# Using molecular orbital for determining leaving group trend and nucleophilicity

Here is the question.

Which one has better leaving group, cyclohexanol or cyclohexane that has $$\ce{OH_{2}^{+}}$$ substituent? Explain.

I know the answer is the latter because if both compounds are subjected in substitution reaction, the latter will yield a weak conjugate base ($$\ce{H2O}$$). There are some interesting facts from there which wants me to prove the answer with molecular orbital approach.

Here, I cite the part from it.

The leaving group itself does play a slight role here in determining the energy of the $$\unicode{x3c3}^*$$ orbital and thus whether it is energetically favourable for the nucleophile to attack.

Based on the cited part, looks the orbital plays the role in nucleophilicity although the iodide example gives counterexample of nucleophilicity vs leaving group.

1. Does HOMO orbital play role in determining nucleophilicity? Based on molecular orbital of $$\ce{OH-}$$ and $$\ce{H2O}$$, the HOMO of $$\ce{H2O}$$ is higher than the former, thus is it correct to interpret $$\ce{H2O}$$ is more nucleophilic than $$\ce{OH2+}$$?

2. If nucleophilicity criteria is solely based on energy of HOMO orbital, is there a bound where the energy of HOMO orbital becomes electrophilic (or the LUMO orbital becomes nucleophilic)?

3. If leaving group is a thermodynamic factor (as noted by first source), would molecular orbital alone (including frontier molecular orbital) suffice to predict the leaving group order?

4. Can molecular orbital describes the iodide case (good leaving group but good nucleophile)?

The leaving group trend is supposed to be for similar weak bases (halides, $$\ce{OTs}$$, and $$\ce{OMs}$$).

## 1 Answer

MO theory can certainly be used to explain substitution reactions from a mechanistic perspective, and to help rationalize trends of relative reactivity. However, I think that the best way to build a mental model that gives you practical predictive power is to also incorporate the simpler ideas that mostly reduce to Lewis acid/base theory that most introductory textbooks teach.

Remember that "nucleophile" and "leaving group" are roles that molecules take when they participate in a specific type of reaction. They are not intrinsic molecular properties. MO theory can fill in part of the picture, but other factors (such as steric hindrance in the nucleophile or electrophile), solvent selection, heat, concentration, etc. can all affect reaction rates and how well a given compound takes up the role of being a nucleophile or electrophile in a given reaction context.

So, to answer your questions:

1) The energy of the HOMO does play a role in determining nucleophilicity since, from a mechanistic perspective, a substitution reaction occurs when the HOMO of a nucleophile overlaps with (and populates electrons into) the LUMO of the electrophile. However, HOMO energy levels are difficult to predict off-hand. A better explanation for the hypothetical difference in nucleophilicity between H2O and H2O+ is that H2O+ is strongly electron-deficient, with the oxygen lacking an octet. Therefore, it is extremely unlikely that the oxygen will act as a nucleophile. This is just Lewis acid/base theory.

2) The LUMO is, by definition, unoccupied. There are no electrons in the LUMO that could act in a nucleophilic manner towards another orbital. On the flip side, if the HOMO is occupied then that orbital is "full" and there is no way that it could act as an electrophile by accepting more electrons.

3) No - see the second paragraph of my answer. The energy of the electrophile's LUMO in question -- the sigma-antibonding orbital -- is certainly relevant, but many other factors are at play: sterics, the bond length between the leaving group and the electrophilic carbon, solvent effects, etc. The most useful, generalizable approach to predicting LG ability is Lewis basicity; the weaker the base, the better the leaving group. Thus, H2O is a better leaving group than OH-, Br- is better than Cl-, etc.

4) From a Lewis/acid base standpoint, one would predict I- to be a good leaving group since it is a very weak base. MO theory predicts that I- will be a better nucleophile than the other halides because its valence electrons reside in 5p orbitals, which are much higher in energy and more close in energy to the sigma-antibonding LUMO of the electrophile.