CPP FIITJEE ISOMERISM

Single correct choice type

1. Which of the following compounds exhibits tautomerism?

(A) chloroethane (B) ethanol
(C) ethoxyethane (D) nitroethane

2. Maleic acid and fumaric acid are

(A) epimers (B) anomers
(C) enantiomers (D) geometrical isomers

3. The dihedral angle between two methyl groups in gauche conformation of n-butane is
(A) 109º 28' (B) 120º
(C) 180º (D) 60º

4. Which of the following compounds can exist in its meso form?

(A) 1, 2dichlorobutane (B) 2, 3dichloropentane
(C) 2, 3dichlorobutane (D) 1, 2dichloropentane

5. For the compound,

total number of optical isomers will be

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

6. Isopentane can form four monobromo isomeric derivatives. How many of them are optically active?
(A) 1 (B) 2
(C) 3 (D) None of these

7. The number of isomers of compounds with molecular formula C2H2Br2 is:

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

8. Which of the following molecule has 'S' configuration?

F CH2Cl

(A) Cl D (B) H3C CH2OH

H H
NH2 COOH
(C) H3C C COOH (D) H2N C H

H CH3
9. Select the optically inactive compound among the following.

(A) CH3 NH2 CH3 NHAc

(B)
NH2 CH3 AcNH CH3

HN NH
(C) HOOC NHAc (D)
AcNH COOH OC CO

10. Which of the following structures are superimposable?

H Br Br Br
Br Me H Me Me H H Me
(1) HO Et (2)Et OH Et Me (4) HO Et
Me Me (3)
OH Me

(A) (1) and (2) (B) (2) and (4) (C) (1) and (4) (D) (1) and (3)

11. Which of the following is a meso compound?

CH3 CH3 CH3
H Cl H Cl Cl H
(1) (2) (3)
Cl H H Cl Cl H
CH3 CH3 CH3

(A) Only (1) (B) Only (2)

(C) Only (3) (D) Both (2) and (3)

12. The two compounds shown in the figure below are

O O

and

H H

(A) diastereomers (B) enantiomers

(C) epimers (D) regiomers
13. Among the following, the Newmann projections of meso–2,3–butanediol are
Me Me Me Me
OH H OH Me Me H
H HO

HO H H OH HO H H OH
Me Me H OH
P Q R S
(A) P, Q (B) P, R
(C) R, S (D) Q, S
14. Which of the following compounds will exhibit geometrical isomerism?
(A) 1-Phenyl-2-butene (B) 3-Phenyl-1-butene
(C) 2-Phenyl-1-butene (D) 1, 1-Diphenyl-1-propene

15. An optically active alkene (A) has the molecular formula C6H12. The catalytic hydrogenation of (A)
gives an achiral product. The structure of (A) would be
(A) CH3CHCH=CHCH3 (B) CH3CHCH2CH=CH2

CH3 CH3
(C) CH3CHCH=CH2 (D) CH3CHCCH
C2 H 5 C2 H 5

16. If specific rotation of glucose solution is 52° and that of fructose solution is –92° then what will be the
specific rotation of invert sugar?
(A) –20° (B) +20° (C) –72° (D) +72°

CHO CHO
17. CH3—CH2—CH2—CH2 and CH3—CH2—CH—CH3 are …………… .
(A) positional isomers (B) functional isomers
(C) tautomers (D) chain isomers

18. Tautomerism will be shown by

(A) CH2==CH—CHO (B) CH3—CH == CH—CHO
OH O
(C) CH2==CH—CH2 (D) C6H5CH

19. Which of the following compounds do not show tautomerism?

O

O
(A) H—O—CH=CH2 (B)

O O

NH O
(C) (D)

20. Which of the following would be optically active?

(A) 1,1–dimethyl ethanol (B) butan–2–ol
(C) butanol (D) 4chloro1hydroxybutane

21. The number of stereoisomers possible for the following compound

CH=CH-CH2CH3
CH 2 CH 3

Me
would be
(A) 4 (B) 6
(C) 8 (D) 16
22. The configuration of chiral carbons in the compound
OH
H CH3

H CH3
OH
would be
(A) Front C–(R), back C–(S) (B) Front C–(S), back C–(R)
(C) Front C–(R), back C–(R) (D) Front C–(S), back C–(S)

23. Which energy profile diagram would you expect for anticlockwise rotation of back side carbon about
the central bond of 2-methyl butane?

