Answer the question on basis of passage given below:
A ketone [A] which undergoes haloform reaction gives compound [B] on reduction. Compound [B] on heating with sulphuric acid gives compound [C] which forms mono-ozonide [D]. The compound [D] on hydrolysis in the presence of zinc dust gives compound [E]. Compound [E] on oxidation gives \(C_2H_4O_2\) [F].
The compound [C] formed can be prepared from reaction of alc. \(KOH\) with:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → CH3–CH(Cl)–CH2–CH3 (2-chlorobutane)
[A] is the methyl ketone butan-2-one. Its reduction gives butan-2-ol [B], and heating [B] with sulphuric acid gives but-2-ene [C] as the main alkene. But-2-ene is the alkene that fits the rest of the passage: its mono-ozonide gives a single compound, acetaldehyde [E], which is oxidised to \( \ce{C2H4O2} \) [F].
Alcoholic KOH removes HX from a haloalkane (dehydrohalogenation). 2-Chlorobutane has β-hydrogens on C1 and C3; by Saytzeff’s rule the more substituted alkene is the main product:
\( \ce{CH3CHClCH2CH3} \xrightarrow[\Delta]{\text{alc. }\ce{KOH}} \underset{\text{major}}{\ce{CH3CH=CHCH3}} + \underset{\text{minor}}{\ce{CH2=CHCH2CH3}} \)
Why the other options are incorrect
Option (1) CH3CH2CH2Cl — Incorrect. 1-chloropropane gives propene, \( \ce{CH3CH=CH2} \); its ozonolysis gives HCHO as well as \( \ce{CH3CHO} \), not a single compound [E].
Option (2) 2-Chloro-2-methylbutane — Incorrect. Gives 2-methylbut-2-ene (major), whose ozonolysis gives acetone and acetaldehyde.
Option (4) 2-Chloro-2-methylpropane — Incorrect. Gives 2-methylpropene, \( \ce{(CH3)2C=CH2} \); its ozonolysis gives acetone and HCHO.
NCERT (2026-27 edition):
Elimination Reactions – “… If there is possibility of formation of more than one alkene due to the availability of more than one β-hydrogen atoms, usually one alkene is formed as the major product. These form part of a pattern first observed by Russian chemist, Alexander Zaitsev (also pronounced as Saytzeff) who in 1875 formulated a rule which can be summarised as “in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.” Thus, 2-bromopentane gives pent-2-ene as the major product.”