X:01 JEE 19-2002 CHEM 201901

JEE (MAIN+ADVANCED)

STEREOISOMERISM

S.No Pages

1. Stereoisomerism 01 – 21

2. Exercise-1 (Subjective Questions) 22 – 25

3. Exercise-2 (Objective Questions) 26 – 38

4. Exercise-3 (Section-A) 39 – 40

5. Exercise-3 (Section-B) 41 – 44

6. Exercise-4 45

7. Exercise-5 46 – 52

8. Answer Key 53 – 55

CONTENT

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STEREOISOMERISM

ISOMERISM

INTRODUCTION

(i) The compoundwhich have the same molecular formula but differ in physical and chemicalproperties arecalled as Isomer and the phenomenon is called Isomerism.

(ii) The term ‘isomer’was given byBerzellius.(iii) The isomer was derived from Greek word meaning ‘equal or like part’ (isos= equal; meros = parts)

Isomerism

Structural Isomerism

Configurational Conformational

Stereoisomerism

Chain isomerism

Position isomerism

Ring chain isomerism

Functional isomerism

Metamerism

TautomerismGeometricalIsomerism

OpticalIsomerism

STEREOISOMERISM / SPACE ISOMERISM

Compounds having same molecular formula, same connectivityand structural formula but differ due tospatial arrangement of group or atom are said to be stereo isomers and phenomenon is termed asstereoisomerism.It is divided into two parts : (1) Configuration isomerism (2) Conformational isomerism

1. Configuration isomerism : Stereoisomers which are not interconvertible at room temperature areknown as configurational isomers

Configurational isomerism is further divided into two parts

(A) Geometrical isomerism

(B) Optical isomerism

GEOMETRICAL ISOMERISM

It is type of configurational isomerism which arises due to restricted rotation of atoms or groupsaround a double bonded system or cyclic system.

Conditions

(i) Restricted rotation(ii) Different functional grouparound system which is showing geometrical isomerism.


STEREOISOMERISM

For example:

(a) C = C

X Y

X Y

Not showing G.I.

(b) C = C

X Y

Y X

Showing G.I.

(Because same group X and Ypresent on each carbon)

Geometrical isomers are named as(a) cis- trans isomers (b) E and Z isomers (c) Syn-anti isomers

(a) Cis- trans isomers. When like atoms or groups attached at the same side of double bondedC-atom-called as cis. isomers. When like atoms or groups are on the opposite sides ofdoubly bonded carbon, are called. trans isomers.

a

b b

aC = C

cis isomers

a

b a

bC = C

trans isomers

C = C

Cl Cl

H H

cis isomer

C = C

Cl

ClH

H

trans isomer

C = C

HOOC COOH

H H

Maleic acid(cis isomer)

C = C

HOOC

COOHH

H

Fumaric acid(trans isomer)

(b) E and Z isomers,(i) The above system is used for derivatives of alkenes in which all the four substituents

should be different

aC = C

b

(ii) Following a set of rules (Cahn - Ingold-Prelog rules) the substituents on a double

bond are assigned priorities.(iii) The double bond is assigned the configuration E (From entgegen, the german word

for opposite) if the two groups of higher priority are on the opposite sides of thedouble bond.

C = C1

12

2

E form

(iv) On the other hand, the double bond is assigned the configuration Z (From zusamenn,the German word for together) if the two groups of higher priority are on the sameside of the double bond.


STEREOISOMERISM

C = C1 1

2 2Z form

Priority rule : cahn, ingold & prelog proposed a sequence rule.Rule-1 When atom or group of atom which are directly attach to the stereogenic

centre have higher atomic number will have higher proirity. Example

C = CF

I

Cl

Br(1)

(2) (2)

(1)Z-form

C = CNH2 NO2

CH3F(1) (2)

(1)(2)

E-form

Rule-2 When the atomic number will be same, priority assigned on basis of atomicweight.

C = CNH2 H

DCH – CH3 2

(1) (2)

(2) (1)E-form

Rule-3 When both atomic number and atomic weight are same then proirity will bedecided by the next joining atom.

C = CCH – CH – CH3 2 2

CH – CH3

CH3

CH – CH2 3

CH3

(E-form)

(1) (2)

(2) (1)

C = CC

CH3

CH3

CH3CH2

OH

(2)

(1)

CH – CH – CH2 3

CH3

(2)CH – CH3

CH3(1)

(Z-form)

Rule-4 If multiple bonded group attach to the double bonded carbon, then they areconsidered in following manner.

C =O C –O

O C

– C A

A

– C – A – C

CA


STEREOISOMERISM

For example :

Ex.-1 Cl – CH2

NH2C = C

CH = O

CH – OH2

(2)

(2)(E-form)

(1)

(1)

Ex.-2

NC

CH

C = CC CH

(2)

(Z-form)

(1) (1)

(2)CH = CH2

O

(c) Syn-anti isomers This type of isomerism exhibit by oximes and Azo compound. Oximesare the compounds formed by the reactions of aldehydes or ketones with hydroxyl amine.The products obtained have all the necessary conditions for Geometrical isomerism. i.e.restricted rotation they can be represented by the general formula

C = O + H – N – OH2

C = N – OH

Oxime

Oxime of aldehydeand oxime of unsymmetricalketone also showGeometrical Isomerism

C

N OH

HCH3

Syn form

Aldoximes

C

NHO

HCH3

Anti formWhen –OH group and H atom is same side, then it is syn form otherwise anti form

C

NOH

CH3CH – CH2 2

Anti form

C

NOH

CH3 CH – CH2 2

Syn form

In unsymmetrical Ketoxime, if-OH and the alphabetically alkyl present on the same side ofdouble bond, then it is called as syn form and other isomer is anti form

Geometrical Isomerism in Azo compound :

Eq. Ph – N = N – Ph

N

N

Ph

Ph

Syn form

N

N

Ph

Ph

Anti form


STEREOISOMERISM

Geometrical Isomerism in cyclo alkane :

In cyclic compound the rotation about C – C single bond is not free because of the regidity causedby the presence of other carbon of the ring which keep them tightly held, thus a disubstituted cycliccompound (having the two substitution at the separate carbon) will also show GeometricalIsomerism.The substituents on the same side are cis-isomers while the substituents on oppositesides represent trans-isomers.

H

Cl HH

H

Cl

Cl

Cl

H

H

H

Cl Cl

H

Cl Cl

H H

Trans form Trans form Cis form Cis form

No. of geometrical isomers in polyenes :

Case-1 Incase of unsymmetric alkene. If R1 R2 (R1 – CH = CH – CH = CH – R2)

n2.I.Gof.no

n number of double bond showing geometrical isomerism

Ex.- CH3 – CH = CH – CH = CH – CH2 – CH3

N = 2n = 22 = 4.

Case-2 Incase of symmetric alkene. If R1 = R2 (R1 – CH = CH – CH = CH – R2)

1P1n 22.I.Gof.no

if n is even no. then , P =2

n

if n is odd no. then, P =2

1n

Ex.- CH3 – CH = CH – CH = CH – CH = CH – CH3

here2

13P

3n

N = 22 + 21 = 6

OPTICAL ISOMERISM

Configurational isomers which are differ in their optical activity.

Optical Activity : The ability of optically active substances to cause rotation in the plane ofoscillations of polarized light is called optical activity. The substances which do not have anyinteraction with plane polarized light are called optically inactive substances.


STEREOISOMERISM

Following experiment was conducted to determine the optical activity of a substance

(a) Under ordinaryconditions, the light wavesoscillate in infinite number of planes passing through the lineof propagation at right angle.

(b) Plane polarized light is a light whose vibrations take place in only one of these possible planes.(c) Ordinary light can be turned into plane polarized light by passing it through Nicol prism (made up of

calcite, a special crystalline form of CaCO3)(d) When plane polarize light is passed through the liquid or dissolved state of such substances.(e) The plane of oscillation gets rotated through some angle towards left or right of the original plane of

oscillations.The substanceswhich rotate theplaneofpolarized light are calledopticallyactive substances.(f) The substances which rotate the plane of polarized light in the clockwise direction, i.e., towards right are

called dextrorotatory substances (Latin : dextro means right). This is indicated by putting a better d or(+) sign before the name of the substances.

(g) The substances which rotate the plane of polarized light in the anticlockwise direction, i.e., towards leftare called leavorotatorysubstances (Latin : laevus means left). This is indicated by putting letter ‘ ‘ or(–) sign before the name of the substance.

(h) The angle through which the plane of polarized light is rotated is represented byand is called observedangle of rotation.

Polarimeter

(i) The amount of rotation caused by an optical active compound depends on various factors-(a) Wavelength of light beam(b) Temperature(c) Densityor concentration(d) Length of the solution through which light beam has been passed.

Specific Rotation : The specific rotation of optically active compound can be defined as the amountof optical rotation observed when plane polarised light is passed through a solution of 1 gm per mlconcentration solution in a 1 dm long tube.

Specific rotation =C

][ t

l

Cause of optical activity:-(a) In order to exhibit optical activity, an object or molecule must be chiral.(b) Any molecule or object is said to be chiral if does not have anyelement of symmetry i.e. a plane of

symmetry or a centre of symmetry.


STEREOISOMERISM

Plane of symmetry:- A 'Plane' of symmetry' is a plane which divides an object in such a way that thepart of it on one side of the plane is the mirror image of that on the other side for eg. a ball issymmetrical while a hand is asymmetric

A chiral molecule or object is non-superimposable on its mirror images.(c) Hence, chiral objects or molecules are also called dissymmetric objects or molecules the word chiral

in fact is derived from the Greek word cheir, meaning hard.

