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The structure of cyanidin is as follows:

Cyanidin

There are eight double bonds in the entire compound, and hence $8 \times 2 = 16$ π-electrons, which does not satisfy the $4n + 2$ rule. However, the molecule is usually depicted as aromatic in literature. MarvinSketch (a software) also reports the entire molecule as aromatic.

  1. Is this molecule aromatic? Why?
  2. Are all anthocyaninidins aromatic?
orthocresol
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    I guess you have to count the electrons for the upper benzene ring and lower fused rings seperately. Then the lower one has 10 electrons (aromatic) and upper one has 6 (again aromatic) – Kartik May 02 '17 at 04:28
  • Probably that's true. Upper ring is aromatic and lower ring is also aromatic if counted separately. – Suraj S May 02 '17 at 04:38

1 Answers1

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No doubt it is aromatic.

The $4n+2$ rule is intended for a single cycle of π-electrons, which may possibly include multiple rings but can't have any interior atoms or pendant π- bonds (which would not fit in a cycle). Here, as Kartik indicates in a comment, there are evidently two such cycles and you have to consider them separately. One cycle has six π-electrons and the other has ten. You have a molecule (or more accurately, a molecular ion) with two aromatic cycles.

orthocresol
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Oscar Lanzi
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  • Thanks for the reply. What do u mean by interior atoms? –  May 02 '17 at 12:01
  • Look at pyrene. The presence of two conjugated atoms surrounded by the others means you can't fit them all into a single cycle. Thus the $4n+2$ rule is not reliable. And true to form it doesn't work. – Oscar Lanzi May 02 '17 at 12:12
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    Hückel's rules can only be applied to the compounds they were derived from: monocyclic, planar hydrocarbons. For most aromatic compounds they are a lot less reliable and only a very rough approximation. https://goldbook.iupac.org/html/H/H02867.html – Martin - マーチン May 07 '17 at 12:07