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

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: \(\text{CH}_3\text{CHO}\)