11.42Cyclooctatetraene has eight pelectrons and thus does not satisfy the (4n12) pelectron require-ment of the Hückel rule.All of the exercises in this problem involve counting the number ofpelectrons in the various speciesderived from cyclooctatetraene and determining whether they satisfy the (4n12)pelectron rule.(a)Adding 1 pelectron gives a species (C8H82) with 9 pelectrons. 4n12, where nis a wholenumber, can never equal 9. This species is therefore not aromatic.(b)Adding 2pelectrons gives a species (C8H822) with 10pelectrons. 4n12510 whenn52.The species C8H822is aromatic.(c)Removing 1 pelectron gives a species (C8H81) with 7 pelectrons. 4n12 cannot equal 7.The species C8H81is not aromatic.(d)Removing 2pelectrons gives a species (C8H821) with 6pelectrons. 4n1256 whenn51.The species C8H821is aromatic.(It has the same number of pelectrons as benzene.)11.43(a, b)Cyclononatetraene does not have a continuous conjugated system of pelectrons. Conjuga-
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This note was uploaded on 09/26/2011 for the course CHM 2210 taught by Professor Reynolds during the Fall '01 term at University of Florida.