Have you ever wondered why the Huckel's rule number of pi electrons, 2, 6, 10, 14, make molecules so stable? Okay. Maybe you weren't wondering that, maybe it wasn't keeping you up at night, but in this video, I'm going to finally explain why these numbers are so important, why they are so special, and we're going to learn a visualization technique that will help us to understand molecular orbital theory better when it comes to aromatic molecules. Okay? So this is called the inscribed polygon method. It also goes by a lot of other names. It's also called the polygon and circle method or a Frost Circle, okay? If you see any of those terms, it's all referring to the same exact thing. It's a method that helps us to visualize the identities of pi electrons, okay? And this is going to explain to us. It's going to be a visual representation of why our Huckel's rule numbers are so special, why they are so important. Okay? So what we are going to do is we are going to do a worked example of 3 different molecules and then we'll go ahead and, you know, do some practice problems. Okay? So here we have 3 different molecules. It says use the polygon and circle method to predict the stability of the following molecules. I'm going to go ahead and add some stuff here. So for example, with this nitrogen, let's go ahead and make this a hydrogen. Okay? So let's go ahead and make it a nitrogen with one lone pair and a hydrogen. The cyclobutadiene, we are keeping just as is. The triangle, it's got a longer name that I don't want to complicate you with. But, the 3-membered ring has a negative charge. That's fine. So, what we're going to do is we are just going to do kind of steps and I am going to do all the steps with all 3 molecules so you can see how they work. The very first step of this method, whether you call it a Frost Circle or polygon and circle, your first method is your first step is always going to be to draw the polygon, whatever shape it is, with one corner facing down. So as you can see, I've already done that step. I've already drawn it every single shape with one of these corners facing down. Now, if you were given the shape like this, then it would simply be your responsibility to redraw it in a way that one corner is at the bottom. Think of it like it's standing on its very tip. Okay? Now, what we're going to do in the next step is you have to draw molecular orbitals on all corners of the ring. Guys, that sounds a lot harder than it is. It just means draw a line next to every atom of the ring. Okay? So, a 5-member ring gets 5 molecular orbitals. 4-member ring gets 4. 3-member ring gets 3. It's that easy. Okay? Now, we're going to draw a line that splits the polygon down the middle in a basically, you know, a horizontal line that's going to split it into 2 different halves. Okay? So, now for the square, that happens to be easy because a halfway line would just go right through the middle of both of those corners. Now, for a triangle, that's also easy because you're just going to put the dotted line somewhere in the middle, but now for a 5-membered ring or sometimes bigger rings, sometimes it can get confusing where to put the line. Obviously, when it's drawn like this, the line is going to go above 3 molecular orbitals and below 2 molecular orbitals. But some students because they're really bad at drawing, they just suck at drawing. I don't blame you. I used to be one of you. I had to get good because this is like my job now. Some students make the mistake of doing this. They like go below that molecular orbital. And you'd be surprised if you draw, let's say you draw your 5-member ring like this, No, like this. Okay? No, I'm messing up. But there's a way to draw it. There you go. So let's say you draw your 5-number ring like this. Well then when you draw the halfway point, you're going to think that it's actually below 4 orbitals and above only 1. But that's not the way it should be. You should If you're splitting with an uneven number of orbitals, you should make it so that they're as close together as possible. So instead of being 4 orbitals on top and 1 orbital on the bottom, it should be 2 on the top and 3 on the bottom like we have here. So hopefully, that kind of helps. If you ever see a completely unequal number of orbitals, that means you probably drew it wrong. You should have a relatively even number of orbitals on both sides. Okay? Excellent. So, we drew the line. Now you're going to insert the number of pi electrons that you have into your orbitals starting from your lowest energy orbital and working your way up, guys. This is called the Aufbau principle, right? Aufbau principle was the building up principle, just means that you have to always fill your lowest energy orbitals first. And we're saying that energy goes up. So energy increases the higher you go. So, let's just look at the 1st molecule. The 1st molecule is one that we already learned how to solve. It's a heterocycle. How many pi electrons does that molecule have? You have to think, will the nitrogen donate? Does it want to donate its lone pair? Yes, it does. Okay? Because we've got 2, 4 electrons, 4 pi electrons, that lone pair is going to make it 6. It's sp3, so that works out. So that means I have 6 electrons to add. How do we add them? 12 here. Then my 3rd electron goes here. My 4th electron goes here. Okay. That's a whole other rule. Okay. That was actually Hund's rule. Hund's rule says that you can't, if you have orbitals of the same energy level, you have to fill them evenly. Okay? So, your 3rd and your 4th electrons go 1 a piece. But we have 6 total. So then my 5th goes here and my 6th goes there. That's it. So I'm done with that step. Let's move on to cyclobutadiene. How many pi electrons do I have for that one? Four. Okay. So it's going to be 1, 2, 3, 4 according to Aufbau principle and then according to Hund's rule I have those 2 even, energy orbitals, so I have to fill them evenly. Okay? Now, just so you guys know, there's a more technical term for orbitals that have the same energy level. Do you guys remember that name? That term, it's actually from chapter 1 of organic chemistry. They're called degenerate orbitals. Okay? So if you have degenerate orbitals, that means that you have to fill them evenly. Okay? That's it. You can't just put 2 on one side and 0 on the other. That doesn't make a lot of sense. Finally, what do we get for the triangle? Okay. The cyclopropenyl anion. So what we would get is 1, 2, 3, 4, right? Because again we've got 4 pi electrons and so we've got to have that even distribution at the top. Perfect. So what did we just do? You might be wondering, okay, Johnny, we drew all these arrows, but where is this going? Well, it