See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: SN2 reactivity depends on the degree of substitution at the carbon bearing the leaving group. Under SN2 conditions, the reaction proceeds faster at less hindered (lower substitution) carbon centers, and Br- is the nucleophile provided by LiBr in DMSO (a polar aprotic solvent that enhances nucleophilicity of Br-). Step 1 - Identify the two electrophilic sites in the starting material: - Site 1: The tertiary carbon of the cyclopentane ring bearing Cl (tertiary alkyl chloride, C1 of ring). - Site 2: The terminal carbon of the -CH2CH2Cl side chain bearing Cl (primary alkyl chloride). Step 2 - Apply SN2 selectivity: SN2 reactions strongly favor primary over secondary over tertiary carbons because steric hindrance disfavors attack at hindered centers. The tertiary C-Cl on the ring is essentially inert to SN2 due to extreme steric crowding (tertiary + part of a ring). The primary C-Cl at the end of the -CH2CH2Cl chain is a typical primary alkyl chloride, highly reactive toward SN2. Step 3 - Predict the product: Br- (from LiBr) attacks the primary carbon of the -CH2CH2Cl group via SN2, displacing Cl-. The tertiary C-Cl on the ring remains unreacted. This gives a product where the ring still bears Cl at C1, and the side chain is now -CH2CH2Br. Step 4 - Match to options: - (a) Br on ring, Cl on chain end - wrong (tertiary SN2 does not occur preferentially) - (b) Cl on ring, Br on chain end - correct (primary SN2 only) - (c) Br on ring and Br on chain - wrong (both substitutions would require tertiary SN2) - (d) Cl on ring, vinyl group - wrong (this would be elimination, not SN2 under these conditions, and an unlikely pathway) Therefore, the correct answer is B.