Halogenation of ketones is an interesting reaction. For example, it is involved in the preparation of haloforms. Halogenation using molecular halogen takes place at the $\alpha$ position to the carbonyl group in the presence of an acid or a base. In a basic medium, the stability of carbanion is important. In an acidic medium stability of enol form is important, as the reaction goes through enol.


An $\alpha$-halogenated ketone undergoes subsequent halogenation to form polyhalogenated ketone. Chloroform is prepared by the action of sodium hypochlorite on acetone. In this process trichloroacetone is formed as an intermediate. Trichloroacetone can react with hydroxide ion in two possible ways as shown below.
(i) $\mathrm{CCl}_3 \mathrm{COCH}_3+\mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{HO}^{-}} \mathrm{CH}_3 \mathrm{COOH}+\mathrm{CHCl}_3$
(ii) $\mathrm{CCl}_3 \mathrm{COCH}_3+\mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{HO}^{-}} \mathrm{CCl}_3 \mathrm{COOH}+\mathrm{CH}_4$
The correct statement(s) is/are:

• A

  An $\alpha$-chloroketone reacts rapidly by $\mathrm{S}_{\mathrm{N}} 1$ mechanism.
• B

  Reaction of 2-bromo-3-pentanone with bromine in an alkaline medium would give mainly 2,4-dibromo-3-pentanone.
• Correct answer
  C

  Methyl ethyl ketone reacts with chlorine in the presence of an acid to form 3-chloro-2-butanone.
• Correct answer
  D

  In an alkaline medium, reaction (i) is favoured over reaction (ii) due to stabilization of carbanion.

Hence, the correct answer is options: 'C', 'D'.