CHC
3
H3

H
HH CH3
Conformation at dihedral angle = 0°

PE PE
(A) (B)

0 60 120 180 0 60 120 180

Dihedral angle Dihedral angle

PE PE
(C) (D)

0 60 120 180 0 60 120 180

Dihedral angle Dihedral angle

24. The minimum number of carbon for an alkane (acyclic form) to show optical isomerism is
(A) 3 (B) 5
(C) 7 (D) Not possible

25. The compound which is not isomeric with diethyl ether is

(A) methyl n-propyl ether. (B) 1-butanol.
(C) 2-methyl propan-2-ol. (D) butanone.

26. Geometrical isomerism is shown by

H I H H3 C I
(A)
C=C C
H Br
=
(B) C
B
r
H3C Cl H ClC
(C) C=C l
H3C Br H3C
C
(D) =
C
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27. The molecules
H3C H H CH3
C=C=C and C=C=C
H CH3 H3C H
are :
(A) enantiomers (B) diastereomers (C) structural isomers (D) none of these

28. The most stable conformation of 1,2–diphenyl ethane is

H C6H5 C6 H 5 C6 H 5
CH H C6H5 H
6 5 H H
(A) (B) H (C) H (D)
C6H5 H H H
H H H
H H C6 H 5
H C6H5

29. Which of the following can form geometrical isomer?

Cl
(A) (B) CH3–CH=N–OH (C) (D) all of the above
Cl
30. The following pair of compounds represents
CH3 CH3
H OH H H

H H H H
CH3 CH2OH
(A) same compound (B) chain isomers (C) position isomers (D)enantiomers

CHO CHO H
H OH H OH HO CHO
31. HO H HOH2 C OH OH
H
CH2OH H CH2OH
(I) (II) (III)
Choose the correct statement.
(A) (I) and (II) are identical; (I) and (III) are diastereomers.
(B) (I) and (II) are enantiomers; (I) and (III) are diastereomers.
(C) (I) and (II) are identical; (I) and (III) are enantiomers.
(D) (I) and (II) are enantiomers; (I) and (III) are identical.

32. The functional isomers of 1butyne are

(1) CH3CCCH3 (2) CH3C=CH2
CH3
(3) CH2=C=CHCH3 (4) CH2=CHCH=CH2
(A) (1), (3) and (4) only (B) (2) and (3) only
(C) (2) and (4) only (D) (3) and (4) only

33. Which of the following compound does not show geometrical isomerism?
Me Me Me
(A) (B) C=C=C=C
Et Et
Me
Me Me
(C) MeCH=CHCH=CHEt (D) C=C=C
Et Et

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34. Which one of the following compound will be chiral ?
..
S (B) H3C CH3 H3CC CH2
C (C) (D)Both (A) and (B)
(A) Me Et H H
O HC CH2

35. Which of the following group of compounds are not functional isomers?
(A) Nitro alkane and alkyl nitrite. (B) Aldehydes and ketones.
(C) Alcohols and aldehydes. (D) Alcohols and ether (both saturated).

36. The number of structural isomers for C6H14 is

(A) 3 (B) 4
(C) 5 (D) 6

37. The relationship between the given stereoisomers

CH3 CH3
H Br Br H
H OH and HO H
H Cl Cl CH3
CH3 H
would be
(A) They are identical. (B) They are diastereomers.
(C) They are enantiomers. (D) They are epimers.

38. Total number of geometrical isomers of the compound CH3CH=CHCH=CHC2H5 is

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

39. Which of the following is not a meso compound?

CH3
H CH3 H OH
(A) (B) H H (C) (D) H Cl
H CH3 H OH
CH3
Cl Cl Cl H
40. Which of the following does not show tautomerism?
O O
(A) C6H5CCH3 (B) C6H5CNH2
O O O
CH3
(C) C6H5CCH2CH (D) C6H5CN
CH3

41. Which of the following compound does not show cistrans isomerism?
CH3
(A) 2methyl2heptene (B)
CH3
CHCH3

(C) 2heptene (D)

CH3

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42. Cyclohexanone tautomerises in presence of catalyst (H+)
O OH
H
H+
H
Identify the product of the following tautomerisation reaction
O D+
TCH –C–CH H
2 2
OH OD
(A) TCH2–C=CH2 (B) TCH2–C=CH2
OT
(C) CH2=C–CH3 (D) all of these

