See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: SN2 reactivity depends on the nucleophilicity of the attacking species. A stronger/better nucleophile leads to a faster SN2 reaction. Step 1: Identify what changes between the first and second reaction in each option. Option (a): First reaction uses EtO^- (ethoxide, a strong nucleophile); second reaction uses EtOH (ethanol, a neutral, weak nucleophile). The second reaction is LESS reactive than the first toward SN2. This does not satisfy the condition. Option (b): First reaction uses EtO^- (oxygen nucleophile); second reaction uses EtS^- (sulfur nucleophile). In SN2 reactions, nucleophilicity follows polarizability trends down a group. Sulfur is more polarizable than oxygen, making EtS^- a better nucleophile than EtO^- in SN2 reactions (especially in protic or aprotic solvents). Therefore, the second reaction (with EtS^-) is MORE reactive toward SN2 than the first (with EtO^-). This satisfies the condition. Option (c): Both reactions use the same nucleophile (CH3O^-) and the same substrate (Et-Cl); only the labels (1m) and (2m) are swapped. The substrate and nucleophile are essentially the same in both, so one is not inherently more reactive than the other in a meaningful distinction for SN2 purposes as described. This does not clearly satisfy the condition. Option (d): First reaction uses Ph3P (phosphorus nucleophile, very good SN2 nucleophile due to high polarizability); second reaction uses Ph3N (nitrogen nucleophile, less polarizable, sterically hindered, poorer SN2 nucleophile). The second reaction is LESS reactive than the first. This does not satisfy the condition. Step 2: Conclude. Only option (b) correctly shows that the second reaction (using EtS^-, a softer, more polarizable, better SN2 nucleophile) is more reactive toward SN2 than the first reaction (using EtO^-). Therefore, the correct answer is B.