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Topic: Problem of the Week - 08/31/09  (Read 14367 times)

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Offline azmanam

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Re: Problem of the Week - 08/31/09
« Reply #15 on: September 01, 2009, 11:51:44 AM »
yup, last step is hetero-ene

yup, stereochem is wrong in final product.

what to do, what to do... :)
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Offline Dan

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Re: Problem of the Week - 08/31/09
« Reply #16 on: September 01, 2009, 12:25:18 PM »
Hmmm, by the power of ring flipping...
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Offline azmanam

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Re: Problem of the Week - 08/31/09
« Reply #17 on: September 01, 2009, 01:27:36 PM »
Yes, the ring flip is the key.  Good work.

As fun as pericyclic reactions are, this was a conformational analysis problem at heart.  The oxy cope needed the six-membered chairlike transition state to get the desired stereochem, and the ring flip was important for the ene.

I'm not sure why the shown diastereomer is the product of this exercise.  Perhaps just to get people to think conformational-ly.  In the event, the diastereomer that Dan drew (and I drew when I first attempted the mechanism) is the favored mechanism in the reaction.  In the paper, dr's range from 2.2:1 (alcohol up) to > 25:1 (alcohol up). 

http://dx.doi.org/10.1021/ja066830f

In any event, since we have time left this week, here's a follow up question

Follow Up QUESTION:  In a non-tandem process, the oxy-cope is known for similar compounds with just the free alcohol (that is, without the allyl group - see figure).  If the rate for the free-alcohol version of the oxy-cope is set at K1 = 1, the anionic oxy-cope benefits from a significant rate enhancement, K2 = 1010-1017

Why?
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Offline Dan

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Re: Problem of the Week - 08/31/09
« Reply #18 on: September 01, 2009, 01:29:18 PM »
Hang on, I've been scribbling away and I'm still not convinced. What is wrong with the following scheme, in which all the pericyclic reactions proceed by chair transition states?
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Offline azmanam

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Re: Problem of the Week - 08/31/09
« Reply #19 on: September 01, 2009, 01:34:19 PM »
nothing.  see the paper.  it's substrate dependent, but the 'wrong product' is the major diastereomer in the systems tested.  I'm guessing the diastereomer given as the 'right product' is intended to make the student take a second look.  Just a guess though.
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Offline blind

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Re: Problem of the Week - 08/31/09
« Reply #20 on: September 01, 2009, 03:46:03 PM »
Follow Up QUESTION:  In a non-tandem process, the oxy-cope is known for similar compounds with just the free alcohol (that is, without the allyl group - see figure).  If the rate for the free-alcohol version of the oxy-cope is set at K1 = 1, the anionic oxy-cope benefits from a significant rate enhancement, K2 = 1010-1017

Why?


Does it have something to do with the metal?

Offline azmanam

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Re: Problem of the Week - 08/31/09
« Reply #21 on: September 01, 2009, 03:49:49 PM »
potassium is better than sodium, but any counter cation will increase the rate.
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Offline sjb

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Re: Problem of the Week - 08/31/09
« Reply #22 on: September 02, 2009, 03:33:13 AM »
In any event, since we have time left this week, here's a follow up question

Follow Up QUESTION:  In a non-tandem process, the oxy-cope is known for similar compounds with just the free alcohol (that is, without the allyl group - see figure).  If the rate for the free-alcohol version of the oxy-cope is set at K1 = 1, the anionic oxy-cope benefits from a significant rate enhancement, K2 = 1010-1017

Why?


If I recall correctly, it's to do with the enol  ::equil:: ketone / enolate equilibrium, and so indirectly pKas etc.

S

Offline azmanam

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Re: Problem of the Week - 08/31/09
« Reply #23 on: September 02, 2009, 10:49:32 AM »
Yes, the enolate is much more stable than the alkoxide, and that will keep K-2 small relative to K-1.  It wasn't the answer I was thinking of, but it is correct.

There's another phenomenon at play that makes K2 larger than K1.  Anyone see what else might be going on?
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Offline azmanam

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Re: Problem of the Week - 08/31/09
« Reply #24 on: September 07, 2009, 10:13:12 AM »
The answer lies in the bond dissociation energy.  The alkoxide results in ground state destabilization compared to the neutral alcohol.  Below are a couple of ways of depicting it.

http://dx.doi.org/10.1016/S0040-4020(97)00679-0
Knowing why you got a question wrong is better than knowing that you got a question right.

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