Lecture 23. Aromaticity
Vranken, David Van
This is the second quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. This video is part of a 26-lecture undergraduate-level course titled "Organic Chemistry" taught at UC Irvine by Professor David Van Vranken. 
 I saw this rather horrific story on the news last night. Dad was watching his kid in the garage and suddenly this rattlesnake jumped out of nowhere and bit the child on the finger. And I can't imagine how scared they both were. I can't imagine how scared that little child was going to be of snakes far into the future. Fortunately, this child was doing all right. They took him to a specialist at Loma Linda University and they were able to inject 22 vials of anti-venom in order to stop the action of the snake venom. And so let's go ahead and talk about this anti-venom business here. What's going on with anti-venom that they're giving to people in the hospital? So if you go for a walk in the woods and get bit by a snake, what's your prognosis? So this is a Southern Pacific rattlesnake. That's the most common rattlesnake that you find in the Lake Elsinore region. And if they take you in with a rattlesnake bite, they're probably going to give you this commercial anti-venom. It's produced commercially. They don't make it there in the hospital. And so let's talk about this particular venom called CroFab. If you look at the prescribing information or just the title in the box, it says, Crotolidiae Polyvalent Immune Fab, Ovine. And what that tells you is that it's an antigen binding fragment that is produced from monoclonal antibodies that come from sheep. That's what Ovine means. It means it comes from sheep blood. So what they do is they typically collect venom from many snakes and then they use that venom to immunize sheep. So you immunize sheep, you combine it with something called Croin's Adjuvant, typically, to get the immune system going. And then a couple of weeks later, you give it another boost and the immune system freaks out and is totally ready and starts cranking out tons and tons of antibodies to all the different things that you find in snake venom, in this particular snake venom. And there are lots of nasty things in snake venom. Snake venoms are designed to mess with the cardiovascular system, at least these types of snakes. So they contain enzymes and proteins that degrade tissue, that break down proteins, that lyse blood cells, and that prevent clotting. So you can imagine what a bad mix of toxins this is. So when you inject this into a sheep, the sheep's immune system produces all these different antibodies to all these different bad things that you find inside of the snake venom. And so then you come along and you collect blood from the sheep, doesn't kill the sheep, right? You can go give blood and it doesn't kill you. And then they isolate the antibodies, they use affinity chromatography to isolate all the antibodies and leave all the blood junk behind. And so what do they do with all these antibodies? The last thing in the world you want to do is take a sheep antibody and inject it into your human being. See these parts right here on the antibody called the antigen binding pheromones bind specifically to all those snake toxins. But this part right here doesn't help you. It's just a little bullseye that helps your immune system home in on this and say, hey, that's not human, I'm going to attack it. So what they have to do to prepare this for a human being is to chop off this constant fragment down here and they use an enzyme to do that. And they use an enzyme that's derived from papayas called papain. And that papain comes along and cleaves a single amide bond on each arm here. It's just a single carbon nitrogen bond, but it's extremely specific. And so the enzyme papain cleaves off these little arms, they throw away the bottom part, don't need that. And these top parts look a lot like the antigen binding fragments from your antibodies. Your immune system can't tell the difference for all intents and purposes. So what they're doing is they're collecting these antigen binding fragments that bind to all these different snake venom components. They put it into a vial and those things sit in the refrigerator waiting for somebody to get bitten by a snake. So it's a simple use of an enzyme to cleave very specifically amide bonds here on the arms of immunoglobulins of antibodies. Say again. How effective is that? Is this venom? I'm not an expert. How effective is the anti-venom? I'm not an expert on anti-venoms, but after reading a couple of these stories, it's very effective. If you can get to the hospital very quickly, you're in good shape. If you're stuck out on the woods on a six-day hike, that's a problem. And this is one of the nastiest toxins. You notice they had to use a lot of ant, whatever they said, 22 vials of anti-venom and the kid's doing fine. So you're a lot larger than a kid, so it would be a lot harder to hurt you. But I had a mistake on one of my slides during the last lecture on the Diels-Alder reaction. And I was pointing out this nomenclature for XO versus endo. And what I said was the substituent down here that's on the same side as the alkene bridge is referred to as endo. But then I had a typographical, I mean just a mistake. I said over here substituents on the same side and I again said alkene. I should have said alkane. That should be alkane bridge there, not alkene bridge. They can't both be on the same side as the alkene bridge. So this is the alkane bridge here and that's XO. I'll just write alkane. So please replace the letter E there with the letter A on the bottom. And if you're not exactly sure, look at the book because I just had the same word twice and that's because I was copying and pasting too much. Okay so we moved on from chapter 16. Chapter 16 was a lot of dienes and conjugation and we moved on to chapter 17 to talk about aromaticity. And one of the main points about aromaticity is that you learned all these different reactions of double bonds. Lots and lots of different reactions of double bonds and what's going on there. In chapter 10 you learn more reactions of pi bonds in alkynes in chapter 11. But for some reason the double bonds of benzene are resistant to the typical reagents that you know. Bromine doesn't add across the double bonds of benzene rings. HBR doesn't add across the double bonds of benzene rings. You can't hydroborate the double bonds of benzene rings. And so it seems like it's pretty important for us to be able to recognize when double bonds have this property of aromaticity. When is a double bond part of an aromatic ring system? We need to know that so we know which double bonds have low reactivity towards typical electrophilic reagents. So I'm going to give you just one moment here. I'm going to give you, if I can get my act together here. Yes, sir. It is amazing how quickly I have now mixed up all of my little sheets that I had here. So you will have to just pardon me. You can stare at the chalkboard behind me while I'm sitting here trying to sort through this. I want to note that we have extra office hours next week. Next week is the last week of lectures in this class. So of course what that means to you is you've got one week left to study for the final exam which is on Monday of final exams week. So please make this last week count as we are going through that kind of material. So I want to talk about aromaticity and how to recognize aromatic ring systems when they're there in front of you. So I'll just do the best to sort through my disarray of notes here. I want to introduce you to some, oh that's not the, here sorry about that. That's not what I wanted to give you. Here's what I wanted to show you. It's to try to give you the ability to recognize aromatic and anti-aromatic compounds. So probably you have your notes in front of you and know where they go. Okay, so let's talk about this. How do you recognize an aromatic compound? This property of aromaticity that causes double bonds to be unreactive. That causes protons in your NMR spectrum to be shifted way downfield to 7 to 8 parts per million. That's super particular to benzene rings in the proton NMR. So it's important for us to recognize aromaticity and we're going to use a set of three rules in order to decide whether something is aromatic or not. Here are those three rules. And the first of those rules is kind of a mathematical rule but I'll have to explain how we do this accounting for this mathematical rule. It's called Huckel's Rule. And it's very simple math. It's 4N plus 2 where N is a whole number that allows you to calculate how many electrons should you be counting up when you try to add up the electrons in a potentially aromatic ring system. And because N is always a whole number like 1, well 0, 1, 2, 3 or 4, when you do the math that means it's telling us that all aromatic ring systems have either 2 electrons, 6 electrons, 10 electrons, etc. And in practice you typically don't get things with more than about 14 electrons. It takes a lot of synthetic work to try to design systems that have more than that many electrons but we still give you the mathematical formula in case you want to calculate it out to infinity. So let's take the canonical aromatic compound and that's benzene. When I think of aromaticity, benzene is always the first thing that I think of. And if I count the electrons here in these pi bonds, it's 2, 4, 6 electrons. And there's a special requirement for those electrons that I'm adding up. First of all, they must be in a planar cyclic array. So if I look at these pi bonds in benzene, they are in a planar cyclic array. Benzene is planar, benzene is cyclic and they're all connected to each other and conjugated. If I come over here and I look at this 6-carbon chain that is not connected here, there's no bond here. This is not aromatic because it's not cyclic. So it does have the 6 electrons. It satisfies that Huckel's 4n plus 2 rule but it's not cyclic. So obviously that can't be an aromatic ring. In fact, this is just a really reactive trying. And then the final point here is that the pi bonds and or p orbitals, you can have 2 electrons in a p orbital, they have to all overlap with each other. And so when I come over here and I look at this compound, cycloheptotriene, this CH2 group prevents this end of this double bond here from interacting with this end of this double bond over here. So those two ends are not in conjugation. They are not part of a continuous conjugated cyclic array. And I'll show you this business about p orbitals. None of these actually have an example of electrons in a p orbital. Okay, so that would be an example of an aromatic ring system. And so let's go ahead and extend that idea. So I'm going to compliment that by telling you about something that would be the opposite of aromaticity. And that would be anti-aromaticity. So what would be an example of something that is anti-aromatic? And I think I'm messing this up again. So let me show you, just give me the ability to write on here. Okay, so anti-aromaticity is the opposite of aromaticity. Aromatic compounds are stable. There's a thermodynamic driving force for organic compounds to want to form aromatic ring systems. Those confer stability. And conversely, it is thermodynamically unfavorable for systems to form anti-aromatic compounds. So what makes a compound anti-aromatic? And again, there's going to be three sets of rules that I'm going to apply, and they're almost identical to the rules for deciding aromaticity. And the only difference is, instead of 4n plus 2 electrons, it's 4n electrons. So whenever I see 4 electrons, like 2 pi bonds in a cyclic array, or 8 electrons in a cyclic array, or 12 electrons, etc., that tells me that that's probably going to be a very difficult compound to isolate. Probably that compound is going to be filled with self-loathing and try to self-destruct or react in some way. So everything else is the same, if I see 4 electrons or 8 electrons in a planar cyclic array, all conjugated together, and they're all in overlapping pi bonds and p orbitals, so they're all conjugated, then that tells me there's going to be problems. So let me take an example, and whenever I think of anti-aromatic, here's the prototype. Let me go ahead and use a black pen for this so I can draw this more clearly. Cyclobutadiene is the canonical anti-aromatic compound. You cannot just make cyclobutadiene and put it in a flask. It's not stable. Yes, question? You listed 14 electrons before, and the dot was all before. Yeah, thank you. 16 electrons. Thanks for asking that. I guess I'm not so good at math here with 4n. Once you get to those larger numbers, it's very atypical. I'm sorry about that math mistake there. Okay, so of course the problem here is I've got 2 double bonds, pi bonds in a cyclic array, and I add up the electrons here, and this is definitely anti-aromatic. And that's why nobody has ever made this on its own. Chemists have managed to trap cyclobutadiene in little cage-like molecules so it can't escape. That's one way to stabilize it. The problem is that when you try to make cyclobutadiene, it's not like we don't have reactions or elimination reactions that can make that. Is that it starts to react like this resonance structure. It hates being anti-aromatic so much. You can imagine if I take the electrons in this bond here and I give one electron to each carbon, I'll draw a resonance structure that's a diradical. And that's a better picture for cyclobutadiene. It's like each electron is saying, forget that other thing away from me. I do not want to acknowledge the existence of that other electron. I don't want to have a pi bond. I don't want to have conjugation. So it acts like this. This is how much cyclobutadiene hates being an anti-aromatic molecule. Now there are other apparent anti-aromatic molecules that have been synthesized that somehow manage to avoid being anti-aromatic. So for example, chemists have made cyclo-octatetraene, a cyclo-octane ring with four double bonds and it looks kind of aromatic. It looks kind of like a benzene ring. The problem is that it's got eight electrons. So there's two electrons in each pi bond, two electrons here, two electrons here, two electrons here and two electrons on top here. That's eight