[azmanam]

yup, last step is hetero-ene

yup, stereochem is wrong in final product.

what to do, what to do...

[Dan]

Hmmm, by the power of ring flipping...

[azmanam]

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?

[Dan]

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?

[azmanam]

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.

[blind]

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?

[azmanam]

potassium is better than sodium, but any counter cation will increase the rate.

[sjb]

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  ketone / enolate equilibrium, and so indirectly pKas etc.

S

[azmanam]

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?

[azmanam]

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