See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: For SN1 and SN2 reactions to give the same constitutional product (excluding stereoisomers), the carbocation intermediate in SN1 must not rearrange to give a different carbon skeleton or a different connectivity than what SN2 would give directly. Step 1: Analyze each option. Option (a): 1-chloro-2-methylcyclohexane. The Cl is on a secondary carbon adjacent to a secondary carbon bearing a methyl group. In SN1, the secondary carbocation formed at C2 could potentially shift the methyl group or hydride to give a rearranged carbocation, leading to a different product than SN2 (which gives direct substitution). So SN1 and SN2 products may differ. Option (b): 1-chloro-1-methylcyclohexane. This is a tertiary alkyl chloride. SN2 is essentially impossible at tertiary carbons due to steric hindrance, so this does not satisfy the condition of both reactions giving the same product (SN2 barely proceeds). Option (c): 1-chloro-4-methylcyclohexane. The Cl is on a secondary carbon at C1, and the methyl is at C4 (para position on the ring). In SN1, the secondary carbocation at C1 forms. There is no adjacent group capable of migrating to give a rearranged carbocation of different connectivity because the methyl at C4 is too far away and no hydride/alkyl shift leads to a more stable or differently connected carbocation. Therefore, the SN1 carbocation is attacked by the nucleophile at the same carbon where Cl left, giving the same constitutional product as SN2 (just different stereochemistry). Excluding stereoisomers, both SN1 and SN2 give the same product: 1-nucleophile-4-methylcyclohexane. Option (d): Ph-CH(CH3)-CHCl-CH3. The Cl is on a secondary carbon adjacent to a benzylic carbon. In SN1, the carbocation could form at the secondary carbon or rearrange/shift to the benzylic position (which is more stable), giving a different constitutional product than SN2 direct substitution. So products differ. Step 2: Why option (c) is correct. In 1-chloro-4-methylcyclohexane, the leaving group carbon (C1) is a simple secondary carbon with no possibility of rearrangement to a more stable or differently connected carbocation. Both SN1 (via symmetrical secondary carbocation at C1, no rearrangement) and SN2 (direct backside attack at C1) give substitution at C1, yielding the same constitutional product (excluding stereoisomers: SN2 gives inversion, SN1 gives racemization, but same connectivity). Therefore, the correct answer is C.