Solutions Slot – 1 (Chemistry) Page # 1

ISOMERISM
EXERCISE – I SINGLE CORRECT (OBJECTIVE QUESTIONS)
Q.1 [C]
C2H5 – O – C2H 5 & CH3 – O – CH2 – CH2 – CH3
R R'
R & R′ are different so these are metamers

Q.2 [A] C7H7Cl

CH3
CH2 – Cl CH3 CH3
Cl

Cl Cl

Q.3 For positional isomerism in alkene 4C-atom required →

4 3 2 1 4 3 2 1
CH3 – CH2 – CH = CH2 & CH3 – HC = CH = CH 3
1-Butene 2-Butene
For geometrical isomerism in alkene also 4c–atom required →
CH3 CH3 CH3 H
C=C C=C
H H H CH3
Cis - alkene Trans-alkene
* * * * *
Q.4 * Cl
*
Cl Cl

Q.5 All (A) (B) & (D) can be written in other isomeric form.
(A) CH3 – CH – CH2 – CH3 CH2 – CH2 – CH2 – CH3
OH OH
2-butanol 1-Butanol
(B) CH3 – CHO
CH2 – C – H CH2 = C – H
O OH
H
Cl
(D) Cl – CH2 – CH2 – Cl and CH3 – CH
1,2-dichloro ethane Cl
1,1–dichloro ethane
But (C) option CH2 = CH – Cl can not be written in an isomeric form.

H
Q.6 (A)
Cis
Trans

Q.7 Molar mass of Br = 80 [Dibromo Derivative]

Br = 80 ↓
160 Molar mass = 186
(B) C2H2 = 12 × 2 + 2 = 26
186 = 160 + x

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 Page # 2 Solutions Slot – 1 (Chemistry)

x = 186 – 160 = 26 ⇒ C2H2

H H
(1) The compound is C2H2Br2 C=C
Br Br
H Br H Br
(2) C=C (3) C=C
Br H H Br

Q.8 [C] CH2 - CH 2

OH OH
OH
OH OH
H OH
H H OH

H H H H
H H HH
Gauche form OH H
Staggered/Anti Eclipsed form
Hydrogen Bonding Present
Stability order ⇒ Gauche > Anti > Eclipse

Q.9 C4H 11N

(1) H3C – H2C – CH2 – CH2 – NH2 (2) CH3 - CH2 – CH – CH3
NH2
CH3 CH3
(3) H3C – CH (4) H 3C – C – NH2

H 2C - NH 2 CH3

d−
87.5 − 12.5 75

Q.10 [B] Optical purity = = =
d+
87.5 + 12.5 100
α mix
Optical purity = α
pure

75 α mix
=
100 30°
75 × 30 45
αmix = = = 22.5°
100 2

Q.11
= 10 m θ = 30° dilute by 1 litre = 1000 m

20 gm
C= θ = α.C.
.
200 m

20
θ=αC
Angle of rotation = 3θ × × 10 = 6°
1000
20
30° = α × × 10
200
200
30° = α ×
200
30 × 200
specific angle of rotation α = 30° α=
200

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 Solutions Slot – 1 (Chemistry) Page # 3

4 1
1
R 4 NH2 R S
HO H
3 CF3 2 HS
Q.12 [C] Reverse because (4) group at
3 COOH
2

wedge.
3
4
H CH3
CHO R → S (C) R,S,S
2 1

Q.13 [C]

(1) Vertical plane (2) Horizontal plane (3) Molecular Plane

CH3 Cl

H Cl H Br
Q.14 [D]
H H CH3
Br
COOH COOH
O O
CH3 – CH – CH – COOH CH3 – CH – CH – COOH
Br Cl Cl Br

Position isomers

Q.15 [A] Geometrical Isomers can be Diastereomers or Enantiomers.

CH3 CH 3 CH3 H
C=C C=C
H H H CH3

R S R S

E Z

H Cl Cl H

Enantiomers
Because geometrical isomers are around restricted rotation which can not be rotate or
interconvertable.

Q.15 [D]
OH CH2 – OH

Phenyl alcohol Benzyl alcohol None of these

No Relation

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 Page # 4 Solutions Slot – 1 (Chemistry)

EXERCISE – II MULTIPLE CORRECT (OBJECTIVE QUESTIONS)

O OH O
Q.1 [B] *S C = CH – C*– C – CH3
.. H H

Number of chiral centers = 2

OH O

Q.2 [B] *
R * R'
R
A symmetric = 2n = 22 = 4 → 1) RR Enantiomer
2) SS
3) RS 
Enantiomer
4) SR
1,3 1,4 2,3 2,4 are Diastereomers.
Ans. 2,4
*
*
Q.3 [B]
* *
(K) (L) (M)
Meso Meso
CH3

Q.4 [D] Plane not present so optically active ⇒ Resolvable

CH3
Trans-1,3-dimethyl cyclo hexane.

OH OH OH
OH C2
OH OH OH
Q.5 (I) (II) (III) (IV)
O O O
OH OH O
OH
OH By C2
On Rotating by C2
same molecule
same molecule obtained.
Identicle compounds are = II & IV

OH OH OH OH
OH OH OH
OH
Q.6 (I) O OH (II) (III) (IV)
O O O
OH OH OH
I & III are diastereomers because they are not mirror image.
CH3 CH3
H OH H OH
Q.7 (A) H OH
HO H
CH3
CH3
Plane present
Newmann Fisher
meso compound

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 Solutions Slot – 1 (Chemistry) Page # 5

CH3 OH
H
OH H CH3
(B) H OH H OH Plane not present
CH3 CH3

C2H5
C2H5 H OH
HO H H H
(C) H OH (D) H OH
C2H5
Plane not present CH3
Plane not present

HO OH
HO OH
HO OH
Q.8 (I) (II) HO OH (III) Rotate out of (IV)
Plane present Plane present plane 180°

rotate out of plane

180º

HO OH HO OH

En antiomer

Q.9 (B) & (C) compounds are R-configuration.

CH3 2
1 (3) (4)
(2) CH3
2 1
2
3 (4) 5
(A) 4 Cl (b) Cl R S
3
5 1 3
(1) Cl 4
4
Cl
(S) -4-chloro-1-methyl cyclohexene (S) -5-chloro-1-methyl cyclohexene.

Cl O
Cl Cl
O *
H
CH2
Q.10 (A) (B) H2C (C) CH2 CH2 (D)
* H
Cl Cl CH2 CH3 CH2

Two asymmetric
centre present.

Q.11 (C) They are conformers as well as enantiomers.

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 Page # 6 Solutions Slot – 1 (Chemistry)

CH3
*
CH – CH = CH – CH2 – CH – CH2
Q.12
CH3 OH Br
asymmetric T.S.I = 2n = 22 = 4
n = no. of double bond + no. of chiral center = 1 + 1 = 2

Q.13 When optically pure amine react with racemic mixture of acid it gives pair of diastereomers
were which can be superated by Fractional distillation followed by Hydrolysis gives acid which
are optically pure and present in different fraction.

