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Aromatic stability V

Aromaticity of polycyclic compounds, such as naphthalene. Created by Jay.

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Video transcript

Huckel's rule can only be applied to monocyclic compounds. However, there are some polycyclic compounds that seem to have some form of aromatic stability. And one of those compounds is naphthalene. So over here, on the left, we have the dot structure for naphthalene. Naphthalene is a white solvent that is traditionally the component of moth balls. And so it has a very distinctive smell to it. Now naphthalene is aromatic. However, it's not as stable as benzene. But we could think about it as two fused benzene-like rings. And so if I were to analyze a naphthalene molecule using our criteria for aromaticity, I could look at each carbon in naphthalene. And I could see that each carbon has a double bond to it. So each carbon is sp2 hybridized. And therefore each carbon has a p orbital, so an unhybridized p orbital. And so there are a total of 10 carbons in naphthalene. And so if I go over here to the drawing on the right, each of those carbons has a p orbital. So I could draw in the p orbitals on each one of my carbons in here like that. Now, these p orbitals are right next to each other, which means they can overlap. And so if you think about the criteria for a compound to be aromatic, this would sort of meet that first criteria, there right? You can see that you have overlap of these p orbitals. And in this case, we have delocalization of electrons across the two rings. Now, when we think about the second criteria, which was Huckel's rule in terms of number of pi electrons our compound has, let's go ahead and analyze naphthalene, even though technically we can't use Huckel's rule. But if we look at it, we can see that there are 2, 4, 6, 8, and 10 pi electrons. So naphthalene has 10 pi electrons. Think about Huckel's rule, 4n plus 2. If n is equal to 2, 4 times 2, plus 2 is equal to 10 pi electrons. And so 10 pi electrons is a Huckel number. And so it looks like naphthalene fulfills the two criteria, even though again technically we can't apply Huckel's rule to polycyclic compounds. But those 10 pi electrons are fully delocalized throughout both rings. And one way to show that would be using resonance structures. So if we were to draw a resonance structure for naphthalene, I could take these electrons and move them in here. And then these electrons would go over here. And then these would go over there. So that would give me another resonance structure. So if I go ahead and draw the results of the movement of those electrons, I would now have my pi electrons over here like this. So it's a benzene-like ring on the left. And then on the right, we still have these pi electrons in here like that. Now, in this case, I've shown these pi electrons right here. I've shown them on the left side. But in reality, those pi electrons are above and below our single bond, in terms of the probability of finding those electrons. And so I don't have to draw it the way I did it here. I could draw it like this. So let me go ahead and put this is going to be equivalent to this structure. So these aren't different resonance structures. I'm just drawing a different way of representing that resonance structure over here. So I could show those electrons in blue over here on this side like that. So let me go ahead and highlight those electrons. So the electrons in blue are right here. And these two drawings are equivalents after I put in my other electrons right there. So the dot structures are just an imperfect way of representing the molecule. So I could show those pi electrons on the left, I could show them on the right. It's really the same thing. Once I draw this picture, I'm now able to draw another resonance structure. So I can draw another resonance structure from this one right here. I could move these electrons over here, move these electrons over here, and then finally, move these electrons over here. And so when I go ahead and draw the resulting dot structure, now I would have, let's see, these pi electrons are still here. And then going around my ring, it would look like this. So there are a total of three resonance structures that you can draw for naphthalene. Again, showing the delocalization of those 10 pi electrons. And showing you a little bit about why naphthalene does exhibit some aromatic stability. It's not quite as aromatic as benzene. All of benzene's bonds have the exact same length. But naphthalene is shown to have some aromatic stability. Naphthalene is the simplest example of what's called a polycyclic aromatic hydrocarbon. And there are several different examples of polycyclic aromatic hydrocarbons. Another example would be something like anthracene. And so there are many, many examples of ring systems that contain fused benzene-like rings throughout the system. But instead of focusing on those, I wanted to do another example which is an isomer of naphthalene. And it's called azulene. And azulene is a beautiful blue hydrocarbon, which is extremely rare in organic chemistry to have a hydrocarbon that's blue. It also has some other interesting properties. It has an increased dipole moment associated with the molecule. There's also increased electron density on the five-membered ring. So over here on the left, we have azulene. And here's the five-membered ring over here on the left. And it turns out there are more electrons on the five-membered ring than we would expect, giving it a larger dipole moment. And if we think about a possible resonance structure for azulene, we can figure out why. So if I took these pi electrons right here and moved them in here, that would push these electrons off onto this carbon. So if I go ahead and draw the resulting resonance structure, I would have an ion that looks like this. So if I think about my formal charges, if I think about these electrons in blue right here, those are going to go off onto that top carbon. So there's that top carbon is going to get a lone pair of electrons, which gives that top carbon a negative 1 formal charge. So it's a negative formal charge on that carbon. And then if I think about this carbon over here, this carbon lost a bond. And so that's going to end up with a positive charge. So we have a carbocation right here like that. And if I analyze this resonance structure, it has two formal charges in it. But if I look over on the right, I can see on the right there, this is a seven-membered ring on the right. And if I look at it, I can see there are six pi electrons. And then right here, I have a carbocation. And so this is one of the examples we did in the last video. So every carbon is sp2 hybridized. And so we have overlapping p orbitals. And we have a total of 6 pi electrons. And so this seven-membered ring is aromatic. If I look over here on the left, I can see that I have a five-membered ring. And I have some pi electrons right here. And the pi electrons right here, as we saw in the example of naphthalene are actually being shared by both rings. So I could pretend like those electrons are right here on my ring. And then this ring also has electrons like that with a negative 1 or more charge. And again in the last video, we saw that this ion is aromatic. It has a total of six pi electrons. So go ahead and highlight those. So these, these, and these are all pi electrons when you think about resonance structures. And so 6 pi electrons. And all the carbons turn out to be sp2 hybridized. Again, look at the previous video for a much more detailed explanation as to why these two ions are aromatic. And so since these ions are aromatic they have some aromatic stability. And this resonance structure, over here on the right, is a much greater contributor to the overall picture of the molecule. And the negative charge is delocalized throughout this five-membered ring over here. And the positive charge is delocalized or spread out throughout this seven-membered ring. And that is what gives azulene its larger dipole moment. So there's a larger dipole moment in azulene than expected because of the fact that this would give us two aromatic rings, which confers, of course, extra stability. And so once again, this ion down here was the cyclopentadienyl anion. And then this cation over here was the cycloheptatrienyl cation from the previous video. The redistribution of these electrons allows azulene to absorb in the orange region, which is difficult for most organic molecules because it's a longer wavelength. And that allows it to reflect in the blue region, which is again the rare, especially for a hydrocarbon. And the fact that it's blue is where this part of the name comes in there, like azure, as in blue. So these are just two examples of some ring systems that also exhibit some form of aromatic stability.