The major product obtained in the photobromin | vedclass

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Question

(A) The photobromination of $2-$methylbutane involves the formation of a free radical intermediate.$CH_3-CH(CH_3)-CH_2-CH_3 \xrightarrow[Br_2]{hv} CH_

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$2-$bromo$-2-$methylbutane

Vedclass · Answer

(A) The photobromination of $2-$methylbutane involves the formation of a free radical intermediate.$CH_3-CH(CH_3)-CH_2-CH_3 \xrightarrow[Br_2]{hv} CH_3-C(Br)(CH_3)-CH_2-CH_3$ ($2-$bromo$-2-$methylbutane).The reactivity of hydrogen atoms towards free radical substitution follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.Since $2-$methylbutane has one tertiary $(3^{\circ})$ hydrogen atom at the $C-2$ position,the abstraction of this hydrogen leads to the most stable tertiary free radical.Therefore,$2-$bromo$-2-$methylbutane is the major product.

The major product obtained in the photobromination of $2-$methylbutane is

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