See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: SN2 reactivity depends on steric accessibility at the carbon bearing the leaving group, the nature of the leaving group, and electronic effects. Less hindered (primary > secondary > tertiary) carbons react faster in SN2. Also, allylic/benzylic systems are more reactive than simple alkyl halides due to partial positive charge stabilization in the transition state. Vinyl and aryl halides are extremely unreactive toward SN2 due to the sp2 carbon and back-lobe inaccessibility. Analysis of each option: (a) (CH3)2CHBr (secondary) vs CH3-CH2-CH2-Br (primary): The primary alkyl halide is less hindered and MORE reactive in SN2 than the secondary. So the FIRST compound is LESS reactive, not more. This option does NOT satisfy the condition. (b) CH3-CH2-CH2-Br (primary bromide) vs CH3-CH2-CH2-I (primary iodide): Iodide is a better leaving group than bromide (I- is more stable/polarizable), so the propyl iodide is MORE reactive in SN2. The first compound is LESS reactive. This option does NOT satisfy the condition. (c) Ph-Br (aryl halide) vs CH3-CH2-CH2-Br (primary alkyl bromide): Aryl halides are essentially inert toward SN2 (sp2 carbon, back-side attack impossible). So Ph-Br is LESS reactive. This option does NOT satisfy the condition. (d) CH2=CH-CH2-Cl (allyl chloride, primary allylic) vs H2C=CH-Cl (vinyl chloride): Allyl chloride has the leaving group on an sp3 carbon that is primary and allylic, making it highly reactive toward SN2 (allylic activation lowers activation energy). Vinyl chloride has the leaving group on an sp2 carbon; back-side attack is geometrically impossible and the C-Cl bond has partial double bond character due to resonance, making it extremely unreactive toward SN2. Therefore, the first compound (allyl chloride) is MORE reactive than the second (vinyl chloride) in SN2. Therefore, the correct answer is D.