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(B) The reaction proceeds in two steps:$1$. Aniline reacts with bromine water $(Br_2/H_2O)$ to form $2,4,6$-tribromoaniline as product $(A)$ due to th

Vedclass · Accepted Answer

$1,3,5-$tribromobenzene

Vedclass · Answer

(B) The reaction proceeds in two steps:$1$. Aniline reacts with bromine water $(Br_2/H_2O)$ to form $2,4,6$-tribromoaniline as product $(A)$ due to the strong activating effect of the $-NH_2$ group.$2$. $2,4,6$-tribromoaniline is then treated with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ to form the diazonium salt,which is subsequently reduced by $H_3PO_2$ and $H_2O$ to remove the $-NH_2$ group,yielding $1,3,5$-tribromobenzene as product $(B)$.

Product $(B)$ in this reaction is

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