H

C

Cl

Cl

C

H

CH3 (I)

CH3 (II)

CH3 (III)

CH3 (IV )

The forms I and II are optical enantiomers and so are forms III and IV. These two pairs of enantiomers will give rise to two possible racemic modifications. The I and III are not mirror images or enantiomers and yet they are optically active isomers. Similarly, the other 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 diastereoisomers or diastereomers. Diastereomers have different physical properties and differ in specific rotation. They may have same or opposite signs of rotations.


... 38

Isomerism in Tartaric Acid The two asymmetric carbon atoms in tartaric acid are

* C H (O H )C O O H * C H (O H )C O O H The possible arrangements are

COOH H

C

OH

HO

C

H

COOH

COOH

COOH

HO

C

H

HO

C

H

H

C

OH

H

C

OH

HO

C

H

H

C

OH

COOH COOH COOH COOH (I) (II) (III) (IV ) 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) (II) (III)

Structure I will rotate the plane of polarized light to the right and will represent (+) – tartaric acid. Structure II will rotate the plane of polarized light to the left and will represent (–) – tartaric acid. 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. COOH P lane of sym m e try H C OH

H

C

OH

COOH Such arrangement is also termed as meso forms. The physical properties of the four tartaric acids are tabulated below. Name (+) – tartaric acid (–) – tartaric acid (±) – tartaric acid meso – tartaric acid

Melting point (°C) 179 170 206 140


Density 1.760 1.760 1.697 1.666

[∝]2 +12° –12° 0° 0°

... 39

Number of Possible Stereoisomers in Compounds Containing Different 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, then the 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 the number of asymmetric carbon atoms is even, then The number of d- and l- forms, a = 2n–1, The number of meso- forms, m =

n   −1 2 2  .

n

Total number of optical isomers = a + m = 2n−1 + 2 2 3.

−1

.

When the molecule can be divided into two equal halves and the number (n) of asymmetric carbon atom is odd, then n 1 − 2,

the number of d- and l- forms, a = 2n−1 − 2 2 The number of meso- forms, m = 2

n 1 − 2 2

.

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

Resolution of Racemic Mixture The separation of a racemic mixture into its (+) or (–) components is termed resolution. Since the optical isomers have identical physical properties, they cannot be separated by ordinary methods such as fractional crystallization or fractional distillation. The following methods can be used for resolution. 1. Chemical methods 2. Mechanical methods 3. Biochemical methods 4. Kinetic methods 5. Selective adsorbtion In general, we use chemical method of resolution in which ± mixture is mixed with another suitable optically active 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. 2.

(+) – Acid (–) – Base (–) – Acid (–) – Base


... 40

Inspection of the configuration of the two salts shows that they are not enantiomorphic. Therefore, they have 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 necessarily chiral (as in the case of alkyne), but in which the reaction produces predominantly or exclusively one stereoisomeric 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 necessarily true. For example synthesis of trans-2-butene from 2-butyne.

CH3

C

C

CH3

(1) L i, E tN H 2 (2) N H 4 C l

H

CH3 C

C

H

CH3 Trans - 2 - bu tene

2-B utyne

Here the possible products can be trans-2-butene as well as cis-2-butene. But trans-2-butene is formed predominantly thus showing that it is a stereo selective reaction. Example of stereospecific reaction Reaction 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 gives racemic mixture.

H 3C C

Br

C C l4 CH3

tra ns - 2 - butene

H C

B r2

C H

CH3

H

C H CH3 2, 3 - D ibrom obuta ne (m eso com pou nd) Br


... 41

H 3C

CH3

H

CH3

Br

C

H

Br C

C

B r2

+

C C l4

C

H

C Br

H 3C H cis - 2 - bu ten e

C Br

H

Br

CH3

CH3

(R acem ic m ixture) 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.

