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11-oxabicyclo(4.4.1)undeca-1,3,5,7,9-pentaene

I have been told by my instructors that for a compound to be aromatic, it must be planar. In this given molecule the oxygen seems to be out of the plane of the conjugated system: how can it still be aromatic, then?

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    $\begingroup$ Aromaticity is a product of the conjugated bonds in a system, not of the molecule as a whole. $\endgroup$ – matt_black Jan 12 at 13:34
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    $\begingroup$ Please use titles that are descriptive, not cryptic. "This compound" is 1,6-oxido[10]annulene (traditional name) or 11-oxabicyclo(4.4.1)undeca-1,3,5,7,9-pentaene (PIN) and not only the oxygen bridge, but also the ring is not planar. Read about aromaticity of 1,6-methano(10)annulene, the ideas are applicable here as well. $\endgroup$ – andselisk Jan 12 at 13:56
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    $\begingroup$ Related; chemistry.stackexchange.com/questions/86734/… $\endgroup$ – user55119 Jan 12 at 15:44
  • $\begingroup$ I point out that the Wikipedia pages on [10]-annulene and methano annulene contain conflicting info about the ring planarity. Still the less distorted and thus aromatic is the second one. $\endgroup$ – Alchimista Jan 13 at 8:24
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The aromatic system should be planar. The oxygen is not part of the aromatic system.

Please note, of course, that your drawing does not capture the 3-dimensional geometry of the system. You would still have to analyze the carbon skeleton to see if it satisfied the planarity constraint enough to be aromatic.

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