Orbital Diagram:4-atoms- 1,3-butadiene - Online Tutor, Practice Problems & Exam Prep

In 4-atom π-conjugated systems, such as dienes, molecular orbital diagrams illustrate the arrangement of electrons in bonding and antibonding orbitals. For 1,3-butadiene, the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) can be identified, with 4 electrons filling the bonding orbitals. The notation for these orbitals can express them as combinations of π orbitals, aiding in understanding their structure. Recognizing these frontier orbitals is crucial for grasping molecular stability and reactivity in organic synthesis.

Drawing MO Diagram for Dienes

Hey guys. In this video we're going to learn how to draw the molecular orbital diagrams for 4-atom pi conjugated systems. So guys, 4-atom pi conjugated systems are usually in the form of diene, so 2 double bonds next to each other. And just so you know, a diene can also be more generalized to be called a polyene. So I'm going to be referring to anything that's conjugated with 4 or more atomic orbitals as being some form of polyene because that's usually the way that it works. Okay? And we know that polyenes can resonate. Okay? But what we're going to do here is we're going to learn how to draw the molecular orbitals for a diene, a 4-atom conjugated system using the rules of molecular orbitals that I've already shown you guys before. So nothing's going to change, this is just an application. Okay? So let's go ahead and do this example. Okay? So it says predict the LCAO model of 1,3-butadiene, identify the HOMO and LUMO orbitals. Very cool. So, let's go ahead and just start off with the basics here, which is that we have 4 atomic orbitals and we have 4 out of 4 electrons in those orbitals because we have 2 pi bonds. That makes sense, right? And we know that according to the Aufbau principle, that means that 2 electrons are going to sit in Ψ₁ and 2 electrons are going to sit in Ψ₂, okay? What we might not remember is how to actually draw the molecular orbital. So let's go ahead and do that now. So, what would be the next thing that we do if all of our orbitals are already drawn for us? We just have to draw them in the phases, what's the next thing to do? Let's go ahead and shade the first orbital. So the first orbital should not change. I should do this, this, and this. Cool. Awesome. So that's our first step. The next step is let's go to the last orbital and start flipping it because you know that it has to flip every time, so this one has to go up and then down and then up again. Cool. Awesome. And now we need to increase our nodes. Okay? So now we have to increase our nodes. So we have we're starting off with 0 nodes here, then it needs to be 1, 2, 3. So the way we're going to do this is try to add these nodes in the most symmetrical way possible. So that means that I'm going to put 1 node in the middle here, a node here, a node here for 2, and then finally for 3, it's just a node in between every single orbital. Cool and that means that now I'm ready to shade the other orbitals in. And what that means is that I would get one phase change here, right, because that's just the only node, then I would get 2 phase changes for this one. So I get the first phase change and the second phase change and then the last one I would just get all of them are changing. Very cool. Awesome. So now we know what our molecular orbitals look like, we know how many electrons there are, and now we just have to indicate which ones are the HOMO and LUMO orbitals, okay. And, what we would have, first of all, what I have for this section here is just how many pi electrons we have in each one. So then this one would be 2 pi electrons, 2 pi electrons, 0 pi electrons, and 0 pi electrons, okay. Now you don't have to do this every time, but since we're new at looking at HOMO and LUMO I just want to make it very clear which of these have electrons and which of these don't and we know that according to the ordering it would only be up to Ψ₂, right. So that means that my HOMO orbital or the highest occupied one must be Ψ₂ and that means that my LUMO orbital must be Ψ₃ because that is the lowest energy one that doesn't have any electrons in it. Once again, these are collectively known as your frontier orbitals. You may not know what they do yet, but it's important that you're able to identify them. Cool? So that's it for this video. In the next video, I want to talk about a specific type of notation that's used for dienes.

