The haloform reaction is a chemical reaction where a haloform (CHX3, where X is a halogen) is produced by the exhaustive halogenation of a methyl ketone (a molecule containing the R–CO–CH3 group) in the presence of a base. R may be H, alkyl or aryl. The reaction can be used to produce chloroform (CHCl3), bromoform (CHBr3), or iodoform (CHI3).
1) When methyl ketones are treated with the halogen in basic solution, polyhalogenaton followed by cleavage of the methyl group occurs.
2) The products are the carboxylate and trihalomethane, otherwise known as haloform.
3) The reaction proceeds via successively faster halogenations at the α-position until the 3 H have been replaced.
4) The halogenations get faster since the halogen stablises the enolate negative charge and makes it easier to form.
5) Then a nucleophilic acyl substitution by hydroxide displaces the anion CX3 as a leaving group that rapidly protonates.
6) 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.
Mr. Lalit Sardana(IIT-JEE AIR 243) will help you to discover more through this video. He will explain haloform reaction to you in detail and if you find something difficult then you may drop your comments or questions in the comment box or Ask module respectively. This topic is asked in 11th CBSE, 12th CBSE, JEE Mains, JEE Advanced, VITEEE, AIPMT, AFMC, MPPET, CET, KCET,etc.
