# C=O and C=C antibonding orbitals

Why is the pi star orbital of the C=C bond higher in energy than the antibonding pi orbital of the C=O bond?

I thought it was the other way around; it's relatively easy to add something to a C=C alkene bond (i.e. halogenation) but halogenation with diatomic halogens doesn't occur with C=O bonds.

• With halogenation it is not the $\pi^{*}$ but the $\pi$ orbitals that are involved in the reaction with e.g. $\ce{Br2}$ because the alkene is acting as a nucleophile here. And because $\ce{O}$ is more electronegative than $\ce{C}$ it should be logical that $\pi$ is lower in energy for $\ce{C=O}$ than for $\ce{C=C}$. – Philipp Oct 27 '14 at 16:45