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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4$\begingroup$ Aromaticity is a product of the conjugated bonds in a system, not of the molecule as a whole. $\endgroup$– matt_blackCommented Jan 12, 2021 at 13:34
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4$\begingroup$ 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 ♦Commented Jan 12, 2021 at 13:56
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3$\begingroup$ Related; chemistry.stackexchange.com/questions/86734/… $\endgroup$– user55119Commented Jan 12, 2021 at 15:44
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$\begingroup$ 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. $\endgroup$– AlchimistaCommented Jan 13, 2021 at 8:24
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1 Answer
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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.
