CHAPTER
1
STEREOISOMERISM
Recap of Early Classes
In Previous chapters we have studied about basic principles of organic chemistry,electronic
displacement effects and their applications & structural isomerism in organic compounds. In this
chapter we will discuss stereo isomerism in organic compounds.
1.0
2.0
INTRODUCTION
GEOMETRICAL ISOMERISM
2.1 Geometrical Isomerism in Alkenes
2.2 Nomenclature Systems of Geometrical isomers
2.3 Geometrical Isomerism in Oximes [>C = N–OH]
2.4
3.0
Geometrical Isomers In Azo Compounds (
N N
)
2.5 Geometrical Isomers in Cycloalkanes
2.6 Physical Properties of Cis–Trans Isomers
2.7 Number of Geometrical isomers in polyenes
OPTICAL ISOMERISM
3.1 Optical activity
3.2 Types of Symmetry
3.3 Chiral Compound
3.4 Asymmetric carbon (or) Chiral Carbon
3.5 Projection Formula of Chiral Molecules
3.6 Enantiomers
3.7 Diastereomerism
3.8 Meso isomers
3.9 Calculation of number of optical isomers
3.10 Pseudo Chiral Centre
3.11 D - L System (Relative configuration) : Application on correct Ficher Projection Formula
3.12 Absolute Configuration (R, S configuration)
3.13 Optically active compounds having no asymmetric carbon
3.14 Optical Purity
EXERCISE-1
EXERCISE-2
EXERCISE-3
EXERCISE-4(A)
EXERCISE-4(B)
EXERCISE-5
sdfsfggg
Stereoisomerism
1.0 INTRODUCTION
SL AL
Stereoisomers
Configurational
Conformational
(Interconvertible
non-resolvable)
(non-interconvertible
resolvable)
Geometrical
Isomers
Optical
Isomers
Diastereomers
Enantiomers
(Mirror-image (non-mirror image
optical isomers)
stereoisomers)
Two or more than two compounds having same molecular formula, same structural formula but different
arrangements of atoms or groups in space.
Configurational Isomerism
SL AL
Non interconvertable Stereo isomers are known as Configurational Isomers.
2.0 GEOMETRICAL ISOMERISM
SL AL
2.1
Stereo isomer which cannot interconvert in each other at room temperature due to restricted rotation known
as Geometrical isomerism.
Geometrical Isomerism in Alkenes
SL AL
Reason : Restricted rotation about double bond : It is due to overlapping of p–orbital.
Restricted Rotation
Free Rotation
CH3
CH3
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H
CH3 CH
CH
sp2
sp 2
Example :
CH3
H
Note : cis
l
H
CH3
C
C
CH 3
H
CH 3
H
C
C
H
CH 3
cis–2–butene
trans–2–butene
trans is possible only when p bond break.
Condition for Geometrical isomerism
Only those alkenes show G. I. in which "Each sp2 carbon have different atoms or groups"
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a
x
b C C y
Geometrical isomerism present
a
b C
C a
b
Geometrical isomerism present
p
a
a
C
C
y
C
C
y
q
No Geometrical isomerism
2.2
p
q
No Geometrical isomerism
Nomenclature Systems of Geometrical isomers
SL AL
(a)
Cis–Trans System : If same groups at same side then cis and if same groups at different side then
trans.
a
b C C a
a
b C C b
a
b
[Same groups, same side]
[Same groups different side]
cis
trans
Example :
a
C
x
C
a
z
y
x
C
cis
Example :
CH 3
H
z
y
cis
C
C
Cl Cl
C
H
CH 2 CH 3
C
Br
trans-2–pentene
(b)
C
Br
It does not show Geometrical isomers So no cis–trans
E – Z System :
E (Entgegen) : When high priority groups are opposite side.
Z (Zussaman) : When high priority groups are same side.
LP
HP
C C HP
C C HP
LP
LP
HP
LP
'E'
'Z'
HP – High priority and LP – Low priority
Priority Rules : (Cahn Ingold Prelog Rule)
Rule I : Priority is proportional to atomic number of atom which is directly attached to sp2 carbon.
Br
C
[HP] I
'
C
[HP]
F
Cl
Cl
Cl C
Cl Cl
[HP]
Z'
C
C
CH 3
N H 2 [HP]
'E'
Rule II : If rule-I is failed then consider next atom
CH 3
[HP]
F
[LP] CH3 H2C
C
C
C
CH 3
C CH
3
H
'Z'
2
CH 3
CH 3
[C, C, C]
[HP]
[C, C, H]
[LP]
decreasing order of atomic number
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l
Stereoisomerism
Rule III :– If multiple bond is present then consider them as :-
C
C
C
O
O
Example :
[HP]
N C
C
H
N
H
C
C
C
C
C
O
O
C
C
C
C
N
C
C
C
C
C
N
C
C
N
N
C
[O, O, H]
C
H
C
O
C
C
OH
[O, O, O]
[HP]
Z'
Prioity order for some groups is :
Br > Cl > OH>NH2>COOH>CHO>CH2OH>CN>C6H5>C = CH>C(CH3)3>CH
CH2>CH(CH3)2
Rule IV : If isotopes are present then consider atomic weight.
Rule V : Higher priority is assigned to bond pair then lone pair.
Example :
(i)
(ii)
(iii)
(iv)
H
[HP]
D
CH 3
[HP]
Cl
CH3
H
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(v)
C
C
C
C
CH3
C
CH2
CH
N
C
C
C
CH3 C
CH3 CH
C
CH3
CH2 CH3
'Z'
[HP]
H
CH3
'Z'
[HP]
CH3
'E'
Cl
C
C
C
CH2 Cl
[Cl, H, H]
[HP]
COOH
[O, O, O]
[LP]
CBr3
;'E'
[Br, Br, Br ]
CHI2
[I, H, H]
'E'
3
CH2 = CH
(vi)
HC
C
C(CH3)3
C == C
CH – CH = CH2
'E'
NºC
(vii)
C=N
HO – C
CH3
'Z'
O
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N
C
(viii) H2C = C = CH
CH2 – C(CH3)3
C == C
'E'
C
CH
O
Cl – C
(ix)
Br – C
C =C
C º CH
'Z'
Ph
O
O
CH3O – C
(x)
Cl – C
COOH
C =C
'E'
CHO
O
CH3
(xi)
CH3 – C – CH2
HO – CH2
C =C
CºN
'E'
CH = N – OH
E-Z Configuration
1.
Assign E & Z configuration?
O
(I)
O
O
O
(II)
O
E
Z
CH2CH2CH3
Ph
(V)
F
|
C–C–C
F
C–C–C
HO
(VII)
4
OHC
Br
(IV)
NC
(VI)
Cl
CH = CH2
HOOC
CHO
Z
O
C º CH
CH3
C – CH3
|
CH3
Cl
O
O
(VIII)
O
O
O
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CH – CH3
|
CH3
(III)
Stereoisomerism
Me
(IX)
Et
+
OH
H
(X)
O¯ Li
16
OH
D
CH3
|
CH – CH2 – CH3
Me
(XI)
18
OMe
F
Cl
C º CH
(XII)
Me
CH3
H
2.3
Geometrical Isomerism in Oximes [>C = N–OH]
SL AL
(i)
(ii)
Oximes show G. I. due to restricted rotation about double bond.
Only those oximes show Geometrical isomerism in which sp2 carbon have two different groups.
[CH3—CH O + H2N—OH] ¾¾® CH3—CH N—OH (oxime)
Example : Acetaldoximes has two Geometrical isomers –
CH3 C
H
CH3 C
N OH
syn
When H and OH are on the same side
Example :
Ph—CH
Ph
HO
N
anti
When H and OH are on the opposite side
C
H
N
OH
Ph
C
H
HO
N
N—OH
H
[syn.]
[Anti]
Benzaldoxime
l
Ketoxinme
H3C
H5C2
C=N
H3C
OH
H5C2
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syn methyl ethyl ketoxime
or anti ethyl methyl ketoxime
2.4
C=N
OH
syn ethyl methyl ketoxime
or anti methyl ethyl ketoxime
Geometrical Isomers In Azo Compounds (
N N
)
SL AL
N
N
Ph
Ph
(syn)
Ph—N
N
Ph
N
Ph
(Anti)
N—Ph (Azo benzene)
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2.5
Geometrical Isomers in Cycloalkanes
SL AL
Cycloalkanes show Geometrical isomers due to restricted rotation about single bond. Only those cyclo alkanes
show Geometrical isomers in which atleast two different carbons have two different groups.
2.6
Me
Me
H
H
Two Geometrical isomers
Me
Me
H
H
Me
H
H
Me
cis
trans
Physical Properties of Cis–Trans Isomers
SL AL
S.No.
