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38 Organic Chemistry Solutions Manual Problem 2 Draw a full orbital diagram for all the bonding and antibonding orbitals in the three- membered cyclic cation shown here. The molecule is obviously very strained. Might it survive by also being aromatic? Purpose of the problem Revision of MO diagrams for conjugated systems with aromaticity. Suggested solution There are only two electrons in this simple cation but we need to mix the π bond (π and π* orbitals) with the empty p orbital to give the MOs. One MO will be bonding all round the ring and this is the only one that matters to the structure. The others may have given you problems. We can mix the p- AO with the π-MO as they have the same symmetry in a three-membered ring but we cannot mix P with π*. So our three MOs are + - and π*. π-MOs of the alkene p-AO of the cation = = π - p 88 p π 88 = + P The cyclic cation is stable and can be made because of the gain in energy in populating only the lowest all-bonding orbital. As it happens and are degenerate. If you count the number of bonding and antibonding interactions in each you will see that the net result is one antibonding interaction in both orbitals. It is also aromatic, having 4n + 2 π-electrons where n = 0. You get the same result if you mix three p-AOs, one on each carbon atom. Then it is easier if we look down on the ring showing the top lobe of the P orbital at each atom. The lowest, all-bonding orbital has no nodes (except the plane of the ring) and the two degenerate, antibonding, and unoccupied orbitals have one node each. and one node P no nodes