Which of the following compounds will not undergo aldol condensation?

The correct answer is Option (1) → Formaldehyde

Aldol condensation needs an α-hydrogen, a hydrogen on the carbon next to the C=O group. The α-hydrogen is acidic because the carbonyl group withdraws electrons and the carbanion (enolate) formed is stabilised by resonance. Dilute alkali removes it, and the carbanion adds to the carbonyl carbon of a second molecule. The β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol) formed loses water on heating to give an α,β-unsaturated carbonyl compound:

    \( \displaystyle \ce{2CH3CHO} \xrightarrow{\text{dil. NaOH}} \ce{CH3-CH(OH)-CH2-CHO} \xrightarrow[\ce{-H2O}]{\Delta} \ce{CH3-CH=CH-CHO} \)

Formaldehyde, HCHO, has no carbon other than the carbonyl carbon, so it has no α-hydrogen and cannot form the carbanion. It does not undergo aldol condensation.

• Option (1) Formaldehyde — Does not undergo aldol condensation (no α-hydrogen). HCHO has no carbon next to the carbonyl carbon, so no carbanion can form.

• Option (2) Ethylmethyl ketone — Undergoes aldol condensation (has α-hydrogens). Butanone, \( \ce{CH3COCH2CH3} \), has five α-hydrogens (on the \( \ce{CH3} \) and \( \ce{CH2} \) groups next to C=O), so it gives a ketol and then the condensation product.

• Option (3) Acetaldehyde — Undergoes aldol condensation (has α-hydrogens). Ethanal has three α-hydrogens and is the standard example of aldol condensation (equation above).

• Option (4) Propanal — Undergoes aldol condensation (has α-hydrogens). \( \ce{CH3CH2CHO} \) has two α-hydrogens on the \( \ce{CH2} \) group and undergoes aldol condensation.

NCERT (2026-27 edition):
Aldol Condensation