Chapter 18: Enols and Enolates

Summary

• When methyl ketones are treated with the halogen in basic solution, polyhalogenaton followed by cleavage of the methyl group occurs.
• The products are the carboxylate and trihalomethane, otherwise known as haloform.
• The reaction proceeds via successively faster halogenations at the α-position until the 3 H have been replaced.
• The halogenations get faster since the halogen stablises the enolate negative charge and makes it easier to form.
• Then a nucleophilic acyl substitution by hydroxide displaces the anion CX₃ as a leaving group that rapidly protonates.
• This reaction is often performed using iodine and as a chemical test for identifying methyl ketones. Iodoform is yellow and precipitates under the reaction conditions.

A yellow precipitate indicates a positive result in the iodoform test (centre tube).

QUESTION Could a methyl aldehyde undergo this reaction ? ANSWER

Related Reactions

• α-halogenation under acidic conditions
• α-halogenation under basic conditions

MECHANISM OF THE HALOFORM REACTION OF METHYL KETONES
Step 1: First, an acid-base reaction. Hydroxide functions as a base and removes the acidic α-hydrogen giving the enolate.
Step 2: The nucleophilic enolate reacts with the iodine giving the halogenated ketone and an iodide ion.
Step 3: Steps 1 and 2 repeat twice more yielding the trihalogenated ketone.
Step 4: The hydroxide now reacts as a nucleophile at the electrophilic carbonyl carbon, with the C=O becoming a C-O single bond and the oxygen is now anionic.
Step 5: Reform the favourable C=O and displace a leaving group, the trihalomethyl system which is stabilised by the 3 halogens. This gives the carboxylic acid.
Step 6: An acid-base reaction. The trihalomethyl anion is protonated by the carboxylic acid, giving the carboxylate and the haloform (trihalomethane).

© Dr. Ian Hunt, Department of Chemistry