How will you carry out the following conversion?
$Aniline \rightarrow m-Bromonitrobenzene$

(N/A) The conversion of $Aniline$ to $m-Bromonitrobenzene$ can be carried out in the following steps:
$1$. $Aniline$ is treated with $HNO_2$ at $273-278 \ K$ to form $Benzene$ $diazonium$ $chloride$.
$2$. $Benzene$ $diazonium$ $chloride$ is treated with $HBF_4$ to form $Benzene$ $diazonium$ $fluoroborate$ $(C_6H_5N_2^+BF_4^-)$.
$3$. $Benzene$ $diazonium$ $fluoroborate$ is treated with $NaNO_2$ in the presence of $Cu$ and heat to form $Nitrobenzene$.
$4$. $Nitrobenzene$ is then treated with $Br_2$ in the presence of $CH_3COOH$ to form $m-Bromonitrobenzene$.