Which of the following does not give iodoform reaction?

Question

Accepted Answer

The iodoform reaction is a chemical test for the presence of methyl ketones or alcohols that can be oxidized to methyl ketones. The reaction involves the formation of a yellow precipitate of iodoform (CHI₃) when the compound is treated with iodine and a base.

Let's analyze each option to determine why Option 1 is the correct answer:

1. **HCHO (Formaldehyde)**: Formaldehyde does not give the iodoform reaction. The iodoform test is specific for compounds containing the CH₃C=O group (methyl ketones) or alcohols that can be oxidized to such ketones. Formaldehyde lacks the necessary methyl group adjacent to the carbonyl group, as it is an aldehyde with the structure H-C=O. Therefore, it cannot form iodoform, making Option 1 the correct answer.

2. **$CH_3CH(OH)CH_3$ (Isopropyl alcohol)**: Isopropyl alcohol can give a positive iodoform test. It is a secondary alcohol with the structure CH₃-CHOH-CH₃, which can be oxidized to acetone (CH₃COCH₃), a methyl ketone. Acetone readily undergoes the iodoform reaction, producing iodoform. Therefore, this option is incorrect.

3. **$CH_3COCH_3$ (Acetone)**: Acetone is a classic example of a compound that gives a positive iodoform test. It is a methyl ketone with the structure CH₃-CO-CH₃. The presence of the CH₃C=O group allows it to react with iodine and a base to form iodoform. Thus, this option is incorrect.

4. **3-Methylbut-3-en-2-one**: This compound has the structure CH₃-C(=CH₂)-C(=O)-CH₃. It contains a methyl ketone group (CH₃C=O), which can participate in the iodoform reaction. Despite the presence of a double bond, the methyl ketone group is sufficient for the formation of iodoform. Therefore, this option is incorrect.

In conclusion, Option 1 (HCHO) is the correct answer because formaldehyde does not have the structural requirements to undergo the iodoform reaction, unlike the other options which contain or can be converted to methyl ketones.