Which of the following compound(s) on being warmed with \(I_2\) solution and \(NaOH\), will give iodoform?

A. \(CH_3CH_2OH\)

2. \(CH_3-CO-CH_3\)

3. \(CH_3-CHOH-CH_3\)

4. \(CH_3OH\)

Choose the correct option from the ones given below:

A, B, C only

The correct answer is option 3. A, B, C only.

The iodoform test is a chemical test used to identify the presence of methyl ketones or secondary alcohols. The reaction involves the use of iodine (\(I_2\)) in the presence of sodium hydroxide (\(NaOH\)).

The general reaction for the iodoform test can be represented as follows:

\[ R-CHOH-R' \, \text{or} \, R-CO-CH_3 + 3I_2 + 4NaOH \rightarrow R-COO^-Na^+ + 3I^- + 3H_2O \]

If \(R\) is a methyl group, then the compound will react positively in the iodoform test, forming a yellow precipitate of iodoform (\(CHI_3\)).

Now, let's analyze the given compounds:

1. \(CH_3CH_2OH\) (Ethanol) - This is a primary alcohol. Ethanol is the only primary alcohol to give the triiodomethane (iodoform) reaction.

2. \(CH_3-CO-CH_3\) (Acetone) - This is a methyl ketone. Methyl ketones react positively in the iodoform test, forming iodoform.

3. \(CH_3-CHOH-CH_3\) (Isopropanol) - This is a secondary alcohol. Secondary alcohols react positively in the iodoform test, forming iodoform.

4. \(CH_3OH\) (Methanol) - This is a primary alcohol. Primary alcohols do not give a positive iodoform test.

Based on the analysis, compounds 2 (acetone) and 3 (isopropanol) will give a positive iodoform test. Therefore, the correct option is 3. A, B, C only. These compounds will produce a yellow precipitate of iodoform when warmed with \(I_2\) solution and \(NaOH\).