I was searching on the internet whether cyclopropene is aromatic or not. I found out that the electrons are not delocalized, causing the molecule to be non-planar and so forth, a lot of reasons.

But I don't know why I just don't seem to fall for it. My question is, if there are 4n+2 pi electrons, why aren't they delocalized? And what exactly stops them from being delocalized? I have tried but failed to come to a conclusion about the reasoning. Any help would be greatly appreciated.

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    $\begingroup$ Your question is hinged on a false premise. There are two pi electrons, and of course they are delocalized between two atoms. This is a phenomenon known as double bond. You must have heard of it before. $\endgroup$ – Ivan Neretin Oct 30 '20 at 6:52
  • $\begingroup$ I might be making a mistake, but I'll just repeat what my instructor told me. He told me that if the delocalized pi electrons of a cyclic compound can be denoted as 4n+2 where "n" is an integer, then the compound can be called aromatic. And in the case of cyclopropene, if the pi electrons were delocalized then there would be 2=4n+2, giving n=0, an integer. Hence the question, $\endgroup$ – Habib Oct 30 '20 at 8:55
  • $\begingroup$ Well, there is an important condition missing here. Cyclopropene is weird, let's put it aside. Imagine cyclohexene: a thing made of four -CH2- and one -CH=CH-. It is surely cyclic. It surely contains 4n+2 electrons (namely, 2). Is it aromatic? What do you think? $\endgroup$ – Ivan Neretin Oct 30 '20 at 9:08
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    $\begingroup$ You seem to be asking why there is no delocalisation of the pi electrons in the 3-membered ring. The answer to this is very obvious: There is no free continuous path for the delocalisation of electrons over the ring to occur. Yes, there are 2 carbon atoms with overlapping p orbitals but the third one is bonded to two hydrogen atoms and does not have an available p orbital to overlap with the p orbitals of the other two atoms. $\endgroup$ – Tan Yong Boon Oct 30 '20 at 9:23

Yup, they aren't. With the saturated carbon atom in the ring, you can't get a delocalized ring of pi electrons that you would need to apply Huckel's Rule.

A more rigorous analysis in Ref. [1] give us a curious conflict between theoretical and experimental results for this molecule. The theoretical calculations suggest that delocalization of the pi bond occurs through overlap with the $\ce{C-H}$ antibonding orbitals, making the molecule somewhat aromatic after all. But the experimental results referenced in the paper do not support thys prediction. With delocalized pi bonding the theory predicts that the $\ce{C-C}$ to the methylene carbon should be shorter than the $\ce{C-C}$ bonds in cyclopropane, but the experimental data show essentially the same bond lengths. Moreover, the quoted experimental dipole measurements do not agree with the modeled electron shift towards the methylene carbon. The experimental evidence, therefore, is that even with the possibilities offered by molecular orbitals pi delocalization fails in cyclopropene.


1. L. Radom, W. A. Lathan, W. J. Hehre, and J. A. Pople, "Molecular Orbital Theory of the Electronic Structure of Organic Compounds. VIII. Geometries, Energies, and Polarities of C3 Hydrocarbons", Journal of the American Chemical Society 93:21, 5339-5341 (1971).


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