Chiral centre:-(i)Acarbon atom bonded to four different atoms / groups in the molecule is called Chiral centre.

(ii) The chiral centre in the molecule is represented by asterisk (*). For example, the second carbon inlactic acid is chiral centre because it is bonded to four different groups(– H, – CH3, – OH and – COOH).

CH – C – COOH3

H

OH

*

Chiral centre

(iii) Some more examples of molecules having one chiral centre are

CH – CH – C H3 52

Br

*

2-Bromobutane

CH – CH – C H3 52

OH

*

2-Butanol

CH CH3 2CH – CH – C H2 52

CH3

*

3-Methylhexane

(d) Compounds which are mirror images of each other and are not superimposable are termedenantiomers and the phenomenon is described as enantiomerism.

Enantiomers : Enantiomers are molecules which are mirror images of each other i.e. they should benon-superimposable.

d d

c cb ba aMirror

Characteristics of Enatiomers :Some of the important characteristics of enantiomers are as given below :

(i) Enantiomers have identical physical properties such as melting point, boilingpoint, density, refractiveindex etc.


STEREOISOMERISM

(ii) Enantiomers are opticallyactive substances. They rotate the plane of polarized light in oppositedirections but to the equal extent.

(iii) Enantiomers have identical chemical properties. This means that they form same products as a resultof chemical combination. However, their reactivity, i.e., rates of reaction with other opticallyactivesubstances are different.

(iv) Enantiomers have different biological properties. For example (+) -sugar plays significant role inanimal metabolism. On the other (–)-sugar does not playanyrole in metabolism.

Representation of Enantiomers :(a) Perspective formulae (b) Fischer Projection Formulae

(a) Perspective formulae :(i) In this method, the four groups bonded to the chiral centre are represented by different means.

(ii)Anormal line represents the bond lying in the plane of paper.

(iii)Abroken line represents the bond going behind the plane of the paper and a solid wedgerepresents the bond projected out towards the viewer.

C

W

X

Y

Z

in the planebelow the plane

above the plane

(b) Fischer Projection Formulae(i) EmilyFischer devised a most simple and convenient method to represent the three dimensionalarrangement of groups bonded to chiral centre.

(ii) He used the point of intersection of two perpendicular lines to represent the chiral centre.

(iii) Horizontal lines represent the bonds projected out of the plane of the paper towards viewer.

(iv) Vertical lines on the other hand, represented the bonds projected back from the plane of thepaper away from the viewer.

(v) The Fischer projection formulae of enantiomers of 2-butanol and lactic acid are as under

Enantiomers of 2-butanol Enantiomers of lactic acid

CH3 CH3CH3 CH3

C H52 COOH COOHC H52

H HHO HOOH OHH H

Chiral centre


STEREOISOMERISM

Important PointsAbout Fischer Projection Formula :(1) Fischer projection of a stereoisomer must not be lifted from the plane of the paper and turned over.

Such an operation would result into an arrangement which is enantiomer of the original stereoisomer.

W W

Y Y

X ZZ X

(A) Enantiomer of A

Lift andturn over

(2) Fischer’s projection can be rotated in the plane of the paper about the chiral centre through 180º orits whole number multiple. Such an operation produces the same arrangement

W

WY

Y

X ZZ X

(A) (Same as A)

Rotate in the paperplane through 180°

(3) Fischer projection should not be turned in the plane of the paper through angle of 90º about the chiralcentre. Such an operation also produces enantiomer of the original compound.

W

W

Y Z

X YZ

X

(A) Enantiomer of A

Turn through 90°in the plane of paper

(4) Keeping one group as steady, the other groups in the Fischer projection can be rotated clockwise oranticlockwise simultaneously. Such operation would give same arrangement as the original.

W W

Y Z

X YZ X

(A) (Same as A)

Clockwise movementof groups other than W

Fixed

Diastereomers :The stereoisomers which are non-superimposable and do not bear mirror image relationship are calleddiastereomers. For Example, a compound having two asymmetric carbon atoms can have fourstereoisomers as shown below in the case of tartaric acid :

COOH

(I) (II)Mirror

One pair of enantiomers

OHH

HHO

CH3

COOH

HO H

H OH

CH3

COOH

(III) (IV)Mirror

Second pair of enantiomers

OH

OH

H

H

CH3

COOH

H

H

HO

HO

CH3


STEREOISOMERISM

(I) is mirror image of (II); similarly (III) and (IV) are mirror images of each other. Thus, the fourisomers are two pairs of enantiomers. Now compare (I) with (III); theyare neither superimposablenor they are mirror images. They are called diastereomers. (I) and (IV) are also diastereomers, asare (II) and (III) and (II) and (IV).

Characteristics of Diastereomers are :(i) Theyshow similar but not identical chemical properties. The rates of reactions are different.

(ii) Theyhave different physical properties, such as melting points, boilingpoints, densities, solubilities,refractive indices, etc.

(iii) Theycan be easilyseparated through fractional crystallization, fractional distillation, chromatography,etc.

(iv) Diastereomers are also encountered in the case of geometrical isomers :

C = C C = CC = C C = C

H3C H3CCH3

CH3H HH

H

Cis Trans

MESO COMPOUND

(a) The compounds containing two or more chiral centres but possessing achiral molecular structurebecause of having plane of symmetry, are called meso compounds.

CH3 CH2CH3CH2CH3

CH2CH3CH3

H H

H H

Br

Br OH

OH

Meso Compound

(2S, 3R) (2S, 4R)

* *

* *

CH2

Plane of symmetry is represented bydotted line

(b) Meso compounds do not rotate the plane of polarized light in any direction, i.e., they are opticallyinactive.

(c) This is because of achiral nature of their molecules. Because of the present of plane of symmetry theoptical rotation caused by half of the molecule is compensated by the rotation caused by the otherhalf.

(d) This cancellation of rotation within the molecule is referred to as internal compensation.(e) In short, the meso compounds are optically inactive due to internal compensation.


STEREOISOMERISM

CONFIGURATION(i) Three dimensional arrangement of groups about the chiral centre is called configuration(ii) There are two methods for assigning configuration to a molecule :

(a) Relative method (b)Absolute method.

(1) Relative method of configuration (D, L System) :(a) It uses D–Glyceraldehyde and L–Glyceraldehyde as the basis for the configuration determination.

(b) The stereochemical descriptor D refers to the arrangement in which – OH group attached to thechiral centre is on the right side of Fischer projection, whereas descriptor L refers to arrangement inwhich – OH group is on the left side of the Fischer projection of glyceraldehyde.

CHO

CH OH2

OHH C3

CHO

CH OH2

CH3HO

(D) (L)

(d) This method was found suitable for the study of optically active sugars as the sugars are defined aspoly hydroxy aldehydes and ketones.(e) Glyceraldehyde also contains hydroxy and aldehyde groups but this method cannot be used forthose molecules which do not process hydroxy aldehyde groups like CFCl BrI.(f) If two or more than two –OH groups are present then D, Lconfiguration is decided on basis of –OHgroup of lowest chiral in the Golden rule following fischer projection

Golden Ruleusually, the Fischer projection is drawn, so that the longest carbon chain in the molecule is vertical withhe highlyoxidised function at the top.Examples :

COOH

Et

OHH

COOH

CH3

OH

OH

H

H

D-form D-form

COOH

Et

OH

H

H

HO

L-form

(i) (ii) (iii)

R–S system (Absolute configuration)

R Rectus (Right)

S Sinister (Left)

R-S nomenclature is assigned as follow :

Step-I :- By the set of sequence rule, we give the priority order of atom or group connectedthrough the chiral carbon.

Step-II :- If atom/group of minimum priority present on the verticle line, then

Movement of eyes in clockwise direction = R


STEREOISOMERISM

Movement of eyes in anticlockwise = S

Movement of eyes taken from 1 2 3 through Low molecular weight group (if needed)

CWR

1

2

3

4

1 2

S

ACW

3

4

R

Step III :- If minimum proirity group present on the horizontal line, then

clockwise rotation S

anticlockwise rotation R1

2

ACW

3

4

R

1

2

ACW

34

R

Example:-

(1)

COOH

2

3

CH3

ACW

OH1H

4

R

(2)

COOH

2

3

OH

Br1

NH2

4

R

Optical Isomerism in compound containing no chiral carbon atom

Various types of compounds belonging to this group are allenes, alkylidene cycloalkanes, spirocompounds (spiranes) and properlysubstituted biphenyls.

(i) Allenes These are the organic compounds of the following general formula.> C = C = C <

Allenes containing even number of double bonds exhibit optical isomerism provided the two groupsattached to each terminal carbon atom are

C = C = C C = C = Ca aa x

b bb yor

(ii) Alkylidene, cycloalkanes and spiro compounds :When one or both of the double bonds in allenes are replaced by one and two rings, the resultingsystems are respectively known as alkylidene cycloalkanes and spiranes.

C

H

COOH

H C3

H

1-Methyl cyclo hexylidene-4-acetic acid(Alkylidene cycloalkane)

orC C C

a a

b b

CH2 CH2

CH2 CH2

Spiranes


STEREOISOMERISM

(iii) Biphenyls :Suitablysubstituted diphenyl compounds are also devoid of individual chiral carbon atom,but the molecules are chiral due to restricted rotation around the single bond between the twobenzene nuclei and hence theymust exist in two non-superimposable mirror images of each other.

