Orbital Diagram:5-atoms- Allylic Ions - Online Tutor, Practice Problems & Exam Prep

Understanding molecular orbital diagrams for a 5-atom conjugated system is crucial for predicting reactivity. The linear combination of atomic orbitals (LCAO) model helps identify bonding, non-bonding, and antibonding orbitals. Nodes must be symmetrically placed, with the number of nodes increasing with each orbital. The non-bonding orbital is often the most reactive, as it neither stabilizes nor destabilizes the molecule. Recognizing these patterns aids in grasping concepts like resonance and molecular stability, essential for advanced organic chemistry.

Drawing MO Diagram

Alright guys, in this video I'm going to show you how to draw the molecular orbital diagrams for a 5 atom conjugated system. Let's go ahead and get started. So like propionyl ions, those 3 atom atom conjugated systems, 5 atom conjugated systems have the ability to resonate reacting in multiple locations. Okay? Same idea. And regardless of the identity of the ion, no matter whether it's a positive charge or a negative charge or anything in between, you can explain the reactivity of the molecule using molecular orbitals. Okay? So what we're going to do here is the number one thing we're going to do is we're going to draw the molecular orbitals for a 5 atom conjugate system because guess what? It's a little bit confusing, and if we don't go through it together, you may not know how to do it, okay? So predict the LCAO model of the 5 carbon system, identify the bonding, non-bonding, and anti-bonding orbitals. Cool.

So guys, what we're basically saying here is that we don't know what type of molecule or what type of ion this is. It could be 0 electrons, 1 electron, or 2 electrons. Okay? But that is going to go into the 5th atomic orbital, and we're going to figure out what to do with that later. For right now, our main job is to fill in what the atomic orbitals look like. Okay. So I've already given you the 5 molecular orbitals; now we just have to fill in the phases, okay. So what do you think is a good place to go from here? Like what's a good starting point? Let's go ahead and fill in the 1st atomic orbital in each and keep it exactly the same. By the way, the rules have not changed at all. I'm just showing you how to apply them to a 5 atom system. Cool.

So then over here, we should probably start flipping the last one back and forth. Flip, flip, flip, and flip. Not too bad. So the only ones we really have to decide are these 3 in the middle, okay. Now what we need to do is start putting in our nodes. Just to remind ourselves, we should have 0, 1, 2, 3, and 4 nodes. Okay? They have to keep increasing. So let's start off with our first one. Where do you think is the right place to put a node if I can only put 1 and it needs to be symmetrical? It needs to be in the middle. Okay? So that's going to be our first node; that means we're going to ignore that orbital; it's going to get deleted. Cool, so let's go to 2. Okay, so where can we put 2 nodes in a place that's symmetrical? Where can we do it? So guys, this is a little bit tricky, but it's going to make sense when I say it. You have to put it on orbital 2 and orbital 4, right? That's the only way to keep it symmetrical, which means that in this one, we're going to delete 2 orbitals. Okay? Cool. Let's go to the next one. So for the next one, we need to put in 3. What do you think?

So the place to put in 3 would be, first of all, we have to put 1 in the middle, right? Because the middle is an odd number, so you have to put 1 in the middle for sure. Okay? And then the other places that you would put them would be here and here, okay? Now those are the only logical places to put them. And the reason is if you were to put one next to the one in the middle, well then, that wouldn't be very symmetrical. That would just be nodes in one place, right? You want to keep them evenly spaced. And let's say that you were to cover up these nodes here, well then what would happen once again is that you have many nodes next to each other and then not many nodes here. So notice that what we're trying to do is keep our nodes evenly spaced. So by evenly spacing it out, we would have a node delete one, and then a node again. Okay. Cool. And then finally, the last one should always be easy, guys, because it means that every single position gets a node. Okay? So see how, I hope, the rules didn't change, but I hope it helped seeing how I think about it, okay? And doing it together.

So what that means is that we're now going to fill in our other nodes; this one is down and this one is up. That's our one phase change. For this one, it's going to be up as the other 2 are deleted, right? So for this one, I do a phase change up, ignore the middle one, and then a phase change down, and, finally, for this one, they're all flipping. Cool. Awesome guys. So now we just filled in our molecular orbitals, and now we just have to put in the pi electrons into the orbitals. So we know that for sure you would get 2 in Psi 1, you'd get 2 in Psi 2, and whatever the identity of that 5th atom is, whatever's in here is going to go right here.

Now, what do we see that's interesting about Psi 3? What's the shape of Psi 3? Now by the way, this is going beyond what the question was asking, but I just think it's a really cool application of what we know about molecular orbitals. What are the places that Psi 3 could react? It has orbitals only at position a, position c, and position e. Right? Notice that position b doesn't exist and position d doesn't exist. Okay? And that explains why whatever ion this is, let's say it's a positive charge, it can only resonate here and here because those are the only places where electrons could actually interact with other electrons since those are the only orbitals that do not have nodes. Isn't that cool? Awesome.

