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.

  • $\begingroup$ 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. $\endgroup$ Commented Mar 24, 2016 at 14:07
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    $\begingroup$ 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... $\endgroup$
    – jerepierre
    Commented Mar 24, 2016 at 17:17
  • $\begingroup$ @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? $\endgroup$ Commented Mar 24, 2016 at 17:24
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    $\begingroup$ 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. $\endgroup$
    – StevieD
    Commented Mar 25, 2016 at 7:56
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    $\begingroup$ @EashaanGodbole, in the appropriate circumstances, you could have some kind of trapping agent present that will react more quickly than interconversion. $\endgroup$ Commented Jan 29, 2017 at 23:15

2 Answers 2


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).


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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