Organic compound ‘D’ is in the reaction seque | vedclass

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(C) Step $1$: Nitrobenzene reacts with conc. $HNO_3$ and conc. $H_2SO_4$ at $323-333 	ext{ K}$ to form $m$-dinitrobenzene $(A)$.Step $2$: $m$-dinitro

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Nitrobenzene

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(C) Step $1$: Nitrobenzene reacts with conc. $HNO_3$ and conc. $H_2SO_4$ at $323-333 	ext{ K}$ to form $m$-dinitrobenzene $(A)$.Step $2$: $m$-dinitrobenzene is reduced by $Fe/HCl$ to form $m$-nitroaniline $(B)$.Step $3$: $m$-nitroaniline reacts with $NaNO_2/HCl$ at $273 	ext{ K}$ to form $m$-nitrobenzenediazonium chloride $(C)$.Step $4$: $m$-nitrobenzenediazonium chloride reacts with ethanol $(C_2H_5OH)$ to form nitrobenzene $(D)$, acetaldehyde $(CH_3CHO)$, $N_2$ gas, and $HCl$. Thus, $D$ is nitrobenzene.

Organic compound ‘$D$’ is in the reaction sequence: Nitrobenzene $\xrightarrow[323-333 \text{ K}]{\text{Conc. } HNO_3, \text{Conc. } H_2SO_4} A \xrightarrow{Fe/HCl} B \xrightarrow[273 \text{ K}]{NaNO_2/HCl} C \xrightarrow[ -N_2]{C_2H_5OH} D + CH_3CHO + HCl$. Identify ‘$D$’.

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