Molecular Perturbations
Within the frame of LCAO, we search to express an
unknown system from an known system.
The unknown system are in general real systems:
molecule or supermolecule.
Known systems may be real systems, simple
molecules, fragments of real systems or partners
within a reaction (fragments of a supermolecule:
reactants).
1
Molecular Perturbations
i= Sr crfr and Ei are solutions for a known system.
We like to find solutions 'i= Sr c'rfr and E'i for an
unknown system that is resembling (whose
Hamiltonian, H', does not differ by much from H). If
so, solutions (E'i,y'i) are close to (Ei,yi) and P=H’-H is
an operator whose terms Pij=[<yi|P|yj>]* are weak. P
is the operator defining the perturbation.
*data usually characterize atoms pij=<fi|P|fj> which are few
terms but no so small (a bond formation between fi and fj).
Pij=<yi|P|yj> are derived from the pij. Pij are significantly
weaker than pij due to delocalization of AOs within MO
expressions. In general, results remain significant even though
some pij are important.
2
Molecular Perturbations
'I = S ai I = Sr crfr unknown MOs, 'i, are linear
combinations of known MOs, i, that are linear
combinations of AOs, fi; they are hence linear
combinations of AOs, i.
If we were to find exact solutions 'i= Sr c'rfr it we be
easier to solve the secular determinant directly.
The advantage of using perturbations is to make it
within approximations but simply.
3
Molecular Perturbations
“Simply” means not rigorously that is somehow negative.
However
° understanding (as opposed to modeling) is always a
work of simplification.
° perturbation is a direct comparison. We search for the
difference between 2 large numbers (total energies)
with a large error bar each. The difference calculating
them independently and making the difference may be
large. A direct estimation of the difference even using
simple theory may be more precise.
4
Molecular Perturbations
Secular determinant without perturbation:
If known, the secular equation is solved (diagonal)!
E1-E
0
0
0
E2-E
0
=0
0
0
Ej-E
5
Molecular Perturbations
Secular determinant with perturbation:
E1+P11-E
P12
P1j
P21
E2+P22-E
P2j
=0
Pj1
Pj2
Ej+Pjj-E
We search to develop the determinant, retaining only the largest terms.
One term is larger than any other: the diagonal.
6
First order Energy solution
Since perturbations are small, all the Pij are small and the largest
terms are those from the diagonal:
Searching for E’j close to Ej, every term of the diagonal is large
except one: E’i - Ej= Ei - Ej which is not small (assuming no
degeneracy).
Only one term is small; that for j: E’i - Ej= Pjj is a “first order” small
term.
Expressing the determinant, the product of the terms of the
diagonal is contains the smallest number of small terms and is
the greatest.
The secular determinant is
P (E’i+Pii-E) = 0 → E’j = Ej+ Pjj
7
Orbital associated with the First order Energy solution (Zero
order)
Searching for
y’ close to y
j
j
Secular equation: Pj1 a1 + Pj2 a1 + …+ (Ej+Pjj-E) aj + …+ Pjn an = 0
All the terms are small except one.
All the ai are zero except aj. aj=1
The orbital is unperturbed to this order
y’ =y
j
j
8
Second order Energy solution
Developing the determinant is:
the diagonal
- all the terms obtained by a single permutation of two terms in the diagonal
+ negligible terms.
