chapter 5 – organic molecules: electronic structure and...
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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)
∑ ϕ=i
iikk cψ
and use this ansatz in the Schrödinger equation:
0=− EψψH
and obtain
( ) 0=∑ ϕ−i
ikik EHc
Multiplication by jϕ from left and integration:
( ) ( ) 0=∑ −=∑ −=∑ −i
jikjiiki
kiki
kik SEHcijEiHjciEHjc ˆˆ
with the Hamilton matrix iHjH jiˆ=
and the overlap matrix ijS ji = .
Secular equations:
( ) 0=∑ −i
jikjiik SEHc or ( ) 0=− kk cSEH
Solution: Secular determinant: 0=− SEH ⇒ eigenvectors kc , eigenvalues kE .
(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:
==
else ; 0neighborsnext ji,for ;
jifor ; βα
H ji
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 2 -
=
=else ; 0
ji ; 1jiS
(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 ∑i
il basis functions jϕ which can be transformed to a i sets of il functions,
i.e. ijlk i
χ ....
11
== is the k-th function belonging to irrep j. As '' jjS jj ≠= for 0 and '' jjH jj ≠= for 0 ,
the ''
jjkkH and '
'jj
kkS matrices are blockdiagonal:
0
00
00
00
00
00
00
22221
221
2211
11111
111
1111
22221
221
2211
11111
111
1111
111
1
111
1
111
1
111
1
=
−
cSS
SSSS
SS
EHH
HHHH
HH
lll
l
lll
l
lll
l
lll
l
O
L
MOM
K
L
MOM
K
O
L
MOM
K
L
MOM
K
,,
,,
,,
,,
,,
,,
,,
,,
,,
,,
,,
,,
,,
,,
,,
,,
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-1AB, 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, Cn
2, ..., Cnn = E
j-th irreducible representation: ( ) jmmn
j εC =Γ with nπi
eε2
=
For the benzene case we obtain the character table:
C6 C6 C62 C6
3 C64 C6
5 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 C6
5
Α Γ6 1 1 1 1 1 1
Β Γ3 1 −1 1 −1 1 −1
Γ1 1 ε −ε∗ −1 −ε ε∗
Ε1 Γ5 1 ε∗ −ε −1 −ε∗ ε
Γ2 1 −ε∗ −ε −1 −ε∗ −ε
Ε2 Γ4 1 −ε −ε∗ −1 −ε −ε∗
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 / C3
2
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 1ϕ yields automatically the orthogonal set of
SALCs. After normalization:
6Γχ = 1/√6( +1 1ϕ +1 2ϕ +1 3ϕ +1 4ϕ +1 5ϕ +1 6ϕ )
3Γχ =1/√6( +1 1ϕ -1 2ϕ +1 3ϕ -1 4ϕ +1 5ϕ -1 6ϕ )
1Γχ =1/√6( +1 1ϕ +ε 2ϕ −ε∗3ϕ -1 4ϕ −ε 5ϕ +ε∗
6ϕ )
5Γχ =1/√6( +1 1ϕ +ε∗2ϕ −ε 3ϕ -1 4ϕ −ε∗
5ϕ +ε 6ϕ )
2Γχ =1/√6( +1 1ϕ −ε∗2ϕ −ε 3ϕ -1 4ϕ −ε∗
5ϕ −ε 6ϕ )
4Γχ =1/√6( +1 1ϕ −ε 2ϕ −ε∗3ϕ -1 4ϕ −ε 5ϕ −ε∗
6ϕ )
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):
6ΓΑ = χχ ; 3ΓΒ = χχ ; 51 ΓΓΕ1,1 += χχχ ; ( )5Γ1ΓΕ1,2 −= χχiχ ; 4Γ2ΓΕ2,1 += χχχ ; ( )42 ΓΓΕ2,2 −= χχiχ
Αχ = 1/√6( +1 1ϕ +1 2ϕ +1 3ϕ +1 4ϕ +1 5ϕ +1 6ϕ )
Βχ = 1/√6( +1 1ϕ -1 2ϕ +1 3ϕ -1 4ϕ +1 5ϕ -1 6ϕ )
Ε1,1χ = 1/√12( +2 1ϕ +1 2ϕ −1 3ϕ -2 4ϕ −1 5ϕ +1 6ϕ )
Ε1,2χ = 1/2( +1 2ϕ +1 3ϕ -1 4ϕ -1 5ϕ )
Ε2,1χ = 1/√12( +2 1ϕ −1 2ϕ -1 3ϕ +2 4ϕ -1 5ϕ -1 6ϕ )
Ε2,2χ = 1/2( +1 2ϕ −1 3ϕ +1 4ϕ −1 5ϕ )
(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:
∑=Ηi
ihˆ
yielding wave functions which are the product of the single electron wave functions:
iiϕΠ=Ψ .
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 -
( )( ) ( )
( )n
N
N
NN
N ϕϕ=ϕ
ϕϕ=Ψ KMN
K
1
1 11121!
,...,,
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 multi-
occupation 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: { { { { {
444 3444 {rg
g
u
g
u
g
g
g
g
g
u
AAa
Ab
Ab
Ab
Ab
1
1
2
1
21
1
23
1
22
1
21 1 2 1 1 1
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
221
23
22
210 12111 uuggu abbbb=Ψ Symmetry: A1g
guuggu babbbb 22
123
22
211 212111=Ψ B2u
guuggu babbbb 32
123
22
212 212111=Ψ B3u
guuggu babbbb 22
123
22
213 212111=Ψ B3u
guuggu babbbb 32
123
22
214 212111=Ψ 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 = B2u
b3g 1 -1 -1 1 1 -1 -1 1
au×b3g 1 -1 -1 1 -1 1 1 -1 = B3u
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
Dipole activity of transitions:
10 ΨΨ μr : gu
u
u
u
g ABBBB
A 12
1
2
3
1
≠=≠
⊗
⊗ ⇒ el. dipole active, y-polarisation
20 ΨΨ μr : gu
u
u
u
g ABBBB
A 13
1
2
3
1
≠≠=
⊗
⊗ ⇒ el. dipole active, x-polarisation
Chapter 5 – Organic Molecules: Electronic Structure and Chemical Reactions – p. 9 -
30 ΨΨ μr : gu
u
u
u
g ABBBB
A 13
1
2
3
1
≠≠=
⊗
⊗ ⇒ el. dipole active, x-polarisation
40 ΨΨ μr : gu
u
u
u
g ABBBB
A 12
1
2
3
1
≠=≠
⊗
⊗ ⇒ el. dipole active, y-polarisation
5.6 Symmetry and configuration interaction
We consider a set of many electron states iΨ . 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
∑ Ψ=Ψi
iikk a
which again leads to a set of secular equations
( ) 0=∑ −i
jikjiik SEHa or ( ) 0=− kk aSEH .
Here, jiH and jiS are matrix elements over the many electron functions ji ,Ψ . Again 0≠jiH
for ji Γ=Γ and 0=jiH for ji Γ≠Γ , 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:
22
20 ugba=Ψ Symmetry: A1g
13
12
21 uug bba=Ψ B1g
23
22 ugba=Ψ 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.