Lecture 14 February 7, 2011
Reactions O2, Woodward-Hoffmann
Nature of the Chemical Bond
with applications to catalysis, materials
science, nanotechnology, surface science,
bioinorganic chemistry, and energy
William A. Goddard, III, wag@wag.caltech.edu
316 Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry,
Materials Science, and Applied Physics,
California Institute of Technology
Teaching Assistants: Wei-Guang Liu <wgliu@wag.caltech.edu>
Caitlin Scott <cescott@caltech.edu>
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
1
Last time
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
2
Bond H to O2
Bring H toward px on Left O
Overlap doubly
occupied (pxL)2
thus repulsive
Overlap singly
occupied (pxL)2
thus bonding
Get HOO bond angle ~ 90º
S=1/2 (doublet)
Antisymmetric with respect to plane:
A” irreducible representation (Cs
group)
2A”
state
Ch120a-Goddard-L14
Bond weakened by ~ 51 kcal/mol
due
toA.loss
inIII,Oall2 rights
resonance
© copyright 2011
William
Goddard
reserved
3
Bond 2nd H to HO2 to form hydrogen peroxide
Bring H toward py on right O
Expect new HOO bond angle ~ 90º
Expect HOOH dihedral ~90º
Indeed H-S-S-H:
HSS = 91.3º and HSSH= 90.6º
But H-H overlap leads to steric effects for HOOH,
net result:
HOO opens up to ~94.8º
HOOH angle  111.5º
trans structure, 180º only 1.2 kcal/mol higher
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
4
Compare bond energies (kcal/mol)
O2 3Sg-
119.0
50.8
HO-O
68.2
17.1
HO-OH
51.1
67.9
H-O2
HOO-H
51.5
85.2
Interpretation:
OO s bond = 51.1 kcal/mol
OO p bond = 119.0-51.1=67.9 kcal/mol (resonance)
Bonding H to O2 loses 50.8 kcal/mol of resonance
Bonding H to HO2 loses the other 17.1 kcal/mol of resonance
Intrinsic H-O bond is 85.2 + 17.1 =102.3
compare CH3O-H: HO bond is 105.1
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
5
Bond O2 to O to form
ozone
Require two OO s bonds get
States with 4, 5, and 6 pp
electrons
Ground state is 4p case
Get S=0,1
but 0 better
Goddard
et al Acc. Chem. Res.
6, 368 (1973)
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
6
Bond O2 to O to form ozone
lose O-O p resonance, 51 kcal/mol
New O-O s bond, 51 kcal/mol
Gain O-Op resonance,<17 kcal/mol,assume 2/3
New singlet coupling of pL and pR orbitals
Total splitting ~ 1 eV = 23 kcal/mol, assume ½
stabilizes singlet and ½ destabilizes triplet
Expect bond for singlet of 11 + 12 = 23 kcal/mol, exper = 25
Expect triplet state to be bound by 11-12 = -1 kcal/mol,
probably
between +2
and -2
Ch120a-Goddard-L14
© copyright
2011 William A. Goddard III, all rights reserved
7
Alternative view of bonding in ozone
Start here with 1-3 diradical
Transfer electron from central doubly
occupied pp pair to the R singly occupied
pp.
Now can form a p bond the L singly
occupied pp.
Hard to estimate
strength
of bond
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
8
Ring ozone
Form 3 OO sigma bonds, but pp pairs overlap
Analog: cis HOOH bond is 51.1-7.6=43.5
kcal/mol. Get total bond of 3*43.5=130.5 which
is 11.5 more stable than O2.
Correct for strain due to 60º bond angles = 26
kcal/mol from cyclopropane. Expect ring O3 to
be unstable with respect to O2 + O by ~14
kcal/mol,
But if formed it might be rather stable with
respect various chemical reactions.
Ab Initio Theoretical Results on the Stability of Cyclic Ozone
L. B. Harding and W. A. Goddard III
J. Chem. Phys. 67, 2377 (1977) CN 5599
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
9
More on N2
The elements N, P, As, Sb, and Bi all have an (ns)2(np)3
configuration, leading to a triple bond
Adding in the (ns) pairs, we show
the wavefunction as
This is the VB description of N2, P2, etc. The
optimum orbitals of N2 are shown on the next slide.
