Energia de Deslocalização em Compostos

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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 311
(b) �e delocalisation energy is given by Edeloc = Eπ − Nπ(α + β), where
Nπ is the number of π electrons. In the ground state of benzene, two
electrons occupy the orbital with energy E0 and four electrons occupy the
degenerate pair of orbitals with energy E±1. Hence the π-electron binding
energy of benzene is Eπ = 2E0 + 4E±1 = 2(α + 2β) + 4(α + β) = 6α + 8β.
�erefore the delocalization energy is Edeloc = 6α + 8β − 6(α + β) = 2β .
To calculate the delocalization energy of hexatriene, the energies of the
occupied molecular orbitals are required. �e six electrons in the π sys-
tem of hexatriene occupy the three lowest energy orbitals, therefore the
π-electron binding energy is
Eπ = 2E1 + 2E2 + 2E3
= 2[α + 2β cos(π/7)] + 2[α + 2β cos(2π/7)] + 2[α + 2β cos(3π/7)]
= 6α + (6.98...)β
Hence the delocalization energy of hexatriene is Edeloc = 6α+(6.98...)β−
6(α + β) = 0.99β . �e lowering in energy as a result of delocalization
in benzene in greater than in hexatriene, consistent with the ‘aromatic’
system in benzene.
(c) �e π-electron binding energy in cyclooctatetraene is
Eπ = 2E0 + 4E±1 + 2E±2 = 2(α + 2β) + 4(α + 1.41β) + 2α = 8α + 9.64β
�e delocalization energy is therefore Edeloc = 8α + 9.64β − 8(α + β) =
1.64β .
�e π-electron binding energy in octatetraene is
Eπ = 2E1 + 2E2 + 2E3 + 2E4
= 2[α + 2β cos(π/9)] + 2[α + 2β cos(2π/9)]
+ 2[α + 2β cos(3π/9)] + 2[α + 2β cos(4π/9)]
= 8α + (9.51...)β
�erefore the delocalization energy is Edeloc = 8α+(9.51...)β−8(α+β) =
1.52β . �e di�erence in delocalization energies between cyclooctate-
traene and octatetraene is much smaller than the di�erence in the values
for benzene and hexatriene. �is is taken as an result of benzene being
‘aromatic’ in contrast to cyclooctatetraene which is sometimes described
as ‘antiaromatic’.
P9E.4 Within the Hückel approximations, the secular determinant of ethene is
∣α − E β
β α − E ∣
�e hamiltonian matrix is of the same form, but with diagonal elements α
H = ( α β
β α ) = α1 + β ( 0 1
1 0 )