turns out that we can use this diagram to understand why molecules are more stable or less stable, and we can understand the identities of the electrons and the identities of the orbitals. Because it turns out that that halfway point actually represents what we call the nonbonding line in the molecular orbital theory, where everything below that line represents a bonding molecular orbital. Okay? So bonding molecular orbitals are the first ones to get filled. They're the ones that contribute to bonding. They're the ones that contribute to things wanting to stay together. And then our antibonding orbitals are the ones that get filled up after all the bonding ones are full. Okay? You would never put something in the top orbital until all of your bottom ones are full. Okay? Well, it turns out that when you have filled molecular orbitals, remember molecular orbital would just be one of these. Right? If all of your molecular orbitals are filled, that's going to contribute to unique stability because remember that we learned a long time ago from chapter 1 of organic chemistry and from gen chem gen chem that orbitals love to be have 2 electrons. If an or if an orbital has 2 electrons, it's 2 electrons, it's happy. Okay? Now, what if all of the bonding orbitals have 2 electrons? That's going to make it uniquely stable because that means that basically all of your bonding orbitals are perfectly filled. Okay? Now, what if you have partially filled molecular orbitals? What if you have a weird number of electrons and you have some molecular orbitals that are hanging out with 1 electron apiece, that's going to contribute to unique instability, guys. That's going to make it unstable because now you have these unfilled orbitals that are trying to get filled with something. No orbital likes to only carry 1 electron. They always want to carry 2. Okay? So what does this mean back to our diagrams? Well, look what's going on. Normally, we would predict that what would be the aromatic of these molecules. Well, the first one would be aromatic. The second one is supposed to be antiaromatic. The last one is also supposed to be antiaromatic. Now are you seeing a pattern here? Notice, look what's going on. The aromatic molecule, the one with 6 pi electrons, Huckel's rule number happens to have all of its bonding orbitals perfectly filled, right? That's going to make it stable. That's going to make it really stable, right? Whereas the ones that are antiaromatic, the ones that have a non-Huckel's rule number or what we call Breslow's rule number, notice that they have this. What is that? They have these partially filled orbitals. Partially filled. Do you think that's going to make it stable? That's going to make it uniquely unstable. Look at these numbers. The electrons here are 4 electrons, the electrons over there for the first one is 6. So guys, the reason that the Huckel's rule numbers of 2, 6, 10, 14 are so stable is because those are the combinations of electrons that are always going to be required to perfectly fill your bonding orbitals. Any number off of that, an odd number. Right? Or instead of an odd number, a 4n number. These are going to kick electrons up to the antibonding orbitals and make unfilled partially filled orbitals that are going to contribute to instability. Okay? So hopefully, this helps you guys get a better grip on, wow, this wasn't just like a memory game. This actually made sense. These numbers are important. There's a reason that benzene is so stable and it's because this example here is benzene. It's because benzene has 6 pi electrons so all of its bonding orbitals are perfectly filled. Alright? So, you guys are going to do some practice so you guys can get a better idea of this, but hopefully that makes sense. Let me know if you have any questions. Let's move on to the next topic.
Frost Circle - Online Tutor, Practice Problems & Exam Prep
Inscribed Polygon Method
Video transcript
Inscribed Polygon Method
Video transcript
Hey guys, let's take a look at the following question. So here, apply the polygon circle method, which we also call Frost Circles, to the following compound. Does it show any special stability? If yes, why? So basically, we have to determine if this molecule is aromatic, non-aromatic, or anti-aromatic. Aromatic shows very high stability. Non-aromatic just means your normal compound. Anti-aromatic means you're super unstable. So if we take a look at the name, we have tropylium cation. Now, another name for tropylium just means that we have cycloheptatriene. The tropylium just means cycloheptatriene. Cyclo means ring. Hepta means 7 carbons. So, we have 7 carbons and triene means 3 double bonds. Now, I say the word cation, which means this carbon up here which is not double bonded like the others is positive. That will be our molecule. Now, here we're going to draw the frost circle. Now, we're going to draw an apex down the molecule, so the point part down. The larger these rings get, the harder it becomes to draw them. It takes practice, guys, to make sure you draw it correctly. So not the straightest but good enough. Now, we're going to cut this in half. So here, this is your nonbonding region. Down here is your bonding. Up here is your antibonding. Everywhere a carbon touches becomes a molecular orbital. Now, we're going to say how many pi electrons do we have? We have 2, 4, 6 pi electrons. So 1 up, 1 down. Up, up, down, down. We're going to say this molecule shows special stability because all the molecular orbitals in the bonding region are completely filled in, which would indicate that we have an aromatic compound. So tropylium cation is aromatic. Here we know it from the old rules that we talked about, but we also know it because we just did a frost circle and it proves it. The molecular orbitals in the bonding region are all completely filled in.
Do you want more practice?
More setsYour Organic Chemistry tutors
- Draw a Frost circle for the cyclopropenyl anion and compare it to the Frost circle for the cyclopropenyl catio...
- Draw a Frost circle for the cyclopentadienyl cation and compare it to the Frost circle for the cyclopentadieny...
- (•) Use a Frost circle diagram to construct the molecular orbital diagram for the molecules shown. Would you e...
- (••) Using a Frost circle, a student drew the molecular orbital picture shown for the cyclopentadienyl anion, ...
- Following the instructions for drawing the energy levels of the molecular orbitals for the compounds , draw th...
- Following the instructions for drawing the energy levels of the molecular orbitals for the compounds , draw th...
- Following the instructions for drawing the energy levels of the molecular orbitals for the compounds <IMAGE...