See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: SN2 reactivity depends on steric hindrance at and around the carbon bearing the leaving group. Primary carbons react fastest, secondary slower, tertiary essentially do not react. Even primary carbons can be severely hindered if bulky groups are nearby (beta-branching effect). Step 1 - Analyze option (a): First compound is n-butyl chloride (primary, unhindered) and second is 2-chloro-2-methylbutane (secondary, more hindered). The second is LESS reactive. Option (a) is incorrect. Step 2 - Analyze option (b): First compound is (chloromethyl)cyclohexane - primary carbon attached to cyclohexane ring, relatively accessible. Second compound is chlorocyclohexane - secondary carbon, more hindered. The second is LESS reactive. Option (b) is incorrect. Step 3 - Analyze option (c): First compound is isobutyl bromide (1-bromo-2-methylpropane) - primary carbon with one beta-methyl branch, moderately hindered. Second compound is neopentyl bromide (1-bromo-2,2-dimethylpropane) - primary carbon but with two methyl groups on the beta carbon (gem-dimethyl), severely hindered. The second is LESS reactive. Option (c) is incorrect. Step 4 - Analyze option (d): First compound has Cl on a carbon that is directly adjacent to a quaternary/heavily branched center (secondary or primary carbon with extreme beta-branching), making it very hindered. The second compound is 1-chloro-3,3-dimethylbutane (primary Cl on C1, with the gem-dimethyl quaternary center at C3, one carbon further away - gamma position). Although still hindered, the bulky group is one carbon further from the reaction site compared to the first compound, meaning the second compound experiences less steric hindrance than the first toward SN2. Therefore the second compound is MORE reactive than the first in SN2. Option (d) is correct. Why other options fail: In (a), (b), and (c) the second compound is more sterically hindered than the first, making it less reactive in SN2. Only in (d) does moving the bulky neopentyl-type group farther from the electrophilic carbon result in the second compound being more reactive than the first. Therefore, the correct answer is D.