electrons. And the way that this avoids being anti-aromatic is it bends into this boat-like shape that prevents the double bonds from being in conjugation. How clever of that molecule that it figured out that it can bend into this kind of shape so that the pi bonds are literally orthogonal to each other. So in fact, cyclo-octatetraene with four double bonds is not anti-aromatic simply because it's not planar. If you tried to somehow flatten it to make it planar, it would freak out and probably start to act like a diradical or do Diels-Alder reactions. It would find some way to not be to react and turn into a stable molecule. Okay, so try to recognize aromatic compounds because they don't do addition of the reagents that you're familiar with, bromine, HBR, hydroboration, hydration. They have special properties in the NMR spectrum. And don't expect that you'll be able to make anti-aromatic compounds because you won't be able to synthesize those. They'll find some way to finagle themselves out of being aromatic. They won't be aromatic. Okay, so let's talk about this issue of planarity. How do you know if an atom is planar? Now I'll show you why this is important in a moment. Not every aromatic compound has only carbon atoms in there. Sometimes you have to count lone pairs as part of your 4n plus 2 calculation or your 4n calculation. And we have to decide how we're going to calculate that. We have to decide which pairs of electrons we're going to count. And I want to start off by addressing this issue of how do you know whether a pair of electrons is in a p orbital or not in a p orbital? I'm going to start off by looking at ammonia and I think all of you know. Did we? Let me, does somebody have a set of notes that I can look at so I can try to get back on track? Plainer aromatic ions. Okay, perfect, thank you. And I'll get to this, I'll come back to this issue in a second. Okay, how do you know if something is planar? Before we look at this issue of lone pairs, how do you know whether an ion is planar? So let me start off by taking a look at a series of ions. So for most of these compounds here, there's very little opportunity for these to bend out of planarity. These are pretty much forced to be planar because they are so small, the ring size is so small. There's no opportunity for these to bend out of planarity like that cycloactotetraene. Kind of the largest ring size I'll talk about here is a seven-membered ring. And so let's talk about what are our options here? I obviously can't have a double bond here on this last carbon atom, the top carbon atom of this three-membered ring. And so what I'm going to do is I'm going to put a positive charge here. If I add up how many electrons are in this empty p orbital, and I'm going to struggle to draw that empty p orbital there, there's zero electrons. So when I add up all the electrons in this array, this array of conjugated double bonds, then what I'll find is that there's two electrons. And because there's two electrons here, that's aromatic. That's 4n plus 2 where n equals 0. So this would be a much more stabilized carbocation than you would expect. And if it truly is aromatic, then we ought to be able to draw a bunch of resonance structures where we delocalize the pi electrons all over that ring. And let's just practice that just to make sure that it has this property of delocalization like benzene. I'm trying to draw an arrow here where I shift that double bond. Let's start to draw some of these resonance structures where I shift, and I'll have to draw this a little bigger here, where I shift that double bond around. If I take the electrons away from this carbon and move them here, now my cation is over here, and I can draw another resonance structure where I move those electrons to here and take the electrons away from this other bottom carbon. And so this has this property of delocalization just like a benzene ring. It has 4n plus 2 electrons. It's planar. In fact, I don't think that could be anything but planar again. So this very strange looking, even with all that strain, cyclopropenium ions are stable and have been made in the laboratory. It's not like you can put it in a flask, but they have been made, and you can get spectra of those kinds of species. Okay, so what do we need to do to a 5-membered ring in order to make that aromatic? If I put a carbocation here like I did with this cyclopropenium ion, if I put a carbocation there, I'll only have 4 electrons, and I definitely do not want to do that. What I'd really like to have here is electrons 5 and 6. So what I really want to have for this to have this property of aromaticity is I want that to be a carbanion at that center. And so I'd really like for these electrons to be in a p-like orbital. And I'll address that. I was just getting to that. Is that p or sp3? So now if I add up my electrons, I've got a planar array, and I have 2, 4, 6 electrons. So let's just add that up 2, 2, and 2. That's 6 electrons. That is an aromatic anion. That carbanion is amazingly easy to make. It's very hard to make carbanions. It's hard to pluck protons off of carbon atoms. But in this system, because this gets to be aromatic, it's very easy to make that by deprotonating carbon. Okay, so now let's come back to the last. The 7-membered ring kind of takes practice to learn how to draw 7-membered rings. That will come with experience. So now if I want to have 6 or 4n plus 2 electrons in this system, do I want to have a carbocation here or a carbanion there? Yeah, we want 0 electrons. We've already got 6 electrons. We don't want to add any more. So we don't want to have a carbanion here, we want to have a carbocation. That would be a stable aromatic ion. So if I add up all my electrons, there's 0 in this p orbital. Let me try to draw that p orbital here. And I'm not going to draw all the different resonance structures. There's lots. But if I add up my electrons, there's 2 in this pi bond, 2 in this pi bond, 2 in this pi bond. There's 0 electrons in that empty p orbital. That's a 6-electron system. That cation would be aromatic and exceptionally stable. So what are the implications here? Implications for you with respect to these aromatic ions is that if you're ever confronted with cyclopentadiene, it is so easy to pull this proton off that you can use regular bases like alkoxide bases. And no other simple alkane allows you to do that. You can just take that and pull that proton off because the product here, I should have used a red arrow there, because the product here is aromatic. The product here is aromatic and heavily stabilized relative to other types of organic molecules. OK. So you can't use alkoxide to pull protons off of hexane or cyclopentane. Or even if you had just one double bond in this ring instead of two, you couldn't use alkoxide to deprotonate that. So those CHs are exceptionally acidic. Somewhere on the order of 34 orders of magnitude more acidic than a regular alkane. Let's come back over here and look at this 7-membered ring. And now it's different. I'm not talking about deprotonating this, but what I've done is I've put an alcohol on here that's been protonated. If you just touch a proton onto that alcohol to put a potential leaving group, this leaving group will fly off. And again, the reason why this is so fast, the reason why that is so exceptionally fast is because it generates an aromatic ion. So reactions that generate aromatic ions, whether it's deprotonation or ionization, those reactions are amazingly fast. And the situation would be totally different now if I did this, just to be clear here. If I only had one double bond in this ring, that's no longer fast. There's no base or simple base that you know of or will have heard of in the Chem 51 series that could ever pull this proton off. So I'm just going to put a dashed line here because this is insane. There's no way you could pull that proton off. That's not going to be aromatic. So you have to look for systems that have the potential to be aromatic. Those reactions will be exceptionally fast if they generate an aromatic ion, either a carb anion or anion. And let me just sort of remind you here that this is fast and this is fast. And this other one is just, there's no way, inconceivable. So gee, isn't it great to become aromatic? You get to enjoy all of that stability. Okay so I drew empty p orbitals with carbocation ions. And then I drew this sort of carb anion where the lone pair was in a p-like orbital. And how do you decide the hybridization of an atom? How do you decide whether electrons are in a p-like orbital? Because only if the electrons are in a p-like orbital do you add them up as part of your 4n plus 2 calculation. So let me just remind you that we'll use nitrogen, not carbon as our example here. I think all of you know that simple amines are pyramidal. Ammonia adopts a pyramidal geometry. That's the lowest energy hybridization for this nitrogen atom. So in this sort of a scheme here, these lone pairs exist in this sp3-like orbital. And that has differential sizes on each end of the orbital. So one end of the orbital is big, the other end of the orbital is small. So that would be an sp3 form of hybridization. If you were to shrink yourself down to the molecular scale and grab these R groups and bend them to the lower planar, you could force this nitrogen atom to adopt an sp2 hybridized geometry. That means each of these bonds have sp2 hybridization, but the electrons in the lone pair, they're in a p-like orbital. And a p-like orbital has the same size on the top and the bottom. And that would be a requirement for us to add this into our 4n plus 2 rule. So let's see how that comes into play. How do you decide in this 5-membered ring system here? What I'm doing is I'm drawing edge on, this is really just the molecule parole. And it's got substituents on all these. I drew substituents on all these. And so the question is, is parole aromatic? Does this 5-membered ring with that nitrogen atom and its lone pair, is that aromatic? Do I count that lone pair in my 4n plus 2 rule? If that lone pair can contribute to aromaticity, it will contribute to aromaticity. So there's two possible hybridizations for this. Most amines don't exist in a planar form, but when I give this lone pair the opportunity to participate in an aromatic ring system, in this 4n plus 2 ring system, it will rehybridize. So flat in this case is more stable because it gets to be aromatic. And that's not normal for amines. Amines don't do that. By being flat, these electrons get to participate in a p orbital. So there's my p orbital there. And when those electrons are in our p orbital, we have equal hybridization there on the top and the bottom face. Or equal, I guess the p orbital has equal size on the top and bottom face. And that allows it to effectively overlap with the pi bonds, which have electron density on the top and the bottom face. So you'll count lone pairs in this 4n plus 2 rule if they can exist in p-like orbitals. And I'm making a big deal about this because there will be cases where lone pairs cannot exist in these p-like orbitals. So you would expect those to have less aromatic resonance energy. Okay, so let's take a look at some systems that have nitrogen in them. Some rings that have nitrogen or oxygen. And try to decide should we count the lone pairs. And I'm going to give you a simple rule to help, I say simple. Maybe it's very wordy, but once you use it a few times, it becomes very user friendly. So I'm sorry, maybe the rule is not, doesn't sound simple. So here's a whole set of heterocycles. Heterocycle means there's a heteroatom in the ring. Usually oxygen or nitrogen, but you can have lower row heteroatoms like sulfur in the ring. So for heteroatoms, how do I decide which of these pairs of electrons to add in to my 4n plus 2 rule to decide whether these are aromatic or anti-aromatic? And here's our simple rule. And I'll read this out to you. If the heteroatom is not part of a double bond, if it's not part of a double bond, then you can count one lone pair towards the 4n plus 2 rule. Now that's a whole string of logic there, so let's practice using that. So if a heteroatom is not part of a double bond, okay, well that nitrogen is part of a double bond. That means I shouldn't count that lone pair. I only count lone pairs if they're attached to atoms that are not part of a double bond. So I'm going to go ahead and cross this out. No, I don't count that. The nitrogen's part of a double bond. So now how many electrons do I have to count? I've got just the double bonds left over. That has six electrons and that's aromatic. So I think you'll be as good as I am at figuring out which of these heteroatoms are part of a double bond. Okay, so let's take this next heterocycle. This is called furan. Okay, if the heteroatom is not part of a double bond, well that oxygen is not part of a double bond. There's only single bonds to that oxygen. Okay, if the heteroatom is not part of a double bond, you count one lone pair towards the 4n plus 2 rule. So I've got two lone pairs here. I'm going to cross one of them out and I'm only going to count one of them. So how many electrons in this system? Six, two, four, six. There's two in this lone pair and then two in each of the pi bonds. That's a total of six electrons. That's aromatic. Okay, let's try to apply it to this heterocycle here that has the two nitrogens next to each other. So if I look at this, again my rule is if the heteroatom is not part of a double bond, well neither of those nitrogens is part of a double bond, then I get to count one lone pair towards the 4n plus 2 rule. So I count both of those lone pairs. Both of those nitrogens are not part of a double bond. So when I count up these electrons here, that would be eight electrons and that would be anti-aromatic. If you could somehow synthesize a molecule with that pattern of bonding, that molecule would be very reactive. It would try to find some way to distort out of planarity or it would try to find some way to react and so it wouldn't have double bonds in that arrangement. That molecule would be extremely unhappy. Okay, so here's a five-membered ring. This is the side chain of the amino acid histidine. It's called a mitazole. And so let's use our little accounting system here. If a heteroatom is