Searching for E’j close to Ej, The secular determinant is
P(E’i+Pii-E) -
S[(E’ +P -E) (E’ +P -E)]
i
(P (E’i+Pii-E)P*ijPji
i
ii
j
+e=0
jj
Secular determinant without perturbation:
E1+P11-E
P12
P1j
P21
E2+P22-E
P2j
=0
Pj1
Pj2
Ej+Pjj-E
9
Second order Energy solution
P(E’i+Pii-E)
-
S [(E’ +P -E) (E’ +P -E)]
i
(P (E’i+Pii-E)P*ijPji
i
P(E’i+Pii-E)
[1 S
–
i
ii
j
S
E – Ej – Pjj = -
i
S
]
[(E’ +P -E) (E’ +P -E)]
ii
P*ijPji
(E’i+Pii-E)
i
jj
P*ijPji
i
-(E’j+Pjj-E)= -
+e=0
j
=0
jj
]
P*ijPji
(Ei-Ej)
10
Second order Energy solution
E = Ej + Pjj +
S
E’j = Ej + Pjj +
For real terms
P*ijPji
i
S
(Ej-Ei)
i
P*ijPji
(Ej-Ei)
Dimension of an energy:
Square of an energy over
an energy
E’j = Ej + Pjj +Si Pij2/ DE
Second order term (2 orbital mixing)
If Ej< Ei stabilization of the orbital of lower energy
If Ej>Ei destabilization of the orbital of higher energy
11
Increases the energy gap
Orbital associated with the second order Energy solution (first
order)
Secular equation: Pi1 a1 + Pi2 a2 + …+ (Ei+Pii-E) ai+ Pij aj + …+ Pjn
an = 0
Product of 2
small terms
not small
terms if ij
not small
terms if ij
Second order
First order
First order
no dimension
Searching a solution close to j, the ith line has small terms of
the first order. ai = Pij/DE aj
The orbital is
y’ = y S
j
j
+
ij
Pij/(Ei-Ej) y i
Such expression is not orbital is not normalized and has to be
multiplied by 1/√N
In phase for the lowest combination
12
Combination of 2 AOs of same symmetry
Ei’-Ei = Pij2/ DE
y’ = y S
j
j
+
ij
Pij/DE y i
Two orbital mixing: a bonding combination and an antibonding
one.
The bonding orbital
is the in-phase combination
Antibonding
Niveau Antiliant
is issued from the orbital of lowest energy
A
(larger coefficient of mixing)
B
A
The antibonding orbital
is issued from to the orbital of highest energy
The interaction widens the gap;
It is more important when S is large and DE small
• 10 eV rule
• Frontier orbitals
B
A
B
Niveau
liant
Bonding
13
2 OA interaction
degenerate or
close in energy
-D/2
D/2
b
f1
f2
f1
D/2 -E
b
f2
b
-D/2 -E
=0
(D/2 -E) (-D/2 -E) = b2  E2 - (D2 /4) = b2
E2 = b2 + (D2 /4)
E = ±√[b2 + (D2 /4)]
general
• If D = 0, E+ - E = b
√ (b2+D2/4)
The geometric mean
of b and D/2
• If b <<D,
E+ = (D/2) (1 + 4 b2/ D2) 0.5
E+ = (D/2) (1 + 2 b2/ D2) = D/2 (1 + 2 b2/ D)
E+ - E = b2/ D
 2nd order Perturbation term
14
Frontier Orbitals
• due to perturbation
formula: Ei’-Ei = Pij2/ DE
• SOMO, HOMO and
LUMO
• The least stable but the
most mobile electrons
• The largest amplitude
on the weakly bound
atoms that are reactive
sites.
Kenichi Fukui
Japan
Nobel 1981
15
3rd order perturbation
Case of two fragments interacting.
