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(D) The photobromination of alkanes involves the formation of free radicals. The stability of free radicals follows the order: $3^\circ > 2^\circ > 1^

Vedclass · Accepted Answer

$2$-bromo-$2$-methylbutane

Vedclass · Answer

(D) The photobromination of alkanes involves the formation of free radicals. The stability of free radicals follows the order: $3^\circ > 2^\circ > 1^\circ$. In $2$-methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$,there are four types of hydrogen atoms. Abstraction of the hydrogen atom from the $C-2$ position leads to a $3^\circ$ free radical,which is the most stable. Therefore,the reaction proceeds via the $3^\circ$ free radical intermediate to form $2$-bromo-$2$-methylbutane as the major product.

Which product is the major product in the photobromination of $2$-methylbutane?

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