not part of a double bond, well this nitrogen atom on top here is part of a double bond. So I don't count that lone pair. The nitrogen's part of a double bond. Okay, let's look at the bottom nitrogen. And our rule says if the heteroatom is not part of a double bond, you count one lone pair towards 4n plus 2. Oh, well we count that. So now we can count up our electrons. There's two in this pi bond, two in this pi bond, and two in that lone pair. That's six electrons. The side chain of histidine, a mitazole, is aromatic. It's an aromatic ring. So every protein in your body is filled, histidine is a common amino acid. It's filled with aromatic rings on the surface of your proteins and the inside. Okay, here's one of the bases of DNA and RNA. This is adenine. And so now there's two sets of rings here. And I can draw different resonance structures. Whenever you've got a double bond between two different rings that are fused together, you double count that double bond. You can look at each ring independently and count that double bond towards each ring. So let's take each of these rings independently. I think you can see that the ring over here on the left hand side is a mitazole. We already did that one. Right? Let's apply our rule. If the lone pair is not part of a double bond, well this lone pair is part of a double bond. I'm not going to count that. And I come over here and this nitrogen atom, well that's part of a double bond. That means I'm not going to count that. And what about this nitrogen atom over here? Well that's part of a double bond. I can't count that. So the only lone pair I count is on the nitrogen that's not part of a double bond. So now when I start off and first count up this five-membered ring here and I add up those electrons. There's two in the lone pair, two in the pi bond, two in the pi bond. This ring here has six electrons and that's aromatic. Now I come over here to the six-membered ring. I don't count any of those lone pairs. And there's a lone pair on the nitrogen but that's not the amino substituent. That's not inside the ring. That's not part of this ring. So don't even worry about the substituents. So I have two, four, six electrons. This ring is aromatic. Both the rings of the DNA base adenine or aromatic. And you would draw the same conclusions about guanine, cytosine, depending on which resonance structure you draw for those. You could consider the other DNA bases to have either partial or full aromaticity. Okay. So let's just come back to why does this rule work? The reason why this rule works. I mean it's very empirical. You don't gain any insight about the geometry of the molecule simply by deciding whether an atom is part of a double bond. All those electrons that we were counting as part of that rule will exist in p-like orbitals. So when we were counting lone pairs in our 4n plus 2 rule, that's because they were able to adopt geometries that were flat so that the lone pair could be in a p-like orbital. So it could interact effectively both and I'm not sure I want to try to draw this in here but with pi bonds that have electron density top and bottom. What about the electron lone pairs that we crossed out? In every case where we crossed out a lone pair, when a heteroatom was part of a double bond like here in pyridine, what that means is that lone pair existed in an sp2-like orbital sticking out of the side of the aromatic ring. So when you didn't count the electrons, they're not sticking up and down in a p-orbital, they're sticking off to the side of the aromatic ring or of the benzene ring or pyridine ring. The implications are these lone pairs that are involved in aromaticity, they're busy. Don't try to mess with those because you'll disrupt aromaticity. So what you would find is that lone pairs like this are simply not reactive. It's very difficult to protonate them. You couldn't protonate that. In fact, if you try to protonate, if you struggle to protonate pyridine by throwing in a powerful acid, it protonates on carbon even. So don't try to mess with that lone pair because it's doing something. It's trying to make the system aromatic. In contrast, the lone pairs here that are sticking out to the side of an aromatic ring, no problem reacting with those. And you know that already. Pyridine is a base. You've already done reactions where you used pyridine as a base. So you can do all kinds of reactions with these lone pairs that are not part of the aromatic 4n plus 2 rule. So no problem reacting with the lone pair on pyridine or this crossed out lone pair on histidine. If you look inside the active sites of enzymes, one of the most common lone pairs used in catalysis is this lone pair that's sticking off to the side of a histidine side chain. If you look at every single amide bond that is produced in every single protein in your body, every single amide bond is created because this lone pair on the bottom of this adenine ring, a single adenine, a single lone pair inside the ribosome is responsible for the synthesis of every single protein, every single amide bond in the human body. So lone pairs that stick off to the side are completely reactive whereas lone pairs that are part of aromatic systems are not reactive at all. Okay. So I think you can see from adenine that you confuse benzene rings together and have much larger aromatic ring systems. So I'm going to give you an example of some fused systems here. So you can see that there's more than one way to account for this 4n plus 2 rule. There's more than one way to do the addition there. So this is a molecule called naphthalene. If you go to the hardware store and buy moth balls, it turns out that crystalline naphthalene just by solid balls of crystalline naphthalene and for whatever reasons moths seem to find that repulsive and will tend to stay out of your closets. So there's multiple resonance structures that I can draw for this. If I count around, if I just ignored this ring, this bond in the middle of those two rings, that sigma bond, just pretend that it's not there, I can count around the outside edges of this flat molecule. And if I count around the outside edges of that flat molecule, what I would find is that there's 2, 4, 6, 8, 10 electrons. That satisfies our 4n plus 2 rule and that's aromatic. There's another way I could have done the accounting for this, that's slightly different, another way to describe this is I could have drawn a resonance structure for naphthalene in which I put a double bond in the middle. There's a different way I can shift around those electrons. And so remember what I said is if there's a double bond at the ring fusion, you can count that double bond for both rings. So in other words, the other way I could have accounted this is I could have said in this ring over here on the left hand side, there's 2, 4, 6 electrons and that's aromatic. And then I look over here at this ring on the other side and there's 2, 4, 6, I double count, I get to count that double bond again and that's aromatic. So I can count that middle, that double bond in the center twice and I reach the same conclusions either way and that is that all of the carbons in this ring system are part of an aromatic ring. I won't reach a different conclusion. So you can fuse these things in all kinds of different ways. Let me give you another example of another aromatic ring that's very common that's fused and this time it's a parole ring fused onto a benzene ring. This is the side chain of the amino acid tryptophan so I'll just write trip here and there's a lone pair and there's nitrogen. So again in this ring system both rings, both of these rings in this heterocyclic, this is called an indole, this bicyclic ring system, both of these rings are aromatic. This ring over here which obviously is benzene, that's aromatic, it has 6 electrons. If I look at this 5-membered ring over here there's 2 electrons in the nitrogen, 2 electrons on this side bond and then 2 electrons over here that's also aromatic. So the side chain of tryptophan is fully aromatic as is the side chain of tyrosine and phenylalinane. You guys are probably learning about those in Bio98 right now. Okay so you can, it can go endlessly. You can fuse benzene ring after benzene ring after benzene ring onto each other. You may have heard of carbon nanotubes or graphene, that's basically what those molecules are. They're just infinite tubes made out of benzene rings all fused together. Okay so the last parts of this chapter are interesting but they're not going to be useful to you for anything else in this quarter. They have to do with molecular orbital base of aromaticity and I think you ought to read that. I would never discourage you from reading anything but it's not the kind of thing that I can test on. People learn about those powerful concepts if you take upper division classes and I would love to teach those to you there but we just can't fit those in with the time that we've got. So be able to recognize aromatic rings, be able to recognize anti-aromatic rings. You need to be able to know when ionization reactions will generate aromatic anions or aromatic cations. Those ionizations or deprotonations will be fast and be able to recognize when heterocycles with nitrogen and oxygen are aromatic. So be ready when we come back on Monday we're going to do chemical reactions of benzene rings and we're going to do a lot of them.
 I saw this rather horrific story on the news last night.  Dad was watching his kid in the garage and suddenly this rattlesnake jumped out of nowhere  and bit the child on the finger.  And I can't imagine how scared they both were.  I can't imagine how scared that little child was going to be of snakes far into the future.  Fortunately, this child was doing all right.  They took him to a specialist at Loma Linda University and they were able to inject 22  vials of anti-venom in order to stop the action of the snake venom.  And so let's go ahead and talk about this anti-venom business here.  What's going on with anti-venom that they're giving to people in the hospital?  So if you go for a walk in the woods and get bit by a snake, what's your prognosis?  So this is a Southern Pacific rattlesnake.  That's the most common rattlesnake that you find in the Lake Elsinore region.  And if they take you in with a rattlesnake bite, they're probably going to give you  this commercial anti-venom.  It's produced commercially.  They don't make it there in the hospital.  And so let's talk about this particular venom called CroFab.  If you look at the prescribing information or just the title in the box, it says, Crotolidiae  Polyvalent Immune Fab, Ovine.  And what that tells you is that it's an antigen binding fragment that is produced from monoclonal  antibodies that come from sheep.  That's what Ovine means.  It means it comes from sheep blood.  So what they do is they typically collect venom from many snakes and then they use that  venom to immunize sheep.  So you immunize sheep, you combine it with something called Croin's Adjuvant, typically,  to get the immune system going.  And then a couple of weeks later, you give it another boost and the immune system freaks  out and is totally ready and starts cranking out tons and tons of antibodies to all the  different things that you find in snake venom, in this particular snake venom.  And there are lots of nasty things in snake venom.  Snake venoms are designed to mess with the cardiovascular system, at least these types  of snakes.  So they contain enzymes and proteins that degrade tissue, that break down proteins, that  lyse blood cells, and that prevent clotting.  So you can imagine what a bad mix of toxins this is.  So when you inject this into a sheep, the sheep's immune system produces all these different  antibodies to all these different bad things that you find inside of the snake venom.  And so then you come along and you collect blood from the sheep, doesn't kill the sheep,  right?  You can go give blood and it doesn't kill you.  And then they isolate the antibodies, they use affinity chromatography to isolate all  the antibodies and leave all the blood junk behind.  And so what do they do with all these antibodies?  The last thing in the world you want to do is take a sheep antibody and inject it into  your human being.  See these parts right here on the antibody called the antigen binding pheromones bind  specifically to all those snake toxins.  But this part right here doesn't help you.  It's just a little bullseye that helps your immune system home in on this and say, hey,  that's not human, I'm going to attack it.  So what they have to do to prepare this for a human being is to chop off this constant  fragment down here and they use an enzyme to do that.  And they use an enzyme that's derived from papayas called papain.  And that papain comes along and cleaves a single amide bond on each arm here.  It's just a single carbon nitrogen bond, but it's extremely specific.  And so the enzyme papain cleaves off these little arms, they throw away the bottom part,  don't need that.  And these top parts look a lot like the antigen binding fragments from your antibodies.  Your immune system can't tell the difference for all intents and purposes.  So what they're doing is they're collecting these antigen binding fragments that bind  to all these different snake venom components.  They put it into a vial and those things sit in the refrigerator waiting for somebody to  get bitten by a snake.  So it's a simple use of an enzyme to cleave very specifically amide bonds here on the arms  of immunoglobulins of antibodies.  Say again.  How effective is that?  Is this venom?  I'm not an expert.  How effective is the anti-venom?  I'm not an expert on anti-venoms, but after reading a couple of these stories, it's very  effective.  If you can get to the hospital very quickly, you're in good shape.  If you're stuck out on the woods on a six-day hike, that's a problem.  And this is one of the nastiest toxins.  You notice they had to use a lot of ant, whatever they said, 22 vials of anti-venom and the  kid's doing fine.  So you're a lot larger than a kid, so it would be a lot harder to hurt you.  But I had a mistake on one of my slides during the last lecture on the Diels-Alder reaction.  And I was pointing out this nomenclature for XO versus endo.  