COOH

COOHHOOC

HOOCF

FNO2

O2N

Number of optical isomers :Case - 1 When the molecule is unsymmetrical. (It cannot be divided into two halves)

Number of d and l isomers = 2n

Number of meso form = 0

Total number of optical isomers = 2n

Where n is the number of chiral carbon atoms.

For eg. 2, 3 - Pentane diol

CH

H C OH

H C OH

C H

3

2 5

|*— —

|*— —

| d and l isomers = 22 = 4

Case - 2 When the molecule is unsymmetrical, number of chiral carbon = even number

Number of d and l forms =2(n – 1)

Number of meso form =2(n/2 – 1)

Total number = addition of the above

= 2n – 1 +1–

2n

2

For eg.: Tartaric acid

H C OH

H C OH

|*— —

|*— —|

COOH

COOH

Number of d and l forms = 2(n/2 – 1) =2

Number of meso form = 2(2/2 – 1) =2º = 1

Total optical isomers = 3

Case- 3. When the molecule is symmetrical.

number of d and l form = 2(n – 1) – 2 (n/2 – ½)

Number of meso form = 2(n/2 – ½)

Total number of isomers = 2n–1

RACEMIC MIXTURE(a) An equimolecular mixture of a pair of enantiomers is called racemic mixture or racemic modification.(b) A racemic mixture is optically inactive. This is because of the fact that in equimolecular mixture of

enantiomeric pairs, the rotation caused by the molecules of one enantiomer is cancelled bythe rotationcaused by the molecules of other enantiomer.


STEREOISOMERISM

(c) This type of compensation of optical rotation in a racemic mixture is referred to as externalcompensation. Thus, racemic mixture becomes optically inactive because of external compensation.

(d) Representation of Racemic mixture : The racemic mixture of a particular sample is indicated byusingthe prefix (d, l ) or (±). For example, racemic mixture of lactic acid is represented as (±) lactic acid.

RACEMIZATION

It is a process of conversion of an opticallyactive compound into the racemic modification. Both (+)and (–) forms of the compound are capable of racemizations under the influence of heat, light orchemical reagents.

RESOLUTIONThe process of separation of constituent enantiomeric forms from the racemic mixture is known asresolution.

1. Chemical method : This is probably the best method of resolution. The racemic mixture is to combinewith another opticallyactive compound and the resulting products (salt - formation) differ in properties,particularly in solubilityin various solvents. Byfractional crystallization from a suitable solvent, theycanbe separated.

2. Mechanical method : If the d and l-forms of a substance exist in well defined crystalline forms, theseparation can be done byhand picking with the help of magnifying lens and a pair of tweezers.

3. Biochemical method : In this method, the resolution is done byuse of micro-organisms, when certainbacteria or moulds are added to a solution of a racemic mixture, they decompose one of the opticallyactive forms more rapidly than the other.

2. CONFORMATIONAL STEREOISOMERISMS

Different non-identical arrangement of atoms or group in a molecule that result by the rotationabout a single bond and that can easily be reconverted at room temperature are known asconformational sterio isomers of conformers.

Conformation isomerism :The different arrangement of atoms is space that result from the free rotation of group about C–C bondaxis are called conformers, and this phenomenon is called conformation isomerism. The basic structureof the moleculevarious bond length andbond angle remain thesame. There are infiniteno. of conformersof any molecule two out of them are defined as staggered and eclipsed.

Condition of conformation :There should be three-bond present in a molecule.

Projection of Tetrahedral Carbon Atom :

Newman projection : In this method the molecule is observed along the central carbon-carbonbond. a circle is drawn and centre of the circle represents the front carbon. the bonds of the frontcarbon are drawn from the centre of the circle while the bonds at the hack carbon are drawn fromthe periphery.

H H

HH

HH

C-H bond of back carbon

C-H bond of front carbon


STEREOISOMERISM

Saw horse projection : In this method, central carbon-carbon bond of the molecule is representedby a straight line written in bond of the molecule is represented by a straight line written in slightlytilted manner and the molecule is observed from the right side.

H

H

H

H

H

HC

C

Dihedral angle :– Angle between valencies of two adjacent atoms

Conformation in ethane (CH3–CH3)

Saw horse formula :H H

H

H

HH

Eclipsed

60°

H

H

H

HH

H

Straggered

Newman's projection formula :

HH

H HH H

60°

H

HH H

H H

Eclipsed form : When H-atoms of one carbon are directly behind the other is called eclipsed form.

Staggered form : The hydrogen of two atoms are maximum distance with respect to one another.

Skew form : The different forms which exist between 0° to 60°.

Stability : The eclipsed form is less stable than staggered due to Vander Waal repulsion and torsionalstrain.

Eclipsed < Skew < Staggered

VanderWaal repulsion : Repulsion between atoms or group of atoms.

Torsional strain : Bond pair-bond pair repulsion in eclipsed form.Potential energy curve : Eec – Est = 3 Kcal / mole or 12.5 kJ / mole

3 Kcal

Pot

enti

alen

ergy

0 60° 120° 180°Staggered Eclipsed Staggered

Rotation


STEREOISOMERISM

Conformation in propane :

H HHCH3

CH3

CH3

HH HHH HHH H

H

H

H

60° 60°

Eclipsed EclipsedStaggered

13.8 kJ

Pot

enti

alen

ergy

0 60° 120° 180°Staggered Eclipsed Staggered

Rotation

Eec – Est = 13.8 kJ/mole

Conformation in n-butane :

3223 CHCHCHCH4321

(C2–C3 bond rotation)

CH3CH3

H HH H

Fully Eclipsed

CH3

CH3

HH H

H

Partially staggered(Gauche form)

CH3

CH3H HH

H

Partially Eclipsed

CH3

CH3

H

H H

H

Fully staggeredor Anti form

CH3

CH3

HH

H

H

Partially Eclipsed

CH3

CH3

H

H HH

Partially staggered(Gauche form)

8Kcal

3.4Kcal


STEREOISOMERISM

Stability order : Anti form > Gauche > Partial eclipse > fully eclipsed

Exercise : (i) n-pentane (about C2–C3)(ii) 3-methyl pentane (about C2 – C3)

Some important example :

Ethylene glycol :H

O–H

O–H

H HH

Gauche form is most stable due to intramolecular H-bonding.

***

OH|CHCHZ 22 Z= –OH, –NH2, –F, –CHO' –COOH, OCH3

Gauche in all cases due to H-bonding.

Gauche effect : In a lone pair containing compound bulkier group should be placed in Gauche (60°)from l.p.As l.p. has minimum steric repulsion.

(i) CH3CH2 2HN

CH3

H H

H H

N <CH3H

H

H H

N

(ii)

3

3

CH|

HOCHCH

CH3

CH3

H

H

O>

CH3CH3

H

H

O

Draw most stable conformation of following compound.(i) n-pentane (C2–C3) bond rotation.

C–C–C–C–C

H H

H HC H52

CH3

H

H H

H


STEREOISOMERISM

(ii) n-hexane (C2–C3) rotation(C3–C4) rotation

C–C–C–C–C–C

C H73

CH3

H

H H

H

C H52

C H52

H

H H

H

(iii) 3-methyl pentane (C2–C3)

C

C–C–C–C–C

C H52

CH3

CH3H

H H

(iv) 3,3-dimethyl hexane (C3–C4) :

C|

CCCCCC|C

C H52

C H52

CH3CH3

HH

(v) 2,2,3,4,5,5-hexamethyl hexane (C3–C4) :

CCCC||||

CCCCCC||CC

CH3

CH3H

H

C–C–C

C–C–C

C

C

Conformation in cyclo alkane :

Baeyer's strain theory : According to Baeyer's strain theory, the amount of the strain is directlyproportional to the angle through which a valency bond has deviated from its normal position. i.e.

Amount of deviation d =2

1(109° 28' – Valency angle)

in cyclopropane d =2

1(109° 28' – 60°) = 24°44'

in cyclobutane d =2

1(109° 28' – 90°) = 9°44'

in cyclopentane d =2

1(109° 28' – 108°) = 0°44'

in cyclohexane d =2

1(109° 28' – 120°) = –5°16'


STEREOISOMERISM

Heat of combustion :Cyclohexane > Cyclopentane > Cyclobutane > cyclopropane

Heat of combustion per –CH2–Cyclopropane > Cyclobutane > Cyclopentane > Cyclohexane

Orbital picture of angle strain :

C

C

C

109°28’C

C

C

In cyclopropane however, the C–C–C bond angle cannot be 109° 28' but instead must be 60°.As aresult the C-atom cannot be located to permit their sp3 orbitals to point toward each other. there is lessoverlap and the bond is weaker than the usual C–C bond.

Conformation of cyclohexane : Cyclohexane exist in different form, such on chair form, boat formtwist boat form, half chair form.Stability order : Chair form > boat form

Boat conformation :

H H

Flag polehydrogen

Flag polehydrogen

HH

due to intraction between flag pole hydrogens boat conformation is unstable.Chair form of cyclohexane :

HaHe

H H

H HH H

H H

Newman’s projectionThere are two type of hydrogen in cyclohexane:(i) Axial hydrogen (Ha)(ii) Equatorial hydrogen (He)

HaHa

Ha

Ha

HaHa

Axial bondEquatorial bond

He

He

He

HeHe

He

He

In cyclohexane all carbon mentain tetrahedral geometryso that cyclohexane is most stable cycloalkane.At room temperature one chair form flips to another chair form during flipping axial bonds converts toequatorial & equatorial bonds converts to axial.

Ha

He

flippings

ringHa

He


STEREOISOMERISM

K = 1Both chair form are equallystable.

