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Topic: Problem of the Week - 11/16/09  (Read 18163 times)

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Offline azmanam

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Re: Problem of the Week - 11/16/09
« Reply #15 on: November 17, 2009, 09:06:42 AM »
Yes, tmartin, that is the correct transition state.  tBu equatorial, rotate the cyclopentenone to minimize A1,3 strain with the OTMS.  There are four possible transition states, two with tBu axial, two with OTMS axial.  OTMS axial, without A1,3 strain is the best of the 4.

Act II.  Alas, the major diastereomer (formed in >25:1 dr) was not the desired diastereomer.  The desired diastereomer arises from the other tBu equatorial transition state (the one that originally suffered from A1,3 strain).  We can't change any of the existing stereocenters in the starting material (they're needed in the natural product), but how can we modify the structure of the 2+2 starting material to aid in reversing the diastereoselectivity toward the desired diastereomer?
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Offline stewie griffin

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Re: Problem of the Week - 11/16/09
« Reply #16 on: November 17, 2009, 09:13:24 AM »
Take off the TMS group to relieve some of the sterics... maybe you could even lactonize the CO2Et with the (now) free alcohol to form a rigid 5 membered ring which should force the desired TS that azmanam has drawn on the right.

Offline tmartin

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Re: Problem of the Week - 11/16/09
« Reply #17 on: November 17, 2009, 09:35:09 AM »
I agree with Stewie.  Seems to be a simple way to modify the reaction if all stereocenters must remain intact.

Heory:  Sorry, I did not see your earlier posts when I responded before.  I was just trying to get at what azmanam said, the enolate would be more reaction than a general run of the mill organocuprate.  The references I posted were from Carey and Sundberg part A (one of the orgo textbooks Ive used), and really amount to something along the following (see figure). 

I've also seen a few examples of allenoates (from conjugate addition) adding into an ester.  I'm not sure if the mix of the zinc/copper will increase the reactivity, they are less basic than say a grignard reagent though, and are mild nuclephiles.  They can add to aldehydes with added BF3 according to the notorious text (A Knochel paper, Tetrahedron Lett., 29, 3887).

Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #18 on: November 17, 2009, 09:44:57 AM »
Thank you. I also much agree with stewie griffin and am waiting for ACT 3, but why aznamam didn't say something about it?

Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #19 on: November 17, 2009, 09:50:02 AM »
 :)

Offline azmanam

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Re: Problem of the Week - 11/16/09
« Reply #20 on: November 17, 2009, 10:06:35 AM »
You have the right idea with deprotecting the secondary alcohol, but I'm looking for a more specific reason. The oxygen atom of the alcohol is the '3' in the A1,3 interaction, so there's still a non-trivial A1,3 interaction.  But deprotecting the alcohol doesn't just remove some of the negative steric strain of the TS, it also adds some positive, stabilizing features to the transition state to actually help make it actively favored (in spite of the still present A1,3 strain)

Heory: yes, this reaction has precedent.  The scheme you do could theoretically be possible, but all examples I've seen have had the intramolecular Claisen (Diekmann) as the acylation step:
http://dx.doi.org/10.1055/s-2006-949653
http://dx.doi.org/10.1021/jo00051a058 (Title: Six-membered cyclic .beta.-keto esters by tandem conjugate addition-Dieckmann condensation reactions)
Organic Syntheses, Coll. Vol. 8, p.112 (1993); Vol. 66, p.52 (1988).
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Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #21 on: November 17, 2009, 10:12:21 AM »
Deprotecting the OH will form a hydrogen bond, and making a lacton is also a good way.
« Last Edit: November 17, 2009, 10:32:59 AM by Heory »

Offline Dan

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Re: Problem of the Week - 11/16/09
« Reply #22 on: November 17, 2009, 11:36:40 AM »
Yeah, it's got to be lactone formation, I think Stewie already said this a few posts earlier.
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Offline azmanam

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Re: Problem of the Week - 11/16/09
« Reply #23 on: November 17, 2009, 03:40:31 PM »
I like the lactone suggestion a lot (I'll actually have more to say about that later), but surprisingly the authors don't form the lactone.  Rather, like stewie mentioned, the TMS group is removed to give the free OH.  As Heory mentioned, in this manner the structure is set nicely for an intramolecular hydrogen bond. 

