Question

When ethyl acetoacetate (CH3COCH2CO2CH2CH3) is treated with one equivalent of CH3MgBr, a gas is evolved from the reaction mixture, and after adding aqueous acid, ethyl acetoacetate is recovered in high yield. Identify the gas formed and explain why the starting material was recovered in this reaction. Be sure to answer all parts.

Answer 1

Answer:Methane gas is evolved from the reaction mixture.

Explanation:

When ethyl acetoacetate is treated with grignard reagent a carbanion is generated.

There are two acidic hydrogens which are present on the carbon which is in between the ester and the ketone group in ethylacetoacetate.

These two protons are also called active methylene protons and they are very acidic in nature due to the presence of two electron withdrawing substituents that is an ester and ketone.

CH₃MgBr is grignard reagent and it is an organo-metallic copmpound  . Carbon here in CH₃MgBr exists as carbanion  CH3⁻ which is basic enough to abstract the acidic protons present on ethylacetoacetate.

As CH3⁻ abstracts an acidic  proton from ethylacetoacetate it become CH₄ which is methane. As methane is a gas so it is methane gas which is evolved from the reaction mixture.

As  the acidic proton is abstracted from ethylacetoacetate which leads to generation of carbanion and this carbanion is very stable as it can be delocalized on to the two carbonyl groups . As we add aqueous acid to the reaction mixture the carbanion can again be protonated and its protonation would lead to the generation of ethylacetoacetate again.