The perturbation is the interaction between fragments
(without perturbation within each fragment)
Permutation between 3 lines of the determinant:
’1 = ’1 +
P1b
E1-Eb
b +
P1b Pb2
(E1-Eb) (E1-E2)
2
2
1
A Fragment :
The orbital 1 and 2 belong to
the same fragment and are
orthogonal inside A
b
B Fragment : A single orbital
1 and 2 both interact with b; they mix, the phase relation being
imposed by their relation with B. This causes polarization
16
S. Imagaki, H. Fujimoto K. Fukui J. Amer. Chem. Soc 98 (1976) 4054 L. Libit R. Hoffmann J. Amer. Chem. Soc 96 (1974) 1370
rd
3
order perturbation
another view point:excitation
’1 = ’1 +
P1b
E1-Eb
b +
P1b Pb2
(E1-Eb) (E1-E2)
2
2
1
A Fragment :
The orbital 1 and 2 belong to
the same fragment and are
orthogonal inside A
b
B Fragment : A single orbital
1 and 2 mix. 1 is partially depopulated and 2 becomes partially
populated. This corresponds to mix with an excited state: IC
17
S. Imagaki, H. Fujimoto K. Fukui J. Amer. Chem. Soc 98 (1976) 4054 L. Libit R. Hoffmann J. Amer. Chem. Soc 96 (1974) 1370
3rd order perturbation
’1 = 1 + P1b/(E1-Eb) b
At variance with 1, ’1 is not orthogonal to 2
’’1 = ’1 + P1’2/(E1-E2) 2
Let express P1’2 by developing ’1
P1’2 = < ’1 IPI 2 > = < 1 IPI 2 > + P1b/(E1-Eb) < b IPI 2 >
P1’2 =
0
+
P1b Pb2/(E1-Eb)
’’1 = ’1 + P1b Pb2/ [(E1-Eb)(E1-E2)] 2
’’1 = 1 + P1b/(E1-Eb) b + P1b Pb2/ [(E1-Eb)(E1-E2)] 2
2
1
b
B Fragment :
A Fragment :
The third order terms account for the polarization of the fragment
18
Eb below E1
y2
-
+
y1
-
1
+
1
2
b
y' 1
b
y' 2
Increases decreases
’1 = ’1
2
-
b +
2
Decreases increases
’2 = ’1
-
b + 2
19
Eb intermediate between E1 and E2
y2
+
y1
-
+
+
1
+
1
2
y' 1
b
y' 2
Decreases increases
’1 = ’1
2
b
+
b +
Decreases increases
2
’2 = ’1
-
b+2
20
Eb above E1
+
+
y2
-
+
y1
-
+
1
2
1
b
y' 1
b
y' 2
Decreases increases
’1 = ’1
2
+
b +
Increases decreases
2
’2 = ’1
-
b-2
21
Examples
To judge the method, we will search for known systems (easily calculated
without perturbation)
• Cyclobutadiene from butadiene
22
Cyclobutadiene from Butadiene
-.3717
-.6015
-1.618
0.3717
0.6015
0.6015
C2
C4
C1
C2
C1
C3
-.3717
0.6015
C4
-.3717
-.618 C2
-.6015
0.3717
0.618
-.3717
0.6015
0.3717
0.6015
1.618
0.6015
0.3717
23
MO Energies from perturbation
What we know is <f1IPIf4> = 1: there is
a bond between atoms 1 and 4 that did
not exist for butadiene
A Mirror symmetry is preserved
S
S
A
A
Only orbitals of the same symmetry mix together
24
Conservation of Orbital
Symmetry
H C Longuet-Higgins E W Abrahamson
The Molecular orbitals are solution of the symmetry operators of the molecule.
MOs from different symmetry groups do not mix.
Hugh Christopher Longuet-Higgins
1923-2004
25
Orbitals ’1 from perturbation
26
New lowest orbital
’1= (fA + fB+ fC+ fD ) + < 1 IPI 3>/(E1-E3) (fA -fB-fC+ fD )
+ 0.2
The + sign means in phase
between 1 and 4
The new orbital looks like 1
but with modulation of
amplitude indicated by red
arrows
27
Exact solutions First set
1/2
-2 b
1/2
1/2
0
1/2
2b
28
Exact solutions Second set
1/2
-2b
1/√2
1/√2
0
1/2
2b
29
Solutions are very accurate even though
<fiIPIfj> is large (introducing a bond as
strong as others). < iIPIj> MOs are less
large due to delocalization of AOs with
MOs.