Conformation of mono substituted cyclohexane :

CH3

Methyl cyclohexane

CH3

4

5 6

3

2

1 HH

H

CH3

HH

H

K > 1(1,3 – 1,5 intraction)

Whenmethylgropuisaddaxialposition than theirwillbemore1,3–1,5 intractionsothat thisconformationwill be least stable. Byring flipping methyl group occupied equatorial position so that new reverse chairform will be more stable.

Write the orders of equilibrium constant for following equilibrium.

(i)

CH3

CH3

K1

(ii)C H52

C H52

K2

(iii) K3

C

C

C

C

C C

(iv) K4

C CC

C

C

C

C

C

order K1 < K2 < K3 < K4Asthe sizeofalkylgroup increases1,3– 1,5–intractionalso increases, so that bulkyalkyl group preferablyoccupied equatorial position.


STEREOISOMERISM

Draw most stable chair conformation of following compound.

(a) 1,2 - dimethyl cyclohexaneCH3

CH3

(b) 1-ethyl-2-methyl cyclohexaneCH3

CH3C H52

C H52

Unstable Stable

(c) 1,3,5-trimethyl cyclohexane

CH3

CH3

CH3

(d) 1,2,3,4,5,6-hexa methyl cyclohexane

CH3

CH3

CH3 CH3

CH3CH3

(e) 1-ethyl-2,3-dimethyl cyclohexane.CH3 CH3

C H52

STEREOISOMERISM

BANSAL CLASSES Private Ltd. ‘Bansal Tower’, A-10, Road No.-1, I.P.I.A., Kota-05, Tel.(0744) 2791000 Page # 22

EXERCISE-1 (Subjective Questions)

Q.1 How many pair(s) of geometrical isomers are possible with C6H12 (only in open chain structures)

Q.2 How many geometrical isomers are possible by

Br

BrBrBr

Q.3 Considering rotation about the C-3 – C-4 bond of 2-methylhexane:(a) Draw the Newman projection of the most stable conformer.(b) Draw the Newman projection of the least stable conformer.

Q.4 Assign the priority order number to the following atoms or groups.(a) –CHO, –CH2OH, –CH3 , –OH

(b) –Ph, –CH(Me)2, –H, –NH2

(c) –COOH, –Ph , –CHO, –CH = CH2

(d) –CH(Me)2, –CH=CH2, –CCH, –Ph

(e) –CH3, –CH2Br, –CH2OH, –CH3Cl

(f) –H, –N (Me)2, –Me, –OMe

(g) –CH = CH2, –Me, – Ph, –Et

(h) –CH2–CH2–Br, –Cl, –CH2–CH2–CH2–Br, (Me)2CH–

(i) –Cl, –Br, –I, –NH2(j) NH2 , NO2 , CH2NH2 , CN

Q.5 Write the structure of :(i) (E) penta-1,3-diene (ii) (2Z, 4E)hexa-2,4-diene(iii) (2E, 4E)-3-ethylhexa-2,4-diene (iv) (R)-2-Bromopentane(v) (S)-3-bromo-3-chlorohexane (vi) (2S, 3R)-2,3-dibromobutane

Q.6 Write the E / Z configuration of each compound.

(i)C = C

F Br

H C3O–CH3

(ii)C = C

O N2 CH –CH2 3

H C–NH3CH3

(iii) C = C

Et

CH –I2Me

Br

(iv) C = C

CH –CH2 3

CH=CH2

Cl

Cl

STEREOISOMERISM


Q.7 Draw the two chair conformers of each compound and indicate which conformer is more stable.(a) cis-1-ethyl-3-methylcyclohexane(b) trans-1-ethyl-2-isopropylcyclohexane(c) trans-1-ethyl-2-methylcyclohexane(d) trans-1-ethyl-3-methylcyclohexane(e) cis-1-ethyl-3- isopropylcyclohexane(f) cis-1-ethyl-4-isopropylcyclohexane

Q.8 Write the configuration of each compound?

(i)

Cl

Cl(ii)

Cl

Cl(iii) Br

Br(iv) Br

Br

Q.9 Draw the most stable conformer of N-methylpiperidine.

Q.10 How many monochlorinated products of methyl cyclohexane are optically active.

Q.11 How many enantiomers are possible on monochlorination of isopentane.

Q.12 Calculate the total number of chiral carbon atoms in.

(i)O

H

H H

H

Progesterone(human female sex hormone)

H C3

H C3

CH3

C=O

(ii)HO

H

H H

HH C3

H C3

CH3

CH3

H C3

H

Cholesterol(important component of membranes:principal component of gallstones)

(iii)

O

O

H

H OH

H

Cortisone(antiinflammatory hormone)

H C3

H C3

CH2OH

C=O

(iv)O

H

H

OH

H

H

Testosterone(human male sex hormone)

H C3

H C3

STEREOISOMERISM


Q.13 In what stereoisomeric forms would you expect the following compounds to exist?(a) EtCH(CO2H)Me (b) MeCH(CO2Et)CO2H

(c) (d)

(e) (f) Et(Me)C=C=C(Me)Et

(g) (h)

(i) (j)

(k)

Q.14 How many stereocenter and pseudochirality center present in the following compound?

O

O

OH N2 NH2

O

Q.15 Calculate the total number of stereoisomers in the following compounds.

(I) (II) (III)

Q.16 The number of stereoisomers (excluding enantiomers) for1-bromo-2-chloro-3-iodocyclopropane

STEREOISOMERISM


Q.17 Calculate the total number of stereoisomers possible for

(i) (ii) C = C = C

H

Cl

CH = CH – CH3

CH = CH – CH3

Q.18 How many stereoisomers are possible for Inositol,

OH

OH

OH

OHHO

HO

Q.19 Comment on the relationship among the following compounds.

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

Q.20 When 20 gm optically active compound is placed in a 10 dm tube, in a 200 ml solution rotates thePPL by 30°.(a) What is the angle of rotation if above solution is diluted to 1 Litre?(b) What is the specific angle of rotation if above solution is diluted to 1 Litre?Write answer of part (a) and (b) in the same order and present the four digit number as answerin OMR sheet. For example : If ans of (a) is 9 and (b) is 99 then fill 0999 in OMR sheet.

STEREOISOMERISM


EXERCISE-2 (Objective Questions)

[SINGLE CORRECT CHOICE TYPE]

Q.1 How many minimum no. of C-atoms are required for position & geometrical isomerism in alkene?(A) 4, 3 (B) 4, 4 (C) 3, 4 (D) 3, 3

Q.2Cl C—O

O

C=CH CH3

H andClC—O

O

C=C

HH

CH3

Shows which type of isomerism(A) Functional group isomerism (B) Geometrical isomerism(C) Metamerism (D) Position isomerism

Q.3 Which of the following does not show geometrical isomerism?(A) 1, 2-dichloro-1-pentene (B) 1, 3-dichloro-2-pentene(C) 1, 1-dichloro-1-pentene (D) 1, 4-dichloro-2-pentene

Q.4 Which of the following will not show geometrical isomerism?

(A) (B)

Me

Me(C)

Me

Cl (D)

Q.5 Geometrical isomerism is possible in:(A) isobutene (B) acetone oxime(C) acetophenone oxime (D) benzophenone oxime

Q.6 Which of the following will show geometrical isomerism?

(A) (B) (C) (D)

Q.7 Which of the following will show geometrical isomerism?

(A) (B)

Me

Me

Br

Br

(C)Br

Me

(D)

STEREOISOMERISM


Q.8 Total number of geometrical isomer of following compound is:

(A) 2 (B) 3 (C) 4 (D) 5

Q.9 Tautomer of which compound can show geometrical isomerism ?

(A)O

(B)

N=O

(C)

O

(D)NO2

Et

Q.10 What characteristic is the best common to both cis-2-butene and trans-2-butene?(A) B.P. (B) Dipole moment(C) heat of hydrogenation (D) Product of hydrogenation

Q.11 The number of cis-trans isomer possible for the following compound

(A) 2 (B) 4 (C) 6 (D) 8

Q.12 Which of the following have zero dipole moment?(A) benzene 1,4- diol (B) trans-1,2-dichloro ethene(C) cis-1,2-dichloro ethene (D) 1,1-dichloro ethene

Q.13 Increasing order of stability among the three main conformation (i.e. eclipse, anti, gauche) ofethylene glycol is :(A) Eclipse, gauche, anti (B) Gauche, eclipse, anti(C) Eclipse, anti, gauche (D) Anti, gauche, eclipse

Q.14 Which of the followings is most stable conformation of 3-Hydroxy propanal ?

(A)

H

H

H

H

CHO

OH

(B)

HH

HH

CHOOH

(C)

HO

H

H

H

CHO

H

(D)OHH

HHCHO

H

Q.15 How many stereoisomers can exist for the following acid

HCO).OH(CH|

HCO.CH|

HCO).OH(CH

2

2

2

(A) Two (B) Four (C) Eight (D) Sixteen

STEREOISOMERISM


Q.16 Which of the following have asymmetric carbon atom?

(A)

HH||

HCCH||BrlC

(B)

HH||

lCCCH||

lCH

(C)

HH||

HCCH||

lCH

(D)

OHBr||

CHCCH||HH

3

Q.17 The number of optically active compounds in the isomers of C4H9Br is(A) 1 (B) 2 (C) 3 (D) 4

Q.18 The number of optically active isomers observed in 2,3-dichlorobutane is:(A) 0 (B) 2 (C) 3 (D) 4

Q.19 The total number of isomeric optically active monochloro derivative of isopentane is:(A) two (B) three (C) four (D) one

Q.20 Select the optically active compound among the following :

(A) HNNH

(B) COOH

COOHH C3

H C3

(C)

CH3H3C

HOOC COOH (D) OC

HN NH

CO

Q.21 How many stereoisomers of the following molecule are possible?HOOC.CH=C=CH.COOH

(A) two optical isomers (B) two geometrical isomers(C)two optical and two geometrical isomers (D) None

Q.22 Which of the following is not chiral ?