The MO description of N2 is
Which we can
draw as
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
10
Ground state of C2
MO configuration
Have two strong p bonds,
but sigma system looks just like Be2 which leads to a bond of ~ 1
kcal/mol
The lobe pair on each Be is activated to form the sigma bond.
The net result is no net contribution to bond from sigma
electrons. It is as if we started with HCCH and cut off the Hs
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
11
Low-lying states of C2
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
12
Van der Waals interactions
For an ideal gas the equation of state is given by
pV =nRT
where p = pressure; V = volume of the container
n = number of moles; R = gas constant = NAkB
NA = Avogadro constant; kB = Boltzmann constant
Van der Waals equation of state (1873)
[p + n2a/V2)[V - nb] = nRT
Where a is related to attractions between the particles,
(reducing the pressure)
And b is related to a reduced available volume (due to
finite size of particles)
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
13
Noble gas dimers
No bonding at the VB or MO level
Only simultaneous electron
correlation (London attraction) or
van der Waals attraction, -C/R6
s
Ar2
Re
De
Ch120a-Goddard-L14
LJ 12-6 Force Field
E=A/R12 –B/R6
= De[r-12 – 2r-6]
= 4 De[t-12 – t-6]
r= R/Re
t= R/s
where s = Re(1/2)1/6
=0.89 Re
© copyright 2011 William A. Goddard III, all rights reserved
14
London Dispersion
The weak binding in He2 and other noble gas dimers was
explained in terms of QM by Fritz London in 1930
The idea is that even for a spherically symmetric atoms such
as He the QM description will have instantaneous fluctuations
in the electron positions that will lead to fluctuating dipole
moments that average out to zero. The field due to a dipole
falls off as 1/R3 , but since the average dipole is zero the first
nonzero contribution is from 2nd order perturbation theory,
which scales like
-C/R6 (with higher order terms like 1/R8 and 1/R10)
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
15
London Dispersion
The weak binding in He2 and other nobel gas dimers was
explained in terms of QM by Fritz London in 1930
The idea is that even for a spherically symmetric atoms such
as He the QM description will have instantaneous fluctuations
in the electron positions that will lead to fluctuating dipole
moments that average out to zero. The field due to a dipole
falls off as 1/R3 , but since the average dipole is zero the first
nonzero contribution is from 2nd order perturbation theory,
which scales like
-C/R6 (with higher order terms like 1/R8 and 1/R10)
Consequently it is common to fit the interaction potentials to
functional forms with a long range 1/R6 attraction to account
for London dispersion (usually referred to as van der Waals
attraction) plus a short range repulsive term to account for
short
Range Pauli Repulsion)
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
16
MO and VB view of He dimer, He2
MO view
ΨMO(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2
Net BO=0
VB view
ΨVB(He2) = A[(La)(Lb)(Ra)(Rb)]= (L)2(R)2
Substitute sg = R + L and sg = R - L
Get ΨMO(He2) ≡ ΨMO(He2)
Ch120a-Goddard-L14
Pauli  orthog of R
to L  repulsive
© copyright 2011 William A. Goddard III, all rights reserved
17
Remove an electron from He2 to get He2+
MO view
Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2
Two bonding and two antibonding  BO= 0
Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su)  BO = ½
Get 2Su+ symmetry.
Bond energy and bond distance similar to H2+, also BO = ½
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
18
Remove an electron from He2 to get He2+
MO view
Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2
Two bonding and two antibonding  BO= 0
Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su)  BO = ½
Get 2Su+ symmetry.
