JEE Chemistry Isomerism MCQs Set 09

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Multiple Choice Questions

Question 1: Identify the compound that exhibits tautomerism:
(a) 2-butene
(b) lactic acid
(c) 2-pentanone
(d) 2-methylpropane
Show Answer & Explanation

Answer: (c) 2-pentanone

Explanation:
1. Keto-enol tautomerism needs a carbonyl compound with an α-hydrogen.
2. 2-Pentanone has α-hydrogens and interconverts with its enol form; the other compounds have no such C=O with α-H.

Teacher's Note:
1. Tautomers are rapidly interconverting constitutional isomers, e.g. keto and enol forms.

Question 2: Out of the following, the alkene that exhibits optical isomerism is:
(a) 3-methyl-2-pentene
(b) 4-methyl-1-pentene
(c) 3-methyl-1-pentene
(d) 2-methyl-2-pentene
Show Answer & Explanation

Answer: (c) 3-methyl-1-pentene

Explanation:
1. Optical isomerism requires a chiral carbon with four different groups.
2. In 3-methyl-1-pentene, C3 carries H, CH₃, C₂H₅ and CH=CH₂, so it is chiral.

Teacher's Note:
1. Look for a carbon bonded to four different groups.

Question 3: The alkene that exhibits geometrical isomerism is:
(a) propene
(b) 2-methylpropene
(c) 2-butene
(d) 2-methyl-2-butene
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Answer: (c) 2-butene

Explanation:
1. Geometrical isomerism needs two different groups on each doubly bonded carbon.
2. Only 2-butene (CH₃–CH=CH–CH₃) satisfies this; the others have two identical groups on one carbon.

Teacher's Note:
1. Each C of the C=C must carry two different groups.

Question 4: The number of stereoisomers possible for a compound of the molecular formula CH₃–CH=CH–CH(OH)–CH₃ is:
(a) 3
(b) 2
(c) 4
(d) 6
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Answer: (c) 4

Explanation:
1. The C=C has different groups on each carbon, so there is cis and trans geometry.
2. C4 (bearing OH) is a chiral centre, giving R and S forms.
3. Total = 2 × 2 = 4 stereoisomers.

Teacher's Note:
1. Count stereo-elements: one double bond and one chiral centre give 2² isomers.

Question 5: Which type of isomerism is shown by 2,3-dichlorobutane?
(a) Structural
(b) Geometric
(c) Optical
(d) Positional only
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Answer: (c) Optical

Explanation:
1. CH₃–CHCl–CHCl–CH₃ has two chiral carbons (C2 and C3).
2. It therefore shows optical isomerism (a pair of enantiomers and a meso form).

Teacher's Note:
1. Chiral carbons give optical isomerism.

Question 6: Which of the following compounds is not chiral?
(a) 1-chloropentane
(b) 2-chloropentane
(c) 1-chloro-2-methylpentane
(d) 3-chloro-2-methylpentane
Show Answer & Explanation

Answer: (a) 1-chloropentane

Explanation:
1. In 1-chloropentane no carbon has four different groups.
2. 2-Chloropentane, 1-chloro-2-methylpentane and 3-chloro-2-methylpentane each have a chiral carbon.

Teacher's Note:
1. A chiral carbon needs four different substituents.

Question 7: Which of the following will have a meso-isomer also?
(a) 2-chlorobutane
(b) 2,3-dichlorobutane
(c) 2,3-dichloropentane
(d) 2-hydroxypropanoic acid
Show Answer & Explanation

Answer: (b) 2,3-dichlorobutane

Explanation:
1. A meso form needs two similar chiral centres and an internal plane of symmetry.
2. 2,3-Dichlorobutane has two identical chiral carbons, so it has d, l and meso forms. The others have only one chiral centre or two dissimilar ones.

Teacher's Note:
1. Meso compounds contain identical chiral centres related by a symmetry plane.

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Isomerism Objective Questions & Solutions for JEE Chemistry

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FAQs

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