Physical properties
Comparison
Remarks
1
Dipole moment
cis > trans
cis-isomer has resultant of dipoles while in trans isomer dipole
moments cancel out
2
Boiling point
cis > trans
Molecules having higher dipole moment have higher boiling
point due to larger intermolecular force of attraction
3
Solubility (in H2O)
cis > trans
More polar molecules are more soluble in H2O
4
Melting point
trans > cis
More symmetric isomers have higher melting points due to
better packing in crastalline lattice & trans isomers are more
symmetric than cis.
5
Stability
trans > cis
The molecule having more vander wall strain are less stable.
In cis isomer the bulky group are closer they have larger
vander waal strain.
Dipole moment [m] :
CH3 C
H
CH 3 C H
cis m ¹ 0
Example :
Illustration 1.
Soltuion
CH3 C
H
H C CH 2 CH3
trans m ¹ 0
H C
CH3
CH3 C H
trans m = Zero
CH3 C
H
Cl C H
cis m ¹ 0
Which of the following show Geometrical isomerism –
(A) 1,1–diphenyl–1–butene
(B) 1,1–diphenyl–2–butene
(C) 2,3–dimethyl–2–butene
(D) 3-phenyl–1–butene
(B)
Cl
Illustration 2.
If dipole moment of chlorobenzene is m, then dipole moment of
is –
Cl
Solution
6
Zero
Cl
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Example :
Stereoisomerism
Illustration 3.
Which of the following show Geometrical isomerism –
(B) H2C
(A) CH3—CH2—CH N—OH
Solution
(C) CH3 C
N
(A), (D)
CH3
Illustration 4.
Which of the following show Geometrical isomerism –
(D) CH 3 C
OH
N
H
H Cl
H
Cl
(A) H
Cl
Solution
2.7
N—OH
(B) Cl
H
(C) Cl
Cl
H
CH2CH3
OH
Me
H
Me
(D)
Br
Cl
Br
Me
Me
(A), (B) and (D)
Number of Geometrical isomers in polyenes
SL AL
R1
HC
(a)
CH
n
R2
If R1 ¹ R2 then number of Geometrical isomers = 2n
[n = number of double bonds.]
Example : CH3—CH=CH—CH=CH—CH=CH—CH2CH3
As n = 3 ® number of Geometrical isomers = 23 = 8
(b)
If R1 = R2 then number of Geometrical isomers = 2n – 1 + 2p – 1
n
n +1
(n is odd)
(when n is even)
and p =
2
2
Example : CH3—CH=CH—CH=CH—CH=CH—CH3 [n = 3]
where p =
Number of Geometrical isomers
= 2n – 1 + 2p – 1
[ p=
3 +1
]
2
= 22 + 21= 4 + 2 = 6
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Geometrical Isomerism
1.
Calculate the total number of open chain isomeric carbonyl compounds of molecular formula C5H8O which
can't show geometrical isomerism.
2.
The number of geometrical isomers in case of a compound with the structure :
CH3–CH=CH–CH=CH–C2H5 is (A) 4
(B) 3
(C) 2
(D) 5
3.
Which one of the following will show geometrical isomerism (A) CH2Cl
C
H
(C) CH2
CH
H
4.
CH3
CH
C
CH3
(B)
C(CH3)2
H
CH
C(CH3)2
(D) CH3CH2CH=CHCH2CH3
CH2Cl
CH
CH2Cl
C
CH2
Which of the following will show geometrical isomerism –
(A) CH3CH = CH2
Br Br
| |
(B) CH3 — C = C — CH2 CH3
(C) CH3CH2CH2CH = CHCH3
(D) CH2 = CH — CH2 — CH3
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5.
Which of the following compounds can exist as geometrical isomers –
(A) CH2Cl2
(B) CH2Cl — CH2Cl
(C) CHBr = CHCl
(D) CH2Cl — CH2Br
6.
Which of the following compounds does not have geometrical isomers (A) 2-Pentenoic acid
(B) 2-Butenoic acid
(C) 3-Pentenoic acid
(D) 3-Butenoic acid
7.
Among the following compounds, the one which does not show geometrical isomerism is (A) C6H5N = NC6H5
(B) C6H5CH = CHC6H5
(C) C6H5 – C = N – OH
CH3
8.
2-Butene exhibits geometrical isomerism due to (A) rotation about the double bond
(C) restricted rotation about the double bond
(D) C6H5 – C = N – CH3
C6H5
(B) rotation about C3–C4 bond
(D) rotation about C1–C2 bond
3.0 OPTICAL ISOMERISM
SL AL
Compounds which have same molecular and structural formula but have different optical activity are known
as optical isomers.
3.1
Optical activity
SL AL
Certain organic compounds, when their solutions are placed in the path of a plane polarized light, have the
remarkable property of rotating its plane through a certain angle which may be either to the left (or) to the right.
This property of a substance of rotating the plane of polarized light is called optical activity and the substance
possessing it is said to be optically active.
The observed rotation of the plane of polarized light [determined with the help of polarimeter] produced by a
solution depends on :
(a)
The amount of the substance in tube ;
(b) On the length of the sample tube;
(c)
The temperature of the experiment and
(d) the wavelength of the light used.
The instrument used to measure angle of rotation is called polarimeter. The measurement of optical rotation
is expressed in terms of specific rotation [a]Dt ; this is given by the following relation :
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Certain substances possess the property to rotate the plane of polarized light. Such substances are called optically
active substances and this phenomenon is called optical activity.
Light has vibrations occur in all planes at right angles to the line of propagation. In plane polarized light the
vibrations take place only in one plane. Plane polarized light can be obtained by passing ordinary light through
a Nicol prism.
Stereoisomerism
[a]Dt =
a obs
[where a = observed angle of rotation]
l´C
[a ]Dt = specific rotation determined at t°C, using D-line of sodium light.
l = length of solution in decimeters
C = concentration of the active compound in grams per millilitre.
The sign attached with the angle of rotation signifies the direction of rotation. Negative sign (–) indicates that
the rotation is towards anti-clockwise, while positive (+) sign means that the direction of rotation is toward right
clockwise.
If the substance rotates plane-polarised light to the right i.e. in clockwise direction it is called dextrorotatory and
indicated by ‘d’ or (+)
If the substance rotates plane-polarised light to the left i.e. in anti clockwise direction it is called laevorotatory
and indicated by ‘l’ or (–)
l
3.2
Condition of Optical Activity
Compound must be unsymmetrical.
Types of Symmetry
SL AL
(i)
Plane of symmetry (POS) : An imaginary plane which bisects any object or molecule into two equal
parts which are mirror images of each other is known as POS.
H
Me
Cl
H
H
Cl
OH
OH
Me
Plane of symmetry
H
Me
Me
Plane of symmetry
Plane of symmetry
(ii)
H
H
Center of Symmetry (COS): It is a point inside a molecule from which on travelling equal distance
in opposite directions one takes equal time.
COOH
CH3
Cl
H
H
COOH
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(iii)
•
R
H
C=C
H
H
H
R
H
Centre of symmetry
CH3
H
Cl
Centre of symmetry
Axis of Symmetry (AOS) (Cn) : Axis of symmetry is an axis such that if one rotates the molecule
by
360 °
, the new position of molcule is superimposable with the original one.
n
H3C
H
C
C
CH3
H
H3C
¾¾®
H
C
C
CH3
H
180°
(iv)
Alternate axis of Symmetry (AAOS) : An imaginary plane around which rotation by angle x gives
its mirror image is known as AAOS. It is represented by Sn where n =
360
x
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JEE-Chemistr y
H3C
H
H
C
C
H
¾¾®
CH3
CH3
C
CH3
C
H
180°
Illustration 5.
Which of the following compounds have POS & COS.
(i)
(ii)
(iii)
(iv)
H3C
(v)
(vi) CCl2H2
H
C
(vii)
C
H
C
CH3
F
(viii)
(ix)
F
(x)
(xi)
F
F
H
(xiii)
(xiv)
F
F
H
F
(xvi)
F
H
Solution
3.3
F
H
F
H
H
H
H
(xv)
F
H
(xvii) H
C
H
(xviii)
H
POS Þ (i), (ii), (iii), (iv), (vi), (viii), (ix), (xi), (xii), (xiii), (xiv), (xv), (xvi), (xvii), (xviii)
COS Þ (iv), (xi), (xiii), (xiv), (xviii)
Chiral Compound
SL AL
Compound which is not super imposable on its mirror image is known as Chiral Compound. All optically active
compounds are Chiral.
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(xii)
F
Stereoisomerism
3.4
Asymmetric carbon (or) Chiral Carbon
SL AL
If all the four bonds of carbon are satisfied by four different atoms/groups, it is chiral. Chiral carbon is designated
by an asterisk (*).