Bond energy and bond distance similar to H2+, also BO = ½
VB view
Substitute sg = R + L and sg = L - R
Get ΨVB(He2) ≡ A[(La)(Lb)(Ra)] - A[(La)(Rb)(Ra)]
= (L)2(R) - (R)2(L)
-
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
19
He2+
+
2S +
g
(sg)1(su)2
2S +
u
(sg
)2(s
u)
BO=0.5
MO good for discuss spectroscopy,
VB good for discuss chemistry
Check H2 and H2+ numbers
Ch120a-Goddard-L14
He2 Re=3.03A
De=0.02 kcal/mol
No bond
H2 Re=0.74xA
De=110.x kcal/mol
BO = 1.0
H2+ Re=1.06x A
De=60.x kcal/mol
BO = 0.5
© copyright 2011 William A. Goddard III, all rights reserved
20
Woodward-Hoffmann rules
orbital symmetry rules
Frontier Orbital rules
Certain cycloadditions occur but not others
Roald
Hoffmann
2s+2s
2s+4s
4s+4s
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
21
Woodward-Hoffmann rules
orbital symmetry rules
Frontier Orbital rules
Certain cyclizations occur but not others
Ch120a-Goddard-L14
conrotatory
disrotatory
disrotatory
conrotatory
© copyright 2011 William A. Goddard III, all rights reserved
22
2+2 cycloaddition – Orbital correlation diagram
ground state
Start with 2 ethene in GS
Occupied orbitals have SS
and SA symmetries
Now examine product
cyclobutane
Forbidden
Occupied orbitals have SS
and AS symmetry
Thus must have high energy
transition state: forbidden
reactions
GS
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
23
2+2 cycloaddition – Orbital correlation diagram
excited state
Start with 1 ethene in GS and one
in ES
Open shell orbitals have SA and AS
symmetries
Now examine product cyclobutane
Allowed
Open shell orbitals have AS and SA
symmetry
Thus orbitals of reactant correlate
with those of product
ES
Thus photochemical reaction
allowed
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
24
Consider butadiene + ethene
cycloaddition; Diehls-Aldor
2+4 Ground State
A
A
Ground state has S,
S, and A occupied
S
Product has S, A,
and S occupied
Allowed
A
Thus transition state
need not be high
Allowed reaction
S
S
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
25
WH rules – 2 + 4
Excited State
A
A
S
Forbidden
A
S
S
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
26
Summary WH rules cycloaddition
2n + 2m
n+m odd: Thermal allowed
Photochemical forbidden
n+m even: Thermal forbidden
Photochemical allowed
n=1, m=1: ethene + ethene
n=1, m=2: ethene + butadience (Diels-Alder)
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
27
A
S
Allowed
A
S
A
S
A
Rotation, C2
Ch120a-Goddard-L14
WH rules – cyclization-GS
S
A
A
Forbidden
A
S
A
S
S
S
Reflection, s
© copyright 2011 William A. Goddard III, all rights reserved
28
Summary WH rules cyclization
2n
n odd: thermal disrotatory
Photochemical conrotatory
n even: Thermal conrotatory
Photochemical disrotatory
n=2  butadiene
n=3  hexatriene
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
29
2D Reaction Surface for H + CH4  H2 + CH3
Product: H2+CH3
H--C
Reactant: H+CH4
Ch120a-Goddard-L14
H--H
© copyright 2011 William A. Goddard III, all rights reserved
30
reaction surface of H + CH4  H2 + CH3 along reaction pat
30.00
H + CH4  H2 + CH3
HF
25.00
Energy (kcal/mol)
(kcal/mol)
Energy
HF
HF_PT2
XYG3
20.00
CCSD(T)
B3LYP
15.00
CCSD(T)
XYG3
BLYP
SVWN
HF_PT2 SVWN
B3LYP
10.00
BLYP
5.00
SVWN
0.00
-2.00
-1.50
-5.00
Ch120a-Goddard-L14
-1.00
-0.50
0.00
0.50
1.00
1.50
2.00
2.50
ReactionR(CH)-R(HH)
coordinate (in Å)
Reaction Coordinate:
© copyright 2011 William A. Goddard III, all rights reserved
31
GVB view reactions
Reactant HD+T
H
D
T
Product H+DT
Ch120a-Goddard-L14
32
Goddard and Ladner, JACS 93 6750 (1971)
© copyright 2011 William A. Goddard III, all rights reserved
GVB view reactions
Reactant HD+T
H
D
T
During reaction, bonding orbital on D stays on D,
Bonding orbital on H keeps its overlap with the orbital on D but
delocalizes over H and T in the TS and localizes on T in the
product. Thus highly overlapping bond for whole reaction
Nonbonding Orbital on free T of reactant becomes partially
antibonding in TS and localizes on free H of product, but it
changes sign
Product H+DT
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
33
GVB view reactions
Reactant HD+T
H
D
T
Bond pair keeps
high overlap while
flipping from reactant
to product
Transition state
nonbond orbital
keeps orthogonal,
hence changes sign
Product H+DT
Ch120a-Goddard-L14
H
D
T
© copyright 2011 William A. Goddard III, all rights reserved
34
GVB analysis of cyclization (4 e case)
4 VB orbitals:
A,B,C,D reactant
φB
φA
φC
Move AB bond;
Ignore D; C
changes phase as
it moves from 3 to 1
φD
2 3
φφB φφA
B
A
2 3
4
1
φφC
C
φφD
D
4
1
φB
Now ask how the CH2 groups
1 and 4 must rotate so that C
and D retain positive overlap.