Cl
Q.14 (D) A single C2 rotational axis but no mirror plane.
Cl

CH3 CH3
H3C H
Q.15 (D) All four substituent are different so it is chiral compound.
CH3
H3C
CH3
H

CH3
CH3 *
O
H3C
Q.16 Br * CH3
CH3 *
CH3
3
Asymmetric = 2 = 8

Cl
Br * * CH = CH *
CH3

Q.17 Chiral = 3 asymmetric = 2 = 16

4

double Bond for G.I. = 1

n=3+1=4

Q.18 (B)
1
1-Heptanol ⇒ CH2 – CH2 – CH2 – CH2 – CH2 – CH2 – CH3

OH
1 2 3 4 5 6 7
2-Heptanol ⇒ H3C – CH – CH2 – CH2 – CH2 – CH2 – CH3

OH
1 2 3 4 5 6 7
3-Heptanol ⇒ H3C – CH2 – CH – CH2 – CH2 – CH2 – CH3

OH

1 2 3 4 5 6 7
4-Heptanol ⇒ H3C – H2C – H2C – CH – CH2 – CH2 – CH3

OH

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 Solutions Slot – 1 (Chemistry) Page # 7

1-Heptanol & 4-Heptanol are not chiral

2-Heptanol & 3-Heptanol are chiral compound.

*
CHBrMe
NO2

Q.19
CH = CH I
Me
Asymmetric 23 = 8

H H
C=C
CH2 CH3

Q.20 (A) or Ph Different

H2C – CH = CH – CH 3
Different
Ph show g.I.

H
(B) 3-Phenyl-1-butene ⇒ H2C = CH – C – CH3
Ph

H H
C=C
Not show g.I. H CH – CH3
Same Ph

H CH2– CH 3
C=C
(C) 2-Phenyl-1-butene. H2C = C – CH2 - CH3 = H Ph
Ph Same
Not show g.I.

Ph Ph H
H C=C
(D) 1,1-di phenyl-1-propene C = C – CH3 = Ph CH3
Same
Ph
Not show g.I.

Q.21 IInd compound is meso compound so θ = 0°

3 2
CH3 εt
Q.22 (A) 4H OH (B) 1HO H
1 4
εt CH3
2 3
R→S R→S
Identicle

O CH3 H
H H H
120°
Q.23 O = –CH3 ≡
H H H
H HH H HH H H CH3

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 Page # 8 Solutions Slot – 1 (Chemistry)

OH OH
Q.24 (A) CH3 – CH2 – CH2 – CH2 and CH3 – C – CH 3
4C-chain
CH3
3-C chain
chain isomers.

α mix 30.8
Q.25 Optical purity = α × 100 = (D) 51.3 × 100 = 60.03
pure

60.03 + 100 160

% of D-enantiomers = = = 80%
2 2
% of L-enantiomer = 100 – 80 = 20%

Major − min or 75 − 25
Q.26 Optical purity = × 100 = × 100 = 50%
Total 100
α mix
O.P. = α × 100
pure

α mix
50 = × 100
(+)158
50 × 158
αmix = 79 = +79°
100

−x (1)H
C4H7Cl → C H   → C H
2
Q.27 +H 4 8 DBE 4 10

(1) H3C – H2C – HC = CH – Cl (2) CH3 – CH – CH = CH2

Cl
(3) CH2 – CH2 – CH = CH2 (4) C – C – C = C
Cl Cl
CH2 – Cl
(5) (6) Cl

CH3

(7) (8)
Cl Cl
CH3 Cl CH2 Cl
(9) (10)

(11) (12)
Cl
Cl

Q.28 (C) Stereoisomer differ from each other in configuration.

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 Solutions Slot – 1 (Chemistry) Page # 9

O
1CHO
O–C–H
HO 2a
H O 2 H
C
2b
Q.29 (C) H OH 1
a R→S
HO H4
4c
H OH H – C – OH
3

CH2OH
5

2 CHO
HO – C – H
2

b
H OH c
4 1 4 H OH
S→R 1
S→R
H – C – OH
CHOH
2
3
CH2OH
3
2S, 3R, 4R

Q.30 Two pure chiral center must be optically active.

Q.31 Physical properties like M.P., B.P., solubility are different for diastereomers.

O
HO CH2 – OH
* *
Q.32
HO * * CH = CH – CH = CH – CH2 – CH2 – CH3
n = 6 asymmetric = 26 = 64

α mix
Q.33 Optical purity = α × 100
pure

10
(D)/x = × 100 = 50%
20
150
% of D(x) compound = = 75%
2
(y) compound = 100 – 75 = 25%
(x) : (y)
75 : 25
3:1

Q.34 C3H6Br2
Br
Br
C–C–C C–C–C
(A) Gem (B) Gem
Br Br
C–C–C C–C–C
(C) (D)
Br Br Br Br
Two gem dibromide.

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 Page # 10 Solutions Slot – 1 (Chemistry)

Q.35 C5H10  1 H2
→ C5H12 = 1 D.B.E.
(1) C – C – C – C = C
H H
C=C
C–C CH3
(2) C – C – C = C – C
H CH3
C=C
C–C H

(3) C – C – C = C (6)
C
(4) C – C = C – C (7)
C
C=C
(5) C – C (8) (9)

C
CH3 (10)
*
CH3
CH3
* (11)
CH3
CH3
(12)
CH3

COOH COOH
R R
H OH H OH

Q.36 H
S
OH HO
R
H ⇒ diastereomers

COOH COOH
Meso-Tartric acid d-Tartric acid

Q.37 C6H5Br3
* Br
Br C–C–C Br
(1) C – C – C Br (2) Br (3) C – C – C
Br Br Br
2 isomers Br
Br
(4) C – C – C – Br (5) C – C – C
Br Br Br Br
Total = 6 isomers.

Q.38 C4H9Br : C – C – C – CH2 – Br C – C – CH2 – Br

C
*
C–C–C–C C
Br C – C – C – Br
Chiral → Optical active
⇒2-isomers C

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 Solutions Slot – 1 (Chemistry) Page # 11

3 CH3
CH3 CH3
1
2 4 a
H 2 Cl
H Cl H Cl 1
R→ S
Q.39 HO H R→
S
3 1 4
HO H HO H
C2H5
2
CH
2 5
3
C2H5
(2s, 3s)4 -2-chloro-3-Hydroxy
5
Pentane

Br Br diequatorial

Q.40(B) Br
equatorial-equatorial
is more stable.
Br
axial-axial

Q.41 HOOC – CH = C = CH – COOH

H h v H
symmetric = C=C=C cis
HOOC COOH
h → v = non planar ⇒ check optical activity
H COOH
C=C=C Trans
HOOC H
Two optical isomers

H Br CH3 Br
CH3 C CH3 C
Q.41 H
C CH3 C
Br H Br H

Br CH 3 Br
H
Br Br
H H
CH3 H
CH3 CH3

3 4 3
CH3 H CH3
4 1 3 1 1 4
H Br R S CH 3 2
Br Br H S R
2
H Br S R H Br H Br S R
4 1 4 1

CH3 CH3 CH3

3 3

SR
Diaste.
RR

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 Page # 12 Solutions Slot – 1 (Chemistry)

Q.43 (C) CH2 = CH – Cl can not be written in an isomeric form.