(A)

Me

Me

COOH

HOOC

(B)

HO

HO

OH

OH

OH

OH

(C) H3C – CH = C = C = C= CH–CH3 (D)

Cl

Cl

F

F

STEREOISOMERISM


Q.23 Which species exhibits a plane of symmetry?

(A) (B)

(C) (D)

Q.24 Which of the following has centre of symmetry?

(A)C

H

HH

H

(B)

F

F

(C)

Cl

Cl

H

HBr

Br

(D) PCl5

Q.25 Which of the following will not show optical isomerism?(A) Cl–CH=C=C=CH–Cl (B) Cl–CH=C=C=C=CH–Cl

(C) (D)

Q.26 Meso-tartaric acid and d-tartaric acid are(A) positional isomers(B) enantiomers (C) diastereomers (D) racemic mixture

Q.27 The structures shown here are related as being:

CH3 CH3

Br

Br

H

H

CH3

Br

Br

H

H

CH3

(I) (II)(A) conformers (B) enantiomers(C) geometrical isomers (D) diastereoisomers

Q.28 What is relationship between

Me

BrMe

H

O

and Me

Me

Br

HO

(A) Diastereomer (B) Enantiomer (C) Conformer (D) None of these

STEREOISOMERISM


Q.29 What is relationship betweenCH CH2 3

Me

BrEtand

CH CH2 3H3C

Br

Et

(A) Enantiomer (B) Identical (C) Diastereomer (D) Epimer

Q.30 The two compounds given below are

(A) enantiomers (B) identical (C) optically inactive (D) diastereomers

Q.31 Which of the following is incorrectly matched?

(A) Enantiomer

(B)

HH CH3H3C

Enantiomer

(C) H

H

H

H

CH3

H3C

CH3

CH3

Identical

(D)

Br Br

Br Br

Identical

Q.32 Which of the following combinations amongst the four Fischer projections represents the sameabsolute configurations?

(A) (II) and (III) (B) (I) and (IV) (C) (II) and (IV) (D) (III) and (IV)

STEREOISOMERISM


Q.33

L -form of the given compound is :

(A)

H

CH2OH

CHO

HOHOH

HHO (B)

CH2OH

CHO

H OHHHHO

HO(C)

CH2OH

CHOH OH

HHHO

HO (D)

CH2OH

CHOH

HHHO

HO

HO

Q.34 Which of the following compound has D-configuration ?

(A)

CHO

H

H

H

HO

HOH C2

OH

OH

H

OH(B)

CHO

H

H

HO

CH OH2

OH

OH

H

H OH

(C)

CHOH

H

HO

CH OH2

OH

OH

H

H

HO(D)

CHO

HO

H

H

HO

CH OH2

H

OH

OH

H

Q.35 Number of possible stereoisomers of glucose are(A) 10 (B) 14 (C) 16 (D) 20

Q.36 The S-ibuprofen is responsible for its pain reliveing property. Which one of the structure shown isS-ibuprofen.

(A)

Me

Me CCH3

C–OH

O

H

(B)C

H

CH3

C – OH

O

Me

Me

(C)C H

CH3

C – OH

O

Me

Me(D) C H

CH3

C – OH||O

Me

Me

STEREOISOMERISM


Q.37

The compound with the above configuration is called:(A) (2S, 3S)-2-chloro-3-pentanol (B) (2S, 3R)-2-chloro-3-pentanol(C) (2R, 3R)-2-chloro-3-pentanol (D) (2R, 3S)-2-chloro-3-pentanol

Q.38 The full name of the compound is

(A) (2R,3R)-3-chloro-2-pentanol (B) (2R,3S)-3-chloro-2-pentanol(C) (2S,3R)-3-chloro-2-pentanol (D) (2S,3S)-3-chloro-2-pentanol

Q.39 The R/S configuration of these compounds are respectively.

CF3

HHO

H NH2

HSCOOH

CH3CHO

H

(A) R,R,R (B) R,S,R (C) R,S,S (D) S,S,S

[ONE OR MORE THAN ONE CORRECT CHOICE TYPE]

Q.40 Which of the following statements is/are not correct?(A) Metamerism belongs to the category of structural isomerism(B) Tautomeric structures are the resonating structures of a molecule(C) Keto form is always more stable than the enol form(D) Geometrical isomerism is shown only by alkenes

Q.41 Tautomer of which of the following can show geometrical isomerism

(A) CH3–CHO (B) CH3CH2–CHO

(C)O

(D)

OO

Q.42 Which will form geometrical isomers ?

(A)

Cl

Cl

(B) CH CH NOH3 (C) (D) All of these

STEREOISOMERISM


Q.43 Which will show geometrical isomerism ?

(A) CH CH NOH3 (B)

Me Me Me Me

Me MeEt Et

(C) NOHH C3 CH C3

(D) HO—N N—OH

Q.44 The isomerism observed in alkanes is :(A) metamerism (B) chain isomerism(C) position isomerism (D) geometrical isomerism

Q.45 Which of the following compound can form geometrically isomeric oxime on reaction with NH2OH

(A)

O

(B)O

(C)

O

(D)O

Q.46 Which of the following can exist in 'syn' and 'anti' form ?

(A) C H6 5—N N—OH (B) C H6 5—N N—C H6 5

(C) C H6 5—CH N—OH (D) (C H )6 5 2C N—OH

Q.47 Which of the following have zero dipole moment?(A) p-Dichlorobenzene (B) Benzene-1, 4-diol(C) Fumaric acid (D) Maleic acid

Q.48 The Z-isomer among the following are :

(A)

C H3 7—C—C H2 5

CH3—C—H (B)

C H2 5—C—C H3 7

CH3—C—H

(C) Cl—C—Br

H—C—F

(D) Cl—C—Br

F—C—H

Q.49 The IUPAC name of the compound :

(A) (2E, 4E, 6Z)-octa-2,4,6-triene (B) (2E, 4E, 6E)-octa-2,4,6-triene(C) (2Z, 4E, 6Z)-octa-2,4,6-triene (D) (2Z, 4Z, 6Z)-octa-2,4,6-triene

Q.50 Which of the following compound has E configuration?

(A)

CHO

H (B)

CH OH2

NH2 (C)

CHO

H

Cl (D)

Cl

Br

H

STEREOISOMERISM


Q.51 Which of the following will show optical isomerism as well as geometrical isomerism.

(A) (B)

(C) (D)

Q.52 Which of the following statements is/are correct?(A) A meso compound has chiral centres but exhibits no optical activity(B) A meso compound has no chiral centres and thus are optically inactive.(C) A meso compound has molecules which are superimposable on their mirror images eventhough they contain chiral centres(D) Ameso compound is optically inactive because the rotation caused by any molecule is cancelledby an equal and opposite rotation caused by another molecule that is the mirror image of the first.

Q.53 Which of the following statements are correct?(A) Any chiral compound with a single asymmetric carbon must have a positive optical rotationif the compound has the R configuration(B) If a structure has no plane of symmetry it is chiral(C) All asymmetric carbons are stereocentres.(D) Alcohol and ether are functional iosmers

Q.54 Which of the following compounds are optically active?

(A) CH3.CHOH.CH2.CH3 (B) H2C=CH.CH2.CH=CH2

(C) (D)

NO2

NO2HOOC

COOH

Q.55 Which out of the following are resolvable pair?

(A)N

HD

T+ en

(B) + en

(C)P

Me

EtPh

O

+ en(D)

H

Me

Et NH2 + en

STEREOISOMERISM


Q.56 Which of the following is/are resolvable pair?

(A)O

+ en (B) N

EtH

D+ en

Me

(C) N

EtD

+ enMe

(D) SEtO

+ en

Me

Q.57 Which conformation of n-Butane has both plane of symmetry and centre of symmetry absent ?(A) fully eclipsed (B) Gauche (C) Partially eclipsed (D) Anti

Q.58 Which of the following operations on the Fischer formula does not change its absolute

configuration?(A) Exchanging groups across the horizontal bond(B) Exchanging groups across the vertical bond(C) Exchanging groups across the horizontal bond and also across the vertical bond(D) Exchanging a vertical and horizontal group

Q.59 Which of the following pairs of compound is/are identical?

(A) (B)

(C) (D)

Q.60 Concider the following structure and pick by the right statement :

CH

CH3

NH2

CH O2 CH3

I

C

HOH

C

CH O2 CH3

II

OH

F OH

III

(A) I and II have R-configuration (B) I and III have R-configuration(C) only III has S-configuration (D) I and III have S-configuration

STEREOISOMERISM


Q.61 Which of the following is a 'threo' isomer?

(A)H

H

OH

OH

CHO

CH OH2

(B)Br

H

H

OH

CH3

CH3

(C) H

ClH

OH

CH3

COOH

(D)H

H

OH

CH3

NH2

COOH

Q.62 Which of the following statements is/are not correct for D-(+) glyceraldehyde?(A) The symbol D indicates the dextrorotatory nature of the compound(B) The sign (+) indicates the dextrorotatory nature of the compound(C) The symbol D indicates that hydrogen atom lies left to the chiral centre in the Fischer projectiondiagram(D) The symbol D indicates that hydrogen atom lies right to the chiral centre in the Fischerprojection diagram

Q.63 Which of the following are correct representation of L–amino acids?

