chem 373- lecture 32 : the hückel method

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.