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(C) To obtain $3,4,5-$Tribromoaniline,we start with $p-$nitroaniline. $1.$ The amino group $(-NH_2)$ is a strong activating group,so treatment with ex

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$p-$Nitroaniline $\xrightarrow[(ii) NaNO_2, HCl, CuBr]{(i) Br_2(excess), acetic acid}$ $\xrightarrow[(iii) Sn, HCl]{}$

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(C) To obtain $3,4,5-$Tribromoaniline,we start with $p-$nitroaniline. $1.$ The amino group $(-NH_2)$ is a strong activating group,so treatment with excess $Br_2$ in acetic acid leads to bromination at the $2$ and $6$ positions relative to the $-NH_2$ group (which are the $3$ and $5$ positions relative to the $-NO_2$ group),yielding $2,6-$dibromo$-4-$nitroaniline. $2.$ The amino group is then converted to a diazonium salt using $NaNO_2/HCl$. $3.$ The diazonium group is replaced by a bromine atom using $CuBr$ (Sandmeyer reaction),resulting in $3,4,5-$tribromonitrobenzene. $4.$ Finally,the nitro group $(-NO_2)$ is reduced to an amino group $(-NH_2)$ using $Sn/HCl$ to yield $3,4,5-$tribromoaniline.

The sequence from the following that would result in giving predominantly $3,4,5-$Tribromoaniline is $:$

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