In butadiene, four atomic p orbitals interact to make four molecular orbitals; in hexatriene (and you will soon see benzene too) six atomic orbitals interact to make six molecular orbitals. We are now going to consider some common conjugated systems made up of three interacting p orbitals. We’ll start with the structure we get from treating propene with a very strong base— one strong enough to remove one of the protons from its methyl group. H+ is removed, so the product must have a negative charge, which formally resides on the carbon of what was the methyl group. That carbon atom started off sp3 hybridized (i.e. tetrahedral: it had four substitu ents), but after it has been deprotonated, it must become trigonal (sp2), with only three substitu ents plus a p orbital to house the negative charge.

We could work out the orbitals of the allyl anion by combining this p orbital with a ready-made π bond, but instead this time we will start with the three separate p atomic orbitals and combine them to get three molecular orbitals. At fi rst we are not concerned about where the electrons are—we are just building up the molecular orbitals. The lowest energy orbital (ψ1) will have them all combining in phase. This is a bonding orbital since all the interactions are bonding. The next orbital (ψ2) requires one node, and the only way to include a node and maintain the symmetry of the system is to put the node through the central atom. This means that when this orbital is occupied there will be no electron density on this central atom. Since there are no interactions between adjacent atomic orbitals (either bonding or antibonding), this is a non-bonding orbital. The final molecular orbital (ψ3) must have two nodal planes. All the interactions of the atomic orbitals are out of phase so the resulting molecular orbital is an antibonding orbital.

We can summarize all this information in a molecular orbital energy level diagram, and at the same time put the electrons into the orbitals. We need four electrons—two from the alk-ene π bond and two more for the anion (these were the two in the C–H bond, and they are still there because only a proton, H+, was removed). The four electrons go into the lowest two orbit-als, ψ1 and ψ2, leaving ψ3 vacant. Notice too that the energy of two of the electrons is lower than it would have been if they had remained in unconjugated p orbitals: conjugation lowers the energy of filled orbitals and makes compounds more stable.

Where is the electron density in the allyl anion π system? We have two fi lled π molecular orbitals and the electron density comes from a sum of both orbitals. This means there is electron density on all three carbon atoms. However, the coefficients of the end carbons are of a significant size in both orbitals, but in ψ2 the middle carbon has no electron density at all—it lies on a node. So overall, even though the negative charge is spread over the whole molecule, the end carbons carry more of the electron density than the middle one. We can represent this in two ways—the fi rst structure below emphasizes the delocalization of the charge over the whole molecule, but fails to get across the important point that the negative charge resides principally at the ends. Curly arrows do this much better: we can use them to show that the negative charge is not localized, but principally divided between the two end carbons.

The problem with these structures carrying curly arrows is that they seem to imply that the negative charge (and the double bond for that matter) is jumping from one end of the molecule to the other. This, as we have seen, is just not so. Another and perhaps better picture uses dotted lines and partial charges. But the structure with the dotted bonds, as with the representation of benzene with a circle in the middle, is no good for writing mecha nisms. Each of the representations has its strong and weak points: we shall use each as the occasion demands.