This question involves the iodoform test, a specific chemical reaction used to identify methyl ketones (like acetone) and compounds that can be oxidized to methyl ketones (like ethanol).

Step 1: Understanding the Reaction
When a methyl ketone (a ketone with the structure R-CO-CH₃) is treated with iodine (I₂) and a base (like potassium hydroxide, KOH), it undergoes a haloform reaction. The base first facilitates the alpha-halogenation of the ketone. Since acetone has three equivalent alpha-hydrogens, all three can be successively replaced by iodine atoms.

Step 2: The Mechanism
The reaction proceeds through the following key steps:

1. The base (OH^-) abstracts an acidic alpha-proton from acetone, forming an enolate ion. ${\mathrm{CH}}_3\mathrm{COCH}3+{\mathrm{OH}}^{-}\to {\left[{\mathrm{CH}}_2\mathrm{COCH}3\right]}^{-}+H_{2}O$
2. The nucleophilic enolate ion attacks an I₂ molecule, leading to the substitution of an alpha-hydrogen with an iodine atom. This step repeats until all three hydrogens are replaced, forming triiodoacetone (CI₃COCH₃).
3. The hydroxide ion then attacks the carbonyl carbon of the triiodo ketone. The unstable intermediate undergoes cleavage, ejecting a triiodomethyl anion (CI₃^-), which is a good leaving group.
4. The triiodomethyl anion rapidly protonates and decomposes to form iodoform (CHI₃), a yellow precipitate, and the carboxylate ion (in this case, acetate ion). $\mathrm{CI}3C(=O)\mathrm{CH}3+{\mathrm{OH}}^{-}\to {\mathrm{CH}}^3{\mathrm{COO}}^{-}+\mathrm{CHI}3$

Final Answer: The major product is a yellow precipitate of iodoform (CHI₃). The other product is acetate ion, which would form acetic acid in an acidic workup.

Why the other options are incorrect:

• Iodoacetone: This is a mono-substituted intermediate (CH₂ICOCH₃), not the final product of the reaction.
• Acetophenone: This is a different compound (C₆H₅COCH₃) and is not related to this reaction.
• Acetic acid: This is a co-product (from the acetate ion), but it is not the characteristic, easily identifiable yellow solid that is the major focus of this test.

Related Topics & Formulae

Haloform Reaction General Equation:
For a methyl ketone: R-CO-CH₃ + 3X₂ + 4NaOH → R-COO^-Na⁺ + CHX₃ + 3NaX + 3H₂O
(Where X can be Cl, Br, or I)

Compounds that give a positive iodoform test: Acetaldehyde, Ethanol, and any methyl ketone.