Question:medium

Isobutane upon bromination under the influence of ultraviolet light at 127 °C affords the following major product:

Updated On: Mar 5, 2026
  • N‐Butyl bromide
  • Isobutyl bromide
  • sec‐Butyl bromide
  • ter‐Butyl bromide
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The Correct Option is D

Solution and Explanation

Bromination of isobutane in the presence of ultraviolet light is a classic example of a free radical halogenation reaction. The process involves replacing a hydrogen atom with a bromine atom.

Let's break down the reaction step-by-step:

  1. Formation of Bromine Radicals: Ultraviolet light promotes the homolytic cleavage of bromine molecules, creating bromine radicals.
    Br2 → 2 Br•
  2. Hydrogen Abstraction: The bromine radical can abstract a hydrogen atom from isobutane, forming an isobutane radical.
    (CH3)3CH + Br• → (CH3)3C• + HBr
  3. Recombination: The isobutane radical can then react with another bromine molecule to form the brominated product.
    (CH3)3C• + Br2 → (CH3)3CBr + Br•

The main site of hydrogen abstraction in isobutane involves the tertiary carbon atom, as tertiary free radicals (those formed at the tertiary carbon center, i.e., where the carbon is attached to three other carbons) are more stable than primary or secondary radicals. This is due to hyperconjugation and greater alkyl group stabilization.

Therefore, the major product of bromination of isobutane is tert-Butyl bromide. The options can be justified and other choices ruled out due to the stability of the tertiary radical that forms the major product.

Conclusion: The correct answer is terr-Butyl bromide as it involves bromination at the most stable tertiary position in isobutane.

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