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During workup of a conjugate addition reaction, it is likely that protonation occurs at the oxygen of the enolate formed. See discussion here: Do enolates get protonated at the carbon or oxygen atom? In that discussion, it is essentially said "The oxygen is probably protonated, but it really doesn't matter." However, I think this could be important in select instances, specifically in the protonation of enolates with diastereotopic faces (which isn't all that rare). But this mostly depends on if tautomerization can occur intramolecularly.

An example of what I'm getting at:

This is an example I just made up. This reaction likely would not actually happen like this. In actually, I could see some γ-deprotonation happening.

In this reaction, one chiral center is formed, and an enolate is formed, and the most likely atom to be protonated upon addition of an acid is the oxygen:

Michael addition and O-protonation

That enol could go on to form one of two products. The intramolecular reaction, I think, would likely give much more of the thermodynamic product (shown below) than an intermolecular reaction, though I'm not certain on that. The reactions here are simplified; I realize both would most likely make some mixture of the two products:

enter image description here

My thoughts are that the four-membered intermediate required for the intramolecular reaction will just be too strained for this to occur. I realize that in basic or acidic solution, catalysis will occur, as the Wikipedia page mentions. I guess the best way to phrase my question is this: In the hypothetical situation that a single ester enol molecule (like that shown as the product of O-protonation above) is dissolved in pure hexane, would that molecule be stuck in in the enol form or would it be able to rearrange? I realize this is more feasible with an dienol, but I'm specifically trying to figure out if the enol can do this.

For the sake of this question, ignore the basicity of the nitrogen. Maybe I should have used the diphenylmethane anion or something. Doesn't matter, it wouldn't reach the proton, either.

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SendersReagent
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  • I guess it is also possible that intermolecular would lead to more thermo product because a smaller acid could fit in on the less accessible side. – SendersReagent Mar 24 '16 at 14:07
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    Instead of hexane, I think gas phase is more what you're suggesting. And if it's gas phase, then someone can probably calculate it... – jerepierre Mar 24 '16 at 17:17
  • @jerepierre That would be fine. I considered that, but I'm an organic chemist. Haha, I don't do anything in the gas phase. Essentially, I was thinking "in a vacuum" but gas phase is to chemists what a vacuum is to physicists, right? – SendersReagent Mar 24 '16 at 17:24
  • I'm inclined to agree with you about ring-strain. However, we are back to the 'do the experiment' approach at this point, I can't really give an answer – Lighthart Mar 24 '16 at 18:49
  • @Lighthart I can't really think of how to come up with an experiment for this one. Or is that what you meant? – SendersReagent Mar 24 '16 at 19:45
  • I mean synthesize compound and see what you get. – Lighthart Mar 24 '16 at 20:47
  • @Lighthart I think I'd have to adjust the setup a bit. Plus, I'm not sure that there would be a significant difference in product distribution unless I used a particularly hindered acid. – SendersReagent Mar 24 '16 at 21:37
  • Can you specify what you mean by intramolecular reaction? Do you refer to a dimerisation kind of mechanism? How do you O-protonate without the lactone getting cleaved? The re-site of the alpha carbon is pretty much shielded with the bulky NPh2 group. I would suspect that any reaction happening there is virtually impossible. That would leave the si-site for an external attack. The intramolecular reaction is probably impossible, too. At least I cannot imagine any four-membered transition state for that where the proton has access to the pi-bond. – Martin - マーチン Mar 24 '16 at 23:41
  • @Martin-マーチン If you look at the top image, the middle image is the lithium enolate of a lactone. There is no doubt in my mind that this would be protonated in the presence of some acid. The lactone wouldn't be cleaved by forming the enol form of the ester unless you think the it is likely to form a ketene and an alcohol. Also, as far as protonation of the re vs si sides, si should, I believe, be the side we are viewing, so I would expect that to be the harder side to protonate, despite its protonation leading to the kinetic product. – SendersReagent Mar 25 '16 at 01:14
  • Also, I specifically mention several times and show that the reaction that I am talking about being inter- or intermolecular here is the tautomerization, and I show and mention right in the title: carbonyl-enol tautomerization. I don't think it is unreasonable to think that, if the enolate is quenched with something like aqueous ammonium chloride, that the small $\ce{NH4^+}$ could lead to some of the cis-bicyclic isomer being formed, especially because the side opposite the one we are seeing is somewhat concave. – SendersReagent Mar 25 '16 at 01:23
  • So I guess, to specify, the intramolecular reaction I'm referring to is just that - an intramolecular migration of the enol proton to the alpha carbon. There are instances of these things being hypothesized, though I'm not sure what has come of those hypotheses (see this discussion). – SendersReagent Mar 25 '16 at 01:37
  • @DGS I'm sorry, the first question should have been: Can you specify what you mean by inter-molecular reaction? I admit I failed to see the cleaved products, so thanks for clearing that up. Also I messed up the faces. I think protonation on the face we are looking at, si, is quite a challenge, probably due to the bulkyness of the phenyl rings. However, if you add NH4Cl to that mix, you can't ignore the amine. I don't know how the last link relates to this question though. – Martin - マーチン Mar 25 '16 at 03:22
  • @Martin-マーチン Oh, okay, that makes sense. Yeah, a dimerization could occur in the way it does with a carboxylic acid -- this would be hard with a bulky enolate, but makes sense to me with a more accessible enolate. Using ammonium chloride was just one example (how I usually protonate my enolates), but yes, I agree that the amine would have to be considered in that case. The link provided deals with gamma-interaction between a proton and a negatively-charged atom, which I feel would be similar to the proton on an enolate being attracted to the alpha-carbon of the enolate. – SendersReagent Mar 25 '16 at 03:46
  • @DGS I don't want to extend that into a discussion in the comments sections, so if you would like to continue to talk about it, I invite you to come to [chat]. Just very briefly now, how I think the case is here very different. You have to access the pi-bond which is perpendicular to the OCC-plane. In HONO you have an in-plane lone pair which could be accessed by the proton, hence giving the four membered transition state. If you consider the amine, then a nice six membered transition state could lead to the thermo product. – Martin - マーチン Mar 25 '16 at 04:00
  • @Martin-マーチン Good point on the HONO thing stereoelectronics. That makes perfect sense. I think this molecule is too rigid for the hydroxyl proton to reach the amine, but it certainly would in a less rigid molecule. – SendersReagent Mar 25 '16 at 04:11
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    Nice thought experiment but I might suggest using something like trimethylsilyl chloride. That won't hope off the oxygen, and Id bet you can induce stereo-selectivity that way. – StevieD Mar 25 '16 at 07:56
  • I understood that letting the tautomerisastion step run for a longer time would give more of the thermodynamically controlled product. What would one do if s/he wants to get the kinetic product? Preferrably without chiral catalysts? – Eashaan Godbole Dec 25 '16 at 07:41
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    @EashaanGodbole, in the appropriate circumstances, you could have some kind of trapping agent present that will react more quickly than interconversion. – SendersReagent Jan 29 '17 at 23:15
  • Trialkyl boranes tend to not add to aldehydes due to the strain of the 4-membered TS, though allyl boranes do, due to the presence of a 6-membered TS, in this case, the intramolecular addition is no different, so I would expect it not to occur. – AS_1000 Jul 26 '17 at 11:58

