In aldehydes, the carbonyl group is bonded to a carbon and hydrogen, while in ketones it is bonded to two carbon atoms. The compounds in which the carbonyl group is bonded to oxygen are known as carboxylic acids and their derivatives.
1,3-Diphenylprop-2-en-1-one is produced by the following
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → C6H5CHO and C6H5COCH3: Cross aldol condensation
1,3-Diphenylprop-2-en-1-one is \( \ce{C6H5-CH=CH-CO-C6H5} \) (benzalacetophenone). Cutting it at the C=C bond gives the two partners: the \( \ce{C6H5CH} \)= part comes from benzaldehyde and the =\( \ce{CH-CO-C6H5} \) part from acetophenone.
Acetophenone has α-hydrogens; with dilute NaOH its carbanion adds to the C=O of benzaldehyde (which has no α-hydrogen), and the β-hydroxy ketone formed loses water:
\( \ce{C6H5CHO + CH3COC6H5} \xrightarrow{\text{dil. NaOH}} \ce{C6H5CH(OH)CH2COC6H5} \)
\( \ce{C6H5CH(OH)CH2COC6H5} \xrightarrow[-\ce{H2O}]{\Delta} \ce{C6H5CH=CHCOC6H5} \)
Two different carbonyl compounds take part, so this is a cross aldol condensation.
Why the other options are incorrect
Option (2) C6H5COOH and C6H5COCH3 — Incorrect. Benzoic acid has no aldehyde/ketone C=O for the carbanion to add to; with NaOH it only forms sodium benzoate.
Option (3) C6H5CHO, self condensation — Incorrect. Benzaldehyde has no α-hydrogen, so it cannot form a carbanion and gives no aldol product (with concentrated alkali it undergoes the Cannizzaro reaction instead).
Option (4) C6H5COCH3, self condensation — Incorrect. Two acetophenone units would give \( \ce{C6H5C(CH3)=CHCOC6H5} \) (1,3-diphenylbut-2-en-1-one), which has a methyl group on C-3.
NCERT (2026-27 edition):
Cross Aldol Condensation
