[faust]

Hello,

I'm looking for the organometallic mechanism of the addition of a copper hydride on a enone by a Michael type addition.

I have the catalytic mechanism


Cu-H + enone --> Copper enolate -->Transmetallation by sigma metathesis with an hydrosilane--> Silyl ether enol + Cu-H.

but I don't have any information about the first step which leeds to copper enolate. I think it's to simplist to consider the attack of H^- in a 1,4 way....

Thank you for your help

[bodegas]

I don't know if I can help, but this kind of reaction resembles the ethylene polimerization with vanadium (Zieggler-Natta catalysis). The orbitals of the double bond estabilish a pi connection with the metal. Then the bond between the metal and the hidrogen attack the double bond with the deslocalization of a pair of electrons to the Oxygen (I think?). I can't be more precise because the chemistry of this reaction is determined by the electronic configuration of the metal center (meaning that it works with vanadium and titanium but copper could have more electrons depending of the other ligands it has (if it has any)).

[FeLiXe]

I have another suggestion, but I am not sure either:

In my eyes The oxygen would form the komplex. Regular head on coordination. That makes the alpha-carbon even more electrophilic, so it is attacked by the hydride. Then you have the enolate.

The same reaction in the other direction, is called beta-elimination and common with enolates. In this case it seems the coordination number of copper is too high. Therefore the hydrogen is taken away from the copper.

[faust]

And what do you think about this :

1- Oxidative Addition of L-Cu-H on the enone with one of the lone pair of the Copper (I) --> forming a Cu(III) complex

2- Reductive elimination by furnishing an hydride to the 4 position and decoordination of the copper forming a Cu(I) specy

[FeLiXe]

I don't know what bond you mean for oxidative addition
do you mean you open up the C=C so you'd get a cupra-cyclo-propane ring?

or do you attack the C=O side-on and get a oxa-cupra-cyclo-propane ring?

I guess in both cases the hydride could move to the C=O carbon and reduce it. In the second case the bond could just move back to the copper and you'd have copper (I) again.

Why do you say 4 position. I'd say the hydride moves to the 1 (i.e. C=O) position because that's where you need to reduce.

[faust]

No; copper hydride is known to be a "soft" hydride (HSAB concept)

If we have the ethyl acrylate, I think the Copper hydride (which have 3 lones pairs) does a 1,4 addition (oxidative addition) forming a Cu(III) complex then reductive elimination (transfering the hydride).

[FeLiXe]

now I understand what you're talking about 

I like organometallic chemistry, but I don't really get what's happening yet.

I know I can't tell you anything new but now that I found out what's going on, I want to post it.
your intermediate looks like this:
http://stud4.tuwien.ac.at/~e0425252/rest/Kupferkomplex.wmf

I don't know which mesomeric structure is more important. Cu(III) sounds kind of exotic but I guess everything is possible in organometallic chemistry ...

[movies]

The reduction step is not and oxidative addition!!  The Cu stays in the same oxidation state the whole time (usually I).  Think about it: you go from a hydride ligand and Cu(I), therefore a net neutral complex, and you add to an enone to give an enolate (anionic ligand) and Cu(?), since the complex must still be neutral (nothing charged was lost) then the Cu must still be Cu(I).

This mechanistic step is most like a "migratory instertion" sometimes called "olefin insertion."  A good example is in the Heck reaction using Pd.  The easiest way to draw this step is through a four-membered ring transition state with the H adding to the beta carbon and the copper ending up on the alpha carbon in a C-bound Cu enolate structure that can isomerize to the eta-3 enolate and the O-bound enolate.

[faust]

Yes! You're totaly right! I have made confusion with the addition of R2CuLi on an enolate!

Thank you again for your help