Lecture 32 : The Hückel Method
The material in this lecture covers the following in Atkins.
14.0 The Hückel approximation (a)The secular determinant (b) Ethene and frontier orbitals (c) Butadiene and p-electron binding energy (d) benzene and aromatic stability
Lecture on-line Huckel Theory (PowerPoint) Huckel Theory (PDF)
Handout for this lecture
The Hückel method
The Hückel approximationcan be used for conjugatedmolecules in which there is an alternation of single and double bonds along a chain
One example is the conjugated - systems π
C2H4
allyl cation
ButadieneCyclo-Butadiene
Benzene
Conjugated systems
We shall also use it for the - bond inethylene
π
Molecular orbital theory In molecular orbital theory we write our orbitals as linear combinations of atomic orbitals :
ψ χk ii
i niC( ) ( )1 1
1= ∑
=
=
We next require that the corresponding orbital energies
W C CH dv
dvnk k
k k(C1, ,.. )
( ) ˆ
( ) ( )211 1
= ∫∫
ψ ψψ ψ
be optimal with respect to {C1, ,.. }C Cn2
or : WC
i = 1,..,ni
δδ
= 0;
This leads to theset of homogeneous equations :
C H WS
i = 1,2,...,n.
k ik ikk=1
k=n−∑ = 0
For which the secular determinantmust be zero
H WS i = 1,2,...,n; k = 1,2,...,n
ik ik− = 0
In order to obtainnon - trivial solutions
The Hückel method
We are going to use as our atomic orbitals a single p function on each carbon atom.
π
The p orbital is the p - function perpendicular to the molecular plane
π
For ethylene we have two p orbitalsπ
pAπ pB
π
pAπ
pBπ
pCπ
pDπ
For butadiene we have four p orbitalsπ
Conjugated systems
The Hückel method For ethylene we have two p orbitalsπ
pAπ pB
πFor ethylene we canuse linear variationtheory to obtain a setof linear homogeneous equations
H E H -ESH -ES H E
11 12 1221 21 11
−− = 0
C H WS
i = 1,2 and n = 1,2
k ik ikk=1
k=n−∑ = 0
Non - trivial solutionsare only possible ifthe secular determinantis zero
Each of the two roots W = E i = 1,2 can be substituted into the set of linear equations to obtain orbital coefficients
i
ψ πi
ik k
k=1
k=nC p
k = 1,2 ; i = 1,2
= ∑
Ethylene
The Hückel method
For butadiene we canuse linear variationtheory to obtain a setof linear homogeneous equations
C H WS
i = 1,4 and n = 1,4
k ik ikk=1
k=n−∑ = 0
pAπ
pBπ
pCπ
pDπ
For butadiene we have four p orbitalsπ
Butadiene
The Hückel method
Each of the two roots W = E i = 1,4 can be substituted into the set of linear equations to obtain orbital coefficients
iψ π
iik k
k=1
k=nC p
k = 1,4 ; i = 1,4
= ∑
H E H -ES H -ES H -ESH -ES H E H -ES H -ESH -ES H -ES H E H -ESH -ES H -ES H -ES H E
11 12 12 13 13 14 1421 21 22 23 23 24 2431 31 32 32 33 34 3441 41 42 42 43 43 44
−−
−−
= 0
Non - trivial solutionsare only possible ifthe secular determinant is zero
Butadiene
The Hückel method
H E H -ES H -ES H -ESH -ES H E H -ES H -ESH -ES H -ES H E H -ESH -ES H -ES H -ES H E
11 12 12 13 13 14 1421 21 22 23 23 24 2431 31 32 32 33 34 3441 41 42 42 43 43 44
