# When is donation into an anti-bonding MO stabilising?

This has been puzzling me for quite a while and I'm not sure if it could be a misconception on my part. When is interaction of a filled orbital with a antibonding MO stabilising and when is the interaction destabilising? I will now detail a few examples of the "stabilising interactions" and the "destabilising interactions". I will use Clayden's Organic Chemistry (2nd ed.) as my primary source of examples.

Destabilising interactions

• Nucleophilic attack at the carbonyl carbon (p. 127)

Here, interaction of the nucleophile's HOMO with the $$\ce {C=O \pi ^*}$$ LUMO destabilises the $$\pi$$ bond, resulting in its cleavage. This is thus a destabilising interaction.

• Nucleophilic attack in the $$\ce {S_N2}$$ reaction (p. 340)

Here, again we see the interaction between a filled HOMO and an empty antibonding MO ($$\ce {C-X \sigma ^*}$$ MO) destabilising the $$\ce {C-X}$$ bond.

Stabilising interactions

• Stabilisation of cis configuration in esters (p. 805, explained in detail here)

Here, we see that the interaction of a filled, nonbonding MO centred on the $$\ce {O}$$ interacts with the $$\ce {C=O \pi ^*}$$ LUMO to stabilise the molecule overall.

• Unreactivity of dichloromethane (p. 804, see here also)

As can be seen in diagram, interaction of nonbonding electrons centred on $$\ce {Cl}$$ with the $$\ce {C-Cl \sigma ^*}$$ stabilises the molecule, as it decreases is reactivity, by raising the level of the new resultant LUMO of the molecule.

• Explaining Zaitsev's rule

This is a classic explanation for the increasing stability of alkenes as substitution increases. Interaction of $$\ce {C-H \sigma}$$ MO with the $$\ce {C=C \pi ^*}$$ MO increases the stability of the alkene.

I am sure there are likely to be more examples (e.g. various other instances of hyperconjugation) but I believe these examples are sufficient to prove my point. Is this inconsistency in the use of frontier MO theory or am I merely misinterpreting its use?

Update

After pondering over this for a while, I was thinking, perhaps we should make use of an MO diagram to more clearly describe the interaction. Below is a simple, typical diagram (from p. 111, with edits). Based on my understanding, all of the above examples I have presented can in fact be described by this straightforward MO diagram. Of course, the energy gap between the two interacting MOs may be different and thus, the strength of interaction would also differ.

In no way does the MO diagram indicate any sort of destabilisation of the molecule.

Such an interaction between a filled MO and an unfilled MO only produces stabilisation, as seen in the diagram. Not only is the filled orbital of a lower energy and hence less likely to interact with LUMOs of other molecules, the empty orbital's energy has also been raised, making it less energetically accessible from the point-of-view of a nucleophile. In other words, the molecule is both less reactive to an electrophile, due to a energetically-lowered filled orbital, and also less reactive towards a nucleophile, due to a energetically-raised empty orbital.

If this is the case, may I question further:

What is the basis of destabilisation in the first 2 examples I have presented above?

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

The difference between the "stabilising" and "destabilising" examples that you have chosen are that the "stabilising" examples are static molecules, whereas the "destabilising" examples are chemical reactions, in which a lot is happening. We'll look at the "stabilising" cases first, as these are easier to understand.

As with any question involving "stability", it is important to consider what we are measuring stabilisation with respect to. In the example of the anomeric effect (e.g. in esters), we are comparing the true electronic state of the ester against a hypothetical scenario, where the lone pair and σ* do not overlap, but everything else is kept the same. In such cases, it is nearly always true that allowing overlap, or delocalisation, leads to a lower energy. This is captured quite well in the last picture in your question body. (See also: Physical reason for why delocalisation leads to stability?)

In a chemical reaction, as we move along the reaction coordinate (RC), we are creating overlap between two different orbitals, just as in the previous case. Naively we might think that this necessarily leads to stabilisation. But this is not physically sensible: if the reaction proceeds via a transition state (TS), this TS must be higher in energy than the reactants (at infinite separation).[1] The problem is that when we bring two molecules together in space, it is not just overlap that is affected: the energies of every orbital will change as nuclear positions are moved, electron-electron repulsions are different, etc. This means that all the orbital energies change in a way which depends not only on overlap with other orbitals (as was the case before), but also a number of other factors.

Instead of going from reactants at infinite separation to the TS directly along the RC, as would be physically realistic, it may be instructive to consider hypothetically doing it in two stages. In the first step, we bring the reactants together from infinite separation to the TS geometry while forbidding any orbital overlap; secondly, we stay at the TS geometry, but turn on the overlap. In real life, you cannot separate these two things: changing the geometry comes with formation of overlap. But thinking about it this way allows us to link it to the "stabilising" cases talked about before.

Let's use the addition of a nucleophile to a carbonyl group as an example to make this more concrete. In the reactants, there are three key frontier orbitals we need to consider: the nucleophile lone pair ($$n$$), and the π-type orbitals of the carbonyl group ($$\pi$$ and $$\pi^*$$).

