Ionization of Aromatics: Videos & Practice Problems

Aromatic compounds, like fovalene and azulene, exhibit unique stability due to resonance structures and the presence of exocyclic double bonds. In fovalene, resonance allows for the formation of strong dipoles, enhancing aromaticity. Azulene, lacking exocyclic double bonds, requires sharing double bonds between rings to achieve aromatic stability. Understanding the role of electrophiles and dipoles is crucial, as the most nucleophilic atom reacts with electrophiles, while dipole direction indicates molecular polarity. These concepts are essential for grasping the behavior of aromatic hydrocarbons and their applications in chemistry.

Resonance of Fulvalenes

In this video, we're going to discuss how aromaticity will drive some molecules to ionize on their own and form very strong dipoles for the sake of remaining stable. Let's look into this. Double bonds can be viewed as a loose pair of electrons that can undergo resonance movement and ionization if it helps to create an aromatic compound. One of the most famous examples of this type of molecule is called a fobalene. Now a fobalene is a hydrocarbon that's composed of 2 fully conjugated rings joined by what we call this is a big word, an exocyclic double bond. It just means that it's a double bond that's not really inside of any ring. You see this a lot when you talk about carbonyls being on the outside of the ring, something like this. Exocyclic would describe that, how it's not inside the ring. How do you count it? Do you count it as 2? Do you count it as 0? That's what this topic is about. With these exocyclic double bonds, we find that you can use them as a loose pair of electrons. Meaning, you could draw a resonance structure and you could split them into charges of positive and negative to help create aromaticity. Here we have 2 questions about this molecule that I'd like to answer. One question is, which of the following atoms, notice there are 8 atoms on this molecule, which of them would you expect to most readily react with an electrophile E⁺? That's already a huge question. You might be like, I have no idea how to answer that. Second of all, does this molecule possess a net dipole? If so, indicate the direction and draw it. Okay? So this is definitely a little bit beyond our level right now because we haven't discussed it. But this turns out to be a pretty easy question. Like I said earlier, these double bonds, if they're exocyclic, they can be seen as a loose pair of electrons. Meaning that I should ionize them in a way that's going to make the most aromatic compounds as possible. I'm going to actually erase this example so I can get a little bit more space. I'm going to draw another version of this molecule here. What I want to do is draw both possibilities and have you guys be the judge of which one's better. In the first possibility, I'm going to split up the lone the double bond so that it becomes a negative charge on one side here and a positive charge on this side here. We're going to ionize this double bond into a resonance structure of just ions. For the blue one, I'm going to do the opposite. I'm going to put the negative charge here and the positive charge here. What we're ending up with is a single bond. If I could erase that other double bond, I would. But a single bond with ions. Does it really matter which direction I ionize? Are both of these equally as stable? Notice that if this does happen, they're going to have very different dipoles. The one that I drew on the left would have a dipole pointing to the right because you always go towards the negative thing. The one that I drew on the right would have a dipole going to the left because the negative's on the other side. Which one is right? Are any of them right? Have you guys figured it out yet? The answer is definitely if this is A and this is B, it's definitely molecule B. Can you help me understand why? Can you explain it? Can you explain it to your friend? Why would molecule B be so much more stable than molecule A? Because we split the exocyclic double bond into ions to make both of the rings stable. We have ring 1 and we have ring 2. Notice that once I place the negative charge on that 6 membered ring, I get 6 pi electrons. Once I place a positive charge on the second ring, I get 2 pi electrons. Are those numbers good? Hell yeah. Those numbers are great. Remember that those are the Huckel's rule numbers. They predict that you're going to have unusual stability in those rings. Now let's look at the other option. What if I put the plus charge on the fiber ring? I would get 4 pi electrons. And what about putting the negative charge on the 3 member ring? I would also get 4 pi electrons. Does this look stable to you? This would have 2 anti-aromatic rings. It would suck terrible stuff. So you're definitely going to go with molecule B, and this is going to help us answer our questions. So now that we have this dipole drawn, the answer is yes. It does possess a net dipole. It will go to the left. So we could put in this case, we could say to the left. But you could just draw it too. Your professor would see it. But then the first question, which of these atoms is most likely to react with an electrophile E⁺? You can answer this question. Now that I drew that resonance it should be pretty clear which atom likes to react with electrophiles the most. Yeah. So any amount of deduction would be that if there's an E⁺ around, my negative charge would be the one attacking the E. So the answer is this atom right here. It would be the one that's most likely to attack the electrophile because of the fact that it's the most nucleophilic atom on the entire molecule because of that resonance structure. Interesting. Fovalenes are definitely like a weird kind of molecule but this entire idea of an exocyclic double bond is actually going to come up quite a bit. I want you guys to know that you can always split an exocyclic double bond into ions into the direction that's going to help you make your compound aromatic. If you don't need electrons, Awesome guys. Now we're going to go ahead and we're going to move on to the next molecule.