−−
−−
= 0
H E H -ESH -ES H E
11 12 1221 21 11
−− = 0
The Hückel approximation 1 0. all overlaps are set to zero S ij =
H E HH H E
11 1221 11
−− = 0
H E H H HH H E H HH H H E HH H H H
11 12 13 1421 22 23 2431 32 33 3441 42 43 44
−−
− = 0
The Hückel method
H E H H HH H E H HH H H E HH H H H
11 12 13 1421 22 23 2431 32 33 3441 42 43 44
−−
− = 0
H E HH H E
11 1221 11
−− = 0
The Hückel approximation 1 0. All overlaps are set to zero S ij =
2. All diagonal terms are set equal to Coulomb integral α
αα
αα
−−
− =
E H H HH E H HH H E HH H H -E
12 13 1421 23 2431 32 3441 42 43
0
αα
−− =E H
H E12
210
The Hückel method
αα
αα
−−
− =
E H H HH E H HH H E HH H H -E
12 13 1421 23 2431 32 3441 42 43
0
αα
−− =E H
H E12
210
The Hückel approximation 1 0. All overlaps are set to zero S ij =2. All diagonal terms are set equal to Coulomb integral α
1
2
3
4
3. All resonance integrals between non - neighbours are set to zero
αα
αα
−−
− =
E H 0 0H E H 00 H E H0 0 H -E
1221 23
32 3443
0
αα
−− =E H
H E12
210
The Hückel method
αα
αα
−−
− =
E H 0 0H E H 00 H E H0 0 H -E
1221 23
32 3443
0
αα
−− =E H
H E12
210
The Hückel approximation 1 0. All overlaps are set to zero S ij =2. All diagonal terms are set equal to Coulomb integral α
1
2
3
4
3. All resonance integrals between non - neighbours are set to zero4. All remaining resonance integrals are set to β
α ββ α β
β α ββ α
−−
− =
E 0 0 E 0
0 E 0 0 -E
0
α ββ α
−− =E
E 0
The Hückel method
α ββ α β
β α ββ α
−−
− =
E 0 0 E 0
0 E 0 0 -E
0
α ββ α
−− =E
E 0
The Hückel approximation 1 0. All overlaps are set to zero S ij =
2. All diagonal terms are set equal to Coulomb integral α
1
2
3
4
3. All resonance integrals between non - neighbours are set to zero4. All remaining resonance integrals are set to β
The Hückel method Solutions for ethylene
α ββ α α β−
− = − − =E E E)2( 2 0
C H WS
i = 1,2 and n = 1,2
k ik ikk=1
k=n−∑ = 0
E ( and negative)1 = +α β α β
E ( and negative) 2 = −α β α β
ψ π π11 2p p= +1
212
ψ π π212
12
= − +p p1 2
ψ πi
ik k
k=1
k=nC p
k = 1,2 ; i = 1,2
= ∑( ;(α β α β− = − ±E) E) = 2 2
The Hückel method
α ββ α β
β α ββ α
−−
− =
E 0 0 E 0
0 E 0 0 -E
Solutions for butadiene
( )αα β
β α ββ α
−−
−−
EE
EE
0
0− −
−=β
β βα β
β α
000
EE
( ) ( )( )
α α ββ α α ββ β
α− −− − − −E
EE
EE
20
− −− + − =β α β
β α β βα
2 2 00
EE E( )
( ) ( )α α β− − −E E4 2 2 − −( )α βE 2 2 − −( )α βE 2 2 +β4
( ) ( )α α β β− − − + =E E4 2 2 43 0
The Hückel method Solutions for butadiene
Thus
E = 1.62 and 0.62α β α β± ±
( ) ( )α α β β− − − + =E E4 2 2 43 0
Dividing by : 2β αβ
αβ
( ) ( )− − − + =E E4
4
2
23 1 0
Introducing : x =( -E)2
2α
β
x2 − + =3 1 0xx = 2.62 ; x = 0.38
The Hückel molecular orbital energy levels of butadiene and thetop view of the corresponding π orbitals. The four p electrons (one supplied by each C)occupy the two lower π orbitals.Note that the orbitals are delocalized.
