Aniline undergoes diazotization with NaNO_2 /HCl to give compound...

Aniline reacts with $\text{NaNO}_2$ and $\text{HCl}$ at $0-5^{{}^{\circ}}\text{C}$ to undergo diazotization, forming benzenediazonium chloride (Compound A).

$\text{Ar}-\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{Ar}-\text{N}_2^+\text{Cl}^-$ (A)

Benzenediazonium chloride reacts with fluoroboric acid ($\text{HBF}_4$) to yield benzenediazonium fluoroborate (Compound B).

$\text{Ar}-\text{N}_2^+\text{Cl}^- \xrightarrow{\text{HBF}_4} \text{Ar}-\text{N}_2^+\text{BF}_4^-$ (B)

When benzenediazonium fluoroborate is heated with aqueous $\text{NaNO}_2$ in the presence of copper, the diazonium group is replaced by a nitro group, forming nitrobenzene (Compound C).

$\text{Ar}-\text{N}_2^+\text{BF}_4^- \xrightarrow{\text{NaNO}_2, \text{Cu}, \Delta} \text{Ar}-\text{NO}_2$ (C)

Thus, Compound C is $\text{Ar}-\text{NO}_2$.

Answer: $\text{Ar} - \text{NO}_2$