A carbonyl compound A (C4H8O), which undergoes the haloform reaction, gives compound B on reduction. B on heating with sulphuric acid gives a compound C, which forms a mono-ozonide D. D on hydrolysis in the presence of zinc dust gives only acetaldehyde (E).
Identify the compound A in the reaction sequence given in the passage.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Butan-2-one
Only aldehydes and ketones with a methyl group joined to the carbonyl carbon (a CH3CO group) give the haloform reaction.
Among the C4H8O carbonyl compounds, the only one with a CH3CO group is butan-2-one, \( \ce{CH3COCH2CH3} \). With sodium hypoiodite it gives iodoform and the salt of an acid with one carbon atom less:
\( \displaystyle \ce{CH3CH2COCH3 + 3NaOI -> CH3CH2COONa + CHI3 + 2NaOH} \)
The rest of the sequence fits: butan-2-one is reduced to butan-2-ol (B), which loses water to give but-2-ene (C); ozonolysis of but-2-ene gives only acetaldehyde.
Why the other options are incorrect
Option (2) 2-Methylpropanal — Incorrect. \( \ce{(CH3)2CHCHO} \) has an H and a (CH3)2CH group on its carbonyl carbon, not a CH3 group, so it gives no haloform reaction.
Option (3) Butanal — Incorrect. \( \ce{CH3CH2CH2CHO} \) also has no CH3CO group, so it gives no haloform reaction.
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. The methyl group is converted to haloform.”