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 C in the reaction sequence given in the passage.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → But-2-ene
Ozonolysis followed by zinc dust and water cuts an alkene at its C=C bond and turns each doubly bonded carbon into a C=O group. An alkene that gives only acetaldehyde must have two identical CH3CH= halves.
A: a C4H8O carbonyl compound that gives the haloform reaction must be a methyl ketone: butan-2-one, \( \ce{CH3COCH2CH3} \).
B: reduction of a ketone gives a secondary alcohol, butan-2-ol:
\( \displaystyle \ce{CH3COCH2CH3} \xrightarrow{\ce{NaBH4}} \ce{CH3CH(OH)CH2CH3} \)
C: heating with sulphuric acid removes a molecule of water and gives an alkene. Butan-2-ol can lose H from C1 or C3, so C is but-1-ene or but-2-ene.
\( \displaystyle \ce{CH3CH(OH)CH2CH3} \xrightarrow[\Delta]{\text{conc. }\ce{H2SO4}} \ce{CH3CH=CHCH3 + H2O} \)
Ozonolysis decides: only but-2-ene splits into two molecules of acetaldehyde:
\( \displaystyle \ce{CH3CH=CHCH3} \xrightarrow[\text{(ii) Zn, }\ce{H2O}]{\text{(i) }\ce{O3}} \ce{2CH3CHO} \)
So C is but-2-ene.
Why the other options are incorrect
Option (1) But-1-ene — Incorrect. \( \ce{CH2=CHCH2CH3} \) is cut into methanal and propanal, not acetaldehyde alone.
Option (2) 2-Methylpropene — Incorrect. \( \ce{CH2=C(CH3)2} \) gives methanal and propanone; it also has a branched chain, which cannot come from butan-2-ol.
Option (4) Butane — Incorrect. Butane is an alkane with no C=C bond, so it is not formed by dehydration and does not undergo ozonolysis.
NCERT (2026-27 edition):
8.2.1 Preparation of Aldehydes and Ketones – “(i) By ozonolysis of alkenes: As we know, ozonolysis of alkenes followed by reaction with zinc dust and water gives aldehydes, ketones or a mixture of both depending on the substitution pattern of the alkene (Unit 9, Class XI).”
7.4.4 Chemical Reactions – “3. Dehydration: Alcohols undergo dehydration (removal of a molecule of water) to form alkenes on treating with a protic acid e.g., concentrated H2SO4 or H3PO4, or catalysts such as anhydrous zinc chloride or alumina.”