The Hückel method C H WS
i = 1,4 and n = 1,4
k ik ikk=1
k=n−∑ = 0
ψ πi
ik k
k=1
k=nC p
k = 1,4 ; i = 1,4
= ∑
Solutions for butadiene
The Hückel method
0.372p p p 0.372p1 2 3 4π π π π+ + +0 602 0 602. .
1
2
3
4
− − +0 602. pp 0.372 0.372p + 0.602p1 2 3 4π π π π
0 602. pp 0.372 0.372p + 0.602p1 2 3 4π π π π− −
1
2
3
4
0.372p p p 0.372p1 2 3 4π π π π− + −0 602 0 602. .
1
2
3
4
1
2
3
4
Solutions for butadiene
The Hückel method Solutions for butadiene
E1 = h2
8mL2 ,
E2 =h
2
8mL2 ,4
E3 =h
2
8mL2 ,9
FMOHuckel
α+1.62β
α+.62β
α−.62β
Comparison between PIB and Huckel treatment ofbutadiene
Same nodal structure
The Hückel method Solutions for butadiene
We can write butadieneas two localized - bondsπ
Or two delocalized - bondsπ 4π
α+1.62β
α+.62β
α−.62β
Butadiene
α+β
Ethylene
α−β
E Bu E Etπ π( ) ( )− 2 ∴ delocalization energy
E Buπ α β α β( ) ( . ) ( . )= + + +2 1 62 2 62
2 4 4 4E Etπ α β α β( ) ( )= + = +
= +4 4 48α β.
Delocalization energy = .48β
( )−36 kJ mol-1
The Hückel method Cyclobutadiene
1
2
4
3
α β ββ α β
β α ββ β α
−−
− =
E 0 E 0
0 E 0 -E
0
α β+ 2
α β− 2
α α
0.5p p p 0.5p1 2 3 4π π π π+ + +0 5 0 5. .
0.5p p p 0.5p1 2 3 4π π π π− + −0 5 0 5. .
0.5p p p 0.5p1 2 3 4π π π π− − +0 5 0 5. .
0.5p p p 0.5p1 2 3 4π π π π+ − −0 5 0 5. .
α β+ 2
α α
α β− 2
α β+
α β−
12
43
12
43
The Hückel method Cyclobutadiene
No delocalization energy
The framework of benzene is formed by the overlap of Csp2 hybrids, which fit without strain into a hexagonal arrangement.
The Hückel method Benzene
Benzene
The C - C and C - H - orbitalsσ
Benzene The Hückel method
α β ββ α ββ α β
β α ββ α β
β β β α
−−
−− =
E 0 0 E 0 0 E 0 0
0 0 E 0 0 0 -E 0 0 -E
0
0
0
C H WS
i = 1, 6 and n = 1, 6
k ik ikk=1
k=n−∑ = 0 ψ π
iik k
k=1
k=nC p
k = 1, 6 ; i = 1, 6
= ∑
E = 2 ; ; α β α β α β± ± ±
Benzene The Hückel method
α β+ 2
α β− 2
α β+
α β−
Delocalization energy =
2( + 2 ) + 4( + +α β α β α β β) ( )− =6 2
−150 kJ mol-1
What you should learn from this lecture
1. Be able to construct the seculardeterminant for a conjugated - system within the Huckel approximation
π
2. Be able to calculate orbital energies for simple systems
3. You will not be asked to find the molecular orbitals. However, you should be able to discuss provided orbitals in terms of bonding and anti - bonding interactions
The Hückel method
2
3
allyl cationα ββ α β
β α
−−
−
E 0 E
0 E = 0
The allyl system
α α+ 1 42.
α α− 1 42.
α
0.5p p p1 2 3π π π+ +0 707 0 5. .
0.5p p p1 2 3π π π− +0 707 0 5. .
0.707p p1 3π π− 0 707.
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