The product C of the following reaction seque | vedclass

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(D) $1$. Aniline reacts with $Br_2/H_2O$ to form $2,4,6$-tribromoaniline as the major product $(A)$.$2$. $2,4,6$-tribromoaniline reacts with $NaNO_2/H

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$1,3,5-$Tribromo$-2-$nitrobenzene

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(D) $1$. Aniline reacts with $Br_2/H_2O$ to form $2,4,6$-tribromoaniline as the major product $(A)$.$2$. $2,4,6$-tribromoaniline reacts with $NaNO_2/HCl$ at $273-278 \ K$ to form the diazonium salt,$2,4,6$-tribromobenzenediazonium chloride $(B)$.$3$. The reaction of the diazonium salt with $HBF_4$ followed by $NaNO_2/Cu, \Delta$ is a variation of the Balz-Schiemann or Sandmeyer-type reaction where the diazonium group $(-N_2^+Cl^-)$ is replaced by a nitro group $(-NO_2)$.$4$. Thus,the final product $C$ is $1,3,5$-tribromo$-2-$nitrobenzene.

The product $C$ of the following reaction sequence is:

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