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. Stability order of carbocations: tertiary > secondary > primary. Additionally, for the same degree of substitution, a better leaving group (I > Br > Cl > F) increases SN1 rate. Analysis of each option: (a) CH3CH2CH2CH2Br (1° alkyl bromide) vs CH3CH2CH(Br)CH3 (2° alkyl bromide): The second compound (2°) would form a more stable carbocation and react faster in SN1. So the first compound does NOT react faster. This option is incorrect. (b) CH3-C(CH3)(Br)-CH2-CH2 — this is a neopentyl-type structure but actually the first compound is CH3-C(CH3)(Br)-CH2-CH2, which upon closer examination is a tertiary bromide (the carbon bearing Br has: CH3, CH3, CH2CH2, and Br — wait, re-reading: CH3-C(CH3)(Br)-CH2-CH2 means the carbon bearing Br has two CH3 groups and a CH2CH2 chain, making it tertiary). The second compound CH3-CH(Br)-CH(CH3)-CH3 is a secondary bromide. A tertiary carbocation is more stable than a secondary carbocation, so the first compound (tertiary bromide) reacts faster in SN1 than the second compound (secondary bromide). This is consistent with the first reacting faster. (c) Bromocyclohexane (secondary cyclic) vs 3-bromocyclohex-2-ene (allylic secondary): The allylic bromide would form an allylic carbocation which is resonance-stabilized, making it MORE stable and faster in SN1 than the simple secondary cyclohexyl bromide. So the second compound reacts faster, not the first. This option is incorrect. (d) CH3-C(CH3)(Br)-CH3 (tertiary bromide) vs CH3-C(CH3)(I)-CH3 (tertiary iodide): Both are tertiary, but iodide is a better leaving group than bromide. Therefore, the iodide (second compound) reacts faster in SN1. The first compound does NOT react faster. This option is incorrect. Conclusion: Only in option (b) does the first compound (tertiary bromide) react faster than the second compound (secondary bromide) in an SN1 reaction, because the tertiary carbocation intermediate is more stable than the secondary carbocation intermediate. Therefore, the correct answer is B.