Which of the following gives iodoform test?

\(CH_3CH_2OH\)

The correct answer is option 2. \(CH_3CH_2OH\).

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.

2. \(CH_3CH_2OH\):  The Iodoform test for ethanol involves three consecutive stages of reaction. Ethanol initially undergoes an oxidation reaction that produces \(CH_3CHO\). In step 2 hydroxide ion abstracts the alpha hydrogen from \(CH_3CHO\) forming 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 \(CH_3CHO\) is completely abstracted and replaced with iodine atoms to form \(CI_3CHO\). In the third and final step, the hydroxide ion attacks the carbonyl carbon of \(CI_3CHO\) to 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.

1. \(CH_3OH\): Methanol \((CH_3OH)\) 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.

3. \(C_2H_5COC_2H_5\): The given molecule is diethyl ether and structurally it is represented as follows:

For a compound to show iodoform test, it is mandatory to have a carbonyl group with a methyl group present in the compound at the position adjacent to carbonyl group i.e., compound must be of type \(R−COCH_3\) to show positive haloform test. As there is no carbonyl group present in the given compound, the given compound will not give an iodoform test. Therefore, \(C_2H_5−O−C_2H_5\) will not give an iodoform test.

4. \(HCHO\): The given molecule is formaldehyde and it does not give a positive iodoform test. The iodoform test is used to identify the presence of a methyl ketone group (\( \text{-CO-CH}_3 \)) or a compound that can be oxidized to form such a group, like ethanol or acetone.

Formaldehyde lacks the necessary \( \text{-CO-CH}_3 \) group or its derivatives, so it does not react with iodine in the presence of a base to form iodoform (\( \text{CHI}_3 \)). Instead, the iodoform test is positive for compounds such as ethanol, acetone, or any other compound that can be oxidized to a methyl ketone.