30
Butadiene from 2 ethylenes
1/√2 (fA - fB)
1/√2 (fA +fB)
1/√2 (fC - fD)
1/√2 (fC +fD)
31
Butadiene from 2 ethylenes
first order terms
-1.5 b
1/√2 (fA - fB)
1/√2 (fC - fD)
-0.5 b
0.5 b
1/√2 (fA +fB)
1.5 b
1/√2 (fC +fD)
< 1/√2 (fA +fB)IPI1/√2 (fc +fD)> = 0.5 b
32
Butadiene from 2 ethylenes
second order terms
-1.5 b - 0.125 b = -1.625 b
1/√2 (fA - fB)
1/√2 (fC - fD)
-0.5 b- 0.125 b = -0.625 b
0.5 b+ 0.125 b= 0.625 b
1/√2 (fA +fB)
1.5 b + 0.125 b = 1.625 b 1/√2 (fC +fD)
[< 1/√2 (fA +fB)IPI1/√2 (fc +fD) >]2 / (b-(-b))= 0.125 b
33
P has to be calculated on the unperturbed MO expressions (before taking into account
1st order perturbation)
New lowest orbital
’1= (fA + fB+ fC+ fD ) + < 1 IPI 3>/(E1-E3) (fA -fB-fC+ fD )
+ 0.5/2.0
The + sign means in phase
between 2 and 3
1.25 0.75 0.75 1.25 without normalization
0.606 0.364 0364 0.606 normalized
34
The total energy stabilization comes only from
second order terms
(interaction between occupied and vacant orbitals)
E = 4 * 0.125 = 0.5 b
This is the energy due to the delocalization.
35
Butadiene from 2 ethylenes
minimizing the repulsion
0.5 b
1/√2 (fA +fB)
1/√2 (fC +fD)
1.5 b
It is better to reduce the overlap: not making bond between atoms 1 and 4 !
Making trans rather than cis butadiene
Avoiding closing ring to cyclobutadiene
2
1
3
4
1
2
3
4
36
Cyclobutadiene from 2 ethylenes
Only one mirror symmetry is preserved by perturbation
37
cyclobutadiene from 2 ethylenes
first order terms
-2 b
1/√2 (fA - fB)
1/√2 (fC - fD)
-0.0 b
0.0 b
1/√2 (fA +fB)
1/√2 (fC +fD)
2b
<1/√2 (fA +fB)IPI1/√2 (fc +fD)> = b
No terms from second order
38
Conservation of Orbital
Symmetry
H C Longuet-Higgins E W Abrahamson
The Molecular orbitals are solution of the symmetry operators of the molecule.
MOs from different symmetry groups do not mix.
Hugh Christopher Longuet-Higgins
1923-2004
39
Non-crossing rule
Energy
The potential energy curves of two MOs do
not cross unless they have different
symmetry.
If the MOs are of the
same symmetry, they
interact. The
interaction increases
when the energy
levels are close.
This opens a gap and
prevents crossing.
Reaction coordinate
or any transformation
40
Non-crossing rule
The potential energy curves of two
electronic states of a diatomic molecule do
not cross unless the states have different
symmetry.
Energy of states
If 2 states are of the
same symmetry, they
interact. The
interaction increases
when the energy
levels are close.
This opens a gap and
prevents crossing.
distance
41
Non-crossing rule
ionic
Atomic or covalent
ionic
atomic
Mixed ionic and
covalent
42
Benzene from pentadienyl + C
radicals
pentadienyl radical without calculations
1. What are the coefficients of the SOMO?
2. What are the coefficients of the bonding
MO antisymmetric relative to the mirror?
3. What is the corresponding Energy level?
4. What are the coefficients of the bonding
MO symmetric relative to the mirror?
5. Deduce from them the MO Energy level?
43
pentadienyl radical using only symmetry and
Normalization.
E = -√3 b
What are the coefficients of the SOMO?
What are the coefficients of the bonding MO
antisymmetric relative to the mirror?
What is the corresponding Energy level?
What are the coefficients of the bonding MO
symmetric relative to the mirror?
E = -b
E=0
Deduce from them the MO Energy level?
E=b
E = √3 b
44
pentadienyl radical using only symmetry and
Normalization.
E = -√3 b
What are the coefficients of the SOMO?
1/√12
Use alternant property
-1/2
What are the coefficients of the bonding MO -1/2
antisymmetric relative to the mirror?
1/2
1/√3
a double bond, symmetrized
What is the corresponding Energy level?
What are the coefficients of the bonding MO
symmetric relative to the mirror?
1/√12: 1/3 +2 (1/2)2 + 2 c2 =1
1/2: 2 (1/2)2 + 2 c2 + 0 =1
1/√3: 1/3 +2 c2 + 0 =1
Deduce from them the MO Energy level?
