C_6H_6 xrightarrow{HNO_3, H_2SO_4} X xrightar | vedclass

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(B) $1$. The reaction of benzene $(C_6H_6)$ with nitrating mixture $(HNO_3 + H_2SO_4)$ is an electrophilic aromatic substitution reaction that yields

Vedclass · Accepted Answer

$3-$nitrochlorobenzene

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(B) $1$. The reaction of benzene $(C_6H_6)$ with nitrating mixture $(HNO_3 + H_2SO_4)$ is an electrophilic aromatic substitution reaction that yields nitrobenzene $(X)$.$2$. Nitrobenzene $(X)$ contains a $-NO_2$ group,which is a strong electron-withdrawing group and a meta-directing group.$3$. When nitrobenzene reacts with chlorine $(Cl_2)$ in the presence of a Lewis acid catalyst $(FeCl_3)$,the incoming electrophile $(Cl^+)$ is directed to the meta-position.$4$. Therefore,the product $Y$ is $3-$nitrochlorobenzene (also known as $m-$nitrochlorobenzene).

$C_6H_6$ $\xrightarrow{HNO_3, H_2SO_4} X$ $\xrightarrow{Cl_2, FeCl_3} Y$. In the above sequence,$Y$ is:

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