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

5,6-dimethylidenecyclohexa-1,3-diene

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.

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    $\begingroup$ Not a conjugated ring system. With the pendant pi electrons we can't apply the 4n+2 rule to it. $\endgroup$ – Oscar Lanzi May 18 '18 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 '18 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. $\endgroup$ – Oscar Lanzi May 18 '18 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. $\endgroup$ – Shoubhik Raj Maiti May 25 '18 at 14:15
  • $\begingroup$ 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! $\endgroup$ – Oscar Lanzi May 25 '18 at 14:26
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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:

Aromaticity

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  • $\begingroup$ What about tropolone? The pi electron system is not closed, but the molecule is aromatic. $\endgroup$ – Shoubhik Raj Maiti May 25 '18 at 14:32
  • $\begingroup$ Sorry, not tropolone, it should be tropone. $\endgroup$ – Shoubhik Raj Maiti May 25 '18 at 14:39

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