Develop <IHI>
4 [(1/√12)(1/2)+(1/2)(1/√3)]= √3
E = -b
1/√3
E=0
1/2
-1/√3
1/2
1/√12
1/2
E=b
E = √3 b 1/√3
45
Benzene from pentadienyl + C radicals
First order term.
E = -√3 b
-1.155 b
1/√12
-1/2
-1/2
1/2
1/√3
E = -b
E=0
1/√3
2/√3 = 1.155 b
E=0
1/2
-1/√3
1/2
1.155 b
1/√12
1/2
E=b
E = √3 b 1/√3
46
Radical chain + C radical atom
comparing the chain with the ring: Aromaticity
First order term.
S
E=0
E=0
4/√(N-1) for the ring
2/√(N-1) for the chain
A
E=0
E=0
0 for the ring
2/√(N-1) for the chain
47
Radical chain + C radical atom
comparing the chain with the ring: Aromaticity
Aromaticity according to Dewar
S
A
When the SOMO is symmetric
The ring is more stable than the chain
The polyene is AROMATIC
N-1 is even
N = 4n +2
When the SOMO is antisymmetric
The ring is less stable than the chain
The polyene is ANTIAROMATIC
N-1 is odd
N = 4n
48
Benzene from pentadienyl + C radicals
Second order terms.
(2/√12)2/√3=0.1924
-1.925 b
E = -√3 b
1/√12
-1/2
-.962 b
1/√3
E=0
0.962 b
1/√12
1/2
1.925 b
E = √3 b 1/√3
49
Benzene from pentadienyl + C radicals
(2/√12)2/√3=0.1924
Second order terms.
-1.925 b
0.1924
-1.155 b
0.1924
E = -√3 b
E = -√3 b
1/√12
-1/2
-1/2
-b
1/2
-0.962 b
1/√3
E = -b
E=0
0.962 b
1/2
b
0.1924
1.155 b
-1/√3
1/2
E = √3 b
0.1924
1.925 b
1/√3
1/√12
1/2
E=b
E = √3 b 1/√3
50
Benzene from butadiene+ethylene
-.3717
-.6015
-1.618
0.3717
0.6015
0.6015
C3
+
-.3717
-.618
0.6015
-.3717
C4
C4
C2
C1
C5
C6
C3
C5
C2
C6
C1
-.6015
0.3717
0.618
-.3717
0.6015
0.3717
0.6015
1.618
0.6015
0.3717
51
Benzene from butadiene+ethylene
Whire arrow:
[2(1/2)0.6015)2/1.618 = 0.447 b
Black arrow:
[2(1/2)0.3717)2/0.618 = 0.447 b
-2.065 b
A
-1.618 b
-0.447 b
-1.065 b
S
-0.618 b
A
0.618 b
S
-0.447 b
0.447 b
1.618 b
-0.447 b
A
-1.0 b
S
1.0 b
0.447 b
-0.447 b
1.065 b
0.447 b
0.447 b
2.065 b
52
Fulvene
0 . 75
-0 . 36
-0 . 19
0 . 66
-0 . 44
-0 . 44
0.15
Here are 4 MOs
Find the missing ones
(coefficients and Energy)
-0 . 35
-0 . 35
0.28
0.15
E = -0.25 b
E = -1.86 b
0 . 25
-0 . 5
Justify the energy 1b
for the orbital at the bottom left
0.52
-0 . 5
Is fulvene more likely a donor 0.0
or an acceptor?
0.0
0.5
0.5
E = 1. b
0.28
0.43
0.43
0.39
0.39
E = 2.12 b
53
The two antisymmetric orbitals are
from a butadiene (no contribution on
the C that belong to the mirror plane)
0.618 b
-1.618 b
54
A non bonding
orbital made of two
pCC
Is fulvene more likely a donor or an acceptor?
