See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The compound is 1-(4-methylphenyl)cyclopentane, which has a para-methyl group on the benzene ring and a cyclopentane ring attached at the ipso carbon. Step 2 - Reaction with SO2Cl2 / hv (free radical chlorination): SO2Cl2 under photochemical conditions acts as a free radical chlorinating agent. Free radical halogenation occurs preferentially at the most stable radical site. The cyclopentane ring is attached to the benzene ring at a tertiary-like benzylic/tertiary carbon. The 1-position of cyclopentane (the carbon bearing the phenyl group) is both tertiary AND benzylic, making it the most stable radical site. Therefore, SO2Cl2/hv chlorinates the cyclopentane at C1 (the carbon attached to the ring). Product (A) = 1-chloro-1-(4-methylphenyl)cyclopentane. Step 3 - Reaction with NBS (N-bromosuccinimide): NBS is a benzylic/allylic brominating agent under radical or ionic conditions. In (A), the methyl group on the para position of the benzene ring is a benzylic CH3. NBS brominates the benzylic methyl group to give a benzyl bromide (CH2Br). Product (B) = 1-chloro-1-(4-(bromomethyl)phenyl)cyclopentane. Step 4 - Reaction with KSH (potassium hydrosulfide): KSH is a nucleophile (HS-). It will undergo SN2 substitution at the more reactive/less hindered site. The benzyl bromide (CH2Br, primary benzylic) is much more reactive toward SN2 than the tertiary C-Cl at the cyclopentane ring (which would be highly hindered for SN2). Therefore, KSH selectively displaces the benzylic bromide to give CH2SH, while the C-Cl at the tertiary cyclopentane position remains intact. Product (C) = 1-chloro-1-(4-(mercaptomethyl)phenyl)cyclopentane, which corresponds to option (b): para-substituted benzene with CH2SH on the methyl-derived position and Cl at the cyclopentane quaternary carbon. Why other options fail: - (a) has Br at cyclopentane and SH at benzylic: this would require KSH to react at the tertiary C-Cl (SN2 disfavored) and leave the primary benzyl bromide intact - incorrect. - (c) has CH2Br remaining and SH at cyclopentane: this would require SN2 at tertiary carbon over primary benzylic - incorrect. - (d) shows elimination product - not consistent with KSH as nucleophile under these conditions - incorrect. Therefore, the correct answer is B.