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Author Topic: Problem of the Week - 11/16/09  (Read 2280 times)
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azmanam
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« on: November 16, 2009, 03:12:22 AM »

(to be read very dramatically:)

After last week's epic, we open this week with a conformational analysis saga in three acts.  We begin... with Act I.  Our hero - lonely - sits on stage, waiting for his cue.  Annnnd ACTION.

QUESTION: Mechanism and product.  You know the drill Smiley
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Heory
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« Reply #1 on: November 16, 2009, 05:25:04 AM »

 Smiley
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jinclean
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« Reply #2 on: November 16, 2009, 05:40:02 AM »

Yes ,i agree with Heory the CuBr&SMe2 make a 1,4 addition to the carbonly of the ester
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tmartin
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« Reply #3 on: November 16, 2009, 06:34:05 AM »

I agree with the conjugate addition, however I think the reaction will proceed a step further, with the resultant enolate attacking the ester of the original nucleophile.  And that diene will then undergo the photocycloaddition.  Hopefully I attached the chemdraw correctly...

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stewie griffin
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« Reply #4 on: November 16, 2009, 07:01:29 AM »

I agree with tmartin. I propose the following 2+2:
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nj_bartel
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« Reply #5 on: November 16, 2009, 10:03:52 AM »

PotW attracts new members!
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azmanam
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« Reply #6 on: November 16, 2009, 11:40:23 AM »

tmartin is correct that the interesting part of step 1 is the allenolate collapsing back down, displacing ethoxide, to give the cyclopentenone as product X.  Stewie is correct that the product of step 2 is the 2+2 (without retro 2+2, heory) to give the cyclobutane, but stewie gets the diastereoselectivity wrong Sad  You've got the right idea in the transisiton state, but your conformational analysis argument is off, and gives the wrong diastereomer.  I meant to post it in the original problem, but the actual major diastereomer from that photocycladdition is formed in greater than 25:1 dr.
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Heory
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« Reply #7 on: November 16, 2009, 05:43:17 PM »

 Smiley
As I've learned, oganocopper reagents don't react with ester groups. At first I thought that the new oganocopper compound that resulted from Micheal Addition would gain a proton at the work-up step to form an alkene.
But here, the ester group did react, why? Huh
Was that because of HMPA, which bond with copper thus activated the oganocopper reagent? Without HMPA, we would not obtain the desired product, would we?

BTW, are there any mavericks fans? Cheers for today's victory!
« Last Edit: November 16, 2009, 06:22:03 PM by Heory » Logged
stewie griffin
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« Reply #8 on: November 17, 2009, 01:29:28 AM »

Dang. I was debating what was going to be worse to place pseudoaxial... the tert-butyl or the TMSO group. Well I guess silicon is quite fat, and thus I have to agree with Heory's transition states.
Heory: After the organocuprate does the michael addition we're left with essentially a fancy metallated enolate. The metal just happens to be copper here, but I don't see any reason why this metallated enolate shouldn't react similarly to others (potassium, sodium, magnesium, titanium, boron, etc.). So I may be missing something but I didn't think it was a problem at all to imagine the copper enolate attacking another carbonyl functionality.
So what's the others parts to this problem???
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tmartin
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« Reply #9 on: November 17, 2009, 02:02:39 AM »

At first I had the t-butyl group pseudoaxial, however after seeing it drawn out I will agree with Heory that the TMS group will go in that position to avoid some unfavorable interactions.

Along the lines of what Stewie said, while esters are "borderline" in terms of reactivity towards general organocuprates (quote from one of my textbooks),  the result of the mixed cuprate-zinc reagent addition is an enolate, and still a nucleophilic species (same textbook).  If allowed to sit around long enough with an electrophile there would be reaction.  The subsequent reaction generally involves alkyl halides but have been shown with other species.

Some references from said texbook on "tandem reactions of enolates from conjugate addition of organocuprates":

http://dx.doi.org/10.1021/ja00222a034
http://dx.doi.org/10.1021/ja00411a063
http://dx.doi.org/10.1021/jo00179a043  
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Heory
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« Reply #10 on: November 17, 2009, 02:04:43 AM »

organo Cu and Zn compounds are not so active as organo Li, Mg etc. ones, so usually ester groups keep intact in presence of them.
For example, Darzen reaction. We have to wait for azmanam to explain it to us.
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Heory
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« Reply #11 on: November 17, 2009, 02:16:26 AM »

tmartin, sorry I've no access to the data base. Could you explain in more detail?
Do you mean that mixing Zn with Cu will increase activity of either of the two? In addition, long reaction time leads to the further claisen condensation?
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azmanam
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« Reply #12 on: November 17, 2009, 02:18:56 AM »

The point for step 1 is that after conjugate addition, the reactive nucleophile is an enolate.  It's not a standard 'organocuprate' as one things of an alkyl organocuprate.  Enolates react just fine with esters - the net reaction is a conjugate addition followed by a Diekmann condensation.  Think of the second half of step 1 as an enolate nucleophile rather than a cuprate nucleophile.

As for the other transition states, they're the same.  The other two transition states you've drawn are rotations of each other (180o rotation about the horizontal axis).  While heory correctly noted the A1,3 strain in the original TS, having t-butyl axial is an even bigger penalty.  So we need to find another transition state that both minimizes the noted A1,3 strain, and is not penalized for having t-butyl axial.
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tmartin
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« Reply #13 on: November 17, 2009, 02:51:39 AM »

How about this?  Sloppy Chemdraw-ing aside, the t-butyl remains equatorial, and there is no steric interactions between the -OTMS and the ester, instead it is -OTMS and H.  And I don't think the t-butyl should be bumping into the H or ester.

