$\xrightarrow{\large{\text{Conc.HNO}_3/\text{Conc.H}_2\text{SO}_4,\ 323-333\text{K}}} A \xrightarrow{\large{\text{Fe/HCl}}} B \xrightarrow{\large{\text{NaNO}_2/\text{HCl},\ 273\text{K}}} C \xrightarrow{\large{\text{C}_2\text{H}_5\text{OH},\ -\text{N}_2}}$ $+ D + \text{HCl}$

Organic compound 'D' is

Benzene undergoes nitration with concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ at $323-333\text{ K}$ to form nitrobenzene (A).

$\text{C}_6\text{H}_6 \xrightarrow{\text{Conc. HNO}_3 / \text{Conc. H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2 \text{ (A)}$

Nitrobenzene (A) is reduced by $\text{Fe/HCl}$ to yield aniline (B).

$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{C}_6\text{H}_5\text{NH}_2 \text{ (B)}$

Aniline (B) undergoes diazotization with $\text{NaNO}_2$ and $\text{HCl}$ at $273\text{ K}$ to form benzene diazonium chloride (C).

$\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl},\ 273\text{K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \text{ (C)}$

Benzene diazonium chloride (C) is reduced by ethanol ($\text{C}_2\text{H}_5\text{OH}$) to benzene. In this process, ethanol acts as a reducing agent and gets oxidized to acetaldehyde ($\text{CH}_3\text{CHO}$).

$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{C}_6\text{H}_6 + \text{CH}_3\text{CHO} + \text{N}_2 + \text{HCl}$

Therefore, the organic compound D is acetaldehyde ($\text{CH}_3\text{CHO}$).

Answer: $\text{CH}_3\text{CHO}$