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].
Compound [F] on treatment with chlorine, in presence of a small amount of red phosphorus gives:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(ClCH_2COOH\)
[F] is the acid \( \ce{C2H4O2} \) obtained by oxidising acetaldehyde [E], that is acetic acid, \( \ce{CH3COOH} \). It has three α-hydrogens on the \( \ce{CH3} \) group next to –COOH.
With chlorine and a small amount of red phosphorus, a carboxylic acid is chlorinated at the α-carbon (Hell-Volhard-Zelinsky reaction):
\( \ce{CH3COOH} \xrightarrow[\ce{(ii) H2O}]{\ce{(i) Cl2/red P}} \ce{ClCH2COOH} \)
The product is chloroacetic acid, \( \ce{ClCH2COOH} \).
Why the other options are incorrect
Option (1) \(CH_3CH_2-Cl\) — Incorrect. Chloroethane has no carboxyl group; the HVZ reaction keeps the –COOH group and only replaces an α-hydrogen.
Option (2) \(CH_3COCl\) — Incorrect. Acetyl chloride is formed when the –OH of the acid is replaced (with \( \ce{PCl5} \), \( \ce{PCl3} \) or \( \ce{SOCl2} \)), not by \( \ce{Cl2} \)/red P.
Option (4) \(ClCH_2CH_2Cl\) — Incorrect. 1,2-dichloroethane is the product of adding \( \ce{Cl2} \) to ethene; it has no –COOH group.
NCERT (2026-27 edition):
Halogenation – “Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give α-halocarboxylic acids. The reaction is known as Hell-Volhard-Zelinsky reaction.”