Edit: Sorry for the huge Chemdraw
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Heory
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« Reply #14 on: November 17, 2009, 03:06:21 AM »

I agree with tmartin.
boat-like TS
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azmanam
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« Reply #15 on: November 17, 2009, 03:06:42 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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stewie griffin
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« Reply #16 on: November 17, 2009, 03:13:24 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.
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tmartin
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« Reply #17 on: November 17, 2009, 03:35:09 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).
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Heory
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« Reply #18 on: November 17, 2009, 03:44:57 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?
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Heory
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« Reply #19 on: November 17, 2009, 03:50:02 AM »

 Smiley
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azmanam
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« Reply #20 on: November 17, 2009, 04:06:35 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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« Reply #21 on: November 17, 2009, 04:12:21 AM »

Deprotecting the OH will form a hydrogen bond, and making a lacton is also a good way.
« Last Edit: November 17, 2009, 04:32:59 AM by Heory » Logged
Dan
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« Reply #22 on: November 17, 2009, 05:36:40 AM »

Yeah, it's got to be lactone formation, I think Stewie already said this a few posts earlier.
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azmanam
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« Reply #23 on: November 17, 2009, 09:40:31 AM »

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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stewie griffin
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« Reply #24 on: November 17, 2009, 12:07:58 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 Smiley. 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??
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tmartin
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« Reply #25 on: November 17, 2009, 02:38:26 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.

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Heory
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« Reply #26 on: November 17, 2009, 05:18:01 PM »

 Smiley
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Dan
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« Reply #27 on: November 17, 2009, 09:39:10 PM »

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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Heory
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« Reply #28 on: November 17, 2009, 10:44:56 PM »

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?
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Heory
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« Reply #29 on: November 18, 2009, 03:15:04 AM »

 Angry
Waiting for the answer
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tmartin
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« Reply #30 on: November 18, 2009, 04:00:17 AM »

Something else I was thinking of... we're focusing on the H-bonding ability, but with the pronounced solvent effects, could it be possible that even if we put a better hydrogen bonding partner there, that the ratio would only increase to 2:1 favored?  If we made the acid, would it even be soluble enough in hexanes (the solvent giving the best ratio previously) to run the reaction?

I think the addition of a Lewis acid could help, but wouldn't that be acting similar to what the H-bonding would accomplish?  With less of a pronounced solvent effect?

Just to throw out another different idea while we wait for a hint or confirmation from azmanam... could we differentiate another spot on the molecule.  Put another substituent somewhere that would tweak the transition state and make the one we want more favorable, and would be easily removed or incorporated into the natural product?

Maybe formation of the lactone would mess up the arrangement in the TS or cause the loss of some favorable overlap?  
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Heory
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« Reply #31 on: November 18, 2009, 04:19:00 AM »

Maybe formation of the lactone would mess up the arrangement in the TS or cause the loss of some favorable overlap? 

I cann't agree more with Mr. Cock, for it's also my idea. I can't express clearly due to my poor English. I will think of another way, according to your hints.
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azmanam
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« Reply #32 on: November 18, 2009, 05:04:33 AM »

Quote
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?

Quite true.  We'd still have A1,3 strain in that case.

Quote
I would think if there was a group better able to H-bond than the ester

On the right track, now.

Quote
If we made the acid, would it even be soluble enough in hexanes (the solvent giving the best ratio previously) to run the reaction?

Good thinking.  I suspect with the methylene chain, t-butyl group, and furan, that it could possible go into hexanes, but that would certainly be something to think about.  In fact, they did not convert to the free acid.

Quote
How about increasing the potential of the ester to H-bond - convert it to a carboxylic acid, amide or carboxylate?

Dan for the win.  Everyone circled the answer, but Dan was the first to mention amide formation.  Amide oxygen atoms are much more Lewis basic than ester oxygen atoms.  The authors did in fact swap the ester for the amide, and ran the reaction again with the free alcohol.  In hexanes, with the new, more Lewis basic amide, the selectivity finally favored the desired diastereomer.  Only a modest 3:1 dr, but overcoming a 1:25 dr in favor of the wrong diastereomer, I think the authors felt OK about 3:1.  Good teamwork and problem solving, all Smiley

Quote
I’m thinking why the authors didn't make a lactone. They didn't know the method?

Oh, I'm quite sure they knew about lactonization.  In fact, that is how they determined the diastereoselectivity of the 2+2.  After the photocycloaddition, one of the diastereomers is quite susceptible to lactonization, the other is geometrically impossible.  So conversion of one diastereomer to the lactone confirmed the orientation of the stereocenters in that product.


(edited to correct image)

As to why they didn't lactonize before?  I really don't know.  The step after 2+2 is in fact lactonization, and the lactone remains until like 5 steps from the end.  Perhaps the fused 5,5 system can't form with the unsaturation at the fused position?  I don't know if I buy that argument, because I can build a model of the fused lactone, and it doesn't seem too strained.  I can also fold that into the desired transition state with ease and no major steric restrictions.  Tis a mystery, I suppose.

This work comes from Crimmins synthesis of ginkgolide B.  Quite a fun read all the way through, but I thought this conformational analysis problem that also got us thinking about what to do when reactions don't go the way you plan was worth talking about in detail.
« Last Edit: November 18, 2009, 05:33:20 AM by azmanam » Logged

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azmanam
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« Reply #33 on: November 18, 2009, 07:41:10 AM »

forgot to include the references...

http://dx.doi.org/10.1021/ja001747s
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Heory
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« Reply #34 on: November 19, 2009, 05:45:03 PM »

 Grin
« Last Edit: November 19, 2009, 06:22:29 PM by Heory » Logged
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