To understand why option D is incorrect, let's first discuss the synthesis of iodoform. Iodoform is a yellow crystalline solid that is commonly used as an antiseptic. It can be prepared from compounds containing a methyl ketone group (RCOCH3) through a reaction called the iodoform reaction.

The iodoform reaction involves the oxidation of the methyl ketone to a carboxylic acid and the subsequent halogenation of the carboxylic acid. The reaction proceeds in the presence of a strong base, such as sodium hydroxide (NaOH), and an iodine source, such as iodine (I2) or sodium hypoiodite (NaOI).

Let's now analyze each option:

a) Ethyl methyl ketone (CH3COCH2CH3): This compound contains a methyl ketone group and can undergo the iodoform reaction. Therefore, option A is a possible precursor for the synthesis of iodoform.

b) Isopropyl alcohol (CH3CHOHCH3): This compound does not contain a methyl ketone group and cannot undergo the iodoform reaction. Therefore, option B is not a possible precursor for the synthesis of iodoform.

c) 3-methyl-2-butanone (CH3COCH2CH(CH3)2): This compound contains a methyl ketone group and can undergo the iodoform reaction. Therefore, option C is a possible precursor for the synthesis of iodoform.

d) Isobutyl alcohol (CH3CH2CH(CH3)OH): This compound does not contain a methyl ketone group and cannot undergo the iodoform reaction. Therefore, option D is not a possible precursor for the synthesis of iodoform.

In summary, iodoform can be prepared from ethyl methyl ketone and 3-methyl-2-butanone, but not from isopropyl alcohol and isobutyl alcohol. Therefore, the correct answer is option D.