Which of the following set of reagents will convert aniline to p-nitroaniline?

$(CH_3CO)_2O/\text{Pyridine}, HNO_3/H_2SO_4$, and $NaOH$

The correct answer is Option (1) → $(CH_3CO)_2O/\text{Pyridine}, HNO_3/H_2SO_4$, and $NaOH$

The $-NH_2$ group of aniline is strongly activating, but in nitration conditions (acidic medium) it becomes protonated to $-NH_3^+$, which is meta-directing.

To obtain para-nitroaniline, the amino group must first be protected (usually as an amide), then nitrated, and finally deprotected.

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Option 1: $(\text{CH}_3\text{CO})_2\text{O}$ / Pyridine, $\text{HNO}_3$ / $\text{H}_2\text{SO}_4$, and $\text{NaOH}$

Acetic anhydride converts aniline into acetanilide (protection step).

The $-\text{NHCOCH}_3$ group is ortho/para directing but less activating, so nitration mainly gives p-nitroacetanilide.

Treatment with $\text{NaOH}$ hydrolyzes the amide, regenerating $-\text{NH}_2$ to form p-nitroaniline.

This is the correct sequence.

Option 2: $\text{Sn}$ / $\text{HCl}$, $\text{NaNO}_2$ / $\text{HCl}$, $\text{H}_3\text{PO}_2$, and $\text{H}_2\text{O}$

$\text{Sn}/\text{HCl}$ reduces nitro groups to amines.

$\text{NaNO}_2/\text{HCl}$ converts amine to diazonium salt.

$\text{H}_3\text{PO}_2$ replaces the diazonium group with hydrogen (deamination).

This sequence removes $-\text{NH}_2$ rather than introducing $-\text{NO}_2$. Not suitable.

Option 3: $\text{HNO}_3$, $\text{H}_2\text{SO}_4$, and $\text{NaOH}$

Direct nitration of aniline in acidic medium forms the anilinium ion $(-\text{NH}_3^+)$, which is meta-directing.

Main product is m-nitroaniline, not para.

$\text{NaOH}$ only neutralizes acid.

Option 4: $\text{HNO}_3$, $\text{H}_2\text{SO}_4$, $\text{H}_3\text{PO}_2$, and $\text{H}_2\text{O}$

Again starts with direct nitration (meta product).

$\text{H}_3\text{PO}_2$ is used to replace diazonium groups with hydrogen, not to achieve para nitration.

This sequence does not yield p-nitroaniline.