Talk:Wacker process

May I suggest a few possible references. The initial stoichiometric reaction was first reported by Phillips:

F.C. Phillips, Am. Chem. J., 1894, 16, 255-277.
F.C. Phillips, Z. Anorg. Chem., 1894, 6, 213-228.

The Wacker reaction was first reported by Smidt et al.:

J. Smidt, W. Hafner, R. Jira, J. Sedlmeier, R. Sieber, R. Rüttinger, and H. Kojer, Angew. Chem., 1959, 71, 176-182.
W. Hafner, R. Jira, J. Sedlmeier, and J. Smidt, Chem. Ber., 1962, 95, 1575-1581.
J. Smidt, W. Hafner, R. Jira, R. Sieber, J. Sedlmeier, and A. Sabel, Angew. Chem., Int. Ed. Engl., 1962, 1, 80-88.

(Only the last article is in English). Hope this is useful. Joseph Wright

really good article guys, great job :)

Spuddddddd 8Mar08 —Preceding unsigned comment added by 193.60.90.97 (talk) 20:37, 8 March 2008 (UTC)[reply]

The Reaction mechanism figuere

if keto-enol tautomerization is not a possible mechanistic step, where does the 3rd H at the final carbon atom of the last step come from? Because in the figure it seems to be the hydrogen that was bond to the oxygen before and this one is from the solvent/water. If one would use D2O there would be a D in the product. So if I'm not misstaken either the explanation or the mechanism-drawing is wrong....nevertheless you guys have a quite goog article on that topic I suppose. lg phil --132.230.20.102 (talk) 13:36, 17 July 2010 (UTC)[reply]

In the figuere Pd is in oxidation state II all the way, it should change to 0... ChristianB (talk) 15:00, 23 May 2009 (UTC)[reply]

Really? If you look at the structure each intermediate, you'll see that Pd is always bonded to at least two chlorides.

Which intermediate do you think is wrong and should be Pd(0)? Please give a reference if possible.

Ben (talk) 18:25, 23 May 2009 (UTC)[reply]

It almost certainly does not go to 0 at any point during the reaction, because it would then be lost to the process. A strong oxidising agent would be required to get it back in solution.

On a sidenote, I don't think the summary description of the process is correct. The whole point of the Cu(II) is that the Pd(II) does not ever go to Pd(0), so the first and second line should probably be combined to give something akin to:

C2H4 + H2O + 2CuCl2 ---> CH3CHO + 2CuCl + 2HCl (in the presence of [PdCl4]2-, but I have no idea how to display it above the arrow)

The 2nd line of the description is not a chemical reaction that takes place, as easily demonstrated by taking some palladium metal and exposing it to a cupric chloride / hydrochloric acid mixture. In fact I believe it is possible to reduce Pd(II) to Pd(0) by exposing it to Cu(0), so going the other way. Harting (talk) 12:37, 2 July 2009 (UTC)[reply]

I did actually have a look and found that three step summary in a book I trust. It's on page 1322 in Comprehensive Inorganic Chemistry Volume 2, 1973 Pergamon Press Ltd. Perhaps I was wrong then, even though it still seems unlikely to me that an actual oxidation of Pd(0) to Pd(II) physically takes place. Harting (talk) 08:46, 3 July 2009 (UTC)[reply]

What do you think is the oxidation state of [PdHCl₂(CH₂=CHOH)]^- ? for me it is 0. --Stone (talk) 13:41, 2 July 2009 (UTC)[reply]

I reckon it's Pd(II). The Cl- are both -1, Hydrogens bonded to Palladium are generally hydrides so -1, the CH2CHOH looks like 0 and the overall charge is -1. 2 * -1 (Cl) - 1 (H) + 0 (organic) + 2 (Pd) = -1 Harting (talk) 08:46, 3 July 2009 (UTC)[reply]

(SO4)2- = 4 * -2 (O) + 6

The H is H+ and will undergo a HCl elimination if possible and then you end up with Pd(0). The whole cycle only makes sence when you go from Pd(2) to Pd(0) without it the process would not need a reoxidation of the Pd by the copper.--09:10, 3 July 2009 (UTC)

The point I was trying to make is that cupric chloride is not a sufficiently strong oxidising agent to oxidise Pd(0) to Pd(II). In fact, copper metal reduces Palladium in solution! If at any point in the industrial Process Pd metal was formed it would not react any further. It would coat the walls of the reactor and the process would soon come to an end. Therefore it is very likely that an intermediate is formed that consists of Pd(II), organic acid and Cu(II) which results in Pd(II), a ketone and Cu(I). I think the Cu(II) is used to "catch" the free electron, so that the Pd(II) doesn't snap it up. In a way the Cu(II) acts as an oxidation state buffer for the Pd(II). Of course that's just my opinion and not a peer reviewed journal article, but I doubt there are any peer reviewed journal articles that mention oxidation of Palladium metal by cupric chloride. 78.105.104.189 (talk) 22:54, 4 July 2009 (UTC)[reply]

Basically you are right, but you do not deal with solid Pd, but with a stable Complex in which the electrochemical potential could be a lot different from what is expected in water.--Stone (talk) 23:17, 4 July 2009 (UTC)[reply]

alternative drawing] might help.--Stone (talk) 09:11, 3 July 2009 (UTC)[reply]

I like the several possible mechanism from some books:

--Stone (talk) 11:58, 3 July 2009 (UTC)[reply]

catalytic cycle is mad scuffed

The catalytic cycle has a few errors in it, likely due to copy/paste errors. The pictured catalytic cycle is inconsistent with the text and with chemical literature.

Numerous palladium species should not be bearing a formal negative charge as the oxidation state of Pd does not change from the +2 oxidation state until the very end after reductive elimination. Some species are drawn as being in the +1 or +3 oxidation state. The palladium species in question are the two alkylpalladium complexes on the bottom, the palladium aqua olefin complex on the middle right, and the alkypalladium hydroxide complex on the middle left.

Additionally, many of the palladium species shift between being aqua and hydroxide complexes, which is also inconsistent with literature. — Preceding unsigned comment added by ChugJugWithU (talk • contribs) 06:59, 19 March 2021 (UTC)[reply]