[sundberg]

Hi!
I've done a Suzuki coupling reaction recently and currently are writing up a report.
I've understood most of the mechanistical cycle, however I am stuck on how my Pd(II) species is reduced to the active Pd(0).
I have found several references and explanations on how Pd(OAc)2 is reduced (and how other palladium catalysts are reduced by organometallics, amines etc). However, we used PdCl2(PPh3)2 with the only additives being PPh3 and K2CO3. How does this work?

My only thoughts so far is that a ligand exchange between Cl and PPh3 works, followed by addition of H2O to the phosphine on the new ligand which gets deprotonated in two steps by Cl- and K2CO3 and gets expelled as triphenylphosphine oxide. In order for our Pd(PPh3)2 species to form this has two occur twice. Am I on the right track at all? Thanks!

[Dan]

That sounds about right to me. The Pd(II) is definately reduced by P(III).

ie. Pd(II) + P(III) ----> Pd(0) + P(V)

I would suggest the following mechanism, but be aware I am by no means an expert!

                             -Cl^-
Cl-Pd-Cl + PPh₃ -------> Cl-Pd-P⁺Ph₃

H₂O
-----------> Cl-Pd-P(OH)Ph₃ -----------> Pd  + POPh₃
-H⁺                                        -HCl

Which I think is pretty much what you said.

[movies]

That's clever, I've never seen it drawn that way before.  I don't know that I agree with all the arrow pushing, but I see where you are going.  I'm not really sure of the proposal for how phosphines reduce Pd(II) to Pd(0); it's much more common with amines really.  Out of curiosity, what was the solvent for your reaction?  It's known that you can reduce Pd(II) to Pd(0) with solvents like DMF, but that's another kind of bizzarre mechanism.

I guess I would draw it first by transferring a phenyl group from triphenylphosphine to the Pd, so that you get a Pd-phenyl bond and displace one of the Cl atoms.  Then you can draw a reductive elimination to make chlorobenzene and Pd(0), or you could transfer another phenyl group and reductively eliminate to make biphenyl and Pd(0).  I guess I would favor the latter, although it does seem kinda strange.  The phosphines that give up the phenyl groups would probably end up as phosphinic acids (I think).

[Dan]

Yeah the arrow pushing was dodgy, my bad, I have edited it.

I did a course on Pd chemistry this year, and we were told that the mechanisms for these Pd(II) -> Pd(0) reductions are a bit of a grey area - and pretty much to live with it. That was the first way I came up with that's all.

[sundberg]

We used THF (the synthesis was performed with microwave heating in a small vial).
I've been looking all night through books and journals for any kind of hints but all I have found so far is reductions by triethyl amine, organometallics etc. Seems like everyone is pretty much clueless when it comes to this specific system.

Maybe I'm getting too tired here but isn't it possible that 2eq of K2CO3 just snatches of the chlorides to make our Pd(0) species and KCl+KCO3 ?
Thanks for your input!

[movies]

I doubt the potassium carbonate has much to do with the reduction.  In THF you can draw a mechanism similar to what you would draw with an amine: coordination, beta-hydride elimination, and ultimately reductive elimination of HCl.

As Dan said, all of these are more than a little iffy, for sure.

Dan, I like your mechanism more and more each time I look at it!  The only thing that bugs me is the 5-coordinate phosphorus intermediate.  It just seems like those three phenyl groups would be too bulky to allow that.  I wonder if there could be an internal mechanism where you had a hydroxide and a PPh₃ ligand cis to one another on the Pd, and then the two reductively eliminate to form (protonated) triphenylphosphine oxide.  In that case you wouldn't have a full-fledged 5-coordinate P, since you would break the Pd-P bond while you formed the P-O bond.  I like it!