In the reaction A xrightarrow{SnCl_2/HCl} B x | vedclass

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(A) The reaction sequence is as follows:$1$. $A$ is Nitrobenzene $(C_6H_5NO_2)$.$2$. Reduction of Nitrobenzene with $Sn/HCl$ gives Aniline $(B = C_6H_

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Nitrobenzene,Aniline,Benzene diazonium chloride

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(A) The reaction sequence is as follows:$1$. $A$ is Nitrobenzene $(C_6H_5NO_2)$.$2$. Reduction of Nitrobenzene with $Sn/HCl$ gives Aniline $(B = C_6H_5NH_2)$.$3$. Diazotization of Aniline with $NaNO_2/HCl$ at $0^{\circ}C$ gives Benzene diazonium chloride $(C = C_6H_5N_2^+Cl^-)$.$4$. Hydrolysis of Benzene diazonium chloride with $H_2O/\Delta$ yields Phenol $(C_6H_5OH)$.Thus,$A$ is Nitrobenzene,$B$ is Aniline,and $C$ is Benzene diazonium chloride.

In the reaction $A$ $\xrightarrow{SnCl_2/HCl} B$ $\xrightarrow{NaNO_2/HCl, 0^{\circ}C} C$ $\xrightarrow{H_2O/\Delta} C_6H_5OH$,what are $A, B$,and $C$?

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