And what I said was the substituent down here that's on the same side as the alkene  bridge is referred to as endo.  But then I had a typographical, I mean just a mistake.  I said over here substituents on the same side and I again said alkene.  I should have said alkane.  That should be alkane bridge there, not alkene bridge.  They can't both be on the same side as the alkene bridge.  So this is the alkane bridge here and that's XO.  I'll just write alkane.  So please replace the letter E there with the letter A on the bottom.  And if you're not exactly sure, look at the book because I just had the same word twice  and that's because I was copying and pasting too much.  Okay so we moved on from chapter 16.  Chapter 16 was a lot of dienes and conjugation and we moved on to chapter 17 to talk about  aromaticity.  And one of the main points about aromaticity is that you learned all these different reactions  of double bonds.  Lots and lots of different reactions of double bonds and what's going on there.  In chapter 10 you learn more reactions of pi bonds in alkynes in chapter 11.  But for some reason the double bonds of benzene are resistant to the typical reagents that  you know.  Bromine doesn't add across the double bonds of benzene rings.  HBR doesn't add across the double bonds of benzene rings.  You can't hydroborate the double bonds of benzene rings.  And so it seems like it's pretty important for us to be able to recognize when double  bonds have this property of aromaticity.  When is a double bond part of an aromatic ring system?  We need to know that so we know which double bonds have low reactivity towards typical  electrophilic reagents.  So I'm going to give you just one moment here.  I'm going to give you, if I can get my act together here.  Yes, sir.  It is amazing how quickly I have now mixed up all of my little sheets that I had here.  So you will have to just pardon me.  You can stare at the chalkboard behind me while I'm sitting here trying to sort through  this.  I want to note that we have extra office hours next week.  Next week is the last week of lectures in this class.  So of course what that means to you is you've got one week left to study for the final exam  which is on Monday of final exams week.  So please make this last week count as we are going through that kind of material.  So I want to talk about aromaticity and how to recognize aromatic ring systems when they're  there in front of you.  So I'll just do the best to sort through my disarray of notes here.  I want to introduce you to some, oh that's not the, here sorry about that.  That's not what I wanted to give you.  Here's what I wanted to show you.  It's to try to give you the ability to recognize aromatic and anti-aromatic compounds.  So probably you have your notes in front of you and know where they go.  Okay, so let's talk about this.  How do you recognize an aromatic compound?  This property of aromaticity that causes double bonds to be unreactive.  That causes protons in your NMR spectrum to be shifted way downfield to 7 to 8 parts  per million.  That's super particular to benzene rings in the proton NMR.  So it's important for us to recognize aromaticity and we're going to use a set of three rules  in order to decide whether something is aromatic or not.  Here are those three rules.  And the first of those rules is kind of a mathematical rule but I'll have to explain  how we do this accounting for this mathematical rule.  It's called Huckel's Rule.  And it's very simple math.  It's 4N plus 2 where N is a whole number that allows you to calculate how many electrons  should you be counting up when you try to add up the electrons in a potentially aromatic  ring system.  And because N is always a whole number like 1, well 0, 1, 2, 3 or 4, when you do the math  that means it's telling us that all aromatic ring systems have either 2 electrons, 6 electrons,  10 electrons, etc.  And in practice you typically don't get things with more than about 14 electrons.  It takes a lot of synthetic work to try to design systems that have more than that many  electrons but we still give you the mathematical formula in case you want to calculate it out  to infinity.  So let's take the canonical aromatic compound and that's benzene.  When I think of aromaticity, benzene is always the first thing that I think of.  And if I count the electrons here in these pi bonds, it's 2, 4, 6 electrons.  And there's a special requirement for those electrons that I'm adding up.  First of all, they must be in a planar cyclic array.  So if I look at these pi bonds in benzene, they are in a planar cyclic array.  Benzene is planar, benzene is cyclic and they're all connected to each other and conjugated.  If I come over here and I look at this 6-carbon chain that is not connected here, there's  no bond here.  This is not aromatic because it's not cyclic.  So it does have the 6 electrons.  It satisfies that Huckel's 4n plus 2 rule but it's not cyclic.  So obviously that can't be an aromatic ring.  In fact, this is just a really reactive trying.  And then the final point here is that the pi bonds and or p orbitals, you can have 2 electrons  in a p orbital, they have to all overlap with each other.  And so when I come over here and I look at this compound, cycloheptotriene, this CH2  group prevents this end of this double bond here from interacting with this end of this  double bond over here.  So those two ends are not in conjugation.  They are not part of a continuous conjugated cyclic array.  And I'll show you this business about p orbitals.  None of these actually have an example of electrons in a p orbital.  Okay, so that would be an example of an aromatic ring system.  And so let's go ahead and extend that idea.  So I'm going to compliment that by telling you about something that would be the opposite  of aromaticity.  And that would be anti-aromaticity.  So what would be an example of something that is anti-aromatic?  And I think I'm messing this up again.  So let me show you, just give me the ability to write on here.  Okay, so anti-aromaticity is the opposite of aromaticity.  Aromatic compounds are stable.  There's a thermodynamic driving force for organic compounds to want to form aromatic  ring systems.  Those confer stability.  And conversely, it is thermodynamically unfavorable for systems to form anti-aromatic compounds.  So what makes a compound anti-aromatic?  And again, there's going to be three sets of rules that I'm going to apply, and they're  almost identical to the rules for deciding aromaticity.  And the only difference is, instead of 4n plus 2 electrons, it's 4n electrons.  So whenever I see 4 electrons, like 2 pi bonds in a cyclic array, or 8 electrons in a cyclic  array, or 12 electrons, etc., that tells me that that's probably going to be a very difficult  compound to isolate.  Probably that compound is going to be filled with self-loathing and try to self-destruct  or react in some way.  So everything else is the same, if I see 4 electrons or 8 electrons in a planar cyclic  array, all conjugated together, and they're all in overlapping pi bonds and p orbitals,  so they're all conjugated, then that tells me there's going to be problems.  So let me take an example, and whenever I think of anti-aromatic, here's the prototype.  Let me go ahead and use a black pen for this so I can draw this more clearly.  Cyclobutadiene is the canonical anti-aromatic compound.  You cannot just make cyclobutadiene and put it in a flask.  It's not stable.  Yes, question?  You listed 14 electrons before, and the dot was all before.  Yeah, thank you.  16 electrons.  Thanks for asking that.  I guess I'm not so good at math here with 4n.  Once you get to those larger numbers, it's very atypical.  I'm sorry about that math mistake there.  Okay, so of course the problem here is I've got 2 double bonds, pi bonds in a cyclic array,  and I add up the electrons here, and this is definitely anti-aromatic.  And that's why nobody has ever made this on its own.  Chemists have managed to trap cyclobutadiene in little cage-like molecules so it can't escape.  That's one way to stabilize it.  The problem is that when you try to make cyclobutadiene, it's not like we don't have reactions or elimination  reactions that can make that.  Is that it starts to react like this resonance structure.  It hates being anti-aromatic so much.  You can imagine if I take the electrons in this bond here and I give one electron to each  carbon, I'll draw a resonance structure that's a diradical.  And that's a better picture for cyclobutadiene.  It's like each electron is saying, forget that other thing away from me.  I do not want to acknowledge the existence of that other electron.  I don't want to have a pi bond.  I don't want to have conjugation.  So it acts like this.  This is how much cyclobutadiene hates being an anti-aromatic molecule.  Now there are other apparent anti-aromatic molecules that have been synthesized that somehow  manage to avoid being anti-aromatic.  So for example, chemists have made cyclo-octatetraene, a cyclo-octane ring with four double bonds  and it looks kind of aromatic.  It looks kind of like a benzene ring.  The problem is that it's got eight electrons.  So there's two electrons in each pi bond, two electrons here, two electrons here, two electrons  here and two electrons on top here.  That's eight electrons.  And the way that this avoids being anti-aromatic is it bends into this boat-like shape that  prevents the double bonds from being in conjugation.  How clever of that molecule that it figured out that it can bend into this kind of shape  so that the pi bonds are literally orthogonal to each other.  So in fact, cyclo-octatetraene with four double bonds is not anti-aromatic simply because  it's not planar.  If you tried to somehow flatten it to make it planar, it would freak out and probably  start to act like a diradical or do Diels-Alder reactions.  It would find some way to not be to react and turn into a stable molecule.  Okay, so try to recognize aromatic compounds because they don't do addition of the reagents  that you're familiar with, bromine, HBR, hydroboration, hydration.  They have special properties in the NMR spectrum.  And don't expect that you'll be able to make anti-aromatic compounds because you won't be  able to synthesize those.  They'll find some way to finagle themselves out of being aromatic.  They won't be aromatic.  Okay, so let's talk about this issue of planarity.  How do you know if an atom is planar?  Now I'll show you why this is important in a moment.  Not every aromatic compound has only carbon atoms in there.  Sometimes you have to count lone pairs as part of your 4n plus 2 calculation or your  4n calculation.  And we have to decide how we're going to calculate that.  We have to decide which pairs of electrons we're going to count.  And I want to start off by addressing this issue of how do you know whether a pair of  electrons is in a p orbital or not in a p orbital?  I'm going to start off by looking at ammonia and I think all of you know.  Did we?  Let me, does somebody have a set of notes that I can look at so I can try to get back  on track?  Plainer aromatic ions.  Okay, perfect, thank you.  And I'll get to this, I'll come back to this issue in a second.  Okay, how do you know if something is planar?  Before we look at this issue of lone pairs, how do you know whether an ion is planar?  So let me start off by taking a look at a series of ions.  So for most of these compounds here, there's very little opportunity for these to bend  out of planarity.  These are pretty much forced to be planar because they are so small, the ring size is  so small.  There's no opportunity for these to bend out of planarity like that cycloactotetraene.  Kind of the largest ring size I'll talk about here is a seven-membered ring.  And so let's talk about what are our options here?  I obviously can't have a double bond here on this last carbon atom, the top carbon atom  of this three-membered ring.  And so what I'm going to do is I'm going to put a positive charge here.  If I add up how many electrons are in this empty p orbital, and I'm going to struggle  to draw that empty p orbital there, there's zero electrons.  So when I add up all the electrons in this array, this array of conjugated double bonds,  then what I'll find is that there's two electrons.  And because there's two electrons here, that's aromatic.  That's 4n plus 2 where n equals 0.  So this would be a much more stabilized carbocation than you would expect.  And if it truly is aromatic, then we ought to be able to draw a bunch of resonance structures  where we delocalize the pi electrons all over that ring.  And let's just practice that just to make sure that it has this property of delocalization  like benzene.  I'm trying to draw an arrow here where I shift that double bond.  Let's start to draw some of these resonance structures where I shift, and I'll have to  draw this a little bigger here, where I shift that double bond around.  If I take the electrons away from this carbon and move them here, now my cation is over  here, and I can draw another resonance structure where I move those electrons to here and take  the electrons away from this other bottom carbon.  And so this has this property of delocalization just like a benzene ring.  It has 4n plus 2 electrons.  It's planar.  In fact, I don't think that could be anything but planar again.  So this very strange looking, even with all that strain, cyclopropenium ions are stable  and have been made in the laboratory.  It's not like you can put it in a flask, but they have been made, and you can get spectra  of those kinds of species.  Okay, so what do we need to do to a 5-membered ring in order to make that aromatic?  If I put a carbocation here like I did with this cyclopropenium ion, if I put a carbocation  there, I'll only have 4 electrons, and I definitely do not want to do that.  What I'd really like to have here is electrons 5 and 6.  So what I really want to have for this to have this property of aromaticity is I want  that to be a carbanion at that center.  And so I'd really like for these electrons to be in a p-like orbital.  And I'll address that.  I was just getting to that.  Is that p or sp3?  So now if I add up my electrons, I've got a planar array, and I have 2, 4, 6 electrons.  