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Aromatic stability II

How the geometry of a molecule plays into aromaticity. Why cyclooctatetraene is not aromatic like benzene. Created by Jay.

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Video transcript

Last video, we observed that benzene exhibits aromatic stabilization. And when chemists first made cyclooctatetraene, which is this molecule right here, they assumed it would react like benzene. Because it looks like it has alternating single, double bonds, and it has a ring. And so they just assumed that it would behave like benzene. It turns out cyclooctatetraene does not react like benzene. So benzene did not react the way cyclohexene did up here. So cyclohexene will give us a mix of enantiomers when the bromine adds across a double bond. Benzene does not do that. Cyclooctatetraene does, as a matter of fact. So it will give you a mixture of enantionmers and the bromine will add across one of those double bonds. It turns out cyclooctatetraene isn't even conjugated. So it looks like it has alternating single and double bonds in this dot structure, but it behaves like isolated double bonds, like four different isolated double bonds. And it's not aromatic. So let's see if we can analyze the reasons for this reaction here. And so we're going to just real quickly review the criteria to determine if a compound is aromatic or not. So a compound or ion is aromatic If it contains a ring of continuously overlapping p orbitals, and also has 4n plus 2, where n is an integer, pi electrons in the ring, which is Huckel's rule. So here is my cyclooctatetraene molecule. And if I count my pi electrons, I can see there are a total of eight pi electrons in this molecules. So eight pi electrons. And each carbon in cyclooctatetraene is sp2 hybridized. So each carbon has a free p orbital. Now the fact that cyclooctatetraene has eight atoms in the ring, means that there's a little bit of angle strain in this molecule. So one possible confirmation for the molecule to adopt is what's called a tub confirmation. So you can see these two carbons can swing up on either side to adopt a tub confirmation. Now if each carbon in the tub confirmation is sp2 hybridized, I could draw a p orbital on each carbon in the tub confirmation here. So when I do that, I'll get this as a picture. Now the problem is, it's a little difficult for the p orbitals to overlap. So it might be easy for these guys to overlap right here, but because of the tub confirmation, it'd be hard to get overlap of these orbitals. And so it turns out that this is the reason why the molecule acts like it's not conjugated. It's because it does have p orbitals, but they can't really overlap in the tub confirmation. And so this violates the first criteria for a compound to be aromatic. And therefore we say that cyclooctatetraene is non-aromatic. So it's not an aromatic because it violates the first criteria. It does not have a range of continuously overlapping p orbitals. The p orbitals don't really overlap very well in the tub confirmation. What if cyclooctatetraene adopted a planar confirmation. So we're just going to pretend like cyclooctatetraene adopts a planar confirmation down here. And once again, each carbon is sp2 hybridized. So each carbon is going to get a p orbital like that. And so we have a total of eight carbons. So eight p orbitals. So eight atomic orbitals. So let me go ahead and write that here. So we have a total of eight atomic orbitals for cyclooctatetraene. And according to MO theory, those eight atomic orbitals are going to give us eight molecular orbitals. So the atomic orbitals are going to cease to exist, and give us eight molecular orbitals. Drawing those eight molecular orbitals would be way too much for this video. So we're not going to actually draw them, but we are going to show where they are in terms of energy using our frost circle. And so this is what we saw in the last video. So how to draw a frost circle. I'm going to go ahead and put a line through the center of my frost circle to help me draw the polygon in here. So what kind of a polygon do I draw? Well, I have an eight-membered ring, so I'm going to draw an eight-sided polygon here. I'm going to start from the bottom here. So I'm going to attempt to draw an eight-sided figure in here. So it would be something like this. Now remember, when you are drawing your frost circle and you're inscribing your polygon, the important thing is where the polygon intersects with your circle. And every point where the polygon intersects with your circle represents a molecular orbital. And so I can see that I would have a total of eight molecular orbitals, because I have eight points of intersection between my polygon and between my circle. And again, the nice thing about a frost circle is it shows