Study of butadiene

A study of the p-p* transitions of butadiene

Here we consider the symmetries and energies of the p and p* orbitals of butadiene.

The highest energy p* orbital has a_u symmetry.

The LUMO is a p* orbital with b_g symmetry.

The HOMO has a_u symmetry.

The lowest p orbital is

As we have seen with ethylene we can model the molecular orbitals by forming linear combinations of the individiual pp orbitals on the four carbon atoms of butadiene.

Formulate linear combinations of f₁, f₂, f₃, and f₄. Use projection operators.

P^Au = 1/2(f₁ + f₂ + f₃ + f₄)

P^Bg = 1/2(f₁ + f₂ - f₃ - f₄)

P^Au = 1/2(f₁ - f₂ - f₃ + f₄)

P^Bg = 1/2(f₁ - f₂ + f₃ - f₄)

The configuration of the ground state is A_1g. This is always the case for a closed shell molecule since the combination of the two spin paired electrons yields a direct product of the symmetry terms that is always totally symmetric. The symmetry of the first p-p* excited state is given by the direct product of the irreducible symmetry representations for those states. For butadiene, we have:

ground state (a_u)²(b_g)²

first excited state (a_u)²(b_g)(a_u)

Thus, we consider the direct product of (b_g)(a_u)

E	C₂	s_v	s_v'
1	-1	-1	1

This has B_u symmetry and since the Cartesian dipole operators transform as

A_u à z

B_u à x,y

this transition is dipole allowed.

According the Huckel theory we have:

Where a' = a - l. The determinant expands to:

a⁴- 4a³l + 6a²l² - 3a²b² - 4al³ + 6ab²l + l⁴- 3b²l² + b⁴= 0

The energies are:

a -(1+Ö5)/2 b , a +(1-Ö5)/2 b , a -(1-Ö5)/2 b , a +(1+Ö5)/2 b

Thus, we have the energy level diagram