Benzene xrightarrow[{{H_2}S{O_4}}]{{HN{O_3}}} | vedclass

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(B) $1$. The reaction of benzene with a mixture of concentrated $HN{O_3}$ and $H_2SO_4$ (nitrating mixture) undergoes electrophilic aromatic substitut

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Nitrobenzene with a chlorine atom at the meta position

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(B) $1$. The reaction of benzene with a mixture of concentrated $HN{O_3}$ and $H_2SO_4$ (nitrating mixture) undergoes electrophilic aromatic substitution to form nitrobenzene $(A)$.$2$. The nitro group $(-NO_2)$ is a strongly deactivating group and is meta-directing for further electrophilic substitution reactions.$3$. When nitrobenzene $(A)$ reacts with $Cl_2$ in the presence of a Lewis acid catalyst like $FeCl_3$,the chlorine atom is directed to the meta position.$4$. Therefore,the major product $(B)$ is $m$-chloronitrobenzene.

$Benzene$ $\xrightarrow[{{H_2}S{O_4}}]{{HN{O_3}}}(A)$ $\xrightarrow[{FeC{l_3}}]{{C{l_2}}}(B)$
The major product $(B)$ is

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$Benzene$ $\xrightarrow[{{H_2}S{O_4}}]{{HN{O_3}}}(A)$ $\xrightarrow[{FeC{l_3}}]{{C{l_2}}}(B)$The major product $(B)$ is