The starting substance for the preparation of iodoform is any one of the following, except (a) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} \quad\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{HCH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\)

Methyl ketones are organic compounds characterized by the presence of a carbonyl group (C=O) with a methyl group (CH₃) on one side. Identifying these compounds is crucial in organic chemistry, and the iodoform test is a well-known method for this purpose. A key feature of methyl ketones is the structure represented by the formula CH₃CO-.

In the iodoform test, the keto group within the methyl ketone reacts with iodine in the presence of a base (usually sodium hydroxide), leading to the formation of a yellow precipitate known as iodoform (CHI₃). This distinct color change is a qualitative indication that a methyl ketone is present. In the context of the given exercise, compound (d) CH₃COCH₃, also known as acetone, has the required methyl ketone structure to produce a positive iodoform test result.

Secondary alcohols are characterized by the hydroxyl group (-OH) attached to a carbon atom that is also connected to two other carbon atoms. The iodoform test serves as a tool to detect secondary alcohols that have at least one methyl group in the alpha position (the carbon atom adjacent to the -OH group).

The reaction mechanism involves the halogenation of the alpha methyl group followed by the formation of iodoform, indicated by the yellow precipitate. In our exercise, compound (a) CH₃CH(OH)CH₃ is a secondary alcohol with a methyl group next to the hydroxyl group, hence it would give a positive iodoform test. In contrast, compound (b) lacks this structure and will not react to produce iodoform.

Halogenation reactions involve the introduction of halogen atoms into an organic compound. In the context of the iodoform test, the specific reaction is the halogenation of methyl ketones or secondary alcohols in the presence of excess iodine and a base. This process is stepwise and initiates with the formation of a halogenated intermediate, which upon further reaction forms the iodoform.

The excess iodine reacts with the methyl group adjacent to the carbonyl or hydroxyl group, resulting in a series of substitutions that eventually lead to the triiodo compound (CHI₃). This compound precipitates out of the reaction mixture as a yellow solid, serving as a visual confirmation of the halogenation reaction and, consequently, the presence of a methyl ketone or suitable secondary alcohol.