See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

The starting material in all cases is 1-(4-methylphenyl)-1-cyclopentane, which has two reactive C-H sites: (1) the benzylic CH3 group on the para position of the ring (a primary benzylic position), and (2) the tertiary benzylic C-H at the cyclopentane carbon directly attached to the ring (a tertiary benzylic position). Key concepts: - NBS (N-bromosuccinimide) is a selective benzylic brominating agent under radical conditions. - SO2Cl2 under hv (light) is a radical chlorinating agent. - Selectivity: When two benzylic positions are present, a tertiary benzylic position is more reactive than a primary benzylic position toward radical halogenation. However, NBS is specifically used for allylic/benzylic bromination and tends to react at the more reactive (tertiary) benzylic site preferentially with 1 equivalent, or both sites with 2 equivalents. - Actually, NBS with 2 equivalents brominates both benzylic positions (the CH3 → CH2Br and the tertiary C-H → C-Br). - SO2Cl2 with 2 equivalents chlorinates both benzylic positions (CH3 → CH2Cl and tertiary C-H → C-Cl). - When two different reagents are used in sequence, the first reagent reacts at the more reactive site (tertiary benzylic), and the second reagent reacts at the remaining site (primary benzylic CH3). Analysis of each conversion: (a) Product has CH2Br (from CH3) and Br at tertiary cyclopentyl carbon → both positions brominated → requires NBS (2 equivalent). Match: (q) (b) Product has CH2-Cl (from CH3) and Cl at tertiary cyclopentyl carbon → both positions chlorinated → requires SO2Cl2/hv (2 equivalent). Match: (p) (c) Product has CH2Br (from CH3) and Cl at tertiary cyclopentyl carbon → tertiary position chlorinated first, then primary benzylic brominated. SO2Cl2/hv first (chlorinates the more reactive tertiary benzylic site), then NBS (brominates the remaining primary benzylic CH3). Match: (s) (d) Product has CH2-Cl (from CH3) and Br at tertiary cyclopentyl carbon → tertiary position brominated first (NBS is selective for the more reactive tertiary benzylic C-H), then SO2Cl2/hv chlorinates the remaining primary benzylic CH3. NBS first then SO2Cl2/hv. Match: (r) Why other options fail: - Using NBS (2 eq) for (b) would give dibromide, not dichloride. - Using SO2Cl2 (2 eq) for (a) would give dichloride, not dibromide. - The sequential reagents differ in order: (r) NBS then SO2Cl2/hv gives Br at tertiary + Cl at primary = product (d); (s) SO2Cl2/hv then NBS gives Cl at tertiary + Br at primary = product (c). Therefore, the correct answer is {"a": "Q", "b": "P", "c": "S", "d": "R"}.