# Ring contraction in a carbocation due to ring strain and back bonding

Today our teacher told us that the following carbocation rearrangement occurs due to back bonding. I could not really follow what he meant. Can someone please explain what is actually happening during the following carbocation rearrangement, and why it happens?

The reason that cyclobutyl carbocations generally rearrange to cyclopropylcarbinyl carbocations is due to resonance stabilization. Here are some drawings that may help, if you build a model and look at the model that will help even more. The $\ce{C-C}$ bonds in a cyclopropane ring are approximately $\ce{sp^{3.74}}$ hybridized (see here), said differently, they have a lot of p-character in the bond. In fact, the p-character is so high that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.
Typically we discuss two conformations of the cyclopropane ring with the carbocation p-orbital, a bisected and a perpendicular conformation. In the bisected conformation the cyclopropane $\ce{C-C}$ bond lies in the same plane as the carbocation p-orbital. The carbocation p-orbital and the $\ce{C-C}$ orbital which resembles a p-orbital (because it is high in p-character) overlap! In the perpendicular conformation, the cyclopropane $\ce{C-H}$ bond is in the same plane as the carbocation p-orbital, the $\ce{C-C}$ bond is not - there is no carbocation stabilization by the cyclopropane ring in this conformation. Molecules where the bisected conformation can be attained show stabilization due to the stabilizing overlap between the carbocation p-orbital and the p-like $\ce{C-C}$ cyclopropane ring bond. Sometimes we use resonance structures such as the following to denote this stabilization.