CO2H CO2H
H OH HO H
Q.44 (B) (I) H OH (IV) HO H
H OH HO H
CO2H CO2H
Identicle Rotate by 180° in the plane.
CO2H
H OH
H OH
H OH
CO2H

CO2H CO2H
HO H H OH
H OH HO H
Q.45 H OH HO H
CO2H CO2H

Mirror image to each other So III & V are evantiomers.

H Cl H Cl

Q.46
H Cl Cl H
The above compounds are different in structure.
Cl
CH2 = C CH = CH
(1) Cl (2)
1,1-dichloro, 1-ethene Cl Cl
1,2-dichloro ethene
Q.47 Only (D) compound has plane of symmetry.

Ph Ph

H H
H H

COOH COOH
Plane present

CHO
1
H OH
2
HO H
3
H OH
Q.48 4 chiral centers are four so 24 = 16 stereoisomers.
H OH
5
CH2OH

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 Solutions Slot – 1 (Chemistry) Page # 13

Q.49 C2H5 – O – C2H5 & CH3 – O – CH2 – CH2 – CH3

R R'
R & R' are different so they are metamers.

Q.50 (A) CH2 = CHBr (B) CH2 = CBr2 (C) ClCH = CHBr (D) Br2C = CCl2

H H H Br H H Br Cl
C=C C=C C=C C=C
H Br H Br Cl Br Br Cl
g.I.x g.Ix g.I. g.Ix
diff. diff.
Ch3 – C – O – CH2 – CH2 – CH3
Q.51 (A) & CH3 – CH2 – C – O – CH2 – CH3 (3) Metamers
O
O
CH 3 CH3
CH 3
(B) and (2) Position isomers

CH3
O OH

(C) (4) Tautomers

O O HO OH
2 CHO
4H
H – C – OH
4 1
(D) and CH3 – C – CHO (1) Enantiomer
CHO 3 2
3
R OH
1
S
(A) A B C D
3 2 4 1
*
Q.52 H2C – CH – CH2 – CH3 21 = 2
Cl Cl
Q.53 CH3 – CH2 – CH3
(1) CH2 – CH2 – CH3 (2) CH3 – CH – CH3

Cl Cl
It is also not chiral It is also not chiral
So only these two isomers.
Cl
Cl2
Q.54 CH3 - CH2 – CH3  → (1) H3C – H2C - HC
Cl
Cl

(2) CH3– C – CH 3 (3) CH2 – CH2 – CH2 (4) CH3 – CH – CH2

Cl
Cl Cl Cl

H
CH3– C *– CH 2

Cl Cl

It is chiral so 2-isomers of this

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Total = 5 isomers
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 Page # 14 Solutions Slot – 1 (Chemistry)

D Cl
S
H Br I H
S
Q.55 H Cl D H
R R
I Br
This are enantiomers

h v h
Q.56 (A) Cl – CH = C = C = CH – Cl
h → h planar show g.I. NOT optical isomerism
(B) Cl – CH = C = C = C = CH – Cl
h v h v

h → v Non planar show optical activity.

COOH
H OH
(C)
H OH

COOH
H
Me
Cl H Cl H
(D) Cl Me
Me Cl
Me H

O O
Q.57 S O and S O
O O
R R'
R & R' are different so show Metamerism.
O
O
C O Cl
Q.58 Cl C O and H3C
C=C
}

H H H
R C=C
H CH3
R'
R & R' are different, So show metamerism.

Q.59 Twisted Boat form is chiral & optically active.

CO2H
H OH COOH COOH
H OH H OH
Q.60 (A)
HO H H OH H OH
COH
2
COOH COOH

Identicle

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 Solutions Slot – 1 (Chemistry) Page # 15

CH3 H CH3
R H
C S H
(B) CH3 C
C=C C=C
Br Br
H H H CH3

cis [diastereomers] Trans

Et Et
R
R
CH3 HS H HO H
HO H HS H
H OH S
(C)   →
Rotate at 180° in the plane
S
H SH CH3 CH3

Et Identicle

Cl Cl
Cl Cl
(D)

Cl Cl
Rotate by C2
Cl e
ic l
Cl ent
Id

Cl

Q.61 Cl Cl

(1) + (1) = 2
(a)

Cl

(1) (1) = 3
Cl (1)
a+b=2+3=5
(b)

Et
Keq.
Et
Q.62 Bulky grap. at equatorial.
Me
Me
kf
>1 So k>1
kb

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 Page # 16 Solutions Slot – 1 (Chemistry)

Me
Me
Q.63 H
H
→ This compound show g.I.
→ This compound show plane of symm.
→ This compound show optical isomers because.
This shows different orientation in 3D.
→ But this compound not posses center of symmetry.

Q.64 Any compound which show g.I. no C atom required.

.. ..
eq. N = N
H H

CH3 CH 3– N – CH – C2H6
Q.65 (I) C2H5 – NH – C – CH3 (II) CH3 CH3
CH3 tert. amine
Sec. amine

F.I
CH3
CH3– CH 2– C – NH – C2H5
(III)
CH3
sec. amine
For conformational isomers M.F. should be same. So II & III are not C.Isomers.
So none of these are correct.

Q.66

(i) Vertical plane (ii) Horizontal plane (iii) Molecular plane

Q.67 C3 F Cl Br I
F F
C – C ≡ C – Br C – C ≡ C – Cl
Cl Br
I I
Cl F
C–C≡C–F C — C ≡C – I
Cl
Br
I Br

O O O O
Q.68 n = 2, 22 = 4

O O O

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 Solutions Slot – 1 (Chemistry) Page # 17

n=2
Q.69 2
2 =4

H CH3 CH3 CH3

Anti H Br Br H
Q.70 + Br2 Addition
  → +
Br H H Br
H CH3 CH3 CH3
Cis-2-Butene
Racemic Mixture
(Threo form) 2 stereoisomers

H CH3
C Me Me Me
Br2 Anti H Br Br H H OH
Q.71 +  
→ + + +
C H2O Addition Br H H Br Br H
H CH3
Me Me Me
Me
HO H
H Br

Me
4 Stereoisomers

H CH3
C H
Anti
Br Me
Q.72 + Br2  
→
C Br Me
CH 3 H
H
Trans-2-Butene Meso
1-stereoisomer

H
*
Q.73 C = C - C – C + Br2 C–C–C–C
Br Br
Optically acive
↓
2 stereoisomer

Br
C – C = C + Br2 C–C–C

Q.74 C C Br
No chiral, no plane
Only 1-steraoisomer

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 Page # 18 Solutions Slot – 1 (Chemistry)

EXERCISE – III SUBJECTIVE QUESTIONS

Q.1 A, D
CH2
Me Me
COOMe Cl
CH2
(I) (II)
Trans (Chiral) Trans (Achiral)

Q.2 (A) 3 isomers

3 isomers
(B)