Example : H
3.5
Br
C*
Cl
I
Projection Formula of Chiral Molecules
SL AL
(i)
Wedge-Dash Projection formula
down
up
CH3
Example : (i) Butan-2-ol
C
H
Ficher Projection formula
Rules of writing Fisher Projection formula
(a)
Groups at Vertical line are away from observer.
(b) Groups at Horizontal line are towards the observer.
(c)
Central ‘C’ atom of the cross is chiral.
(d) High priority group lies at the top of vertical line (Numbering starts from top).
CH3
H
OH
CH2CH3
3.6
Enantiomers
///////////////////////
(ii)
OH
C2H5
CH3
OH
H
CH2CH3
SL AL
Stereoisomers which are mirror-image of each other are called enantiomers (or) enantiomorphs. Thus (i) and
(ii) are enantiomers. All the physical and chemical properties of enantimoers are same except two :
(i)
They rotate PPL to the same extent but in opposite direction. One which rotates PPL in clockwise
direction is called dextro-rotatory [dextro is latin word meaning thereby right] and is designated by d (or)
(+). One which rotates PPL in anti-clockwise direction is called laevo rotatory [means towards left] and
designated by l (or) (–).
(ii)
They react with optically active compounds with different rates.
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3.7
Diastereomerism
SL AL
The optical isomer which are not mirror images to each other are called diastereomers.
Properties
Me
H
H
(i) Dipole moment
(ii) Melting point
(iii) Boiling point
(iv) Solubility & Density
(v) Specific rotation
Different
Different
Different
Different
Different
Et
I
Me
Cl
H
Cl
Cl
Cl
H
Et
II
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JEE-Chemistr y
Example :
CH3
CH3
Cl
Cl
H
H
Cl
Cl
H
H
Cl
Cl
H
Cl
H
H
Cl
CH2CH3
(I)
CH2CH3
(II)
(III)
I, II = Enantiomers,
I, III = Diastereomers,
II, III = Diastereomers,
III, IV = Enantiomers
II, IV = Diastereomers
I, IV = Diastereomers
COOH
COOH
Example :
H
OH
HO
H
H
OH
HO
H
COOH (I)
Achiral
CH2CH3
(IV)
COOH (II)
I and II are identical
Stereoisomerism in Tartaric Acid
Tartaric acid exists in only three stereomeric forms. Two of them (1 and 2) are non superimposable mirror images
of each other so they are enantiomers of each other. The 3 rd isomer is non mirror image stereomers of (1) and
(2) so it is diastereomer of (1) and (2). 3rd isomer containes plane of symmetry so, it is optically inactive. It
is known as meso tartaric acid.
CO2H
HO—C—H
H—C—OH
3.8
CH3
H
CH2CH3
l
CH3
CO2H
H—C—OH
OH—C—H
CO2H
CO2H
(1)
(2)
CO2H
H—C—OH
H—C—OH
CO2H
(3)
Meso tartaric acid
Meso isomers
3.9
Achiral compound which have chiral centers is known meso compound.
They are achiral (optical rotation = 0).
They have [a] = 0 due to internal compensation of optical rotation.
They are diastereomer of d – l pair. So, it has different physical properties than d – l -pair..
Presence of more than one asymmetric ‘C’ atoms.
They are non resolvable.
Calculation of number of optical isomers
SL AL
The number of optical isomers of an organic compound depends on its structure and number of asymmetric
carbon atoms. Thus, the number of optical isomers may be determined from the knowledge of the structure
of the compound as follows :
(a)
When the molecule is unsymmetrical
No. of optically active isomers, a = 2n
Number of meso forms (m) = 0
Number of racemic mixtures, r = a/2
\ Total no. of optically active isomers = (a + m) = 2n
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SL AL
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Stereoisomerism
(b)
When the molecule is symmetrical and has even no. of asymmetric carbon atoms.
No. of optically active isomers, a = 2(n –1)
n
No. of meso forms, m =
-1
22
No. of racemic mixtures, r = a/2
\ Total no. of optically active isomers = a + m
(c)
When the molecule is symmetrical and has an odd no. of asymmetrical carbon atoms.
no. of optically active isomers, a =
2(n–1)
–
n -1
2 2
n -1
No. of meso forms, m = 2 2
\ Total no. of optically active isomer = (a + m) = 2(n–1)
Example Structural formula
CH3 – CH—CH – CH2 – CH3
Cl
Cl
Optical isomer = 2 = 2 = 4
n
l
2
Stereochemical Formula
CH3
CH3
CH3
CH3
H
Cl
Cl
H
H
Cl
Cl
H
H
Cl
Cl
H;
Cl
H
H
Cl
C2H5
I
[a] = +xº
Analysis
I, II – Enantiomers
I, III – Diasteromers
II, III – Diasteromers
C2H5
C2H5
C2H5
II
–xº
III
+yº
IV
–yº
III, IV – Enantiomers
I, IV – Diasteromers
II, IV – Diasteromers
3.10 Pseudo Chiral Centre
AL
A chiral center which becomes achiral on changing the configuration of one of its substituents and vice versa
CH3
OH
CH
Pseudo chiral center ¾® HC
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HC
Br
OH
CH3
3.11 D - L System (Relative configuration) : Application on correct Ficher Projection Formula
SL AL
This method is used to relate the configuration of sugars and amino acids to the enantiomers of glyceraldehyde.
The configuration of (+)-glyceraldenyde has been assigned as D and the compounds with the same relative
configuration are also assigned as D, & those with (-) glyceraldehyde are assigned as L.
CHO
CHO
H
(R)
OH
HO
H
(S)
CH2OH
D-(+)-glyceraldehyde
– OH group on right side
CH2OH
L-(–)-glyceraldehyde
– OH group on left side
13
JEE-Chemistr y
Examples :
H2N
COOH
COOH
COOH
H
H
NH2
NH2
CH2CH2COOH
CH2OH
CH3
D(+)-serine
L-Alanine
H
L-(+)-Glutamic acid
Sugars have several asymmetric carbons. A sugar whose highest numbered chiral centre (the penultimate carbon) has the same configuration as D-(+)-glyceraldehyde (– OH group on right side) is designated as a D-sugar,
one whose highest numbered chiral centre has the same configuration as L-glyceraldehyde is designated as an Lsugar.
CH2OH
|
C=O
CHO
e.g.
H
HO
H
H
OH
H
OH
OH
HO
H
CHO
H
HO
H
H
OH
CH2OH
D-(-)-Threose
CH2OH
CHO
HO
HO
H
H
H
OH
OH
CH2OH
CH2OH
D-(+)-Glucose
H
H
OH
OH
D-Mannose
D-Fructose
3.12 Absolute Configuration (R, S configuration)
SL AL
The actual three dimensional arrangement of groups in a molecule containing asymmetric carbon is termed
absolute configuration.
System which indicates the absolute configurastion was given by three chemists R.S. Cahn, C.K. Ingold and V.
Prelog. This system is known as (R) and (S) system or the Cahn–Ingold system. The letter (R) comes from the
latin rectus (means right) while (S) comes from the latin sinister (means left).
(R) (S) nomenclature is assigned as follows :
Step 1 : Assign priority to the groups which are attached with chiral carbon.
Step 2 : Bring the lowest priority group to dash by even simultaneous exchanges.
Step 3 : Draw an arrow from first priority group to second priority group till third priority group.
Step 4 : If the direction of arrow is clockwise the configuration is R and if anticlockwise it is S.
RS Nomenclature in ficher formula :
Step 1 : Assign priority to the groups which are attached with chiral carbon.
Step 2 : Bring the lowest priority group to vertical line.
Step 3 : Draw an arrow from first priority group to second priority group till third priority group.
Step 4 : If the direction of arrow is clockwise the configuration is R and if anticlockwise it is S.
(1)
(2)C2H5
H
(4)
l
C
C2H5
H
CH3
CH3
(3)
OH
C
S
C2H5
S
OH
CH3
H
Important
Note that the designation of a compound as R or S has nothing to do with the sign of rotation. the Cahn-Ingold
rule can be applied to any three dimensional representation of a chiral compound to determine whether it is R or
S only. For example in above case (i.e. lactic acid), R configuration is laevo rotatory is designated as R-(–)-lactic
acid. Now the other configuration of it will have opposite sign of rotation i.e. S-(+)-lactic acid.
14
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OH
Stereoisomerism
3.13 Optically active compounds having no asymmetric carbon
AL
(i)
Allenes : An sp-hybridized carbon atom possess one electron in each of two mutually perpendicular
p orbitals. When it is joined to two sp2-hybridized carbon atoms, as in allene, two mutually perpendicualr
p-bonds are formed and consequently the p-bonds to the sp2-carbons are in perpendicular planes. Allenes
of the type abC=C=Cab (a ¹ b) are therefore not superimposable on their mirror images and despite
the absence of any asymmetric atoms, exist as enantiomers and several optically active compounds have
been obtained.
