|Ch. 1 - A Review of General Chemistry||4hrs & 47mins||0% complete||WorksheetStart|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete||WorksheetStart|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete||WorksheetStart|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 18mins||0% complete||WorksheetStart|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete||WorksheetStart|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete||WorksheetStart|
|Ch. 8 - Elimination Reactions||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete||WorksheetStart|
|Ch. 10 - Addition Reactions||3hrs & 8mins||0% complete||WorksheetStart|
|Ch. 11 - Radical Reactions||1hr & 57mins||0% complete||WorksheetStart|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 34mins||0% complete||WorksheetStart|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete||WorksheetStart|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete||WorksheetStart|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 18mins||0% complete||WorksheetStart|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete||WorksheetStart|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete||WorksheetStart|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 26mins||0% complete||WorksheetStart|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete||WorksheetStart|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 59mins||0% complete||WorksheetStart|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete||WorksheetStart|
|Ch. 23 - Amines||1hr & 43mins||0% complete||WorksheetStart|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete||WorksheetStart|
|Ch. 25 - Phenols||15mins||0% complete||WorksheetStart|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete||WorksheetStart|
|Aromaticity||8 mins||0 completed|
|Huckel's Rule||10 mins||0 completed|
|Pi Electrons||5 mins||0 completed|
|Aromatic Hydrocarbons||15 mins||0 completed|
|Annulene||17 mins||0 completed|
|Aromatic Heterocycles||20 mins||0 completed|
|Frost Circle||15 mins||0 completed|
|Naming Benzene Rings||13 mins||0 completed|
|Acidity of Aromatic Hydrocarbons||10 mins||0 completed|
|Basicity of Aromatic Heterocycles||11 mins||0 completed|
|Ionization of Aromatics||19 mins||0 completed|
|Physical Properties of Arenes|
|Resonance Model of Benzene|
|Cumulative Aromaticity Problems|
|Polycyclic Aromatic Hydrocarbon Nomenclature|
Concept #1: Inscribed Polygon Method
Have you ever wondered why the Huckel's Rule number of pi electrons (2, 6, 10, 14) make molecules so stable? 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 are they so special. We're going to learn a visualization technique that will help us to understand molecular orbital theory better when it comes to aromatic molecules.
This is called the inscribed polygon method. It also goes by a lot of other names. It's also called the polygon-in-circle method or a Frost Circle. 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. This is going to explain to us. It's going to be a visual representation of why are Huckel's Rule numbers are so special, and why they’re so important.
What we're going to do is we're going to do a worked example of three different molecules. Then we'll go ahead and do some practice problems. Here we have three different molecules. It says “Use the polygon-in-circle method to predict the stability of the following molecules.” I'm going to go ahead and add some stuff here.
For example, with this nitrogen, let's go ahead and make this a hydrogen. Let's go ahead and make it a nitrogen with one lone pair and a hydrogen. The cyclobutadiene, we’re keeping just as is. The triangle, it's got a longer name that I don't want to complicate you with. The three-membered ring has a negative charge. That's fine.
What we're going to do is we're just going to do kind of steps and I'm going to do all of the steps with all three molecules so you can see how they work. The very first step of this method whether you call it a Frost Circle or polygon-in-circle, your first step is always going to be to draw a polygon, whatever shape it is, with one corner facing down.
As you can see, I've already done that step. I've already drawn every single shape with one of these corners facing down. 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. Now what we're going to do in the next step is you have to draw molecular orbitals on all corners of the ring. That sounds a lot harder than it is. It just means draw a line next to every atom of the ring. A five-membered ring gets five molecular orbitals. Four-membered ring gets four. Three-membered ring gets three. It’s that easy.
Now we're going to draw a line that splits the polygon down the middle basically in a horizontal line that's going to split into two different halves. For the square, that happens to be easy because the halfway line would just go right through the middle of both of those corners. 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 five-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 three molecular orbitals and below two 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 my job now. Some students make the mistake of doing this. They go below that molecular orbital. You'd be surprised if you draw, let’s say you draw your five-membered ring like this. But there's a way to draw it. There you go. Let's say you draw your five-membered ring like this. Well then when you draw the halfway point, you're going to think that it's actually below four orbitals and above only one. But that's not the way it should be.
If you're splitting with an uneven number of orbitals, you should make it so that they’re as close together as possible. Instead of being four orbitals on top and one orbital in the bottom, it should be two on the top and three on the bottom like we have here. 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.
