Excess of isobutane on reaction with Br_{2} i | vedclass

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Question

(B) The bromination of alkanes is highly selective. The stability of the free radical intermediate formed during the reaction determines the major pro

Vedclass · Accepted Answer

$2-$bromo$-2-$methylpropane

Vedclass · Answer

(B) The bromination of alkanes is highly selective. The stability of the free radical intermediate formed during the reaction determines the major product. In isobutane,the tertiary radical $((CH_{3})_{3}C^{\bullet})$ is more stable than the primary radical $((CH_{3})_{2}CH-CH_{2}^{\bullet})$. Thus,$2-$bromo$-2-$methylpropane is formed as the major product.

Excess of isobutane on reaction with $Br_{2}$ in presence of light at $125^{\circ} C$ gives which one of the following,as the major product?

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