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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

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

Solution

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.

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