|Chapter 18: Enols and Enolates|
QUESTION Could a methyl aldehyde undergo this reaction ? ANSWER
|MECHANISM OF THE HALOFORM REACTION OF METHYL KETONES|
First, an acid-base reaction. Hydroxide functions as a base and removes the acidic α-hydrogen giving the enolate.
The nucleophilic enolate reacts with the iodine giving the halogenated ketone and an iodide ion.
Steps 1 and 2 repeat twice more yielding the trihalogenated ketone.
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.
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.
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|