What Is the MO Diagram for the Allyl Cation?

The allyl cation is a three-carbon, conjugated system with two π electrons. Its molecular-orbital (MO) diagram shows how the three parallel p orbitals combine and where those electrons go. The key result is that both electrons occupy the lowest-energy π orbital, which extends across the entire carbon chain.

Count the orbitals and electrons

The allyl cation has a planar, conjugated carbon framework. Each carbon is approximately $sp^2$-hybridized and contributes one p orbital perpendicular to the plane of the sigma bonds. Three p orbitals combine to form three π molecular orbitals.

There are two π electrons in the cation. You can count them from either common resonance form, $\mathrm{CH_2{=}CH{-}CH_2^+}$ or ${}^{+}\mathrm{CH_2{-}CH{=}CH_2}$: each has one π bond, containing two electrons. The positive charge does not mean that the π system has no electrons; it means there is one fewer π electron than in the corresponding allyl radical, which has three.

The allyl-cation MO diagram

The three orbitals are conventionally labeled $\pi_1$, $\pi_2$, and $\pi_3^*$. With the usual Hückel convention that the resonance integral $\beta$ is negative, their energies are ordered as follows:

                         Energy increases

       π₃*     α − √2β          empty
               antibonding

       π₂      α                empty
               nonbonding

       π₁      α + √2β          ↑↓
               bonding

                         Energy decreases

Because $\beta<0$, $\alpha+\sqrt{2}\beta$ is the lowest of these energies, and $\alpha-\sqrt{2}\beta$ is the highest. The two electrons pair in $\pi_1$, following the Aufbau principle and the Pauli exclusion principle. The other orbitals remain unoccupied in the ground-state configuration:

$$(\pi_1)^2(\pi_2)^0(\pi_3^*)^0$$

What the orbitals look like

The lowest-energy orbital is bonding across both neighboring carbon pairs. Its p-orbital lobes have the same phase along the chain, so electron density can be shared over all three carbons. In a simple Hückel model, its normalized coefficients, listed from one end of the chain to the other, are

$$\left(\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\right)$$

The nonbonding orbital has no coefficient on the middle carbon. Its two terminal p orbitals have opposite phases, giving a node at the central carbon. Its coefficients are

$$\left(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}\right)$$

The antibonding orbital has a phase change between the middle p orbital and each terminal p orbital. Its coefficients are

$$\left(\frac{1}{2},-\frac{1}{\sqrt{2}},\frac{1}{2}\right)$$

The signs describe orbital phase, not positive and negative electric charge. In an orbital sketch, shading can distinguish the two phases. Each p orbital also has lobes above and below the molecular plane.

What the diagram tells you

Since the occupied bonding orbital is spread over all three carbon atoms, the π electrons are delocalized rather than confined to one C=C bond. The two terminal carbons are equivalent by symmetry. In the simple Hückel picture, each terminal carbon carries half of the π-electron density found at the central carbon; this helps explain why the positive charge is shared between the ends of the allyl system rather than fixed at just one carbon.

The model also gives equal π bond-order contributions for the two adjacent carbon pairs. This does not mean the molecule has two ordinary, fully localized double bonds: the values describe only the π contribution in this simplified model, in addition to the sigma framework.

The empty nonbonding orbital is especially relevant to reactions. It can accept electron density from a nucleophile, and its terminal lobes have amplitude on both ends of the allyl group. This is one reason reactions of allylic cations can form bonds at either terminal position.

Common mistakes

• Putting two electrons in the nonbonding orbital: That is not the allyl-cation ground state. The bonding orbital is lower in energy and is filled first.
• Drawing only two π orbitals: Three p atomic orbitals produce three π molecular orbitals, even though only one π bond is present in either localized resonance form.
• Treating the cation as a three-electron system: The allyl radical has three π electrons; removing one gives the two-electron allyl cation.
• Putting the positive charge only on one end: A resonance structure can place it on either terminal carbon. The delocalized description makes the two ends equivalent.
• Reading phase as charge: Opposite orbital phases indicate a sign change in the wavefunction. They are not opposite electric charges.

How it differs from the allyl radical

The orbital energy diagram is the same for the allyl radical, but its electron count is three instead of two. The first two electrons fill the bonding orbital, and the third occupies the nonbonding orbital singly. The cation therefore has an empty nonbonding orbital, while the radical has a singly occupied one.