The haloform reaction: conversion of methyl ketones to carboxylic acids
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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this reaction is also occur- case1. when we take reagent as bleaching powder instead of NaOH+Cl2 and.Case 2. is we take substrate as secondary alcohol.can you explain both of these
primary as well as seconday alcohol
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
Hi, in the Description I can read “iodoform” but in the picture of reaction I can see “HCl3”. Thx, P.
Sorry… it is not “H CI3 = H CL3” but “H C I3” :-)