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A little quiz...


Dan_Ny
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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...:

post-40949-0-89136600-1300597178_thumb.gif

Now, it took me a while to see, why it proceeded totally regioselective. Who knows the answer? (I mean, besides me :P )

(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 by Dan_Ny
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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.

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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 by Dan_Ny
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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.

 

:rolleyes: 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.

 

post-40949-0-55517700-1300642716_thumb.gif

 

post-40949-0-52631300-1300642948_thumb.gif

 

Now look at our substrate. the only side from which this peroxide can possibly come from is from the bottom:

 

post-40949-0-31767000-1300642694_thumb.gif

 

Now our intermediate looks like this:

 

post-40949-0-41070400-1300642701_thumb.gif

 

 

Now, look at he next picture.

 

post-40949-0-60333000-1300642709_thumb.gif

 

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 by Dan_Ny
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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! tongue.gif

 

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 -

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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?

 

post-40949-0-80245700-1300765213_thumb.gif

 

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?

 

post-40949-0-78967800-1300765682_thumb.gif

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  • 1 year later...

 

 

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 by adianadiadi
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