Certain Aldehydes, ketone and certain alcohol undergo iodoform test. Certain Aldehydes, ketones undergo aldol condensation. While certain aldehydes undergo Cannizaro's reaction. Aldehydes responds to tollen reagent and fehling reagent test.

Which amongst the following undergo iodoform test

Choose the correct option:

B and C only

The correct answer is option 2. B and C only.

Let us look at each of the given compounds:

A. The compound is methanol. 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. 2-Propanol and 1-Propanol are also distinguished using this test.

B. The compound is ethanol. 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 \(Cl_3CHO\). In the third and final step, the hydroxide ion attacks the carbonyl carbon of \(Cl_3CHO\) 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.

C. The given compound is propan-2-ol. Propan-2-ol (also known as isopropanol) can undergo the iodoform reaction. Propan-2-ol contains a methyl group adjacent to a carbonyl group, making it suitable for the iodoform test. The iodoform reaction occurs when propan-2-ol reacts with iodine \((I_2)\) in the presence of a base, typically sodium hydroxide \((NaOH)\). Here's the balanced chemical equation for the reaction:

The reaction forms a yellow precipitate of iodoform (\(CHI_3\)), which confirms the presence of a methyl ketone or a compound containing a methyl group adjacent to a carbonyl group. So, propan-2-ol is suitable for the iodoform test, and the appearance of a yellow precipitate confirms the presence of the necessary functional group.

D. The given compound is propan-1-ol. Propan-1-ol (also known as n-propanol) cannot undergo the iodoform test because it lacks the necessary functional group for the reaction. The iodoform reaction requires a compound containing a methyl ketone group \((CH_3COCH_3)\) or a methyl group adjacent to a carbonyl group. Propan-1-ol does not have these groups; instead, it has a primary alcohol functional group \((–OH)\) at the terminal carbon. The iodoform reaction typically involves compounds like acetone (propanone), which has a methyl ketone group, or compounds like propan-2-ol (isopropanol), which have a methyl group adjacent to a carbonyl group. Since propan-1-ol does not contain these functional groups, it cannot undergo the iodoform reaction.

Thus the correct answer is option 2. B and C only.