Chem3x11 Lecture 12
Incomplete Saturday Jun 2
This lecture is about pericyclic reactions.
(Back to the main teaching page)
- Pericyclic reactions involve cyclic transition states
Some Opening Experimental Observations
Here are some reactions we've not seen before. The outcomes are, on the face of it, difficult to understand, particularly the stereochemical outcomes which are very specific. There are no apparent intermediates. Solvent polarity has little influence on the outcome. The reactions are equilibria. Imagine you were presented with these reactions and asked to explain the mechanism.
This was the situation in the 1960's when an explanation for these reactions, and several other types, finally emerged. The key was to look at the orbitals involved (an approach taken by Woodward and Hoffmann) and this was later simplified just to a consideration of the Frontier Orbitals (i.e., the HOMO and LUMO) (an approach pioneered by Fukui). To understand this, let's make sure we can draw the relevant frontier orbitals for molecules like this.
Frontier Orbitals of Polyenes
The σ framework of an organic molecule contains the strong (low energy) bonds, meaning the corresponding antibonding orbitals are high in energy. Our analysis of frontier orbitals in reactions of this kind typically focusses on the π system, where the higher energy bonding interactions are, as well as the lowest-lying antibonding molcular orbitals. If we therefore forget about the σ bonds and just focus on the π bonds, we can easily sketch the HOMO and LUMO for ethene.
(Remember that the MO's will actually look like clouds spread over the whole molecule, rather than just separate atomic AOs, but there will be nodes for every MO above the HOMO, as we'll see, where the constituent AO's are not in phase - as for the LUMO above)
In the case of butadiene, there are four π molecular orbitals, two bonding and two antibonding. The way to evaluate how they are drawn is to generate an increasing number of nodes.
Based on the number of electrons in the π system, the HOMO and LUMO must be as shown. In the ground state of the molecule, the electrons will just populate the bonding molecular orbitals, as expected. For hexatriene, the analysis is just the same:
So being able to sketch all of these means that we know we have identified the sign of the HOMO and LUMO on each carbon. This will be important for reactivity. Now let's go back and look at one of the reactions in Scheme 1 in a little more detail.
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