Aspirin is a commonly used non-steroidal anti-inflammatory drug which reduces fever, relieves mild pain and decreases swelling. It is prepared by numerous synthetic methods; of which the acetylation of Compound A with acetyl chloride is one of the easiest methods. However, Compound A itself needs to be prepared from Compound B under basic conditions by using carbon dioxide as a reagent. Interestingly, Compound B along with Compound C can be obtained by the air oxidation of Compound D. Compound C is a commonly used industrial solvent and gives a positive iodoform test and a negative Tollen’s test. Compound C easily forms 2,4-DNP derivative (Compound E). Compound C on reaction with Ba(OH)2 under heating conditions produces Compound F whose molecular formula is C6H10O.
The structures of Compound C and Compound E are
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → propanone (Compound C) and propanone 2,4-dinitrophenylhydrazone (Compound E)
The clues in the passage identify Compound C:
It is formed with phenol in the cumene process, which gives acetone.
It gives a positive iodoform test, so it contains a \( \ce{CH3CO} \)– group.
It gives a negative Tollens’ test, so it is a ketone, not an aldehyde.
Two molecules of it give F, \( \ce{C6H10O} \), so C has three carbon atoms.
So C is propanone (acetone), \( \ce{CH3COCH3} \), a common industrial solvent.
Aldehydes and ketones react with 2,4-dinitrophenylhydrazine (2,4-DNP) by addition–elimination: the C=O becomes C=N–NH–Ar and water is lost. The product is a coloured solid, which is used to identify carbonyl compounds. In the product the two \( \ce{NO2} \) groups stay at the 2- and 4-positions of the ring, relative to the NH.
\( \displaystyle \ce{(CH3)2C=O + H2N-NH-C6H3(NO2)2 -> (CH3)2C=N-NH-C6H3(NO2)2 + H2O} \)
Only Option (2) shows propanone together with its correct 2,4-dinitrophenylhydrazone.
Why the other options are incorrect
Option (1) — Incorrect. Compound C is drawn as butanone (\( \ce{C4} \)). It is not formed in the cumene process, and two molecules of it could not give \( \ce{C6H10O} \).
Option (3) — Incorrect. C is right, but in E the \( \ce{NO2} \) groups are at the 3- and 4-positions, so it is not the 2,4-DNP derivative.
Option (4) — Incorrect. C is drawn as a larger ketone, not propanone, and in E the \( \ce{NO2} \) groups are again at the 3- and 4-positions.
NCERT (2026-27 edition):
Chemical Reactions – …
