You Spin My Methyl Round-by-Round
After a [1,4] shift, what change you will observe in the H-RMN spectra compared to that of the original molecule?
Suppose the solution is sufficiently hot: .
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1 solution
To explain this reaction we must use Molecular Orbital Theory. Breaking the cyclopropane bond closest to the positive charge will give us the components necessary to deduce what happens. Two systems of π orbitals rise ; one with only one orbital and the second formed by four, identically to butadiene. We assume that the 2 electrons from the broken bond go in the 4-orbital system. The course of any reaction is dictated by the HOMO-LUMO interaction. In this case, the LUMO will be the unoccupied orbital of the first system and the HOMO will be the second MO of the butadiene systems, as it has 4 electrons( so there are two filled MOs and we take only the last one). The interaction is shown in the figure below. It's easy to see that the methyl group stays behind the cyclopropane ring during the transition. This means that it will stay there after infinitely many [1,4] shifts. Moreover, we may assume that this molecule is in a superposition of all it's valence isomers and the real molecule looks different, more symmetrical. This can also be explained as a supra-antara interaction due to the 4 electrons system.