See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: In SN1 reactions, the rate-determining step is the formation of a carbocation intermediate. The more stable the carbocation formed, the faster the SN1 reaction proceeds. Analysis of Compound A: Compound A is 3-bromocyclopentene (an allylic bromide). When the Br leaves, it generates an allylic carbocation stabilized by resonance over the cyclopentene double bond. This provides moderate stabilization. Analysis of Compound B: Compound B is a 2-bromo-2H-pyran (or similar structure - a bromide on the sp3 carbon of a dihydropyran ring containing an oxygen atom). When Br leaves from this position, it generates a carbocation that is BOTH allylic (stabilized by the adjacent double bond via resonance) AND adjacent to the ring oxygen. The oxygen lone pairs can donate electron density directly into the carbocation center (oxocarbenium ion stabilization), providing an additional resonance contributor. This means the carbocation from B is stabilized by BOTH allylic resonance AND lone pair donation from oxygen (forming an oxocarbenium ion), making it far more stable than the purely allylic carbocation from A. Comparison: The carbocation intermediate from B is stabilized by two resonance effects (vinyl/allylic + oxygen lone pair donation), while the carbocation from A is stabilized only by allylic resonance. Therefore, B forms a more stable carbocation intermediate, lowering the activation energy for the rate-determining ionization step, and B undergoes SN1 hydrolysis faster than A. Why other options fail: - (a) A reacts faster than B: Incorrect, because A forms only an allylic cation while B forms an oxocarbenium-allylic cation (more stable). - (c) Same rate: Incorrect, because the two carbocations have different stabilities. - (d) Neither reacts: Incorrect, both are capable of SN1 reaction as they can form resonance-stabilized carbocations. Therefore, the correct answer is B.