To convert aniline into p-Nitroaniline, three steps are taken. Arrange following given steps in correct sequence.
A. N-Phenyl ethanamide is treated with HNO3, H2SO4 and heated at 288 K.
B. p-Nitro acetanilide undergoes acidic or basic hydrolysis to give p-Nitroaniline.
C. Aniline is treated with acetic anhydride in pyridine.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → C, A, B
Direct nitration of aniline is not used: the –\( \ce{NH2} \) group is easily oxidised by \( \ce{HNO3} \), and in the acid much of the aniline is protonated to the meta-directing anilinium ion. The –\( \ce{NH2} \) group is therefore protected by acetylation, which also reduces its activating effect so that a single nitro group enters the para position:
C. Aniline is acetylated with acetic anhydride in pyridine:
\( \displaystyle \ce{C6H5NH2 + (CH3CO)2O} \xrightarrow{\text{pyridine}} \ce{C6H5NHCOCH3 + CH3COOH} \)
A. N-Phenylethanamide (acetanilide) is nitrated with \( \ce{HNO3} \)/\( \ce{H2SO4} \) at 288 K, mainly at the para position:
\( \displaystyle \ce{C6H5NHCOCH3 + HNO3} \xrightarrow[288\ \mathrm{K}]{\ce{H2SO4}} \text{p-}\ce{O2N-C6H4-NHCOCH3 + H2O} \)
B. p-Nitroacetanilide is hydrolysed (acid or base) to p-nitroaniline:
\( \displaystyle \text{p-}\ce{O2N-C6H4-NHCOCH3 + H2O} \xrightarrow{\ce{H+} \text{ or } \ce{OH-}} \text{p-}\ce{O2N-C6H4-NH2 + CH3COOH} \)
Shortcut: Protection has to come first, so the sequence starts with C; only Option (2) does.
NCERT (2026-27 edition):
Chemical Reactions – “… The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their very high reactivity. Substitution tends to occur at ortho- and para-positions. … This can be done by protecting the -NH2 group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.”