Which test can be used to distinguish between pentan-2-one and pentan-3-one?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Iodoform test
The iodoform test detects a CH3CO– group (a methyl ketone). With sodium hypoiodite (I2 and NaOH), a methyl ketone gives a yellow precipitate of iodoform, CHI3.
Pentan-2-one, \( \ce{CH3COCH2CH2CH3} \): it has a CH3CO– group, so it gives iodoform.
\( \displaystyle \ce{CH3COCH2CH2CH3 + 3NaOI -> CHI3 v + CH3CH2CH2COONa + 2NaOH} \)
Pentan-3-one, \( \ce{CH3CH2COCH2CH3} \): both groups on the carbonyl carbon are ethyl, so there is no CH3CO– group and no yellow precipitate.
Why the other options are incorrect
Option (2) Lucas test — Incorrect. It distinguishes primary, secondary and tertiary alcohols; both compounds here are ketones.
Option (3) Lassaigne’s test — Incorrect. It detects nitrogen, sulphur and halogens in an organic compound; neither ketone contains these elements.
Option (4) Victor Meyer test — Incorrect. It is a test for the classes of alcohols, not for ketones.
NCERT (2026-27 edition):
8.4 Chemical Reactions – “(iii) 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.”
“Iodoform reaction with sodium hypoiodite is also used for detection of CH3CO group or CH3CH(OH) group which produces CH3CO group on oxidation.”