Question

Compute the delocalization energy and pi-bond formation energy of (a) the allyl radical, (b) the cyclobutadiene cation.

Answer 1

1. The delocalization energy and pi-bond formation energy of allyl radical (C3H5)

In the allyl radical, there are 3 π electrons. 3 carbon 2p orbitals form 3 π molecular orbitals

The energies of the three molecular orbitals are (α + β√2), a and (α – β√2) (bonding, nonbonding and antibonding)

The total p electron energy of allyl radical is therefore: 2´( α + β √2) + α

                                                                                                 = 3 α + 2 β√2

                                                                                                      3 α + 2.83 β

In a model allyl radical with a wall to stop the p electrons delocalizing:

The total π electron energy is

2´(α + β) + α = 3 α +2 β

(Remember a is the energy of an electron in a carbon 2p orbital – you need to add a for the energy of the odd electron)

So the delocalization energy of allyl radical is

3 α + 2.83 β – 3 α +2 β

= 0.83 β

B) The delocalization energy and pi-bond formation energy of the cyclobutadiene cation

For cyclobutadiene the theory predicts that the two high-energy electrons occupy a degenerate pair of MO's that are neither stabilized nor destabilized. Hence the square molecule would be a very reactive triplet diradical (the ground state is actually rectangular without degenerate orbitals). In fact, all cyclic conjugated hydrocarbons with a total of 4nπelectrons share this MO pattern, and this forms the basis of Hückel's rule.

Cyclobutadiene	E1 = α + 2β
	E2 = α
	E3 = α
	E4 = α − 2β

Cyclobutadiene contains four sp²hybridized carbons, which leaves four p orbitals for the π bonding.

The four p orbitals produce 4 MO's, as shown below.

As usual, the lowest energy MO (π ₁) has all bonding interactions between the p orbitals, and zero nodes.

The next highest energy MO's are π ₂ and π ₃. They are degenerate and contain one node.

Their overall energy is zero, which is non-bonding (π ₂ and π ₃ both have two bonding and two anti-bonding interactions).

The highest energy MO is π ₄, and comprises solely of antibonding interactions (and two nodal planes).

The four electrons which have to be accommodated are arranged putting 2 electrons in π ₁, and one each in π ₂ and π ₃ (Hund's rule).