[qw098]

Hi Colleagues,

I was wondering what the geometry of the enolate would have to be in the diagram below for the elimination of the hydroxide.

Most importantly, I'd like to understand why

Thanks!

[discodermolide]

Hi Colleagues,

I was wondering what the geometry of the enolate would have to be in the diagram below for the elimination of the hydroxide.

Most importantly, I'd like to understand why

Thanks!

Try drawing a six membered ring with the metal coordinating between the enolate oxygen and the OH group. Examine the possible conformers.

[qw098]

Try drawing a six membered ring with the metal coordinating between the enolate oxygen and the OH group. Examine the possible conformers.

I just did that below. Not quite sure what you mean/what do I do now?! And where did this coordinating metal come from?!

[Dan]

where did this coordinating metal come from?!

You have hydroxide in the scheme in your first post. You can't get a bottle of ^-OH, you need a positive counterion. That cation is your coordinating metal ion.

[discodermolide]

Try drawing a six membered ring with the metal coordinating between the enolate oxygen and the OH group. Examine the possible conformers.

I just did that below. Not quite sure what you mean/what do I do now?! And where did this coordinating metal come from?!

I don't agree with your drawing of the enolate, look up the conformation of cyclohexenes.

[qw098]

You have hydroxide in the scheme in your first post. You can't get a bottle of ^-OH, you need a positive counterion. That cation is your coordinating metal ion.

Awesome Dan, thanks a lot! My coordinating metal would be Na in this case because we used NaOH in the lab!

I don't agree with your drawing of the enolate, look up the conformation of cyclohexenes.

Ok, thanks discodermolide. I never knew cyclohexenes had a different conformation than cyclohexanes. I just did a little bit of reading up on, and my cyclohexene would look like the drawing I have posted below.

In my "3" position I would have my enolate, in the "4" position I would have the coordinating metal, Na in my case, and finally at "5" I would have by hydroxide!

So when I am looking for the geometry of the enolate, what would I say? The enolate and hydroxide are opposite? There must be a nice word that describes this geometry! Or would I simply draw this cyclohexene conformation and that would be a good enough description of the geometry!

Thanks discodermolide for pushing me in the right direction, it was awesome!

[discodermolide]

You have hydroxide in the scheme in your first post. You can't get a bottle of ^-OH, you need a positive counterion. That cation is your coordinating metal ion.

Awesome Dan, thanks a lot! My coordinating metal would be Na in this case because we used NaOH in the lab!

I don't agree with your drawing of the enolate, look up the conformation of cyclohexenes.

Ok, thanks discodermolide. I never knew cyclohexenes had a different conformation than cyclohexanes. I just did a little bit of reading up on, and my cyclohexene would look like the drawing I have posted below.

In my "3" position I would have my enolate, in the "4" position I would have the coordinating metal, Na in my case, and finally at "5" I would have by hydroxide!

So when I am looking for the geometry of the enolate, what would I say? The enolate and hydroxide are opposite? There must be a nice word that describes this geometry! Or would I simply draw this cyclohexene conformation and that would be a good enough description of the geometry!

Thanks discodermolide for pushing me in the right direction, it was awesome!

They should be trans and antiperiplanar for the OH to eliminate. i.e. the enolate  must be a trans enolate, if my memory serves me correctly!

[qw098]

Ok awesome!

And the enolate and the hydroxide are trans/antiperiplanar which makes sense with the conformation I drew up above... right?

[discodermolide]

Ok awesome!

And the enolate and the hydroxide are trans/antiperiplanar which makes sense with the conformation I drew up above... right?

Yes
I should add that what you have here is a movement of electrons, but the same principles apply.
This is the beauty of the aldol reaction, a six membered cyclic transition state, where the stereochemistry of the product can easily be predicted. Look at some of the work of Evans and also Paterson.

[qw098]

Sweet! I'll definitely take a look at the work by Evans and Patterson! Thanks again for the great explanation!

[discodermolide]

Sweet! I'll definitely take a look at the work by Evans and Patterson! Thanks again for the great explanation!

Np, hope it is correct and helped:))