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Topic: Problem of the Week - 11/16/09  (Read 18162 times)

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Offline azmanam

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Problem of the Week - 11/16/09
« on: November 16, 2009, 09:12:22 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 :)
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #1 on: November 16, 2009, 11:25:04 AM »
 :)

Offline jinclean

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Re: Problem of the Week - 11/16/09
« Reply #2 on: November 16, 2009, 11:40:02 AM »
Yes ,i agree with Heory the CuBr&SMe2 make a 1,4 addition to the carbonly of the ester

Offline tmartin

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Re: Problem of the Week - 11/16/09
« Reply #3 on: November 16, 2009, 12:34:05 PM »
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...


Offline stewie griffin

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Re: Problem of the Week - 11/16/09
« Reply #4 on: November 16, 2009, 01:01:29 PM »
I agree with tmartin. I propose the following 2+2:

Offline nj_bartel

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Re: Problem of the Week - 11/16/09
« Reply #5 on: November 16, 2009, 04:03:52 PM »
PotW attracts new members!

Offline azmanam

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Re: Problem of the Week - 11/16/09
« Reply #6 on: November 16, 2009, 05:40:23 PM »
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 :(  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.
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #7 on: November 16, 2009, 11:43:17 PM »
 :)
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? ???
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 17, 2009, 12:22:03 AM by Heory »

Offline stewie griffin

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Re: Problem of the Week - 11/16/09
« Reply #8 on: November 17, 2009, 07:29:28 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???

Offline tmartin

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Re: Problem of the Week - 11/16/09
« Reply #9 on: November 17, 2009, 08:02:39 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  

Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #10 on: November 17, 2009, 08:04:43 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.

Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #11 on: November 17, 2009, 08:16:26 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?

Offline azmanam

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Re: Problem of the Week - 11/16/09
« Reply #12 on: November 17, 2009, 08:18:56 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.
Knowing why you got a question wrong is better than knowing that you got a question right.

Offline tmartin

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Re: Problem of the Week - 11/16/09
« Reply #13 on: November 17, 2009, 08:51:39 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

Offline Heory

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Re: Problem of the Week - 11/16/09
« Reply #14 on: November 17, 2009, 09:06:21 AM »
I agree with tmartin.
boat-like TS

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