So let's just add that up 2, 2, and 2.  That's 6 electrons.  That is an aromatic anion.  That carbanion is amazingly easy to make.  It's very hard to make carbanions.  It's hard to pluck protons off of carbon atoms.  But in this system, because this gets to be aromatic, it's very easy to make that by deprotonating  carbon.  Okay, so now let's come back to the last.  The 7-membered ring kind of takes practice to learn how to draw 7-membered rings.  That will come with experience.  So now if I want to have 6 or 4n plus 2 electrons in this system, do I want to have a carbocation  here or a carbanion there?  Yeah, we want 0 electrons.  We've already got 6 electrons.  We don't want to add any more.  So we don't want to have a carbanion here, we want to have a carbocation.  That would be a stable aromatic ion.  So if I add up all my electrons, there's 0 in this p orbital.  Let me try to draw that p orbital here.  And I'm not going to draw all the different resonance structures.  There's lots.  But if I add up my electrons, there's 2 in this pi bond, 2 in this pi bond, 2 in this  pi bond.  There's 0 electrons in that empty p orbital.  That's a 6-electron system.  That cation would be aromatic and exceptionally stable.  So what are the implications here?  Implications for you with respect to these aromatic ions is that if you're ever confronted  with cyclopentadiene, it is so easy to pull this proton off that you can use regular bases  like alkoxide bases.  And no other simple alkane allows you to do that.  You can just take that and pull that proton off because the product here, I should have  used a red arrow there, because the product here is aromatic.  The product here is aromatic and heavily stabilized relative to other types of organic  molecules.  OK.  So you can't use alkoxide to pull protons off of hexane or cyclopentane.  Or even if you had just one double bond in this ring instead of two, you couldn't use  alkoxide to deprotonate that.  So those CHs are exceptionally acidic.  Somewhere on the order of 34 orders of magnitude more acidic than a regular alkane.  Let's come back over here and look at this 7-membered ring.  And now it's different.  I'm not talking about deprotonating this, but what I've done is I've put an alcohol  on here that's been protonated.  If you just touch a proton onto that alcohol to put a potential leaving group, this leaving  group will fly off.  And again, the reason why this is so fast, the reason why that is so exceptionally fast  is because it generates an aromatic ion.  So reactions that generate aromatic ions, whether it's deprotonation or ionization, those  reactions are amazingly fast.  And the situation would be totally different now if I did this, just to be clear here.  If I only had one double bond in this ring, that's no longer fast.  There's no base or simple base that you know of or will have heard of in the Chem 51 series  that could ever pull this proton off.  So I'm just going to put a dashed line here because this is insane.  There's no way you could pull that proton off.  That's not going to be aromatic.  So you have to look for systems that have the potential to be aromatic.  Those reactions will be exceptionally fast if they generate an aromatic ion, either a  carb anion or anion.  And let me just sort of remind you here that this is fast and this is fast.  And this other one is just, there's no way, inconceivable.  So gee, isn't it great to become aromatic?  You get to enjoy all of that stability.  Okay so I drew empty p orbitals with carbocation ions.  And then I drew this sort of carb anion where the lone pair was in a p-like orbital.  And how do you decide the hybridization of an atom?  How do you decide whether electrons are in a p-like orbital?  Because only if the electrons are in a p-like orbital do you add them up as part of your  4n plus 2 calculation.  So let me just remind you that we'll use nitrogen, not carbon as our example here.  I think all of you know that simple amines are pyramidal.  Ammonia adopts a pyramidal geometry.  That's the lowest energy hybridization for this nitrogen atom.  So in this sort of a scheme here, these lone pairs exist in this sp3-like orbital.  And that has differential sizes on each end of the orbital.  So one end of the orbital is big, the other end of the orbital is small.  So that would be an sp3 form of hybridization.  If you were to shrink yourself down to the molecular scale and grab these R groups and  bend them to the lower planar, you could force this nitrogen atom to adopt an sp2 hybridized  geometry.  That means each of these bonds have sp2 hybridization, but the electrons in the lone pair, they're  in a p-like orbital.  And a p-like orbital has the same size on the top and the bottom.  And that would be a requirement for us to add this into our 4n plus 2 rule.  So let's see how that comes into play.  How do you decide in this 5-membered ring system here?  What I'm doing is I'm drawing edge on, this is really just the molecule parole.  And it's got substituents on all these.  I drew substituents on all these.  And so the question is, is parole aromatic?  Does this 5-membered ring with that nitrogen atom and its lone pair, is that aromatic?  Do I count that lone pair in my 4n plus 2 rule?  If that lone pair can contribute to aromaticity, it will contribute to aromaticity.  So there's two possible hybridizations for this.  Most amines don't exist in a planar form, but when I give this lone pair the opportunity  to participate in an aromatic ring system, in this 4n plus 2 ring system, it will rehybridize.  So flat in this case is more stable because it gets to be aromatic.  And that's not normal for amines.  Amines don't do that.  By being flat, these electrons get to participate in a p orbital.  So there's my p orbital there.  And when those electrons are in our p orbital, we have equal hybridization there on the top  and the bottom face.  Or equal, I guess the p orbital has equal size on the top and bottom face.  And that allows it to effectively overlap with the pi bonds, which have electron density  on the top and the bottom face.  So you'll count lone pairs in this 4n plus 2 rule if they can exist in p-like orbitals.  And I'm making a big deal about this because there will be cases where lone pairs cannot  exist in these p-like orbitals.  So you would expect those to have less aromatic resonance energy.  Okay, so let's take a look at some systems that have nitrogen in them.  Some rings that have nitrogen or oxygen.  And try to decide should we count the lone pairs.  And I'm going to give you a simple rule to help, I say simple.  Maybe it's very wordy, but once you use it a few times, it becomes very user friendly.  So I'm sorry, maybe the rule is not, doesn't sound simple.  So here's a whole set of heterocycles.  Heterocycle means there's a heteroatom in the ring.  Usually oxygen or nitrogen, but you can have lower row heteroatoms like sulfur in the ring.  So for heteroatoms, how do I decide which of these pairs of electrons to add in to my  4n plus 2 rule to decide whether these are aromatic or anti-aromatic?  And here's our simple rule.  And I'll read this out to you.  If the heteroatom is not part of a double bond, if it's not part of a double bond, then you  can count one lone pair towards the 4n plus 2 rule.  Now that's a whole string of logic there, so let's practice using that.  So if a heteroatom is not part of a double bond, okay, well that nitrogen is part of  a double bond.  That means I shouldn't count that lone pair.  I only count lone pairs if they're attached to atoms that are not part of a double bond.  So I'm going to go ahead and cross this out.  No, I don't count that.  The nitrogen's part of a double bond.  So now how many electrons do I have to count?  I've got just the double bonds left over.  That has six electrons and that's aromatic.  So I think you'll be as good as I am at figuring out which of these heteroatoms are part of  a double bond.  Okay, so let's take this next heterocycle.  This is called furan.  Okay, if the heteroatom is not part of a double bond, well that oxygen is not part of a double  bond.  There's only single bonds to that oxygen.  Okay, if the heteroatom is not part of a double bond, you count one lone pair towards the  4n plus 2 rule.  So I've got two lone pairs here.  I'm going to cross one of them out and I'm only going to count one of them.  So how many electrons in this system?  Six, two, four, six.  There's two in this lone pair and then two in each of the pi bonds.  That's a total of six electrons.  That's aromatic.  Okay, let's try to apply it to this heterocycle here that has the two nitrogens next to each  other.  So if I look at this, again my rule is if the heteroatom is not part of a double bond,  well neither of those nitrogens is part of a double bond, then I get to count one lone  pair towards the 4n plus 2 rule.  So I count both of those lone pairs.  Both of those nitrogens are not part of a double bond.  So when I count up these electrons here, that would be eight electrons and that would be  anti-aromatic.  If you could somehow synthesize a molecule with that pattern of bonding, that molecule  would be very reactive.  It would try to find some way to distort out of planarity or it would try to find some way  to react and so it wouldn't have double bonds in that arrangement.  That molecule would be extremely unhappy.  Okay, so here's a five-membered ring.  This is the side chain of the amino acid histidine.  It's called a mitazole.  And so let's use our little accounting system here.  If a heteroatom is not part of a double bond, well this nitrogen atom on top here is part  of a double bond.  So I don't count that lone pair.  The nitrogen's part of a double bond.  Okay, let's look at the bottom nitrogen.  And our rule says if the heteroatom is not part of a double bond, you count one lone  pair towards 4n plus 2.  Oh, well we count that.  So now we can count up our electrons.  There's two in this pi bond, two in this pi bond, and two in that lone pair.  That's six electrons.  The side chain of histidine, a mitazole, is aromatic.  It's an aromatic ring.  So every protein in your body is filled, histidine is a common amino acid.  It's filled with aromatic rings on the surface of your proteins and the inside.  Okay, here's one of the bases of DNA and RNA.  This is adenine.  And so now there's two sets of rings here.  And I can draw different resonance structures.  Whenever you've got a double bond between two different rings that are fused together,  you double count that double bond.  You can look at each ring independently and count that double bond towards each ring.  So let's take each of these rings independently.  I think you can see that the ring over here on the left hand side is a mitazole.  We already did that one.  Right?  Let's apply our rule.  If the lone pair is not part of a double bond, well this lone pair is part of a double  bond.  I'm not going to count that.  And I come over here and this nitrogen atom, well that's part of a double bond.  That means I'm not going to count that.  And what about this nitrogen atom over here?  Well that's part of a double bond.  I can't count that.  So the only lone pair I count is on the nitrogen that's not part of a double bond.  So now when I start off and first count up this five-membered ring here and I add up  those electrons.  There's two in the lone pair, two in the pi bond, two in the pi bond.  This ring here has six electrons and that's aromatic.  Now I come over here to the six-membered ring.  I don't count any of those lone pairs.  And there's a lone pair on the nitrogen but that's not the amino substituent.  That's not inside the ring.  That's not part of this ring.  So don't even worry about the substituents.  So I have two, four, six electrons.  This ring is aromatic.  Both the rings of the DNA base adenine or aromatic.  And you would draw the same conclusions about guanine, cytosine, depending on which resonance  structure you draw for those.  You could consider the other DNA bases to have either partial or full aromaticity.  Okay.  So let's just come back to why does this rule work?  The reason why this rule works.  I mean it's very empirical.  You don't gain any insight about the geometry of the molecule simply by deciding whether  an atom is part of a double bond.  All those electrons that we were counting as part of that rule will exist in p-like orbitals.  So when we were counting lone pairs in our 4n plus 2 rule, that's because they were able  to adopt geometries that were flat so that the lone pair could be in a p-like orbital.  So it could interact effectively both and I'm not sure I want to try to draw this in  here but with pi bonds that have electron density top and bottom.  What about the electron lone pairs that we crossed out?  In every case where we crossed out a lone pair, when a heteroatom was part of a double  bond like here in pyridine, what that means is that lone pair existed in an sp2-like orbital  sticking out of the side of the aromatic ring.  So when you didn't count the electrons, they're not sticking up and down in a p-orbital, they're  sticking off to the side of the aromatic ring or of the benzene ring or pyridine ring.  The implications are these lone pairs that are involved in aromaticity, they're busy.  Don't try to mess with those because you'll disrupt aromaticity.  So what you would find is that lone pairs like this are simply not reactive.  It's very difficult to protonate them.  You couldn't protonate that.  In fact, if you try to protonate, if you struggle to protonate pyridine by throwing in a powerful  acid, it protonates on carbon even.  So don't try to mess with that lone pair because it's doing something.  It's trying to make the system aromatic.  In contrast, the lone pairs here that are sticking out to the side of an aromatic ring,  no problem reacting with those.  And you know that already.  Pyridine is a base.  You've already done reactions where you used pyridine as a base.  So you can do all kinds of reactions with these lone pairs that are not part of the aromatic  4n plus 2 rule.  So no problem reacting with the lone pair on pyridine or this crossed out lone pair  on histidine.  