(1)Lecture 32 : The Hückel Method The material in this lecture covers the following in Atkins

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 approximation

can be used for conjugated

molecules in which there is an alternation of single and

double bonds along a chain One example is the

conjugated - systems π C₂H₄

allyl cation Butadiene

Cyclo-Butadiene

Benzene

Conjugated systems We shall also use it for the - bond in ethylene

π

Molecular orbital theory In molecular orbital

theory we write our orbitals as linear

combinations of atomic orbitals :

ψ_{k i}χ

i i n

C i

( )1 ( )1

1

= ∑=

=

We next require that the corresponding orbital energies

W C C H dv

n k k dv

k k

(C₁, ,.. ) ( ) ˆ

( ) ( )

2 1

1 1

= ∫∫

ψ ψ

ψ ψ

be optimal with respect to {C₁,C₂,..C_n}

or : W

C i = 1,..,n

i

δ

δ = 0;

This leads to the

set of homogeneous equations :

C H WS

i = 1, 2,...,n.

k ik ik

k=1

k=n∑ − = 0

For which the

secular determinant must be zero

H WS

i = 1, 2,...,n; k = 1, 2,...,n

ik − ik = 0

In order to obtain

non - 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 orbitals_π

p^A_π p^B_π

p^A_π

p^B_π

p^C_π

p^D_π

For butadiene we have four p orbitals_π

Conjugated systems

The Hückel method For ethylene we have

two p orbitals_π

p^A_π p^B_π

For ethylene we can use linear variation

theory to obtain a set of linear homogeneous equations

H E H - ES

H¹¹₂₁ - ES₂₁¹² H₁₁ ¹²E

− − = 0

C H WS

i = 1, 2 and n = 1, 2

k ik ik

k=1

k=n∑ − = 0

Non - trivial solutions are only possible if

the secular determinant is 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

ψⁱ _i^{k k}_π

k=1 k=n

C p k = 1, 2 ; i = 1, 2

= ∑

Ethylene

The Hückel method

For butadiene we can use linear variation theory to obtain a set of linear homogeneous equations

