Iodoform test is used to check the presence of carbonyl compounds with the structure R-CO-CH3 or alcohols with the structure R-CH(OH)-CH3 in a given unknown substance.
The reaction of iodine, a base and a methyl ketone gives a yellow precipitate along with an “antiseptic” smell. It also tests positive for a few specific secondary alcohols that contain at least one methyl group in the alpha position.
Iodoform Test Description
When Iodine and sodium hydroxide are added to a compound that contains either a methyl ketone or a secondary alcohol with a methyl group in the alpha position, a pale yellow precipitate of iodoform or triiodomethane is formed. It can be used to identify aldehydes or ketones. If an aldehyde gives a positive iodoform test, then it must be acetaldehyde since it is the only aldehyde with a CH3C=O group. Given below are a few example reactions for positive iodoform tests.
Compounds That Give Positive Iodoform Test
- Methyl Ketones
- Secondary Alcohols that contain Methyl Groups in Alpha Position
Iodoform Test Mechanism
First, the Hydroxide ion removes an acidic alpha hydrogen. This results in the formation of an enolate ion. The enolate anion then goes on to displace an iodide ion from the iodine molecule. This process repeats twice to give R-CO-CI3. Now, a hydroxide ion forms a bond with the carbonyl carbon. This leads to the reformation of the carbonyl group and the elimination of the CI3- anion. An R-COOH group is also formed. The carboxylic acid group and the basic CI3- ion neutralize each other. Thus the iodoform is precipitated. The mechanism for this reaction is illustrated below:
Thus, the pale yellow precipitate of iodoform is formed, which can be identified by its characteristic “antiseptic” smell. The presence of the methyl ketone is confirmed. The iodoform test is a very useful method to identify the presence of these methyl ketones or acetaldehyde in an unknown compound.