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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?
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    $\begingroup$ 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) $\endgroup$
    – Kartik
    May 2, 2017 at 4:28
  • $\begingroup$ Probably that's true. Upper ring is aromatic and lower ring is also aromatic if counted separately. $\endgroup$
    – Suraj S
    May 2, 2017 at 4:38

1 Answer 1

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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.

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  • $\begingroup$ Thanks for the reply. What do u mean by interior atoms? $\endgroup$
    – user16096
    May 2, 2017 at 12:01
  • $\begingroup$ 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. $\endgroup$ May 2, 2017 at 12:12
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    $\begingroup$ 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. goldbook.iupac.org/html/H/H02867.html $\endgroup$ May 7, 2017 at 12:07

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