2 Answers2

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tl;dr: The intramolecular variant is an impossible rearrangement. Only intermolecular proton transfer can be observed.

This is actually a very nice example to discuss sigmatropic rearrangements, because the desired [1,3]-rearrangement would be one. In general, these can proceed in four possible ways, depending on the configuration of the migrating group and whether it remains on the same side of the π system or not. We would have to draw orbital schemes to determine whether or not the orbital of the migrating group is able to interact both with the point of origin and the point of arrival favourably.

With regards to the configuration, the two possibilities are:

  • migration under retention
  • migration under inversion

A migration under retention is always possible because the corresponding orbital’s lobe is already pointing towards the π system it is migrating along. A migration under inversion is only possible if the (atomic) orbital whose bond is broken and later reformed has lobes on two different sides; typically for migrating atoms, this means a $\mathrm p$ orbital.

In the case of a proton shift, we must consider a hydrogen atom. Its only occupied orbital through which it bonds to the carbon chain is $\mathrm{1s}$. This orbital does not have two lobes, therefore a rearrangement mechanism under inversion is impossible; we need only consider the retention mechanisms.

With respect to the π system, the two possibilities are:

  • suprafacial (remaining on the same side)
  • antarafacial (switching sides)

Again, a suprafacial migration is practically always possible. For an antarafacial migration to be considered, the chain must be flexible enough so that both sides can interact with the migrating group; ideally at the same time.

Since the thought experiments are rather hard to perform, here are some depictions to make it easier.

suprafacial migration antarafacial migration
Figure 1: Suprafacial [1,5] (left) and antarafacial [1,7] (right) hydrogen migrations. The depicted π orbital corresponds to the HOMO of a full π system including the $\ce{C-H}$ bond.

Generally, the full π system and the neighbouring $\ce{C-H}$ bond that is to be broken must be considered as I have done in figure 1. This π system can be drawn up with the standard rules for linear π systems. In the [1,5]-rearrangement, the system is penta-2,4-dien-1-yl while in the [1,7] case it is hepta-2,4,6-trien-1-yl. The $\ce{C-H}$ bond is considered part of the π system with the proton being the migrating group. Therefore, the HOMO (the orbital of importance) will be the one in question. (If the $\ce{C-H}$ bond is considered part of hydrogen, we must consider the π system’s LUMO, meaning that the orbital will remain the same. We must always have an occupied and an unoccupied orbital for consideration.)

The image shows us that the [1,5]-migration — over two double bonds — can only occur in a suprafacial manner with hydrogen. If we had a group capable of inversion, an antarafacial manner would be possible given the correct geometry. In the [1,7] case — over three double bonds —, hydrogen can only migrate in an antarafacial manner under retention; the only possible suprafacial manner would be under inversion which hydrogen cannot do.

This has been turned into a general rule of thumb: If your migration is [1,$4n+1$], it may proceed suprafacial under retention but if it is [1,$4n+3$] it must proceed in an antarafacial manner. In the [1,3]-case, only antarafacial migration would be possible, but there is no way that the π system can twist so it would allow for antarafacial migration. The π orbital in question is shown in figure 2 below.

The pi orbital for a 1,3 migration
Figure 2: The π orbital relevant for a [1,3]-migration.

Therefore, an intramolecular proton transfer is impossible and an external source would have to protonate.

One thing to note: the molecule itself is able to deprotonate/reprotonate another molecule of its type intermolecularly. So the compound would not be stuck in the enol form but a second molecule would provide the much desired proton to allow the keto form to be regenerated, even in pure hexane (if it is even soluble in that).

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This is a very hypothetical situation, but let's say that you obtain that enol, and you dissolve it in an aprotic solvent, and that the enol is higher energy than the ketone, I would think a second molecule would act as the base and send your product to the lower energy state. There is no need for this to be intra-molecular, as the Pka of one enol is exactly the same as the Pka of the other enol, they will freely exchange protons up to the point where you will get protonation at the carbon.

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