Which simple chemical test is used to distinguish between ethanal and propanal?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Iodoform test
Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon (\( \ce{CH3CO} \)– group) are oxidised by sodium hypoiodite (NaOI) to give a yellow precipitate of iodoform (\( \ce{CHI3} \)).
Ethanal, \( \ce{CH3-CHO} \), has the \( \ce{CH3CO} \)– group: positive iodoform test.
\( \ce{CH3CHO + 3NaOI -> CI3CHO + 3NaOH} \)
\( \ce{CI3CHO + NaOH -> CHI3 v + HCOONa} \)Propanal, \( \ce{CH3CH2-CHO} \), has an ethyl group on the carbonyl carbon: no iodoform.
Why the other options are incorrect
Option (2) Tollen’s test — Incorrect. Both are aldehydes, so both give a silver mirror.
Option (3) Fehling’s test — Incorrect. Both are aliphatic aldehydes, so both give a red-brown precipitate.
Option (4) Lucas test — Incorrect. It distinguishes primary, secondary and tertiary alcohols; it is not a test for aldehydes.
NCERT (2026-27 edition):
Oxidation of Methyl Ketones by Haloform Reaction – “Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (methyl ketones) are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound. The methyl group is converted to haloform. This oxidation does not affect a carbon-carbon double bond, if present in the molecule.”