[blaisem]

Hi there, sorry for the ambiguous topic name.  I have two questions from a set of notes in which the answer is not supplied, and the notes were submitted online for a course at a different university (so I can't ask the professor).  I was hoping someone here would be able to assist me.

The first question asks which salt will form from a reaction (see image).  I chose product A with the reasoning that the lone pair on former carbonyl (trans to the nitrogen) could delocalize into the antibonding sigma orbital of the N-H sigma bond.  This stabilization is otherwise known as the Anomeric Effect.  I have attempted to depict this interaction in my image.

The second question asks which conformation for two different molecules is preferable.  The two molecules in question are:

Molecule 1:
H₃COCH₂OH

Molecule 2:
FCH₂OH


I answered (also in my image) that the gauche-conformation along the O-C bond is correct for both molecules.  My reasoning is again based on the Anomeric Effect, namely that one Oxygen lone pair would prefer to delocalize into an antibonding sigma orbital of C-H, while the second lone pair would prefer to delocalize into the antibonding sigma orbital of either O-H (Molecule 1) or C-F (Molecule 2).  The preference of the latter case for O-H or C-F over delocalizing into a second C-H bond is that the antibonding sigma orbital of the more electronegative O-H and F-H bonds is lower in energy and therefore a better acceptor.

Are my answers and reasoning correct? Thank you for reading and for any help you can provide!

[synthnick]

I will attempt to solve the first problem.
The anomeric effect is better when the ΔE of the donor and the acceptor is as small as possible. Both compounds have 2 secondary electronic effects (shown in the picture)
 C-O bonds are stronger than C-N so the σ* of C-O are higher in energy("bad" anomeric effect) than σ* of C-N ("good" anomeric effect)
So I think B is more stable.

PS1: i don't think that either A or B is formed under kinetic control
PS2:i think that the σ* of N-H lies on the H
hope I helped

[blaisem]

I will attempt to solve the first problem.

So I think B is more stable.

PS1: i don't think that either A or B is formed under kinetic control
PS2:i think that the σ* of N-H lies on the H
hope I helped

Hi there! Thanks for taking the time to respond.  Unfortunately, I don't have access to a scanner for a few days, so I can't draw a picture back for you   Hopefully, my comments make sense without a picture to help with the details.

C-O bonds are stronger than C-N so the σ* of C-O are higher in energy("bad" anomeric effect) than σ* of C-N ("good" anomeric effect)

The more electronegative Oxygen lowers the energy of both the bonding and antibonding orbitals, even more than Nitrogen.  The bonding orbital having a lower energy makes it more "bound," so it is a worse donor and a stronger bond.  The lower energy of the antibonding orbital makes it a better acceptor.  That's my understanding of how it works.

So I think B is more stable.

Nice picture.  It seems though that the interactions you drew were the same for both pictures, except that in B, two lone pairs are "competing" for the same σ* orbital.  Wouldn't this be less favorable?

I also noticed you drew sp² next to both Oxygens, along with a conformational equilibrium in B.  The Oxygens should be sp³ with two lone pairs on each of them.  I only drew one lone pair in my picture, but the other one is still there.  Maybe this is some of the confusion?

PS2:i think that the σ* of N-H lies on the H

What do you mean by this? I did not know that antibonding orbitals were localized on atoms?

Thanks!

[synthnick]

Hi
First of all , carbamates (i am not 100% sure about the name) are planar and have an extensive pi system so N,C,O,O are sp2 hybridized (just like esters)
B can have two conformations the upper or the lower, the lower is higher in energy due to electron-electron repulsion. So we have to compare A with upper B.Both have one n(Osp2) :rarrow:σ*(C-O) but A has one more but upper B has n(Osp2) :rarrow:σ*(C-N).
So whether σ* is lower on energy , the "better" the anomeric effect ,the preferred product.
With the N-H comment i meant that for the σ* the bigger coefficient would be on H (more electropositive) so the overlap at the anomeric effect that you proposed would be small.
Hope I helped
PS I am not sure which σ* is lower in energy C-N or C-O.

[blaisem]

Hmm, that's a good point on the planarity.  I hadn´t thought of that.  So where does the second lone pair go, just 90 degrees above or below the plane?

You mentioned before you did not think either product would form.  What do you think would form? (You can just give me the first step of the reaction so you don't have to draw it out).

Also, do atomic orbital coefficients for the σ* antibonding orbital of H-N play a role in orbital overlap?  In energy diagrams, we always write molecular orbitals at a discrete (or relative) energy without discerning between the atoms involved, so I figured overlap with the σ* orbital would correspond to one energy no matter if it's at the side of the Nitrogen or the Hydrogen.

Your discussion has definitely caused me to reconsider what I know on the topic, so I would say you are helping a good amount

[azmanam]

I've written extensively about the anomeric effect before[1]. I think the answer to the first question has more to do with the starting material (which is in fact a carbamate) than with the product. I think the pertinent question is: which carbonyl lone pair is most nucleophilic?

The answer to this question has to do with the anomeric stabilization of the starting carbamate, not the anomeric stabilization of the product. I think. I'm willing to be wrong, but this is a fairly classical organic chemistry question, and while I can't recall absolutely, I'm pretty sure the answer makes a starting material argument, rather than a product argument.

Your argument for the acyclic anomeric effect sounds good to me

[1] http://www.chemicalforums.com/index.php?topic=31137
http://www.chemistry-blog.com/2009/03/18/the-anomeric-effect/

[blaisem]

Thanks azmanam for all the info!  Sorry for my late reply; it was a good bit of info to take in, and I was in the middle of moving locations.  At any rate, after reading through it, I naturally have a few questions if you have time to answer any at all.

