[azmanam]

Thought I'd give this board some life and get us to exercise our mental muscles a bit   So I'm initiating a Problem of the Week.  I'll grab a challenging problem (usually from Evans' site*, but not necessarily).  If people seem to like it, I'll try to post one every Monday or so.  But I don't have to be the only one posting problems.  If anyone else wants to try to challenge the others, go ahead (but lets try to only have one unsolved problem at a time, k?)





So, the inaugural problem.  A bit of aldol chemistry, something near and dear to my heart:

QUESTION:  Provide the transition state which accounts for the observed syn selectivity in the following racemic aldol addition (eq 1).  For eq 2, ponder the transition state for the analogous "imine-aldol" addition.  Draw the transition state and illustrate the predicted stereochemical outcome of the addition reaction.  Assume the imine geometry cannot isomerize under the reaction conditions.




Good luck.  I look forward to discussing the answer to the problem, because I came up with a different transition state which gives the wrong answer... but I think I can justify my answer so someone's going to have to talk me out of it.

*http://www2.lsdiv.harvard.edu/labs/evans/problems/index.cgi

[James Newby]

My answer to part 1.  Aldol reactants line up next to each other to maximise orbital overlap and allow reaction to occur.  This leads to a pseudo 6-membered ring transition state (hopefully in the pic)

Sorry in the picture it looks like ive drawn the Z enolate, supposed to be the E enolate

[azmanam]

Cool.  Right track... but not quite the right transition state.  You still have the E(O) enolate (and your text describes the E(O) enolate) in the transition state.  That does give Me up and Ph back as you've drawn in cpd 3, but rotation puts Ph in the plane, and OH back to give the anti product.

*You call it the Z enolate, which is right by IUPAC rules, but for enolates, I've always named them with respect to the former carbonyl oxygen... as the R group (iPr here) changes, the IUPAC naming may change.  By keeping consistent with naming enolates with respect to oxygen, there's no naming confusion.  Thus the SM in the question shows what I would call the Z(O) enolate.

Try the TS again with the Z(O) enolate.  extra credit: wanna name the transition state.  that 6-membered ring has a special name for these types of reacitons.

[James Newby]

ahh! Im supposed to know this for my exam on 3rd june so not a good sign! Transition state is Zimmerman traxler transition state.

[azmanam]

yup, Z-T it is.  If you just redraw with the right enolate geometry I think you'll be on the right track.

[James Newby]

ah so Me goes axial in Z-T transition state, leads to syn Me and Ph.

I drew the Ph pointing the wrong way in my pic as well!  Rotation of the Ph back into the plane of the board moves OH towards us ie solid wedge.  Cheers Azmanam

[azmanam]

so Me goes axial in Z-T transition state, leads to syn Me and Ph.

Yup.  Here's how I drew the transition state, and the on-paper way I unfold Z-T transition states into linear molecules to confirm relative stereochemistry.

I think Ph is fine - it should be pseudoequatorial... but note the isopropyl group is pseudoaxial.  Why is it important that isopropyl is pseudoaxial?

Also: why does enolization of esters of the type shown in the SM give predominantly Z(O) enolates?

[azmanam]

*Ignore me, I am impatient*

anyone wanna take a shot at part b?

[spirochete]

I think Ph is fine - it should be pseudoequatorial... but note the isopropyl group is pseudoaxial.  Why is it important that isopropyl is pseudoaxial?

My guess: In the lowest energy conformation you have a methyl group in the axial position, but it contributes almost zero strain because nothing else is axial in that conformation.  The axial iPr interacts with just one hydrogen.

In the alternate conformation there is an equatorial iPr-methyl interaction.  A semi-rare case where equatorial strain is actually worse than anything contributed by axial. 

I can't remember the specific name for that kind of strain though.

[azmanam]

You're thinking of A(1,3) strain.  That's part of the reason why you get the Z(O) enolate preferentially, but once it's locked into the Z(O) enolate, the methyl group and the iPr group will always be 'trans' to each other.  If you had the E(O) enolate, you'd have interaction b/t iPr and methyl, but in the chair transition state, the iPr pretty much has to be pseudo axial due to the orientation of the enolate geometry.  What else is happening with the iPr in the Z-T transition state which makes it important that iPr is psuedoaxial?  And what do you think might change in the 'imine aldol' case?

[aldoxime_amine]

*http://www2.lsdiv.harvard.edu/labs/evans/problems/index.cgi

I would only like to mention that that site is one of the best I have ever seen . Finally a 'math like' problems database!! I would love to discuss them!

Thanks a lot azmanam..

[azmanam]

It is pretty awesome.  Very challenging problems, but very awesome.  Wanna give the 'imine aldol' a try?

[aldoxime_amine]

Even though I understand the reactions, I must admit I am not much aware of the matter which you guys are discussing...

[Dan]

Interesting. I have drawn the Z-T type transition states, but I can't decide...

In the first one there is 1,3-diaxial clashing of the ⁱPr-Ph, and the second would have to proceed in a boat conformation due to the R group on the E-imine.

What's worse, 1,3-diaxial or boat vs chair? Or am I barking up the wrong tree?

[azmanam]

Nope, that's the right tree.  The salient point is that because the imine can't isomerize, for the reaction to go through the chair TS, the phenyl group is forced into the pseudo axial position in the chair-like Z-T transition state.  To put the phenyl group in the less sterically demanding pseudo-equatorial positoin means the lone pair on the imine can't interact with the metal.  The only way to get the phenyl group in the pseudo-pseudo-equatorial position is to go through the boat-like transition state. 

When I was working through the problem, I said the phenyl/isopropyl steric demands in the chair like TS were too large for the reaction to proceed through the chair-like transition state.  I said the boat-like transition state was the better TS.  It seems however, that the chair-like transition state was the answer they were looking for.  Maybe in this case, you can put the H of the isopropyl group over the chair and rotate the phenyl group such that the steric demands aren't so large?

I don't know.  I still like the boat TS leading to the same syn product, but the correct answer is the chair TS leading to the anti product.