\(C_3H_6O\) does not form a silver mirror with Tollens’ reagent but forms an oxime with hydroxylamine. It can give positive:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Iodoform test
Formation of an oxime with hydroxylamine shows that \( \ce{C3H6O} \) has a carbonyl group. It does not reduce Tollens’ reagent (no silver mirror), so it is not an aldehyde (propanal). The compound is therefore the ketone propanone (acetone), \( \ce{CH3COCH3} \):
\( \ce{(CH3)2C=O + NH2OH} \xrightarrow{\ce{H+}} \ce{(CH3)2C=N-OH + H2O} \)
Acetone has a CH3CO– group, so it gives the iodoform test: with \( \ce{I2} \) and NaOH (sodium hypoiodite) it forms a yellow precipitate of iodoform.
\( \ce{CH3COCH3 + 3NaOI -> CHI3 v + CH3COONa + 2NaOH} \)
Why the other options are incorrect
Option (2) Fehling’s test — Incorrect. Given by aliphatic aldehydes (red-brown \( \ce{Cu2O} \)); ketones such as acetone do not reduce Fehling’s solution.
Option (3) Schiff’s test — Incorrect. Aldehydes restore the pink colour of Schiff’s reagent; ketones do not.
Option (4) Carbylamine test — Incorrect. A test for primary amines (with \( \ce{CHCl3} \) and alcoholic KOH they give foul-smelling isocyanides); acetone has no nitrogen.
NCERT (2026-27 edition):
Oxidation – “(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. The methyl group is converted to haloform.
… Iodoform reaction with sodium hypoiodite is also used for detection of CH3CO group or CH3CH(OH) group which produces CH3CO group on oxidation.”