CHO
H * OH
COOH
H * OH
(C) H OH (D) H * OH 3
2 =8
H OH
meso = 1 isomer CH2OH
COOH
Ans. A, B, C
Cl Cl
Q.3 (i) C=C=C
H h v H
(i) ;g ;kSfxd T;kferh; leko;ork iznf'kZr djrk gaSA
⇓
vryh; ⇒ T;kferh; lfØ;rk iznf'kZr ugha djrk gaSA
(ii) vfUre dkcZu lewg ,d nwljs ds yEcor~ gaSA
;g lgh dFku gaSA
Q.4 B, C
NO2 HOOC

(A)

NO2 HOOC
pos optically. Inactive

CH3 H
(B)
H COOH

Pos X optically active

NO2 HOOC

(C)

NO2 O2N
pos × optically active

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 Solutions Slot – 1 (Chemistry) Page # 19

COOH
H OH
(D) HO H
H OH Pos optically inactive

COOH
Q.5 C4H6O2
First find D.B.E C4H6O2 → C H 2H2
C4H10
4 6
DB.E = 2
(A) ,d MkbdhVksu = CH3 — C — C — CH3
O O
(B) ,d ;kSfxd ftlesa nks ,fYMgkbM lewg gaSA = H — C — CH2 — CH2 — C — H
O O
(C) ,d ,Ydhuksbd vEy = CH3 — CH = CH — C — OH
O
Ans. A, B, C

Q.6 (B), (C) o (D) xyr gaSA

(B) VkWVksesfjd lajpuk,¡ vuquknh lajpuk,a ugha gksrh gaSA
(C) dhVks voLFkk ges'kk bukWy voLFkk esa vf/kd LFkk;h ugha gksrh gaSA
(D) T;kferh; leko;ork dsoy ,Ydhu ds }kjk iznf'kZr ugha dh tkrh gaSA
Q.7 (A) ,d ehtksa ;kSfxd fdjSy dsUnz j[krk gS ysfdu izdkf'kd lfØ;rk iznf'kZr ugha djrk gaSA
(C) ,d ehtksa ;kSfxd og ijek.kq j[krk gS tks ,d nwljs dh niZ.k izfrfcEc ij izfrLFkkfir fd;k tk ldrk gaAS tcfd og
fdbZy dsUnz j[krk gaSA
Q.8 (A) ladsr D ;kSfxd dh izd`fr dks ugha crkrk cfYd d ladsr ;g crkrk gaSA
(D) ladsr D ;g ugha crkrk fd fQ'kj iz{ksi.k esa gkbMªkstu v.kq fdjSy dsUnz ds nk;h rjQ nh;s gaSA tcfd –OH ijek.kq
ds ckjs esa crkrk gaSA

Q.9 (A) C – C – C – C – Br 1-czkseksY;wVsu

H

(B) H3C — H2C — C*— CH3 2-czkseksY;wVsu (fdjSy blfy, izdkf'kd lfØ;)
Br
H2C — C — CH3
(C) 1-czkseks-2-esfFky izksisu (fdjSy ugha gS)
Br CH3
Br
(D) H3C — C — CH3 2-czkseks-2-esfFky izksisu (fdjSy ugh gaS)
CH3
(3) CH3
(4)
H OH
Q.10 (1)
R → S
(2) C2H5

(C) lewgksa dh {kSfrt ca/k ds lgkjs cnyus ij o Å/okZ/kj ca/k ds lgkjs cnyus ij
(2) C2H5

HO H
(4)
(1)

(3) CH3
R
(5)

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 Page # 20 Solutions Slot – 1 (Chemistry)

CH=CH2 CH3 OH H

Q.11 CH3 OH H OH H CH3 HO CH=CH2

H CH=CH2 CH=CH2 CH3 R

s
R s s R
(III) (I) (II) (IV)
II o IV dk leku foU;kl gaSA

Q.12 (A) ,d ehtks ;kSfxd ,d ryh; ,efefr j[krk gaSA

(B) ,d ehtks ;kSfxd de ls de ,d leku ;qXe dk f=kfoe dsUnz j[krk gaSA
(C) The meso comd. is achiral

Me S.C
*
H OH
H *
OH
Me
S.C
achiral compound
Ans. A,B,C

Q.13 (A) CH3 — CH — CH2 — CH3 Pos X izdkf'kd lfØ;

OH
H Cl
(C) C=C=C Pos X izdkf'kd lfØ;
Cl H

COOH NO2

(D) Pos X izdkf'kd lfØ;

NO2 COOH

(B) CH2 = CH — CH2 — CH = CH2 Plane present izdkf'kd lfØ;

Ans. A,C,D

CH3
CH 3 CH3
Q.14 (A) C (B) C
H CH3
g.I (ry vuqifLFkr) g.I X
O.I

CH3 H
CH3 H
CH3 CH3
C C
H H
CH3
(C) H (D) CH3 H g.I
g.I Plane o.I
Plant not present, o.I not

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 Solutions Slot – 1 (Chemistry) Page # 21

(1) NH2

COOH C (2)

COOH
H
H2N H CH2OH
(3)

Q.15 (A) CH2OH (B) (4) Clockwise –R

L-amino acid
D

NH2 (1)
C (2)
H2C COOH
(3)
(C) H
(4)

R s
Anti clockwise
is L-amino acid

OH (2)
COOH (3)
C (3) (2)
* R (S)
H C*
(D) (4)
CH3 H NH2 (1)
(1)
(4)
L-amino acid
Ans. A, C, D

Q.16 A, C, D
1 CH2OH
1 CHO 1 CHO OH 2 C=O
2 2 3
H OH H OH CHO H HO H
3 3 4
HO H HO H HO H H OH
4 4 5
H OH H OH H OH H OH
(A) 5 (B) 5 (C) (D)
H3C H H CH2OH OH CH2OH CH2OH
OH OH H
D-sugar
1 CHO
2
H OH
3
HO H
4
5 5 H OH
H OH HO H 5
H OH
CH3 CH2OH
CH2OH
D-sugar L-sugar
D-sugar

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 Page # 22 Solutions Slot – 1 (Chemistry)

I III
aI X
IV d
d b R
CR IV
Q.17 (A) II

X b a
II
III I

I a

X b R
II
(C) III
d
IV
Ans. A, C

5 7 7
1 3 5

6 8
8
2 4 4
6

Q.18 2

4,5-di ethyl-3-methyl 4,5-di ethyl-2-methyl

octane octane

Constitutional isomerism
⇓
Positional isomerism

H H COOH
H OH
Q.19 (A) (B) H OH
HO OH COOH

Br
H H
Cl H C
C=C
(C) H Cl (D)
Molecular Plane

lHkh ds ry mifLFkr gaSA

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 Solutions Slot – 1 (Chemistry) Page # 23

EXERCISE – IV ADVANCED SUBJECTIVE QUESTIONS

(1) (1) (1)
(4)
(4) OH OH HO (4) (4) OH
(1)

Q.1 (i)
(3) (2) (2) (3) (3) (2) (2) (3)

