See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: Both reactions involve 3-bromocyclohex-1-ene (an allylic bromide). Allylic substrates are reactive in both SN1 and SN2 pathways because the allylic carbocation (for SN1) is stabilized by resonance, and the allylic position is also accessible for backside attack (SN2). Step 2 - Analyze Reaction I: The reagent is CH3O(minus) at HIGH concentration in CH3OH solvent. A strong nucleophile (methoxide) at high concentration favors a bimolecular mechanism. The rate depends on both nucleophile and substrate concentration. This points to SN2. Even though the substrate is secondary (allylic), with a strong, high-concentration nucleophile, SN2 is preferred over SN1. Step 3 - Analyze Reaction II: The reagent is CH3OH (methanol) alone, which is a weak nucleophile and a polar protic solvent. Weak nucleophiles in polar protic solvents favor SN1. The allylic bromide can form a stabilized allylic carbocation intermediate, making SN1 feasible. The absence of a strong nucleophile and presence of a polar protic solvent (CH3OH) supports SN1. Step 4 - Match to options: Reaction I = SN2, Reaction II = SN1. This corresponds to option (c) SN2, SN1. Step 5 - Why other options fail: - (a) SN1, SN1: Reaction I uses strong nucleophile at high concentration, which is inconsistent with SN1. - (b) SN1, SN2: Reaction I should be SN2 (strong nucleophile), and Reaction II should be SN1 (weak nucleophile), so this is reversed. - (d) SN2, SN2: CH3OH alone is too weak a nucleophile to drive SN2 in Reaction II. Therefore, the correct answer is C.