φA
2 3
φC
Clearly 4n© copyright
is conrotatory
2011 William A. Goddard III, all1rights reserved
Ch120a-Goddard-L14
φD
4
35
GVB analysis of cyclization (6 e case)
φB
φA
φC
2
1
Ch120a-Goddard-L14
3
φD
4
© copyright 2011 William A. Goddard III, all rights reserved
36
Apply GVB model to 2 + 2
4 VB orbitals:A,B,C,D reactant
Transition state: ignore C
φB
φA
φB
φC
φD
φD
φC
Ch120a-Goddard-L14
φA
φD
φB
φA
\
4 VB
orbitals
product
φC
Nodal
plane
© copyright 2011 William A. Goddard III, all rights reserved
37
1
2
1
2
Transition state for 2 + 2
3
4
3
4
Transition state: ignore C
Orbitals A on 1 and B on 2
keep high overlap as the
2
1
bond moves from 12 to 23
φA
with B staying on 2 and A
moving from 1 to 3
Orbital D must move from 3 to 1
3
4
but must remain orthogonal to
the AB bond. Thus it gets a
nodal plane
φ
φB
D
The overlap of D and C goes
from positive in reactant to
Nodal
negative in product, hence goingφ
plane
C
through 0. thus break CD bond.
Reaction Forbidden
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
38
GVB model fast analysis 2 + 2
4 VB orbitals:A,B,C,D reactant
1
2
φA
φB
φC
3 φD
4
Move A from 1 to 3 keeping overlap with
B
Simultaneously D moves from 3 to 1 but
must change sign since must remain
orthogonal to A and B
φD
C and D start with positive overlap
and end with negative overlap.
