In the series of reactions: C_6H_5NH_2 xright | vedclass

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Question

(C) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2/HCl$ at $0 - 5\ ^\circ C$ to form benzene diazonium chloride $(C_6H_5N_2^+ Cl^-)$,which is $X

Vedclass · Accepted Answer

$C_6H_5N_2^+ Cl^-$,$C_6H_5NO_2$

Vedclass · Answer

(C) Step $1$: Aniline $(C_6H_5NH_2)$ reacts with $NaNO_2/HCl$ at $0 - 5\ ^\circ C$ to form benzene diazonium chloride $(C_6H_5N_2^+ Cl^-)$,which is $X$.Step $2$: Benzene diazonium chloride $(C_6H_5N_2^+ Cl^-)$ reacts with $HNO_2$ in the presence of $H_2O$ to form nitrobenzene $(C_6H_5NO_2)$,which is $Y$,along with the evolution of $N_2$ gas and formation of $HCl$.Therefore,$X$ is $C_6H_5N_2^+ Cl^-$ and $Y$ is $C_6H_5NO_2$.

In the series of reactions: $C_6H_5NH_2$ $\xrightarrow{NaNO_2/HCl}_{0 - 5\ ^\circ C} X$ $\xrightarrow{HNO_2}_{H_2O} Y + N_2 + HCl$. $X$ and $Y$ are respectively:

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