Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 1 5. Organic Molecules: Electronic Structure and Chemical Reactions
5.1 Brief reminder: LCAO approximation
We describe the molecular wave functions as a Linear Combination of Atomic Orbitals
(LCAO)
ψ k = ∑ cik ϕi
i
and use this ansatz in the Schrödinger equation:
Ĥψ − Eψ = 0
and obtain
∑ cik (H − Ek )ϕi = 0
i
Multiplication by ϕ j from left and integration:
(
)
∑ cik j Hˆ − Ek i = ∑ cik j Hˆ i − Ek j i = ∑ cik (H ji − Ek S ji ) = 0
i
i
i
with the Hamilton matrix H ji = j Hˆ i
and the overlap matrix S ji = j i .
Secular equations:
∑ cik (H ji − Ek S ji ) = 0 or (H − Ek S )ck = 0
i
Solution: Secular determinant: H − E S = 0 ⇒ eigenvectors ck , eigenvalues Ek .
(5.1 Example: secular equation for butadien, C-2p orbitals only).
5.2 Brief: reminder: Hückel molecular orbital model (HMO)
We consider only the π-system of a conjugated hydrocarbon using the approximations:
 α ; for i = j

H ji =  β ; for i, j next neighbors
 0 ; else

Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 2 1 ; i = j
S ji = 
0 ; else
(5.2 Comments to HMO theory, example: butadiene).
5.3 The connection to symmetry
The LCAO ansatz in a basis set of n orbitals leads in general to a couples set of n secular
equations. With increasing number of basis function the numerical effort to solve the system
becomes becomes large quite rapidly. Even if some of the matrix elements of the H or S
matrix are zero, this does not necessarily simplify the problem.
The only way to drastically reduce the (numerical) effort for solving the system is to find a
basis transformation which blockdiagonalizes the H, S matrizes. A basis of symmetry adapted
fulfils this requirement:
We assume a set of ∑ li basis functions ϕ j which can be transformed to a i sets of li functions,
i
i.e. χ kj==11....li is the k-th function belonging to irrep j. As S jj' = 0 for j ≠ j' and H jj' = 0 for j ≠ j' ,
i
the H kkjj'' and S kkjj'' matrices are blockdiagonal:
  H11,,11 K H11,,l1

 M O
M
  1,1
1 ,1
  H l ,1 L H l ,l


0



0


 

1
1

 S11,,11 K S11,,l1


0
 M O M

 S 1 ,1 L S 1 ,1
l ,l

 l ,1



0 − E
0






O
0








1
0
1 1
1
H12,1,2 K H12,l, 2
M
O
M
2 ,2
H l ,1 L H l2,,l2
0
1
1
1 1
0
1 1
S12,1, 2 K S12,l,2
M O M
Sl2,,12 L Sl2,,l2
0
1
1
1 1


0 




0  c = 0


O 


 