43. Which of the following depicts the same stereoisomer?

CH2CH3 CH3 CH2CH3
Br Cl CH3CH2 Cl H3 C Br

CH3 Br Cl
1 2 3
(A) 1 and 2 (B) 1 and 3
(C) 2 and 3 (D) 1, 2 and 3

44. Which of the following conformational isomers is most stable?

H3C CH3
H3C
H CH3

(A) (B)
H
H HH H H
H
H3CH H3 C
H H
(C) (D)
H H
H
H H3 C H H3 C

45. HCN and HNC exhibit

(A) functional isomerism. (B) tautomerism.
(C) metamerism. (D) none of these.

46. Choose the correct statement about the compounds, 1, 2 and 3:

COOC 2H COOCH3 COOCH3

5 H Br H Br
H Br
H Br H Br Br H
COOCH3 COOC2 H5 COOC2 H5
(1) (2) (3)
(A) (1) and (2) are identical (B) (1) and (2) are diastereoisomers
(C) (1) and (3) are enantiomers (D) (1) and (2) are enantiomers

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47. 1,3Dichloroallene (CHCl=C=CHCl) shows
(A) geometrical isomerism. (B) optical isomerism.
(C) both (A) and (B). (D) none of these.

48. Which of the following compound has zero dipole moment?

(A) Cis but2ene (B) Trans but2ene
(C) But1ene (D) 2methylprop1ene

49. The configurations of two polyhydric alcohols having same molecular formula are given below. How
are they related to each other?
CH3 CH3 H—
HO—C—H C—OH
HO—C—H H—C—OH
H—C—OH H—C—
CH3 CH3 OH
(I) (II)
(A) Enantiomers (B) Diastereomers
(C) Identical (D) Constitutional isomers

50. The enol form of acetone after treatment with D2O gives
O OH OD
(A) CH3C=CH2 (B) CH3CCD3 (C) CH2=CCH2D (D) CD2=CCD3
OD

51. Number of sterocentres and stereoisomers of the following compound will be respectively
CH3
OH

(A) 1 and 2 (B) 2 and 4

(C) 3 and 8 (D) 3 and 6

52. Which of the following compound can exhibit both geometrical isomerism and enantiomerism?
(A) CH3CHOHC2H5 (B) CH3CHOHCOOH
CH3
(C) CH–CH=CHCH3 (D) All of them
C2H5

53. Which of the following substituted benzenes would furnish three isomeric compounds when one more
substituent is introduced?
Cl Cl Cl Cl
Cl
1. 2. 3. 4.
Cl

Cl
Select the correct answer using the codes given below:
(A) 1, 2 and 3 (B) only 1
(C) 2 and 4 (D) 1 and 3

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54. Which one of the following is the least stable conformer?
CH3 CH3 CH3 OH
HO H CH H OH H CH3
H 3

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

H OH H OH H OH H CH3
CH3 OH CH3 OH

55. Number of isomers obtained on monobromination of mdimethyl benzene is

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

56. The number of isomeric optically active isomers of monochloro isopentane is:
(A) 2 (B) 3
(C) 4 (D) 5

57. The maximum number of optically active isomers possible for the following compound is
H H
HO + COO
N
H H
4hydroxy proline
(A) 2 (B) 4
(C) 6 (D) 8

58. The maximum number of optically active isomers possible for the following compound is
HOOCCH(OH)CH(OH)CH(OH)COOH
(A) 2 (B)4 (C) 6 (D) 8

One or more options may correct :

59. Which of the following compound(s) is/are optically active?

CH3 CCl3
CH2OH COOH
(A) H Cl (B) (C) (D) H OH
H Cl HO H H 3C CH2Cl HO H
CH3 CH3 COOH CCl3

60. Keto-enol tautomerism is observed in

(A) C6H5CHO (B) C6H5COCH3 (C) C6H5COC6H5 (D) C6H5COCH2COCH3

61. Only two structural isomeric monochloro derivatives are possible for:
(A) n-Butane (B) 2,4-Dimethyl pentane (C) Benzene (D) 2-Methyl propane

62. Which of the following is/are the pair of isomer(s):

(A) (CH3)2CHOC2H5 ; CH3(CH2)2OC2H5 (B) CH3CH2NO2 ; CH2(NH2)COOH
(C) (CH3)2CH(CH2)2CH3 ; CH3(CH2)2CH(CH3)2 (D) CH3CH2CO2H ; CH3CO2CH3