The LUMO level -0.25 b is low in
energy
Fulvene is an acceptor (interacting
with an electron rich dienophile such
as : CH2=C(OMe)2
E= b
LUMO: E=-0.25 b
55
Hexatriene
Let call 1= af1 + bf2 + cf3 + c4f4 + c5f5 + c6f6 the lowest orbital
1. Identify a, b and c to the numbers 0.232, 0.418 and 0.521.
2. Express c4, c5 and c6 in terms of a, b and c
3. What relationship connects a, b and c?
4. Give energy E1 of 1 as a function of a, b and c. Calculate E1.
5. The coefficients of the other orbitals are obtained by the following
permutations:
c1
c2
c3
3
c
a
b
2
b
c
a
1
a
b
c
Give c2 et c3 the sign of the coefficients (c1 will be conventionally chosen
as positive).
We give E2=1.247 b et E3=0.445 b.
56
Hexatriene
Let call 1= af1 + bf2 + cf3 + c4f4 + c5f5 + c6f6 the lowest
orbital
1.
a=0.232 b=0.418 c=0.521
2. Express c4, c5 and c6 in terms of a, b and c: c6=c1 c5=c2 and c4=c3
3. What relation link a, b and c? a2+b2+c2 = 0.5
4. Give the energy E1 of 1 as a function of a, b and c. Calculate E1.
4ab+4bc+2c2 = E1 = 1.802 b.
5. The coefficients of the other orbitals are obtained by the following
permutations:
3
2
1
c1
c2
c3
c
a
b
b
c
a
a
b
c
2 c2 and c3 are positive
3 c2 is positive an c3 is negative.
E2=1.247 b and E3=0.445 b.
57
Ring C14 and pyrene
by perturbation from hexadienes
y
A
1'
1
2'
2
3'
3
4'
4
5'
x
5
6'
6
B
pyrène
2 hexatrienes + atoms A and B
58
The antisymmetric combinations
remain unperturbed
-2.
- 1.802
A
S
- 1.247
A
- 0.445
S
A
y
A
1'
A+S
1
2'
2
3'
3
4'
0.445
A
4
5'
6'
S
1.247
x
5
6
B
A
1.802
S
2
HEXATRIENES
Hexatriene
A,B
EN C14
CANNULENE
A,B
14 ring
en pointillé les correspondances entre orbitales
antisymétriq ues par rapport au plan contenant A et B
59
The Symmetric combinations
interact
y
-2.
- 1.802
- 1.247
A
S
1
2'
2
3'
A
3
4'
5
6'
S
x
4
5'
- 0.445
S
A
1'
6
B
A+S
0.445
1.247
1.802
y
A
1'
1
2'
A
2
3'
S
2
3
4'
4
5'
HEXATRIENES
ANNULENE EN C14
A
A
S
A,B
x
5
6'
6
B
en pointillé les correspondances entre orbitales
antisymétriq ues par
rapport au the
plan contenant
degenerate
(except
lowestA et B
Orbitals are
and highest levels); correlation jumps from
one energy level to the next.
60
Interaction with C=C
2 stabilizing interactions between
Frontier Orbitals
HOMO S
y
LUMO
S
y
pyrène
A
1'
1
2'
A
2
3'
3
4'
4
5'
A
x
5
6'
1'
1
2'
2
3'
3
4'
4
5'
6
B
S
x
5
6'
6
B
LUMO
HOMO
61
CH2
H2C
C
Trimethylenemethane
CH2
√(2/3)
1/√3
-1/√6
From symmetry, 4 degenerate orbitals; 2 of them remain non bonding,
The two other mix (3(1/√3)b). This is a diradical
1/√6
-1/√2
-√3 b
1/√3
1/√6
1/√2
√3 b
62
Exercise
•What are Mo’s coefficients energy level of the p orbital
of lowest energy level for the cyclopentadienyl radical?
•Two MOs from cyclopentadienyl are MOs from
butadiene? Which ones?
• Give all the energy levels for the cyclopentadienyl
radical without any calculation.
•
Considering Ag6. Each Ag will be described by a
single valence eorbital, 5s , occupied by a single electron.
Among the 3 below which one is the most stable when
fragments are located far from each other (Ag5 will always
be assumed with 5-fold symmetry). (Ag5 +Ag); (Ag5+ +Ag-)
et (Ag5- +Ag+).
