See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: The rate of an S_N1 reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group. The more stable the carbocation, the faster the S_N1 reaction. Step 1 - Analyze each option: (a) Bromocyclohexane: Ionization gives a secondary cyclohexyl carbocation (2° carbocation). Moderately stable. (b) 3-Bromocyclohex-1-ene: Ionization gives an allylic carbocation. The carbon bearing Br is allylic (adjacent to the C=C double bond). Loss of Br- generates a secondary allylic carbocation, which is stabilized by resonance delocalization over the double bond. This makes it significantly more stable than a simple secondary carbocation. (c) (Bromomethyl)cyclohexane: Ionization gives a primary carbocation (-CH2+), which is the least stable. Primary carbocations are very unstable and S_N1 is disfavored. (d) 1-Bromonorbornane (bridgehead): Ionization would give a bridgehead carbocation. Due to Bredt's rule, bridgehead carbocations in small bicyclic systems are extremely unstable because the geometry does not allow the required planar sp2 hybridization at the bridgehead. This compound is the least reactive in S_N1. Step 2 - Compare stabilities: - Option (b) gives allylic (resonance-stabilized) secondary carbocation — most stable among these options. - Option (a) gives simple secondary carbocation — moderately stable. - Option (c) gives primary carbocation — very unstable. - Option (d) gives bridgehead carbocation — essentially cannot form (Bredt's rule violation). Step 3 - Conclusion: The allylic stabilization in option (b) makes its carbocation intermediate the most stable, so it undergoes S_N1 most rapidly. Therefore, the correct answer is B.