JEE Advanced Chemistry: Stereochemistry / Isomerism questions with solutions

Q1. Among the following compounds, how many can exhibit geometric (cis-trans) isomerism? (i) HO-C(CH3)=CH-C(F)(CH3)2 (ii) CH3-C(CH3)=C(F)(CH3) (iii) CH3-N(=O)-OH (iv) H-N-C(CH3)=CH-CH(CH3)2 (v) CH3-C(=O)Cl (vi) (CH3)2C=CH-CH3 (vii) A cyclohexane ring bearing two different substituents on the two carbons of an endocyclic C=C bond (viii) CH3-C(CH3)=C(CH2CH3)2 (ix) CH3-C(Cl)=C(CH3)2

1. 1
2. 2
3. 3
4. 4

Answer: 4

(i) C(CH3)(OH)=CH-: left C bears OH and CH3 (different); right C bears H and C(F)(CH3)2 (different) -> YES. (ii) C(CH3)(CH3)=: two identical CH3 on left C -> NO. (iii) CH3-N(=O)-OH: N=O with N carrying CH3 and OH (different) -> shows syn/anti isomerism -> YES. (iv) C(CH3)=CH-: left C bears CH3 and N-group (different); right C bears H and CH(CH3)2 (different) -> YES. (v) C=O: one side is only O -> NO. (vi) C(CH3)2=: two CH3 on one carbon -> NO. (vii) Two different substituents on each double-bond carbon in the ring -> YES. (viii) C=C(CH2CH3)2: right C has two identical ethyl groups -> NO. (ix) C=C(CH3)2: right C has two CH3 -> NO. Count: (i), (iii), (iv), (vii) = 4.

Q2. Which of the following compounds will show geometrical (cis-trans) isomerism?

1. (A) A cyclopentane ring with identical T substituents on two adjacent upper carbons and a geminal dichloro group (Cl, Cl) on the topmost carbon.
2. (B) A cyclohexene ring with identical D substituents on the two carbon atoms adjacent to and on the same side as the double bond.
3. (C) A spiro arrangement of four cyclopropane rings with a central cyclopropane connected to three others at its vertices.
4. (D) A cyclohexadiene ring with identical D substituents on two adjacent carbons and two double bonds within the ring.

Answer: (B) A cyclohexene ring with identical D substituents on the two carbon atoms adjacent to and on the same side as the double bond.

In cyclohexene (option B), the two sp3 carbons each carry one D and one H substituent; since neither carbon has two identical groups, restricted rotation in the ring fixes their relative orientation, giving rise to cis and trans isomers. The geminal dichloro group in option A prevents G.I. at that center; the spiro polycyclopropane in option C has no appropriate stereocenters; the diene in option D places the D-substituted carbons on sp2 centers that cannot show classical cis-trans ring isomerism.

Q3. Match the following cyclohexane-based compounds (P-S) with their isomerism properties (1-4). Compound P: 1,4-disubstituted cyclohexane bearing a formyl group (-CHO) at C1 and a phenyl group (Ph) at C4. Compound Q: Cyclohexane bearing CH3 and CHO on the same carbon (C1), and a phenyl group (Ph) on another ring carbon. Compound R: Cyclohexane bearing a formyl group (-CHO) at C1 and a phenyl group (Ph) at C3. Compound S: 4-phenylcyclohexan-1-one (cyclohexanone with Ph at the para position). Isomerism types: (1) Shows tautomerism and optical isomerism but not geometrical isomerism (2) Shows tautomerism and geometrical isomerism but not optical isomerism (3) Shows geometrical isomerism and optical isomerism but not tautomerism (4) Shows tautomerism, geometrical isomerism, and optical isomerism

1. (A) P-2, Q-1, R-3, S-4
2. (B) P-2, Q-3, R-4, S-1
3. (C) P-3, Q-2, R-4, S-1
4. (D) P-2, Q-3, R-1, S-4

Answer: (D) P-2, Q-3, R-1, S-4

P: 1,4-positions give cis/trans (geometrical) isomers and CHO allows tautomerism; both cis and trans forms have planes of symmetry, so no optical isomers -> matches (2). Q: gem-disubstituted C1 (CH3+CHO) with Ph on another carbon; no enolisable alpha-H at C1 for typical tautomerism consideration here; geometrical (cis/trans Ph) and optical (chiral C with 4 different groups) -> matches (3). R: 1,3-substitution gives both geometrical and optical isomers; CHO gives tautomerism -> matches (4). S: 4-phenylcyclohexanone has tautomerism (alpha-H) and a chiral centre at C4 bearing Ph (if considered with ring asymmetry) giving optical isomers; symmetric para relationship means no classical geometric isomerism -> matches (1). Best answer: D.