The compound A when treated with HNO_3 (in pr | vedclass

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(B) The reaction sequence is as follows:$1$. The nitration of benzene $(C_6H_6)$ with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$ y

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Benzene

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(B) The reaction sequence is as follows:$1$. The nitration of benzene $(C_6H_6)$ with concentrated $HNO_3$ in the presence of concentrated $H_2SO_4$ yields nitrobenzene $(C_6H_5NO_2)$,which is compound $B$.$2$. Nitrobenzene $(C_6H_5NO_2)$ is then reduced using $Sn$ and $HCl$ to form aniline $(C_6H_5NH_2)$.$3$. Therefore,the starting compound $A$ is benzene.

The compound $A$ when treated with $HNO_3$ (in presence of $H_2SO_4$) gives compound $B$,which is then reduced with $Sn$ and $HCl$ to aniline. The compound $A$ is

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