63
Exercise
• Using PMO theory based on Frontier orbitals, tell which
among the 3 structures which is the most stable. (every
Ag-Ag distance will be supposed equal).
6A
6B
6C
•Show (using properties of alternant systems) that this
structure has an energy close to that of an hexagonal
structure.
•Knowing that this structure is more stable guess another
structure encore plus structure even more stable?
64
cyclopentadiene
-1.618
-.618
0.618
1.618
65
C10H8 bicyclic compounds
with a common CC bond
Naphtalene vs. Azulene
(22-8)/2=7 insaturations: 2 rings 5 double bonds
Are these structures stable?
Which one has the lowest energy?
Why azulene is named azulene?
66
Comparison with C10H10 :
a ring closure
1/√5=0.4472
0.3717/√2= 0.2628
√(1/5-0.26282)=0.3618
0.6015/√2= 0.4253
√(1/5-0.42532)=0.1383
67
Naphtalene vs. Azulene
Naphtalene formation is better than azulene
formation
Between 1 and 6 large lobes
Between 1 and 5 small (zero considering the
rule for alternants: 1 and 6 are both
starred atoms)
68
HOMO and LUMO of azulene
A
S
LUMO
HOMO
High HOMO, Low LUMO: this is why it is blue!
69
Polarization of the Frontier orbitals
mixing with the orbital with the closest
HOMO
energy level
On C5
HOMO mixes out of phase with blue
LUMO in phase with blue
On C7
LUMO
70
4n+2 e :Aromaticity C7+ C5-
3 bonding orbitals to accommodate 6e
71
The anion is stable (aromatic)
The cation is not (antiaromaticJahn-Teller situation)
Thorium+4
72
Azulene is not an alternant
hydrocarbon
C7 is positively charged and C5 is
negatively charged.
There is a large dipole moment!
73
H3
1/√2
-1/√2
-1/√6 -1/√6
-b
2b
√(2/3)
2b
1/√3
74
Benzene built from 3 double bonds
75
First Order pertubation: Construction of symmetry orbitals
From
and
One MO generates a set of 3 MOs
-
-
+
+
-
-
+
+
+
-
+
-
+
The interaction is b’= (1/√2)(1/√2) = 0.5 b
76
-2b
b’= -b’’= b/2
-
1/√6
2b’’
-b
1/√3
-b’’
-0.5b
-
+
1/√12+
-1/2
+
-1/√12
-b’
b
2b’
1/√3
-1/√12
0.5b
-1/√12
2b
1/2
-
+
-
-1/2
-1/2
+
+
+
-
1/√6
+
77
Energies: 2nd order perturbation
Each p orbital interacts with two p* orbitals
DE’ = 2 P2/DE
The interaction P is b’= (1/√2)(1/√2) = 0.5 b
DE’ = b/2
b’= b/2
P2/DE = b’2/(+1(-1))= b/4
P has to be calculated on the unperturbed MO expressions
(before taking into account 1st order perturbation)
78
-
-2b
-b/4
-b/4
-b
-
-0.5b
b/4
b/4
b
+
+
S
A
-
+
0.5b
-
-
S
A
+
-
+
+
2b
+
79
Mixing of SS orbitals
After solving 3x3 determinant for degenerate MOs
-b (1/√3)(1/2)=1/√12
-0.5b
(1/√3)(1/2)=1/√12
S
1/2
-1/2
2(1/√12)(1/2)=1/√12
1/√3
0.5b
b
-1/√12
S
-1/√12
P/DE =(3/√12)/[0.5-(-0.5)]= √3
80
Mixing of SS orbitals
After solving 3x3 determinant for degenerate MOs
1/√12
1/√3
1/2
S
-1/2
-1/√12
S
2/√3→1/√3
S
 * - √3 
1/√12 -1/√3→ -1/√12
Before normalization → After normalization
 + √3 *
1/√3
-1/√12
1→1/2
0
1/2
S
0
-1/2
-1→ -1/2
Before normalization → After normalization
81
The perturbation is large
We are building half of the bonds which is
not a small perturbation.
P/DE = √3 /1
Formula are accurate because there is no
other terms beyond 2nd order.