(Ex. a = phenyl, b=1-naphthyl)
b
a
C
C
C
a
(ii)
a
b
b
b
C
C
C
a
Alkylidene cyclo alkanes : The replacement of one double bond in an allene by a ring does not alter
the basic geometry of the system and appropriately substituted compounds exist in optically active forms.
H
H3C
CO2H
H
Related compounds in which sp2-carbon is replaced by nitrogen have also been obtained as optical
isomers.
H
OH
N
CO2H
l
l
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l
l
Difference between Racemic mixture and Meso compound
A racemic mixture contains equimolar amounts of enantiomers. It is optically inactive due to external
compensation.
External compensation
If equimolar amounts of d and l-isomers are mixed in a solvent, the solution is inactive. The rotation of each
isomer is balanced (or) compensated by the equal but opposite rotation of the other. Optical inactivity having
this origin is described as due to external compensation. Such mixtures of (+) and (–) isomer (Racemic mixtures)
can be separated into the active components.
A meso compound is optically inactive due to internal compensation
Internal compensation
In meso tartaric acid the inactivity is due to effects within the molecule and not external. The force of rotation
due to one of the molecule is balanced by the opposite and equal force due to the other half. The optical
inactivity so produced is said to be due to internal compensation. It occurs whenever a compound containing
two (or) more asymmetric carbon atoms has a plane (or) point of symmetry. Since the optical inactivity of such
a compound arises within the molecule, the question of separating into active components does not arise.
Special Point
(i)
Chiral nitrogen containing tetra alkyl ammonium ion show optical isomerism.
R4
R1
R1
NÅ
NÅ
R3
(I)
R2
R2
R3
R4
(II)
I and II enatiomers
15
JEE-Chemistr y
(ii)
Chiral nitrogen containig tertiary amine do not show optical isomerism
Reason :- Rapid umbrella inversion.
N
R1
R2
(I)
N
Room temperature
R3
R3
R2
(II)
R1
(iii)
Energy required for this interconversion is available at room temperature so I and II are identical
(iv)
Ortho substituted biphenyl compounds do not have any chiral carbon but due to chiral molecule, they
are optically active.
NO2 HOOC
SO3H H3C
COOH O2 N
Horizontal
plane
CH3 HO3 S
Optically active
Vertical
plane
3.14 Optical Purity
SL AL
Some times we deal with mixtures that are neither otpically pure nor racemic mixture. In these cases we
specify the optical purity of the mixture. It is defined as the ratio of its rotation to the rotation of pure enantiomer.
observed optical rotation
Þ Optical purity = optical rotation of pure enantiomer × 100
e.g. If we have some 2-butanol with observed rotation of +9.72, we compare this rotation with +13.5 rotation
of the pure (+) enantiomer.
Optical purity =
9.72
´100 = 72%
13.5
That means 72% is pure (+) 2-Butanol and 28% is (± mixture)
Total (+) isomer = 72 + 14 = 86%, (–) isomer = 14%
Mark meso isomers among following
(1)
COOH
OH
H
(2)
OH
COOH
H
H
(4)
OH
H
*
OH
Solution
16
(1), (2), (4) and (5)
(5)
Cl
Cl
Me
Me
(3)
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
Illustration 6.
Stereoisomerism
Illustration 7.
Which of the following is optically active –
(A) CH3—CH C CH—CH3
Cl
(C)
C
C C
(B) CH2
Cl
Me
(D)
Br
Solution
Br
(A), (C) and (D)
Illustration 8.
Give number of Stereoisomers following compounds:
(i) Cl
Et
C
CH2
C
C C
Cl
Br
(ii) F
Cl
Cl
Cl
(iii) Me
(iv)
CH=C=C=CH–Me
Cl
(v)
(vi)
(vii)
(viii)
Cl
Cl
H
H3C—C=C=C=C
(ix) Cl
(x)
Cl
(xi) Cl
HC
CH
Me
CH=CH–Cl
Cl
(xii)
Cl
Cl
Solution
(i) = 3, (ii) = 4, (iii) = 4, (iv) = 3, (v) = 4, (vi) = 8, (vii) = 10, (viii) = 6, (ix) = 32
(x) = 2, (xi) = 6, (xii) = 2
Illustration 9.
Give (R), (S) & (E), (Z) configuration of following compounds.
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
CH3
H3C
H
Br
Me
Cl
(a)
(b) H
H
Cl
CH3
(c) F
H3C
H
CHO
H
Me
COOH
(d) F
OH
CH3
Solution
Me
(e)
Et
Cl
H
(f) Cl
H
(a) = (Z, R), (b) = (Z,S), (c) = (R), (d) = (S), (e) = (S), (f) = (S)
H
H
Me
17
JEE-Chemistr y
Optical Isomerism
CH3
1.
H
Cl
HO
H
C2H5
The compound with the above configuration is called:
(A) (2S, 3S)-2-chloro-3-hydroxypentane
(B) (2S, 3R)-2-chloro-3-hydroxypentane
(C) (2R, 3R)-2-chloro-3-hydroxypentane
(D) (2R, 3S)-2-chloro-3-hydroxypentane
2.
Find out the total number of cyclic isomers of C6H12 which are optically active?
3.
How many enantiomers are possible on monochlorination of isopentane.
4.
Write the structure of:
(i) (E) penta-1,3-diene
(iii) (2E, 4E)-3-ethylhexa-2,4-diene
(v) (S)-3-bromo-3-chlorohexane
What are the relationships between the following pairs of isomers?
(a)
and
(c)
and
(b)
(d)
and
and
(e)
and
(f)
(g)
and
(h)
(i)
18
and
and
and
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5.
(ii) (2Z, 4E)hexa-2,4-diene
(iv) (R)-2-Bromopentane
(vi) (2S, 3R)-2,3-dibromobutane
Stereoisomerism
ANSWERS
BEGINNERS BOX-1
1.
(Z) – I, II, III, VI, VII ; (E)– IV, V, VIII, IX, X, XI, XII
BEGINNERS BOX-2
1.
6.
(8)
(D)
BEGINNERS BOX-3
1.
(A)
4.
(i)
2.
7.
(A)
(D)
3.
8.
(D)
(C)
2.
8
3.
4
4.
(B)
5.
(C)
(ii)
Et
Et
C2H5
(iii)
(iv)
H
Br
CH3
C3H7
(v)
Cl
Br
H
Br
Br
(vi) H
C2H5
5.
CH3
CH3
(a) Enantiomers
(b) Enantiomers
(c) Geometrical isomers & Diastereomers
(d) Positional
(e) Optical (Diastereomers)
(f) Diastereomers,
(g) Enantiomers,
(h)Identical,
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
(i) Geometrical isomers (Diastereomers)
19
JEE-Chemistr y
SINGLE CHOICE CORRECT QUESTIONS
H3C
1.
C
C
H
H
Exhibits :H3C
COOH
(A) Tautomerism
(C) Geometrical isomerism
H3C
2.
C
(B) Optical isomerism
(D) Geometrical and optical isomerism
Given compound has x geometrical isomers and y optical active isomers
H3C
H
C
C
H
OH
CH2—CH2—C—CH3
H
The value of x and y respectively are :(A) 4 and 4
(B) 2 and 2
3.
(C) 2 and 4
(D) 4 and 2
The correct structure of trans–2–hexenal is (A)
CHO
(B)
(C)
CHO
(D)
4.
Which of the following compounds is chiral ?
(A) CH3CH2CH2Cl
(B) CH3CHDCH3
5.
In the structure :
CHO
CHO
(C) CH3CHClCH3
(D) CH3CH2CHDCl
CH3
OH the configurations at the chiral centres are :
Br
H
H
CH3
(A) 2R, 3R
6.
(B) 2S, 3R
(C) 2R, 3S
(D) 2S, 3S
Which of the following compound contains a pseudo-asymmetric carbon atom
(A) CH 3CHCHCH 2
(B) CH3CH–CH–CHCH3
Br Br Br
Br OH Br
(D) CH 3CHCHCHCH3
(C) CH3CHCHCHCH3
OH Br Br
Consider the following structures (A), (B), (C) and (D) -
CH3
C2H5
Br
(A) Cl
C2H5
Br
(B) Cl
CH3
Which of the following statements is not correct
(A) B and C are identical
(C) A and C are enantiomers
8.
Cl
(C) CH3
Cl
Br
(D) C2H5
C2H5
(B) A and B are enantiomers
(D) B and D are enantiomers
In which of the following properties do enantiomers differ from each other
(A) Solubility in an achiral solvent
(B) Reactivity with a chiral reagent
(C) Melting point
(D) Dipole moment
20
CH3
Br
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
7.
