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(C) The reaction sequence is as follows:$1$. Nitration of benzene with $HNO_3/H_2SO_4$ gives nitrobenzene.$2$. Bromination of nitrobenzene with $Br_2/

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$4-$bromo$-1-$chloro$-2-$iodobenzene

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(C) The reaction sequence is as follows:$1$. Nitration of benzene with $HNO_3/H_2SO_4$ gives nitrobenzene.$2$. Bromination of nitrobenzene with $Br_2/FeBr_3$ gives $m$-bromonitrobenzene because the $-NO_2$ group is meta-directing.$3$. Reduction of $m$-bromonitrobenzene with $H_2/Pd/C$ gives $m$-bromoaniline.$4$. Chlorination of $m$-bromoaniline with $Cl_2/FeBr_3$ gives $2$-chloro-$5$-bromoaniline (the $-NH_2$ group is strongly ortho/para directing,and the para position is occupied by $-Br$,so it directs to the ortho position).$5$. Diazotization with $NaNO_2/HCl$ converts the $-NH_2$ group to a diazonium salt $(-N_2^+Cl^-)$.$6$. Treatment with $KI$ replaces the diazonium group with an iodine atom,resulting in $1$-chloro-$2$-iodo-$4$-bromobenzene (or $4$-bromo-$1$-chloro-$2$-iodobenzene).

Give the product of the following reaction sequence:

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