Identify the major product (99 %) formed when | vedclass

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(B) The reaction of $(CH_3)_3CH$ (isobutane) with $Br_2$ in the presence of $UV$ light is a free radical substitution reaction.Bromination is highly s

Vedclass · Accepted Answer

$(CH_3)_3CBr$

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(B) The reaction of $(CH_3)_3CH$ (isobutane) with $Br_2$ in the presence of $UV$ light is a free radical substitution reaction.Bromination is highly selective compared to chlorination.The stability of the intermediate free radical determines the major product.The tertiary free radical $(CH_3)_3C^{\bullet}$ is more stable than the primary free radical $(CH_3)_2CHCH_2^{\bullet}$.Therefore,the bromine atom replaces the hydrogen atom at the tertiary carbon to form $2-$bromo$-2-$methylpropane as the major product $(99 \%)$.

Identify the major product $(99 \%)$ formed when $(CH_3)_3CH$ is treated with $Br_2$ in $UV$ light.

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