|WikiProject Chemistry||(Rated C-class, Low-importance)|
The diagram of the acetoxy group is obviously wrong (the carboxyl group erroneously has an oxygen double-bonded to oxygen, rather than to carbon).
I was going to say the same thing, somebody fix this!!
Other than that, in the Regioselectivity and Stereochemistry subsection, the autor is trying to explain resonance stabilization by telling us the following:
"By inspection of these structures, it is seen that the positive charge of the mercury atom will sometimes reside on the more substituted carbon (approximately 4% of the time)"
First of all, the resonance structures are not shown. Therefore, it might be nesessary to draw them. Second, there is no such thing like "4% of the time etc", the resonance sturctures show mere resonance contributors, which is a model for understanding and visualizing the reactive sites of a molecular species. The molecule or ion does not "switch" between those resonance structures. I shall try to correct those errors.
- D'oh, I can't believe I missed that. I've corrected the acetoxy image. I also agree that 4% of the time is vague. I merely mean that the structure has little weight and is a minor resonance contributor because it is comparatively less stable than the other resonance structures. The structure is meaningful only in the sense that it gives the carbon its electropositive character. I'll get to these corrections shortly, with a complete treatment of the resonance effects, if you don't beat me to it. -D. Wu 05:25, 4 April 2006 (UTC)
I've added a description of the mercury double bond interaction in greater detail. --Nemsen 19:41, 6 January 2007 (UTC)
What about the first structure with the OH and the Hg attached to the same carbon? That should be fixed. 18.104.22.168 16:36, 9 October 2007 (UTC)jeff
The time-resolved description of resonance given in the article obviously needs to be corrected. The structures are not shown, but it is suggested that the "carbocation" formed via resonance is tertiary, whereas the structure given in the mechanism would yield a secondary cation. Furthermore, please note that relative rates of nucleophilic attack should be explained in terms of the respective transition states rather than charge distribution in the substrate.
Third step in the mechanism for Oxymercuration
According to the book: Organic Chemistry, fifth edition by Paula Yurkanis Bruice the third step of the mechanism is that another molecule of water attacks a hydrogen of the water group forming an alcohol and the electrons from de bond O-H (the attacked H) are passed to Oxygen.
Either way is correct. The hydronium will get deprotonated by that acetate ion.--Nemsen 01:41, 5 December 2006 (UTC)
I've posted the mechanism for the demercuration reaction. If you have any questions, please refer to the literature that I've cited.--Nemsen 01:44, 5 December 2006 (UTC)
Are you sure about this mechanism for demercuration? I have McMurray Organic Chemistry 1996 and it says that the mechanism is uncertain but is thought to occur by radicals.
The paper that I have cited was a thorough study on the demercuration reaction mechanism. McMurray is right; the reaction does in fact follow a radical reaction pathway for three main reasons:
1)Demercuration can be accomplished by employing a sodium amalgam.
2)Yields of oxymer/demer products are low when the reactants contained substituents that are sensitive to radicals (i.e. cyclopropanes, olefins, halides, etc.)
3)Side products that are isolated could have only formed when a radical is generated.
Nevertheless, the mechanism that I have drawn also involves a radical pathway (notice the half arrow I drew ;) ). The hydride bridges both the boron and the mercury atom by an 'electron deficient bond.' Thus, the lines connecting the boron and the mercury to the hydride only represent one electron rather than the classical two. These types of bonds behave like radicals. Sorry about the confusion. I hoped that helped. --Nemsen 19:26, 6 January 2007 (UTC)
How does that mechanism explain the scrambling, though? Looks like the C-Hg bond stays intact until the reductive elimination and I would think the red. elim. would retain the stereochemistry. 22.214.171.124 16:34, 9 October 2007 (UTC)jeff
Shall we buy the reagent BH3-THF in the stores?
I have only BF3-aether solution. I would like to do the experiment. Shall we buy the reagent BH3-THF in the stores? The reagent seems to be too dear to a student. —Preceding unsigned comment added by Me3CBr (talk • contribs) 14:33, 23 July 2008 (UTC)
- I presume that you are referring to Hydroboration-oxidation reaction, a similar reaction? If you have BF3.ether, you can prepare BH3 (as B2H6) from it by reaction with NaBH4, then trap it in THF. See J. Am. Chem. Soc. 1958, 80, 1552, and Organic Reactions 13, 1 (1963) describes the apparatus needed. However, diborane catches fire spontaneously in air, so this reaction should NOT be done anywhere except a proper lab hood/fume cupboard, and it should only be performed by an experienced chemist. I'd recommend saving up for some borane.THF, which is safer to work with! One even safer alternative would be 9-BBN, which you may be able to buy more cheaply than borane from a company like Strem. Walkerma (talk) 03:45, 31 July 2008 (UTC)