(A) (B)

(C) (D)

Q.64 Which of the following are D sugars:

(A) (B)

(C) (D)

STEREOISOMERISM


[MATCH THE COLUMN]

Q.65 Column I Column II

(A) (P) Position isomers

(B) (Q) Geometrical isomers

(C)C=N C=N

OH

OH

Me Me

(R) First is E and second is Z form

(D)

OH

OH

OH

OH

(S) Structural isomer


Compound Total number of stereoisomer

(A) (P) 2

(B) (Q) 4

(C) (R) 6

(D) (S) 8

STEREOISOMERISM



(A)

CH3

CH3

Cl

ClH

HOH

H&

CH3

CH3

Cl

ClHO

HH

H(P) Identical

(B)

Et

H

HHSMe

OH &

OH

Et

Me

SHH

H (Q) Diastereomers

(C)

H2C

MeH

C CH

Et

&

H2C Me

H

C CH

Et

(R) Conformer

(D)

HH

H HH H

ClCl

&

HH

H

H

H H

Cl

Cl

(S) Positional

Q.68 Match List-I, List-II & List-III :List-I List-II List-III

(a)

CH3

CH3

HO H

Br H

(1) C – CHO

H

Br

CH3

H

CH3

(i) (2R, 3R)

(b)

CH3

CH3

BrHHO

H (2) C – C

H

Br

H

OH

CH3

CH3

(ii) (2S,3S)

(c)

CH3

CH3

BrH OH

H (3) C – C

H

HO

H

Br

CH3

CH3

(iii) (2S,3R)

(d)

CH3

CH3

Br

HO HH (4) C – CBr

H

OH

CH3

HCH3

(iv) (2R, 3S)

Q.69 Column I Column IICompound (Open chain) Number of optically active isomer

(A) Unsymmetrical compound with ‘n’ chiral carbon (P) 2n–1

(B) Symmetrical molecule with ‘n’ chiral carbon when n is even (Q) 2n–1 – 2n–1/2

(C) Symmetrical molecule with ‘n’ chiral carbon when n is odd (R) 2n

STEREOISOMERISM


EXERCISE-3

SECTION-A

(JEE Main Previous Year's Questions)Q.1 Stereo - Isomerism includes - [AIEEE-2002]

(A) Geometrical isomerism only (B) Optical isomerism only(C) Both geometrical & optical isomerism (D) Position& Functional isomerism

Q.2 Which of the following does not show geometrical isomerism - [AIEEE-2002](A) CH3 – CH = CH – CH3 (B) CH3 – CH2 – HC = CH2

(C) CH3 –

Cl|C CH – CH3 (D) ClHC = CH – CH2 – CH3

Q.3 Racemic mixture is formed bymixing two – [AIEEE-2002](A) Isomeric compounds (B) Chiral compounds(C) Meso compounds (D) Enanitiomerswith chiral carbon

Q.4 Geometrical isomerism is not shown by– [AIEEE-2002](A) 1, 1– dichloro–1–pentene (B) 1, 2– dichloro–1–pentene(C) 1, 3– dichloro–2–pentene (D) 1, 4– dichloro–2–pentene

Q.5 Among the following four structures I to IV

(i) C H –CH–C H2 5 3 7

CH3

(ii) C H – C – CH – C H2 5 2 5

CH3O

(iii) H–C–CH3

H

H

(iv) C H –CH–C H2 5 2 5

CH3

It is true that – [AIEEE-2003](A) Only (iii) is a chiral compound (B) Only(ii) and (iv) are chiral compounds(C)All four are chiral compounds (D) Only(i) and (ii) are chiral compounds

Q.6 Which of the following compounds is not chiral ? [AIEEE-2004]

(A) 1 – chloropentane (B) 2 – chloropentane

(C) 1 – chloro –2–methyl pentane (D) 3 – chloro – 2 – methyl pentane

Q.7 Which of the following will have a mesoisomer also - [AIEEE-2004]

(A) 2 - Chlorobutane (B) 2, 3 - Dichlorobutane

(C) 2,3 - Dichloropentane (D) 2-Hydroxypropanoic acid

Q.8 Which types of isomerism is shown by2,3–dichlorobutane ? [AIEEE-2005]

(A) Optical (B) Diastereo (C) Structural (D) Geometric

Q.9 Increasing order of stability among the three main conformations (i.e. Eclipse, Anti, Gauche) of

2-fluoroethanol is [AIEEE 2006]

(A) Gauche, Eclipse,Anti (B) Eclipse,Anti, Gauche

(C)Anti, Gauche, Eclipse (D) Eclipse, Gauche,Anti

STEREOISOMERISM


Q.10 Which of the following molecules is expected to rotate the plane of plane-polarised light?

[AIEEE-2007]

(A) (B) (C) (D) H

COOH

H N2

H

Q.11 Which one of the following conformations of cyclohexane is chiral ? [AIEEE-2007]

(A) Twist boat (B) Rigid (C) Chair (D) Boat

Q.12 The absolute configuration of

HO C2 CO H2

HO HOH

H

is [AIEEE-2008]

(A) R, R (B) R, S (C) S, R (D) S, S

Q.13 Out of the following, the alkene that exhibits optical isomerism is [AIEEE-2010](A) 3-methyl-1-pentene (B) 2-methyl-2-pentene(C) 3-methyl-2-pentene (D) 4-methyl-1-pentene

Q.14 Identifythe compound that exhibits tautomerism. [AIEEE-2011](A) 2-Butene (B) Lactic acid (C) 2-Pentanone (D) Phenol

Q.15 How manychiral compounds are possible on monochlorination of 2-methyl butane ? [AIEEE-2012](A) 4 (B) 6 (C) 8 (D) 2

Q.16 For which of the following molecule significant µ 0? [JEE Main 2014]

(a)

Cl

Cl

(b)

CN

CN

(c)

OH

OH

(d)

SH

SH

(A) (a) and (b) (B) only (c) (C) (c) and (d) (D) only (a)

Q.17 Which of the followingcompounds will exhibit geometrical isomerism? [JEE Main 2015]

(A) 2 -Phenyl -1 -butene (B) 1, 1 -Diphenyl -1 -propane

(C) 1 -Phenyl -2 -butene (D) 3 -Phenyl -1 -butene

Q.18 The absolute configuration of [JEE Main 2016]CO H2

CH3

H

H

OH

Cl

is :(A) (2R,3R) (B) (2R, 3S) (C) (2S,3R) (D) (2S, 3S)

STEREOISOMERISM


SECTION-B

(JEE ADVANCED Previous Year's Questions)

Q.1 True or False:2, 3, 4-Trichloropentane has three asymmetric carbon atoms. [JEE 1990]

Q.2 Isomers which can be interconverted through rotation around a single bond are [JEE 1992](A) Conformers (B) Diastereomers (C) Enantiomers (D) Positional isomers

Q.3 The optically active tartaric acid is named as D–(+)–tartaric acid because it has a positive(A) optical rotation and is derived from D-glucose(B) pH in organic solvent(C) optical rotation and is derived from D–(+)–glyceraldehyde(D) optical rotation onlywhen substituted bydeuterium [JEE 1992]

Q.4 The molecules that will have dipole moment are : [JEE 1992](A) 2,2-dimethyl propane (B) trans 2-pentene(C) cis 3-hexene (D) 2,2,3,3-tetramethyl butane

Q.5 The shows: [JEE 1995(Scr.)]

(A)geometrical isomerism (B) optical isomerism(C) geometrical & optical isomerism (D) tautomerism

Q.6 How manyoptically active stereoisomers are possible for butane-2,3-diol? [JEE 1997](A) 1 (B) 2 (C) 3 (D) 4

Q.7 The number of possible enantiomeric pairs that can be produced during monochlorination of 2-methylbutane is [JEE 1997](A) 2 (B) 3 (C) 4 (D) 1

Q.8 When cyclohexane is poured on water, it floats, because: [JEE 1997](A) cyclohexane is in 'boat' form (B) cyclohexane is in 'chair' form(C) cyclohexane is in 'crown' form (D) cyclohexane is less dense than water

Q.9 Which of the followingcompounds will show geometrical isomerism? [JEE 1998](A) 2-butene (B) propene (C) 1-phenylpropene (D) 2-methyl-2-butene

Q.10 Identify the pairs of enantiomers and diastereomers from the following compounds I, II and III.[JEE 2000]

STEREOISOMERISM


Q.11 Which of the followingcompounds will exhibit geometrical isomerism? [JEE 2000 (Scr.)](A) 1-Phenyl-2-butene (B) 3-Phenyl-1-butene(C) 2-Phenyl-1-butene (D) 1,1-Diphenyl-1-propene

Q.12 The number of isomers for the compound with molecular formula C2BrClFIis(A) 3 (B) 4 (C) 5 (D) 6 [JEE 2001 (Scr.)]

Q.13 Which of the followingcompounds exhibits stereoisomerism? [JEE 2002 (Scr.)](A) 2-methylbutene-1 (B) 3-methylbutyne-1(C) 3-methylbutanoic acid (D) 2-methylbutanoic acid

Q.14 Which of the following hydrocarbons has the lowest dipole moment? [JEE 2002](A) cis-2-butene (B) 2-butyne (C) 1-butyne (D) H2C = CH – C CH

Q.15 In the given conformation, if C2 is rotated about C2–C3 bond anticlockwise by an angle of 120° thenthe conformation obtained is [JEE 2004 (Scr.)]