Recipe:
1. The AO basis is a n dimensional reducible representation of the symmetry group.
Determine the number and symmetry type of contained irreps (section 3.13).
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 3 2. Construction of a set of orthonormal SALC using projection operators (section 4.6).
3. Reformulate secular equation in SALC basis.
(5.3 Example: butadiene).
5.4 Cyclic electron systems and cyclic groups
As an example we investigate the electronic structure of the π-system of benzene:
The symmetry group of the molecules is
D6h:
E
2C6
2C3
C2
3C2’
3C2’’
i
2S3
2S6
σh
3σd
3σvT
The characters can be determined by applying the operations and counting the number of
unaffected (+1) and inverted (-1) AOs (only those AOs which are moved in space contribute
to the character):
Γπ :
6
0
0
0
-2
0
0
0
0
-6
2
0
The representation can be analysed according to section 3.13, yielding:
Γπ = A2u + B2g + E1g + E2u
Now the SALCs can be constructed applying the projection operators. However, there is a
simpler procedure using the properties of cyclic groups.
We do not necessarily have to take into account all symmetry elements of the group, but can
choose any subgroup as well (accepting only partial symmetry adaptation if the choice of
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 4 subgroup is bad). A good choice for cyclic electron systems is the corresponding pure
rotational group (C6 in this case).
The uniaxial rotational groups are cyclic groups (section 3.4). All cyclic groups are Abelian
(section 3.4). In an Abelian group each elements belongs to its own class (AB=BA,
A=B-1 AB, i.e. all elements are only self-conjugated). Therefore, the order of the group is
equal to the number of classes and all representations must be one-dimensional (section 3.12).
There is a general formulation for the j-th irrep of a cyclic group:
Cyclic group Cn: Cn1, Cn2, ..., Cnn = E
j-th irreducible representation: Γ j (Cnm ) = ε jm with ε = e n
2 πi
For the benzene case we obtain the character table:
C6
C6
C62
C63
C64
C65
E=C66
Γ1
ε1
ε2
ε3
ε4
ε5
ε6
Γ2
ε2
ε4
ε6
ε8
ε10
ε12
Γ3
ε3
ε6
ε9
ε12
ε15
ε18
Γ4
ε4
ε8
ε12
ε16
ε20
ε24
Γ5
ε5
ε10
ε15
ε20
ε25
ε30
Γ6
ε6
ε12
ε18
ε24
ε30
ε36
It can be shown that the representations fulfil the orthogonality requirements in section 3.12.
The character table can be rearranged as follows:
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 5 C6
E
C6
C3
C2
C32
C65
Α
Γ6
1
1
1
1
1
1
Β
Γ3
1
−1
1
−1
1
−1
Γ1
1
ε
−ε∗
−1
−ε
ε∗
Γ5
1
ε∗
−ε
−1
−ε∗
ε
Γ2
1
−ε∗
−ε
−1
−ε∗
−ε
Γ4
1
−ε
−ε∗
−1
−ε
−ε∗
Ε1
Ε2
There are 6 1-dimensional irreps. Γ1 and Γ5 (Γ2 and Γ4) can be combined to form two
dimensional reducible representations with non-complex characters. If some additional
symmetry is introduced such as e.g. mirror planes, which makes C6 / C65 and C3 / C32
conjugate, Γ1 / Γ5 and Γ2 / Γ4 merge to two 2-dimensional irreps E1 and E2.
An analysis of the benzene AO’s in terms of the group C6 yields:
Γπ = Γ6 + Γ3 + Γ1 + Γ5 + Γ2 + Γ4
Application of the projection operators of C6 to ϕ1 yields automatically the orthogonal set of
SALCs. After normalization:
χ Γ 6 = 1/√6(
+1 ϕ1 +1 ϕ2 +1 ϕ3 +1 ϕ4 +1 ϕ5 +1 ϕ6 )
χ Γ 3 =1/√6(
+1 ϕ1 -1 ϕ2
χ Γ1 =1/√6(
+1 ϕ1 +ε ϕ2 −ε∗ ϕ3 -1 ϕ4
−ε ϕ5 +ε∗ ϕ6 )
χ Γ 5 =1/√6(
+1 ϕ1 +ε∗ ϕ2 −ε ϕ3 -1 ϕ4
−ε∗ ϕ5 +ε ϕ6 )
χ Γ 2 =1/√6(
+1 ϕ1 −ε∗ ϕ2 −ε ϕ3 -1 ϕ4
−ε∗ ϕ5 −ε ϕ6 )
χ Γ 4 =1/√6(
+1 ϕ1 −ε ϕ2 −ε∗ ϕ3 -1 ϕ4
−ε ϕ5 −ε∗ ϕ6 )
+1 ϕ3 -1 ϕ4
+1 ϕ5 -1 ϕ6
)
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 6 If real coefficients are preferred, one may choose suitable linear combinations (although this
is not necessary in principal):
χ Α = χ Γ 6 ; χ Β = χ Γ 3 ; χ Ε1,1 = χ Γ1 + χ Γ 5 ; χ Ε1,2 = i( χ Γ1 − χ Γ 5 ) ; χ Ε2,1 = χ Γ 2 + χ Γ 4 ; χ Ε2,2 = i( χ Γ 2 − χ Γ 4 )
χΑ =
1/√6( +1 ϕ1 +1 ϕ2 +1 ϕ3 +1 ϕ4 +1 ϕ5 +1 ϕ6 )
χΒ =
1/√6( +1 ϕ1 -1 ϕ2
+1 ϕ3 -1 ϕ4
χ Ε1,1 = 1/√12( +2 ϕ1 +1 ϕ2 −1 ϕ3 -2 ϕ4
χ Ε1,2 = 1/2(
+1 ϕ2 +1 ϕ3 -1 ϕ4
χ Ε2,1 = 1/√12( +2 ϕ1 −1 ϕ2 -1 ϕ3
χ Ε2,2 = 1/2(
+1 ϕ5 -1 ϕ6
)
−1 ϕ5 +1 ϕ6 )
-1 ϕ5
+2 ϕ4 -1 ϕ5
)
-1 ϕ6
+1 ϕ2 −1 ϕ3 +1 ϕ4 −1 ϕ5
)
)
(5.4 Energy levels of benzene in HMO model)
(5.5 Energy levels of cyclic unsaturated hydrocarbons in HMO model, geometric construction)
(5.6 Example: naphthalene, C60)
5.5 Symmetry of many electron states
How can we derive the symmetry of many electron states from the symmetry of the single
electron wave functions discussed so far? In the simplest approximation we assume that the
Hamiltonian is the sum of single electron Hamiltonians:
ˆ = ∑ ĥ
Η
i
i
yielding wave functions which are the product of the single electron wave functions:
Ψ = Πi ϕi .
The permutation symmetry for fermions cab be taken into account by generating the product
wavefunction of N electrons in N spin orbitals in the form of the Slater determinant:
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 7 -
Ψ (1,2 ,..., N ) =
1
N!
ϕ1 (1) K
N
ϕ N (1)
M
ϕN (N )
= ϕ1 K ϕ n
We consider the spatial symmetry of the many electron wave function. In general the
symmetry is obtained as the direct product of the irreps of the participating one electron
functions. The may however arise limitations from the Pauli principle, excluding multioccupation of spin orbitals. This will be further discussed in chapter 6.
Example: Naphtalene ground state
3b3g
2au
3b1u
2b3g
2b2g
1au
||
2b1u
||
1b3g
||
1b2g
||
1b1u
||
ground state: 1b12u 1b22g 1b32g 2b12u 1au2
{{{{{
A1g4A4
1 g A1 g A1 g A1 g
1
42444
3
A1g
Important: A fully occupied set of space orbitals belonging to one arbitrary irrep of the group
belongs to the totally symmetric irrep (easy to see for 1-dim. irreps., more complicated to
proof for more-dimensional irreps.)
5.5 Symmetry of excited states and dipole selection rules
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 8 Example: Lowest excited states in naphthalene
Ψ0 = 1b12u 1b22g 1b32g 2b12u 1au2
Symmetry:
A1g
Ψ1 = 1b12u 1b22g 1b32g 2b12u 1au 2b2 g
B2u
Ψ2 = 1b12u 1b22g 1b32g 2b12u 1au 2b3 g
B3u
Ψ3 = 1b12u 1b22g 1b32g 2b1u 1au2 2b2 g
B3u
Ψ4 = 1b12u 1b22g 1b32g 2b1u 1au2 2b3 g
B2u
D2h
E
C2x
C2y
C2z
i
σxy
σxz
σyz
au
1
1
1
1
-1
-1
-1
-1
b2g
1
-1
1
-1
1
-1
1
-1
au×b2g
1
-1
1
-1
-1
1
-1
1
b3g
1
-1
-1
1
1
-1
-1
1
au×b3g
1
-1
-1
1
-1
1
1
-1
b1u
1
1
-1
-1
-1
-1
1
1
b1u×b2g
1
-1
-1
1
-1
1
1
-1
= B3u
b1u×b3g
1
-1
1
-1
-1
1
-1
1
= B2u
= B2u
= B3u
Dipole activity of transitions:
r
Ψ0 μ Ψ1 :
 B3u 
≠ 
 
