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Molecular Orbitals of The Allyl Cation, Allyl Radical, and Allyl Anion

In the last post, we showed how to build a molecular orbital (MO) diagram for a typical C-C pi bond.

We saw that:

The molecular orbitals for a pi bond look like this:

Drawing out the molecular orbitals for a single pi bond seems simple enough. But what happens to the molecular orbital diagram if we add a third contributing p-orbital? Or if we have an adjacent double bond, contributing 2 further p-orbitals, giving us 4 conjugated p orbitals in total? Understanding what happens to the molecular orbitals will allow us to understand their energies, and (as we’ll see later) their reactivity.

Generalizing The Rules For Building Up Pi Molecular Orbitals

We can use the lessons learned during building up the molecular orbitals of the pi bond to generalize toward building up larger (linear) pi systems containing N molecular orbitals. [note 1]

Two key lessons, really. First, the number of p orbitals tells you the number of pi orbitals. Second, the lowest and highest energy orbitals are the easiest to draw: the lowest energy one will have phases all aligned, and the highest energy one will have phases all alternating. We’ll be using this generalization a lot.

So let’s take this to the next level. Three molecular orbitals. We call this the allyl system. Let’s go!

The Allyl Cation, Allyl Radical, and Allyl Anion

You’ve heard of methyl, ethyl, propyl, and butyl; our name for the three carbon unit H2C=CH-CH2  is “allyl“. Allyl alcohol, for example, is H2C=CH-CH2OH .

By now you’ve likely encountered the allyl cation, the allyl radical, or the allyl anion?  Maybe you’ve learned that the allyl cation is a “resonance-stabilized” carbocation, or the allyl radical is a “resonance stabilized” radical?

In each of these “pi systems” the allyl carbon has an available p-orbital that is conjugated with the adjacent pi bond. Another way of saying the same thing is that the allyl carbon is in resonance with the pi bond, such that we can draw resonance structures of each of these molecules. [We won’t do that again here: we covered it two posts ago, in this post on  conjugation and resonance – if you need a refresher,  go back and read it to re-familiarize yourself with conjugation, resonance, and orbital overlap. ]

Since the allyl system has three p-orbitals that are in conjugation, how many Pi molecular orbitals will it have?

Three. (N=3)

Next question. What will those three pi molecular orbitals look like? Say you’re asked the following question:

“Draw the molecular orbitals of the allyl cation, the allyl radical, and the allyl anion.”

(a common enough exam question, believe it or not!).

How would you approach that?

The allyl cation, radical, and anion all utilize the same framework of molecular orbitals. They only differ in the number of pi electrons they possess (2, 3, and 4, respectively).

Knowing that, we can divide this problem into two stages.

Let’s do this.

The Molecular Orbitals For The Allyl System (N=3)

We’ll start by drawing the orbitals. The two easiest to draw are the lowest-energy and highest-energy orbitals, so let’s do those first.

What about the middle orbital, π2 ? It should have one node, and be intermediate in energy between π1 and π3

Where to put that single node? If you’re like I was, your first temptation for drawing the orbitals will likely be to put a node between one of the carbons, giving orbitals that look something like this:

It’s actually incorrect!

Why? That’s a good question, and the answer isn’t obvious. It turns out that because of the way the math works in the Schrodinger equation, the nodes are always “balanced” with respect to the centre of the orbital. The node can’t just be placed anywhere; when there is one node, it must be in the centre. [By the way, this is a general feature of MO’s where the number of contributing orbitals is odd, e.g. N=3, N=5, N=7… the first node will be on the central carbon.]

That means that the node is directly on that middle carbon.

Weird, right? What does that actually mean? It means that in that molecular orbital there is zero electron density on that central carbon. We don’t need to go into further detail for our purposes, but I’m including a note where it gives us some clues about reactivity. [note 2]

Now, if we arrange these orbitals together by energy (most stable at bottom) we get the following orbital diagram for the allyl system.

This is just the framework for the orbitals in the allyl system. It’s not specific to the allyl cation, radical, or anion yet, because those three species differ in the number of Pi electrons they possess.

Now, let’s populate the orbitals with electrons.


Molecular Orbitals of The Allyl Cation (2 pi electrons)

In the allyl cation we have 2 pi electrons in total: in the resonance form below, we see two electrons in the pi bond, and zero on the allyl carbon.

When we populate our allyl orbitals with two electrons, building from the ground floor up, we get the following:

Note that the “highest occupied” molecular orbital of the allyl cation is π1 – the lowest-energy orbital in the system. The “lowest unoccupied” molecular orbital is π.

We generally abbreviate the terms “highest occupied molecular orbital”  as HOMO and “lowest-unoccupied molecular orbital ” as LUMO. They are often called the “frontier” molecular orbitals  and are where most of the action happens in reactions, as we’ll see in future posts.

Molecular Orbitals of the Allyl Radical (3 pi electrons)

What about the allyl radical? Here we have 3 pi electrons: two in the pi bond, and a third one in that half-filled p orbital.

Where do we put that third electron? In the next-highest orbital, of course. Into π2 it goes!

That gives us a half-filled π2, which is our new “HOMO”, and πnow becomes our “LUMO”. [I don’t want to add to the confusion, but sometimes half-filled molecular orbitals like this are referred to as a “singly occupied molecular orbital”, or SOMO. Not super-important at this stage]. 

Molecular Orbitals of the Allyl Anion (4 pi electrons) 

What about the allyl anion? Here we have 4 pi electrons: two in the pi bond, and two more as a lone pair on the terminal carbon.  This fills up π.

Our HOMO remains πand our LUMO remains π3 .

This satisfactorily shows the molecular orbitals for the allyl cation, radical, and anion.


Here, we’ve shown how to extend the general principles of molecular orbitals we learned in building up the molecular orbitals of a pi bond (N=2)  toward the allyl system (N=3). The key points are as follows:

In the next post, let’s look at the molecular orbitals for the butadienyl system (N=4).

Note 1. The rule of thumb that for “the Nth molecular orbital will have N-1 nodes” applies best to linear molecules. Cyclic molecules (such as benzene) follow the same general principles, but due to symmetry will have “nodal planes”. For example the lowest energy MO of benzene has zero nodes, but the next-highest energy level of benzene is “doubly degenerate” meaning that there are two ways to draw a single nodal plane. More here. We’ll cover this in a separate post.

[An advanced note: one consequence of the node being in the centre of π2 is that  the radical has zero electron density in the middle carbon (C-2). This is consistent with everything we’ve ever observed about the reactivity of allyl radicals – they react at the termini (C-1 or C-3), and never the middle carbon (C-2). ]

Final note. Here’s a final (advanced) footnote, that only the extremely curious should look at. It’s showing how the 3 molecular orbitals of the allyl system are formed through interaction of two Pi molecular orbitals with a p orbital. This results in lowering of the energy of the  π orbital (resulting in a stabilization of energy E) and raising of the energy of the  π* orbital. The energy of the p orbital remains unchanged; intermediate in energy between bonding and antibonding, it is referred to as a “nonbonding” molecular orbital.

Adapted from Ian Fleming’s excellent “Frontier Orbitals and Organic Chemical Reactions”.


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