1. Since you mentioned in your linked forum thread that the delocalizing effect of conjugation involves orbitals delocalizing with antibonding orbitals as opposed to electrons delocalizing into antibonding orbitals, does this mean that the only important consideration is the energy level of the antibonding orbital and that factors like molecular coefficients aren't important?  For example, above in this thread was a little discussion about how viable the N-H σ* orbital was for the cis-Oxygen lone pair to conjugate with.  It was mentioned that the N-H σ* orbital was not as effective for conjugation because the molecular orbital coefficient was centered on the hydrogen, which is far away from the cis-Oxygen.  Is consideration of molecular orbital coefficients for the empty antibonding orbital relevant for these conjugation effects?

2. Since electrons in bonding orbitals also delocalize in part with nearby empty antibonding orbitals, does this mean that this type of conjugation is present in just about every molecule to some extent, even if the energy difference is so small that there are no conformational or structural implications?  I'm just trying to wrap my head around the idea of hyperconjugation, because it was hardly mentioned in my undergraduate studies, at least not in proportion to how potentially prevalent it really is.  I guess this might explain how the idea of an "electron cloud" exists as opposed to localized electrons.

3. In your blog, you open the discussion with Glucose.  It was asked in the forum thread as well as by Mitch in your blog why the OH group would prefer to be equatorial while the methoxy group is axial.  Technically, the O-H antibonding orbital should behave similarly to a O-C antibonding orbital with regards to conjugation facilitation.

I had a theory of my own to explain this:

The 2 O-H bond in the axial position does interact with the 1-Oxygen via the Anomeric Effect; however, in the equatorial position, it also interacts with a lone pair from the 3-Hydroxy group.  Thus, the Anomeric Effect is present in both axial and equatorial conformations.  The energy difference between the diastereomers is either equivalent or small enough that steric effects dominate and the equatorial position is preferred.

When the 2 O-H bond is replaced with a 2-methoxy group, the preference is axial. The axial 2 O-C σ* orbital participates as expected in the Anomeric Effect with the 1-Oxygen.  In addition to this, though, a σ* orbital from C-H bond on the methoxy group is conjugated with by a lone pair from the 3-hydroxy group as well.  In the equatorial position, the lone pair from the 1-Oxygen is too far from the C-H bond to conjugate appreciably with it, while the 3-hydroxy group is assisted by being one bond removed from the pyrane ring, so it can conjugate with a neighboring axially-oriented methoxy group.

Summary: In the case of 2-hydroxy, we have one conjugative interaction no matter whether it is axial or equatorial, so steric effects are the deciding factor in prioritizing the equatorial position.  In the case of 2-methoxy, we have two conjugative interactions in the axial position, as opposed to just one in the equatorial position, and the axial position is preferred.

Anyways, that was my attempt at an explanation   It would be interesting to see whether the hydroxyl group in 2-hydroxy-3,4,5,6-tetrahydropyrane is axial or equatorial.  If it's still equatorial, then I am wrong.  If it is changed to axial, though, then maybe there was some interaction with the 3-hydroxy group in the Glucose case.

Thanks again for the discussion!

[azmanam]

NP. I've been away from the forums for a while (a couple of years!) and it's good to get back here, even if it's just pointing people to things I wrote a long time ago

1. My argument does not get sophisticated enough to consider orbital coefficients. I'm admittedly using more of a hybrid-orbital argument than a molecular orbital argument. I don't know if it's a factor here, and I don't know if it's the best argument here, but I know I did not take it into account when working through my answer, and neither did any of my Twitter friends to whom I also posed this question (to make sure I would be giving sound advice... Turns out this is a stumper for a lot of really bright chemists!)

My answer to your question, and the whole thread so far, is that you're looking at the wrong orbital interactions. The answer to this question comes from assessing the starting material, not the product. The C=O of the SM has two lone pairs, one on the 'N' side, and one on the 'O' side. These lone pairs can interact with the C-N σ* and the C-O σ* of the carbamate through the anomeric effect. One lone pair makes a better overlap, which stabilizes that lone pair more, and makes that lone pair a worse nucleophile. The other lone pair makes a worse overlap, which does not stabilize that lone pair as much, and makes that lone pair the better nucleophile, and the one which is active in the cyclization reaction. What you said above "The more electronegative Oxygen lowers the energy of both the bonding and antibonding orbitals, even more than Nitrogen." This is key here.

2. Yes, this type of conjugation is wildly prevalent. You know the Newman projection of ethane prefers staggered, of course. Maybe you learned it was for steric reasons. The hydrogens don't want to be 'occupying the same space' when eclipsed. Some will argue this is an inadequate explanation, as the hydrogens are really too far apart and angled off in the wrong directions to significantly be 'occupying the same space.' A more complete explanation involves the 6 degenerate hyperconjugation interactions of each C-H σ bonding orbital with each syn-coplanar C-H σ* antibonding orbital of the other carbon. One could argue it's why hydrocarbons in general prefer the linear, all staggered conformation.

3. Maybe I'd have to see a drawing of what you're describing, because I don't see it from your description. With the 1-OR group in the equatorial position, the C-O σ* orbital is only syn-coplanar to the O1-C6 σ bond and the C3-C4 σ bond. While they can stabilize the equatorial group through hyperconjugation, I don't see anyway the C3 OH group lone pair can interact with the C2-O σ* orbital.

Keep up the good work