63. Tautomerism is not exhibited by :

CH3
(A) –CH=CH–OH (B) O= =O (C) –C=O (D)
HO–C– –OH
H
O
CH3

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64. Which of the following compounds show tautomerism?
O
O
Me Me
(A) C2H5NO2 (B) Acetoacetic ester (C) (D)
Me Me
65. Which of the compounds will show geometrical isomerism?
Br Br
H H H H
(A) C (B) C (C) C (D) C
Br Br Br Br
Br

66. Which type of isomerism is shown by 2,3dichlorobutane?

(A) Conformational (B) Geometrical
(C) Optical (D) Tautomerism

67. Which of the following compounds exhibit(s) optical isomerism?

HOOC NO2

(A) (B) CH2=C=CH2

NO2 COOH
C6 H5 C6 H5 CH2 CH2
(C) C=C=C (D) C=C
-napthyl -napthyl CH2 CH2

68. Which of the following compound(s) is/are optically active?

CH3
(A) S=O (B)
H5C2
Br CH3 Cl
H3C 3 CH
(C) H
NO2 CH3
H
(D)

69. Which of the following will show optical isomerism? (A)CH3–

CH3
CH=C=CH–CH3 (B) C
Cl
CH3
HOOC O2N
CH3
(C) Cl C (D)
H Cl
O2N Br

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70. Which statements are correct for the following compounds?
CHO CHO

H OH HO H
HO H H OH

C6H5 C6H5
(I) (II)
(A) Both are in threo form. (B) Both are enantiomers.
(C) Both are diastereomers. (D) Both are in erythro form.

71. Two isomers ‘A’ and ‘B’ of same energy reacts with a reagent R to form products C and D respectively
as shown in the graph.

A+R
PE or B+R C
D
POR
Which of the following statement(s) is/are correct?
(A) The two transition states are enantiomers.
(B) The two transition states are diastereomers.
(C) A and B are enantiomers.
(D) C and D are diastereomers.

Comprehension Type

Passage # 1
Broadly speaking, there are four types of stereoisomers, namely conformational, geometrical,
enantiomers and diastereomers. Although conformational isomers, have same configuration, geometrical
isomers have different atoms attached to each of the doubly bonded carbon atom and enantiomers are due
to chirality in the molecule. Enantiomers are also known as optical isomers and diastereomers are those
stereoisomers which are not enantiomers.
Observe the following structure and answer the questions given below:
H3C H
C=C CH3
H C
OH H
72. How many stereoisomers are possible in 3penten2ol, CH3CH=CHCHOHCH3?
(A) 1 (B) 2 (C) 3 (D) 4

73. The structure can have how many other diastereomers?

(A) 1 (B) 2 (C) 3 (D) Nil

74. The compound given above, when treated with D2/Ni at elevated temperature gives
CH3CHDCHDCH(OH)CH3 is capable to show
(A) Optical isomerism only (B) Conformation only
(C) Both optical as well as conformation (D) Conformational and geometrical

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Passage # 2
For most compounds all the molecules have the same structure, whether are not this structure can be
satisfactorily represented by a Lewis formula. But for many other compounds there is a mixture of two or
more structurally distinct compounds that are in rapid equilibrium. When this phenomenon, called
tautomerism, exists, there is a rapid shift back and forth among the molecules. In most cases, it is a proton
that shifts from one atom of a molecule to another. A very common form of tautomerism is that between a
carbonyl compounds containing an hydrogen and its enol form.
O OH
C C C=C
H
The structures of keto forms of four compounds have been shown below.
6
O O O O
5
1

4
OH Cl 3
2 NO 2
(I) (II) (III) (IV)
The compound represented by structure (I) also has an OH functional groups, while compounds
represented by structures (II) and (III) have substituents Cl and NO2 respectively. The last compound
represented by structure (IV) does not have another functional group. The structures represented by (I) to
(IV) are not aromatic but readily acquire aromaticity in presence of an acid or a base.

