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).
The compound (E) of the passage, acetaldehyde, can also be prepared by the addition of water to a hydrocarbon X in the presence of H2SO4 and HgSO4. Under the same conditions, propyne gives the compound Y. Identify X and Y.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → X = Ethyne, Y = Propanone
In the hydration of alkynes, water adds across the triple bond with H2SO4 and HgSO4 as catalysts, and the enol formed changes to a carbonyl compound.
X — Ethyne is the one alkyne that gives an aldehyde, acetaldehyde:
\( \displaystyle \ce{HC#CH + H2O} \xrightarrow{\ce{H2SO4},\ \ce{HgSO4}} \ce{CH3CHO} \)
Y — All other alkynes give ketones. Propyne gives propanone, not propanal:
\( \displaystyle \ce{CH3C#CH + H2O} \xrightarrow{\ce{H2SO4},\ \ce{HgSO4}} \ce{CH3COCH3} \)
Why the other options are incorrect
Option (1) X = Ethyne, Y = Propanal — Incorrect. X is right, but propyne gives a ketone, propanone; only ethyne gives an aldehyde.
Option (2) X = Ethene, Y = Propanone — Incorrect. Y is right, but ethene is an alkene. Addition of water to an alkene in acid gives an alcohol, here ethanol, not acetaldehyde.
Option (4) X = Ethene, Y = Propanal — Incorrect. Both are wrong: ethene gives ethanol, and propyne gives propanone.
NCERT (2026-27 edition):
8.2.1 Preparation of Aldehydes and Ketones – “(ii) By hydration of alkynes: Addition of water to ethyne in the presence of H2SO4 and HgSO4 gives acetaldehyde. All other alkynes give ketones in this reaction (Unit 9, Class XI).”