Michael Reaction

For blog assignment #9, Dr. Mullins explained that we would need to find and work out a problem that deals with anything within chapters 24 and 26.  After reading through both chapters I decided to pick a problem on something that I've had difficulty with trying to understand the concept.  So I decided to work out a problem that involves the Michael reaction from chapter 24.  A Michael reaction involves two carbonyl compounds: the enolate of one carbonyl compound and an alpha, beta-unsaturated carbonyl compound.  The Michael reaction is when the conjugate addition (1,4-addition) of a resonance-stabilized enolate to the beta carbon of an alpha, beta-unsaturated carbonyl system.  In this Michael reaction, the enolate acts as the nucleophile and attacks the beta carbon, which is the key step to this mechanism. The Michael reaction always forms a new carbon-carbon bond on the beta carbon of the Michael acceptor.  The question is below:

Draw the product formed from a Michael reaction with the given starting materials using ^-OEt, EtOH.

In step 1, the enolate is formed by a base, ^-OEt, removing an acidic proton between the two carbonyl groups.  The oxygen atom from the base attacks the hydrogen and the pair of electrons and a negative charge are given to the alpha carbon on the enolate.  This lone pair of electrons and negative charge on this carbon are eventually used as a nucleophile in step 2.  The following reaction depicts step 1 in this Michael reaction:

In step 2, the nucleophilic attack of the enolate to the beta carbon of the alpha, beta-unsaturated carbonyl compound takes place.   After this nucleophilic attack occurs, a new carbon-carbon bond is formed and a stabilized enolate is formed.  Also, after the nucleophilic attack of the enolate on the beta carbon occurs, the electron pair of the double bond and a negative charge are given to the alpha carbon on this compound.  The following reaction depicts step 2 in this reaction:

In step 3, the lone pair and negative charge of the alpha carbon on the compound protonate and attack the hydrogen atom from EtOH.  With this protonation if the enolate and a new carbon-hydrogen bond is formed.  The following step in this reaction depicts step 3:

The Michael reaction involves first the formation of the enolate then the nucleophilic attack at the beta carbon and then protonation.  With these steps being taken correctly the end product should end as such:

I would like to say that I think this was the best idea for a blog yet because we can find something in these chapters that we dont understand then research it and construct a problem over it.  The Michael reaction was a difficult reaction to understand, however after constructing a question and working it out I feel a lot more confident with this reaction.


Sources


1)  Organic Chemistry. Janice Gorzynski Smith. 2nd edition