Which of the following compound would undergo Aldol condensation?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Phenylacetaldehyde
Aldol condensation takes place with aldehydes and ketones having at least one α-hydrogen, in the presence of dilute alkali. Phenylacetaldehyde, \( \ce{C6H5-CH2-CHO} \), has two α-hydrogens on the \( \ce{CH2} \) next to the carbonyl group, so it forms an aldol which loses water on heating to give the α,β-unsaturated aldehyde:
\( \ce{2C6H5CH2CHO} \xrightarrow{\text{dil. NaOH}} \ce{C6H5CH2CH(OH)CH(C6H5)CHO} \)
\( \ce{C6H5CH2CH(OH)CH(C6H5)CHO} \xrightarrow[-\ce{H2O}]{\Delta} \ce{C6H5CH2CH=C(C6H5)CHO} \)
Why the other options are incorrect
Option (1) Methanal — Incorrect. HCHO has no α-carbon, so it has no α-hydrogen; with concentrated alkali it gives the Cannizzaro reaction instead.
Option (2) Benzaldehyde — Incorrect. The carbonyl carbon is attached directly to the ring, so there is no α-hydrogen; it also undergoes the Cannizzaro reaction.
Option (3) 2,2-Dimethylbutanal — Incorrect. \( \ce{CH3CH2C(CH3)2-CHO} \); the α-carbon carries two methyl groups and an ethyl group, so it has no α-hydrogen.
NCERT (2026-27 edition):
Aldol Condensation
