Benzene + HNO_{3(conc.)} xrightarrow{H_2SO_4} | vedclass

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(B) $1$. Benzene reacts with conc. $HNO_3$ in the presence of $H_2SO_4$ to form nitrobenzene $(A)$.$2$. Nitrobenzene is reduced by $Sn/HCl$ followed b

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Phenol

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(B) $1$. Benzene reacts with conc. $HNO_3$ in the presence of $H_2SO_4$ to form nitrobenzene $(A)$.$2$. Nitrobenzene is reduced by $Sn/HCl$ followed by treatment with $OH^-$ to form aniline $(B)$.$3$. Aniline reacts with $NaNO_2 + HCl$ at $0-5 \ ^oC$ to form benzene diazonium chloride $(C)$.$4$. Benzene diazonium chloride on hydrolysis with $H_2O/H^+$ yields phenol $(D)$.

Benzene $+ HNO_{3(conc.)}$ $\xrightarrow{H_2SO_4} A$ $\xrightarrow[{(ii) OH^{-}}]{{(i) Sn/HCl}} B$ $\xrightarrow[{0 - 5 \ ^oC}]{NaNO_2 + HCl} C$ $\xrightarrow{H_2O/H^{+}} D$. What is $D$?

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