See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: NaCN is a strong nucleophile (CN⁻), so it preferentially reacts via SN2 mechanism. The rate of SN2 depends on steric hindrance at the carbon bearing the leaving group: less hindered = faster SN2 rate. Step 2 - Analyze each option: (a) The structure shows Br on a primary carbon (CH2) with the adjacent carbon being a simple CH2, and the chain ending in a gem-dimethyl quaternary carbon. This is a primary alkyl bromide. The carbon bearing Br is primary (two H's), and there is no branching at the alpha carbon itself. This gives relatively low steric hindrance at the reaction site. (b) This is a secondary alkyl bromide (Br on a CH) with a methyl group also on the adjacent carbon — secondary with branching nearby, more hindered than (a). (c) This is a primary (neopentyl-type) bromide: Br-CH2-C(CH3)2-... The alpha carbon is primary (CH2) but the beta carbon is quaternary (gem-dimethyl), creating severe steric hindrance (neopentyl system). SN2 is very slow for neopentyl halides. (d) This is a tertiary alkyl bromide (Br on a carbon bearing two methyl groups and a propyl chain). Tertiary halides are the most hindered and react extremely slowly (essentially not at all) by SN2. Step 3 - Comparison: SN2 rate order is primary (unhindered) > primary (neopentyl, hindered) > secondary > tertiary. Option (a) is a primary alkyl bromide without severe beta-branching, making it the least sterically hindered and therefore fastest in SN2 with CN⁻. Step 4 - Why others fail: (d) is tertiary — no SN2; (c) is neopentyl primary — very slow SN2 due to beta quaternary carbon; (b) is secondary — slower than unhindered primary. Therefore, the correct answer is A.