 
A1 g ⊗ B2 u  ⊗ B2 u =  A1 g ⇒ el. dipole active, y-polarisation
B 
≠ 
 1u 
 
r
Ψ0 μ Ψ2 :
 B3 u 
= 
 
 
A1 g ⊗ B2 u  ⊗ B3u ≠  A1 g ⇒ el. dipole active, x-polarisation
B 
≠ 
 1u 
 
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 9 -
r
Ψ0 μ Ψ3 :
 B3 u 
= 
 
 
A1 g ⊗ B2 u  ⊗ B3u ≠  A1 g ⇒ el. dipole active, x-polarisation
B 
≠ 
 1u 
 
r
Ψ0 μ Ψ4 :
 B3u 
≠ 
 
 
A1 g ⊗ B2 u  ⊗ B2 u =  A1 g ⇒ el. dipole active, y-polarisation
B 
≠ 
 1u 
 
5.6 Symmetry and configuration interaction
We consider a set of many electron states Ψi . Beyond the simple one electron picture
considered so far, there might be electron correlations, which lead to deviations from these
states. In order to describe this effect, we start with an configuration interaction (CI) ansatz,
describing the many electron wave function as a linear combination of the many electron
wave functions over the ground and excited states
Ψk = ∑ aik Ψi
i
which again leads to a set of secular equations
∑ aik (H ji − Ek S ji ) = 0 or (H − Ek S )a k = 0 .
i
Here, H ji and S ji are matrix elements over the many electron functions Ψi , j . Again H ji ≠ 0
for Γi = Γj and H ji = 0 for Γi ≠ Γj , i.e. a configuration interaction (mixing of states) is only
expected for states which belong to the same irrep of the group.
(5.7 Example: CI for excited states in naphthalene)
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 10 5.7 Symmetry and chemical reactions
A well known example are the “Woodward-Hoffmann” for cyclisation reactions. We consider
processes
•
for which the considered step is rate-determining
•
for which symmetry elements are preserved during the reaction
Example: [2+2] cycloaddition
1. Determination of preserved symmetry elements:
The symmetry group D2h. For the determination of the irreps it is sufficient to consider
a set of generating elements of the group, e.g. σ, σ’, σ’’.
2. We identify the orbital basis involved in the reaction:
p1 – p4 in case of 2 C2H4,
sp1 – sp4 (hybrid) in case of C4H8
3. Classification of SALCs according to (1) symmetry and (2) energy
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 11 -
“Orbital correlation diagram”
Construction of ground state and lowest excited states:
Ψ0 = a g2 b22u
Symmetry:
A1g
Ψ1 = a g2 b21u b31u
B1g
Ψ2 = a g2 b32u
Ag
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 12 High activation barrier for [2+2] cycloaddition for molecules in ground state:
thermal activation forbidded.
Low activation barrier for [2+2] cycloaddition for molecules in 1st excited state:
photochemical activation allowed.
Similar models for other cycloaddition and cyclisation reactions.