Which of the following set of reagents will convert aniline to p-nitroaniline?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \((CH_3CO)_2O/\text{Pyridine}, HNO_3/H_2SO_4\), and \(NaOH\)
Direct nitration of aniline gives oxidation products and much m-nitroaniline, so the –\( \ce{NH2} \) group is first protected by acetylation, the ring is nitrated, and the protecting group is then removed.
First, aniline is acetylated with acetic anhydride in pyridine; pyridine removes the acid formed:
\( \displaystyle \ce{C6H5NH2 + (CH3CO)2O} \xrightarrow{\text{pyridine}} \ce{C6H5NHCOCH3 + CH3COOH} \)
The acetamido group is a milder activator and still an ortho/para director; being bulky, it sends the nitro group mainly to the para position:
\( \displaystyle \ce{C6H5NHCOCH3 + HNO3} \xrightarrow[\text{288 K}]{\text{conc. }\ce{H2SO4}} p\text{-}\ce{O2NC6H4NHCOCH3 + H2O} \)
Finally, the amide is hydrolysed by aqueous NaOH, which gives back the free –\( \ce{NH2} \) group:
\( \displaystyle p\text{-}\ce{O2NC6H4NHCOCH3 + NaOH} \xrightarrow{\text{heat}} p\text{-}\ce{O2NC6H4NH2 + CH3COONa} \)
So the set is \( \ce{(CH3CO)2O} \)/pyridine, \( \ce{HNO3}/\ce{H2SO4} \) and NaOH.
Why the other options are incorrect
Option (2) \(Sn/HCl, NaNO_2/HCl, H_3PO_2\), and \(H_2O\) — Incorrect. Sn/HCl reduces nitro groups and does nothing to aniline; \( \ce{NaNO2}/\ce{HCl} \) turns aniline into benzenediazonium chloride, and \( \ce{H3PO2} \) then replaces the \( \ce{N2+} \) group by H. The product is benzene, with no nitro group at all.
Option (3) \(HNO_3, H_2SO_4\), and \(NaOH\) — Incorrect. This is direct nitration of unprotected aniline: oxidation products and a large share of the meta isomer are formed, so p-nitroaniline is not obtained cleanly.
Option (4) \(HNO_3, H_2SO_4, H_3PO_2\), and \(H_2O\) — Incorrect. Again the amino group is not protected, and \( \ce{H3PO2} \) is a reducing agent for diazonium salts that has no role here.
NCERT (2026-27 edition):
Electrophilic Substitution – …
