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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.
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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.
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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
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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
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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.
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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.
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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.
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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
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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
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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.
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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
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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
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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.
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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
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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
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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
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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
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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'
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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[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
Q.66 Column I Column II
Compound Total number of stereoisomer
(A) (P) 2
(B) (Q) 4
(C) (R) 6
(D) (S) 8
STEREOISOMERISM
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Q.67 Column I Column II
(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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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[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
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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
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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
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(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
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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
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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
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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
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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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