lectura de huckel

Lecture 32 : The Hckel Method

The material in this lecture covers the following in Atkins.

14.0 The Hckel 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 Hckel method The Hckel 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

C2H4

allyl cation

ButadieneCyclo-Butadiene

Benzene

Conjugated systemsWe shall also use it for the - bond inethylene

Molecular orbital theory In molecular orbital theory we write our orbitals as linear combinations of atomic orbitals :

k ii

i niC( ) ( )1 1

1=

=

=

We next require that the corresponding orbital energies

W C C H dvdvn

k kk k

(C1, ,.. ) ( )

( ) ( )211 1

=

be optimal with respect to {C1, ,.. }C Cn2or :

WC

i = 1,..,ni

= 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 Hckel method

We are going to use as our atomic orbitals a single p function on each carbon atom.

The p orbital is the p - function perpendicular to the molecular plane

For ethylene we have two p orbitals

pA pB

pApB

pCpD

For butadiene we have four p orbitals

Conjugated systems

The Hckel method For ethylene we have two p orbitals

pA pBFor 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

i ik k

k=1

k=nC p

k = 1,2 ; i = 1,2

=

Ethylene

The Hckel 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

pApB

pCpD

For butadiene we have four p orbitals

Butadiene

The Hckel 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 i i

k kk=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 Hckel 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 Hckel 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 Hckel 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 Hckel approximation 1 0. All overlaps are set to zero S ij =2. All diagonal terms are set equal to Coulomb integral

=

E H H HH E H HH H E HH H H -E

12 13 1421 23 2431 32 3441 42 43

0

=E H

H E12

210

The Hckel method

=

E H H HH E H HH H E HH H H -E

12 13 1421 23 2431 32 3441 42 43

0

=E H

H E12

210


1

2

3

4

3. All resonance integrals between non - neighbours are set to zero

=

E H 0 0H E H 00 H E H0 0 H -E

1221 23

32 3443

0

=E H

H E12

210

The Hckel method

=

E H 0 0H E H 00 H E H0 0 H -E

1221 23

32 3443

0

=E H

H E12

210


1

2

3

4

3. All resonance integrals between non - neighbours are set to zero4. All remaining resonance integrals are set to

=

E 0 0 E 0

0 E 0 0 -E

0

=E

E 0

The Hckel method

=

E 0 0 E 0

0 E 0 0 -E

0

=E

E 0


1

2

3

4

3. All resonance integrals between non - neighbours are set to zero4. All remaining resonance integrals are set to

The Hckel method Solutions for ethylene

= =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 = + E ( and negative) 2 =

1 1 2p p= +12

12

212

12

= +p p1 2

i ik k

k=1

k=nC p

k = 1,2 ; i = 1,2

= ( ;( = E) E) = 2 2

The Hckel method

=

E 0 0 E 0

0 E 0 0 -E

Solutions for butadiene

( )

EE

EE

0

0

=

000

EE

( ) ( ) ( )

E E E E E2

0

+

=

2 2 0

0E

E E( )

( ) ( ) E E4 2 2 ( ) E 2 2 ( ) E 2 2 +4

( ) ( ) + =E E4 2 2 43 0

The Hckel method Solutions for butadiene

Thus

E = 1.62 and 0.62

( ) ( ) + =E E4 2 2 43 0

Dividing by : 2

( ) ( )

+ =E E44

223 1 0

Introducing : x = ( -E)2

2

x2 + =3 1 0xx = 2.62 ; x = 0.38

The Hckel molecular orbital energy levels of butadiene and thetop view of the corresponding orbitals. The four p electrons (one supplied by each C)occupy the two lower orbitals.Note that the orbitals are delocalized.

The Hckel method C H WS

i = 1,4 and n = 1,4

k ik ikk=1

k=n = 0

i ik k

k=1

k=nC p

k = 1,4 ; i = 1,4

=


The Hckel method

0.372p p p 0.372p1 2 3 4 + + +0 602 0 602. .

1

2

3

4

+0 602. pp 0.372 0.372p + 0.602p1 2 3 4

0 602. pp 0.372 0.372p + 0.602p1 2 3 4 1

2

3

4

0.372p p p 0.372p1 2 3 4 + 0 602 0 602. .1

2

3

4

1

2

3

4


The Hckel method Solutions for butadiene

E1 =h2

8mL2,

E2 =h2

8mL2,

4

E3 =h2

8mL2,

9

FMOHuckel

+1.62

+.62

.62

Comparison between PIB and Huckel treatment ofbutadiene

Same nodal structure

The Hckel method Solutions for butadiene We can write butadieneas two localized - bonds

Or two delocalized - bonds 4

+1.62

+.62

.62

Butadiene

+

Ethylene

E Bu E Et ( ) ( ) 2 delocalization energy

E Bu ( ) ( . ) ( . )= + + +2 1 62 2 62

2 4 4 4E Et ( ) ( )= + = += +4 4 48 .

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

The Hckel method Cyclobutadiene

12

43

=

E 0 E 0

0 E 0 -E

0

+ 2

2

0.5p p p 0.5p1 2 3 4 + + +0 5 0 5. .

0.5p p p 0.5p1 2 3 4 + 0 5 0 5. .

0.5p p p 0.5p1 2 3 4 +0 5 0 5. .

0.5p p p 0.5p1 2 3 4 + 0 5 0 5. .

+ 2

2

+

12

43

12

43

The Hckel 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 Hckel method Benzene

Benzene

The C - C and C - H - orbitals

Benzene The Hckel method

=

E 0 0 E 0 0 E 0 0

0 0 E 0 0 0 -E 0 0 -E

0

00

C H WS

i = 1, 6 and n = 1, 6

k ik ikk=1

k=n = 0 i i

k kk=1

k=nC p

k = 1, 6 ; i = 1, 6

=

E = 2 ; ;

Benzene The Hckel method

+ 2

2

+

Delocalization energy = 2( + 2 ) + 4( + + ) ( ) =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

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 Hckel method

2

3

allyl cation

E 0 E

0 E = 0

The allyl system

+ 1 42.

1 42.

0.5p p p1 2 3 + +0 707 0 5. .

0.5p p p1 2 3 +0 707 0 5. .

0.707p p1 3 0 707.

lectura de huckel

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