Just like aromatic compounds, anti aromatic compounds have several resonance structures and some of them are also conjugated ( 1,3-cyclobutene).

Why is it that just having 4n electrons makes them unstable?

  • 2
    $\begingroup$ The intricacies of aromacity are still being researched. From what I recall, it still not clear as to what exactly aromaticity is. There are recurring features along aromatic compounds: electron delocalization, stability, and unreactivity. However, these are not completely understood. This is largely due to the fact that electron delocalization and localization is hard to measure directly (some even say that it is not measurable at all). $\endgroup$ Apr 6, 2016 at 7:24
  • $\begingroup$ chemistry.stackexchange.com/questions/30927/… $\endgroup$
    – Mithoron
    Apr 10, 2016 at 15:08

1 Answer 1


In conjugated-pi ring systems of N atoms, you typically have a single low energy orbital and then paired (degenerate) energy levels until you have N orbitals total. This has to do with the number of nodes in the conjugated pi system, and the number of ways you can lay them out with N atoms.

Now, when you add in electrons you add two electrons per orbital, so your 4n+2 electrons fill 2n+1 orbitals -- the single low energy orbital and then n complete sets of degenerate pairs.

If you only have 4n electrons, you can fill the single low energy orbital, then 2(n-1) complete sets of degenerate pairs ... but with two "extra" electrons which aren't paired up. This allows those two electrons to "spread" out between the pair of equal-energy orbitals. Those unfilled/partially filled orbitals mean that the anti-aromatic compound is quite reactive.

The other issue is that "stability" is not an absolute property. Stability is always in reference to some other state. The typical anti-aromatic systems you'll encounter tend to have those frontier orbitals as non-bonding or anti-bonding orbitals, meaning the electrons in the orbitals aren't really contributing to the molecular stability. Generally, if you build the energy level diagrams for alternate systems (e.g. a bent ring where you don't get localization) you'll find that - while the lowest energy orbital isn't as low energy as in the ring-conjugated system - the sum of the energies across all orbitals is indeed lower. And it's the total energy of the system which is optimized. (For example, in cyclobutadiene the ring-conjugated form would have two electrons in a low-energy bonding orbital and two in non-bonding orbitals. In contrast, linear butadiene bonding orbitals are not as low energy as for the ring-conjugated cyclo form, but there's two of them. And the sum of the energies from two linear bonding orbitals is lower than what you get from a single cyclo bonding orbital.)

(Final note: this is all high-level generalization. There's more sophisticated levels of quantum mechanical theory where things get more complex.)


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