Nitrobenzene on reduction using zinc in alkal | vedclass

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Question

(D) The reduction of nitrobenzene with zinc in an alkaline medium (e.g.,$Zn/NaOH$) yields hydrazobenzene $(C_{12}H_{12}N_2)$ as the product $X$.Struct

Vedclass · Accepted Answer

$27 \sigma, 6 \pi$

Vedclass · Answer

(D) The reduction of nitrobenzene with zinc in an alkaline medium (e.g.,$Zn/NaOH$) yields hydrazobenzene $(C_{12}H_{12}N_2)$ as the product $X$.Structure of hydrazobenzene: $C_6H_5-NH-NH-C_6H_5$.Calculation of bonds:$1$. Each phenyl ring $(C_6H_5)$ contains $6$ $C-C$ bonds (including $3$ $\pi$ bonds),$5$ $C-H$ bonds,and $1$ $C-N$ bond.$2$. Total $\sigma$ bonds = $(6 	imes 2) + (5 	imes 2) + (1 	imes 2) + (N-N \sigma) + (2 	imes N-H \sigma) = 12 + 10 + 2 + 1 + 2 = 27$.$3$. Total $\pi$ bonds = $3$ (in first ring) + $3$ (in second ring) = $6$.Thus,$X$ contains $27 \sigma$ and $6 \pi$ bonds.

Nitrobenzene on reduction using zinc in alkaline medium results in $X$. The number of $\sigma$ and $\pi$ bonds in $X$ is

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