(A) fullyeclisped conformation (B) partiallyeclipsed conformation(C) gauche conformation (D) staggered conformation

Q.16 On monochlorination of 2-methylbutane, the total number of chiral compounds formed is(A) 2 (B) 4 (C) 6 (D) 8 [JEE 2004]

Q.17 (i) obs = i

iix

wherei is the dipole moment of a stable conformer of the molecule, Z–CH2–CH2–Z andxi is the mole fraction of the stable conformer.Given : obs = 1.0 D and x (Anti) = 0.82Draw all the stable conformers of Z–CH2–CH2–Z and calculate the value of(Gauche).

(ii) Draw the stable conformer ofY–CHD–CHD–Y(meso form), whenY= CH3 (rotationabout C2–C3) and Y = OH (rotation about C1 – C2 ) in Newmann projection. [JEE 2005]

Q.18 The number of structural isomers for C6H14 is [JEE 2007](A) 3 (B) 4 (C) 5 (D) 6

Q.19 Statement-1 : Molecules that are not superimposable on their mirror images are chiral.becauseStatement-2 : All chiral molecules have chiral centres.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 isTrue, Statement-2 isTrue; Statement-2 is NOT a correct explanation for Statement-1.(C) Statement-1 is True, Statement-2 is False.(D) Statement-1 is False, Statement-2 is True. [JEE 2007]

STEREOISOMERISM


Q.20 The correct statement(s) about the compound given below is (are) [JEE 2008]

CH3

HCl

H C3

Cl H

(A) The compound is opticallyactive(B) The compound possesses centre of symmetry(C) The compound possesses plane of symmetry(D) The compound possesses axis of symmetry

Q.21 The correct statement(s) concerning the structures E, F and G is (are) [JEE 2008]

OH C3

H C3 CH3

(E)

OHH C3

H C3 CH3

(F)

OH

H C3

H C3

CH3

(G)

(A) E, F and G are resonance structures (B) E, F and E, G are tautomers(C) F and G are geometrical isomers (D) F and G are diastereomers

Q.22 The total numberof cyclic structural aswell as stereo isomerspossible for a compoundwith the molecularformulaC5H10 is [JEE 2009]

Q.23 The correct statement(s) about the compound H3C(HO) HC – CH = CH – CH (OH) CH3 (X) is (are)(A) The total number of stereoisomers possible for X is 6(B) The total number of diastereomers possible for X is 3(C) If the stereochemistryabout the double bond in X is trans, the number of enantiomers possible for Xis 4(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for Xis 2 [JEE 2009]

Q.24 The total number of cyclic isomers possible for a hydrocarbon with the molecular formula C4H6 is[JEE 2010]

Q.25 In the Newman projection for 2,2-dimethylbutane [JEE 2010]

H C3

H H

CH3

X

Y

X and Ycan respectively be(A) H and H (B) H and C2H5 (C) C2H5 and H (D) CH3 and CH3

STEREOISOMERISM


Q.26 Themaximumnumberof isomers (includingstereoisomers) that arepossibleonmono-chlorinationof thefollowingcompound, is [JEE 2011]

Q.27 Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possibleconformations (if any), is (are) [JEE 2011]

(A) (B) H – C

(C) H2C = C = O (D) H2C = C = CH2

Q.28 Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct? [JEE 2012]

HO HO

HO

HO HO

HO

HO OH

OH OH

CH3

CH3

CH3 CH3

CH3

H HH

H H HH H

H HCl

Cl

Cl

Cl Cl

(M) (N) (O) (P) (Q)

(A) M & N are non-mirror image stereoisomers (B) M & O are identical(C) M & P are enantiomers (D) M & Q are identical

Q.29 The total number(s) of stable conformers with non-zero dipole moment for the following compound is(are) [JEEAdvance 2014]

Cl

Br

Br

Cl

CH3

CH3

Q.30 The total number of stereoisomers that can exist for M is : [JEEAdvance 2015]

H C3CH3

OH C3

MQ.31 For the given compound X, the total number of optically active stereoisomers is ____.

[JEEAdvance 2018]HO

HO

HO

HOX

This type of bond indicates that the configuration at thespecific carbon and the geometry of the double bond is fixed

This type of bond indicates that the configuration at thespecific carbon and the geometry of the double bond is fixedNOT

STEREOISOMERISM


EXERCISE-4

(Potential Problems Based on CBSE)

Very Short Answer Type Questions

Q.1 Why eclipsed and staggered forms of ethane cannot be isolated at room temperature?

Q.2 What is the cause of geometrical isomerism?

Q.3 What are conformations?

Q.4 Whichof thefollowingexhibitsgeometrical isomerism?(a) 2-methyl propene (b) 1-butene(c) 2-butene (d) 2,3-dibromo-2-butene

Q.5 Draw the Newman’s projection formula of the staggered form of 1, 2-dichloroethane.

Q.6 How manypossible conformations of ethane are there?

Q.7 What is difference between conformation and isomers?

Q.8 Name the structures obtained by free rotation along the C–C bond.

Q.9 Is geometrical isomerism possible around triple bond?

Short Answer Type QuestionsQ.10 Identify the type of isomerism exhibited bythe following compounds:

(i) CH3– CH = CH – CH3 (ii) CH3 – CH = CH – CH2 – CH3(iii) CH3 – CH2 – CH2 – CH3 (iv) C5H12

Q.11 Draw structures to illustrate.(i) Geometrical isomerism(ii) Positional isomerism

Q.12 Classify the following as cis-isomer or trans-isomer or having no geometrical isomers:

(a) C = C

H

H CH3

CH3

(b)C = C

HH

CH3–CH2 C H52

(c) C = C

HH

ClCl

(d) C = C

CH3CH2

CH2CH3

CH3

CH3

Long Answer Type QuestionsQ.13 Draw main conformations of n-butane obtained by rotation around C–2 and C–3.Also give the names

of these conformations. Which of these conformations is most stable and which is the least stable andwhy?

STEREOISOMERISM


EXERCISE-5 (Rank Booster)

[SINGLE CORRECT CHOICE TYPE]

Q.1 Which of the followingstructure can show geometrical isomerism?

(A)

NO2

HOOCCOOH

NO2

(B) 2,3-Dimethyl but-2-ene

(C) 3,4-Diethyl hex-2-ene (D) 1,4-Dichloro benzene

Q.2 What is the correct IUPAC name of the following compound

(A) 2E, 4E, 6Z 4-methyl oct-2, 4, 6 triene (B) 2E, 4Z, 6Z 5-methyl oct-2, 4, 6 triene(C) 2Z, 4Z, 6Z 5-methyl oct-2, 4, 6 triene (D) 2E, 4Z, 6E 4-methyl oct-2, 4, 6 triene

Q.3 Molecular formula C5H10O can have :(A) 6-Aldehyde, 4-Ketone (B) 5-Aldehyde, 3-Ketone(C) 4-Aldehyde, 3-Ketone (D) 5-Aldehyde, 2-Ketone

Q.4 The number of isomers of dibromoderivative of an alkene (molar mass 186 g mol–1) is(A) 2 (B) 3 (C) 4 (D) 6

Q.5 On chlorination of propane number of products of the formula C3H6Cl2 is(A) 3 (B) 4 (C) 5 (D) 6

Q.6 The correct stabilityorder of the following species is

Me

Me

(a)

H Me

H MeH H

HH

(b)

Me

Me

(c)

(A) c < a < b (B) c = b < a (C) c < a = b (D) a = b = c

STEREOISOMERISM


Q.7 Identifydiastereoisomer in the following are :

(A)Cl

Cl

H

H

CH3

CH3

CH3

H

H

Cl

Cl

CH3

(B)

CH3

H

H

Cl

Cl

CH3

CH3

Cl

H

H

Cl

CH3

(C)

HCH3

Cl

Cl

CH3

H

CH3

H

H

Cl

CH3

Cl (D)

HCH3

Cl

Cl

CH3

HH

Cl

Cl

HCH3

CH3

Q.8 What is relationship between ?

HOOC

COOH

Br

BrH

H COOH

Br

H

H

Br

COOH

(A)Enantiomer (B) Diastereomer (C) Conformer (D) Identical

Q.9 The structure of (2R, 3R)C2H5CH(CH3)CH(D)CH2D is

(A) (B) (C) (D)

Q.10 The structure of (2R, 3S)C2H5CH(CH3)CH(D)CH2D is

(A) (B) (C) (D)

Q.11 Select the correct IUPAC name of following compound :

COOHH

NH2

CONH2H

COOH

(A) 2R,3S-3- amino-2- carbamoyl butane-1,4-dioic acid(B) 2S,3R-2-amino-3- carbamoyl butane-1,4-dioic acid(C) 2S, 3R-3-amino-2-carbamoyl butane-1-4-dioic acid(D) 2R, 3R-2-amino-3- carbamoyl butane-1-4-dioic acid

STEREOISOMERISM


[ONE OR MORE THAN ONE CORRECT CHOICE TYPE]

Q.12 Total numberof geometrical isomer of following compound.

(A) 2 (B) 4 (C) 8 (D) 16

Q.13 In which of the following has minimumtorsional strain and minimumVander waal strain.

CH3

CH3

H

H

I

CH3

CH3 CH3

CH3

CH3

CH3

H

H

II

CH3

CH3

H

IIICH3

CH3

H

CH3

CH3

CH3

CH3

H

H

IV

(A) I (B) II (C) III (D) IV

Q.14 Which of the following statements for a meso compound is correct?(A) The meso compound has either a plane or a centre of symmetry(B) The meso compound is chiral.(C) The meso compound is achiral(D) The meso compound is formed when equal amounts of two enantiomers are mixed.