If you look inside the active sites of enzymes, one of the most common lone pairs used in  catalysis is this lone pair that's sticking off to the side of a histidine side chain.  If you look at every single amide bond that is produced in every single protein in your  body, every single amide bond is created because this lone pair on the bottom of this  adenine ring, a single adenine, a single lone pair inside the ribosome is responsible for  the synthesis of every single protein, every single amide bond in the human body.  So lone pairs that stick off to the side are completely reactive whereas lone pairs that  are part of aromatic systems are not reactive at all.  Okay.  So I think you can see from adenine that you confuse benzene rings together and have much  larger aromatic ring systems.  So I'm going to give you an example of some fused systems here.  So you can see that there's more than one way to account for this 4n plus 2 rule.  There's more than one way to do the addition there.  So this is a molecule called naphthalene.  If you go to the hardware store and buy moth balls, it turns out that crystalline naphthalene  just by solid balls of crystalline naphthalene and for whatever reasons moths seem to find  that repulsive and will tend to stay out of your closets.  So there's multiple resonance structures that I can draw for this.  If I count around, if I just ignored this ring, this bond in the middle of those two  rings, that sigma bond, just pretend that it's not there, I can count around the outside  edges of this flat molecule.  And if I count around the outside edges of that flat molecule, what I would find is that  there's 2, 4, 6, 8, 10 electrons.  That satisfies our 4n plus 2 rule and that's aromatic.  There's another way I could have done the accounting for this, that's slightly different,  another way to describe this is I could have drawn a resonance structure for naphthalene  in which I put a double bond in the middle.  There's a different way I can shift around those electrons.  And so remember what I said is if there's a double bond at the ring fusion, you can  count that double bond for both rings.  So in other words, the other way I could have accounted this is I could have said in this  ring over here on the left hand side, there's 2, 4, 6 electrons and that's aromatic.  And then I look over here at this ring on the other side and there's 2, 4, 6, I double  count, I get to count that double bond again and that's aromatic.  So I can count that middle, that double bond in the center twice and I reach the same conclusions  either way and that is that all of the carbons in this ring system are part of an aromatic  ring.  I won't reach a different conclusion.  So you can fuse these things in all kinds of different ways.  Let me give you another example of another aromatic ring that's very common that's fused  and this time it's a parole ring fused onto a benzene ring.  This is the side chain of the amino acid tryptophan so I'll just write trip here and there's a  lone pair and there's nitrogen.  So again in this ring system both rings, both of these rings in this heterocyclic, this is  called an indole, this bicyclic ring system, both of these rings are aromatic.  This ring over here which obviously is benzene, that's aromatic, it has 6 electrons.  If I look at this 5-membered ring over here there's 2 electrons in the nitrogen, 2 electrons  on this side bond and then 2 electrons over here that's also aromatic.  So the side chain of tryptophan is fully aromatic as is the side chain of tyrosine and phenylalinane.  You guys are probably learning about those in Bio98 right now.  Okay so you can, it can go endlessly.  You can fuse benzene ring after benzene ring after benzene ring onto each other.  You may have heard of carbon nanotubes or graphene, that's basically what those molecules  are.  They're just infinite tubes made out of benzene rings all fused together.  Okay so the last parts of this chapter are interesting but they're not going to be useful  to you for anything else in this quarter.  They have to do with molecular orbital base of aromaticity and I think you ought to read  that.  I would never discourage you from reading anything but it's not the kind of thing that  I can test on.  People learn about those powerful concepts if you take upper division classes and I would  love to teach those to you there but we just can't fit those in with the time that we've  got.  So be able to recognize aromatic rings, be able to recognize anti-aromatic rings.  You need to be able to know when ionization reactions will generate aromatic anions or  aromatic cations.  Those ionizations or deprotonations will be fast and be able to recognize when heterocycles  with nitrogen and oxygen are aromatic.  So be ready when we come back on Monday we're going to do chemical reactions of benzene rings  and we're going to do a lot of them.
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I saw this rather horrific story on the news last night. Dad was watching his kid in the garage and suddenly this rattlesnake jumped out of nowhere and bit the child on the finger. And I can't imagine how scared they both were. I can't imagine how scared that little child was going to be of snakes far into the future. Fortunately, this child was doing all right. They took him to a specialist at Loma Linda University and they were able to inject 22 vials of anti-venom in order to stop the action of the snake venom. And so let's go ahead and talk about this anti-venom business here. What's going on with anti-venom that they're giving to people in the hospital? So if you go for a walk in the woods and get bit by a snake, what's your prognosis? So this is a Southern Pacific rattlesnake. That's the most common rattlesnake that you find in the Lake Elsinore region. And if they take you in with a rattlesnake bite, they're probably going to give you this commercial anti-venom. It's produced commercially. They don't make it there in the hospital. And so let's talk about this particular venom called CroFab. If you look at the prescribing information or just the title in the box, it says, Crotolidiae Polyvalent Immune Fab, Ovine. And what that tells you is that it's an antigen binding fragment that is produced from monoclonal antibodies that come from sheep. That's what Ovine means. It means it comes from sheep blood. So what they do is they typically collect venom from many snakes and then they use that venom to immunize sheep. So you immunize sheep, you combine it with something called Croin's Adjuvant, typically, to get the immune system going. And then a couple of weeks later, you give it another boost and the immune system freaks out and is totally ready and starts cranking out tons and tons of antibodies to all the different things that you find in snake venom, in this particular snake venom. And there are lots of nasty things in snake venom. Snake venoms are designed to mess with the cardiovascular system, at least these types of snakes. So they contain enzymes and proteins that degrade tissue, that break down proteins, that lyse blood cells, and that prevent clotting. So you can imagine what a bad mix of toxins this is. So when you inject this into a sheep, the sheep's immune system produces all these different antibodies to all these different bad things that you find inside of the snake venom. And so then you come along and you collect blood from the sheep, doesn't kill the sheep, right? You can go give blood and it doesn't kill you. And then they isolate the antibodies, they use affinity chromatography to isolate all the antibodies and leave all the blood junk behind. And so what do they do with all these antibodies? The last thing in the world you want to do is take a sheep antibody and inject it into your human being. See these parts right here on the antibody called the antigen binding pheromones bind specifically to all those snake toxins. But this part right here doesn't help you. It's just a little bullseye that helps your immune system home in on this and say, hey, that's not human, I'm going to attack it. So what they have to do to prepare this for a human being is to chop off this constant fragment down here and they use an enzyme to do that. And they use an enzyme that's derived from papayas called papain. And that papain comes along and cleaves a single amide bond on each arm here. It's just a single carbon nitrogen bond, but it's extremely specific. And so the enzyme papain cleaves off these little arms, they throw away the bottom part, don't need that. And these top parts look a lot like the antigen binding fragments from your antibodies. Your immune system can't tell the difference for all intents and purposes. So what they're doing is they're collecting these antigen binding fragments that bind to all these different snake venom components. They put it into a vial and those things sit in the refrigerator waiting for somebody to get bitten by a snake. So it's a simple use of an enzyme to cleave very specifically amide bonds here on the arms of immunoglobulins of antibodies. Say again. How effective is that? Is this venom? I'm not an expert. How effective is the anti-venom? I'm not an expert on anti-venoms, but after reading a couple of these stories, it's very effective. If you can get to the hospital very quickly, you're in good shape. If you're stuck out on the woods on a six-day hike, that's a problem. And this is one of the nastiest toxins. You notice they had to use a lot of ant, whatever they said, 22 vials of anti-venom and the kid's doing fine. So you're a lot larger than a kid, so it would be a lot harder to hurt you. But I had a mistake on one of my slides during the last lecture on the Diels-Alder reaction. And I was pointing out this nomenclature for XO versus endo. And what I said was the substituent down here that's on the same side as the alkene bridge is referred to as endo. But then I had a typographical, I mean just a mistake. I said over here substituents on the same side and I again said alkene. I should have said alkane. That should be alkane bridge there, not alkene bridge. They can't both be on the same side as the alkene bridge. So this is the alkane bridge here and that's XO. I'll just write alkane. So please replace the letter E there with the letter A on the bottom. And if you're not exactly sure, look at the book because I just had the same word twice and that's because I was copying and pasting too much. Okay so we moved on from chapter 16. Chapter 16 was a lot of dienes and conjugation and we moved on to chapter 17 to talk about aromaticity. And one of the main points about aromaticity is that you learned all these different reactions of double bonds. Lots and lots of different reactions of double bonds and what's going on there. In chapter 10 you learn more reactions of pi bonds in alkynes in chapter 11. But for some reason the double bonds of benzene are resistant to the typical reagents that you know. Bromine doesn't add across the double bonds of benzene rings. HBR doesn't add across the double bonds of benzene rings. You can't hydroborate the double bonds of benzene rings. And so it seems like it's pretty important for us to be able to recognize when double bonds have this property of aromaticity. When is a double bond part of an aromatic ring system? We need to know that so we know which double bonds have low reactivity towards typical electrophilic reagents. So I'm going to give you just one moment here. I'm going to give you, if I can get my act together here. Yes, sir. It is amazing how quickly I have now mixed up all of my little sheets that I had here. So you will have to just pardon me. You can stare at the chalkboard behind me while I'm sitting here trying to sort through this. I want to note that we have extra office hours next week. Next week is the last week of lectures in this class. So of course what that means to you is you've got one week left to study for the final exam which is on Monday of final exams week. So please make this last week count as we are going through that kind of material. So I want to talk about aromaticity and how to recognize aromatic ring systems when they're there in front of you. So I'll just do the best to sort through my disarray of notes here. I want to introduce you to some, oh that's not the, here sorry about that. That's not what I wanted to give you. Here's what I wanted to show you. It's to try to give you the ability to recognize aromatic and anti-aromatic compounds. So probably you have your notes in front of you and know where they go. Okay, so let's talk about this. How do you recognize an aromatic compound? This property of aromaticity that causes double bonds to be unreactive. That causes protons in your NMR spectrum to be shifted way downfield to 7 to 8 parts per million. That's super particular to benzene rings in the proton NMR. So it's important for us to recognize aromaticity and we're going to use a set of three rules in order to decide whether something is aromatic or not. Here are those three rules. And the first of those rules is kind of a mathematical rule but I'll have to explain how we do this accounting for this mathematical rule. It's called Huckel's Rule. And it's very simple math. It's 4N plus 2 where N is a whole number that allows you to calculate how many electrons should you be counting up when you try to add up the electrons in a potentially aromatic ring system. And because N is always a whole number like 1, well 0, 1, 2, 3 or 4, when you do the math that means it's telling us that all aromatic ring systems have either 2 electrons, 6 electrons, 10 electrons, etc. And in practice you typically don't get things with more than about 14 electrons. It takes a lot of synthetic work to try to design systems that have more than that many electrons but we still give you the mathematical formula in case you want to calculate it out to infinity. So let's take the canonical aromatic compound and that's benzene. When I think of aromaticity, benzene is always the first thing that I think of. And if I count the electrons here in these pi bonds, it's 2, 4, 6 electrons. And there's a special requirement for those electrons that I'm adding up. First of all, they must be in a planar cyclic array. So if I look at these pi bonds in benzene, they are in a planar cyclic array. Benzene is planar, benzene