Compound III isxrightarrow[{ii),CuCl/HCl}]{{i | vedclass

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Question

(C) The reaction sequence is as follows:$1$. The starting material (not explicitly shown but implied as $4-$nitro$-2,6-$dibromoaniline or similar deri

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$2,6-$dibromochlorobenzene

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(C) The reaction sequence is as follows:$1$. The starting material (not explicitly shown but implied as $4-$nitro$-2,6-$dibromoaniline or similar derivative) undergoes diazotization followed by Sandmeyer reaction with $CuCl/HCl$ to replace the amino group with a chlorine atom,forming compound $I$ $(1-chloro-2,6-dibromo-4-nitrobenzene)$.$2$. Reduction of the nitro group in compound $I$ using $Sn/HCl$ yields compound $II$ $(4-amino-3,5-dibromochlorobenzene)$.$3$. Diazotization of compound $II$ with $HNO_2$ at $0-5\,^oC$ followed by treatment with $H_3PO_2$ and $H_2O$ (reductive deamination) removes the amino group,resulting in compound $III$ $(1-chloro-2,6-dibromobenzene)$.

Compound $III$ is
$\xrightarrow[{ii)\,CuCl/HCl}]{{i)\,HNO_2\,(0 - 5\,^oC)}}I$ $\xrightarrow{{Sn/HCl}}II$ $\xrightarrow[{ii)\,H_3PO_2 + H_2O}]{{i)HNO_2\,(0 - 5\,^oC)\,}}III$

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Compound $III$ is$\xrightarrow[{ii)\,CuCl/HCl}]{{i)\,HNO_2\,(0 - 5\,^oC)}}I$ $\xrightarrow{{Sn/HCl}}II$ $\xrightarrow[{ii)\,H_3PO_2 + H_2O}]{{i)HNO_2\,(0 - 5\,^oC)\,}}III$