So you guys are becoming molecular orbital freaks, you guys are totally getting this. Now we do have to do our last part about the orbitals. So guys, there's a little bit of notation that's missing, right. So, first of all, this is bonding; this is also bonding. Anything that's below the 50% mark is bonding. Anything that's above it is anti-bonding, so that means this is anti-bonding, and this is anti-bonding, right? And that also means we should add stars here because these stars are missing. Okay? And then what is psi 3, because it's right at that halfway point? Anytime that you're at the halfway point, that is considered non-bonding, and it neither adds nor subtracts from stability. The non-bonding one is usually the reactive one. It's the one that can react with other ions and other molecules, and that's really where the fun stuff happens. It doesn't make it more stable to have electrons there, it doesn't make it less stable to have electrons there. It's just basically the place where the reactions happen most of the time at the non-bonding position. Cool? Awesome guys, and there's no special notation for it, okay? Awesome guys, so we're done with this problem, let's move on to the next video.

Predict the molecular orbitals and identify the HOMO and LUMO orbitals of the following cation.

Predict the molecular orbitals and identify the HOMO and LUMO of the following cations. So, guys, this should be pretty fast because we've been practicing this quite a bit. So, what I have is 5 atoms that are conjugated, this happens to be a cation, so that means I'm going to do 1212 with nothing in the middle because that's where my cation is, okay? Cool. Now, we have to fill in all of our orbitals but thankfully we just did this so this should be fresh on your mind. Let's shade in the bottoms for number 1. Let's shade in all the bottoms of orbital 1. Let's flip the last orbital over and over again, flip, flip, flip flip, and let's enter with our nodes. Our nodes would go 1, 2, 3, 4. Cool. And now we can start shading everything else, meaning that it's going to do this and this, just this, this and this, and then all these are going to flip. Not so bad, right, you guys are becoming pros. Let's go ahead and fill in our molecular orbitals. So we know there should be 4 electrons, 1, 2, 3, 4. And that means that I can now identify my HOMO, well, first of all, I just want to point out that means that the positive charge actually resides here in side 3, so these are the orbitals that it will react with and that also corresponds with the positions that it will react with. And, I also want to point out the HOMO and the LUMO. So, this is my HOMO side 2 and this is my LUMO side 3. Not too shabby. Thanks, guys. You're doing great. Let's move on to the next video.

Here’s what students ask on this topic:

What is the molecular orbital diagram for a 5-atom conjugated system?

The molecular orbital diagram for a 5-atom conjugated system involves five molecular orbitals formed from the linear combination of atomic orbitals (LCAO). These orbitals are labeled as Ψ1, Ψ2, Ψ3, Ψ4, and Ψ5. The number of nodes increases from 0 in Ψ1 to 4 in Ψ5. Ψ1 and Ψ2 are bonding orbitals, Ψ3 is a non-bonding orbital, and Ψ4 and Ψ5 are antibonding orbitals. The placement of nodes must be symmetrical, and the non-bonding orbital (Ψ3) is often the most reactive. This diagram helps predict the reactivity and stability of the molecule.

How do you determine bonding, non-bonding, and antibonding orbitals in a 5-atom system?

In a 5-atom conjugated system, bonding orbitals are those with fewer nodes and lower energy, typically Ψ1 and Ψ2. Non-bonding orbitals, like Ψ3, have an intermediate number of nodes and are at the midpoint of the energy scale. Antibonding orbitals, Ψ4 and Ψ5, have the highest number of nodes and energy. The placement of nodes must be symmetrical, and the number of nodes increases with each orbital. Bonding orbitals stabilize the molecule, antibonding orbitals destabilize it, and non-bonding orbitals neither stabilize nor destabilize but are often the most reactive.

Why is the non-bonding orbital in a 5-atom conjugated system often the most reactive?

The non-bonding orbital in a 5-atom conjugated system, typically Ψ3, is often the most reactive because it is at the midpoint of the energy scale. It neither stabilizes nor destabilizes the molecule, making it a neutral site for reactions. This orbital has nodes that do not interfere with electron interactions, allowing it to participate in chemical reactions more readily. The non-bonding orbital's reactivity is crucial for understanding the molecule's behavior in various chemical processes.

How do you place nodes symmetrically in a 5-atom molecular orbital diagram?

To place nodes symmetrically in a 5-atom molecular orbital diagram, start with the lowest energy orbital (Ψ1) with 0 nodes. For Ψ2, place 1 node in the middle. For Ψ3, place 2 nodes symmetrically at positions 2 and 4. For Ψ4, place 3 nodes with one in the middle and the other two evenly spaced. Finally, for Ψ5, place 4 nodes, one at each position. Symmetrical placement ensures the correct energy levels and stability of the orbitals.

What is the significance of the LCAO model in understanding 5-atom conjugated systems?

The Linear Combination of Atomic Orbitals (LCAO) model is significant in understanding 5-atom conjugated systems because it helps predict the formation of molecular orbitals from atomic orbitals. By combining atomic orbitals, the LCAO model identifies bonding, non-bonding, and antibonding orbitals, which are crucial for understanding the molecule's reactivity and stability. This model aids in visualizing how electrons are distributed in the molecule and how they interact during chemical reactions.