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 B in the reaction sequence given in the passage.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Butan-2-ol
A ketone is reduced to a secondary alcohol, and an aldehyde to a primary alcohol.
A: a C4H8O carbonyl compound that gives the haloform reaction must carry a CH3CO group, so it is the methyl ketone butan-2-one, \( \ce{CH3COCH2CH3} \).
B: reduction turns its C=O group into CH–OH, giving butan-2-ol:
\( \displaystyle \ce{CH3COCH2CH3} \xrightarrow{\ce{NaBH4}} \ce{CH3CH(OH)CH2CH3} \)
Check: butan-2-ol loses water with sulphuric acid to give but-2-ene, \( \ce{CH3CH=CHCH3} \), whose ozonolysis gives only acetaldehyde, as the passage says.
Why the other options are incorrect
Option (1) Butanoic acid — Incorrect. A carboxylic acid is an oxidation product; reduction of a ketone never gives one.
Option (3) Butane — Incorrect. Butane has no –OH group, so it cannot lose water on heating with sulphuric acid to give the alkene C.
Option (4) Butan-1-ol — Incorrect. A primary alcohol comes from the aldehyde butanal, which has no CH3CO group and gives no haloform reaction.
NCERT (2026-27 edition):
8.4 Chemical Reactions – “(i) Reduction to alcohols: Aldehydes and ketones are reduced to primary and secondary alcohols respectively by sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4) as well as by catalytic hydrogenation (Unit 7, Class XII).”