Dan_Ny Posted March 20, 2011 Posted March 20, 2011 (edited) So guys I had a look at Corey's 1988 synthesis of Ginkgolide B and came over an innocent-looking step - a Baeyer-Villiger-reaction...: Now, it took me a while to see, why it proceeded totally regioselective. Who knows the answer? (I mean, besides me ) (Note: I is not going to be the "usual" reason of migration preference, since the carbons next to the carbonyl are both secondary alkyls...) Hint: From which side is the peroxide going to come from? Edited March 20, 2011 by Dan_Ny
hypervalent_iodine Posted March 20, 2011 Posted March 20, 2011 At a guess, I would say it is to do with stereoelectronics. The cyclopentane with the t-Bu group on it will be feeding more electron density towards the migratory carbon than will the cyclopentene, which would of course stabilise the transition states to a higher degree during the migration. Perhaps I am wrong on that assumption though.
Dan_Ny Posted March 20, 2011 Author Posted March 20, 2011 (edited) Build a model. It's a steric argument. (Why does he use tritylperoxide? Which side does it come from? And which carbon can only migrate after that and why?) Edited March 20, 2011 by Dan_Ny
hypervalent_iodine Posted March 20, 2011 Posted March 20, 2011 Ah, I see it. I had considered that something along those lines may be implicated, but then I couldn't be bothered to model it in my head.
mississippichem Posted March 20, 2011 Posted March 20, 2011 It looks like like we are looking at a competition between an endo and exo transitions state. Is there a pericyclic transition state?
Dan_Ny Posted March 20, 2011 Author Posted March 20, 2011 (edited) Ok, I'll draw it. might take a few minutes, though Ah, I see it. I had considered that something along those lines may be implicated, but then I couldn't be bothered to model it in my head. good! Well the first thing to realize is, that he used a triphenylmethylperoxide for the baeyer villiger oxidation. why would he do that? Now, important to know is that the trityl group is one of the biggest, sterrically most demanding groups that exist. Now look at our substrate. the only side from which this peroxide can possibly come from is from the bottom: Now our intermediate looks like this: Now, look at he next picture. The two carbons i marked green could theoretically both migrate as shown by the green arrows. But look at the carbon I marked red. It is the spiro carbon between two rings and tetrahedral. If the carbon closer to us would migrate, this tetrahedral spiro carbon would become a planar spiro carbon. Now, spiro carbons are never planar - the strain is simply too high (the electon repulsion betwee four coplanar bonds at one atom is too high). Therefore, let the other carbon, far from us, migrate. When it goes for the peroxide oxygen, the red marked carbon will be a bit deformed, but still be tetrahedral. See it? And that's why this one migrates, not the other one. Because of the spiro carbon geometry. He is a genius, after all, this Corey... (sorry, i did not optimize the last few structures, they look a bit weird actually - hopefully you can see it, though) Edited March 20, 2011 by Dan_Ny
hypervalent_iodine Posted March 21, 2011 Posted March 21, 2011 I have my reservations about him. I mean, if you look at his wiki article and scroll to the 2nd last paragraph (end of the Woodward-Hoffman rules), there is a clear implication that he was somehow involved in Woodward's death. Scandal!
Dan_Ny Posted March 21, 2011 Author Posted March 21, 2011 I have my reservations about him. I mean, if you look at his wiki article and scroll to the 2nd last paragraph (end of the Woodward-Hoffman rules), there is a clear implication that he was somehow involved in Woodward's death. Scandal! He was also involved in Woodward's death? Hm, a dangerous person indeed. But my moral vision is beclouded by the beauty of his syntheses, I'm afraid -
Dan_Ny Posted March 22, 2011 Author Posted March 22, 2011 By the way, in the next step, he oxidates an enolate to a alpha-hydroxylactone under the use of an oxaziridine (an "electrophilic hydroxygroup" so to say) (see below). Enolates are known for the fact that they are configurationally unstable at the alpha-carbon. Why does he get complete stereoselectivity for hydroxylation? The next step is something I don't completely understand. Is this a CH-activation? In a paper from 1984? But it seems to be- What is the mechanism of it? A radical-mechanism? Hypervalent?
adianadiadi Posted May 7, 2012 Posted May 7, 2012 (edited) But look at the carbon I marked red. It is the spiro carbon between two rings and tetrahedral. If the carbon closer to us would migrate, this tetrahedral spiro carbon would become a planar spiro carbon. Now, spiro carbons are never planar - the strain is simply too high (the electon repulsion betwee four coplanar bonds at one atom is too high). Sorry to interfere, after a long time, however, I do not see any bond breaking between red and green labelled carbons so that it will make spiro carbon planar. Could you please clarify this? Please try to mark the bonds instead of atoms. Thank you. Edited May 7, 2012 by adianadiadi
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