Lecture 4 – Worksheet: π Orbitals in Cyclobutadiene
Model 1: Isolated π Bonds
The MO diagram for an isolated C=C π bond is shown opposite. Each electron
in the bond has an energy of α + β.
Critical thinking questions
1. What is the total energy of the two π electrons in H2C=CH2?
2. Assuming that the π bonds do not interact (i.e. there is no conjugation), what is the total
energy of the π electrons in the cyclobutadiene?
Model 2: The π Orbitals of C4H4
The particle on a ring solution to the Schrodinger equation can be used
to work out the π molecular orbitals for organic rings. Solving the
equation gives two solutions:
• ψ = sin(jθ) and ψ = cos(jθ) with j = 0, 1, 2, 3....
These can be used to work out the combinations of the p-orbitals on
each carbon atom in the ring. Each π orbital is a combination of the p-
orbitals:
• ψ = a1C1 + a2C2 + a3C3 + a4C4
The particle on a ring solutions can be used to work out the coefficients
a1 – a4. The picture shows a cyclobutadiene ring:
• C1 is at 90°, C2 is at 180°, C3 is at 270° and C4 is at 360°.
C2
C3
C4
C1
θ
Critical thinking questions
1. Complete the table below to work out the coefficients for the levels with j = 0, 1 and 2 for both the
sin(jθ) and cos(jθ) functions. The j = 0 solutions are shown as an example.
C1
C2
C3
C4
angle
90°
180°
270°
360°
j = 0
sin(jθ)
sin(0 × 90°) = 0
sin(0 × 180°) = 0
sin(0 × 270°) = 0
sin(0 × 360°) = 0
cos(jθ)
cos(0 × 90°) = 1
cos(0 × 180°) = 1
cos(0 × 270°) = 1
cos(0 × 360°) = 1
j = 1
sin(jθ)
cos(jθ)
j = 2
sin(jθ)
cos(jθ)