Resonance of Azulene

So guys, azulene is a molecule that we discussed previously, and it turns out that azulene also has resonance structures similar to what we just learned with covalent. Okay? So first of all, azulene, just to recap, is a polycyclic aromatic molecule with a distinctive blue color. I actually went ahead and included an example of a mushroom that is actually colored by a derivative of azulene. So it just goes to show how in nature, azulene is actually kind of like a natural dye that turns things a brilliant blue. Okay? And, we have the same two questions here that we want to answer. We want to answer which atom would most likely react with the electrophile, and does this possess a net dipole? Well, this one's a little bit harder to figure out because there's no exocyclic double bond. Right? There's no exocyclic double bond. I'm just going to write that here, no exocyclic double bond. So we can't just split any of these double bonds into ions. That would not be right. What is our goal? Our goal with azulene and this actually applies to many polycyclics is going to be, I'm going to put here goal to share a double bond between both rings. Okay? What that's going to allow us to do, if we can share a double bond between both rings, that's going to allow us to figure out an arrangement to have them both be aromatic by themselves. Okay? Awesome. So, let me just, once again show you that we basically have 2 different options of how we can do this. I'm going to have to write the smaller one at the bottom here. I'm running out of room a little bit. But there's really only one way to get azulene to share. There's really only 2 ways to get azulene to share double bond between both. Let me write the original structure again. Just give me one second. You guys should all be writing this as well since I'm going to use that as an example. Okay. There are 2 different ways we could do this. Okay? Either we could take this double bond and use that double bond to make a double bond in the middle. Now, if we do that, that's going to break an octet right here. So if we make that bond, we have to break this bond. And then we would do that. And what that would do is that would give us a product that would now have a double bond in between the rings. Now, the other option would be to go from the small ring to the middle. So then I would go like this and I actually messed up with my double bonds. Oops. Okay. So this is an error guys. This is the kind of stuff I'm human. I actually drew an 8-membered ring. So let's just erase this really quick. This is going to get ugly a little bit. We're going to do this. There we go. Okay. My apologies. So if you want, you can pause the video so you can catch up. But go ahead and draw a 7-member ring, not an 8. Okay? So back to the molecule at hand, we would go ahead and we could go from the small ring to make a double bond. But once again, we have to break a bond so then we break this one here. Okay? So let's kind of look at what you would get in terms of charges, in terms of products from both of these. Okay? I'm actually going to use the space here at the bottom to draw both of the products. Okay? And then, we can see which one looks better. This will also give me a chance to redeem myself with that 7-member ring. I thought I was doing pretty well, but I guess not. 1, 2, 3, 4, 5, 6, 7. Wait. No. I did it again. Ugh. These are hard to draw, man. Okay. So that's 1. And I'll cheat. I'm going to take this, and I'm going to copy it so that I don't have to mess up again. Alright. Cool. Okay. So now we're going to add in what the molecules would look like after these resonance structures have formed. We'll notice that this double bond and this double bond are still the same. But now I have a double bond here and here. Meaning that now what charge should I have where the original double bond left? I should have a positive here and I should put a negative here. Okay? So that's one of the arrangements, one of the possibilities. Another possibility is that these dual ones are still the same. I still have the double one in the middle but now I have the positive here because my bond left and my negative here. We're comparing one resonance structure versus another. And you guys have to tell me which one you think is more stable. Do you think the red version is more stable, or do you think the blue version is more stable? Okay. Hint, I think you should count up pi electrons and see what the aromaticity of both of these molecules would be. So, let's just say that we've got ring 1 and ring 2. So ring 1 has how many pi electrons? Sorry, you couldn't see that. So ring 1 has how many pi electrons? Well, it has 2, 4, 6 from the double bond that's being shared. 2, 4, 6. Positive doesn't count as anything. So this one has 6 pi electrons. So good so far. And then molecule or ring 2, my apologies. So I have ring 1 and ring 2. Ring 2 has how many pi electrons? Well, it appears to have 2, 4, 6. This one also has 6 pi electrons. Right? So that's great. We really can't do better than that but let's just verify that the blue is wrong. So for blue, how many pi electrons do I have with ring 1? Guys, I have 8 pi electrons, right? Because I've got 3 double bonds and then I've got a negative charge at the top. Now, for ring 2, I have how many pi electrons? I have 2, I have 4, I have a positive charge that counts as 0, I end up with 4 pi electrons here. This sucks. Okay? So as you can see, this is a structure that is terrible. This resonance structure would never ever form, whereas this resonance structure is actually highly favored because of the fact that now I have 2 aromatic molecules. Okay? Now, can you see where this is going? Now that I know where my charges are, can I answer the 2 questions? Absolutely. So which atom would you expect to react with electrophile e? Alright, guys. So the answer is it has to be this atom. This is the only atom on the benzene ring. I'm sorry, not on the benzene ring, on the azulene that is going to get pretty much a full negative charge. Okay? Does this molecule possess a dipole? If so, indicate its direction. Hell, yeah. It does. So you could just draw the dipole straight from the positive to the negative. So it would have some kind of slanted dipole like this. Okay? And that would show that there's a dipole going towards the negative. And actually, just so you guys know, one of the biggest reasons for azulene's vivid color is its strong dipole. These chemical properties and physical properties are closely intertwined and actually make it the kind of amazing molecule that it is. Okay? So guys, anyway, so now you know that exocyclic double bonds, you can just split them into ions easily. And you know that for polycyclic systems like this, you have to make sure that when you're drawing resonance that your goal is that you want to share a double bond between both rings to allow it to make them both aromatic. If you don't share a bond, you're never going to make both rings aromatic at the same time. Okay? So I hope that made sense. We're done with this topic. Let's go ahead and move on.