Q.15 Which of the following pairs can be resolved?

(A) (B)

(C)

NO2 NO2

HOOC COOH

NO2 NO2

HOOC COOH

(D)

Q.16 Which of the following is not of pair of enantiomers?

(A)Cl Cl

Cl Cl

(B)Br

BrBr

Br

C = C = C C = C = C

H

HH

H

(C)C

CH3

H3CBr

Br

H H

C(D)

CH3 CH3

CH3 CH3

H Br

H Cl

Cl H

Br H

STEREOISOMERISM


Q.17 Which of the following molecules is/are identical with that represented by

(A) (B) (C) (D)

Q.18 Which of the following is/are the correct Fischer projection of the following :

(A)

OHOHOH

HHH

CH3

C H2 5

(B)HO H

C H2 5

CH3

HOHO

HH

(C)H

C H2 5

CH3

HO HOH

OH

H(D)

H

CH3

Et

HOHOH

HHO

[REASONING TYPE]Q.19 Statement-1 : E-cyclopentadecene is having more HC (Heat of combustion) than

Z isomer.

Statement-2 : E-cyclopentadecene is more stable than Z isomer.

(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

(B) Statement-1 is true, statement-2 is trueandstatement-2 isNOTthecorrectexplanation for statement-1.

(C) Statement-1 is true, statement-2 is false.

(D) Statement-1 is false, statement-2 is true.

[MATCH THE COLUMN]Q.20 Match the column :

Column-I Column-II

(A)NH2

CH OH2

CH3

Hand

OH

CH NH2 2

HH C3

(P) Structural isomer

(B)Et

Cl

CH3

H

andEt

ClCH3

H

(Q) Identical

(C)Et

OH

CH3

H

andEt

OH

HH C3

(R) Enantiomers

(D)OH H

H C5 2

andOHH

H C5 2

(S) Diastereomers

STEREOISOMERISM


Q.21 Match the Column :Column I Column II

(a) A pair of metamer (i)

H H

HH

OH OH

OHHO

COOH COOH

COOH COOH

;

(b) tautomerism (ii) CH3OC3H7 ; C2H5OC2H5

(c) A pair of geometrical isomer (iii)H HOOH H

CH3 CH3

COOH COOH

;

(d) A pair of distereomers (iv)

CH3 H3CCH3

CH3

C=C C=C

H HH

H;

(e) A pair of optical isomer (v) CHCHCHCH||

O

223 ; CH3CH2CH=CH–OH

[SUBJECTIVE]

Q.22 Assign E &Z configuration?

(I)

Z

(II)

E

OO

O

O

O

(III)

CH CH CH2 2 3

CH – CH3

|CH3

(IV) Cl

Ph

(V)C – C – C

|F

C – C – C

FBr

Cl

(VI)

Z

NC

HOOC CHO

CH = CH2

(VII)

HO

OHC C – CH3

C CH

|CH3

CH3

(VIII)O

O

O

O

O

O

(IX)Me

Et

OMe

O¯ Li+

(X)H

D

OH

OH

18

16

STEREOISOMERISM


(XI)Me

Cl

F

Me(XII)

CH – CH – CH2 3

|CH3

C CH

CH3

H

Q.23 Determine whether each of the following compounds is a cis isomer or a trans isomer.

(a) (b)

(c) (d)

(e) (f)

Q.24 Calculate the numberofchiral center in themolecule Ethyl 2,2-dibromo-4-ethyl-6-methoxycyclohexanecarboxylate.

Q.25 With reasons, state whether each of the following compounds I to VII is chiral.

(I) (II) (III)HO C2 CO H2

(IV) (V) (VI) (VII) SPhO

Me

STEREOISOMERISM


Q.26 What are the relationships between the following pairs of isomers?

(a) and (b)

Cl Cl

Cl Cl

and

(c) and

(d) and

(e) and (f) and

(g) and (h) and

(i) and

Q.27 Total number of streoisomers for the following molecule : (including optical)

(i) CH – CH – CH = CH3

|Cl

CH = CH – CH – CH3

|CH3

(ii)Me

COOH

(iii) N

N

NH2

N

N

EtO – P

O

OEt

(iv)

O

OH

(v) (vi)

OH

O

STEREOISOMERISM


ANSWER KEY

EXERCISE-1

Q.1 4 Q.2 6

Q.3 (a)HH

H

H

H

Et

(b)

HH

HH

Et

Q.4 (a) 4,1,2,3 (b) 4,1,2,3 (c) 1,3,2,4 (d) 4,3,2,1 (e) 2,4,3,1 (f) 4,2,3,1 (g) 3,1,4,2 (h) 2,4,1,3(i) 3,2,1,4 (j) 2, 1, 4, 3

Q.5 (i)Me

(ii)

(iii)

C H2 5

(iv)

Pr

CH3

BrH

(v)

C H52

C H73

Br Cl(vi)

CH3

CH3

Br

Br

H

H

Q.6 (i) Z, (ii) Z, (iii) Z, (iv) E

Q.7 Stable are : (a) diequatorial, (b) , (c) (d) ,

(e) , (f)

Q.8 (i)Trans (ii)Cis (iii)Trans (iv)Cis

STEREOISOMERISM


Q.9

Q.10 8 Q.11 4 Q.12 (i) 6, (ii) 8, (iii) 6, (iv) 6

Q.13 Optical : a,b,c,d,f,g,i,j,k ; Geometrical isomer : c,g,j ; None : e,h,

Q.14 3,1 Q.15 (I) 4, (II) 3, (III) 4 Q.16 4

Q.17 (i) 24 (ii) 4 Q.18 9

Q.19 II, III & IV are Identical; I is Enantiomer of these. Q.20 0630

EXERCISE-2

Q.1 B Q.2 C Q.3 C Q.4 A Q.5 C Q.6 D Q.7 A

Q.8 A Q.9 B Q.10 D Q.11 A Q.12 B Q.13 C Q.14 C

Q.15 B Q.16 D Q.17 B Q.18 B Q.19 C Q.20 B Q.21 A

Q.22 D Q.23 D Q.24 C Q.25 A Q.26 C Q.27 D Q.28 C

Q.29 B Q.30 A Q.31 B Q.32 C Q.33 D Q.34 C Q.35 C

Q.36 D Q.37 A Q.38 A Q.39 A Q.40 BCD Q.41 BC Q.42 D

Q.43 ABD Q.44 BC Q.45 BC Q.46 ABC Q.47 AC Q.48 AC Q.49 C

Q.50 AD Q.51 ACD Q.52 AC Q.53 CD Q.54 ACD Q.55 AC Q.56 AD

Q.57 BC Q.58 C Q.59 A Q.60 AC Q.61 B Q.62 AD Q.63 ACD

Q.64 ACD Q.65 (A) P,R,S ;(B) Q; (C) Q,R; (D) P,R,S

Q.66 (A) Q (B) Q (C) S (D) P Q.67 (A) P, (B) S, (C) R, (D) Q

Q.68 (a-4-iii) ; (b-3-iv) ; (c-2-ii) ; (d-1-i) Q.69 (A) R , (B) P, (C) Q

EXERCISE-3

SECTION-A

Q.1 C Q.2 B Q.3 D Q.4 A Q.5 D Q.6 A Q.7 B

Q.8 A Q.9 B Q.10 A Q.11 A Q.12 A Q.13 A Q.14 C

Q.15 A Q.16 C Q.17 C Q.18 C

STEREOISOMERISM


SECTION-B

Q.1 True Q.2 A Q.3 C Q.4 BC Q.5 B Q.6 B Q.7 A

Q.8 D Q.9 AC Q.10 enantiomers–I & III; diastereomers – I & II and II & III

Q.11 A Q.12 D Q.13 D Q.14 B Q.15 C Q.16 B

Q.17 (i)18.0

1D, (ii)Anti form whenY=CH3 & Gauche when Y = – OH Q.18 C

Q.19 C Q.20 AD Q.21 BCD Q.22 7 Q.23 AD Q.24 5 Q.25 BD

Q.26 8 Q.27 BC Q.28 ABC Q.29 3 Q.30 2 Q.31 7

EXERCISE-5

Q.1 C Q.2 D Q.3 B Q.4 B Q.5 C Q.6 C Q.7 D

Q.8 B Q.9 A Q.10 B Q.11 B Q.12 B Q.13 B Q.14 AC

Q.15 BCD Q.16 ABCD Q.17 AD Q.18 AB Q.19 D

Q.20 (A) P , (B) R , (C) Q , (D) R Q.21 (a) ii (b) v (c) iv (d) i, iv (e) i, iii

Q.22 Z – I, II, III, VI, VII ; E – IV, V, VIII, IX, X, XI, XII

Q.23 (a) cis (b) cis (c) cis (d) trans

(e) trans (f) trans

Q.24 3

Q.25 achiral : I, III, IV ; chiral : II,V, VI, VII

Q.26 (a) Enantiomers, (b) Identical, (c) Geometrical isomers & Diastereomers, (d) Positional,

(e) Optical (Diastereomers), (f) Diastereomers, (g) Enantiomers, (h) Identical, (i) Geometrical isomers

(Diastereomers)

Q.27 (i) 25 , (ii) 24 , (iii) 2 , (iv) 4 , (v) 3, (vi) 8

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X:01 JEE 19-2002 CHEM 201901

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Transcript of X:01 JEE 19-2002 CHEM 201901