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(B) The reaction of $1$-ethyl-$4$-nitrobenzene with $Br_2$ in the presence of light $(h
u)$ proceeds via a free radical mechanism. Free radical bromi

Vedclass · Accepted Answer

$1-$($4$-nitrophenyl)$-1-$bromoethane

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(B) The reaction of $1$-ethyl-$4$-nitrobenzene with $Br_2$ in the presence of light $(h
u)$ proceeds via a free radical mechanism. Free radical bromination occurs preferentially at the benzylic position because the resulting benzylic radical is stabilized by resonance with the benzene ring. In $1$-ethyl-$4$-nitrobenzene,the benzylic carbon is the $CH_2$ group attached to the benzene ring. Therefore,the bromine atom replaces one of the hydrogen atoms on the benzylic carbon,leading to the formation of $1$-($4$-nitrophenyl)-$1$-bromoethane as the major product.

The major product obtained in the following reaction is:

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