Question

Ethyl acetoacetate when reacts with one mole methyl magnesium iodide then product of reaction will be:
(a)-

(b)-

(c)-

(d)-

Answer 1

Hint: Ethyl acetoacetate is an organic compound in which there are six carbon atoms, the second carbon atom has the ketone group and the fourth carbon atom has the ester functional group. So, the carbon atom between the ketone and ester group is the acidic methyl group.

Complete answer:
Methyl magnesium iodide is a Grignard reagent in which the halogen atom is iodine and an alkyl group is a methyl group. The formula of methyl magnesium iodide is $C{{H}_{3}}MgI$.
The given compound in the question is ethyl acetoacetate. Ethyl acetoacetate is an organic compound in which there are six carbon atoms, the second carbon atom has the ketone group and the fourth carbon atom has the ester functional group. The formula of Ethyl acetoacetate is given below:

So, the third carbon atom in the compound, i.e., the methyl group between the ketone group and ester group is the most acidic group in the compound. Therefore, when the methyl magnesium iodide attacks the ethyl acetoacetate, then a hydrogen atom will be released from the third carbon atom of the ethyl acetoacetate. Since, after the release of the hydrogen atom, there will be a negative charge on the third carbon atom. In methyl magnesium iodide, magnesium iodide is the positive part and methyl group is the negative part. So, the positive part of the Grignard reagent will attack the negative part of the compound. The reaction is given below:

So, the correct answer is an option (c).

Note:
Don't get confused that, the Grignard reagent will attack the carbonyl group and will form alcohol, but we have to the most acidic part in the compound because the Grignard reagent attacks the acidic part of the compound.