1. 1 + S 12 1 E 1 =   1 = c 1  1 + c 1  2 1 - S 12 1 E 2 = -   2 = c 1  1 - c 1  2 bonding antibonding

2. Let overlap term go to zero Overlap term is non zero

3. Ethylene  system of ethylene

4. H2H2 C2H4C2H4

5.    antibonding empty orbital  bonding fille with 2 e -

6. Now let us look at a more complicated system. Three Orbitals! Allyl anion

7. 1 2 3 How do we calculate the energies and coefficients of the MO’s? We are going to use the LCAO approximation to make three MO’s. Use Symmetry! The ends are the same so they must contibute equally to any MO. So we can make linear combinations where atoms one and three contribute equally.  1 = p z 1 + p z 3  2 = p z 1 - p z 3  3 = p z 2 The second carbon is unique.

8. p z a Ĥ p z a = E =  = 0 Coulomb integral p z a Ĥ p z b = E int =  Interaction integral Only if the two p orbitals are adjacent otherwise Interaction is 0. p z a p z a = 1 normalized p z a p z b = 0 assume overlap = 0 Huckel Approximation

9. 1 2 3 How do we calculate the energies and coefficients of the MO’s? We are going to use the LCAO approximation to make three MO’s. Use Symmetry! The ends are the same so they must contibute equally to any MO. So we can make linear combinations where atoms one and three contribute equally.  1 = p z 1 + p z 3  2 = p z 1 - p z 3  3 = p z 2 The second carbon is unique.

10. 1 2 3  1 = p z 1 + p z 3 We need to normalize our LCAOs  1 2 = 1 N 2 (p z 1 + p z 3 ) (p z 1 + p z 3 ) = 1 N 2 (p z 1 2 + 2 p z 3 p z 1 + p z 3 2 ) = 1 1 + 2 x 0 + 1 N 2 x 2 = 1 So N =

11. 1 2 3 How do we calculate the energies and coefficients of the MO’s? We are going to use the LCAO approximation to make three MO’s. Use Symmetry! The ends are the same so they must contibute equally to any MO. So we can make linear combinations where atoms one and three contribute equally.  1 = ( p z 1 + p z 3 )  2 = ( p z 1 - p z 3 )  3 = p z 2 The second carbon is unique.

12.  1 = ( p z 1 + p z 3 )  2 = ( p z 1 - p z 3 )  3 = p z 2 1 2 3 Same Symmetry 1 2 3 1 2 3 1 3 2 11 22 33

13.  1 = ( p z 1 + p z 3 )  3 = p z 2 11 33 1 2 3 1 2 3 - 11 33 1 2 3 1 2 3 + Bonding MO Antibonding MO

14.  1 = ( p z 1 + p z 3 )  3 = p z 2 ( p z 1 + p z 3 ) Ĥ p z 2 = 22 = 1 2 3 33 +  Energy =  1 2 3 11

15.  1 = ( p z 1 + p z 3 ) -  3 = - p z 2 ( p z 1 + p z 3 ) Ĥ (- p z 2 ) = - 2  = -  Energy =  - 1 2 3 33 - 1 2 3 11

16. 1 2 3 33 + 1 2 3 11 1 2 3  Bonding

17. 1 2 3 33 - Energy = 1 2 3 11 2  - 1 3 Antibonding

18. 1 2 3 What about  2 ? There is no overlap between ends so E = 0

19. 1 3 1 2 3 1 2 3  -  0

20. This is painful! It makes the brain hurt. So use a computer instead. Simple Huckel Molecular Orbital Theory Calculator