R

1

C

C

R

2

HX

R

H C R

1

2

C

HX

R

1

X

H

X

C

C

H

X

R

2

Asymmetric synthesis When a compound containing an asymmetric carbon atom is synthesized by ordinary laboratory methods from a symmetric compound, the product is a racemic mixture. If, however, such a synthesis is carried under the ‘asymmetric influence’ of a suitable optically active reagent, one of the optically active isomers is produced in preference. This kind of process in which an asymmetric compound is synthesized from a symmetric compound to yield 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 acid is first combined with an optically active alcohol, e.g. methanol (R OH) to form an ester, which is then reduced the product upon hydrolysis yields (–) – lactic acid in exess.

CH3 — CO — COOR M enthyl p yru va te

CH3 — CO — COOH + CH3 P yru vic acid CH H 2C

CH2

H 2C

CHOH CH C H (C H 3 ) 2 (M enthol)


+H 2 O H CH3 — C — COOH + ROH OH (–) – lactic acid (exce ss)

... 42

Where

CH3 CH R=

CH2

CH2

CH2

CH — O — CH C H (C H 3 ) 2

Importance of Stereochemistry 1.

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 importance Stereochemistry 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 drugs Chirality is crucial for the effect of drugs. In a majority of cases only one enantiomers is found to have 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 as acidic and basic strengths, boiling point, solubility in water, rates of chemical reactions and stability of intermediates 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


... 43

Resonance Ethylene can be represented by a single Lewis structure. However, many molecules cannot be represented adequately by a single Lewis structure. Two or more structures must then be combined to provide a good description of the molecule. Examples are (C6H6), CO32− ion and SO2, etc.

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

(

)

In case of an ion CO32− 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 called as resonance. O– O– O

O

–

C

O

–

O

–

C O

O

O

C

– O

–

– –O C 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 difference in 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 arbitrary or imaginary. Dispersal (or delocalization) of electrons decreases the potential energy of a molecule and enhances its stability. The resonance energy is a measure of the stability of the molecule. The larger is this energy, the more stable 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) (ii) (iii) (iv) (v)

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


... 44

(vi) (vii)

The electrons always move away from a negative charge. 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

– —C—C

OH

OH

(+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

—C—C

O

CH

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 +M effect, while the substituent with partial or complete positive charge exhibits –M effect, e.g.

+ NH2

NH2

+ NH2 etc.

+R effect of –NH2 group. O

O

O

N

O N

O

O N

+

etc.

+ –R effect of –NO2 group.


... 45

Inductive Effect The electron shift takes place by resonance in a conjugated system. There is an additional way for similar transmission of electrons and this is done through the inductive effect (I). This effect takes place when a group attached to the carbon chain has the tendency to release or withdraw electrons through the chain. It takes place in a saturated carbon chain unlike resonance. It is of two types: + I (i.e. the group attached to the chain is electron-donating) and – I (i.e. the group attached to the chain is electron-withdrawing). CH3, C2H5, CH(CH3)2, C(CH3)3, O–, –NO2, –CN, –N+(CH3)3, –F, –Cl, –Br, –I, –OCH3, –OH, –C6H5

Group showing +I effect: Group showing –I effect:

Chloroacetic acid is a stronger acid than acetic acid because

O Cl - CH2 C

O Cl-CH2 - C

OH

+

-I

H

+

O

Ka = 1.4 × 10-3

O CH3-C

O

+

CH3-C OH

+I

H

+

O

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-donating group (+ 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. Br CH2 COOH F CH2 COOH Ka 2.5 × 10 −3

(ii)

Inductive effect is additive, i.e. more the electron-withdrawing groups attached to the chain, larger is the –I effect. Cl2CHCOOH Cl3CCOOH Ka 2.3 × 10 −1

(iii)

1.3 × 10 −3

5.4 × 10 −2

Since this effect is transmitted through a chain it becomes less effective with distance ClCH2CH2CH2COOH ClCH2CH2COOH Ka 8.32 × 10 −4

3.02 × 10 −5


... 46

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

in presence of reagent X

+ X

Y

Y

in absence of reagent The common example of electromeric effect is the partial charge development in an alkene during addition reaction. R

1

C

2

H

C

R

H

H

Br

R — CH — CH3

–

H

HBr

C—C + –

H R

+ C — CH3

H

H

+

H Br

C1-atom is having positive charge as it has been neutralized by +I effect of R group and has been attacked by 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 of attraction between a hydrogen atom attached by a covalent bond in one molecule and an electronegative atom attached in the same or different molecule is called the hydrogen bond. F

F H

H

F H

The bond is formed due to electrostatic attraction and is represented by a dotted line. It is a weak bond and the 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, acidic strength, basic strength, etc.


