Identify P, Q and R in the correct sequence of reactions shown below:…

The correct answer is Option (4) → A, C and D only

Direct nitration of aniline is not used: it gives tarry oxidation products, and in strong acid the anilinium ion directs to the meta position. So the –\( \ce{NH2} \) group is protected first. Structures B (an N–O linked ester) and E (nitrobenzene, with no N–H group) are not formed in this sequence.

First, aniline is acetylated: P is acetanilide (structure A).

    \( \displaystyle \ce{C6H5NH2 + (CH3CO)2O} \xrightarrow{\text{pyridine}} \ce{C6H5NHCOCH3 + CH3COOH} \)

Then nitration: the moderately activating, bulky –\( \ce{NHCOCH3} \) group directs the nitro group mainly to the para position, so Q is p-nitroacetanilide (structure C).

    \( \displaystyle \ce{C6H5NHCOCH3 + HNO3} \xrightarrow{\ce{H2SO4},\ \text{288 K}} p\text{-}\ce{O2NC6H4NHCOCH3 + H2O} \)

Last, hydrolysis removes the acetyl group, so R is p-nitroaniline (structure D).

    \( \displaystyle p\text{-}\ce{O2NC6H4NHCOCH3 + H2O} \xrightarrow{\ce{H+}\text{ or }\ce{OH-}} p\text{-}\ce{O2NC6H4NH2 + CH3COOH} \)

NCERT (2026-27 edition):
Nitration – “Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.
However, by protecting the –NH₂ group by acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the p-nitro derivative can be obtained as the major product.”