75. Which of the following compound does not have a chiral centre.
(A) (I) (B) (II) (C) (III) (D) (IV)

76. Which of the following compound have maximum rate of enolization?

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

77. The enol form of the these compounds show acidic behaviour in water. Their acidic strength will be
(A) (III) < (II) < (I) < (IV) (B) (III) > (II) > (I) > (IV)
(C) (III) > (II) > (IV) > (I) (D) (II) > (I) > (III) > (IV)

Passage # 3

Type of isomerism occuring by the migration of atom (generally acidic hydrogen) and the movement
of a double bond is called tautomerism. Tautomers are true isomers. The most common tautomerism is
keto–enol tautomerism. This is an example of 1,3–tautomerism. Generally keto form is more stable than
enol form. However some enols are more stable than keto form. Enol form is more stable than the keto
form either due to resonance or due to hydrogen bonding or due to the nature of solvent.

78. Which of the compound(s) will show tautomerism?

O O
O
C
(A) CH3CCH3
H
(B)
3
(C) (D) All of these

79. Which of the following compounds do not exist in the enol form?
O O
O
(A) (B) (C) CH3CCH2–COOC2H5 (D) Both (A) and (B)

O NH

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80. Enol content of ethyl acetate will be maximum in which solvent?
(A) H2O (B) CH3COOH (C) n–hexane (D) aqueous HCl

81. Which of the compound(s) exist mainly in the enol form?

O

O
O
(A) (B) (C) O
(D) All of these
O

Matrix Match Type

82. Column I Column II
(A) 2, 3 di bromo pentane 1. 2 optically active isomers
(B) Tartaric acid 2. 3 optical isomers
(C) Lactic acid 3. 4 optical isomers
(D) HOOC–CHOH–CHOH–CHOH–COOH 4. 2 meso isomers

83. Column I Column II

(A) Optical isomerism CH 3

H
1. D

(B) Geometrical isomerism

2. CH3 D

H H

(C) Decolourisation of Br2 | CCl4 solution 3. CH3C=C–CH3

H H

(D) Reduced to optically inactive by C2H5

catalytic hydogenation 4. CH3–C –CC–H
H

84. Column I Column II

(A) trans-1,4-dimethyl cyclohexane and cis-1,4- 1. Can be isolated in pure forms.

dimethyl cyclohexane.

(B) CH3–CHCl2 and ClCH2–CH2Cl 2. Geometrical isomers.

(C) cis and trans–but-2-ene 3. Different stability.

4. Differ in melting point.

(D) (+) lactic acid (–) lactic acid CH3CHCOOH
OH

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 85. Column I Column II
(A) A pair of optical isomers. O
1. CH3CH2–CH2–C–H ; CH3CH2CH=CHOH
(B) Asymmetric synthesis COOH COOH
H–C–OH ; H–C–OH
2.
HO–C–H H–C–OH
COOH COOH
(C) A pair of diastereomers. O OH
3. CH3–C–COOH ; CH3–C–COOH
H
(D) Enolisation CH3 CH3
4. H–C–OH ; HO–C–H
COOH COOH

86. Column I (Compound) Column II (Type of isomerism exhibited)

(A) CH3CH2CH(CH3)CH2–CH3 1. Functional
(B) CH3CH2CH2Cl 2. Optical
(C) CH3CH2CO2H 3. Chain
(D) CH3CH(OH)CH(Cl)CHO 4. Position

87. Column I (Compounds) Column II (No. of optical isomers)

(A) HOOCCHOHCHOHCHOHCOOH 1. 2

(B) CHOCHOHCHOHCHOHCH2OH 2. 3

(C) HOOCCHOHCHOHCOOH 3. 4

(D) CH3CH(OH)–COOH 4. 8

88. Column I Column II

(A) 2, 3 di bromo pentane 1. 2 optically active isomers

(B) Tartaric acid 2. 4 optical isomers

(C) Succinic acid (butanedioic acid) 3. No optical isomer

CH2–CH3 4. 3 optical isomers

(D) CH3–CH–CH2–CH2–CH3

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Assertion Reasoning Type
(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion.
(B) Assertion is True, Reason is True; Reason is NOT a correct explanation for Assertion.
(C) Assertion is True. Reason is False.
(D) Assertion is False. Reason is True.

89. Assertion : HO2C–CHOH–CHOH–CHOH–CO2H exists in four stereoisomeric forms; two of which

are optically active while the other two are meso-forms.
Reason : Mirror image of meso is another compound.

90. Assertion : CHBr=CHCl exhibits geometrical isomerism but Cl–CH2–CH2–Br does not.
Reason : Presence of C=C is a necessary condition for showing geometrical isomerism.

