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Question

(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

Vedclass · Answer

(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:

Similar Questions

In the following reaction,the product $(Y)$ will be:

The product $(C)$ in the following reaction sequence is:
$4-\text{Bromoethylbenzene}$ $\xrightarrow[\Delta]{Br_2} B$ $\xrightarrow{\text{aq. } KOH} C$

In the following pairs of halogen compounds,which compound undergoes faster $S_N 1$ reaction?

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The correct reactivity order of the following compounds towards alkaline hydrolysis is:-
$I. \ p-CH_3O-C_6H_4-CH_2Cl$
$II. \ C_6H_5-CH_2Cl$
$III. \ p-CH_3-C_6H_4-CH_2Cl$
$IV. \ m-CH_3-C_6H_4-CH_2Cl$

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