In general, as we bring the nucleophile and carbonyl together, mixing of all three orbitals will occur continuously. As the reaction proceeds, we will therefore have three MOs $$\{\psi^{(i)}\}$$ $$(i = 1,2,3)$$ which are linear combinations of the three frontier orbitals listed earlier:

$$\psi^{(i)} = c^{(i)}_n\phi_n + c^{(i)}_\pi \phi_\pi + c^{(i)}_{\pi^*}\phi_{\pi^*}$$

For well-separated reactants, the relevant MOs will (for the most part) be the original three frontier orbitals. So, for example, we could have $$\psi^{(1)} = \phi_\pi$$, $$\psi^{(2)} = \phi_n$$, $$\psi^{(3)} = \phi_{\pi^*}$$. But at the end of the reaction, the MOs of the product will be a complicated mixture of the three orbitals. For example, the $$\ce{C-Nu}$$ $$\sigma$$ orbital might be made up of something that resembles this:

$$\psi^{(1,\text{new})} \approx \frac{1}{\sqrt{3}}(\phi_n + \phi_\pi + \phi_{\pi^*})$$

The reason why these linear combinations (or mixtures) can form is because there is non-zero overlap between all of these orbitals as the reaction proceeds. If there is no overlap between two orbitals, then they cannot mix. So, the way to forbid any overlap from occurring in the first step is to simply not change the coefficients of the MOs as we bring the reactants closer to each other (towards the TS geometry). Of course, this is not physically possible, which is why it is purely a hypothetical scenario.

Even though the MOs themselves do not change in this first step, their energies do change, because the Hamiltonian is changing (by virtue of the nuclear positions changing). So, the energies of the MOs will change. To throw an additional wrench into the works, it is not just the electronic energy that we must consider, but also the nuclear energy (when we bring nuclei closer to each other, they will repel each other, and this is not captured in MO diagrams which only illustrate electronic energy). As you can imagine, it is very hard to make any real analysis as to what happens in this step.

In any case, once we reach the TS geometry, the energy of the system will definitely have changed in some way. We can now carry out the second step by "turning on" the overlap, i.e. finding the appropriate linear combinations of $${n, \pi, \pi^*}$$. Since nothing else here is changing (in particular, no nuclear movement), except for us "allowing" overlap to occur, it is exactly analogous to the "stabilising" case which was discussed earlier. So the second step always leads to a decrease in energy.

The problem is that, from the infinitely-separated starting materials, we have carried out two steps to get to the TS. In the second step, the energy has decreased – a "stabilisation" of sorts. But in the first step, we don't know what has happened. If a TS is to be higher in energy than the reactants, then it follows that there must have been an increase in energy in the first step.[2]

Simple LCAO-MO diagrams don't quite capture the complexity involved in the first step – they can only really properly describe the second step. A better way of drawing it might be to do something like this:

I haven't drawn the orbitals in the TS, but you can imagine that they are starting to morph into what they will be in the products. So $$\psi^{(1)}$$ will be somewhat intermediate between $$\pi_\ce{C=O}$$ and $$\sigma_\ce{C-Nu}$$. Of course, this is purely hypothetical - especially in the middle, the orbitals and their energies have no physical meaning at all. In a real reaction you would go straight from SM to TS, and not via an unphysical intermediate stage as is depicted here.

So, therein lies the difference between the two cases you have described. In the "stabilising" case, the only thing that is happening is a magical "switching on" of overlap. In the "destabilising" case of a chemical reaction, in addition to the overlap (which is stabilising), you also have a glut of other factors which can lower or (more frequently) raise the energy of the system as it move towards the TS. The problem is that we cannot separate the two from each other. Creating overlap comes with all the other strings attached, and so overall as we move along the RC towards the TS, there is (typically) an increase in energy.

Finally, how do we relate this back to the description of nucleophilic addition in Clayden? The wording is definitely somewhat loose (organic chemists frequently do this for ease of use), but it does have a grounding in quantum theory. I'll reproduce the image here and try to relate each phrase:

Electrons in HOMO begin to interact with LUMO

This refers to the developing overlap that we looked at in step 2, i.e. the formation of MOs which are linear combinations of the HOMO ($$n$$) and LUMO ($$\pi^*$$). On its own this seems to be a stabilising effect, but recall that this does not come on its own: we also have to consider the different nuclear positions and its implications on the energy.