91. Assertion: CH3CH=CHCH=CHC2H5 is capable of showing cistrans isomerism and

has 4 geometrical isomers.
Reason: The compound has restricted rotation and each atom having restricted rotation is linked
to 2 different groups and number of geometrical isomers is given by 2n where ‘n’
represents the number of double bonds capable of exhibiting geometrical isomerism in
unsymmetric molecules.

92. Assertion: Compound is optically inactive when it is asymmetric.

Reason: Asymmetry in the molecule is due the absence of plane of symmetry, centre of
symmetry, simple axis of symmetry and alternating axis of symmetry.

93. Assertion : Enol form of cyclohexane1,3,5 trione is less stable than its keto form.
Reason : In enol form, it contains a benzene ring.

94. Assertion : Racemic mixture as well as meso compounds are optically inactive.
Reason : Racemic mixture contains two different type of molecules whereas meso compound
contains only one type of molecule.

95. Assertion : Keto form of phenol is less table than phenol itself.
Reason : Phenol is resonance stabilised by aromatic character

96. Assertion : Meso tartaric acid and (+)tartaric acid are diastereomers.
Reason : Mesotartaric acid and (+)tartaric acid are mirror images of each other.
97. Assertion : Melting point of trans–2–butene is higher than that of cis–2–Butene.
Reason : Dipole moment of trans–2–Butene is zero while that of cis–2–Butene is not zero.

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 ANSWERS

Only one option correct:

O O
1. (D) CH 2  N  O CH 2  N  OH

2. (D) These contain –C = C–

H

H H

3. (D)
H CH3
60º
CH3

CH3

H Cl
4. (C)
H Cl Plane of symmetry

CH3
Meso form

.
. Note: chiral centre are marked.
5. (B)

CH3 CH3 CH3

Br2 *
(B) CH3–CH–CH2–CH3  CH3–CH–CH–CH3 + CH2–CH–CH2–CH3
6. *
Br Br
H H H Br H
7. (A) C=C ; C=C Br ; C=C
Br Br Br H Br H
8. (D)

9. (D) The molecule has a plane of symmetry, so optically inactive.

Br H
Me H  Br Me
10. (D) 
Et Me HO Et
OH Me
2 exchanges on each chiral atom

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 CH3 Cl CH3
H Cl HH Cl H Cl
11. (A)  
Cl H Me
Cl H
CH3 CH3 CH 3
Presence of plane of symmetry and center of symmetry

12. (B) They are non superimposable mirror images

13. (B) Meso compounds have got the plane of symmetry.

CH3 CH2Ph CH3 H

14. (A) C=C and C=C
H H H CH2Ph

15. (C)
H2
CH3 - CH - CH = CH2 CH 3- CH - C H2 5 ( no chiral atom)
C2 H5 C2 H5

16. (A) Let, 1 mole of sucrose be taken. On hydrolysis, one mole of glucose and one mole of fructose
are formed.
Optical rotation of 2 moles of mixture = 1  52° + 1  (–92)° = –40°
40
Hence optical rotation of 1 mole of mixture =  = –20°
2
1CHO
17. (D) CH3—CH2—CH2—CH2 (Pentan-1-al)
5 4 3 2

1CHO
CH3—CH2—CH—CH3 (2-methyl butan-1-al)
4 3 2

O O—H
18. (B) CH2—CH==CH—C—H CH2==CH—CH==C—H
H
O

O
19. (D) does not undergo tautomerism as it does not have acidic hydrogen.

20. (B) Due to presence of one chiral carbon

21. (C) 2 double bond are capable of showing geometrical isomerism and one
chiral center is present
22. (C)
23. (B)
*
24. (C) CH3CH2C H–CH2–CH2–CH 3
CH3

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25. (D) Butanone (CH3COCH2CH3) and diethyl ether (C2H5OC2H5) do not have same molecular
formula and hence are not isomers.
26. (B) Both the carbon have individually two different groups and rotation is restricted due to double
bond.

27. (A) They are non super imposible mirror images

28. (D) Stability order of various conformers follows:

Anti > gauche > partially eclipsed > fully eclipsed
a – gauche
b – fully eclipsed
c – partially eclipsed
d – anti

29. (D) In the given compounds, there is restricted rotation about ‘C–C’ bond C–N bonds.

