When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: CH₃–CH(OH)–CH(CH₃)–CH₃ + HBr → CH₃–CBr(CH₃)–CH₂–CH₃. Give a mechanism for this reaction. (Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
The reaction proceeds through a carbocation rearrangement mechanism involving the following steps:
Step I: Protonation of Alcohol 3-methylbutan-2-ol is protonated by HBr to form an oxonium ion. The hydroxyl group gets protonated, making it a better leaving group.
Step II: Formation of Secondary Carbocation The oxonium ion loses a water molecule to form a secondary carbocation (2° carbocation) at the second carbon atom.
Step III: Carbocation Rearrangement The secondary carbocation is less stable. A hydride ion shift occurs from the 3rd carbon atom to the 2nd carbon atom. This rearrangement converts the secondary carbocation into a more stable tertiary carbocation (3° carbocation).
Step IV: Nucleophilic Attack The bromide ion (Br⁻) attacks the tertiary carbocation to form the final product, 2-bromo-2-methylbutane (CH₃–CBr(CH₃)–CH₂–CH₃).
This rearrangement occurs because tertiary carbocations are more stable than secondary carbocations due to greater hyperconjugation and inductive effects.
Explanation
This question tests understanding of nucleophilic substitution reactions and carbocation stability. The key concept is that carbocations can rearrange to form more stable intermediates. The textbook hint explicitly guides students to show the hydride shift from the 3rd carbon atom, converting the secondary carbocation to a tertiary one before the final product formation.
Solution Steps
Step 1: Protonation of -OH group by H⁺ from HBr
Step 2: Loss of H₂O to form secondary carbocation
Step 3: Hydride shift from C-3 to C-2 forming tertiary carbocation
Step 4: Attack by Br⁻ to form the alkyl bromide product