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.

  • 2
    $\begingroup$ Not a conjugated ring system. With the pendant pi electrons we can't apply the 4n+2 rule to it. $\endgroup$ May 18, 2018 at 12:26
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    $\begingroup$ 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. $\endgroup$
    – porphyrin
    May 18, 2018 at 14:09
  • $\begingroup$ 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. $\endgroup$ May 18, 2018 at 16:29
  • $\begingroup$ @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. $\endgroup$
    – S R Maiti
    May 25, 2018 at 14:15
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    $\begingroup$ @SRMaiti, tropone is formally aromatic since the carbonyl is polarised towards the more electronegative oxygen atom which leads to a charge-separated aromatic resonance form. For more detailed explanation see chemistry.stackexchange.com/questions/15272/is-tropone-aromatic. $\endgroup$
    – Exeplone
    Feb 18 at 0:13

1 Answer 1


In this question, conjugation of p-orbitals in the system is misrepresented. It should be noted that, a π-system to be an aromatic or anti-aromatic, there should be a closed loop(s) of p-orbitals (conjugated), which is planner. What makes it an aromatic or anti-aromatic whether it follows Hückel's rule of (4n+2) π-electrons or not.

Aromatic (most stable): If a planer, closed loop(s) of p-orbital system has (4n+2) π-electrons (in the closed system), it is aromatic. Example:, benzene (a closed p-orbital system with six π-electrons) and naphthalene (a closed p-orbital system with ten π-electrons).

anti-Aromatic (least stable): If a planer, closed loop(s) of p-orbital system has (4n) π-electrons (in the closed system), it is anti-aromatic. Example:, cyclobutadiene (a closed p-orbital system with four π-electrons) and cyclopentadiene carbocation (a closed p-orbital system with four π-electrons).

non-Aromatic (stability depends on the system): Every other systems are non-aromatic. Examples: 5,6-dimethylidenecyclohexa-1,3-diene and heptafulvene (In both cases, each p-orbital system is not closed).

See following diagram:


  • $\begingroup$ What about tropolone? The pi electron system is not closed, but the molecule is aromatic. $\endgroup$
    – S R Maiti
    May 25, 2018 at 14:32
  • $\begingroup$ Sorry, not tropolone, it should be tropone. $\endgroup$
    – S R Maiti
    May 25, 2018 at 14:39

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