![]() | This article is rated B-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||
|
May I suggest a few possible references. The initial stoichiometric reaction was first reported by Phillips:
The Wacker reaction was first reported by Smidt et al.:
(Only the last article is in English). Hope this is useful. Joseph Wright
Spuddddddd 8Mar08 —Preceding unsigned comment added by 193.60.90.97 ( talk) 20:37, 8 March 2008 (UTC)
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)
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)
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)
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)
alternative drawing] might help.-- Stone ( talk) 09:11, 3 July 2009 (UTC)
I like the several possible mechanism from some books:
-- Stone ( talk) 11:58, 3 July 2009 (UTC)
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)
![]() | This article is rated B-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||
|
May I suggest a few possible references. The initial stoichiometric reaction was first reported by Phillips:
The Wacker reaction was first reported by Smidt et al.:
(Only the last article is in English). Hope this is useful. Joseph Wright
Spuddddddd 8Mar08 —Preceding unsigned comment added by 193.60.90.97 ( talk) 20:37, 8 March 2008 (UTC)
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)
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)
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)
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)
alternative drawing] might help.-- Stone ( talk) 09:11, 3 July 2009 (UTC)
I like the several possible mechanism from some books:
-- Stone ( talk) 11:58, 3 July 2009 (UTC)
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)