Back to Directory
NEET CHEMISTRYMedium

Consider the following reaction and identify the product (P): [Specific Reaction Image Missing - Likely Reaction of 3-Methylbutan-2-ol with HBr]

A

2-Bromo-3-methylbutane

B

2-Bromo-2-methylbutane

C

2-Methylbut-2-ene

D

1-Bromo-2-methylbutane

Step-by-Step Solution

The reaction of a secondary alcohol with a hydrogen halide (like HBr) proceeds via an SN1S_N1 mechanism involving a carbocation intermediate:

  1. Protonation: The -OH group accepts a proton (H+H^+) to form water (H2OH_2O), which is a good leaving group.
  2. Carbocation Formation: Water leaves, generating a secondary (22^{\circ}) carbocation (e.g., from 3-methylbutan-2-ol, a 22^{\circ} carbocation is formed).
  3. Rearrangement: A less stable carbocation rearranges to a more stable one if possible. Here, a 1,2-hydride shift occurs, converting the secondary carbocation into a more stable tertiary (33^{\circ}) carbocation.
  4. Nucleophilic Attack: The bromide ion (BrBr^-) attacks the stable tertiary carbocation to form the major product, 2-Bromo-2-methylbutane (Tertiary Alkyl Halide).

Direct attack on the secondary carbocation yields the minor product.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started