Ionization of Aromatics

Concept: Concept: Resonance of Fulvalenes

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Video Transcript

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 hopes to create an aromatic compound. One of the most famous examples of this type of molecule is called a fulvalene. A fulvalene is a hydrocarbon that’s composed of two 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 two? Do you count it as zero? 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 and you could split them into charges of positive and negative to help create aromaticity. Here we have two questions about this molecule that I’d like to answer. One question is which of the following atoms, notice there’s eight atoms on this molecule. Which of them would you expect to most readily react with an electrophile e positive? That’s a really huge question. You might be like “I have no idea how to answer that.” Second of all, does this molecule possess in that dipole? If so, indicate the reaction and draw it.
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 is better. In the first possibility, I’m going to split up 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 chare here and the positive charge here. What we’re ending up with is a single bond. If I could erase that and the 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 is 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 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 six-membered ring, I get six pi electrons.
Once I place a positive charge on the second ring, I get two pi electrons. Are those number 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 five-membered ring? I would get four pi electrons. What about putting the negative charge on the three-membered ring? I would also get four pi electrons.
Does this look stable to you? This is like a molecule from hell. This would have two anti-aromatic rings. It would suck. Terrible stuff. You’re definitely going to go with molecule B. this is going to help us to answer our questions. Now that we have this dipole drawn, the answer is yes it does possess in it. That dipole would go to the left. 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 are most likely to react with an electrophile e plus? You can answer this question. Now that I drew that resonance structure, it should be pretty clear which atom likes to react with electrophiles the most. Any amount of deduction would be that if there’s an e plus around, my negative charge would be the one attacking the e. 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, right?
Fulvalenes are definitely 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 could 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, put the negative charge on the top. If you do need electrons, bring them down. That’s how you work with them.
Awesome guys. Now we’re going to go ahead and we’re going to move on to the next molecule.  

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

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