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 α 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 α-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.
(i) CCl3COCH3+H2OHO−→CH3COOH+CHCl3
(ii) CCl3COCH3+H2OHO−→CCl3COOH+CH4
The correct statement(s) is/are.
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 α 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 α-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.
(i) CCl3COCH3+H2OHO−→CH3COOH+CHCl3
(ii) CCl3COCH3+H2OHO−→CCl3COOH+CH4
The correct statement(s) is/are.
The correct options are
C Methyl ethyl ketone reacts with chlorine in the presence of an acid to form 3-chloro-2-butanone
D In an alkaline medium, reaction (i) is favoured over reaction (ii) due to stabilization of carbanion
The acidity of α-H of carbonyl compound is high, therefore halogenation takes place at α position. Also, the halogen attack follows S2N mechanism via enolization of ketone. In basic medium, polyhalogenation can take place at the same αC due to the higher acidity of its proton. Thus, reaction of 2-bromo-3-pentanone with bromine in alkaline medium will give 2,2-dibromo-3-pentanone.
CH3CH2BrCOC2H5+Br2→CH3CHBr2COC2H5
In acid catalysed reaction, relative rates of formation of the enols determines the product distribution, In ethyl methyl ketone (2-butanone), since but-2-en-2-ol is more stable than but-1-en-2-ol (more substituted alkene is more stable). Therefore will give halogenation product as 3-chloro-2-butanone.
CH3CH2COCH3+H+→CH3CH=C(OH)CH3Br2⟶CH3CHBrCOCH3
During polyhalogenation, the stability of CCl−3 over CH−3 favours reaction (i) over (ii).
Thus option C and D are correct.
C Methyl ethyl ketone reacts with chlorine in the presence of an acid to form 3-chloro-2-butanone
D In an alkaline medium, reaction (i) is favoured over reaction (ii) due to stabilization of carbanion
The acidity of α-H of carbonyl compound is high, therefore halogenation takes place at α position. Also, the halogen attack follows S2N mechanism via enolization of ketone. In basic medium, polyhalogenation can take place at the same αC due to the higher acidity of its proton. Thus, reaction of 2-bromo-3-pentanone with bromine in alkaline medium will give 2,2-dibromo-3-pentanone.
