Identify the final product formed when benzenamine reacts with the given reagents in the sequential order as:
i. (CH$_3$$-$CO )$_2$O / Pyridine.
ii. Conc. HNO$_3$ + H$_2$SO$_4$ & then reacted with H$_3$O$^+$
iii. NaNO$_2$ HCl (273K) & then reacted with H$_3$PO$_2$(aq).

Benzenamine (aniline) reacts with acetic anhydride in the presence of pyridine to form acetanilide. This step protects the amino group and prevents polysubstitution and oxidation.

Acetanilide undergoes nitration with a mixture of concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ to form $p$-nitroacetanilide as the major product due to steric hindrance at the ortho position.

Hydrolysis of $p$-nitroacetanilide with $\text{H}_3\text{O}^+$ removes the acetyl group, yielding $p$-nitroaniline.

$p$-Nitroaniline reacts with $\text{NaNO}_2$ and $\text{HCl}$ at $273\text{ K}$ to form $p$-nitrobenzenediazonium chloride.

Reaction of the diazonium salt with aqueous $\text{H}_3\text{PO}_2$ (hypophosphorous acid) reduces the diazonium group, replacing it with a hydrogen atom (deamination).

The final product formed is nitrobenzene.

Answer: Nitrobenzene