R S S R R S R

SR RR
diastereomer
1
OH
R
2
R 2
3 4 OH
(ii) 3 S
S 3
2

Identicle

CHO CH2OH CH2OH

H OH Rotate at OH H HO H
H OH 180º HO H HO H
(iii) CH2OH CHO CHO

Identicte

4
HO 4 1 HO
H 1
H2N H
Q.2 (i) NH2
HO 3 COOH COOH 3 HO
2 2

R R

Identicle
R CH2OH CH2OH
(ii) H OH H OH R RR
SR Diastereomer
H OH H3C H R
S CHO

(iii)
Meso Pos X o.Active

diastereomers
2 2

CH=O CH=O
1
Q.3 (a) 4 H OH HO H4
1

CH3 CH3
3
3
R S
Enantiomers

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 Page # 24 Solutions Slot – 1 (Chemistry)

CH=O CHO
H OH R HO H S RS
(b) Diastereomer
H OH R H OH R RR
CH3 CH3

Br Br
Br Br Br Br
2 3

(c) 1 Br 3 2 Br 4

4 1

Indenticle

CH3 CH3
S
Q.4 (a) H Br Br H
R
CH2CH3 CH2CH3

Enantiomer

CH3 2
CH2CH3

(b) H Cl 4 H CH3
S 3
S
CH2CH3 1 Cl
Identicle
CH3 CH3

(c) H Br S H Br S SS
RS Diastereomer
H Br Br H S
CH3 R CH3

CH3 CH3
H Br Br H mirror image of meso
(d)
H Br Br H always identicle
CH3 CH3

Br Br
Br Br
(e)
Br
Br

Enantiomer

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 Solutions Slot – 1 (Chemistry) Page # 25

OH H OH H
R S
S CH3 S CH3
Q.5 (a)

RS
SS Diastereomer

OH H OH H CH3

R E CH3 R Z
S S
(b)

Diastereomer

OH H OH
H

CH3 CH3

(c)
different
compound

OH H OH H
3 5 3 5
1
1
2 4 6 CH3 2 4 6 CH3
(d)

Structural

OH H OH H
R S
S CH3 R
CH3
(e)

RS
SR Enantiomers

O O
CH3 CH3
S S

Q.6 (a) CH3 R

CH3 S
(b)
SS
RS Diastereomers

O O
CH3 CH3

CH3 CH3

Not isomers (diff. compunds)

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 Page # 26 Solutions Slot – 1 (Chemistry)

O O
CH3 CH3

(c) CH3
CH3

enantiomer (due to mirror' image)

O O
CH3 CH3 CH3

(d) CH3

Structural isomer (Positional)

O O
CH3 CH3

(e)
CH3 rotate by 180º
CH3
Identicle

CH3 H
HOOC OH
Q.7 (i) (ii) HO H (iii) C=C=C
O HO H

achiral COOH

O
C
H N C6H5 COOH
N N=N
(iv) HOOC (v)
HOOC
a chiral
Chiral

H Cl Cl H

Q.8 (I) cis (polar)

Pos present
optical inactive

H Cl H
Cl
(II)

Pos ugha gS blfy, ;kSfxd izdkf'kd lfØ; gaSA

µ=0

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 Solutions Slot – 1 (Chemistry) Page # 27

Q.9 Column -A Column - B

Br CH3 Br
R
(A) fdjSy (a) Br H CH3 H
H3C H
R
Enamtiomer
H3C H CH3 H CH3
H
R
(B) OH fdjSy (b) HO S HO

Enant.
CH3 CH3 CH3 CH3
CH3 R
R
(C) Meso (C) R S

CH3 CH3 CH3

RR
Dias.
RS
OH OH OH
OH OH OH OH OH OH
(D) fdjSy (d)

Diastereomer

COOH COOH

HOOC H
H H

H COOH

cis - isomer (Trans - isomer

Q.10 (Plane of symm.) (Pos absent)
Polar Non Polar

o. Inactive o. Active
Non resolvable Resolvable

Q.11 Nitrophenol
OH OH OH
NO2

NO2
NO2
Ortho. Nitro Phenol M-Nitro Phenol P-Nitro Phenol
Total = 3 isomers.

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 Page # 28 Solutions Slot – 1 (Chemistry)

Q.12 C4H10O

C—C—C—O—C C—C—C—O—C

F.I Metamers

C — C — C — C — OH C—C—C—C
OH
Positional isomers

CHO

* CHOH
Q.13 (I) 22 = 4 isomers.
* CHOH
CH2OH
(II)

H COOH COOH COOH COOH H

H COOH

HOOC COOH COOH

H H H H H

3 isomers

COOH
H * OH
(III) H OH 22 = 4 isomers.
H *
OH
COOH

Q.14 ;kSfxd tks izdkf'kd leko;ork iznf'kZr djrs gaSA

COOEt
(a) Et — CH — COOH (b) Me — CH — COOH (c) (d)
O
Me Ph COOH
Et–Me Me–Et
(f) (g)
C=C=C
Ph COOH
Cl

(j) (k)
O
;kSfxd tks T;kfefr; leko;ork iznf'kZr djrs gaSA
H
Ph COOH

(c) (g) (j)

O Ph COOH
H
I
H H
Not show G.I (o.I) (e) C=C=C=C (n)
H H
Br

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 Solutions Slot – 1 (Chemistry) Page # 29

Cl Cl

H H
Q.15 (I) (II)
Cl H
H Cl
Pos present so Achiral Chiral
due to Pos. absent

Cl COOH O O
(III) C=C=C (IV)
Cl Me
a chiral Pos absent
(Plane present) So chiral

Ph

:
Me
(V) (VI) S=O
Ph

COOH Chiral
Pos absent
so chiral

. O
H3C — CH — CH2 — CH2— CH2— CH3
Q.16 1 2 3 4 5 6
CH3

CH3 CH3
CH
H H
3

(a) vf/kd LFkk;h la:i.k

H 4 H
CH2–CH3

CH3 CH3
CH
CH2–CH3
(b) de LFkk;h la:i.k

H H
H H

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 Page # 30 Solutions Slot – 1 (Chemistry)

Br
H H
Br H
Cl
Q.17 (a) (b) H3C
Cis H Cis

H H H
Br
CH3
CH3
(c) Br (d)
H
Cis Trans

H
CH3 H
Cl
CH3
CH3
(e) (f)
H
H
Trans Trans

Q.18 dksbZ oLrq tks dksbZ Hkh lefefr ugha j[krk gaSA

Q.19 3 – Pentene – ol
Me Me Me
Me HO
H OH H H OH HO H
H H Me Me

Me Me Me H

OH right OH left D - Cis L - Cis

side = D side = L
Trans Trans
dqy 4 f=kfoe leko;oh

Q.20 bl ;kSfxd ds izdkf'kd lfØ; f=kfoe leko;oh

Cl
Cl Cl

Cl Cl
Cl

Cl
Cl Cl

ry mifLFkr ugha gaS blfy, izdkf'kd lfØ;

Cl Cl
Cl

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 Solutions Slot – 1 (Chemistry) Page # 31

EXERCISE – V(A)
Cl Br

1.

Br Cl

Identical

H H H
2. H3C –* C –* C –* C –CH3

Cl Cl Cl

3. Confermers

CH3 H
C=C H
CH3 G.I. *
4. C Exhibit Optical Isomerism only
X
CH3 COOH

CH3 CH3 CH3

* CH –OH H – C – OH HO – C – H

5. * CH –OH HO – C – H H – C – OH

CH3 CH3 CH3

(1) (2) (3)
Plane not present
Meso (Optically active) (Three Optical isomers)

6. Cyclohexane is less dense than water

7.

