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 condensation reaction for converting Compound C to Compound F is called ........, and the structure of Compound F is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Aldol Condensation, 4-methylpent-3-en-2-one
An aldehyde or ketone with at least one α-hydrogen undergoes aldol condensation in the presence of a base. When two molecules of the same carbonyl compound react, it is called aldol condensation; when two different carbonyl compounds react, it is called cross-aldol condensation. The first product is a β-hydroxy carbonyl compound (the aldol or ketol), which loses water on heating and gives an α,β-unsaturated carbonyl compound.
Compound C is acetone (iodoform positive, Tollens’ negative, from the cumene process). With \( \ce{Ba(OH)2} \), two molecules of acetone give the ketol 4-hydroxy-4-methylpentan-2-one (\( \ce{C6H12O2} \)), which loses water on heating.
The product, 4-methylpent-3-en-2-one (mesityl oxide), has the formula C6H10O, which matches F. Only one carbonyl compound (acetone) is involved, so the reaction is aldol condensation, and F is the α,β-unsaturated ketone.
Why the other options are incorrect
Option (1) Aldol Condensation — Incorrect. The name is right, but the structure is the ketol 4-hydroxy-4-methylpentan-2-one (\( \ce{C6H12O2} \)), not the \( \ce{C6H10O} \) compound formed on heating.
Option (2) Cross-aldol Condensation — Incorrect. Both parts are wrong; only one carbonyl compound reacts, and the structure is the ketol, not F.
Option (3) Cross-aldol Condensation — Incorrect. The structure is right, but the reaction of acetone with itself is not a cross-aldol condensation.
NCERT (2026-27 edition):
Aldol Condensation