... 47

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

H O H O

H

H

H

O

H O 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 in water because they form hydrogen bond with water.

Do You Know? 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 bond intramolecularly and intramolecular H-bonding reduces the boiling point.

OH δ

+

–

δ

H

O

O

N+

O –

δ O o-N itro phenol

δ H

H

N +

+

O

–

O

p-N itro phenol (interm olecular H -bo nding)

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


... 48

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

OH

OH

OH

CH3

CH3

CH3

CH3 NO2

NO2

NO2

2,6-dimethyl-4-nitrophenol has a value of ionization constant comparable to p-nitrophenol. But the acidity of 3,5-dimethyl-4-nitrophenol is almost 10 times lower. This reduced acidity is explained in terms of steric effects. 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 also termed as steric inhibition of resonance.

Illustrative Example Example 1:

N, N-dimethylaniline and its 2,6-dimethyl derivative, which one is readily undergoing azo-coupling reaction? Explain.

Solution:

Me

Me

Me

Me

N

N

N

+ +

Ph

N

–H

N H

Me

Me

+

N

N

+

Ph N


N

Ph ... 49

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

Me

Me

Me

Me +

N H3C

N CH3

H3C

CH3

– Due to the presence of four bulky —CH3 groups —N(CH3)2 cannot be in the same plane with 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.

Me3C– > Me2CH– > MeCH2 – > CH3 – However, in certain cases, particularly in case of conjugated double bond or benzene ring, the reverse order can be found. It can be explained by the extension of resonance, where delocalisation occurs in the follwing manner: H H — C — CH

H CH2

H

H—C

CH — CH2

H (3 similar structures)

H H — C — CH

H CH2

CH3

H—C

CH — CH2

CH3 (2 structures)

H H3C — C — CH

H CH2

CH3

H3C — C

CH — CH2

CH3 (only one structure)


... 50

CH3 H3C — C — CH

CH2

CH3

No such extended structure possible which can explain the stability

This type of delocalization involves σ− and π−bond orbitals and is given the name hyperconjugation or no bond resonance. Like resonance more is the hyperconjugative structures, more stable is the ion or molecule. The number of hyperconjugative structures in an alkene is obtained by the number of C — H bonds 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 both at the ortho and para positions, e.g. H H H H

H—C—H

b.

H—C—H

+

+ H—C H

+ H—C H

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

(2 hyperconjugative structures) H3C — CH

CH — CH3

H3C — CH — CH

CH2 H

+

(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 resultant ion can be stabilized by any means (resonance, inductive effect, H-bonding, etc). If the ion can be stabilized then the parent molecule is always acidic. Phenols are stronger acids than alcohols (the ion cannot be stabilized) but weaker than the carboxylic acids. This is so because after the carboxyl group loses a proton, the carboxylate anion is stabilized by resonance in which the negative charge remains on the oxygen atom unlike in phenol in which it is delocalized over the C-atoms as well. Chemistry / Nomenclature and Isomerism

... 51

O

O

R-C

R-C OH

+

H

O

+

R-C

O

O Resonance structures

Electron-withdrawing groups (− NO 2 , − CN) present on the ring in phenols help to delocalize the negative charge on the oxygen atom and stabilize the anion and hence increases the acidic strength. Electrondonating groups (− OCH3 , − CH3 ) do just the opposite. OH

O

O

O

etc

N O

+

N O

O

N

+

O

O

+

N O

O

+

O

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

.. NH2

NH2

N H2

NH2

NH2

–

– – N o re sonance po ssible

Now if we compare the acidic strength of o-, m- and p-nitrophenols, then – O O O O O O N N N – O O O – o-nitrophe noxide io n

O–

O N

O

O

–

N O


O ... 52

O

O

O

O

NO2

NO2

NO2

NO2

m-nitrophenoxide

O

O

O

– NO2

NO2

N O

O

p-nitrophenoxide

O

O

O

NO2

NO2

N

O O 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 has less tendency to donate proton (H+).

