# Are 2-pyridone and cycloheptatrienide antiaromatic?

Determine which one of the following compounds is not aromatic:

I think 2-pyridone (preferred name: pyridin-2(1H)-one) (1) is non-aromatic. It already contains three double bonds. If I take the lone pairs on the nitrogen atom and consider them π-electrons, this would result in eight π-electrons which results in antiaromaticity.

I am guessing maybe the $$\ce{C=O}$$ double bond doesn't count as part of the ring and therefore if the lone pair on nitrogen is $$\mathrm{sp^2}$$-hybridized, it will result in an aromatic compound.

Similarly, cycloheptatrienide (2) seems to also be non-aromatic as the $$\mathrm{sp^3}$$-hybridized carbanion causes a non-planar structure. And if $$\mathrm{sp^2}$$-hybridized, creating eight π-electrons and therefore becoming antiaromatic.

## 1 Answer

This can be confusing, especially when you are given a molecule that has a pendant carbonyl oxygen like your first molecule.

Strictly speaking, the $$4n+2$$ and $$4n$$ electron rules apply only when the conjugated electrons form exactly a single ring, meaning there can be no pendant carbonyl oxygen atoms (or other double-bonded goodies like sulfur atoms or imino groups).

Therefore your first molecule may indeed be aromatic, as I discuss below, but the electron-counting rule you're supposed to use does not tell you that unless you have been taught the appropriate modification for this type of situation. Absent that condition, the problem is not well-posed.

There is empirical evidence for such a modification in the case of pendant carbonyl oxygens, which conveniently have both a high electronegativity and a high rate of occurrence compared with other double-bonded ligands. If polarizing the carbonyl group to $$\ce{\overset{+}{C}-\overset{-}{O}}$$ leaves the ring fully conjugated with an appropriate $$4n+2$$ count of remaining π-electrons, then the compound shows properties consistent with this “aromatic zwitterion” model (such as an enhanced dipole moment and a more strongly basic oxygen atom than would be expected of a ketone). Trying this modification with your first molecule does indeed leave a conjugated ring with six π-electrons, so the aromatic-zwitterion structure is favored. The nitrogen atom being able to donate an electron pair helps this along, but even without such an atom cyclopropenone and tropone show similar aromatic-zwitterion character.

The second molecule, actually the $$\ce{C7H7^-}$$ ion, is also not quite cut and dried. Of course it cannot be aromatic, as the OP correctly points out, but is it nonaromatic with a nonconjugated, $$\mathrm{sp^3}$$-hybridized carb-anion center, or antiaromatic with all carbons $$\mathrm{sp^2}$$-hybridized and the charge delocalized? For larger rings the energy penalty for antiaromaticity drops off, so it's not as clear-cut as say cyclobutadiene would be. Experimental investigation or sophisticated computation is needed to prove that $$\ce{C7H7^-}$$ is in fact nonplanar and nonaromatic.