Br OHCl
Stereoisomerism
9.
(+) - Mandelic acid (Ph – CH(OH) – COOH) has a specific rotation of + 158º. What would be the observed
specific rotation of a mixture of 25% (–) -mandelic acid and 75% (+) -mandelidc acid :
(A) + 118.5°
(B) –118.5°
(C) – 79°
(D) + 79°
10.
The structures shown here are related as being:
H
(I)
Br
CH3
C
C
CH3
H
Br
Br
C
(II) H3C H3C
C
H
H
Br
(A) confermers
11.
(B) enantiomers
(A)
COOH
H
H
H
(C)
H
(B)
CH3
OH
H
H
H
H
H
Ph
Ph
COOH
H
H
(D)
H
HOOC
H
14.
H
H
H
H
13.
(D) diastereoisomers
Which species exhibits a plane of symmetry?
CH3
12.
(C) geometrical isomers
OH
COOH
H
H
COOH
Recemic mixture is formed by mixing two(A) isomeric compounds
(C) meso compounds
(B) chiral compounds
(D) enantiomers with chiral carbon
Geometrical isomerism is not shown by(A) Pent–1–ene
(C) 1–Chloropent–2–ene
(B) Pent–2–ene
(D) 2–Chloropent–2–ene
Following types of compounds I and II can not show
(I) CH3CH=CHCH3
(II) CH3
CH
OH , are studied in
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
CH2CH3
(A) chain isomerism
(C) stereoisomerism
15.
(B) position isomerism
(D) metamerism
Among the following four structures I to IV
H
C2H5
CH
CH3
(I)
C3H7
C2H 5
C
CH
O
CH3
(II)
It is true that(A) All four are chiral compounds
(C) only III is a chiral compound
C2H5
H
CÅ
H
(III)
C2H5
CH
C2H5
CH3
(IV)
(B) only I and II are chiral compounds
(D) only II and IV are chiral compounds
21
JEE-Chemistr y
16.
Which of the following will have a meso-isomer also(A) Methylcyclopentane
(B) 1,3–Dimethylcyclopentane
(C) 1–Bromo–2–chlorocyclopentane
(D) 3–Bromocyclopent–1–ene
17.
Amongst the following compounds, the optically active compound is
CH3
(A) CH3
C
CH2
CH
CH
CH3
(B) CH3
CH2
(D) CH 3
CH2 CH2 CH3
H
(C) CH3
C
C2H5
18.
Which of following compounds is not chiral
(A) 1–chloropentane
(C) 1–chloro–2–methyl pentane
(B) 2–chloropentane
(D) 3–chloro–2–methyl pentane
19.
Of the five isomeric hexanes, the isomer which can give two structural monochlorinated compounds is(A) 2-methyl pentane
(B) 2,2–dimethyl butane
(C) 2,3–dimethyl butane
(D) n-hexane
20.
Which of the following molecules is expected to rotated the plane of plane-polarised light ?
CHO
(A) HO
H
(B)
CH2OH
(C)
H2N
H
Ph
NH2
H
SH
COOH
H
(D) H2N
Ph
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
H
22
Stereoisomerism
1.
2.
SECTION - 1 : MULTIPLE CHOICE CORRECT QUESTIONS
In which of the following cases, cis-trans nomenclature can not be used :(A) Cl—CH=CH—Cl
(B) CH3—CH=CH—CHO
(C) C6H5—N=N—C6H5
(D) CH3—CH=C(Cl)C2H5
Consider the following compounds :
CH3
CH3
H———OH
CH3
HO———H
HO———H
HO———H
H———OH
HO———H
CH3
(I)
CH3
(II)
CH3
(III)
Choose the correct statements :
(A) I and III are enantiomers
(C) II and III are diastereomers
3.
Which will show geometrical isomerism :-
(A) C6H5–CH=NOH
4.
(B) I and II are diastereomers
(D) I, II and III are all optically active
(B)
(C) C6H5–N=N–C6H5
H3C CH3
CHO
HO———H
H———OH
HO———H
C6H5
(I)
C6H5
(II)
(A) Both are in threo form
(C) Both are diastereomers
(B) Both are enantiomers
(D) Both are in erythro form
Observe the following structures and pick up the correct option (s) mentioned below :H
CH3
H
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
H H
Which statement (s) is/are correct for :-
CHO
H———OH
5.
(D)
CH2Cl
(A)
Cl
H
H
H
CH3
Cl
H
Cl
(B)
(A) The two are position isomers
(B) None of the two shows optical isomerism
(C) Only A shows optical isomerism
(D) The two are not related to each other regarding isomerism
6.
The R and S enantiomers of an optically active compound differ in :(A) their reactivity with chiral reagents
(B) their melting points
(C) their optical rotation of plane polarized light
(D) their solubility in achiral reagents
23
JEE-Chemistr y
7.
Which of the following statements (s) is (are) incorrect :Y
X
(A) X is cis- and Y is trans
(C) X is trans and Y is cis
(B) X is Z and Y is E
(D) X and Y are diastereomers
8.
Which of the following compounds will show geometrical isomerism :(A) 2-butene
(B) propene
(C) 1-phenylpropene
9.
Which of the following notations are correct :
(A)
(C)
10.
Br
F
C=C
Cl
(Z)
I
H3C
H
C=C
(Z)
(B)
CH3
H
(D)
Br
I
C=C
(E)
H 3C
H
Cl
I
C=C
(E)
(D) 2-methyl-2-butene
CH 2CH2CH 3
CH 3
For which of the following pairs of compounds are the correct notation given :-
Ph
Ph
Ph
and
N=N
(A)
Anti-azobenzene
N=N
Ph
Syn-azobenzene
(B)
C=N
OH
H
Syn-acetaldoxime
and
C=N
H
Anti-acetaldoxime
H
H
CO2H
C=C
H
(C)
OH
H3C
H3C
NH2
Trans-o-aminocinnamic acid
H
CO2H
C=C
and
NH2
Cis-o-aminocinnamic acid
ClCH2
H3C
CH2CH2CH3
CH3
C=C
C=C
and
H
CH3
(D)
CH3
BrH2C
Z-isomer
11.
Which can show geometrical isomerism as well as optical activity.
(A) 1,1–dichlorocyclohexane
(B) 1,2–dichlorocyclohexane
(C) 1,3–dichlorocyclohexane
(D) 1,4–dichlorocyclohexan
12.
Which of the following will show optical isomerism as well as geometrical isomerism.
H3C
C
(A)
H3C
CH3
CH3
C
(B)
CH3
H
H3C
C
(C)
H3C
24
H3C
CH3
C
(D)
H
CH3
H3C
CH3
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
E-isomer
Stereoisomerism
CH3
13.
CH3
H———Cl
H———Cl
Cl———H
H———Cl
CH2CH3
(I)
CH2CH3
(II)
(A) I and II are enantiomers
(C) I is 2R, 3R; While II is 2R, 3S
14.
(B) I is 2S, 3S; while II is 2S, 3R
(D) I and II are diastereomers
Which of the following statements are true regarding following structures :COOCH3
COOH
COOH
H———OH
H———OH
H———OH
H———OH
H———OH
HO———H
COOH
COOCH3
(c)
COOCH3
(b)
(a)
(A) A and B are diastereomers
(C) B and C are diastereomers
(B) A and C are diastereomers
(D) A and B are enantiomers
SECTION - 2 : COMPREHENSION BASED QUESTIONS
(SINGLE CHOICE CORRECT QUESTION)
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
Comprehension-1
Geometrical isomerism is a kind of stereoisomerism which is present in the compounds containing a double
bond (C=C, C=N, N=N) and arise due to the restricted or frozen rotation about the double bond. The atoms
or groups attached to the doubly bonded carbons must be different. In aldoximes, the isomer is named as syn if
hydrogen and hydroxyl groups are on the same side of C=N bond and if these are on opposite sides, the isomer
is named as anti. In ketoximes, the prefixes syn and anti indicate which group of ketoxime is syn or anti to
hydroxyl group.
15.
Which of the following does not show geometrical isomerism ?
(A) 1,2-Dichloropent-1-ene
(B) 1,3-Dichloropent-2-ene
(C) 1,1-Dichloropent-1-ene
(D) 1,4-Dichloropent-2-ene
16.
On treating with NH2OH, which can form two products ?
(A) Acetaldehyde
(B) Acetone
(C) Formaldehyde
(D) Benzophenone
17.
Number of stereoisomers of the compound 2-chloro-4-methylhex-2-ene is/are
(A) 1
(B) 2
(C) 4
18.