The hydrogen bonding stabilization turns out to be quite solvent dependent.  The use of nonpolar, aprotic solvents encourage intramolecular hydrogen bonding, but polar protic solvents break up hydrogen bonding.  Thus, even with the OH deprotected, running the photocycoaddition in methanol still gives a >25:1 dr for the undesired product.  THF gives 7:1 dr, and hexanes gives 1.1:1 dr... in all three cases still favoring the undesired diastereomer.

So we're getting less worse selectivity, but still not selective.  This does follow the order the authors describe in the paper.  At this point, the authors modify the structure once more (again, not changing stereocenters) to try to favor the desired diastereomer even more.  Act III: what is one more possible structural variation we could do to try to boost our dr in the right direction? (no, it's not lactone formation)
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Offline stewie griffin

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Re: Problem of the Week - 11/16/09
« Reply #24 on: November 17, 2009, 06:07:58 PM »
Well I'd really like to just make the epimer at the OTMS position. But I can't do that. It seems like that A1,3 strain is being real troublesome. Making the OTMS smaller by converting to just an OH didn't solve the problem, but it helped. And we can't really make the OH any smaller :). So how about we try to make the CO2Et smaller. Here's two suggestions:
1) Reduce the CO2Et all the way down to an alcohol. Not only will it be smaller but the H-bonding interaction will still be present. So that's a win-win..
2) Remove the CO2Et completely. I know that beta-keto esters can be decarboxylated via a Krapcho decarboxylation. However here with the unsaturation present in the dihydrofuran I'm not certain we can do the Krapcho here.
tmartin, Dan, Heory... what do you think??

Offline tmartin

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Re: Problem of the Week - 11/16/09
« Reply #25 on: November 17, 2009, 08:38:26 PM »
I think a Krapcho here would be difficult.  It seems like that anion would be hard to stabilize due to formation of an allene inside the 5-membered ring to make the enolate.  Although I also think it would be great if we could just get rid of the ester.  It worked previously with the TMS group...

Reducing the ester would be a good solution and still allow for H-bonding, but I wonder will it significantly reduce the A1,3 strain?  With the solvent dependence for H-bonding exhibited, I would think if there was a group better able to H-bond than the ester, that would help the selectivity.  The alcohol would probably do it or maybe conversion of the ester to the carboxylic acid (wouldn't help on the reduced A1,3 strain front though), as either could be good H-bonding partners.


Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #26 on: November 17, 2009, 11:18:01 PM »
 :)

Offline Dan

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Re: Problem of the Week - 11/16/09
« Reply #27 on: November 18, 2009, 03:39:10 AM »
I like the stronger Lewis acid idea, but could BCl3 be a bit too powerful? If you activate the carbonyls too much you're asking for side reactions...

How about increasing the potential of the ester to H-bond - convert it to a carboxylic acid, amide or carboxylate? Although, if you tried to hydrolyse the ester in acid, you'd probably get lactonisation anyway so it seems unlikely that this is what they did...

edit: sorry, just realised tmartin beat me to it!
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Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #28 on: November 18, 2009, 04:44:56 AM »
I’m thinking why the authors didn't make a lactone. They didn't know the method? The lactone was difficulty to make due to large stain or 2+2 TS of the lactone is hard to organise also due to large stain? Or they found a better way?

I though hydrolysing the ester is enough. If this till doesn't work well I would use Lewsis acid. How about milder LAs such as AlCl3, FeCl3, TiCl4, ZnCl2?

Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #29 on: November 18, 2009, 09:15:04 AM »
 >:(
Waiting for the answer

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