is cyclic and they're all connected to each other and conjugated. If I come over here and I look at this 6-carbon chain that is not connected here, there's no bond here. This is not aromatic because it's not cyclic. So it does have the 6 electrons. It satisfies that Huckel's 4n plus 2 rule but it's not cyclic. So obviously that can't be an aromatic ring. In fact, this is just a really reactive trying. And then the final point here is that the pi bonds and or p orbitals, you can have 2 electrons in a p orbital, they have to all overlap with each other. And so when I come over here and I look at this compound, cycloheptotriene, this CH2 group prevents this end of this double bond here from interacting with this end of this double bond over here. So those two ends are not in conjugation. They are not part of a continuous conjugated cyclic array. And I'll show you this business about p orbitals. None of these actually have an example of electrons in a p orbital. Okay, so that would be an example of an aromatic ring system. And so let's go ahead and extend that idea. So I'm going to compliment that by telling you about something that would be the opposite of aromaticity. And that would be anti-aromaticity. So what would be an example of something that is anti-aromatic? And I think I'm messing this up again. So let me show you, just give me the ability to write on here. Okay, so anti-aromaticity is the opposite of aromaticity. Aromatic compounds are stable. There's a thermodynamic driving force for organic compounds to want to form aromatic ring systems. Those confer stability. And conversely, it is thermodynamically unfavorable for systems to form anti-aromatic compounds. So what makes a compound anti-aromatic? And again, there's going to be three sets of rules that I'm going to apply, and they're almost identical to the rules for deciding aromaticity. And the only difference is, instead of 4n plus 2 electrons, it's 4n electrons. So whenever I see 4 electrons, like 2 pi bonds in a cyclic array, or 8 electrons in a cyclic array, or 12 electrons, etc., that tells me that that's probably going to be a very difficult compound to isolate. Probably that compound is going to be filled with self-loathing and try to self-destruct or react in some way. So everything else is the same, if I see 4 electrons or 8 electrons in a planar cyclic array, all conjugated together, and they're all in overlapping pi bonds and p orbitals, so they're all conjugated, then that tells me there's going to be problems. So let me take an example, and whenever I think of anti-aromatic, here's the prototype. Let me go ahead and use a black pen for this so I can draw this more clearly. Cyclobutadiene is the canonical anti-aromatic compound. You cannot just make cyclobutadiene and put it in a flask. It's not stable. Yes, question? You listed 14 electrons before, and the dot was all before. Yeah, thank you. 16 electrons. Thanks for asking that. I guess I'm not so good at math here with 4n. Once you get to those larger numbers, it's very atypical. I'm sorry about that math mistake there. Okay, so of course the problem here is I've got 2 double bonds, pi bonds in a cyclic array, and I add up the electrons here, and this is definitely anti-aromatic. And that's why nobody has ever made this on its own. Chemists have managed to trap cyclobutadiene in little cage-like molecules so it can't escape. That's one way to stabilize it. The problem is that when you try to make cyclobutadiene, it's not like we don't have reactions or elimination reactions that can make that. Is that it starts to react like this resonance structure. It hates being anti-aromatic so much. You can imagine if I take the electrons in this bond here and I give one electron to each carbon, I'll draw a resonance structure that's a diradical. And that's a better picture for cyclobutadiene. It's like each electron is saying, forget that other thing away from me. I do not want to acknowledge the existence of that other electron. I don't want to have a pi bond. I don't want to have conjugation. So it acts like this. This is how much cyclobutadiene hates being an anti-aromatic molecule. Now there are other apparent anti-aromatic molecules that have been synthesized that somehow manage to avoid being anti-aromatic. So for example, chemists have made cyclo-octatetraene, a cyclo-octane ring with four double bonds and it looks kind of aromatic. It looks kind of like a benzene ring. The problem is that it's got eight electrons. So there's two electrons in each pi bond, two electrons here, two electrons here, two electrons here and two electrons on top here. That's eight electrons. And the way that this avoids being anti-aromatic is it bends into this boat-like shape that prevents the double bonds from being in conjugation. How clever of that molecule that it figured out that it can bend into this kind of shape so that the pi bonds are literally orthogonal to each other. So in fact, cyclo-octatetraene with four double bonds is not anti-aromatic simply because it's not planar. If you tried to somehow flatten it to make it planar, it would freak out and probably start to act like a diradical or do Diels-Alder reactions. It would find some way to not be to react and turn into a stable molecule. Okay, so try to recognize aromatic compounds because they don't do addition of the reagents that you're familiar with, bromine, HBR, hydroboration, hydration. They have special properties in the NMR spectrum. And don't expect that you'll be able to make anti-aromatic compounds because you won't be able to synthesize those. They'll find some way to finagle themselves out of being aromatic. They won't be aromatic. Okay, so let's talk about this issue of planarity. How do you know if an atom is planar? Now I'll show you why this is important in a moment. Not every aromatic compound has only carbon atoms in there. Sometimes you have to count lone pairs as part of your 4n plus 2 calculation or your 4n calculation. And we have to decide how we're going to calculate that. We have to decide which pairs of electrons we're going to count. And I want to start off by addressing this issue of how do you know whether a pair of electrons is in a p orbital or not in a p orbital? I'm going to start off by looking at ammonia and I think all of you know. Did we? Let me, does somebody have a set of notes that I can look at so I can try to get back on track? Plainer aromatic ions. Okay, perfect, thank you. And I'll get to this, I'll come back to this issue in a second. Okay, how do you know if something is planar? Before we look at this issue of lone pairs, how do you know whether an ion is planar? So let me start off by taking a look at a series of ions. So for most of these compounds here, there's very little opportunity for these to bend out of planarity. These are pretty much forced to be planar because they are so small, the ring size is so small. There's no opportunity for these to bend out of planarity like that cycloactotetraene. Kind of the largest ring size I'll talk about here is a seven-membered ring. And so let's talk about what are our options here? I obviously can't have a double bond here on this last carbon atom, the top carbon atom of this three-membered ring. And so what I'm going to do is I'm going to put a positive charge here. If I add up how many electrons are in this empty p orbital, and I'm going to struggle to draw that empty p orbital there, there's zero electrons. So when I add up all the electrons in this array, this array of conjugated double bonds, then what I'll find is that there's two electrons. And because there's two electrons here, that's aromatic. That's 4n plus 2 where n equals 0. So this would be a much more stabilized carbocation than you would expect. And if it truly is aromatic, then we ought to be able to draw a bunch of resonance structures where we delocalize the pi electrons all over that ring. And let's just practice that just to make sure that it has this property of delocalization like benzene. I'm trying to draw an arrow here where I shift that double bond. Let's start to draw some of these resonance structures where I shift, and I'll have to draw this a little bigger here, where I shift that double bond around. If I take the electrons away from this carbon and move them here, now my cation is over here, and I can draw another resonance structure where I move those electrons to here and take the electrons away from this other bottom carbon. And so this has this property of delocalization just like a benzene ring. It has 4n plus 2 electrons. It's planar. In fact, I don't think that could be anything but planar again. So this very strange looking, even with all that strain, cyclopropenium ions are stable and have been made in the laboratory. It's not like you can put it in a flask, but they have been made, and you can get spectra of those kinds of species. Okay, so what do we need to do to a 5-membered ring in order to make that aromatic? If I put a carbocation here like I did with this cyclopropenium ion, if I put a carbocation there, I'll only have 4 electrons, and I definitely do not want to do that. What I'd really like to have here is electrons 5 and 6. So what I really want to have for this to have this property of aromaticity is I want that to be a carbanion at that center. And so I'd really like for these electrons to be in a p-like orbital. And I'll address that. I was just getting to that. Is that p or sp3? So now if I add up my electrons, I've got a planar array, and I have 2, 4, 6 electrons. So let's just add that up 2, 2, and 2. That's 6 electrons. That is an aromatic anion. That carbanion is amazingly easy to make. It's very hard to make carbanions. It's hard to pluck protons off of carbon atoms. But in this system, because this gets to be aromatic, it's very easy to make that by deprotonating carbon. Okay, so now let's come back to the last. The 7-membered ring kind of takes practice to learn how to draw 7-membered rings. That will come with experience. So now if I want to have 6 or 4n plus 2 electrons in this system, do I want to have a carbocation here or a carbanion there? Yeah, we want 0 electrons. We've already got 6 electrons. We don't want to add any more. So we don't want to have a carbanion here, we want to have a carbocation. That would be a stable aromatic ion. So if I add up all my electrons, there's 0 in this p orbital. Let me try to draw that p orbital here. And I'm not going to draw all the different resonance structures. There's lots. But if I add up my electrons, there's 2 in this pi bond, 2 in this pi bond, 2 in this pi bond. There's 0 electrons in that empty p orbital. That's a 6-electron system. That cation would be aromatic and exceptionally stable. So what are the implications here? Implications for you with respect to these aromatic ions is that if you're ever confronted with cyclopentadiene, it is so easy to pull this proton off that you can use regular bases like alkoxide bases. And no other simple alkane allows you to do that. You can just take that and pull that proton off because the product here, I should have used a red arrow there, because the product here is aromatic. The product here is aromatic and heavily stabilized relative to other types of organic molecules. OK. So you can't use alkoxide to pull protons off of hexane or cyclopentane. Or even if you had just one double bond in this ring instead of two, you couldn't use alkoxide to deprotonate that. So those CHs are exceptionally acidic. Somewhere on the order of 34 orders of magnitude more acidic than a regular alkane. Let's come back over here and look at this 7-membered ring. And now it's different. I'm not talking about deprotonating this, but what I've done is I've put an alcohol on here that's been protonated. If you just touch a proton onto that alcohol to put a potential leaving group, this leaving group will fly off. And again, the reason why this is so fast, the reason why that is so exceptionally fast is because it generates an aromatic ion. So reactions that generate aromatic ions, whether it's deprotonation or ionization, those reactions are amazingly fast. And the situation would be totally different now if I did this, just to be clear here. If I only had one double bond in this ring, that's no longer fast. There's no base or simple base that you know of or will have heard of in the Chem 51 series that could ever pull this proton off. So I'm just going to put a dashed line here because this is insane. There's no way you could pull that proton off. That's not going to be aromatic. So you have to look for systems that have the potential to be aromatic. Those reactions will be exceptionally fast if they generate an aromatic ion, either a carb anion or anion. And let me just sort of remind you here that this is fast and this is fast. And this other one is just, there's no way, inconceivable. So gee, isn't it great to become aromatic? You get to enjoy all of that stability. Okay so I drew empty p orbitals with carbocation ions. And then I drew this sort of carb anion where the lone pair was in a p-like orbital. And how do you decide the hybridization of an atom? How do you decide whether electrons are in a p-like orbital? Because only if the electrons are in a p-like orbital do you add them up as part of your 4n plus 2 calculation. So let me just remind you that we'll use nitrogen, not carbon as our example here. I think all of you know that simple amines are pyramidal. Ammonia adopts a pyramidal geometry. That's the lowest energy hybridization for this nitrogen atom. So in this sort of a scheme here, these lone pairs exist in this sp3-like orbital. And that has differential sizes on each end of the orbital. So one end of the orbital is big, the other end of the orbital is small. So that would be an sp3 form of hybridization. If you were to shrink yourself down to the molecular scale and grab these R groups and bend them to the lower planar, you could force this nitrogen atom to adopt an sp2 hybridized geometry. That means each of these bonds have sp2 hybridization, but the electrons in the lone pair, they're in a p-like orbital. And a p-like orbital has the same size on the top and the bottom. And that would be a requirement for us to add this into our 4n plus 2 rule. So let's see how that comes into