CH3 CH3

(A) C=C (Z)-2- Butene

H Z H

H CH3
C=C
(B) (E)-2-Butene
CH3 E H

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 Page # 32 Solutions Slot – 1 (Chemistry)

CH3 Ph

(C) C=C (Z)-1- Phenyl Propene

H Z H

H Ph
C=C
(E)-1- Phenyl Propene
CH3 E H

CH3 Ph

8. C=C
H Z H

H Ph
C=C
CH3 E H

F Br
C=C
9. 2 isomer
I Cl

Cl Br
C=C
2 isomer
I F

F Br
C=C
2 isomer
Cl I
Total = 6 isomer

10.
(A) CH3 – CH2 – C =CH2 No chiral centre

CH3

(B) CH3 – C – C CH No chiral centre

CH3

(C) CH3 – CH – CH2 – COOH No chiral centre

CH3

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 Solutions Slot – 1 (Chemistry) Page # 33

(D) * chiral centre is present

CH3 – CH2– CH – COOH

CH3

This compound show the O.I. because of chiral carbon

11.

4
CH3
CH3
120°
H H H CH3
120°

H H H
120° H

CH3 H

gauche Conformotion

12. S t r u c t u r a l I s o m e r C H14
6
DBE = 0 = no cyclic structure

(A)

(B)

(C)

(D)

(E)

13. Stateneut - I is correct because all non super imposable mirror images are optical isomer.
Statement - II is incorrect because it is not necessary to have chiral centre for optical
activity

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 Page # 34 Solutions Slot – 1 (Chemistry)

CH3

Cl H
*

14. H Cl Plane absent (optically active)

*

CH3
15.
CH3 O CH3 OH CH3 CH3

(B) & E
Z
CH3 CH3 CH3 CH3 CH3 OH
(E) (F) (G)

CH3 OH CH3 CH3

(C) & E
Z
CH3 CH3 CH3 OH

CH3 OH CH3 CH3

(D) & E
Z
CH3 CH3 CH3 OH
16. H3C(OH) CH – CH = CH–CH (OH) CH3

* *
CH3 – CH –CH = CH – CH– CH3

OH OH
n = 3(odd)
n +1
T.S.I = 2n – 1 + 2 p –1 p=
2
3 +1
=2 3–1
+2 2– 1 p= =2
2
=6

CH3

H3C – CH2 – C – CH3

17.
CH3
H CH3
CH3 CH3
CH3 CH3

H H H H

C2H5 CH3

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 Solutions Slot – 1 (Chemistry) Page # 35

18. In option (B) & (C) All atom are in out plane

H H
C = C= C
19.
H H
2 2
SP SP SP

CH3 H

C C
20. CH3–CH =CH– –CH =CH– –CH =CH–CH3    → CH3 – CH = 0 +
Ozonolysis

H CH3
CH3

C
0 = CH – – CH = 0
H

Products are optically inactive.

21. At room temperature compound B and C are unstable because both of the

Compound are anti aromatic

OH OH

R Cl H R Cl
H

S CH3 R
22. H CH3 H

OH OH

OH OH OH

R S R
H Cl Cl H H Cl

S R R
H CH3 CH3 H CH3 H

OH OH OH
[O] [P] [Q]

M&N Diastereomers
M&O Identical
M&P Enantiomer
M&Q Diatereomers

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 Page # 36 Solutions Slot – 1 (Chemistry)

EXERCISE – V(B)
1. C4 H10 O
(i) CH3 – CH2 – CH2 – OH
(ii) CH3 – CH2 – CH – CH3

OH
(iii) CH3 – CH–CH 3

CH2–OH

CH3

(iv) CH3 – C –OH

CH3

2.

CH3 CH3

H OH OH H

(I) OH H (II) OH H

CH3 CH3

CH3

OH H

(III) H OH

CH3

I & III niZ.k izfrfcEc gaSA = izfrfcEc leko;oh

I & II vkSj II & III niZ.k izfrfcEc ugh gaS o foofje leko;oh gaSA

3. µ obc = Σ µ i Xi Z – CH2 – CH2 – Z

(i) 1 = µg x g Xanti + Xg = 1

1 = µ g × 0.18 0.82 + Xg = 1

1
µg = D Xg = .18
0.18

(ii) Y – CHD – CHD – Y

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 Solutions Slot – 1 (Chemistry) Page # 37

ehtksa voLFkk
When y = CH3

CH3
H OH

1 2 3 4

CH3–CHD–CHD–CH3
H D

CH3
,UVh voLFkk

When Y = OH
OH – CHD – CHD – OH

HO ''''
''''
H ''''
HO

H D

D
xkWp voLFkk
4. C5H10

1 2 3. 4

5 6 7.

2H2

5. C4H6 C4H10
DBE

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 Page # 38 Solutions Slot – 1 (Chemistry)

EXERCISE – VI
1.
G.I. O
H OH

(A) * *

CH3 H
Stereogenic area = 3
chiral centre =2
Total no. of stereoisomers = 23 = 8(even)
Soln. = Q,S

* H
* CH3
H
(B) C
HC
3

CH2
POS is absent
COS is absent
Chiral Centre = 2
Total no. of stereoisomers = 22 = 4 (even)
Sol4. = Q,S

COOH

* OH
H
* OH
(C) H

COOH
POS is present
Chiral Centre = 2
Total no. of stereoisomers = 2n–1+2p–1
n 2
P= = =1
2 2
= 22–1 +21–1 = 3 (odd)
soln. P.S

OH
HO *
(D) CH2 NH2
HO
POS is absent
Chiral Centre = 1
Total no. of stereoisomers = 24
= 21 = 2 (even)
n
Sol . = Q,R

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 Solutions Slot – 1 (Chemistry) Page # 39

2.
O O

Me C–CH2CH2C–OH
N
Me Me
(A)

Br
Me

Me O
Pos & Cos will not present in this molecule. because the group become
– N– C – C – COOH
perpendicular to benzene ring due to ortho effect. there fore it can show optical isomerism.

O O

Me
C–CH2CH2C–OH
N
Me Me
(B)

Br Br
Me
Symmetry(i.e Pos) is present. there fore it can not show optical isomerism.

Cl H
CH – CH = C=C
(C)
Me
Cl
Terminal groups present perpendicular to each other. therefore Pos & Cos will not present .
so it can show optical isomerism and coupound will be resolvable.

Me

(D)

Me
(Cis)-1,2- di methyl cylohexane
Pos is present. Compound will non resolvable. but its trans isomer will be optically active.
there pure it can show optical isomerism.

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 Page # 40 Solutions Slot – 1 (Chemistry)

Cl Cl
Cl Cl

3. (A)

Cl Cl
They are diastereomers because they are not mirror image.

Cl
Cl
Cl

(B)
Cl Cl
Cl
They are position isomer,because the position of chlorine aloms are different.

Cl Cl

(C)

Cl Cl Cl Cl
They are identical because symm. is present and they are mirror image.

