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Ethyl acetoacetate when reacts with one mole methyl magnesium iodide then product of reaction will be :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` |
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Answer» Correct Answer - C `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). |
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