30. (C) The given compounds are CH3CH2CH(OH)CH3 and CH3CH2CH2CH2OH. They are position
isomers.

CHO CHO CHO CHO

H OH H OH First H OH Second H OH
31. (A) HO H HOH2 C OH HO CH2OH interchange HO H
CH2OH H interchange H CH2OH
(I) (II) (I)
Similarly
H CHO
HO * CHO H OH
After two
HO H interchanges at C * H OH
CH2OH CH2OH
(III) (III)
 (I) and (II) are identical; (I) and (III) are diastereomers.
32. (D) Alkyne and diene are different fuctional group

33. (D) 34. (D)

35. (C) Alcohols and aldehydes are not functional isomers due to different molecular formula

36. (C) The various isomers possible are:

nhexane, 2 methylpentane, 3methylpentane, 2,2dimethylbutane, 2,3dimethylbutane.

37. (B) The given stereoisomers are diastereomers.

38. (C) In a compound having n double bonds and two different terminals, the number of geometrical
isomers is 2n. The given compound has n = 2. Therefore, number of geometrical isomers = 22
= 4.
39. (D) No plane of symmetry and no center symmetry

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40. (D) No acidic hydrogen

41. (A) CH3CH2CH2CH2C=CCH3 does not show cis trans isomerism.

H CH3

42. (B) H+ will be replaced by D+ during enolization but not T+. This is because heavier isotope forms
stronger bond

CH2CH3 CH3 C2H5 C2H5

43. (D) Br Cl C2H5 Cl  Cl CH3  Br Cl

CH3 Br Br CH3
(1) (2) (same as 1)

C2H5 C2H5
H3C Br  Br Cl
Cl CH3
(3) (same as 1)

H3C
H H
44. (D)

H H
H3C
Minimum steric repulsion and minimum dihedral strain. Therefore, most stable conformation.
 
45. (B) H–CN C  N H (Shifting of H+ occurs)
46. (D) By rotation on the plane of the paper, (1) produces a mirror image of (2).

47. (B) It shows optical isomerism as the molecule does not possess any element of symmetry.

48. (B) Dipole moment is a vector quantity. It gets cancelled in the case of trans but2ene.

49. (A) They are non superimpasible mirror image of each other

50. (D) All enolizable H+ will be replaced by D+

51. (C) Three chiral atoms. Number of stereoisomers = 23 = 8

H
CH3 CH3
52. (C) C–C=C has got a double bond which can show geometrical isomerism and has got
C2H5 H H
a chiral carbon to show optical isomerism.
53. (D)
Cl Cl
(1) (1) (2) (1)

(2) (2) Cl
(3)
(3) (2)

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54. (D)
OH
H CH3

H CH3
OH
No intramolecular hydrogen bonding and more steric repulsion

55. (C) 56. (C)

57. (B) Due to presence of two chiral atoms
58. (A)

One or more options may correct:

59. (B, D) 60. (B, D) 61. (A, D) 62. (A, B, D) 63. (B, C)

64. (A, B, D) 65. (A, B, C, D) 66. (A, C) 67. (A, C) 68. (A, B, C)

69. (A, B, C, D) 70. (A, B) 71. (B, C, D)

Comprehension Type

72. (D) 73. (B) 74. (C) 75. (D) 76. (C)

77. (C) 78. (D) 79. (D) 80. (C) 81. (D)

Matrix Match Type

82. (A)  3 ; (B)  1, 2 ; (C)  1 ; (D)  1, 3, 4

83. (A)  1, 2, 4 ; (B)  1, 2, 3 ; (C)  3, 4 ; (D)  4
84. (A)  1, 2, 3, 4 ; (B)  1, 3, 4 ; (C)  1, 2, 3, 4 ; (D)  1
85. (A)  2, 4 ; (B)  3 ; (C)  2 ; (D)  1
86. (A)  3, 4 ; (B)  4 ; (C)  1 ; (D)  1,2,3,4
87. (A)  3 ; (B)  4 ; (C)  2 ; (D) – 1
88. (A)  2 ; (B)  1, 4 ; (C)  3 ; (D) – 1

Assertion Reasoning Type

89. (C) Mirror image of meso is the same compound.

90. (C) 91. (A) 92. (D)

93. (B) Energy released due to resonance in benzene is not sufficient to convert trione into enol form.
94. (B) 95. (A) 96. (C) 97. (B)

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