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(A) The reaction of ethylbenzene with $Br_2$ in the presence of $UV$ light is a free radical substitution reaction. The bromine radical abstracts a hy

Vedclass · Accepted Answer

$1-$bromo$-1-$phenylethane

Vedclass · Answer

(A) The reaction of ethylbenzene with $Br_2$ in the presence of $UV$ light is a free radical substitution reaction. The bromine radical abstracts a hydrogen atom from the benzylic carbon because the resulting benzylic radical is resonance-stabilized by the phenyl ring. The radical formed at the $\alpha$-carbon (benzylic position) is more stable than the radical at the $\beta$-carbon. Therefore,the major product is $1$-bromo-$1$-phenylethane $(C_6H_5CH(Br)CH_3)$.

The major product of the following reaction is:

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