Cl
Cl Cl

(D)
Cl
Cl
They are idencal because Pos is present and they are mirror image to each other.

4.
O
*

(A)

Me *
Me Cl
POS is present
COS is absent
Chiral centre / stereo centre = 2 (even)
Total no. of stereo isomers = 2n–1+2p–1
n 2
P= = =1
2 2
= 22–1 + 21–1 = 3 (odd) P
Dipole moment will non zero because both chlorine atoms are same side
Soln. = [P,Q,R]

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 Solutions Slot – 1 (Chemistry) Page # 41

Br

(B)
Br
POS is present
COS is present
chiral centre = 0
No. of stereo isomer = 2 (cis & trans)
Dipole moment = 0 [ Cos is present ]

H Cl

C
Stereo
(C) Centre C

Cl H
POS is absent
COS is absent
stereo centre = 2
Dipole moment = 0 [ cos is present]

(D)
COOH

H OH

H OH

COOH
Chiral centre = 2
Pos X
Cos X
A o s = c
2
No. of stereo isomer = 22–1 + 21–1 = 3

5.
CH3 CH2OH H CH2NH2
H H3C
(A) Position isomer
NH2 OH

2- Amino-1-Propanol 1-Amino-2-Propanol

CH3 Cl
Cl CH 3
H H
S R
(B) Enantiomer

Et Et

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 Page # 42 Solutions Slot – 1 (Chemistry)

CH3 H
OH Et
H H3C
S S
(C) identical
Et OH

CH Me Me
2 5
R H5C2 S
(D) Enantiomer
OH H H OH

CH3

H Br

H Cl

6. H F

CH3
Chiral Centre = 3
POS is absent
COS is absent
Total optical active isomer = 23
=8
op. active isom
No. of enatiomeric pair =
2
8
= =4
2
Cl Cl

7.

Cl Cl Cl Cl
They are not mirror image so they are Diastereomers.

Cl Cl
Cl
Cl
8. (A) (B)

Cl Cl
POS is present POS is present
COS is absent COS is absent

(C) (D)

POS is present POS is present

COS is absent COS is present

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 Solutions Slot – 1 (Chemistry) Page # 43

EXERCISE – VII
Sol4. =1

CH3
CH3
NH2
(i) (ii)
NH2
o-Toludine m-toludine

CH3
CH2–NH2

(iii) (iv)

NH2
P- toludine Benzyle amine

Sol4. =2
O O

NH– C – CH3 C – NH–CH3

(i) (ii)

O O

CH–
2 C – NH2 CH2–NH–C –H
(iii) (iv)

CH3
O
CH3

C – NH2
(v) (vi) C – NH2
O

CH3

O
CH3

NH – C – H
(vii) (viii)
C – NH2
O

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 Page # 44 Solutions Slot – 1 (Chemistry)

CH3

CH3

O
(ix) (x)
NH– C – H
NH – C – H
O

Soln.=3= 5
O
O
(i) (ii)

O
O CH3
CH3

CH3
(iii) (iv)

CH3

O
CH 3

(v)

CH3

Soln.=4=1

CH3
CH3
C2H5 & C=C
C=C C2H5
Cis Trans

Sol4.=5=1
CH3 CH3 CH3

&

CH3 CH3 CH3

cis Trans
1,2 Dimethyl Cyclo Propane

Sol4. = 6 =2

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 Solutions Slot – 1 (Chemistry) Page # 45

Cl Cl

*
CH–CH–CH–Cl
3 CH–CH–CH–CH
2 2 3
*
CH3 CH3

Sol4=7=2
Me Me

Me Me
POS is absent
COS is absent
Both are optically active

Sol4=8=2
CH3 CH–OH
2 CH3 H
CH–CH=CH–CH
3 2 C=C C=C

H H H
OH CHOH
2

Cis Tarns

CH3 CH3 CH3 OH

CH–CH=C–CH
3 3
C=C C=C
H OH H CH3
OH
Cis Trans

Sol4=9

OH OH OH OH

(a)=4
CH3
CH3 CH3 CH3

OH OH OH

(b)=3

OH OH OH

Sol4. =10

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 Page # 46 Solutions Slot – 1 (Chemistry)

Me
Me
*
H CH=C=C
* H
H D
(a)
H
CH=C
Me
Stereogenic area = n =4
Total stereo isomer = 2n = 24 = 16

H
H
NO2 C=C
Me
(b)

COOH Me

Stereogenic area =2
Total stereoisomers = 22 = 4

Sol4. = 11

CH2– COOH

*CH – COOH
* *
(a) CH3– CH – CH – COOH (b)
HO – CH – COOH
OH OH *

S.I = 22 = 4 S.I = 22 = 4

* OH
*
(c) * (d) *

S.I =21 =2 S.I. = 23 = 8

*
* O OH
*
(e) (f)
COOH
* OH

S.I =22 = 4 S.I =22 =4

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 Solutions Slot – 1 (Chemistry) Page # 47

*
(g)
*

S.I =22 = 4

SOl4.= 12

* CH3
HO CH–C–NH * * S

N CH3
O
NH2 *
C=O
HO
Total Chiral Centre = 4

(a) Amoxicillin

HO O
O *
*

O
CH3 CH3
* *
* **

HC
3
*

Total Chiral Centre = 4

(b) Mevacor

Sol =13

CH3

(A) (B)
CH3 CH3
CH3

POS is Present POS is Present

COS is absent COS is absent

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 Page # 48 Solutions Slot – 1 (Chemistry)

CH3

(C) (D)
CH3 CH 3
CH3

POS is Present POS is Present

COS is absent COS is absent

CH3 Cl CH3 Cl

(E) (F)

CH 3 Cl CH3

POS is absent POS is Present

COS is absent COS is Present

Cl CH 3 H 3C Cl

(G) (H)

H3C Cl Cl CH3

POS is absent POS is Present

COS is absent COS is absent

Chiral = C,E
Achiral = A,B,D,F,G,H

Sol4. = 14

a.

POS is absent chiral

COS is absent

b.

POS is present Achiral

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 Solutions Slot – 1 (Chemistry) Page # 49

COS is absent

c.

POS is present Achiral

COS is absent

d.

POS is absent chiral

COS is absent

e.

POS is present Achiral

COS is present

f.

POS is present Achiral

COS is absent

Sol4. = 15

a.

POS is absent chiral

COS is absent

b.

POS is absent chiral

COS is absent

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 Page # 50 Solutions Slot – 1 (Chemistry)

c.

POS is present meso

COS is absent

d.

