# Differences in Gibbs activation energies for rotation around a bond

Question
Why is there a difference in energy in the rotational barrier between these two structures?

My reasoning is such:

Due to conjugation between the O electrons and the pi- double bonds, the bond has actually single bond characteristics and is therefore weaker than a conventional double bond alkene. But that doesnt explain the difference between the molecules? Can someone please explain?

• One possibility is that there is an unstable resonance structure you haven't drawn, where (going from the right-side one you drew) the pi bond between the left Oxygen and Carbon dissociates, putting a positive charge on Carbon. This would be better stabilized by having a more substituted carbon attached to it. However, I'm skeptical that alone is enough to explain this large a difference in energy. Especially since steric effects would favor B. – William Kappler Nov 18 '15 at 3:48

In both molecules, there is π delocalization across the $\ce{C=C}$ bond which extends across the two carbonyl groups on the alkene. In order to maximize the overlap of the p orbitals involved in this conjugated system, it means that this portion of each molecule has to be flat. It is not a problem for the conjugated system to be flat in molecule B, but in molecule C you have three bulky methyl groups near the conjugated system. Steric clashes in molecule C between the t-butyl group and the carbonyl group on the same side of the $\ce{C=C}$ bond mean the carbonyl will rotate out of the plane of the conjugated system to avoid "bumping into" the t-butyl group, lowering the overlap between the p orbitals in the conjugated π system and thus destabilizing the molecule.
Because molecule C does not experience as much favorable pi electron delocalization as molecule B, molecule C sits at a relatively higher energy than molecule B. The transition states of the two molecules, however, are not so different in energy. The Gibbs energy of activation is defined as the difference in Gibbs energy between the transition state of a reaction and the reactants (where our "reaction" is a rotation about the $\ce{C=C}$ bond). As molecule C sits at a higher energy relative to molecule B, but the transition states are relatively similar, it makes sense that there is a lower barrier to rotation about the $\ce{C=C}$ bond in molecule C.