# Why is 5,6-dimethylidenecyclohexa-1,3-diene non-aromatic?

The compound 5,6-dimethylidenecyclohexa-1,3-diene is supposedly non-aromatic:

But I have no idea why this is so. The compound is probably planar. All the atoms on the ring are $\ce{sp^2}$ hybridized. So, the p-orbitals are in conjugation. There are 4-$\ce{\pi}$ electrons, as I suppose the external pi bonds do not participate in resonance. So, it should be antiaromatic. Why is it non-aromatic?

I am a school student, so an detailed description would be helpful.

• Not a conjugated ring system. With the pendant pi electrons we can't apply the 4n+2 rule to it. – Oscar Lanzi May 18 '18 at 12:26
• It is conjugated though, in effect it is electronically similar to a linear polyene. To make a 3rd double bond in the ring would mean making a radical species of unreasonably high energy. – porphyrin May 18 '18 at 14:09
• Or you would need a zwitterionic structure where one methylene carbon is negative and the other, seemingly identical carbon would be positive and short of an octet. This is also not pretty. We are just not getting this thing to be aromatic, period. – Oscar Lanzi May 18 '18 at 16:29
• @OscarLanzi, what is 'pendant pi electrons'? I am guessing it means the pi electrons outside the ring. But if we only consider the pi electrons in the ring, then it should be anti-aromatic (conjugated ring system, 4 pi electron. – Shoubhik Raj Maiti May 25 '18 at 14:15
• Like the "pendant" decoration hanging off a necklace or chain. Pendant pi electrons, or even vacant pendant pi orbitals, modify the ring orbitals and bust the 4n+2 rule no matter how you try to count. DO NOT apply the 4n+2 rule, or any rule, beyond its range of validity! – Oscar Lanzi May 25 '18 at 14:26