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I'm trying to figure out how azulene is an aromatic compound; I understand it is cyclic, flat, and follows Huckel's rule, but I don't understand how it is conjugated.

I have circled what I believe to be an unconjugated area in yellow. Can someone help me figure out how this molecule is conjugated? Thank you.

azulene conjugation attempt

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    $\begingroup$ Azulene is a pretty special hydrocarbon. In particular, the exact linkage you highlight is what makes it special, from a graph-theoretic standpoint. I'm not sure whether Hückel's rule or simple arrow-pushing should be expected to describe it well. $\endgroup$ Aug 6, 2022 at 1:21
  • $\begingroup$ Thank you for responding. I think Huckel's rule explains it well because 4(2) + 2 = 10, but what would be a good way to determine that azulene is conjugated if arrow-pushing can't describe it well? $\endgroup$
    – Eneluza
    Aug 6, 2022 at 1:27
  • $\begingroup$ The $\sigma$-bond circled in yellow keeps the molecule planar. If this bond were replaced with two hydrogens to form a cyclodecapentaene, the molecule would not be planar nor aromatic. $\endgroup$
    – user55119
    Aug 6, 2022 at 1:48
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    $\begingroup$ Hückel's rule likely gives the right answer for the wrong reasons - this is a recurring theme in high-school and undergraduate chemistry. A satisfactory explanation other than "it is experimentally observed to be aromatic" may be quite involved. $\endgroup$ Aug 6, 2022 at 10:57
  • $\begingroup$ What definition of "conjugated" are you using? I consider doubly (or triply) bonded carbon atoms connected by a single (lone) single (sigma) bonds to be conjugated. For example, I'd view all the carbon atoms in this molecule as part of a chain of conjugated bonds, even though the two chains are connected by a single bond: C(C=CC(=CC=CC)C(C=CC=C)=CC=CC)=C So, the "internal" bond of azulene might not be conjugated, but both carbon atoms that are part of that internal bond are conjugated for other reasons... $\endgroup$
    – Curt F.
    Sep 29, 2022 at 20:30

1 Answer 1

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If you insist on neutral atoms then the internal linkage is not conjugated.

BUT ...

The notable thing about azulene is that the seven-and five-membered rings can be aromatic when the former is positively charged and the latter negatively charged. That puts six electrons into each ring, and we all know about the $4n+2$ rule (even if technically it should be applied to single-ring systems).

If you allow a positive charge on the seven-membered ring and a negative charge on the five-membered ring, you can render contributing structures with a double bond in the internal linkage.

enter image description here

http://www.chemspider.com/Chemical-Structure.8876.html

According to Wikipedia, azulene is actually known with a dipole moment having the positive end in the seven-membered ring, and its chemical reactivity (electrophilic in the seven-membered ring, nucleophilic in the five-membered ring) is also consistent with this polar contributing structure.

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  • $\begingroup$ Hi, thank you so much for the response. I understand how a resonant structure of azulene can form a cation on the seven-membered ring and an anion on the five-membered ring; however, I still do not understand how it is conjugated. Can you provide me with a diagram showing how to conjugate it with arrow pushing? If that is not possible, how can I determine that it can be conjugated and is therefore aromatic? $\endgroup$
    – Eneluza
    Aug 6, 2022 at 2:32
  • $\begingroup$ The monocyclic is the first word in the definition. It really is important that this rule is not applied to polycyclic compounds. Therefore Hückel's rules do not apply to azulene. doi.org/10.1351/goldbook.H02867 $\endgroup$ Sep 29, 2022 at 18:03
  • $\begingroup$ Yes I know that, but the OP was asking for contributing structures with an internal pi bond. So I had to go along. The existence of a good contributing structure with this feature differentiates azulene from a 10-pi mono-cycle. With that, however, Hückel calculations indicate the internal pi bond is weaker in azulene than in naphthalene. $\endgroup$ Sep 29, 2022 at 18:23

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