82
Exercice The Bow-tie molecule
Second Mirror
C6
C3
C1
C5
C2
First Mirror
C4
C6H4 molecule is planar with three planes as mirror symmetry, so called
the molecule plan, the first mirror (containing C1 and C2) and the
second mirror (perpendicular to C1-C2) respectively. We call it the bowtie molecule hereafter.
83
Bow-tie
1. Explain what partition of the molecular orbitals the molecule plane performs. (Give the labels
of the molecular orbitals that this mirror symmetry performs)
the s and p separation: s orbitals are symmetric and p orbitals are antisymmetric.
2. Write down the secular determinant for each symmetry group of the p orbitals by considering
the two other mirror symmetries (“first” and “second “ mirror as indicated on the figure
above).
SS
C1
C6
SA
C1
C6
C1
-x+1
2
C1
-x-1
2
C6
1
-x+1
C6
1
-x+1
AA
C6
AS
C6
C6
-x-1
C6
-x-1
3. Two orbitals are pure symmetry orbitals. What are their energies?
x = -1 (unit b)
4. Make a drawing of these two molecular orbitals and give their atomic coefficients.
Coefficients are 0.5.
84
Bow-tie
5. Solve the secular determinant involving orbital symmetric relative to the two mirrors and
give the energy levels.
SS (1-x)2=2 x=1±√2 (2.414 and -0.414)
6. Solve the secular determinant involving orbital antisymmetric relative to the two mirrors
and give the energy levels.
SA (-1-x) (1-x) =2 x=±√3 (1.732 and -1.732)
7. What is the total (p) energy of the molecule?
E= 2 (2.414 + 1.732-0.414)=7.464
8. Show a Molecular diagram
(the energy levels with values in b units and label of symmetry).
85
Bow-tie
II-1. The energy levels for the hexatriene (the linear C6H8 chain) are E1=1.802 b, E2=1.247
b, E3=0.445 b E4=1.802 b, E5=1.247 b and E6=0.445 b. Compare the stability of the bowtie molecule with that of the hexatriene molecule. Is the Bow-tie molecule aromatic?
It is aromatic: Its energy is larger an that of the chain 6.988 b. However 6e does not
correspond to an optimal count since the HOMO is antibonding.
II-2.What would be the ideal charge for the Bow-tie molecule?
Positive (formally 2+ would allow removing the two antibonding electrons but would
induce electrostatic repulsion).
II-3. Remind what are the energy levels and the coefficients of the orbitals of C3H3 (a
single equilateral triangle)
86
Bow-tie
II-4. What would be the ideal electron count for such a ring? What would then be the charge for
the ring?
The ideal number of electrons is 2 (filling the bonding level and not the antibonding one); this
implies a total charge +1.
II-5. We are now building the Bow-tie molecule from two three member rings. Make a schematic
drawing for the energies at the first order considering perturbations.
87
Bow-tie
II-6. Calculate the second order shifts and show the resulting MO diagram
P= (1/√3)(√2/√3) P2/DE=2/27=0.074
88
Bow-tie
III-1. We now search finding the MOs using symmetry orbitals? What are the energy
levels of the symmetry orbitals? Describe them?
These are double bonds whose energy levels are ±1 (the horizontal one and the
combinations of the two vertical ones)
III-2. Use perturbation theory to calculate the MOs from symmetry orbitals of the SS
symmetry.
The interaction P is 4x(1/2)x(1/√2) = √2 = 1.414
The shifts are then ±1.414 for SS (first order term)
The resulting levels are 1±1.414 for SS (first order term)
III-3. Write the secular determinant for the SA orbitals and make the full diagram.
1/√2 (C1-C2)
1/2
(C3+C4+C5+C6)
1/√2 (C1-C2)
-x-1
√2
1/2 (C3+C4+C5+C6)
√2
-x+1
SA
89
Bow-tie
III-3. make the full diagram.
III-4. Why second order perburbation theory fails for the SA interaction?
Because P=√2 is not small relative to DE=1
The application of second order theory for SA would lead to 2 and -2 instead of 1.732 and -1.732
90