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Voiceover: In the last video, we used the criteria for aromaticity to see that heterocycles can be aromatic too. In this video, we're going to look at more aromatic heterocycles, specifically five-membered rings. We'll start with pyrrole right down here. The pyrrole molecule, as you can see, five atoms in the ring, and if we take a look at the carbons in the ring, we can see that those carbons all have a double bond to them. Therefore, each of those carbons is sp two hybridized, meaning each of the carbons has a free p orbital. So I can go ahead and draw a free p orbital on each of those four carbons like that. In terms of the nitrogen on the ring, I need to know the hybridization state of this nitrogen. The best way to do that is to calculate the steric number. We know the steric number is equal to the number of sigma bonds plus number of lone pairs of elections. I can see that here is a sigma bond, here is a sigma bond, and here is a sigma bond, so three sigma bonds plus lone pairs of electrons. There's one lone pair of electrons on that nitrogen there, so the steric number should be equal to four, which would imply an sp three hybridization state for pyrrole. We know that's not the case because pyrrole is an aromatic molecule, and sp three hybridized nitrogen would mean no p orbitals at that nitrogen, which would violate the first criterion for this compound to be aromatic. And so there must be some way to get that nitrogen to be sp two hybridized, and of course, we saw how to do that in the end to the last video. This lone pair of electrons on this nitrogen is actually not localized to this nitrogen. We can take this lone pair of electrons and move them in here so that lone pair of electrons can participate in resonance. If those lone pairs of electrons move into there to form a pi bond, that would kick these electrons off onto this carbon, so the resonance structure will have nitrogen with a pi bond here now and a lone pair of electrons on this carbon, which would give this carbon a negative one formal charge. We still have a pi bond over here like that, and this nitrogen now would have a plus one formal charge. Now when we analyze the hybridization state of this nitrogen, we can see that once again, we're going for sigma bonds, so one sigma bond, two sigma bonds, three sigma bonds, so three sigma bonds, this time no lone pairs of electrons because that lone pair of electrons is now de-localized in resonance, and so three plus zero is of course three, meaning that this nitrogen is now sp two hybridized. Since that nitrogen is sp two hybridized, it has a free p orbital, so we can go ahead and draw the p orbital on that nitrogen. You could think about in terms of dot structure, these two electrons over here, these two electrons are actually de-localized and participate in resonance, so that lone pair of electrons you could think about as occupying a p orbital here and they're actually de-localized. We have all these pi electrons de-localized throughout our ring, and so let's go ahead and check the criteria for aromaticity. Pyrrole does contain a ring of continuously overlapping p orbitals, and it does have four n plus two pi electrons in that ring, so let's go ahead and highlight those. We had these pi electrons, so that's two, these pi electrons, so that's four, and then these pi electrons here in magenta are actually de-localized in the ring, so that gives us six pi electrons. So if n is equal to one, four times one plus two gives me six pi electrons. Pyrrole has six pi electrons and also has a ring of continuously overlapping p orbitals, so we can say that it is aromatic. Let's go ahead and look at another molecule here so similar to it. This is imidazole. For imidazole, once again, we have the same sort of situation that we had for pyrrole with this nitrogen right here, so at first, it looks like that nitrogen might be sp three hybridized, but we can draw a resonance structure for it. We can take these electrons in here and move them in here, which will kick these electrons off onto that top nitrogen. Let's go ahead and draw the resonance structure for that. We would have some pi electrons here. We would have a double bond here. We would now have a plus one formal charge on this nitrogen, and we would have a negative one formal charge on this top nitrogen right here like that. Once again, we've seen that this nitrogen in blue, it's actually sp two hybridized. Let me go ahead and highlight it in blue over here on the right. This nitrogen is actually sp two hybridized, so it's the exact same situation that we saw in pyrrole, which, at first, it looks like those electrons might be localized to that nitrogen. Those electrons are actually not. Those electrons are de-localized in the ring because of the possible resonance structure. And so we've determined that that nitrogen in blue that's bond to that hydrogen is sp two hybridized, so over here, I can go ahead and identify that nitrogen in blue that's bond into the hydrogen. It's sp two hybridized. Therefore, it has a p orbital like that, so I can go ahead and draw the p orbital on that nitrogen. If I look at the carbons, if I look at the carbons in this molecule, let me go ahead and highlight these carbons in red here, so these three carbons of my original dot structure, we can see they have a double bond to them, so those are all sp two hybridized, so I can go ahead and draw in the p orbital on those carbons as well. Then once again, going back to the original dot structure and this time looking at the nitrogen, I'm going to go ahead and draw this one in magenta, this nitrogen right here, if I look at this first dot structure, it's just like the example in pyridine that we saw in the last video. It's actually sp two hybridized. Since it's sp two hybridized, we could think about these electrons in here in magenta as participating in resonance and the electrons in blue here as being localized to that nitrogen atom, localized to an sp two hybridized orbital, and so I can go ahead and identify this