What is the correct sequence of reagents used | vedclass

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(B) The conversion of nitrobenzene to $m$-dibromobenzene involves the following steps:
$1$. Bromination of nitrobenzene using $Br_2/Fe$ gives $m$-bro

Vedclass · Accepted Answer

$\xrightarrow{Br_2/Fe}$ $\xrightarrow{Sn/HCl}$ $\xrightarrow{NaNO_2/HCl}$ $\xrightarrow{CuBr/HBr}$

Vedclass · Answer

(B) The conversion of nitrobenzene to $m$-dibromobenzene involves the following steps:
$1$. Bromination of nitrobenzene using $Br_2/Fe$ gives $m$-bromonitrobenzene.
$2$. Reduction of $m$-bromonitrobenzene using $Sn/HCl$ gives $m$-bromoaniline.
$3$. Diazotization of $m$-bromoaniline using $NaNO_2/HCl$ at $0-5^{\circ}C$ gives $m$-bromobenzenediazonium chloride.
$4$. Sandmeyer reaction of $m$-bromobenzenediazonium chloride with $CuBr/HBr$ replaces the diazonium group with a bromine atom to yield $m$-dibromobenzene.

What is the correct sequence of reagents used for converting nitrobenzene into $m$-dibromobenzene?

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