δ H

+

O

−

Oδ N O

Aliphatic and aromatic carboxylicacids The strength of an organic acid RCOOH depends on the stability of RCOO– (carboxylic ion) as compared to RCOOH in aqueous solution. For an aliphatic acid, the stability of carboxylate ion determines the strength of the acid, e.g. O O CH3 — CH2 — C — OH

H3C


CH2

– + C—O+H

... 53

Since ethyl (CH3CH2) group is electron releasing, it will reduce the stability of carboxylate anion and hence acidic strength decreases. O O – + CF2 C—O+H Here CH3CF2 group is much less electron donating than previous one and hence stabilises carboxylate anion. So the acid strength increases. H3C

CH3 — CF2 — C — OH

Ask Yourself What is the decreasing order of acid strength of following? H3C — CH2 — COOH H2C CH — COOH HC

C — COOH III

II

I

Acidic strength in aromatic acids depends on the stability of the benzoate anion. Electron-withdrawing groups substituted in benzene ring disperse the negative charge of –COO– group and increases the stability of 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. OH HO HO O O O

C

C

C etc.

H—C—H

H — C H+

H — C H+

H

H

H

OH

HO

O C

HO

O C

H

O C

C—H

H+ C—H

H+ C—H

H

H

H

etc.

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


... 54

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

O

C

O

C

etc.

N

N O

O

O

H OH

OH C

O

C

O etc.

N

O

N—O

O

O 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 nature of the group, most of the o-substituted benzoic acid is stronger than benzoic acid. It can be explained by the fact that due to the presence of o- substituent coplanarity of –COOH group with phenyl ring is being disturbed 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 intramolecular H-bonding as follows.

O C O

O H

(Carboxylate ion is stabilised by H-bonding)


... 55

Same effect is observed with phthalic acid.

O C

O

C—O

H

O (Anion is stabilised by intramolecular H-bonding)


Arrange the following compounds according to the decreasing order of acidic strength. COOH COOH COOH

OCH3

(a)

OCH3 OCH3 I

III

II COOH

(b) HCOOH and

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.

HO

HO C

O

O C etc.

OCH3


+ OCH3

... 56

HO

HO C

C

O

O etc.

OCH3

(b)

+ OCH3

Between o- and p-isomers, o-methoxy benzoic acid is more acidic due to ortho effect. Therefore, the decreasing order is II > I > III. In formic acid, the carboxylate ion is stabilised by equivalent resonating structure. Hence, its energy distribution is very regular. – O O HCOOH H—C H+ + H — C – O O Whereas in benzoic acid, due to the +R effect of phenyl ring more negative charge will be accumulated on oxygen atom of –COOH group and hence decreases the acid strength. HO HO – C O C O

etc. COOH

∴ HCOOH >

Relative Strength of Organic Bases Strength of organic bases increases as the ability to donate lone pairs of electrons increases for aliphatic and 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. Tertiary amine has minimum H-bonding stability.


... 57

H

OH2

R

+ R — NH

+

OH2

N

OH2 R

H

H

R

+

R—N—H H

OH2

OH2

OH2

R

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

Electron-donating groups increase the basic strength and electron-withdrawing groups have adverse effect on 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 etc.

N O

N

O

– O

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

NH2

NH2 –

OMe

etc.

+ OMe

–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 on N-atom is involved in resonance or not. Involvement in resonance decreases the basic strength and vice versa.