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. This is called the Aufbau Principle. Aufbau Principle was the building up principle. It just means that you have to always fill your lowest energy orbitals first. We're saying that energy goes up. Energy increases the higher you go.
Let's just look at the first molecule. The first 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 because we've got 4 pi electrons. That lone pair is going to make it 6. It's sp3 so that works out. That means I have six electrons to add. How do we add them? One and two here. Then my third electron goes here. My fourth electron goes here.
That's a whole other rule. That was actually Hund's rule. Hund’s rule says that if you have orbitals of the same energy level, you have to fill them evenly. Your third and your fourth electrons go one apiece. But we have six total so then my fifth goes here and my sixth goes there. That's it. I'm done with that step.
Let's move on to cyclobutadiene. How many pi electrons do I have for that one? Cuatro. That's going to be one, two, three, four according Aufbau Principle and then according to Hund’s rule that I have those two even energy orbitals. I have to fill them evenly.
Just so you guys know, there's a more technical term for orbitals that have the same energy levels. Do you guys remember that name, that term? It’s actually from chapter 1 of organic chemistry. They're called degenerate orbitals. If you have degenerate orbitals, that means that you have to fill them evenly. That's it. You can't just put two on one side and zero on the other. That doesn't make a lot of sense.
Finally, what do we get for the triangle, the cyclopropenyl anion? What we would get is one, two, three, four because again, we've got four pi electrons. We've got to have that even distribution at the top.
What did we just do? You might be wondering, “Johnny, we drew all these arrows but where is this going?” It turns out that we can use this diagram to understand why molecules are more stable or less stable. We can understand the identities of the electrons and the identities of the orbitals. It turns out that that halfway point actually represented what we call the nonbonding line in the molecular orbital theory where everything below that line represents a bonding molecular orbital. 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.
Then our antibonding molecular orbitals are the ones that get filled up after all the bonding ones are full. You would never put something in the top orbital until all of your bottom ones are full. It turns out that when you have filled molecular orbitals, remember molecular orbitals would just be one of these. 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 gen chem that orbitals love to be have two electrons. If an orbital has two electrons, it's happy.
What if all of the bonding orbitals have two electrons? That's going to make it uniquely stable because that means that basically all of your bonding orbitals are perfectly filled. What if you have partially filled molecular orbitals? What if you have a weird number of electrons and you have some molecular orbitals just hanging out with one electron apiece? That's going to contribute to unique instability. 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 one electron. They always want to carry two.
What does this mean back to our diagrams? Look what's going on. Normally, we would predict that what would be the aromaticity of these molecules? The first one would be aromatic. The second one is supposed to be anti-aromatic. The last one is also supposed to be anti-aromatic. Are you seeing a pattern here? Notice. Look what's going on.
The aromatic molecule, the one with six pi electrons, Huckel's Rule number happens to have all of its bonding orbitals perfectly filled. That's going to make it stable. That's going to make it really stable. Whereas the ones that are anti-aromatic, 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.
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 four electrons. The electrons over there for the first one is six. The reason that the Huckel's Rule numbers have 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 or instead of an odd number, a 4n number. These are going to kick electrons up to the antibonding orbitals and make partially filled orbitals that are going to contribute to instability.
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 six pi electrons so all of its bonding orbitals are perfectly filled.
You guys are going to do some practices so you 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.
Example #1: Inscribed Polygon Method
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 is this molecule aromatic not aromatic or anti aromatic, aromatic shows very high stability non aromatic just means your normal compound and anti aromatic means you're super unstable so if we take a look at the name we have tropyllium another name for tropyllium just means that we have cycloheptatriene, So the tropyllium just means cycloheptatriene, cyclo means ring hepta mean seven cabons, so we have seven carbons and tri means three double bonds.
Now I say the word cation which means this carbon up here which is not double bond like the others is positive, so that would be our molecule. Now here we're going to draw the frost circle. Now we're going to draw apex down the molecule so point part down. The larger these rings get the harder it becomes to draw them so it takes practice guys to make sure you draw it correctly, not the straightest but good enough. Now we're going to cut this in half so here this is your non bonding 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 we have we have 2,4, 6 pi electrons so one up one down up up down down. We're going to say this molecule shows special stability because all of the molecule orbitals in the body 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 tools that we talked about but we also know it because we just did a frost circle and it approves it the molecular orbitals in the body region are all completely filled it.
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