Reaction of aniline with NaNO_{2} + dil. HCl | vedclass

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(C) The correct answer is $C$ $(III)$.Step $1$: Aniline reacts with $NaNO_{2} +$ dil. $HCl$ at $0-5^{\circ} C$ to form benzene diazonium chloride $(C_

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$III$

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(C) The correct answer is $C$ $(III)$.Step $1$: Aniline reacts with $NaNO_{2} +$ dil. $HCl$ at $0-5^{\circ} C$ to form benzene diazonium chloride $(C_{6}H_{5}N_{2}^{+}Cl^{-})$.Step $2$: The benzene diazonium chloride then reacts with $CuCN$ (Sandmeyer reaction) to replace the diazonium group with a cyano group $(-CN)$,yielding cyanobenzene (benzonitrile) and releasing $N_{2}$ gas.Structure $III$ represents cyanobenzene.

Reaction of aniline with $NaNO_{2} +$ dil. $HCl$ at $0^{\circ} C$ followed by reaction with $CuCN$ yields:

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