The haloform reaction: conversion of methyl ketones to carboxylic acids

by James

Description: Addition of a dihalogen such as iodine, bromine or chlorine to a methyl ketone in the presence of base results in a carboxylic acid and a haloform (such as iodoform, pictured)
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{ 3 comments… read them below or add one }

Petr Menzel May 25, 2014 at 9:50 am

Hi, in the Description I can read “iodoform” but in the picture of reaction I can see “HCl3″. Thx, P.

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Petr Menzel May 25, 2014 at 9:54 am

Sorry… it is not “H CI3 = H CL3″ but “H C I3″ :-)

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James Briscoe November 4, 2014 at 11:58 pm

So I understand that the haloform reaction when using Iodine provides a nifty way to identify methyl ketones because the LG ‘CI3 precipates as a yellow solid in the form of Iodoform; I also understand that secondary methyl alcohols are oxidized by I2 in a radical-led oxidation rxn to a methyl ketone, which obviously goes through a subsequent reaction with I2. However, why do primary alcohols such as ethanol or propanol not react with I2? Would methanol react? My book does not expand anymore than just that I2 will react with methyl ketones and secondary methyl alcohols.

Thanks!
James

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