C H WS

i = 1, 4 and n = 1, 4

k ik ik

k=1

k=n∑ − = 0

p^A_π

p^B_π

p^C_π

p^D_π

For butadiene we have four p orbitals_π

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

i ψⁱ _i^{k k}_π

k=1 k=n

C p k = 1, 4 ; i = 1, 4

= ∑ H E H - ES H - ES H - ES

H - ES H E H - ES H - ES H - ES H - ES H E H - ES H - ES H - ES H - ES H E

11 12 12 13 13 14 14

21 21 22 23 23 24 24

31 31 32 32 33 34 34

41 41 42 42 43 43 44

− −

− −

= 0 Non - trivial solutions

are only possible if

the secular determinant is zero

Butadiene

The Hückel method

H E H - ES H - ES H - ES H - ES H E H - ES H - ES H - ES H - ES H E H - ES H - ES H - ES H - ES H E

11 12 12 13 13 14 14

21 21 22 23 23 24 24

31 31 32 32 33 34 34

41 41 42 42 43 43 44

− −

− −

= 0

H E H - ES

H¹¹₂₁- ES₂₁¹² H₁₁ ¹²E

− − = 0

The Hückel approximation

1. all overlaps are set to zero S _ij = 0

H E H H¹¹₂₁ H₁₁ ¹²E

− − = 0

H E H H H H H E H H H H H E H H H H H

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

− −

− = 0

The Hückel method

H E H H H H H E H H H H H E H H H H H

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

− −

− = 0

H E H H¹¹₂₁ H₁₁ ¹²E

− − = 0

The Hückel approximation

1. All overlaps are set to zero S _ij = 0 2. All diagonal terms are set equal to Coulomb integral α

α α

α α

− −

− =

E H H H H E H H H H E H

H H H - E

12 13 14

21 23 24

31 32 34

41 42 43

0 α − α

− = E H

H₂₁ ¹²E 0

The Hückel method

α α

α α

− −

− =

E H H H H E H H H H E H

H H H - E

12 13 14

21 23 24

31 32 34

41 42 43

0 α − α

− = E H

H₂₁ ¹²E 0

The Hückel approximation

1. All overlaps are set to zero S _ij = 0 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

α α

α

α

− −

− =

E H 0 0 H E H 0 0 H E H 0 0 H - E

12

21 23

32 34

43

0 α − α

− = E H

H₂₁ ¹²E 0

The Hückel method

α α

α

α

− −

− =

E H 0 0 H E H 0 0 H E H 0 0 H - E

12

21 23

32 34

43

0 α − α

− = E H

H₂₁ ¹²E 0

The Hückel approximation

1. All overlaps are set to zero S _ij = 0 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

4. All remaining resonance integrals are set to β

α β

β α β

β α β

β α

− −

− =

E 0 0 E 0 0 E

0 0 - E

0

α β

β − α

− = E

E 0

The Hückel method

α β

β α β

β α β

β α

− −

− =

E 0 0 E 0 0 E

0 0 - E

0

α β

β − α

− = E

E 0

The Hückel approximation

1. All overlaps are set to zero S _ij = 0 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

4. All remaining resonance integrals are set to β

The Hückel method Solutions for ethylene

α β

β − α α β

− = − − =

E

E ( E)^{2 2} 0

C H WS

i = 1, 2 and n = 1, 2

k ik ik

k=1

k=n∑ − = 0

E₁ = +α β α ( and negative)β E₂ = −α β α ( and negative) β

ψ₁ = 1 p¹_π + p_π² 2

1 2

ψ₂ 1 _{π π}

2

1

= − p¹ + 2 p² ψⁱ _i^{k k}_π

k=1 k=n

C p k = 1, 2 ; i = 1, 2

= ∑ (α − E)² = β²; (α − E) = ± β

The Hückel method

α β

β α β

β α β

β α

− −

− =

E 0 0 E 0 0 E

0 0 - E Solutions for butadiene

(α )α β

β α β

β α

− −

− −

E

E

E

E 0 0

− −

− = ββ β

α β

β α 0 0

0

E

E

( ) ( )

( )

α α β

β α α ββ β

− − α

− − − −

E E

E E

E

2

0

− −

− + − =

β α β

β α β β

2 2 0 α

0 E

E ( E)

(α − E)⁴ − (α − E)^{2 2}β − −(α E)^{2 2}β − −(α E)^{2 2}β +β⁴ (α −E)⁴ − 3(α − E)^{2 2}β + β⁴ = 0

The Hückel method Solutions for butadiene

Thus

E = α ± 1.62 and β α ± 0.62β (α − E)⁴ − 3(α − E)^{2 2}β + β⁴ = 0 Dividing by β² : α

β

α β

( −E)⁴ − ( − E) + =

4

2

3 2 1 0

Introducing : x = ( - E)²

2

α β

x² − 3x + =1 0

x = 2.62 ; x = 0.38

The Hückel molecular orbital energy levels of butadiene and the top 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 ik

k=1

k=n∑ − = 0

ψⁱ _i^{k k}_π

k=1 k=n

C p k = 1, 4 ; i = 1, 4

= ∑ Solutions for butadiene

The Hückel method

0.372p¹_π + 0 602. p²_π + 0 602. p³_π + 0.372p⁴_π

1

2

3

4

−0 602. pp¹_π − 0.372²_π + 0.372p + 0.602p³_π ⁴_π 0 602. pp¹_π − 0.372²_π − 0.372p + 0.602p³_π ⁴_π

1

2

3

4

0.372p¹_π − 0 602. p²_π + 0 602. p³_π − 0.372p⁴_π

1

2

3

4

1

2

3

4

Solutions for butadiene

The Hückel method Solutions for butadiene

E₁ = h² 8mL² ,

E₂ = h² 8mL4 ², E₃ = h² 8mL9 ²,

FMO Huckel

α+1.62β α+.62β

α−.62β

Comparison between PIB and Huckel treatment of

butadiene

Same nodal structure

The Hückel method Solutions for butadiene We can write butadiene

as two localized - bondsπ

Or two delocalized - bondsπ ^4π

α+1.62β α+.62β

α−.62β

Butadiene

α+β

Ethylene α−β

E_π(Bu) − 2E_π(Et) ∴ delocalization energy

E_π(Bu) = 2(α + 1 62. β) + 2(α + .62β)

2E_π(Et) = 4(α β+ ) = 4α + 4β

= 4α + 4 48. β

Delocalization energy = .48β (−36 kJ mol^-1)

The Hückel method Cyclobutadiene

1

2 4

3

α β β

β α β

β α β

β β α

− −

− =

E 0 E 0 0 E

0 - E

0

α + 2β α − 2β

α α

0.5p¹_π + 0 5. p²_π + 0 5. p³_π + 0.5p⁴_π 0.5p¹_π − 0 5. p²_π + 0 5. p³_π − 0.5p⁴_π

0.5p¹_π − 0 5. p²_π − 0 5. p³_π + 0.5p⁴_π

0.5p¹_π + 0 5. p²_π − 0 5. p³_π − 0.5p⁴_π

α + 2β α α

α −2β

α β+ α β−

1 2

4 3

1 2

4 3

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 - orbitals

σ

Benzene The Hückel method

α β β

β α β

β α β

β α β

β 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 ik

k=1

k=n∑ − = 0 ψⁱ _i^{k k}_π

k=1 k=n

C p k = 1, 6 ; i = 1, 6

= ∑ E = 2 ; ; α ± β α β α β± ±

Benzene The Hückel method

α + 2β α − 2β

α β+ α β−

Delocalization energy = 2( + 2 ) + 4( +α β α β) − 6(α β+ ) = 2β

−150 kJ mol^-1

What you should learn from this lecture 1. Be able to construct the secular

determinant 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 cation

α β

β α β

β α

− −

− E 0 E

0 E

= 0 The allyl system

α + 1 42. α α − 1 42. α

α

0.5p¹_π + 0 707. p²_π + 0 5. p³_π 0.5p¹_π − 0 707. p²_π + 0 5. p³_π

0.707p¹_π − 0 707. p³_π