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(C) Step $1$: Reaction of $1-chloro-4-nitrobenzene$ with $aq. NH_3$ undergoes nucleophilic aromatic substitution $(S_NAr)$ to form $4-nitroaniline$ as

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$1,3-$dibromo$-5-$nitrobenzene

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(C) Step $1$: Reaction of $1-chloro-4-nitrobenzene$ with $aq. NH_3$ undergoes nucleophilic aromatic substitution $(S_NAr)$ to form $4-nitroaniline$ as product $(A)$.Step $2$: $4-nitroaniline$ reacts with $Br_2/H_2O$ to form $2,6-dibromo-4-nitroaniline$ as product $(B)$ due to the strong activating effect of the $-NH_2$ group.Step $3$: The reaction of $(B)$ with $(i) NaNO_2 + HCl$ followed by $(ii) H_3PO_2$ involves diazotization of the $-NH_2$ group followed by deamination (reduction of the diazonium salt to a hydrogen atom).Thus,the final product $(C)$ is $1,3-dibromo-5-nitrobenzene$.

The product $(C)$ of the following reaction is:

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