Identify product B in the following sequence | vedclass

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(D) Step $1$: Aniline reacts with $NaNO_2 + HCl$ at $273-278 \ K$ to form $A$,which is Benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
Step $2$: When

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Phenol

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(D) Step $1$: Aniline reacts with $NaNO_2 + HCl$ at $273-278 \ K$ to form $A$,which is Benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
Step $2$: When Benzenediazonium chloride is heated with water $(\Delta, H_2O)$,it undergoes hydrolysis to form $B$,which is Phenol $(C_6H_5OH)$,along with the evolution of Nitrogen gas $(N_2 \uparrow)$ and Hydrochloric acid $(HCl)$.
Therefore,product $B$ is Phenol.

Identify product $B$ in the following sequence of reaction:
Aniline $\xrightarrow[273 \ K]{NaNO_2+HCl} A$ $\xrightarrow[\Delta]{H_2O} B + N_2 \uparrow$

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Identify product $B$ in the following sequence of reaction: Aniline $\xrightarrow[273 \ K]{NaNO_2+HCl} A$ $\xrightarrow[\Delta]{H_2O} B + N_2 \uparrow$