in this video, we're going to discuss how air Metis ity will drive some molecules toe ionized on their own and form very strong die polls for the sake of remaining stable. So let's look into this so 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. Okay, one of the most famous examples of this type of molecule is called a foe of a lien. Okay, now a foe of a lien is a hydrocarbon that's composed of two fully conjugated rings, joined by what we call. There is a big word and XO cyclic 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 like carbon eels being on the outside of the ring. Something like this Exhaust cyclic would describe that how it's not inside the ring. So how do you count it? Do you count as two? Do you counted zero. That's what this topic is about. Okay, so with these eggs, a cyclic the whole bonds, Okay, we find that you can use them as a loose pair of electrons, meaning you could draw resident structure and you could split them into charges of positive and negative to help create air Metis ity. Okay, so 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, ah, eight atoms on this molecule. Which of them would you expect to most readily react with an electric file e positive? So it's already like a huge question like you might be like, I have no idea how to answer that. Second of all, does this molecule possessing that die poll. If so, indicate direction and draw. 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. Okay, Like I said earlier, these double bonds, if their exhaust cyclic, they can be seen as a loose pair of electrons, meaning that I should ionized them in a way that's gonna make the most aromatic compounds as possible. So I'm gonna actually erase this example so I could get a little bit more space. I'm gonna draw another version of this molecule here. Okay? And what I want to do is draw both possibilities. And have you guys be the judge of which one's better. Okay, so in the first possibility, I'm going to split up the loan, The double bond, that it becomes a negative charge on one side here and a positive charge on this side here case, we're gonna ionized this double bond into a resident structure of just ions. Okay, now for the blue. What? I'm gonna do the opposite. I'm gonna put the negative church here and the positive church here. So what we're ending up with is a single bond. Okay, if I could reach that other double one would. But a single bond with Ion it's okay. So it doesn't really matter which direction I ion eyes could. I just Are both of these equally a stable Okay, notice that if this does happen, they're gonna have very different die polls. The one that I drew on the left would have a die pole pointing to the right because you always go towards the negative thing. And the one that I drew on the right would have a typo going to the left because the negatives on the other side Which one is right? Are any of them right? Have you figured out yet? So 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 b so much more stable than molecule? A. Because we split the Exocet click double bond in tow ions to make both of the rings stable, we have ring one way, have ring to and notice that once I placed the negative charge on that six member of ring, I get six pie electrons, right? Once I place a positive charge on the second ring, I get two pi electrons. Are those numbers good? Hell, yeah. Those numbers are great. Remember that those are the Hucles rule number, so they predict that you're gonna have unusual stability in those rings. Now let's look at the other option. What if I put the plus charge on the 500 ring? I would get four pi electrons. And what about on the putting the negative charge on the three member grain. I would also get four pi electrons. Does this look stable to you guys? This is like a molecule from hell. This would have to anti aromatic rings. It would suck terrible stuff. Okay, so you're definitely gonna go with molecule B, and this is gonna help us to answer our questions. So now that we have this di pull drawn, um, the answer is yes, it does possess in a net dipole it would go to the left. Okay, So we could put in this case you could say to the left, but you could just draw it to your professor would see it. Okay, But then, the first question which of these atoms is most likely to react with an electric felt e? Plus, you can answer this question now that I drew that residence structure, it should be pretty clear which Adam likes to react with Elektra files the most. Yeah, So any amount of deduction would be that if there's an e plus around my negative charge would be the one attacking the E. So the answer is this Adam right here would be the one that's most likely to attack the electric file. because of the fact that it's the most nuclear Filic Adam on the entire molecule because of that resonant structure. Interesting. Right? So full of Elaine's air. Definitely. Like a weird kind of molecule. But this entire idea of an exhaust cyclic double bond is actually gonna come up quite a bit. And I want you guys to know that you could always split in exhaust cyclic double bond in tow ion into the direction that's gonna help you make your compound aromatic. Okay, So if you don't need electrons, put the negative charge on the top. If you do need electron to bring them down, okay, that's where you work with them. Awesome guys. So now we're gonna go ahead and we're gonna move on to the next molecule.
