Question:medium

What is the first step and the final product formed in the reaction of HBr with \( CH_3-CH(CH_3)-CH=CH_2 \)?

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Always check for carbocation rearrangement (hydride/methyl shift) in alkene additions!
Updated On: Apr 22, 2026
  • Protonation at more substituted carbon; \( CH_3-C(Br)(CH_3)-CH_2-CH_3 \)
  • Protonation at less substituted carbon; \( CH_3-CH(CH_3)-CHBr-CH_3 \)
  • Radical initiation; \( CH_3-CH(CH_3)-CH_2-CH_2Br \)
  • Protonation followed by hydride shift; \( CH_3-C(Br)(CH_3)-CH_2-CH_3 \)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
Electrophilic addition of H-Br to alkenes proceeds via a carbocation intermediate. Carbocations can undergo rearrangement (like hydride or methyl shifts) to reach a more stable state before the final nucleophilic attack.
Step 2: Detailed Explanation:
The starting material is 3-methyl-1-butene: \( CH_3-CH(CH_3)-CH=CH_2 \).
Step 1: Protonation. \(H^+\) adds to the terminal \(CH_2\) (Markovnikov's rule) to form a secondary (\(2^\circ\)) carbocation: \( CH_3-CH(CH_3)-\overset{+}{C}H-CH_3 \).
Step 2: Rearrangement. This \(2^\circ\) carbocation is adjacent to a tertiary carbon. A 1,2-hydride shift occurs to form a more stable tertiary (\(3^\circ\)) carbocation: \( CH_3-\overset{+}{C}(CH_3)-CH_2-CH_3 \).
Step 3: Nucleophilic Attack. \(Br^-\) attacks the \(3^\circ\) carbocation to form the final product: 2-bromo-2-methylbutane.
The question asks for the "first step" (which produces the stable \(3^\circ\) cation after rearrangement) and the "final product".
Step 3: Final Answer:
The final product is 2-bromo-2-methylbutane, formed via a tertiary carbocation intermediate.
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