Which carbon in the following compound is most likely to react with an electrophile? 

• A. A
• B. B
• C. C
• D. D

Answer: (A)

Here’s what students ask on this topic:

What is the role of exocyclic double bonds in the ionization of aromatic compounds?

Exocyclic double bonds play a crucial role in the ionization of aromatic compounds by providing a loose pair of electrons that can undergo resonance. This resonance can split the double bond into positive and negative charges, enhancing the aromaticity of the molecule. For example, in fovalene, the exocyclic double bond can be ionized to create a resonance structure that stabilizes the molecule by making both rings aromatic. This process helps in forming strong dipoles and contributes to the overall stability of the aromatic compound.

How does aromaticity influence the stability of molecules like fovalene and azulene?

Aromaticity significantly influences the stability of molecules like fovalene and azulene. In fovalene, the ionization of exocyclic double bonds into resonance structures creates aromatic rings, which are highly stable due to Huckel's rule (4n+2 π electrons). Azulene, lacking exocyclic double bonds, achieves stability by sharing double bonds between its rings, ensuring both rings are aromatic. This shared aromaticity stabilizes the molecule. The presence of aromatic rings reduces the molecule's reactivity and increases its stability, making it less likely to undergo chemical reactions that would disrupt its aromatic system.

Why is azulene's vivid blue color related to its dipole moment?

Azulene's vivid blue color is closely related to its strong dipole moment. The molecule's resonance structures create a significant separation of charges, resulting in a pronounced dipole. This dipole affects the electronic transitions within the molecule, particularly the π-π* transitions, which are responsible for its color. The strong dipole moment enhances these transitions, leading to the absorption of light in the visible spectrum and the emission of a blue color. Thus, the dipole moment not only contributes to the molecule's stability but also to its distinctive color.

How do you determine which atom in an aromatic molecule is most likely to react with an electrophile?

To determine which atom in an aromatic molecule is most likely to react with an electrophile, you need to identify the most nucleophilic atom. This is typically the atom with the highest electron density, often indicated by a negative charge in resonance structures. For example, in fovalene, the ionization of the exocyclic double bond creates a resonance structure where one atom carries a negative charge. This negatively charged atom is the most nucleophilic and thus the most likely to react with an electrophile (E+). Identifying these resonance structures helps predict the reactivity of different atoms in the molecule.

What is the significance of Huckel's rule in determining the stability of aromatic compounds?

Huckel's rule is significant in determining the stability of aromatic compounds as it provides a criterion for aromaticity. According to Huckel's rule, a molecule is aromatic if it has (4n+2) π electrons, where n is a non-negative integer. This rule helps predict the unusual stability of aromatic compounds. For instance, in fovalene, the resonance structures create rings with 6 π electrons (n=1), making them aromatic and highly stable. Conversely, structures with 4n π electrons are anti-aromatic and less stable. Thus, Huckel's rule is essential for understanding the stability and reactivity of aromatic compounds.