See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: SN2 reaction rate depends on (1) the leaving group ability and (2) steric accessibility of the electrophilic carbon. Step 1 - Identify leaving groups: (a) OMe on a tertiary carbon - methoxy is a very poor leaving group (MeO- is a strong base/poor leaving group), AND the carbon is tertiary (highly hindered). Extremely slow for SN2. (b) OTs on a secondary cyclopentyl carbon - tosylate is an excellent leaving group, but the carbon is secondary and part of a ring (moderate steric hindrance). (c) OMe on a primary carbon - methoxy is a poor leaving group even on a primary carbon. Very slow. (d) OTs on a primary carbon (n-propyl tosylate) - tosylate is an excellent leaving group AND the carbon is primary (least hindered). This combination is ideal for SN2. Step 2 - Compare options: - Tosylate (OTs) is one of the best leaving groups because the negative charge is delocalized over the sulfonate group, making it a stable, weak base after departure. - Primary substrates are the most reactive in SN2 because there is minimal steric hindrance for backside attack by the nucleophile (CN-). - Option (d) combines both advantages: primary carbon + excellent leaving group (OTs). - Option (b) has OTs but is secondary and cyclic (more hindered than primary). - Options (a) and (c) have OMe, a poor leaving group, so they react very slowly regardless of substrate class. Step 3 - Why other options fail: (a) Tertiary carbon + poor leaving group = negligible SN2 rate. (b) Secondary cyclopentyl + good leaving group = moderate rate, but slower than primary. (c) Primary carbon + poor leaving group = slow due to leaving group inability. (d) Primary carbon + best leaving group = fastest SN2 rate among all options. Therefore, the correct answer is D.