Thus break bond  forbidden
Ch120a-Goddard-L14
φC
© copyright 2011 William A. Goddard III, all rights reserved
φB
φA
\
39
Next examine 2+4
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
40
GVB 2+4
φC
φB
φD
φA
23
1
4
6
5
φF
φD
φC
23
1
4
6
φE
φA
φB
φF
5
φE
1. Move AB bond;
Ignore D; C changes
phase as it moves
from 3 to 1
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
41
GVB 2+4
2. Move EF
bond; C
changes phase
again as it
moves from 1
to 5
φA
φB
φD
φC
23
1
4
φA
φB
φD
23
6
φF
5
φE
3. Now examine overlap of D with
C. It is positive. Thus can retain
bond CD as AB and EF migrate
Ch120a-Goddard-L14
1
4
φE
φC
6
5
φF
Reaction
Allowed
© copyright 2011 William A. Goddard III, all rights reserved
42
GVB 2+4
φC
φB
φD
φA
23
1
4
φA
φB
φD
φC
23
1
4
2. Move EF
bond; C
changes phase
again as it
moves from 1
to 5
φA
φB
φD
23
6
φF
5
6
φE
φF
5
φE
1
4
3. Examine final
φE
1. Move AB bond;
φC
overlap of D with
Ignore D; C changes
C. It is positive.
6
5
phase as it moves
Thus can retain
φF
from 3 to 1
bond CD as AB
Reaction
Allowed
Ch120a-Goddard-L14
© copyright
2011EF
William
A. Goddard III, all
rights reserved
and
migrate
43
2D Reaction Surface for H + CH4  H2 + CH3
Product: H2+CH3
H--C
Reactant: H+CH4
Ch120a-Goddard-L14
H--H
© copyright 2011 William A. Goddard III, all rights reserved
44
reaction surface of H + CH4  H2 + CH3 along reaction pat
30.00
H + CH4  H2 + CH3
HF
25.00
Energy (kcal/mol)
(kcal/mol)
Energy
HF
HF_PT2
XYG3
20.00
CCSD(T)
B3LYP
15.00
CCSD(T)
XYG3
BLYP
SVWN
HF_PT2 SVWN
B3LYP
10.00
BLYP
5.00
SVWN
0.00
-2.00
-1.50
-5.00
Ch120a-Goddard-L14
-1.00
-0.50
0.00
0.50
1.00
1.50
2.00
2.50
ReactionR(CH)-R(HH)
coordinate (in Å)
Reaction Coordinate:
© copyright 2011 William A. Goddard III, all rights reserved
45
GVB view reactions
Reactant HD+T
H
D
T
During reaction, bonding orbital on D stays on D,
Bonding orbital on H keeps its overlap with the orbital on D but
delocalizes over H and T in the TS and localizes on T in the
product. Thus highly overlapping bond for whole reaction
Nonbonding Orbital on free T of reactant becomes partially
antibonding in TS and localizes on free H of product, but it
changes sign
Product H+DT
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
46
GVB view reactions
Reactant HD+T
H
D
T
Bond pair keeps
high overlap while
flipping from reactant
to product
Transition state
nonbond orbital
keeps orthogonal,
hence changes sign
Product H+DT
Ch120a-Goddard-L14
H
D
T
© copyright 2011 William A. Goddard III, all rights reserved
47
GVB analysis of cyclization (4 e case)
4 VB orbitals:
A,B,C,D reactant
φB
φA
φC
Move AB bond;
Ignore D; C
changes phase as
it moves from 3 to 1
φD
2 3
φφB φφA
B
A
2 3
4
1
φφC
C
φφD
D
4
1
φB
Now ask how the CH2 groups
1 and 4 must rotate so that C
and D retain positive overlap.
φA
2 3
φC
Clearly 4n© copyright
is conrotatory
2011 William A. Goddard III, all1rights reserved
Ch120a-Goddard-L14
φD
4
48
Apply GVB model to 2 + 2
4 VB orbitals:A,B,C,D reactant
Transition state: ignore C
φB
φA
φB
φC
φD
φD
φC
Ch120a-Goddard-L14
φA
φD
φB
φA
\
4 VB
orbitals
product
φC
Nodal
plane
© copyright 2011 William A. Goddard III, all rights reserved
49
1
2
1
2
Transition state for 2 + 2
3
4
3
4
Transition state: ignore C
Orbitals A on 1 and B on 2
keep high overlap as the
2
1
bond moves from 12 to 23
φA
with B staying on 2 and A
moving from 1 to 3
Orbital D must move from 3 to 1
3
4
but must remain orthogonal to
the AB bond. Thus it gets a
nodal plane
φ
φB
D
The overlap of D and C goes
from positive in reactant to
Nodal
negative in product, hence goingφ
plane
C
through 0. thus break CD bond.
Reaction Forbidden
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
50
GVB model fast analysis 2 + 2
4 VB orbitals:A,B,C,D reactant
1
2
φA
φB
φC
3 φD
4
Move A from 1 to 3 keeping overlap with
B
Simultaneously D moves from 3 to 1 but
must change sign since must remain
orthogonal to A and B
φD
C and D start with positive overlap
and end with negative overlap.
