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.,$Zn/NaOH$) produces hydrazobenzene $(C_{12}H_{12}N_2)$ as the final product $X

Vedclass · Accepted Answer

$27 \sigma, 6 \pi$

Vedclass · Answer

(D) The reduction of nitrobenzene with zinc in an alkaline medium (e.g.,$Zn/NaOH$) produces hydrazobenzene $(C_{12}H_{12}N_2)$ as the final product $X$.Structure of hydrazobenzene: $Ph-NH-NH-Ph$.Each phenyl ring $(C_6H_5)$ contains $6 \pi$ bonds (from the aromatic ring) and $12 \sigma$ bonds ($6$ $C-C$ and $6$ $C-H$).For two phenyl rings,we have $12 \pi$ bonds and $24 \sigma$ bonds.Additionally,the central $N-N$ linkage has $1 \sigma$ bond,and the two $N-H$ bonds contribute $2 \sigma$ bonds.Total $\sigma$ bonds = $24 + 1 + 2 = 27$.Total $\pi$ bonds = $3 + 3 = 6$ (from the two benzene rings).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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