Which of the following pairs of compounds can be distinguished by the iodoform test?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Acetophenone and benzophenone
The iodoform test (heating with iodine and sodium hydroxide, i.e. sodium hypoiodite) is given by compounds with a \( \ce{CH3CO-} \) group (or a \( \ce{CH3CH(OH)-} \) group, which is oxidised to \( \ce{CH3CO-} \)). A pair can be told apart only if one member gives the test and the other does not.
Acetophenone, \( \ce{C6H5COCH3} \) — a methyl ketone, so it gives a yellow precipitate of iodoform:
\( \displaystyle \ce{C6H5COCH3 + 3NaOI -> C6H5COONa + CHI3 + 2NaOH} \)
Benzophenone, \( \ce{C6H5COC6H5} \) — both groups on the carbonyl carbon are phenyl groups; with no \( \ce{CH3CO-} \) group, it gives no iodoform.
One positive, one negative: the pair is distinguished.
Why the other options are incorrect
Option (1) Benzaldehyde and benzophenone — Incorrect. Neither \( \ce{C6H5CHO} \) nor \( \ce{C6H5COC6H5} \) has a \( \ce{CH3CO-} \) group, so both give a negative test.
Option (2) Benzaldehyde and formaldehyde — Incorrect. \( \ce{C6H5CHO} \) and \( \ce{HCHO} \) have no methyl group on the carbonyl carbon, so both are negative.
Option (3) Acetophenone and acetaldehyde — Incorrect. Acetaldehyde, \( \ce{CH3CHO} \), also has a \( \ce{CH3CO-} \) group, so both give iodoform.
NCERT (2026-27 edition):
8.4 Chemical Reactions: 3. 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. … Iodoform reaction with sodium hypoiodite is also used for detection of CH3CO group or CH3CH(OH) group which produces CH3CO group on oxidation.”