Resonance of Azulene
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So guys, as you lean is a molecule that we discussed previously. And it turns out that as you lean also has resonance structures similar to what we just learned with fall Villines. Okay, So first of all, as you lean 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 a xylene. So just goes to show how in nature as a lien is actually kind of like a natural diet turns things this brilliant blue. Okay. And we have the same two questions here that we wanna answer. We wanna answer which Adam would most likely react with the Electra file. And does this possess a net dipole? Well, this one's a little bit harder to figure out because there's no exhaust cyclic double bond, right? There's no exhaust cyclic double bond. I'm just gonna write that here. No XO, cyclic double wind. So we can't just split any of these double bonds toe ions. Okay, That would not be right. Okay, So what is our goal? Our goal with as you lean and this actually applies to many policy clicks is gonna be I'm gonna put your goal to share a double bond between both rings. Okay, What that's gonna allow us to do if we can share a double one between both rings, that's gonna allow us to figure out an arrangement toe, have them both be aromatic by themselves. Okay, awesome. So let me just once again show you that we basically have two different options of how we can do this. I'm gonna have to write the smaller one at the bottom here. I'm running out of room a little bit, but there's really Onley. One way to get as you lean to share. There's really only two ways to get actually to share double one between both. Let me write the original structure again. Just give me one second. You guys should all be writing. This is well, since I'm gonna use that as an example, okay? There's two different ways we could do this. Okay, Either we could take this double bond and use that Don't want to make Thio make a double one in the middle. Now, if we do that, that's gonna make that's gonna break in octet right here. So if we make that bond, we have to break this bond, and then you 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. Okay, Now, the other option would be to go from this small ring to the adult to the middle. So then I would go like this, okay? And I actually messed up with my double bonds. Oops. Okay, So this isn't This is an error, guys. This is the kind of stuff I'm human. I actually drew an eight member ring, so let's just erase this really quick. This is going to get ugly a little bit. We're gonna do this. There we go. Okay. My apologies. So if you want, you can pause the videos, you can catch up, but go ahead and try. Draw seven member grain, not in eight. Okay, So, back thio 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 off products from both of these. Okay, I'm actually gonna 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 seven member ring. I thought I was doing pretty well, but I guess not. 1234567 Wait, no, I did it again. Oh, these air hard to draw, man. Okay, so that's one, and I'll cheat. I'm gonna take this, and I'm going to copy it. It's that I don't have to mess up again. All right, Cool. Okay, So now we're gonna add in what the molecules would look like after these resident structures have formed. Well, notice that this double bond and this door wonder still the same. But now I have a double one here and here, meaning that. Now what charge should I have? Were the original deal one left. I should have a positive here on. I should put a negative here. Okay, so that's one of the arrangements. One of the possibilities another possibility is that these will wonder Still the same. I still love the dull one in the middle. But now I have the positive here because my bond left and my negative here. Okay, so we're comparing. We're comparing one resident 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. I electrons and see what the authenticity of both of these molecules would be. Okay, so let's just say that we've got ring one and ring, too. So Ring one has how maney pie electrons. Oh, sorry. You couldn't see that. So ring one has how maney pie electrons. What has 246 from the dull one that's being shared 246 positive doesn't count as anything. So this one has six pie electrons so good so far, and then molecule are ring beat. Bring to my apologies. So I bring one and ring too. Ring to Has how many pie electrons? Well, it appears tohave to 46 This one also has six pie electrons, right? So that's great. We really can't do better than that. But it's just verify that blue is wrong. So for blue, how maney pie electrons do I have with with ring one guys, I have ate pie electrons, right? Because I've got three gold ones and then I've got a negative charge of the top now for ring, too. I have how maney pie electrons. I have to. I have four have a positive charge that counts. Zero. I end up with four pi electrons here. This sucks. Okay, So, as you can see, this is a structure that is terrible. This residence structure would never, ever form. Whereas this residence structure is actually highly favored because of the fact that now I have to aromatic molecules. Okay. Now can you see where this is going? Now that I know where my charges are, can I answer the two questions? Absolutely. So which Adam, would you expect to react with? Electrical eat? Alright, guys. So the answer is, it has to be this. Adam, this is the Onley Adam on the benzene ring. I'm sorry on the benzene ring on the as you lean that is going to get pretty much a full negative charge. Okay, Does this molecule possess a die poll? If so, indicated destruction. Hell, yeah, does. So you could just draw the die post straight from the positive to the negative. So would have some kind of slanted die poll like this. Okay. And that would show that there's a die poll going towards the negative. And actually, just you guys know one of the biggest reasons for as Yellen's vivid color is it strong die poll. These chemical properties and physical properties are closely intertwined and actually make it the kind of amazing molecule that it iss. Okay, so, guys, anyway, so now you know that exhaust cyclic double wands, You could just split them in tow ions easily. And you know that for poly cyclic systems like this, you have to make sure that when you're drawing residents, that your goal is that you want to share a double bond between both rings toe, allow it to make them both aromatic. If you don't share a bond, you're never gonna make both rings or 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?