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Studying for a test? Prepare with these 7 lessons on Aromatic compounds.
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We have seen that a compound or ion is aromatic if it contains a ring of continuously overlapping p orbitals, and also if it has 4n plus 2 pi electrons in the ring, where n is an integer. So for example, n could be 0, or 1, or 2, or so on. We can use these criteria to analyze the cyclopropenyl cations. So here is my cyclopropenyl cation right there. And if I see how many pi electrons it has, that would be 2 pi electrons in the ion. So let me go ahead and write 2 pi electrons here. When I look at the carbons in the ion, this carbon has a double bond to it, so it's sp2 hybridized. Same with this carbon. And then this top carbon here, this carbocation is also sp2 hybridized. So we know that sp2 hybridized carbons have a free p orbital. So each carbon in this ion has a p orbital on it. So I can go ahead and show that on this diagram. So I'm sketching in the p orbital on each carbon in my ring. And so that allows me to see that my p orbitals could overlap side by side. And so this ion fulfills the first criteria. It has a ring of continuously overlapping p orbitals. I know that p orbitals are atomic orbitals. And so I have a total of three atomic orbitals. And according to MO theory, those three atomic orbitals are going to combine to give me three molecular orbitals. And I can analyze the relative energy levels of those molecular orbitals using what's called a frost circle. And so we've seen these frost circles before. And I'm just going to go ahead and start by drawing a line here to divide my circle in half, which divides my bonding from my antibonding molecular orbitals. And we always start at the bottom of our frost circle. And we inscribe a polygon to match the number of atoms in my ring. So in this case, I have a three-membered ring. And so I'm going to inscribe a triangle into my frost circle. So I'm going to attempt to draw a triangle in my frost circle here. And not the best triangle, but the important thing is where your polygon intersects with your frost circle represents the energy level of your molecular orbitals. So I can see I have a total of three molecular orbitals, looking at my frost circle. And when I go and represent them over here, I know that I have one molecular orbital right here. And I have two molecular orbitals up here. Bonding molecular orbitals are always lower in energy. So I know that this orbital down here is my bonding molecular orbital. And therefore, I have two antibonding molecular orbitals up here. I know I have two pi electrons to worry about. So it's completely analogous to electron configurations. I'm going to fill my lowest energy orbital first. And so that would be my bonding molecular orbital. And I only have two pi electrons. And so those 2 pi electrons are going to completely fill my bonding molecular orbital. Let's analyze this in terms of Huckel's rule. So I can see that this would be a 2 here, to represent my two pi electrons. And then I have 0 electrons in my antibonding molecular orbitals. So it would be like 4 times 0. So 4 times 0 is, of course, 0. Plus 2, gives me a total of 2, which is Huckel's rule. So I have two pi electrons in the ring. And this ion satisfies the second criteria as well. So the cyclopropenyl cation is aromatic. So if I go ahead and write this ion is aromatic over here. And so it's extra stable. And this observation allows us to explain some of the properties associated with this molecule down here, which is cyclopropenone. Cyclopropenone has a huge amount of angle strain with a three-membered ring here. And it also has increased dipole moment from what we might expect. And we can explain both of those by looking at the resonance structure for the cyclopropenone molecule. So if I think about drawing a resonance structure, I could take these pi electrons and move them off onto my oxygen here. And so the resonance structure would have my three-membered ring like that. And then it would have this oxygen with now three lone pairs of electrons around it, giving it a negative 1 formal charge. Took a bond away from my carbonyl carbon right here. So that carbon is going to get a plus 1 formal charge. And so you can see that we've just formed the cyclopropenyl cation here, which we know is extra stable. This is aromatic. And so while this is an extremely minor resonance structure for most carbonyl compounds, for this one, this contributes a little bit more to the resonance hybrid because of the extra stability associated with the cyclopropenyl cation, the fact that it is aromatic. And so that affords some extra stability to this molecule. And we could also draw it where we show that positive charge is being spread out throughout our three-membered ring like that. And then we have our oxygen over here with a negative charge. And this picture allows us to see the increased dipole moment a little bit more. So because the part on the left is extra stable as a positive charge, it's aromatic. That means we have an increased dipole moment and also increasing the stability despite the significant angle strain associated with the cyclopropenone molecule. So you can use the concept of aromatic stability to analyze the structure of other molecules as well. Let's do one more example. Let's look at this molecule, which is cyclobutadiene. So over here on the left, we have cyclobutadiene. If I look at the pi electrons, here is 2 pi electrons. And here's another 2, for a total of 4 pi electrons. So there are 4 pi electrons in my molecule. When I look at the carbons, each carbon has a double bond. So each carbon is sp2 hybridized. So each carbon has a free p orbital. And so I can sketch in my p orbitals over here on my diagram like that. Makes it a little bit easier to see that these p orbitals could overlap side by side. So the first criteria has been fulfilled. I have a ring of continuously overlapping p orbitals. But when I look at my second criteria, our second criteria was 4n plus 2 pi electrons. And so I don't have that. I have a total of 4 pi electrons. And so that's really 4n, where n is equal to 1. So 4 times 1 is equal to 4 or so. I have 4n pi electrons. I don't have 4n plus 2. So I already know that this compound is not aromatic just by looking at that. But let's go ahead and draw in our molecular orbitals and see where we put those 4 pi electrons in our molecular orbitals. So my four p orbitals here on this diagram. It means I have a total of four atomic orbitals for cyclobutadiene. Once again, MO theory tells me four atomic orbitals are going to give me four molecular orbitals. And I can draw in my frost circle right here. So once again, we're going to draw a line through the center to separate my bonding from my antibonding molecular orbitals. You always start at the bottom of your frost circle. Four-membered ring. So we're going try to draw a four-sided figure in our frost circle. So I'm going to attempt to draw this square in here. And once again, where our polygon intersects with our circle represents the energy level of our molecular orbitals. So I have a total of four molecular orbitals. And if I go over here, I have a molecular orbital below the line. So that's my bonding molecular orbital. I have a molecular orbital above the center line there. So that's my antibonding molecular orbital. And this time I have two molecular orbitals that are right on the line, which represents non-bonding molecular orbitals. When I go ahead and put in my 4 pi electrons, once again I fill lowest energy molecular orbitals first. So that takes care of 2 pi electrons. And now I have two more. And so here I have the non-bonding molecular orbitals are on the same energy level. And so if you remember Hund's rule from electron configurations, you can't pair up these last two pi electrons because the orbitals are of equal energy. And so this is the picture that we get. And you can see that I have two unpaired electrons. And two unpaired electrons implies that this molecule is extremely reactive. And experimentally, it is. So cyclobutadiene will actually react with itself. So it's experimentally extremely reactive which tells you that it's not extra stable. It's not aromatic. So you don't have 4n plus 2 pi electrons. And so the second criteria is not fulfilled, but this compound does satisfy the first criteria. And so the term for this compound is anti-aromatic, which means it fulfills the first criteria, a ring of continuously overlapping p orbitals, but does not satisfy the second criteria. It does not have 4n plus 2 pi electrons. It has 4n pi electrons. And so we say that is anti-aromatic. And there are actually very few examples of anti-aromatic compounds. But cyclobutadiene is considered to be anti-aromatic. In the next video, we're going to look at a few more examples of aromatic stability.