Filling of $$\pi^*$$

I think the way to interpret it is that, at the start of the reaction the two filled orbitals $$\psi^{(1)}$$ and $$\psi^{(2)}$$ are simply $$n$$ and $$\pi$$, so the coefficients $$c_{\pi^*}^{(1)}$$ and $$c_{\pi^*}^{(2)}$$ are zero. As the reaction progresses, these coefficients steadily increase, which in a sense corresponds to a population of the $$\pi^*$$-orbital, even though the $$\pi^*$$-orbital ceases to exist in its own right once we bring the starting materials together.

causes $$\pi$$ bond to break

In a sense we could define a "$$\pi$$ bond order" to be some kind of difference between the coefficients of the $$\pi$$ orbital in the filled MOs, and the coefficients of the $$\pi^*$$ orbital in the filled MOs. At the start these are $$1$$ and $$0$$ respectively, for a bond order of $$1$$. As the reaction proceeds and the orbitals mix, we find that $$c_\pi$$ decreases in the filled MOs, $$c_{\pi^*}$$ increases, and at the end of the reaction we have a bond order of $$0$$. So loosely speaking, it is true that "population of the $$\pi^*$$ orbital" (i.e. increasing $$c_{\pi^*}$$ in the filled MOs) leads to the cleavage of the $$\pi$$-bond.

Electrons from $$\pi$$ bond end up as negative charge on oxygen

This is somewhat true again since the filled MOs morph from $$\pi_\ce{C=O}$$ plus a lone pair to $$\sigma_\ce{C-Nu}$$ and a lone pair, but it's important to remember that it is not a one-to-one correlation, i.e. the $$\pi_\ce{C=O}$$ MO does not become the oxygen lone pair all by itself.[3] More properly, the oxygen lone pair is a linear combination of multiple MOs in the starting material.

It's also important to not ascribe too much importance to exactly which electrons go where, as this violates quantum indistinguishability.

Notes

[1] There are barrierless reactions which do not proceed via a transition state, such as $$\ce{H + H -> H2}$$, and in such cases it is quite true that we always get stabilisation as we move along the RC towards products. However, these reactions don't fully capture the complexity involved in other reactions with TSes and it's better to think of these as being special cases of the other reactions.

[2] Barrierless reactions don't obey this line of thinking, of course. In fact, they don't even have a TS, so there isn't a well-defined geometry for us to distort towards in the first step. You can pick any geometry along the reaction coordinate between reactants and products, and it would probably work.

[3] Under some special circumstances it is possible to have a one-to-one relationship between reactant MOs and product MOs. See e.g. the first correlation diagram in Theoretical basis behind orbital correlation diagram for pericyclic reactions. But note that here there is only one orbital of each symmetry species, so there is no mixing going on and no linear combinations being formed. See also the last section in the answer on "Real pericyclic reactions".

• Ok I think your answer definitely consolidated our discussion from yesterday. But I would like to ask further regarding the change in the nuclear coordinates (i.e. the first step): Firstly, do MOs only depict the electronic energies? Secondly, from the point of view of an electron, when it is brought near to another nuclei, there would be net fall in PE or is it net gain in PE due to the increased inter-electron repulsion? – Tan Yong Boon Dec 27 '18 at 0:49
• (1) Yes, for a better understanding of this look up the Born–Oppenheimer approximation in a textbook. The bottom line is that we separate the Schrödinger equation for the whole molecule into a nuclear component and an electronic component, and the MOs are part of the (approximate) solution for the electronic bit, so they don't reflect nuclear repulsion energies (which fall out of the nuclear component). (2) Not 100% sure, but I don't think this can be qualitatively predicted. You would probably need to work out the actual integrals. – orthocresol Dec 27 '18 at 0:54
• In that case, is it possible to pinpoint what is the exact cause of destabilisation in the "first step"? Can we even confirm that there is destabilisation in the "first step" to begin with? – Tan Yong Boon Dec 27 '18 at 0:56
• I don't think there's much insight that can be gleaned there, and that's why I deliberately refused to say anything about it in the actual answer. There are several components in the Schrodinger equation: (1) nuclear-nuclear repulsions, (2) electron-electron repulsions, (3) electron-nuclear attractions, (4) kinetic energy. It seems intuitive to me that an increase in (1) and (2) causes destabilisation, but an increase in (3) would also be stabilising. For different reactions these factors would contribute to different extents. [The KE should be constant if the wavefunctions don't change.] – orthocresol Dec 27 '18 at 1:02
• Thank you very much for spending so much of your precious time in having this discussion with me! I have definitely gained much from it. Merry Christmas and happy New Year! – Tan Yong Boon Dec 27 '18 at 1:37

You are making the mistake of looking at part of the molecule instead of the whole. Let's look at your example involving esters, or better yet carboxylate ions. You have the carboxyl group and then a second oxygen atom. When the latter donates its pi electrons to the molecular orbital, they go into an antibonding orbital of the carboxyl group. So that part of the bonding is weaker. But this interaction also makes a new pi bonding interaction involving the second oxygen atom. You have to count that, too, to evaluate the bonding in the whole molecule. When the extra bonding outweighs the extra antibonding in the whole molecule, as in pi electron radicalization in an water or carboxylare, the interaction is favored even though not every part of it is.

• Ok I think we are off to a good start. I know exactly what you are saying. The donation to the antibonding MO destabilises the particular bond but brings about a stronger bonding interaction that can overall bring down the overall energy of the molecule. – Tan Yong Boon Dec 26 '18 at 13:26