# Do the π-electrons from the C=O bond in tropone contribute to Hückel 's rule?

This is a continuation of Is tropone aromatic?

I'm looking at the resonance form that does have the carbonyl double-bond, 1a. Can this resonance structure also be considered aromatic? Would the electrons from the C=O double bond also be delocalized through extended conjugation?

To me, this looks like a very similar situation to styrene, when we have an alkenyl sidechain attached to a benzene ring. Apparently, for alkenyl side-chains, Huckel's number is not affected, leaving an already aromatic compound to stay aromatic even though more electrons come into the conjugation.

In the case of tropone, would the C=O double bond interfere with electron counting for Hückel's rule?

In addition to Philipp's answer I'd like to share some orbital pictures which shall make the case rather obvious.

Let me shortly remind you of the key part of the IUPAC definition for aromaticity:

1. In the traditional sense, 'having a chemistry typified by benzene'.

I will therefore follow the same approach as in my answer to "Is phenoxide aromatic?" in comparing the orbital pictures.

The following orbitals are for the π system of benzene, in increasing order of energy (DF-BP86/def2SVP). The two orbitals on the right are degenerate:

Analogously, the π system for cyclohepta-2,4,6-trien-1-one is extensively delocalised and holds eight electrons. Note that the symmetry of this molecule is only $C_\mathrm{2v}$ removing a lot of constraint from the electronic structure. This is also why the Hückel rule breaks down, since there are no more degeneracies observed.

Also note, that the LUMO of this molecule (pictured below) still has considerable bonding character and that reduction would most likely not destroy aromaticity. (The same statement also applies to oxidation.)

Maybe my answer concerning the question why aniline is considered aromatic might help a bit. There I describe that groups that are cross-conjugated with the aromatic ring are not included into the "Hückel-count" and give some arguments why this is so.

A similar argument would apply here: The keto group is cross-conjugated with the ring's π-system but in a different way than the amino group is in aniline because the carbonyl carbon is part of the ring. So, the role of the keto group's π system is only to act as a bridge between the $\ce{C=C}$ π-orbitals so that the ring is "closed". Thus, you wouldn't count the keto group's electrons but only the $\ce{C=C}$ π-electrons. And this would lead to the conclusion that your cyclic ketone will be aromatic if it really is flat (that will have to be checked by NMR or other methods) because there are 6 π-electrons in the ring.

• So para-quinone is antiaromatic because not counting the conjugated pi electrons there are four in the ring. Does that make sense? Para-quinone is usually described as nonaromatic instead, and a look at molecular orbitals (even qualitatively) should convince you that the eight total pi electrons are in fact optimal for pi bonding in this molecule. – Oscar Lanzi Jan 29 '18 at 22:30

You can't apply Huckel's 4n+2 rule except for purely cyclic conjugation. When there are pendant pi bonds like the one in tropone all bets are off on 4n+2. You need to look at the MO structure to see if the total pi electron count, eight in this case, is optimal for the whole molecule, not just the ring. (In tropone it is.)