Orbital Diagram:3-atoms- Allylic Ions - Online Tutor, Practice Problems & Exam Prep

In a three-atom conjugated system, the allylic position, adjacent to a double bond, can resonate, allowing for molecular orbital explanations of reactivity. The molecular orbital theory indicates that the bonding molecular orbital (psi 1) holds the most stable electrons, while the non-bonding orbital (psi 2) has a node at the allylic position, preventing reactivity there. This illustrates why resonance structures cannot depict movement to the middle carbon, emphasizing the importance of understanding molecular orbitals in predicting chemical behavior.

Orbital Diagram

Hey guys, in this video, I'm going to show you how to draw molecular orbital diagrams for 3-atom conjugated systems. Let's get started. So guys, because a 3-atom conjugated system is an odd number of atoms, what we're usually going to end up dealing with is an allylic position. Remember that the allyl position or an allylic position means it's the position next to a double bond, and that's usually the case when you have an odd number of atoms because many times there's a double bond and then there's something next to the double bond. That's like the 3rd atom. Okay? Now what we know about allylic positions from previous lessons is that allylic positions are able to resonate. Remember that the allylic position can resonate from one side of the double bond to the other. Okay? So it turns out that the whole idea of resonating an allylic position can actually be explained through molecular orbitals. You already know how to explain it using resonance. You should be able to draw resonance arrows and be able to show how the allylic position can move from one side to the next. But it actually turns out that the fundamental explanation of an allylic position reaction can actually be explained through molecular orbital theory. So that's what we're going to do right here. I'm going to show you guys how molecular orbitals explain the reactive positions of an allylic ion. So here we go. It says simplified LCAO model of a propanyl ion and here what I've drawn is a general propanyl ion. Basically we've got our 3-atom conjugated system. So we have 3 atomic orbitals, like let's just say a, b, and c, right. And what we see is that we have a double bond that we know is going to contribute 1 pi electron for each atom, so we have 12. And then we have this unknown ion or this unknown non-bonding orbital in position C and that I've labeled with a question mark. And the reason is because what I'm about to explain to you applies to any ion, no matter what the identity of that question mark is. Whether it's empty, whether it's an empty orbital or whether it's a radical or whether it's a lone pair. It doesn't matter. What I'm able to teach you applies in all those situations and what the molecular orbital theory would say about this molecule is that how would you structure the molecular orbitals? Well, it says that you would have, first of all, a molecular orbital with 0 nodes, 1 node and 2 nodes, and that we would put those pi electrons in order of Aufbau principle. So we would put the 2 electrons from that first double bond into psi 1, right, which is the most stable the bonding orbital. And then what we would do is we would put whatever is left over, whatever is here, whether it's 0 electrons, 1 electron, or 2 electrons, they would go into psi 2. Does that make sense so far? No matter what the question mark is, we know it's going into psi 2. Now what is unique about psi 2? Look at psi 2. Notice that psi 2 actually has a node at atom B. If this is atom A, atom B and atom C like we had before, we had a, b, and c, right? Notice that there's a node at atom B. What does that mean? What that means is that regardless of the identity of that ion, it cannot react at position B. It can only react at either position A or at position C, but because no electrons ever pass through atom B, you will never find the allylic position reacting there. So what this does is it explains the theory behind the resonance structures that we've learned how to draw for so long. Remember that an allylic position ion can resonate to the other side of the bond, but it can never resonate to the middle. You've never seen that, for example, this, let's say this is a positive charge, you can't move it to the middle. You can only switch places with the double bond, but you can't move it to the middle. And the reason is that the molecular orbital theory states that there's no way that that ion can react at the middle carbon because that orbital doesn't even have any electrons. It can only react at the orbitals with electrons, which would either be orbital C in this case or orbital A in this case, but never B. Isn't that so cool? So what we're going to do next is an example.

Explaining Resonance through MO Theory

Use both resonance theory and MO theory to predict the reactive sites of the following radical. So guys, first of all, let's do the resonance one because that's the one that we should know the best. You should already know this from prior lessons. So, how does a radical resonate? Do you remember, does it resonate with 1 arrow, 2 arrows, or 3 arrows? 3. Three of the fish hook arrows. So what we would expect is that you have a resonance structure that looks like this. Fish hook, fish hook, fish hook. Does that ring a bell? So let's go ahead and draw our resonance structure and what we would expect is that now I would get a new double bond here and a new radical here. Makes sense? Cool. We would put this in brackets and then using resonance theory, which would be the reactive sites of that radical? Where would be the places that the radical could react? It would be here and here. Okay, those are the 2 positions that can react as a radical. So if we were to call this atom A, atom B, and atom C of the pi conjugated system, right, of the conjugated system, what we would say is that resonance predicts radical character at positions A and C. Right? Resonance predicts, there's an s there if you can't see it, radical character at positions A and C. Those are the positions that could react as radicals.

