The LUMO of CH3OH2+

Question 11a of the first semester final exam for Chem 125a for 2001-2002 was:

Reaction of 1-butanol with HBr involved protonating the OH group to give the molecule a lower LUMO, and then attacking it with the HOMO of Br- to give a transition state like the one on the right, where the colinear dotted lines show the forming Br-C bond and the breaking C-OH2 bond.

a) Explain why protonation lowers the energy of the LUMO, and draw a picture of the LUMO to explain the direction from which it is attacked by Br-, and why the dotted lines should be roughly colinear.

The expected answer for the question was:

Protonation generates a positive charge (on the Oxygen atom), which lowers the electron energy for all MOs in the vicinity, including that of the LUMO.

The LUMO is s*C-O (see right). Best overlap is available from the face of the carbon atom exactly opposite the O, as shown by the arrow.

This answer was given full credit, but it does not contain the whole truth.

We can often gain a very good understanding of molecular structure and reactivity by considering simple localized orbitals, like s*C-O shown above, that involve only two atoms. Gaining this kind of insight was a major goal of Chem 125a, and it should prove very useful in understanding the reactions you encounter in Chem 125b.

But you should remember that such simple orbitals are often "disguised" by mixing with other orbitals of similar energy to give complex MOs that include contributions from many atoms. In such instances the "true" LUMO (or HOMO) that is calculated by a quantum mechanics program, like MacSpartan, shows us more than is convenient to see for the purpose of achieving a simple understanding. ("I wanted to catch a little one!")

The LUMO of CH3OH2+ is a good example. The figure below shows three different structures of this molecule and, below each, the corresponding molecular LUMO.

Structures and LUMOs were determined using STO 6-31G**, the most sophisticated calculation available with MacSpartan. In each structure the methyl group's third H is obscured by the black C atom. A single contour (where the electron density would be 0.001 e/Å3) is shown for each orbital. Red and blue code opposite signs. Unfortunately MacSpartan exchanged red and blue for the orbital shown in the center. For comparison with the flanking MOs, you must pretend that its colors are swapped.

Images in the center are for the cation in its lowest-energy geometry. The flanking structures are distorted, as the molecule would become when it approaches the transition state for reaction.

The structure on the left is the lowest-energy structure possible if the central C-O bond is artificially stretched by 30%, as it would be if Br- were approaching from the bottom, mixing its HOMO with s*C-O, and causing the C-O bond to break. (Note that the CH3 group has become more planar than in the central structure.)

The structure on the right is the lowest-energy structure possible if the O-H bond on the right is artificially stretched by 30%, as it would be if Br- were approaching from the top right, mixing its HOMO with s*C-H, and causing the C-H bond to break.

The flanking LUMOs are relatively simple. Each is dominated by the s* orbital between two atoms whose bond is being broken for that particular structure, with just traces of the other two localized s* orbitals.

The center LUMO, for undistorted CH3OH2+, is formed by mixing these two LUMOs with the right-to-left mirror image of the LUMO on the right, that is, the s* orbital whose occupancy by electrons would weaken the other O-H bond.

Apparently the three localized s* orbitals, involving the individual pairs of atoms, O-C, O-H(1), and H(2)-O, are similar enough in energy that the small overlap among them can cause this confusing mixture.

Not to worry however.

When the structure approaches the transition state (and for understanding rate, the transition state energy is what we need to predict!) the s* orbital of the breaking bond is much lower in energy than the other two (because of reduced overlap), and three localized s* orbitals no longer mix strongly with one another. So a single localized pairwise orbital dominates the LUMO of the important structure.

Thus the simple-minded pair-wise orbital approach we took really does work for understanding this reaction.

The shape of the LUMO for undistorted CH3OH2+ reminds us that three paths are possible for reaction with bromide. The HOMO of Br- can attack either at an H to form H-Br and "displace" CH3OH, or it can attack at C to form CH3Br and displace H2O. In fact, as the shape of the LUMO might suggest, the former reaction, deprotonation, occurs much more easily, but it is a "nothing" reaction, since HBr can just protonate CH3OH again and again, waiting for the opportunity to displace H2O and form the ultimate products.

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copyright 2002 J. M. McBride