In the reaction sequence A xrightarrow{Sn/HCl | vedclass

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(A) The given reaction sequence is: $A$ $\xrightarrow{Sn/HCl} B$ $\xrightarrow{HNO_2, 0^{\circ}C} C$ $\xrightarrow{C_2H_5OH} C_6H_6$. $1$. $A$ is nitr

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$NO_2, NH_2, N_2^+Cl^-$

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(A) The given reaction sequence is: $A$ $\xrightarrow{Sn/HCl} B$ $\xrightarrow{HNO_2, 0^{\circ}C} C$ $\xrightarrow{C_2H_5OH} C_6H_6$. $1$. $A$ is nitrobenzene $(C_6H_5NO_2)$ containing the $-NO_2$ group. $2$. Reduction of $A$ with $Sn/HCl$ gives $B$,which is aniline $(C_6H_5NH_2)$ containing the $-NH_2$ group. $3$. Reaction of $B$ with $HNO_2$ at $0^{\circ}C$ (diazotization) gives $C$,which is benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$ containing the $-N_2^+Cl^-$ group. $4$. Reaction of $C$ with $C_2H_5OH$ reduces the diazonium salt to benzene $(C_6H_6)$. Therefore,the functional groups are $-NO_2, -NH_2,$ and $-N_2^+Cl^-$.

In the reaction sequence $A$ $\xrightarrow{Sn/HCl} B$ $\xrightarrow{HNO_2, 0^{\circ}C} C$ $\xrightarrow{C_2H_5OH} C_6H_6$,identify the functional groups present in $A, B,$ and $C$.

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