Now let's go ahead and actually draw out the MO diagram and see if the MO diagram also predicts it. So, how many atomic orbitals do we have? 3, right? Because we have 3 atoms that are conjugated, that means I'm going to draw 3 energy states. So this is going to be 1, 2, 3. Let me try to make it a little bit more even. There we go. Okay. So we know this is going to be Psi 1, Psi 2, Psi 3. And how many electrons in total are there? There are 3. There's 1 electron from the radical A and then there's 1 pi electron from B and there's 1 pi electron from C. So one of these is a radical because it's not bound, it's not resonating with another lone electron, so it's called a radical, but the other ones are called pi electrons because they're inside of a pi bond. Cool. Awesome. So we have those 3 single electrons. Now we have to draw our energy states. So I'm going to try to do this quickly. And some of you may already have these memorized by now because we've done this a few times, but if not that's fine, we're going to go through the rules again. So how do we predict where our lobes are, the phases? So remember the first one, let's just draw them all with the dark facing down. Remember that your next rule says that you would then draw your first one the same in every single energy state. Remember that the next rule says that you would flip the other one every time, flip flip and then we would increase the number of nodes. So we have 0 nodes on the first one and then one node on the second, meaning that I should put a node right down the middle. And then for the last one I should put a node here and a node here, which would be 2. Lastly, we have one node that needs to be deleted because I'm sorry, one orbital needs to be deleted because a node passes through it. And there you go, that's our MO diagram. Now we just have to fill it in with electrons. And what we would do is according to Aufbau principle start with and then Pauli exclusion says you can only put 2, so we're going to have to put another one up here. So what does this MO diagram tell us? What it tells us is that the radical is only going to react in 2 places and not 3. It's only going to react at orbital A and at orbital C, but it cannot react at orbital B because orbital B is a node. So, by definition, mathematically, no electrons can react at that position because it can't have any electrons. Isn't that cool? So MO theory predicts that the radical can only react at positions A and C but not B. I'm just being extra clear there because, that is a unique learning that you see once you draw the molecular orbital diagram. And really guys, when it comes down to it, Molecular Orbital Theory is the underlying theory that provides us with the knowledge of reactivity of lots of reactions. Resonance was one way to, to kind of, it's a shorthand to draw it, but really the true theory lies in the MO theory. So it's really cool that I'm getting to show this and getting to share this with you guys, and hopefully help you understand it.

Cool. So that's it for this video, let's move on to the next one.

Consider the MO's of following allyl cation. Which of the following are HOMO and LUMO?

• A.
• B.
• C.
• D.
• E.

Here’s what students ask on this topic:

What is an allylic position in a three-atom conjugated system?

An allylic position in a three-atom conjugated system refers to the position adjacent to a double bond. In such systems, the allylic position is crucial because it can participate in resonance, allowing the electrons to delocalize over the system. This delocalization can be explained using molecular orbital theory, where the allylic position can resonate from one side of the double bond to the other, but not to the middle carbon due to the presence of a node in the non-bonding orbital (ψ2).

How does molecular orbital theory explain the reactivity of allylic ions?

Molecular orbital theory explains the reactivity of allylic ions by describing the distribution of electrons in molecular orbitals. In a three-atom conjugated system, the bonding molecular orbital (ψ1) holds the most stable electrons, while the non-bonding orbital (ψ2) has a node at the middle carbon (B). This node means that electrons do not pass through this position, preventing reactivity there. As a result, the allylic ion can only react at the terminal positions (A or C), aligning with the observed resonance behavior.

Why can't the allylic position resonate to the middle carbon in a three-atom conjugated system?

The allylic position cannot resonate to the middle carbon in a three-atom conjugated system because the non-bonding molecular orbital (ψ2) has a node at the middle carbon (B). This node indicates that no electrons are present at this position, preventing any reactivity or resonance there. Consequently, the allylic ion can only resonate between the terminal positions (A and C), but never to the middle carbon.

What is the significance of nodes in molecular orbitals for allylic ions?

Nodes in molecular orbitals are regions where the probability of finding an electron is zero. For allylic ions in a three-atom conjugated system, the non-bonding orbital (ψ2) has a node at the middle carbon (B). This node is significant because it prevents any reactivity at this position, as no electrons can be found there. This explains why the allylic ion can only react at the terminal positions (A or C) and not at the middle carbon, aligning with the observed resonance structures.

How do you draw a molecular orbital diagram for a three-atom conjugated system?

To draw a molecular orbital diagram for a three-atom conjugated system, follow these steps: 1) Identify the atomic orbitals involved (A, B, and C). 2) Combine these orbitals to form molecular orbitals: bonding (ψ1), non-bonding (ψ2), and anti-bonding (ψ3). 3) Fill the molecular orbitals with electrons according to the Aufbau principle, starting with the lowest energy orbital (ψ1). 4) Note that ψ2 has a node at the middle carbon (B), preventing reactivity there. This diagram helps explain the resonance behavior and reactivity of the allylic ion.