Thus break bond  forbidden
Ch120a-Goddard-L14
φC
© copyright 2011 William A. Goddard III, all rights reserved
φB
φA
\
51
Next examine 2+4
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
52
GVB 2+4
φC
φB
φD
φA
23
1
4
6
5
φF
φD
φC
23
1
4
6
φE
φA
φB
φF
5
φE
1. Move AB bond;
Ignore D; C changes
phase as it moves
from 3 to 1
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
53
GVB 2+4
2. Move EF
bond; C
changes phase
again as it
moves from 1
to 5
φA
φB
φD
φC
23
1
4
φA
φB
φD
23
6
φF
5
φE
3. Now examine overlap of D with
C. It is positive. Thus can retain
bond CD as AB and EF migrate
Ch120a-Goddard-L14
1
4
φE
φC
6
5
φF
Reaction
Allowed
© copyright 2011 William A. Goddard III, all rights reserved
54
GVB 2+4
φC
φB
φD
φA
23
1
4
φA
φB
φD
φC
23
1
4
2. Move EF
bond; C
changes phase
again as it
moves from 1
to 5
φA
φB
φD
23
6
φF
5
6
φE
φF
5
φE
1
4
3. Examine final
φE
1. Move AB bond;
φC
overlap of D with
Ignore D; C changes
C. It is positive.
6
5
phase as it moves
Thus can retain
φF
from 3 to 1
bond CD as AB
Reaction
Allowed
Ch120a-Goddard-L14
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2011EF
William
A. Goddard III, all
rights reserved
and
migrate
55
Benzene and Resonance
referred to as Kekule or VB structures
Ch120a-Goddard-L14
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56
Resonance
Ch120a-Goddard-L14
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57
Benzene wavefunction
is a superposition of the VB structures in (2)
benzene as
≡
Ch120a-Goddard-L14
+
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58
More on resonance
That benzene would have a regular 6-fold symmetry is not
obvious. Each VB spin coupling would prefer to have the
double bonds at ~1.34A and the single bond at ~1.47 A (as
the central bond in butadiene)
Thus there is a cost to distorting the structure to have equal
bond distances of 1.40A.
However for the equal bond distances, there is a
resonance stabilization that exceeds the cost of distorting
the structure, leading to D6h symmetry.
Ch120a-Goddard-L14
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59
Cyclobutadiene
For cyclobutadiene, we have the same situation, but here the
rectangular structure is more stable than the square.
That is, the resonance energy does not balance the cost of
making the bond distances equal.
1.34 A
1.5x A
The reason is that the pi bonds must be orthogonalized,
forcing a nodal plane through the adjacent C atoms, causing
the energy to increase dramatically as the 1.54 distance is
reduced to 1.40A.
For benzene only one nodal plane makes the pi bond
orthogonal
to both other
bonds,
leading
to lower
cost
Ch120a-Goddard-L14
© copyright
2011 William
A. Goddard
III, all rights
reserved
60
graphene
Graphene: CC=1.4210A
Bond order = 4/3
Benzene: CC=1.40 BO=3/2
Ethylene: CC=1.34 BO = 2
CCC=120°
Unit cell has 2 carbon atoms
1x1 Unit cell
This is referred to as graphene
Ch120a-Goddard-L14
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61
Graphene band structure
1x1 Unit cell
Unit cell has 2 carbon atoms
Bands: 2pp orbitals per cell
2 bands of states each with N states
where N is the number of unit cells
2 p electrons per cell  2N electrons for
N unit cells
The lowest N MOs are doubly occupied,
leaving N empty orbitals.
The filled 1st band touches
the empty 2nd band at the
Fermi energy
Get semi metal
Ch120a-Goddard-L14
2nd band
1st band
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62
Graphite
Stack graphene layers as ABABAB
Can also get ABCABC Rhombohedral
AAAA stacking much higher in energy
Distance between layers = 3.3545A
CC bond = 1.421
Only weak London dispersion
attraction between layers
De = 1.0 kcal/mol C
Easy to slide layers, good lubricant
Graphite: D0K=169.6 kcal/mol, in plane bond = 168.6
Thus average in-plane bond = (2/3)168.6 = 112.4 kcal/mol
112.4 = sp2 s + 1/3 p
Diamond: average CCs = 85 kcal/mol  p = 3*27=81 kcal/mol
Ch120a-Goddard-L14
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63
energetics
Ch120a-Goddard-L14
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64
Allyl Radical
Ch120a-Goddard-L14
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65
Allyl wavefunctions
It is about 12 kcal/mol
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© copyright 2011 William A. Goddard III, all rights reserved
66
stop
Ch120a-Goddard-L14
© copyright 2011 William A. Goddard III, all rights reserved
67