In the reaction,Aniline xrightarrow[text{Dil. | vedclass

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(C) The reaction sequence is as follows:$1$. Aniline reacts with $NaNO_2$ and $	ext{Dil. } HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $

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$para$-hydroxyazobenzene

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(C) The reaction sequence is as follows:$1$. Aniline reacts with $NaNO_2$ and $	ext{Dil. } HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(P)$: $C_6H_5NH_2 + NaNO_2 + 2HCl ightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$2$. Benzenediazonium chloride $(P)$ then undergoes an electrophilic substitution (coupling reaction) with phenol in the presence of $NaOH$ ($pH$ $9-10$). The coupling occurs primarily at the $para$-position due to steric hindrance at the $ortho$-position,forming $p$-hydroxyazobenzene $(Q)$: $C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{NaOH} C_6H_5-N=N-C_6H_4-OH(p) + HCl$

In the reaction,Aniline $\xrightarrow[\text{Dil. } HCl]{NaNO_2} P$ $\xrightarrow[NaOH]{\text{Phenol}} Q$,identify $Q$.

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