... 58


Arrange the following according to the increasing order of basic strength. NH2 NH2 NH2

OMe OMe OMe I Solution:

II

III

–OMe group has –I effect and also +R effect when it is present to o- and p-positions with respect 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 so that it no longer shows its initial properties. The organic compounds are bonded by covalent bonds and they can undergo reactions only by breaking some bonds and forming new bonds. There is more than one way in which this can happen but all reactions involve electrons. Electrons are unit negative charges and the magnitude and disruption of charge are important consideration in the study of organic reactions, and eventually their mechanisms. A reaction mechanism is simply a detailed description of the sequence of steps 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 is broken and the nature of the resulting fragments. There are fundamentally two different modes of bond cleavage 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. A : B → A + : B −

A:B A– : B+ The species obtained after heterolytic cleavage or heterolysis are charged ions. The carbon containing ions 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 divided equally between the two fragments, known as free radicals. A : B → A ⋅ + B⋅


... 59

Free radicals are uncharged. The species so produced above are called reaction intermediates. Reactions involving heterolytic fission are known as ionic or polar reactions and those involving homolytic fission are 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 atom has 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 inductive effect or hyperconjugation, For example benzyl cation is stabilised through resonance as + CH2 CH2 CH 2

etc

+

+ Benzyl cation

CH2

+ CH CH2

+ CH2 CH CH2 Stabilised through resonan ce

A llyl cation H2C

no resonance hence unstable.

CH

Vinyl cation

Inductive effect: Since a carbocation has a positive charge, an electron-releasing group with +I effect will stabilize it. C H3 CH3 CH 3

C H3 >

C

+

>

C H3

>

C H3

C H

+

>

H

C

+

H 1°

2° 3° 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.


... 60

Do You Know? Carbocations are stabilised by hyperconjugation also.

CH3 H 3C — C

CH3

+

H 3C — C C H 2H+

CH2 — H

C H 2H

CH3 + H CH2

etc.

+

H 3C — C

C

CH3 CH3 Thus, tertiary carbocation is more stable than secondary and so on. Rearrangement of carbocations

CH3

CH3 H 3C — C H — C H 2 O H

C onc. H 2 SO 4 H

+ H 3C — C H — C H 2 — O H 2

+

–H 2 O

A dition – ~C H 3 sh ift

+ H 3C — C H — C H 2C H 3

CH3 + H 3C — C — C H 2

2° (less stab le than 3°)

H

CH3 H 3C — C — C H 3 + (3° m ost stable cation)

~H –

1° cation very le ss stable and w ill go for rearrangem ent for better sta bility

hydride shift

Hence, 3° carbocation is the stable intermediate.

Ask Yourself

Is tertiary cation most stable cation always?


... 61

The Carbanion An ion with a negative charge on the carbon atom is called a carbanion. In a carbanion the carbon atom has 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 carbanion Inductive Effect : C H3 C H3 C H3

H

C

C H3

C

C H3

H

H

C H3

C

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

C H 2−

C H −2

> NO2

O C H3

Illustrative Examples Example 1:

Which one is more stable, C6H5 + or C6H11+ ? Explain.

+

sp

+

s p2

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


... 62

The Free Radical A species which contains an unpaired electron is called a free radical. A free radical has a total of seven electrons. It is considered electron deficient and neutral in nature. Geometrically a free radical may be planar 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

OH− , OCH3− , − CN− , − I− , CH3 COO − , NH2 , CH3− , H2O, NH3 , NH3 — NH2

..

..

..

..

etc.

H 2O , C H 3 — O .. — C H 3 , C 2 H 5 — O.. H , N H 3 , All nucleophiles are in general Lewis bases.

N ..

Electrophilic Reagents (Electrophiles) A reagent which attacks a region of high electron density or electron-loving is known as electrophile. All positively charged or electron deficient species are electrophilic. H+ , CH3+ , NO + , Cl+ , Br + , Ag+ 2

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 two adjacent groups by covalent bonding. A carbene is neutral and possesses two free electrons, i.e. a total of six electrons. It is also considered as electron deficient. Carbene is of two types (i) Singlet carbene: CH2 hybridisation sp2 it is v-shaped

CH2

hybridisation sp it is linear shaped Triplet carbene is more stable than single carbene. (ii)

Triplet carbene:


... 63

Do You Know? Carbene is an electrophile.