The geometrical isomerism is shown by :
CH2
(A)
CH2
(B)
(D) 16
CHCl
CHCl
(C)
(D)
Comprehension-2
The optical isomers rotate the plane of plane-polarised light. A sp3-hybridised carbon atom attached to four
different atoms or groups is called an asymmetrical centre or chiral centre. Chiral molecules do not possess any
of the elements of symmetry. A chiral molecule cannot be superimposed on its mirror image. These stereoisomers
are called enantiomers. Molecules having a plane of symmetry or centre of symmetry are superimposable on
their mirror images and are achiral. The stereoisomers that are not mirror images of each other are called
diastereomers. A mesoisomer has a plane of symmetry and is optically inactive due to internal compensation.
25
JEE-Chemistr y
Which of the following is a meso isomer also ?
(A) 2-Chlorobutane
(C) 2, 3-Dichloropentane
(B) 2-3, Dichlorobutane
(D) 2-Hydroxypentanoic acid
20.
Which of the following compounds is not chiral ?
(A) DCH2CH2CH2Cl
(C) CH3CHDCH2Cl
(B) CH3CH2CHDCl
(D) CH3CHClCH2D
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
19.
26
Stereoisomerism
1.
SECTION - 1 : NUMERICAL ANSWER BASED QUESTIONS
The total number of stereoisomers of the compound
CH 3CH=CH–CH–CH=CH–CH3 is OH
2.
The total number of stereoisomers of 2,3–dibromobutane is -
3.
How many meso stereoisomers are possible for 2, 3, 4-pentanetriol -
4.
How many statements is/are correct ?
(i) Stereo-isomers which are not mirror image of each other are known as diastereoisomers.
(ii) In every case, a pair of enantiomers have a mirror-image relationship.
(iii) If a compound has an enantiomer it must be chiral.
(iv) If a compound has a diastereomer it must be chiral.
(v) Any molecule containing a single asymmetric carbon must be chiral.
(vi) Any chiral compound with a single asymmetric carbon must have a positive optical rotation if the compound
has the R configuration.
5.
How many chiral compounds are possible on monochlorination of 2–methyl pentane ?
6.
How many cyclic isomers are possible for C5H10 molecular formula including stereoisomers ?
7.
How many total isomers are possible for the formula C 2FClBrI.
8.
Give number of chiral carbon atoms in the compound X, Y and Z. Give your answer as XYZ.
Me
Me
Me
Me
(X)
9.
(Y)
(Z)
For the given reaction how many monochloro products will be obtained (all isomers) ?
Cl / hv
2
¾¾¾¾
® products
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
10.
SECTION - 2 : MATRIX - MATCH QUESTIONS
Match the column I with column II.
Column-I (reaction)
Column-II (stereoisomers)
(A) CH3—CH CH—CH N—OH
(P) 2
(B)
(Q)
4
(C) CH3—CH
CH—CH
CH—CH
CH—CH3
(R)
6
(D) CH3—CH
CH—CH
CH—CH
CH—Ph
(S)
8
27
JEE-Chemistr y
1.
The absolute configuration of
(1) S, S
HO2C
CO2H
HO H H OH
(2) R, R
[AIEEE-2008]
is
(3) R, S
(4) S, R
(3) Propene
[AIEEE-2009]
(4) 2–methyl propene
2.
The alkene that exhibits geometrical isomerism is :(1) 2–butene
(2) 2–methyl–2–butene
3.
Out of the following, the alkene that exhibits optical isomerism is :(1) 2-methyl-2-pentene
(2) 3-methyl-2-pentene
(3) 4-methyl-1-pentene
(4) 3-methyl-1-pentene
[AIEEE-2010]
4.
How many chiral compounds are possible on monochlorination of 2–methyl butane ?
(1) 6
(2) 8
(3) 2
(4) 4
[AIEEE-2012]
5.
Which of the following compounds will exhibit geometrical isomerism?
[JEE MAINS - 2015]
(2) 3 - Phenyl - 1 - butene
(3) 2 - Phenyl - 1 - butene
(4) 1, 1 - diphenyl - 1 - propane
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
(1) 1 - Phenyl - 2 - butene
28
Stereoisomerism
1.
SECTION - 1 : SINGLE CHOICE CORRECT QUESTIONS
The number of stereoisomers possible for a compound of the molecular formula
CH3–CH = CH–CH(OH)–Me is :(1) 4
(2) 6
(3) 3
(4) 2
[AIEEE-2009]
2.
Which branched chain isomer of the hydrocarbon with molecular mass 72 u gives only one isomer of mono
substituted alkyl halide ?
[AIEEE-2012]
(1) Neohexane
(2) Tertiary butyl chloride (3) Neopentane
(4) Isohexane
3.
Maleic acid and fumaric acids are :(1) Tautomers
(2) Chain isomers
4.
The absolute configuration of
(3) Geometrical isomers
[AIEEE-2012(Online)]
(4) Functional isomers
[JEE MAINS - 2016]
CO2H
H
OH
H
Cl
CH3
is
(1) (2R, 3S)
(2) (2S, 3R)
(3) (2S, 3S)
(4) (2R, 3R)
1.
SECTION - 2 : NUMERICAL ANSWER BASED QUESTIONS
The number of chiral carbons in chloramphenicol is _______
[JEE MAINS - 2020 - ONLINE]
2.
The number of chiral centres in penicillin is ______
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
[JEE MAINS - 2020 - ONLINE]
29
JEE-Chemistr y
1.
[IIT-2008]
The correct statement(s) concerning the structures E, F and G is (are)
O
H3C
H3C
(E)
OH
H3C
CH3
H3C
(A) E, F and G are resonance structures
(C) F and G are geometrical isomers
2.
H3C
CH3
(F)
CH3
H3C
(G)
OH
(B) E, F and E, G are tautomers
(D) F and G are diastereomers
[IIT-2008]
The correct statement(s) about the compound given below is (are) :-
Cl
H
CH3
H3C
Cl
(A) The compound is optically active
(C) The compound possesses plane of symmetry
H
(B) The compound possesses centre of symmetry
(D) The compound possesses axis of symmetry
3.
The correct statement(s) about the compound H3C(HO)HC – CH = CH – CH(OH)CH3 (X) is (are) :
[IIT-2009]
(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 stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is
4
(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2
4.
Which of the given statement(s) about N,O,P and Q with respect to M is (are) correct ?
HO
Cl
H
HO
HO
H
CH3
OH
H
Cl
H
H
HO
CH3
OH
H
CH3
OH
HO
H
H
HO
H
H
HO
CH3
Cl
CH3
M
N
O
(A) M and N are non-mirror image stereoisomers
(C) M and P are enantiomers
5.
[IIT-2012]
Cl
P
Cl
Q
(B) M and O are identical
(D) M and Q are identical
H3C
CH3
[JEE ADVANCED-2015]
H3C
6.
M
O
Total number of isomers, considering both structural and stereoisomers, of cyclic ethers with the molecular
formula C4H8O is ___
[JEE ADVANCED-2019]
30
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
The total number of stereoisomers that can exist for M is
Stereoisomerism
ANSWERS
EXERCISE-1
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
B
C
B
D
B
B
D
B
D
D
D
D
A
D
B
Que.
16
17
18
19
20
Ans.
B
C
A
C
A
EXERCISE-2
l
l
MULTIPLE CHOICE CORRECT QUESTIONS
1.
(CD)
2.
(ABC)
3.
(ABCD)
6.
(AC)
7.
(AB)
8.
(AC)
11. (BC)
12. (ACD)
13. (BD)
COMPREHENSION BASED QUESTIONS
15. (C)
16. (A)
17. (C)
20. (A)
4.
9.
14.
(AB)
(AC)
(BCD)
5.
10.
(CD)
(BC)
18.
(D)
19.
(B)
4.
9.
(4)
(6)
5.
(6)
EXERCISE-3
l
l
NUMERICAL ANSWER BASED QUESTIONS
1.
(4)
2.
(3)
3.
(2)
6.
(7)
7.
(6)
8.
(021)
MATRIX MATCH QUESTIONS
10. (A) - (Q), (B) - (P), (C) - (R), (D) - (S)
EXERCISE-4(A)
1.
(2)
2.
(1)
3.
(4)
4.
(4)
5.
(1)
5.
(2)
EXERCISE-4(B)
l
l
SECTION - 1 : SINGLE CORRECT CHOICE QUESTIONS
1.
(1)
2.
(3)
3.
(3)
4.
(2)
SECTION - 2 : NUMERICAL ANSWER BASED QUESTIONS
1.
(2)
2.
(3)
EXERCISE-5
(BCD)
(10)
2.
(AD)
3.
(AD)
4.
(ABC)
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Enthusiast\Che\Unit-06\Stereoisomerism
1.
6.
31
Stereoisomerism
Not To Be Discussed in Class
1.
SECTION - 1 : SINGLE CHOICE CORRECT QUESTIONS
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
2.
Which of the following cannot be written in an isomeric form?