you the relative energies of your eight molecular orbitals. So I would have three bonding molecular orbitals, which would be the ones down here. So these three points of intersection will give you the relative energies. And so I have three bonding molecular orbitals. And then up here at the top, I also have three molecular orbitals. But these are my antibonding molecular orbitals. Those are higher energy. And my two points of intersection that are right on the center line here, represent two non-bonding molecular orbitals like that. So when I fill my molecular orbitals, again it's analogous to electron configurations. I have a total of eight pi electrons that I need to worry about for a planar cyclooctatetraene molecule. And so I can go ahead and start to fill in my pi electrons like that. So that takes care of six of them and I have two more. And since this is analogous to electron configurations, I'm going to follow Hund's rule and not pair up my electrons in an orbital here. So that represents my eight pi electrons. And since I have unpaired electrons, I have two unpaired electrons, that predicts a very unstable molecule if it were to adopt a planar confirmation. And if I think about in terms of Huckel's rule, I know it doesn't follow Huckel's rule. Huckel's rule is 4n plus 2, where n is an integer. And the two comes from the fact that I have this orbital down here. And I do have 4n right here. But I don't have 4n, where n is an integer for these two electrons up here. And so this is where it breaks down. And so I have a total of eight pi electrons, which does not follow Huckel's rule. And so because the number of electrons is incorrect, this molecule is definitely not going to adopt a planar confirmation. And so cyclooctatetraene has a tub confirmation, and not planar. It is not aromatic. It is considered to be non-aromatic because of the violation of the first criteria. But it is possible to react cyclooctatetraene. It's possible to oxidize it. And so let's see what happens when we do that. So if we take cyclooctatetraene and we oxidize it. So it's going to lose some electrons. So I'm going to say that these pi electrons are going to stay. And we're going to lose the pi electrons on the left. So if I take away a bond from these two carbons that used to have that double bond there, they're going to be positively charged. And so I could draw a resonance structure for this. I could move these electrons over here. And so if I go ahead and show that resonance structure, then this carbon still has a plus 1 charge. And the other positive charge moves over here to this carbon, like that. And you could continue drawing resonance structures for this molecule. I am not going to do that. I just want to show you that the positive charges are spread out throughout the entire ion here. And so one way to represent that would be to just show the electrons are spread out throughout this entire ion. And the whole thing has a 2 plus charge like that. And so when I analyze this dication that I got from cyclooctatetraene, I realize that all the carbons are sp2 hybridized. So if I look at these, my carbocations those are sp2 hybridized. Everything with a double bond on it is sp2 hybridized. And so I have eight sp2 hybridized carbons. And so each one of those sp2 hybridized carbons has a p orbital. And I also have six pi electrons. I have two, four, and six pi electrons. So six pi electrons follows Huckel's rule. And so this ion turns out to be planar. So it actually does look like this up here. And so there's opportunity for those p orbitals to overlap side by side. And so this ion actually fulfills the first criteria. It contains a ring of continuously overlapping p orbitals. And when you analyze the second criteria, it has six pi electrons. And so it would fill the bonding molecular orbitals. It no longer has these two electrons up here. So it has six pi electrons. It fulfills Huckel's rules. So there's a total of 6 pi electrons for this dication. And so this ion is said to be aromatic. So this is an aromatic ion. It fulfills both of the criteria for it to be aromatic. So it's extra stable. And that's the reason why it turns out to be planar, because it has eight carbons like cyclooctatetraene. And so because of that number, it has some strain that it has to overcome. So it would like to turn into a tub to overcome some of that strain. But the fact that it's a planar dication, means that there must be some sort of extra stability in that ion that's forcing it to be planar. And that extra stability is due to the fact that it is aromatic. And so this ion has been proved to be planar. So this video is just an overview of cyclooctatetraene, why it itself is non-aromatic. But if you turn it into a cation, it can be aromatic, which is further proof for the concept of aromatic stabilization.