Types of organic reactions Organic compounds undergo a large number of interesting reactions. These reactions are broadly classified into following: 1. Substitution

2. Addition

3. Elimination

4. Rearrangement 5. Condensation 6. Isomerisation Substitution 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 reaction is called nucleophilic substitution reaction. R − X + OH − → R OH + X − SN1 Reaction: Unimolecular nucleophilic substitution reaction. It is a two-step process. First step involved the formation of carbocation which is slow and rate determining step. The rate of substitution depends on the concentration of the substrate. CH3 CH3 SN1 + CH3 — C — CH2Cl CH — C — CH2 3 – OH CH3 CH3 slow The carbonium ion formed can undergo rearrangement to give more stable carbonium ion before attack of the nucleophile. CH3 CH3 CH3 – + 1, 2-Methyl anion shift Fast OH CH3 — C — CH2CH3 CH — C — CH – CH CH3 — C — CH2 3 2 3 + OH CH3

In SN1 reaction, there can be racemisation. Order of reactivity of RX as SN1 3° > 2° > 1° > CH3X SN2 Reaction: This is called bimolecular nucleophilic substitution and it is one-step process. SN2 is second order reaction, because substrate and nucleophile both are involved in the rate determining step.

CH3

d– OH

– H — C — Br + OH

CH3

H C

.

d– Br

CH3 F ast

HO — C — H

slow C H 2C H 3

C H 2C H 3

C H 2C H 3

Transition state unsta ble Chemistry / Nomenclature and Isomerism

... 64

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 > C = C <, − C ≡ C−, > C = O or − C ≡ N bond and the π bond is converted into σ bond. It can be electrophilic addition and nucleophilic addition. Cl 2 → Cl CH 2 − CH 2 Cl CH 2 = CH 2  CCl 4

+

H

H2 O

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

Conc. H2SO4 – H2O

CH3 — CH

CH – CH3

H

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 a carbocation (as in SN1 reaction). In second step, carbocation loses a proton from an adjacent carbon atom to form a π bond in presence of base.

First step

L CH3 CH3

C—C

CH3 CH3

– –L Slow

CH3

CH3

–H Fast

H

CH3 C—C + CH3 CH3 H Carbocation

Second step Base +

CH3 CH3

+ C—C

CH3

+

CH3 CH3

C

C

CH3 CH3

H


... 65

The first step is slow and rate determining step E1 reaction is favoured in compounds in which one leaving group 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 the removal of a proton, H+ by a base generating a carbanion (II). In the second step, leaving group releases from carbanion (II) to form alkene. L L + CH Ph CH3 Ph – 3 –H Slow Ph CH3 + Base C—C C—C C = C – Fast H CH3 H CH3 – L H CH 3

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

E2 Reaction This is one step process, which includes breaking of 2 σ bonds and formation of one π bond simultaneously. δ– L L CH 3 CH3 slow – C C C—C B ase + CH 3 CH 3

H

δ–

B

H

Transition state

CH3 CH3

C =C

CH3

+ H -B ase + L

–

CH3 It is a bimolecular since substrate and base are involved in the rate determining step. E2 reaction does not proceed 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 3° > 2° > 1°

The order of reactivity of alkyl halides in E1 and E2 are E 1, E 2


... 66

Take-off 1.

The correct order of electronegativity of carbon atom in different hybrid orbitals is (a) sp > sp2 > sp3

(b) sp3 > sp > sp2

(c) sp3 > sp2 > sp

(d) sp > sp3 > sp

2.

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

(1989) (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) H 2 C CH C N (b) H C C C C C H (c) H 2 C C C C C H 2 (d) H 2 C CH2

4.

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

(1989)

5.