(A) CH3–CH(OH)–CH2–CH3
(B) CH3–CHO
(C) CH2=CH–Cl
(D) Cl–CH2CH2–Cl
3.
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
4.
Which of the following will not show geometrical isomerism.
Me
Me
(A)
5.
6.
(B)
Geometrical isomerism is possible in:
(A) isobutene
(C) acetophenone oxime
(D)
Cl
(B) acetone oxime
(D) benzophenone oxime
Which of the following have asymmetric carbon atom?
C l Br
|
|
(A) H - C - C - H
|
|
H H
7.
(C)
Me
H Cl
|
|
(B) H - C - C - C l
|
|
H H
Meso-tartaric acid and d-tartaric acid are
(A) positional isomers
(B) enantiomers
H Cl
|
|
(C) H - C - C - H
|
|
H H
H H
|
|
(D) H - C - C - CH 3
|
|
Br OH
(C) diastereomers
(D) racemic mixture
CH3
8.
The full name of the compound
HO
H
H
Cl
is
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Nurture\Che\Unit-06\Stereoisomerism
C2H5
(A) (2R,3R)-3-chloro-2-pentanol
(C) (2S,3R)-3-chloro-2-pentanol
9.
(B) (2R,3S)-3-chloro-2-pentanol
(D) (2S,3S)-3-chloro-2-pentanol
The structure of (2R, 3R)C2H5CH(CH3)CH(D)CH2D is
C2H5
(A)
C2H5
H3C
H
H
H
D
(B) H
CH2D
C2H5
CH3
D
CH2D
H
CH3
CH3
(C) D
H
CH2D
(D)
H3C
H
D
H
CH2D
251
JEE-Chemistr y
10.
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
11.
The interchange of two groups (Br and CH3) at the chiral centre of the projection formula (A) yields the formula
(B), while the interchange of another set of two groups (C 2H5 and Cl) of (A) yields the projection formula (C)-
Br
CH3
Br
Cl
CH3
C2H5
(A)
Br
CH3 C2H5
Cl
C2H5
(B)
Cl
(C)
Which of the following statements is not correct about the structures (A), (B) and (C) (A) B and C are identical
(B) A and C are enantiomers
(C) B and C are enantiomers
(D) A and B are enantiomers
12.
Molecular formula C5H10O can have :
(A) 6-Aldehyde, 4-Ketone
(C) 4-Aldehyde, 3-Ketone
Br
COOH Br
COOH
13.
and
HOOC
are
Br
Br
(A) Positional
14.
(B) 5-Aldehyde, 3-Ketone
(D) 5-Aldehyde, 2-Ketone
COOH
(B) Chain
(C) Geometrical
(D) Functional
Which of the following compounds is/are chiral and resolvable :+
(B) [C6H5 N (CH3) (C2H5) (C3H7)] Cl–
(A) C6H5N(CH3) (C2H5)
COOH
(C) CH3–CH2–CH(CH3) N (CH3) (C2H5)
(D)
15.
An organic molecule necessarity shows optical activitiy if it :
(A) Contains asymmetric carbon atoms
(B) is non polar
(C) is non superimposable on its mirror image
(D) is superimposable on its mirror image
16.
Ordinary light can be converted into plane polarized light with the help of a :
(A) Nickel prism
(B) Nicol prism
(C) Diffraction grating
17.
The structure shows :
H3C
H3C
C=C
H
C
CH3
(A) Geometrical isomerism
(C) Geometrical & optical isomerism
252
H
COOH
(B) Optical isomerism
(D) tautomerism
(D) Quartz cell
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Nurture\Che\Unit-06\Stereoisomerism
COOH
Stereoisomerism
18.
Isomers which can be interconverted through rotation around of single bond are (A) Conformers
(B) Diastereomers
(C) Enantiomers
(D) Positional isomers
19.
Rotation of polarised light can be measured by :
(A) Monometer
(B) Galvanometer
(C) Polarimeter
(D) Viscometer
20.
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 an organic solvent
(C) optical rotation and is derived from D–(+)– glyceraldehyde
(D) optical rotation only when substituted by deuterium
21.
Which of the following compound will exhibits geometrical isomerism :
(A) 1–phenyl–2–butene
(B) 3–phenyl–1–butene
(C) 2–phenyl–1–butene
(D) 1, 1–diphenyl–1–propene
22.
Which of the following exhibits stereoisomerism–
(A) 2–Methylbutene–1
(C) 3–Methylbutanoic acid
23.
(B) 3–Methylbutyne–1
(D) 2–Methylbutanoic acid
Hydrogen of the following compound in the presence of poisoned palladium catalyst gives :
Me
H Me
H
Me
(A) optically active compound
(C) a racemic mixture
(B) an optically inactive compound
(D) a diastereomeric mixture
24.
SECTION - 2 : MULTIPLE CHOICE CORRECT QUESTIONS
Which types of isomerism is shown by 2,3–dichloro butane(A) structural
(B) geometric
(C) optical
(D) diastereo
25.
Which of the following has/have P.O.S. (plane of symmetry) ?
H3C
CH3
C
(A)
C
H
26.
(B)
CH3
H
Cl
(C) H
Cl
H
(D) F
Br
I
Compound having optically active stereoisomer
COOH
HOOC
CH3
H 3C
(B)
(A)
Ph
H 3C
CH3
Ph
(C) H 3C
CH–CH3
Which will show geometrical isomerism ?
(A) H2N2
H3C
C = N–OH
(C)
H3C
CH3
CH 3
COOH
(D)
Ph
27.
Cl
CH3
Ph
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Nurture\Che\Unit-06\Stereoisomerism
H
COOH
(B) 1-Chloro-2-phenylethene
(D) HO–N = N–OH
253
JEE-Chemistr y
28.
Which of the following are optically active :-
(A) H2C=C=CH2
(B)
C=C=C
F
H
H
(C) H N
2
F
—CO2H
(D)
NH2
Cl
F
F
SECTION - 3 : COMPREHENSION BASED QUESTIONS
Paragraph-1 for (Q.no. 29 to 30)
R,S configuration is a useful tool for determination of enantiomers, diastereomers and identical compounds. If
configuration of all chiral centres are opposite then structures are enantiomers. If all chiral centres have same
configurational then they are identical and if some have same configuration and some have oppsite configuration then they are diastereomers ?
CH3
H
Cl
H
NH2
Cl
(I)
30.
H
H
H 3C
NH2
H
Cl
H3C
NH2
H
H
(II)
(III)
CH3
29.
CH3
CH3
Cl
CH3
H 3C
H
NH2
(IV)
Among above structures find out enantiomeric structures :
(A) II & III
(B) I & II
(C) I & IV
(D) III & IV
Which of the following is not diastereomer ?
(A) I & III
(B) II and III
(D) II and IV
(C) III & IV
31.
Which of the followings compound is optically inactive ?
CH3
CH3
(A) H
Cl
254
OH
H
C 2H 5
Cl
(B) HO
H
CH3
(C)
CH3
H
Cl
(D) H
Cl
CH3
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Nurture\Che\Unit-06\Stereoisomerism
Paragraph-2 for (Q.no. 31 to 32)
Presence of chiral centre is not an essential condition to show optical isomerism. Essential condition is compound
should show non-superimposable mirror image. Allene do not contain chiral carbon but snow optical isomerism
when different groups are attached on double bonded carbons.
Stereoisomerism
32.
Which of the following allene compounds is optically active ?
H 3C
C
(A)
C
C
H
H 3C
33.
H
CH3
C
(D)
H 3C
C
H
C
C
C
C
H 3C
CH3
C
CH3
C
(B)
H
(C)
H
CH3
C
C
H
H
C
CH3
SECTION - 4 : MATRIX - MATCH QUESTIONS
Match the following column :
Column-I
(Compounds / Orientations)
Column-II
(Relation)
OH
Cl
OH
(A)
&
Cl
NC
Position isomers
(Q)
Diastereomers
(R)
Geometrical isomers
(S)
Enantiomers
(T)
Stereoisomers
CN
CN
(B)
(P)
&
CN
H
CH3
HO
OH
(C)
&
H
OH
CH3
H
HO
H
CH 3
CH3
Me
(D)
&
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Nurture\Che\Unit-06\Stereoisomerism
Me
255
JEE-Chemistr y
Match the following column :
Column-I
CH3
H
Column-II
H
CH2OH
H3C
(A)
CH2NH2
(P)
and
NH2
CH3
H
OH
Cl
H
Cl
(B)
CH3
and
(Q) Identical
Et
CH3
H
(C)
Et
H
H3C
OH
Et
and
H3C2
HO
(R)
Enantiomers
(S)
Diastereomers
OH
Et
(D)
Structural
H3C2
H
and
H
OH
35.
