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.
How the Frontier Orbitals Guide an Example Reaction
The reaction we'll start with is the transformation of that octatriene to the cyclohexadiene shown in Scheme 1. The starting material is a triene with methyl groups on the end. The mechanism is just a cyclic rearrangement of electrons, resulting in one less π bond and one more σ bond (largely explaining the position of equilibrium). This reaction type is called an Electrocyclic Reaction and is one of the reactions we'll consider. The transition state has a cyclic arrangement of electrons, like an electric circuit. We're going to see various types of these reactions with small variations, but if they involve a cyclic arrangement of electrons in the transition state we call them pericyclic reactions generally.
This simple 2D representation does not explain what's really going on, nor the stereochemical outcome. Notice that there is a σ bond being formed from two π bonds. Since the electrons that are moving around are in the HOMO, we just need to consider that frontier orbital. A way to think about what's going on is that the π orbitals at the ends of the triene are "rotating" so they overlap and form a new σ bond, thereby creating the ring. If we look back at our HOMO diagram for the triene (Scheme 4) we see that these end lobes have the same sign. If they are to productively overlap to give a new σ bond, lobes of the same sign (symmetry) must come together. This specifies a direction of rotation for this reaction to work, in this case the direction of rotation is the opposite way for each lobe, and we call this disrotatory. The need for a specific rotation in the bonds mandates that the two methyl groups end up on the same face of the ring.
So in this case the reaction shown is possible because the interacting MO lobes are of the correct sign. We say the reaction is symmetry allowed. Let's test this idea with the second reaction in Scheme 1, generating the 4-membered ring.
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