C_6H_5NH_2 xrightarrow{Br_2/CCl_4} ?What is t | vedclass

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(C) Aniline is highly activated towards electrophilic substitution due to the strong electron-donating effect of the $-NH_2$ group.When aniline reacts

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$o$- and $p$-bromoaniline

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(C) Aniline is highly activated towards electrophilic substitution due to the strong electron-donating effect of the $-NH_2$ group.When aniline reacts with $Br_2$ in a polar solvent like water,it forms $2, 4, 6$-tribromoaniline.However,when the reaction is carried out in a non-polar solvent like $CCl_4$ at low temperature,the activation of the benzene ring is reduced.This allows for mono-substitution,resulting in a mixture of $o$-bromoaniline and $p$-bromoaniline.

$C_6H_5NH_2 \xrightarrow{Br_2/CCl_4} ?$
What is the product?

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$C_6H_5NH_2 \xrightarrow{Br_2/CCl_4} ?$What is the product?