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Idoform Test

Last updated date: 27th May 2024
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What is an Iodoform Test?

The iodoform test is used to determine whether a given unknown chemical substance contains carbonyl compounds with an α-methyl group and the structural formula (R - CO - CH3) or alcohols with the formula R - CH (OH) - CH3. Iodine and aqueous sodium hydroxide (NaOH) solution or potassium iodide (KI) and sodium hypochlorite (NaClO) solution can also be used to conduct the test. In the iodoform test, the unknown is let to react with a mixture of excess iodine and sodium hydroxide.

Iodoform Test Using Iodine and Sodium Hydroxide.

Iodoform tests using l2 and NaOH are more commonly used. A pale yellow precipitate known as triiodomethane or iodoform is produced when iodine is added to the unknown compound that contains aldehydes or ketones in the presence of excess sodium hydroxide, and it has an "antiseptic" odour. Triiodomethane precipitation demonstrates the presence of aldehydes and ketones.

Iodoform Test Using Iodine and Sodium Hydroxide.

Iodine and aqueous sodium hydroxide (NaOH) combine with acetone to produce sodium acetate (CH3 COONa) and iodoform, also known as triiodomethane, which is a yellow precipitate (CH3)

Acetone Showing A Positive Iodoform Test

Acetone Showing A Positive Iodoform Test

Iodoform Test Using Potassium Iodide and Sodium Chlorate (I) Solutions

Sodium hypochlorite is another name for sodium chlorate (I). A small amount of aldehyde or ketone is mixed with a potassium iodide solution, and then sodium chlorate (I) solution is added. The same pale yellow precipitate as before shows the presence of an aldehyde or a ketone.

Iodoform Test Mechanism

Acidic alpha hydrogen from methyl ketone is first removed by the hydroxide ion as a result a water molecule is eliminated. An enolate ion is created as a result of the previously mentioned step. The enolate ion then displaces the iodide ion from iodine. To produce R - CO - l3, this step is repeated twice as methyl ketone contains a total of 3 acidic alpha hydrogen. The carbonyl carbon now forms a bond with a hydroxide ion. As a result, the carbonyl group reforms, and the Cl3 anion is eliminated. Moreover, an R-COOH group is also formed. The basic ion and carboxylic acid group balance each other out. Iodoform precipitates as a result. The following diagram illustrates the reaction's mechanism:

Mechanism of Iodoform Formation

Mechanism of Iodoform Formation

Iodoform is subsequently precipitated into a light yellow substance that has a distinctively "antiseptic" odour (positive test). It is verified that methyl ketone is present. The iodoform test is a fairly effective way to determine whether these methyl ketones or acetaldehyde are present in an unknown substance.

Iodoform Test for Alcohol

Some secondary alcohols that include at least one methyl group in the alpha position also exhibit the Iodoform reaction along with ethanol. Since ethanol is the only major alcohol to produce a positive iodoform test result, the iodoform test is useful in distinguishing ethanol from methanol.

  • Iodoform Test for Ethanol (CH3CH2OH):

The Iodoform test for ethanol involves three consecutive stages of reaction. Ethanol initially undergoes an oxidation reaction that produces CH3CHO. In step 2 hydroxide ion abstracts the alpha hydrogen from CH3CHOforming an enolate ion and water. Enolate ion now abstracts iodide from the iodine molecule. This step is repeated till all the 3 alpha hydrogen atoms originally present in CH3CHOis completely abstracted and replaced with iodine atoms to form Cl3CHO. In the third and final step, the hydroxide ion attacks the carbonyl carbon of Cl3CHOto form a formate ion and iodoform also known as triiodomethane which is a yellow colour precipitate thus confirming the presence of Ethanol. Ethanol hence gives a positive iodoform test.

Iodoform Test Mechanism in Ethanol

Iodoform Test Mechanism in Ethanol

  • Iodoform Test for Methanol:

Methanol (CH3OH) does not answer the iodoform test because there are no alpha hydrogen atoms present in methanol. The same reason holds good for the unreactiveness of tertiary alcohols (alpha hydrogen atoms are absent) to the iodoform test. Hence this test can be used to distinguish between ethanol and methanol. 2-Propanol and 1-Propanol are also distinguished using this test.

Important Questions

  1. Do all the alcohols answer positively to the iodoform test?

All the alcohols that have alpha hydrogen atoms in them i.e primary alcohols and secondary alcohols answer the iodoform test while the tertiary alcohol does not answer this test because tertiary alcohols do not have alpha hydrogen atoms in them.

  1. What does iodoform test for?

The presence of an aldehyde or ketone with a methyl group as one of the groups directly attached to the carbonyl carbon is detected by the iodoform test. Methyl ketone is the name given to such a ketone. In the iodoform test, the unknown is let to react with a mixture of excess iodine and sodium hydroxide.


The following are detected by the iodoform reaction as positive: Methyl Ketones, Acetaldehyde, Ethanol, and Secondary alcohols containing the methyl group in the first position. Triiodomethane precipitate, a yellow substance, is produced successfully. When ethanol tests positive while methanol does not, it is useful to distinguish between the two substances. Tertiary alcohols are not tested with the Iodoform test.

Practice Questions

  1. Which of the following compounds does not answer the iodoform test?

    1. Propanal

    2. Propanone

    3. Ethanol

    4. Acetone

  2. Which of the following is not a reagent used in the iodoform test?

    1. Sodium hydroxide

    2. Iodine

    3. Potassium iodine

    4. Sodium carbonate


  1. (a)

  2. (d)

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FAQs on Idoform Test

1. The chemical molecule C3H6O has two functional isomers, A and B. While isomer A does not produce any iodoform precipitate when heated with NaOH and I2, isomer B does. Write the A and B formula down.

The compound in question has the chemical formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$. The chemical B will be acetone or 2-propanone since one of its functional isomers, B, exhibits an iodoform test that can only be demonstrated by substances containing methyl ketone. Propanal is its functional isomer A.

2. How do you distinguish propan-2-one and pentan-3-one?

The Iodoform test can be used to distinguish between propan-2-one and pentan-3-one. While pentan-3-one does not respond, propan-2-one tests positive and results in the precipitation of pale yellow Iodoform.

3. How do you distinguish benzaldehyde and acetophenone?

The Iodoform test can be used to distinguish between benzaldehyde and acetophenone. Acetophenone, a methyl ketone, answers an iodoform test and yields a yellow precipitate of iodoform, but benzaldehyde does not answer the iodoform test.