play. How do you decide in this 5-membered ring system here? What I'm doing is I'm drawing edge on, this is really just the molecule parole. And it's got substituents on all these. I drew substituents on all these. And so the question is, is parole aromatic? Does this 5-membered ring with that nitrogen atom and its lone pair, is that aromatic? Do I count that lone pair in my 4n plus 2 rule? If that lone pair can contribute to aromaticity, it will contribute to aromaticity. So there's two possible hybridizations for this. Most amines don't exist in a planar form, but when I give this lone pair the opportunity to participate in an aromatic ring system, in this 4n plus 2 ring system, it will rehybridize. So flat in this case is more stable because it gets to be aromatic. And that's not normal for amines. Amines don't do that. By being flat, these electrons get to participate in a p orbital. So there's my p orbital there. And when those electrons are in our p orbital, we have equal hybridization there on the top and the bottom face. Or equal, I guess the p orbital has equal size on the top and bottom face. And that allows it to effectively overlap with the pi bonds, which have electron density on the top and the bottom face. So you'll count lone pairs in this 4n plus 2 rule if they can exist in p-like orbitals. And I'm making a big deal about this because there will be cases where lone pairs cannot exist in these p-like orbitals. So you would expect those to have less aromatic resonance energy. Okay, so let's take a look at some systems that have nitrogen in them. Some rings that have nitrogen or oxygen. And try to decide should we count the lone pairs. And I'm going to give you a simple rule to help, I say simple. Maybe it's very wordy, but once you use it a few times, it becomes very user friendly. So I'm sorry, maybe the rule is not, doesn't sound simple. So here's a whole set of heterocycles. Heterocycle means there's a heteroatom in the ring. Usually oxygen or nitrogen, but you can have lower row heteroatoms like sulfur in the ring. So for heteroatoms, how do I decide which of these pairs of electrons to add in to my 4n plus 2 rule to decide whether these are aromatic or anti-aromatic? And here's our simple rule. And I'll read this out to you. If the heteroatom is not part of a double bond, if it's not part of a double bond, then you can count one lone pair towards the 4n plus 2 rule. Now that's a whole string of logic there, so let's practice using that. So if a heteroatom is not part of a double bond, okay, well that nitrogen is part of a double bond. That means I shouldn't count that lone pair. I only count lone pairs if they're attached to atoms that are not part of a double bond. So I'm going to go ahead and cross this out. No, I don't count that. The nitrogen's part of a double bond. So now how many electrons do I have to count? I've got just the double bonds left over. That has six electrons and that's aromatic. So I think you'll be as good as I am at figuring out which of these heteroatoms are part of a double bond. Okay, so let's take this next heterocycle. This is called furan. Okay, if the heteroatom is not part of a double bond, well that oxygen is not part of a double bond. There's only single bonds to that oxygen. Okay, if the heteroatom is not part of a double bond, you count one lone pair towards the 4n plus 2 rule. So I've got two lone pairs here. I'm going to cross one of them out and I'm only going to count one of them. So how many electrons in this system? Six, two, four, six. There's two in this lone pair and then two in each of the pi bonds. That's a total of six electrons. That's aromatic. Okay, let's try to apply it to this heterocycle here that has the two nitrogens next to each other. So if I look at this, again my rule is if the heteroatom is not part of a double bond, well neither of those nitrogens is part of a double bond, then I get to count one lone pair towards the 4n plus 2 rule. So I count both of those lone pairs. Both of those nitrogens are not part of a double bond. So when I count up these electrons here, that would be eight electrons and that would be anti-aromatic. If you could somehow synthesize a molecule with that pattern of bonding, that molecule would be very reactive. It would try to find some way to distort out of planarity or it would try to find some way to react and so it wouldn't have double bonds in that arrangement. That molecule would be extremely unhappy. Okay, so here's a five-membered ring. This is the side chain of the amino acid histidine. It's called a mitazole. And so let's use our little accounting system here. If a heteroatom is not part of a double bond, well this nitrogen atom on top here is part of a double bond. So I don't count that lone pair. The nitrogen's part of a double bond. Okay, let's look at the bottom nitrogen. And our rule says if the heteroatom is not part of a double bond, you count one lone pair towards 4n plus 2. Oh, well we count that. So now we can count up our electrons. There's two in this pi bond, two in this pi bond, and two in that lone pair. That's six electrons. The side chain of histidine, a mitazole, is aromatic. It's an aromatic ring. So every protein in your body is filled, histidine is a common amino acid. It's filled with aromatic rings on the surface of your proteins and the inside. Okay, here's one of the bases of DNA and RNA. This is adenine. And so now there's two sets of rings here. And I can draw different resonance structures. Whenever you've got a double bond between two different rings that are fused together, you double count that double bond. You can look at each ring independently and count that double bond towards each ring. So let's take each of these rings independently. I think you can see that the ring over here on the left hand side is a mitazole. We already did that one. Right? Let's apply our rule. If the lone pair is not part of a double bond, well this lone pair is part of a double bond. I'm not going to count that. And I come over here and this nitrogen atom, well that's part of a double bond. That means I'm not going to count that. And what about this nitrogen atom over here? Well that's part of a double bond. I can't count that. So the only lone pair I count is on the nitrogen that's not part of a double bond. So now when I start off and first count up this five-membered ring here and I add up those electrons. There's two in the lone pair, two in the pi bond, two in the pi bond. This ring here has six electrons and that's aromatic. Now I come over here to the six-membered ring. I don't count any of those lone pairs. And there's a lone pair on the nitrogen but that's not the amino substituent. That's not inside the ring. That's not part of this ring. So don't even worry about the substituents. So I have two, four, six electrons. This ring is aromatic. Both the rings of the DNA base adenine or aromatic. And you would draw the same conclusions about guanine, cytosine, depending on which resonance structure you draw for those. You could consider the other DNA bases to have either partial or full aromaticity. Okay. So let's just come back to why does this rule work? The reason why this rule works. I mean it's very empirical. You don't gain any insight about the geometry of the molecule simply by deciding whether an atom is part of a double bond. All those electrons that we were counting as part of that rule will exist in p-like orbitals. So when we were counting lone pairs in our 4n plus 2 rule, that's because they were able to adopt geometries that were flat so that the lone pair could be in a p-like orbital. So it could interact effectively both and I'm not sure I want to try to draw this in here but with pi bonds that have electron density top and bottom. What about the electron lone pairs that we crossed out? In every case where we crossed out a lone pair, when a heteroatom was part of a double bond like here in pyridine, what that means is that lone pair existed in an sp2-like orbital sticking out of the side of the aromatic ring. So when you didn't count the electrons, they're not sticking up and down in a p-orbital, they're sticking off to the side of the aromatic ring or of the benzene ring or pyridine ring. The implications are these lone pairs that are involved in aromaticity, they're busy. Don't try to mess with those because you'll disrupt aromaticity. So what you would find is that lone pairs like this are simply not reactive. It's very difficult to protonate them. You couldn't protonate that. In fact, if you try to protonate, if you struggle to protonate pyridine by throwing in a powerful acid, it protonates on carbon even. So don't try to mess with that lone pair because it's doing something. It's trying to make the system aromatic. In contrast, the lone pairs here that are sticking out to the side of an aromatic ring, no problem reacting with those. And you know that already. Pyridine is a base. You've already done reactions where you used pyridine as a base. So you can do all kinds of reactions with these lone pairs that are not part of the aromatic 4n plus 2 rule. So no problem reacting with the lone pair on pyridine or this crossed out lone pair on histidine. If you look inside the active sites of enzymes, one of the most common lone pairs used in catalysis is this lone pair that's sticking off to the side of a histidine side chain. If you look at every single amide bond that is produced in every single protein in your body, every single amide bond is created because this lone pair on the bottom of this adenine ring, a single adenine, a single lone pair inside the ribosome is responsible for the synthesis of every single protein, every single amide bond in the human body. So lone pairs that stick off to the side are completely reactive whereas lone pairs that are part of aromatic systems are not reactive at all. Okay. So I think you can see from adenine that you confuse benzene rings together and have much larger aromatic ring systems. So I'm going to give you an example of some fused systems here. So you can see that there's more than one way to account for this 4n plus 2 rule. There's more than one way to do the addition there. So this is a molecule called naphthalene. If you go to the hardware store and buy moth balls, it turns out that crystalline naphthalene just by solid balls of crystalline naphthalene and for whatever reasons moths seem to find that repulsive and will tend to stay out of your closets. So there's multiple resonance structures that I can draw for this. If I count around, if I just ignored this ring, this bond in the middle of those two rings, that sigma bond, just pretend that it's not there, I can count around the outside edges of this flat molecule. And if I count around the outside edges of that flat molecule, what I would find is that there's 2, 4, 6, 8, 10 electrons. That satisfies our 4n plus 2 rule and that's aromatic. There's another way I could have done the accounting for this, that's slightly different, another way to describe this is I could have drawn a resonance structure for naphthalene in which I put a double bond in the middle. There's a different way I can shift around those electrons. And so remember what I said is if there's a double bond at the ring fusion, you can count that double bond for both rings. So in other words, the other way I could have accounted this is I could have said in this ring over here on the left hand side, there's 2, 4, 6 electrons and that's aromatic. And then I look over here at this ring on the other side and there's 2, 4, 6, I double count, I get to count that double bond again and that's aromatic. So I can count that middle, that double bond in the center twice and I reach the same conclusions either way and that is that all of the carbons in this ring system are part of an aromatic ring. I won't reach a different conclusion. So you can fuse these things in all kinds of different ways. Let me give you another example of another aromatic ring that's very common that's fused and this time it's a parole ring fused onto a benzene ring. This is the side chain of the amino acid tryptophan so I'll just write trip here and there's a lone pair and there's nitrogen. So again in this ring system both rings, both of these rings in this heterocyclic, this is called an indole, this bicyclic ring system, both of these rings are aromatic. This ring over here which obviously is benzene, that's aromatic, it has 6 electrons. If I look at this 5-membered ring over here there's 2 electrons in the nitrogen, 2 electrons on this side bond and then 2 electrons over here that's also aromatic. So the side chain of tryptophan is fully aromatic as is the side chain of tyrosine and phenylalinane. You guys are probably learning about those in Bio98 right now. Okay so you can, it can go endlessly. You can fuse benzene ring after benzene ring after benzene ring onto each other. You may have heard of carbon nanotubes or graphene, that's basically what those molecules are. They're just infinite tubes made out of benzene rings all fused together. Okay so the last parts of this chapter are interesting but they're not going to be useful to you for anything else in this quarter. They have to do with molecular orbital base of aromaticity and I think you ought to read that. I would never discourage you from reading anything but it's not the kind of thing that I can test on. People learn about those powerful concepts if you take upper division classes and I would love to teach those to you there but we just can't fit those in with the time that we've got. So be able to recognize aromatic rings, be able to recognize anti-aromatic rings. You need to be able to know when ionization reactions will generate aromatic anions or aromatic cations. Those ionizations or deprotonations will be fast and be able to recognize when heterocycles with nitrogen and oxygen are aromatic. So be ready when we come back on Monday we're going to do chemical reactions of benzene rings and we're going to do a lot of them.
cyclo-octatetraene, a cyclo-octane ring with four series of ions. So for most of these compounds to make that by deprotonating carbon. Okay, so hybridization there on the top and the bottom neither of those nitrogens is part of a double five-membered ring. This is the side chain of the