The C=O double bond is the most important functional group in organic chemistry. It is present in aldehydes, ketones, acids, esters, amides, and so on. We shall spend several chapters discussing its chemistry so it is important that you understand its electronic structure from this early stage. We’ll use the simplest carbonyl compound, methanal (formaldehyde), as our example. As in alkenes, the carbon atom needs three sp² orbitals to form σ bonds with the two H atoms and the O atom. But what about oxygen? It needs only to form one σ bond to C, but it needs two more hybrid orbitals for its lone pairs: the oxygen atom of a carbonyl group is also sp² hybridized. A p orbital from the carbon and one from the oxygen make up the π bond, which also contains two electrons. This is what the bonding looks like:

For the MO energy diagram, we’ll again just consider the bonding between C and O. First, we hybridize the orbitals of both atoms to give us the 3 × sp² orbitals and 1 × p orbital we need. Notice that we have made the AOs at O lower in energy than the AOs at C because O is more electronegative. Once we have accounted for the non-bonding sp² orbitals at O and the two C–H bonds, we allow the two remaining sp² orbitals to interact and make a σ and a σ* orbital, and the two p orbitals to make a π and a π* orbital.

The fact that oxygen is more electronegative than carbon has two consequences for this diagram. Firstly, it makes the energy of the orbitals of a C=O bond lower than they would be in the corresponding C=C bond. That has consequences for the reactivity of alkenes and carbonyl compounds, as you will see in the next chapter. The second consequence is polarization. You met this idea before when we were looking at NO. Look at the fi lled π orbital in the MO energy level diagram. It is more similar in energy to the p orbital on O than the p orbital on C. We can interpret this by saying that it receives a greater contribution from the p orbital on O than from the p orbital on C. Consequently, the orbital is distorted so that it is bigger at the O end than at the C end, and the electrons spend more time close to O. The same is true for the σ bond, and the consequent polarization of the C=O group can be represented by one of two symbols for a dipole—the arrow with the cross at the positive end or the pair of δ+ and δ– symbols.

Conversely, if you look at the antibonding π* orbital, it is closer in energy to the p orbital on C than the p orbital on O and therefore it receives a greater contribution from the p orbital on C. It is distorted towards the carbon end of the bond. Of course, being empty, the π* orbital has no effect on the structure of the C=O bond. However, it does have an effect on its reactivity—it is easier to put electrons into the antibonding π* orbital at the C end than at the O end.