The enolic form of acetone contains: (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

(1990)

6.

Discuss the hybridization of carbon atoms in allene (C3H4) and show the π − orbital 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)

CH3 H 3C C

CH CH2

CH3 (a) 3,3,3-trimethyl-1-propene (c) 3,3-dimethyl-1-butene


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

... 67

9.

The UPAC name of the compound

(1987)

CH2 = CH – CH (CH3 )2 is :

10.

11.

(a) 1,1-dimethyl-2-propene (c) 2-vinyl propane

(b) 3-methyl-1-butene (d) none of the above

The IUPAC name of C6H5COCl is : (a) benzoyl chloride (c) benzene carbonyl chloride

(b) benzene chloro ketone (d) c hloro phenyl ketone

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

C 6H 5 CH3

C

O CH

C

CH3

CH3 CHCl CH3 12.

Suggest an IUPAC names for the following compounds.

CH3 | (i) CH3 CH CH2 CH2 C CH2 CH3 | | CH2CH3 CH2CH3

(ii)

CH2 CH2 CH3

(iii) C H 3 C H C H CH3 | CH CH3

(iv)

C 2H 5

13.

Suggest an IUPAC name for each of the following. (i) CH =CHCH CH CH N 2 2 2 2

CH3

(ii)

CH3

O CH2 CH2 OH

O

(iii)

CN


(iv)

CN

... 68

14.

Write the IUPAC names of following compounds C H 2C l (i) B r — C H 2 — C H — (C H 2 )5 — C H C lB r

CH2 — CH — CH2

(ii)

OH

OH

OH NO2

(iii)

HOOC

*(iv)

NO2

COOH

(C H 3) 2C C O O C 2 H 5

(v)

NO2 15.

16.

An isomer of ethanol is: (a) methanol (b) diethyl ether

(c) acetone

(d) dimethyl ether

(1986)

The number of isomers of C6H14 is: (a) 4 (b) 5

(c) 6

(d) 7

(1987)

17.

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

18.

How many monochloroderivatives are possible for the following? (i) CH3 — CH2 — CH2 — CH2 — CH2 — CH3

(ii) C H 3

CH

CH2

(1986)

CH2

CH3

CH3

CH3

CH3

(iii) C H 3

CH2

CH

CH2

CH3

(iv) C H 3

C

CH2

CH2

CH3

CH3

CH3 (v) C H 3

C

CH3 C H2

CH3

C H3 Chemistry / Nomenclature and Isomerism

(vi) C H 3

CH

CH

CH3

CH3 ... 69

19.

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

20.

The number of geometric isomers in case of the following compound CH3CH = CHCH = CHCH2CH3 is (a) 3 (b) 2 (c) 4

(d) 5

21.

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

H3C

H C

C

H3C

H C COOH

H3C

The above compound will exhibit (a) geometrical isomerism (c) geometrical and optical isomerism 24.

(1983)

(b) optical isomerism (d) tautomerism

Which one of the following compounds will show geometrical isomerism? (a) CH3 — CH = CH2

(b)

CH3 C

(c)

(d) HO — N = N — OH

N — OH

CH3 CH3

CH3

25.

H

OH

Br

H C2H5

HO

H

Br

H C2H5

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


... 70

26.

Which of the following have asymmetric carbon atom?

Cl Br

H

(a) H C C H H H

(1989)

Cl

(b) H C C C l H H

H Cl

H H

(c) H C C D H H

(d) H C C C H 3 Br OH

27.

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

*28.

Which of the following compound will be optically active?

(1997)

F CH3 CH3 (b)

(a)

H

H F

CH3 (c)

H

H

CH3

COOH

(d)

CH3 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)

O (a)

C 6H 5 C

O H

(b)

O (c)

C 6H 5 C

C 6H 5


(d)

C 6H 5 C C H 3

O

O

C 6H 5 C

CH2 C

CH3

... 71

32.

The compound exhibits

(a) geometrical Isomerism (c) optical isomerism 33.

(b) tautomerism (d) conformational isomerism

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

CH3 H 34.

