Question

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

• A. N‐Butyl bromide
• B. Isobutyl bromide
• C. sec‐Butyl bromide
• D. ter‐Butyl bromide

Answer 1

The Correct Option is D

To determine the major product of isobutane bromination under ultraviolet light at 127 °C, we need to understand the mechanism of free radical halogenation, specifically bromination.

Concept of Free Radical Bromination:

• In the presence of UV light, bromination proceeds via a free radical chain mechanism, involving initiation, propagation, and termination steps.
• In bromination, selectivity is high for highly substituted carbon atoms, meaning tertiary hydrogens have the highest reactivity compared to secondary and primary hydrogens.

Structure of Isobutane:

• Isobutane, also known as 2-methylpropane, has the molecular structure with the formula (\text{CH}_3)_3 \text{CH}.
• The carbon attached to three methyl groups is a tertiary carbon, thus having a tertiary hydrogen that is most susceptible to bromination.

Bromination Reaction:

• Under the influence of UV light, a bromine molecule splits to form bromine radicals.
• The bromine radical reacts with the tertiary hydrogen on isobutane, forming a tertiary radical.
• This tertiary radical quickly reacts with another bromine molecule, resulting in the formation of tert-butyl bromide.

Conclusion:

Given the high selectivity for the most substituted carbon atom in bromination, the major product of this reaction is tert-butyl bromide.

Explanation of Options:

• N-Butyl bromide: Not formed, as there is no linear butane segment in isobutane.
• Isobutyl bromide: Not formed, as there is no primary carbon available in isobutane.
• Sec-Butyl bromide: Not formed, as there is no secondary carbon favored over the tertiary site in isobutane.
• Tert-Butyl bromide: Correct answer, formed by the replacement of the tertiary hydrogen.

Hence, the correct answer is ter‐Butyl bromide.