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Given the following alkyl bromides: (A) CH3–CH2–CH2–CH2–Br (B) CH2=CH–CH(Br)–CH3 (C) CH3–CH(Br)–CH3 Which sequence correctly represents their order of S_N1 reactivity?

1. B > C > A
2. B > A > C
3. C > B > A
4. A > B > C

Correct answer: B > C > A

Solution

SN1 rate follows carbocation stability. (B) CH2=CH-CHBr-CH3 gives a resonance-stabilised allylic secondary cation (most stable), (C) isopropyl is a simple secondary cation, and (A) n-butyl is primary. Order: B > C > A.

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