SECTION - 5 : NUMERICAL ANSWER BASED QUESTIONS
The number of cis-trans isomer possible for the following compound
36.
The number of isomers of dibromoderivative of an alkene (molar mass 186 g mol –1) is
37.
The number of optically active compounds in the isomers of C 4H9Br is
38.
How many optically active stereoisomers are possible for butane –2,3–diol :
39.
The number of possible enantiomeric pairs that can be produced during monochlorination of 2–methyl butane
is :
40.
The number of isomers for the compound with molecular formula C2BrCIFI is :
41.
SECTION - 6 : SUBJECTIVE QUESTIONS
Which of the following compounds can exist as geometric isomers ?
CH2Cl2, CH2Cl—CH2Cl, CHBr = CHCl, CH2Cl—CH2Br.
42.
Why does cyclopentene not exhibit geometric isomerism though it has a double bond.
43.
Why does 2-butene exhibit cis-trans isomerism but 2-yne does not ?
44.
Write structural isomer of C6H14. What is relation between them ?
45.
How many pseudo chiral carbon are present in butane–1,2,3–triol and cyclohexane–1,2,3–triol ?
46.
(+) 2-butanol has specific rotation of + 13.9° when measured in pure form. A sample of 2-butanol was found
to have an optical rotation of –3°. What is the stereomeric composition of this mixture ?
47.
N-methylethenamine as such does not show any stereoisomerism but one of its resonance form exhibit
stereoisomerism. Explain.
256
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34.
Stereoisomerism
48.
Assign Cahn-ingold prelog priorites to the following sets of substituents :
(i) —H, —Br, —CH2CH3, —CH2CH2OH
(ii) —COOH, —COOCH3, —CH2OH, —OH
(iii) —CN, —CH2NH2, —CH2NHCH3, —NH2 (iv) —Br, —CH2Br, —Cl, —CH2Cl
49.
Identify whether the stereogenic centre is present or not :
(i) 2-Cyclo penten-1-ol
(ii) 3-cyclo penten-1-ol
(iii) 2-bromopentane
(iv) 3-bromopentane
50.
Discuss the optical activity of tertiary amines of the type R1R2R3N :
51.
Draw the enantiomer of the following structure :
CH3
H
Cl
C
C2H5
52.
2,4-Hexadiene has three geometrical isomers. Draw their structures.
53.
Assign R and S configuration to the chiral carbons in the following :
CH3
(i)
54.
CH3
HO———H
(ii)
H———OH
CH3
H———OH
H———OH
CH3
Give relation between following compounds.
(A)
(B)
(C)
(D)
(E)
(F)
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Nurture\Che\Unit-06\Stereoisomerism
SECTION - 7 : ASSERTION–REASON QUESTION
These questions contains, Statement-I (assertion) and Statement-II (reason).
(A) Statement-I is True, Statement-II is True ; Statement-II is a correct explanation for Statement-I
(B) Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explanation for Statement-I
(C) Statement-I is True, Statement-II is False.
(D) Statement-I is False, Statement-II is True.
55.
Statement-I : All double bond containing compounds show geometrical isomerism.
Because
Statement-II : Alkenes have restricted rotation about the double bond.
56.
Statement-I : Meso-tartaric acid is optically active.
Because
Statement-II : Optically active molecule is a molecule that cannot be superimposed on its mirror image.
57.
Statement-I : Propadiene is optically inactive.
Because
Statement-II : Propadiene has a plane of symmetry.
58.
Statement-I : Molecules that are not superimposable on their mirror images are chiral
Because
Statement-II : All chiral molecules have chiral centres.
257
JEE-Chemistr y
ANSWERS
SINGLE CHOICE CORRECT QUESTIONS
1.
(B)
2.
(C)
3.
(D)
6.
(D)
7.
(C)
8.
(A)
11. (C)
12. (B)
13. (A)
16. (B)
17. (B)
18. (A)
21. (A)
22. (D)
23. (B)
l
l
l
l
4.
9.
14.
19.
(A)
(A)
(B)
(C)
5.
10.
15.
20.
(C)
(A)
(C)
(C)
MULTIPLE CHOICE CORRECT QUESTIONS
24. (CD)
25. (ABC)
26. (BCD)
27.
(ABD)
28.
(BC)
COMPREHENSION BASED QUESTIONS
29. (B)
30. (D)
31. (D)
32.
(A)
MATRIX MATCH QUESTIONS
33. (A) - (QRT), (B) - (QRT), (C) - (ST), (D) - (P)
34.
(A) - (P), (B) - (R), (C) - (Q), (D) - (R)
38.
(2)
NUMERICAL ANSWER BASED QUESTIONS
35. (4)
36. (3)
37. (2)
40. (6)
l
l
41.
39.
(2)
SUBJECTIVE QUESTIONS
Only CHBr = CHCl can exist as geometric isomers :
Br
H
C
C
Cl
H
and
Br
H
C
C
H
Cl
In CH2Cl— CH2Cl and CH2Cl—CH2Br, the carbon atoms are connected by a single bond about which the
groups can rotate relatively freely. Thus any conformation of the halogen atoms may be converted into any other
simply by rotation about the single bond. In CH2Cl2, the configuration of the molecule is tetrahedral and all
interchanges of atoms yield exactly equivalent configurations.
H
42.
H
This is cis form. Two H atoms on the same side. To get trans, ring must be twisted.
Double bond becomes severely twisted-destabilized. Effective overlap of P orbitals is missing, so does not exist.
The Pz orbitals forming p-bonds and the empty Pz orbital of the carbon with +ve charge are parallel. So the
electrons may be delocalized. The +ve charge is effectively spread out over two carbons; delocalized.
H
H
C
C
H
+
p-orbital
empty orbital
C+ H
+
CH2=CH— C H 2 ¬¾® C H 2 —CH= CH
In n-propyl cation, + I effect of R increases the stability.
In allyl + M effect increases the stability. But + M effect in allyl cation is more effective. So allyl > propyl.
A group with + M effect stabilizes cation; destabilizes anion.
44.
(a)
258
(b)
(c)
(d)
(d)
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Nurture\Che\Unit-06\Stereoisomerism
43.
Stereoisomerism
Number of pseudo chiral carbon in butane–1,2,3–triol is 0.
Number of pseudo chiral carbon in cyclohexane–1,2,3–triol is 1.
46.
Let x is the % of (+) 2-butanol.
13.9 x – 13.9 (100 – x) = – 300.
x = 39.2, % of d form = 39.2, % of l form = 60.8.
:
45.
CH2 =CH — NH—CH3
47.
:
Å
CH2 —CH = NH—CH3 (this shows Geometrical isomerism)
48.
(a) —Br > —CH2CH2OH > —CH2CH3 >— H,
(b) —OH > —COOCH3 >—COOH >—CH2OH
(c) —NH2 >—CN >—CH2NHCH3 >—CH2NH2
(d) —Br >—Cl >—CH2Br >—CH2Cl
49.
(i)
Hydroxyl bearing carbon is stereogenic centre.,
H
OH
4=3
(ii)
3=4
5=2
2=5
H
It has no stereogenic centre.
OH
Br
(iii) CH 3—C —CH2—CH 2— CH 3 bromine bearing carbon is stereogenic centre.
H
Br
(iv) CH3—CH2—CH—CH2—CH3 It has no stereogenic centre.
••
Tertiary amines have pyramideal geometry with sp -hybridization at nitrogen.
It should be a chiral molecule (assuming lone pair to be a substituent). Thus,
R1
N
R3
R2
tertiary amines exist as racemic mixture but they cannot be resolved.
–1
This is due to the reason that the energy difference between the isomer is very small (25 kJ mol ). Hence, rapid
nitrogen or amine inversion takes palce.
2
R1
••
N
sp -hybridized
•
N
•
R1
R2
R3
2
sp
••
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Nurture\Che\Unit-06\Stereoisomerism
50.
3
2
sp
N
R1
R2
R3
R2 R3
Planar transition state
Enantiomers
Tertiary amine N-oxide has four group hence nitrogen inversion is not possible, thus tertiary amine -N-oxide can
be resolved.
259
JEE-Chemistr y
CH3
51.
Cl
H
C
C2H5
52.
53.
54.
,
CH3
CH3
HO—R ——H
S
H——
—OH
(i) H——R—OH
CH3
R
—OH
(ii) H——
CH3
(A), (B) Þ Enantiomers.
(A), (C) Þ Enantiomers.
(A), (D) Þ Identical.
(A), (E) Þ Identical.
(A), (F) Þ Enantiomers.
ASSERTION–REASON QUESTIONS
55. (D)
56. (D)
57.
(A)
58.
(C)
JPR\COMP.251\D\Allen(IIT-JEE Wing)\2020–21\Nurture\Che\Unit-06\Stereoisomerism
l
,
260