Aldehydes and ketones react with halogens under acid conditions, resulting in halogenation at the α-carbon.
The mechanism involves formation of the enol tautomer, which acts as a nucleophile. A halogen atom is bound to the α-carbon and the final step involves loss of a proton.
Treatment of a methyl ketone with excess iodine and sodium hydroxide results in tri-iodination of the methyl group. The resulting CI3 group is a good leaving group and is displaced by the hydroxide ion to form a yellow precipitate (CHI3).
Aldehydes and ketones react with chlorine, bromine or iodine in acidic solution, resulting in halogenation at the α-carbon (Fig. 1).
Since acid conditions are employed, this process does not involve an enolate ion. Instead, the reaction takes place through the enol tautomer of the carbonyl compound. The enol tautomer acts as a nucleophile with a halogen by the mechanism shown (Fig. 2). In the final step, the solvent acts as a base to remove the proton.
α-Halogenation can also be carried out in the presence of base. The reactionproceeds through an enolate ion which is then halogenated (Fig. 3). However, it is difficult to stop the reaction at mono-halogenation since the resulting product is generally more acidic than the starting ketone due to the electron-withdrawing effect of the halogen. As a result, another enolate ion is quickly formed leading to further halogenation.
This tendency towards multiple halogenation is the basis for a classical test called the iodoform test which is used to identify methyl ketones. The ketone to be tested is treated with excess iodine and base and if a yellow precipitate is formed, a positive result is indicated. Under these conditions, methyl ketones undergo α-halogenation three times (Fig. 4). The product obtained is then suscep-tible to nucleophilic substitution whereby the hydroxide ion substitutes the tri-iodomethyl (−CI3) carbanion – a good leaving group due to the three electron-withdrawing iodine atoms. Tri-iodomethane is then formed as the yellow precipitate.
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