H

H 3 2 CH3

H

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

(2004)

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 measurable dipole 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

OH (a) C2H5OH

(b)

NO2 OH NO 2

(c) Chemistry / Nomenclature and Isomerism

(d) CH3COOH ... 72

39.

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

(1986)

40.

Which of the following species has the highest hyperconjugative stability?

CH3 (a) H3C — CH

CH2

(b) H3C — C

CH3 CH3 (c) H3C — C 41.

CH

H3C CH3 (d) H3C — C

C — CH3

Which of the following is the most stable carbocation? +

+

(a) (CH3 )3 C (c) *42.

CH2

+

CH2

(b) (CH3 )2 CH +

(d) CH2CH2NO2

Which one of the following free radicals has the least bond angle? (a) •CF3

(b) •CH3

(c) •CH(CH3 )2

(d) •C(CH3 )3

43.

Inductive effect involves (a) σ-electron (c) Both (a) and (b)

(b) π-electron (d) None of these

44.

In which of the following, the resonance is possible?

45.

(a) CH2 = CH — CH2 — CHO

(b) CH3COCH3

(c) CH2 = CH — CHO

(d) CH2 = CH — CH2 — CH = CH2

The most stable free radical, will be

(a)

(b)

(c)

(d)


... 73

*46.

Account for the basicity order of the following compounds.

NH 2

N

47.

N

H Arrange the following compounds in the increasing order of acidity. OH

OH

OH N O2 ,

, NO 2 (i)

NO2 (iii)

(ii)

48.

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

49.

A nucleophile must possess (a) two lone pairs of electrons (c) an overall positive charge

50.

H3C — CH

CH2

Br2/H2O

(b) an unpaired electron (d) tendency to donate electron pair

H3C — CH — CH2 OH

51.

52. 53.

Br

The above reaction is an example of (a) addition reaction (c) elimination reaction

(b) condensation reaction (d) substitution reaction

Which of the following is an electrophile? (a) RNH2 (c) CN–

(b) : CH2 (d) H2O

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

(d) CH3NH2

Which of the following is not a nucleophile? (b) OH– (a) CN–

(d) BF3


(c) NH3

... 74

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

56.

(d) carbonium ion

Which of the following has the highest nucleophilicity? (a) F−

57.

(c) carbene

(b) OH−

(c) CH3 −

− (d) NH2

Separate the following into electrophiles and nucleophiles.

CH3 OH, NO2+ , AlCl3 , H2S, RSH, Br + , OH− , NH3 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. – – C H 3 – C C , C H 3 – C H 2 – C H 2 a nd C H 3– C H = C H –

62.

Which of the following compounds exhibit geometrical isomerism? (b) (CH)2(COOH)2 (a) C2H5Br (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 (c) Acetophenone-oxime

(b) Isobutane (d) Benzophenone-oxime

*65.

The specific rotation of a pure enantiomer is +10°. If it is isolated from a reaction with 30% racemisation and 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

... 75

67.

Which of the following statements is not correct? (a) Enantiomers are similar chemically, but their rates of reaction with other optically active substances are 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 (c) trans, trans

(b) trans, cis (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 for Statement–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 for Statement–I (c) Statement–I is True, Statement–II is False (d) Statement–I is False, Statement–II is True


... 76

Answer Key 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 H3 C

CH

O

C

C H3

49. d

52. b 56. c

53. d

O H

50. a 54. d

51. b 55. b

57. Electrophiles: NO2+, AlCl3, Br+, Nucleophiles: CH3OH, H2S, RSH, OH–, NH3 61. C H 3 – C C– > C H 3 – C H = CH – > C H 3 – C H –2 CH 2– 62. b

63. d

66. b

67. b

64. c

65. c

68. (a) two geometrical isomers (b) three stereosiomers (c) two enantiomers (d) two conformational isomers Chemistry / Nomenclature and Isomerism

69. b

70. d

71. c ... 77

Nomenclature & Isomerism [1-77]

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