In the iodoform reaction, acetone (CH3COCH3) is treated with iodine and a base to form iodoform (CHI3) and an alkoxide ion. The reaction proceeds through a series of steps, starting with the formation of an enolate ion and followed by halogenation and deprotonation. Let's analyze each option to determine which compound is not formed in the iodoform reaction of acetone.

a) CH3COCH2I
In the iodoform reaction, the enolate ion of acetone is halogenated. In this case, the iodine atom replaces one of the hydrogen atoms, resulting in the formation of CH3COCH2I. This compound can be formed in the iodoform reaction, so option a) is not the correct answer.

b) CH3COCHI2
Similar to option a), in the iodoform reaction, the iodine atom can replace two hydrogen atoms on acetone, resulting in the formation of CH3COCHI2. This compound can also be formed in the iodoform reaction, so option b) is not the correct answer.

c) CH3COCl3
In the iodoform reaction, the iodine atom is required for the halogenation step. Chlorine atoms cannot replace hydrogen atoms in acetone during the reaction. Therefore, CH3COCl3 is not formed in the iodoform reaction. This compound is the correct answer.

d) ICH2COCH2I
This compound is not directly formed in the iodoform reaction of acetone. The iodoform reaction involves the substitution of iodine atoms for hydrogen atoms. In this case, there are no hydrogen atoms adjacent to the carbonyl group. Therefore, this compound is not formed in the iodoform reaction.