Chemistry - Is carbonyl-enol tautomerization only intermolecular or can it be intramolecular?

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).