Alternative MO Notation for Dienes

So guys, it turns out that specifically for 4 atom conjugated systems, for dienes, a lot of textbooks and some professors will actually describe molecular orbitals as the sum of pi orbitals. So they'll think about it as, you know, you have 4 molecular orbitals, and those orbitals are actually the sum of different combinations of 2 orbitals. Okay? This is only true for 4 orbital systems. In case you see it, I want you to know first of all that it's not very important. So this type of notation doesn't supersede what I already taught you. You already know how to do 4 orbital systems or 4 atomic orbital systems, but I just wanted to go ahead and do an example with you so that in case you see it, you know what you're looking at. So this is the way that it works. So basically you know how usually we start with our dark lobes facing down on our first orbital, that's just the way that I like to do it. So that is usually, if that's your starting point then that would be called your positive pi because it's just the way that you started. Okay? If you were to then draw an antibonding orbital from there, remember that the first one has to stay the same, but the second one has to flip, right? So that would be called the positive pi star because it's like you started with a positive pi, but then you made it antibonding, okay? Well, just as correct of a starting point would be the negative pi. Negative pi just means that you flipped it the other way. That's totally fine. Some people want to use the negative pi as their starting point all the time. I just like to put the dark lobe down, but if you're using this one, this would be called negative pi. And then if you wanted to make the antibonding version of it, the antibonding would be negative pi star. Got it so far? Now we know that according, this is just basically looking at 2 different ethane molecules, we know that what would happen is that in you would have 2 electrons here and 2 electrons there, let's go up and look again at this. Your size, let's go up and look again at this, what we want to do is figure out how can we describe these 4 molecular orbitals in terms of the sum of 2 pi molecular orbitals. Okay? So for example, notice that for psi 1, I have 4 lobes all facing down, right? So that means that if I wanted to use this type of notation, what would be the right summation of these orbitals to make that thing? And I'm going to draw it for you right now and hopefully, you'll understand better. And I'll just draw it right here so we have a little bit of room. Should be equal to what? It should be equal to positive pi plus I should put it in brackets, positive pi. Positive pi. Why? Because the way that I drew it had both lobes at the bottom and it was just 2 of those. So it was a positive pi, plus another one of those. So they're next to each other, that would psi 1. Cool? Let's go ahead and do psi 2 now. What is psi 2 going to be? Well let's go ahead and look at it again for reference. So psi 2, it looks like the first two didn't change, but the second one did. So then what this one would be if we look at it, if you look at all the different possibilities, it's actually going to be positive pi again plus negative pi, right? Because in this case what we're adding, the second one that we're adding is the opposite configuration or the opposite phase, okay. The opposite phase of my bonding orbital would be this one. So that means that what Psi 2 is is actually it's you're adding a positive pi plus a negative pi. Cool? Let's keep going. Psi 3, notice that it appears to be 2, well, it appears to be 2 antibondings, right? Because notice that the first one has the phases different, the second one also has the phases different. Notice that the first one is starting off with the lobe down, the second one is starting off with the first lobe up. So let's see what it would be. That means that it would be positive pi star, right? Because that's the first one, plus negative pi star. Right? Because if you add those together, don't you get psi 3? It's just basically adding the top one there and the top one there and you would get psi 3. And then finally Psi 4 would be what? It would be that everything's changing. It actually looks like it's the same thing twice right? It's your antibonding, 1 twice. So what that means is that it would just be plus pi star. I'm so why? I kept saying psi, I meant pi. Positive? Yeah, I should have put a positive around it. So it would be a positive pi star. Does this make sense? Cool. So you might be saying, Johnny, why are you showing me this? Like why is this important? The truth of the matter is it's not important at all. This is just a specific type of notation that certain textbooks and certain professors like to use to represent a 4 atom conjugated system. Sometimes they might say, discuss Psi 3 in terms of pi and then you'd have to actually do this. But for the purposes of actually understanding molecular orbital theory, this doesn't provide any extra insight. It's just showing you all the different combinations of smaller molecular orbitals that could be used to make a bigger one. Okay? Cool guys. So I hope that this made sense. Let's move on to the next video.

Here’s what students ask on this topic:

What is the molecular orbital diagram for 1,3-butadiene?

The molecular orbital diagram for 1,3-butadiene involves four π atomic orbitals combining to form four molecular orbitals: two bonding (Ψ1 and Ψ2) and two antibonding (Ψ3* and Ψ4*). According to the Aufbau principle, the four π electrons will fill the lowest energy orbitals first, resulting in Ψ1 and Ψ2 being fully occupied. Ψ1 has no nodes, Ψ2 has one node, Ψ3* has two nodes, and Ψ4* has three nodes. The highest occupied molecular orbital (HOMO) is Ψ2, and the lowest unoccupied molecular orbital (LUMO) is Ψ3*.

How do you identify the HOMO and LUMO in 1,3-butadiene?

To identify the HOMO (Highest Occupied Molecular Orbital) and LUMO (Lowest Unoccupied Molecular Orbital) in 1,3-butadiene, first draw the molecular orbital diagram. The four π electrons will occupy the two lowest energy orbitals (Ψ1 and Ψ2). The HOMO is the highest energy orbital that contains electrons, which is Ψ2. The LUMO is the lowest energy orbital that does not contain electrons, which is Ψ3*. Recognizing these orbitals is crucial for understanding the molecule's reactivity and stability.

What is the significance of nodes in molecular orbitals of 1,3-butadiene?

Nodes in molecular orbitals are regions where the probability of finding an electron is zero. In 1,3-butadiene, the number of nodes increases with the energy of the molecular orbital. Ψ1 has no nodes, Ψ2 has one node, Ψ3* has two nodes, and Ψ4* has three nodes. The presence of nodes affects the energy and stability of the orbitals: more nodes generally mean higher energy and less stability. Understanding nodes helps in predicting the behavior of electrons in chemical reactions.

How do you draw the molecular orbitals for a 4-atom π-conjugated system?

To draw the molecular orbitals for a 4-atom π-conjugated system, follow these steps: 1) Draw the four atomic p orbitals. 2) Combine them to form four molecular orbitals with increasing energy. 3) Assign the number of nodes: Ψ1 (0 nodes), Ψ2 (1 node), Ψ3* (2 nodes), and Ψ4* (3 nodes). 4) Shade the lobes to indicate the phase changes. 5) Fill the orbitals with the appropriate number of π electrons, following the Aufbau principle. This method helps visualize the electron distribution and energy levels in the system.

What is the LCAO model for 1,3-butadiene?

The Linear Combination of Atomic Orbitals (LCAO) model for 1,3-butadiene involves combining the atomic p orbitals of the four carbon atoms to form molecular orbitals. The resulting molecular orbitals are Ψ1, Ψ2, Ψ3*, and Ψ4*. Ψ1 and Ψ2 are bonding orbitals, while Ψ3* and Ψ4* are antibonding. The four π electrons occupy the two lowest energy orbitals (Ψ1 and Ψ2). This model helps in understanding the bonding and antibonding interactions within the molecule.