nitrogen in magenta over here, so I'm saying this nitrogen is in magenta. I can say that it is sp two hybridized. It's the same situation as pyridine, so I can go ahead and draw a p orbital on it. And that lone pair of electrons in blue that's on that nitrogen, that lone pair of electrons occupies, let me go ahead and put them in blue, that lone pair of electrons is going to occupy an sp two hybridized orbital on that nitrogen, so those electrons are not involved in resonance. Those electrons are localized to that nitrogen. And so when we go ahead and think about the criteria for aromaticity, our first criteria are we have overlapping p orbitals here, and so I can see that that is the case. We have overlapping p orbitals. Everything is sp two hybridized in our ring. And of course, we also need four n plus two pi electrons, and so let me go ahead and label those in magenta. My pi electrons, going back to my original dot structure, here's two, here's four, and then this lone pair on our nitrogen, those are actually pi electrons, when we think about this resonance structure here, so I'm going to go ahead and label them in magenta again. So there actually are six pi electrons in the imidazole molecule, and those pi electrons are de-localized around this ring, around this overlapping p orbitals, and so the imidazole molecule is aromatic as well. The imidazole molecule is actually extremely important in biochemistry, so let's take a look at a famous molecule that contains the imidazole ring, and this molecule is called histamine, which, anyone who has allergies, has heard of histamine. You can see the imidazole ring over here in the left on the histamine molecule. If you want to understand biochemistry, it's very useful to understand these concepts found in organic chemistry. Histamine would be an example of a biological aromatic heterocycle. This portion of the molecule, the imidazole ring, is aromatic. It satisfies the two criteria, as we have seen. Let's do one more example of an aromatic heterocycle here. This time, let's look at an example that has sulfur in it. This molecule is called thiophene. This is really the sulfur analog to the pyrrole molecule that we studied. Let's start by analyzing the sulfur in terms of its steric number. I'm going to look at this sulfur right here. I'm going to calculate the steric number of that sulfur, so number of sigma bonds to that atom. Here is a sigma bond, and here is a sigma bond, so we have steric number is equal to number of sigma bonds plus number of lone pairs of electrons so we have two lone pairs of electrons around that sulfur like that, so that would be two plus two, which is equal to four, which implies that that sulfur is sp three hybridized, but that doesn't work for our concept of aromaticity because in order for something to be aromatic, our atom needs to be sp two hybridized, so it has a free p orbital, just like these carbons right here. Once again, these carbons all have a double bond to them, so those carbons all have a p orbital, so I can go ahead and sketch in those, those p orbitals on those carbons like that. This sulfur looks like it's sp three hybridized, but of course it's going to be a similar example to the pyrrole molecule. I can show a resonance structure for this thiophene molecule. I can take one of these known pairs of electrons, I'm going to say it's the lone pair on the right here, and I could show them moving in to form a pi bond between the sulfur and this carbon, which would of course kick these electrons off onto that carbon. And so when I go ahead and draw this resonance structure, now there is a pi bond between the sulfur and this carbon, lone pair of electrons moved out onto this carbon, giving that carbon a negative one formal charge. There is still a lone pair of electro- still a pi bond, I should say, over here, and there is still a lone pair of electrons left on this sulfur. So now, we can also give this sulfur a plus one formal charge, and we can analyze it in terms of its steric number, so once again, number of sigma bonds. This is a sigma bond, this is a sigma bond, and so the steric number would be equal to two plus how many lone pairs of electrons, now there's only one, so two plus one is equal to three. So now we can see that it's actually sp two hybridized, so it has three sp two hybrid orbitals, and one of those sp two hybrid orbitals is going to contain that lone pair of electrons in blue there, so that lone pair of electrons. Again, it's sp two hybridized, which means it has two other sp two hybrid orbitals, and those hybrid orbitals are forming bonds with those carbons there. Since it's sp two hybridized, it also has an unhybridized p orbital, so I can go ahead and draw in that unhybridized p orbital. I could think about one of those lone pairs of electrons on my original dot structure occupying a p orbital, so let's go back here and look at the thiophene original dot structure I drew. I could think about one of this lone pair of electrons. I'm going to mark it in magenta as occupying a p orbital, so I can go ahead and put it in there like that. And I could think about the other lone pair of electrons on that dot structure as occupying an sp two hybridized orbital out to the side. Whenever you see this situation, think about where those lone pairs of electrons actually are. I also have these as being pi electrons and these as being pi electrons, so two, four, six, a total of six pi electrons, which is Huckel's number, de-localized throughout our ring, throughout our overlapping p orbitals, and so we can say that the thiophene molecule is aromatic. It satisfies both of the criteria for this. There are, of course, analogs to thiophene using oxygen, and that's of course a similar situation. You can draw other resonance structures, but I just wanted to draw this one to show that one of those lone pairs is actually de-localized and involved in resonance, and one of those lone pairs is actually localized to the sulfur atom and localized in an sp two hybridized orbital. So this is how to analyze aromatic heterocycles.