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

Options

A. `CH_(3)-overset(O)overset(||)(C)-CH_(2)-overset(O)overset(||)(C)-CH_(3)`
B. `CH_(3)-underset(CH_(3))underset(|)overset(O^(Θ)Mg^(o+)Br)overset(|)(C)-CH_(2)-overset(O)overset(||)(C)-CH_(3)`
C. `CH_(3)-overset(O)overset(||)(C)-underset(overset(o+)(MgBr))overset(Θ)(CH)-CO_(2)Et`
D. `CH_(2)^(-)overset(O)overset(||)(C)-CH_(2)-CO_(2)Et`

Answer

C

Explanation

`CH_(3)-overset(O)overset(||)(C)-CH_(2)-overset(O)overset(||)(C)-O-Et overset(CH_(3)MgBr)rarrCH_(3)-overset(O)overset(||)(C)-underset(overset(o+)(MgBr))overset(Θ)(CH)-CO_(2)Et`
(because of active methylene group acid-base take place).