POS is present Achiral

COS is absent

Sol4.16
Br

Br Br H CH3
C—C
H H H CH3
(i)
CH3 CH3
Br
POS is present meso
COS is present

Br

Br H H CH3
C—C
H CH3 CH3 H
(ii)
CH3 Br
Br

POS is absent chiral

COS is absent
CH3 CH3
OH OH OH
(iii)
OH
POS is present meso
COS is absent

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Solutions Slot – 1 (Chemistry) Page # 51

CH3
OH OH OH
(iv)
CH3
OH

POS is present meso

COS is absent

CH3 CH3
OH OH OH

(v)

OH

POS is absent chiral

COS is absent
CH2OH
CH2OH

OH H OH
H
(vi) H OH
H OH

CH2OH CH2OH
POS is present meso
COS is present

CHO
HO CHO
H

H OH

(vii) OH H
H OH
CHOH
2
CHOH
2

POS is absent chiral

COS is absent

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 Page # 52 Solutions Slot – 1 (Chemistry)

CH2 OH

CH2 OH CH2 OH CH2OH

HO H HO
H OH

H H OH
(viii)
H H
OH OH OH
CH2OH
CH2 OH CH2 OH

POS is present meso

COS is present

CH2 OH

CH2 OH CH2 OH CH2OH

HO H
H OH

(ix) H OH H
H OH H OH OH
CH2OH
CH2 OH

POS is absent chiral

COS is absent
Sol4.17 (A)

(i) CH3 – CH – CH – CH3 Chiral centre = 0

CH3 CH3

(ii) CH3 – CH – CH– CH2 – CH3 Chiral Centre = 2

Br Br
Symm= x
Meso = x

Br
CH3
(iii)

C.C = 2
Pos/Cos = X
Meso = X

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 Solutions Slot – 1 (Chemistry) Page # 53

(B)
(i) CH3 – CH2 – CH – CH – CH2 – CH3

CH3 CH3
C.C = 2
Pos = present
Meso

(ii) CH3 – CH2 – CH – CH – CH2 – CH3

CH2 CH2

CH3 CH3

C.C = 0
Meso = X

(C)

(i)

C.C = 2
Pos = present
Meso

(ii)

C.C = 0
Meso = X

(iii)

C.C = 2
Pos = present
Meso

sol4.=18

Cl Cl Cl Cl Cl Cl

C – C – C –C C – C– C– C – C C –C – C – C

C C
(i) (ii) (iii)

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 Page # 54 Solutions Slot – 1 (Chemistry)

Pos is present Pos & Cos is absent Chiral Center = 0

Achiral Chiral Achiral

(B)
Cl
Me Br Me
Me

Me Br Cl
(i) (ii) (iii) (iv)
Pos is present Pos is present Pos is present Pos is present
Achiral Achiral Achiral Achiral

CH3
Br Br Br Br

(C) C – C – C –C – C C – C – C –C– C

CH3
(i) (ii) (iii)

Pos is present Pos & Cos is absent Pos is present

Achiral Chiral Achiral

Sol.4=19
Cl H H Cl

Enantiomer
H O
H HO H
(2S , 3R) (2R , 3S)

H Cl HO H

Diastereomers
H HO H Cl
(2S , 3S) (2R , 3S)

H Cl HO H

Diastereomers
HO H Cl H
(2R , 3S) (2R , 3R )

Sol4.20
1.

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 Solutions Slot – 1 (Chemistry) Page # 55

COOH
COOH COOH
H OH COOH
R
H OH
R
OH H
H HO
OH H H OH COOH
COOH
(2R,3R)

2.

COOH COOH
COOH
H COOH
HO
R OH
H
S
H OH
H
H H OH OH
OH
COOH
COOH
(2R,3S)

3.

COOH COOH
HOOC OH COOH COOH

S
HO H
S
H OH
HO H HO H OH H
COOH
H
(2S,3S)

4.

COOH
COOH COOH
COOH COOH
H S
HO H
S
H OH
HO H
HO H OH COOH
OH
(2S,3S)
a. Same Compound 3&4
b. Enantiomer 1&3
C. Meso 2
D. Diastereomers (2,3) (2,4) (1,2)

Sol4. =21

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 Page # 56 Solutions Slot – 1 (Chemistry)

HO H HO H H OH

H OH OH
OH H H
(i) (ii) (iii)
(2R,3R) (2R,3S) (2S,3S)

Enantiomer 1&3
Meso (ii)
Diastereomers (i,ii & (ii,iii)

Sol.4 =22

(i)
Cl

a. C – C – C = C + HCl C – C –* C –C x=2
b. C – C – C = C + HBr Pero
.→ C – C – C – C y =1 (x+y=3)

Br

(ii)
Br
H3C CH 3
a. C=C + HBr CH3 – C – CH2 – CH3 x = 1
H3C H
CH3

H3C CH 3
C=C *
b. + HBr Pero
.→ CH3 – CH – CH– CH3 y = 2
H3C H |
Br
CH3
(X+y=3)

(iii)
Cl

a. CH3 – CH2 CH2 CH = CH2 +HCl C – C – C – C *– C x=2

H H
C=C H+
b. + H2O → C2H5 – CH– CH2 – C2H5 y=2
C2H 5 C2H5
OH

c. + HBr z=1
CH3 H 3C Br
[ x+y+z = 5]

Sol4.=23

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 Solutions Slot – 1 (Chemistry) Page # 57

Cis Anti addition Threo

Trans Anti addition Erythro
(i)

C2H5 C2H5

H Br Br H
CH
2 5 C3H7
C=C
(a) + Br2 Anti
 → Br
add . H + H Br
H H
C3H7 C3H7
Cis-3-heptene Threo (x=2) (b)
C2H5 C2H5

C2H5 H Br Br H
H
C=C Br + Br
H
+ Br2 Anti
 → H
add . H
C3H7
C3H7 C3H7
Trans-3-heptene Erythro (y=2)
C2H5

H Br
C2H5 H
C=C Br
(c)
H
+ Br2 Anti
 → H
add . (z=1)
C2H5
C2H5
meso (x+y+z) = 5

(ii) + Br2 Anti

 →
add . + x=2
CH3 CH 3 CH3 Br Br CH3 Br CH3CH3 Br

+H2 Syn
 add
→ + y=2
CH3 C2H5 CH3 H C2H5H H CH H CH
3
2 5

x+y=4
Sol4.=24
Cis Syn addition Erythro
Trans Syn addition Threo

(i) + Br/CH2Cl Anti

 →
add .

CH3 CH2CH3

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 Page # 58 Solutions Slot – 1 (Chemistry)

CH2CH3 CH2CH3

CH3CH2 CH3 CH3 H H CH3

+ H2 / Pt Syn H CH3 + CH3 H

(a)  add
→
CH3 CH2CH3
CH2CH3 CH2CH3
Trans (x = 2)
CH2CH3

CH3CH2
CH2CH3 H CH3

+ H2 / Pt Syn H CH3
(b)  add
→ y=1
CH3 CH3
CH2CH3
Cis meso
CH2CH3 CH2CH3

CH3 CH3 H3C H CH3 H

H +
(c) + H2 / Pt Syn
 add
→ CH3 H CH3
CH3 CH2CH3 CH3
CH2CH3 CH3 CH2CH3 CH3
(Z = 2)
Trans x+y+z = 2 + 1 + 2 = 5

(ii)

(a) + H2 / Pt Syn
 add
→ (x = 1)
CH3 CH3 CH3 H CH3H

(b) +Br2 Antiadditi on

   → +
CH3 CH2CH3 CH3 H H CHCH H CH CH H
2 3 3 2 5

(y = 2)
x+y = 1 + 2 = 3

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