 Okay, so there's an exam next week and it's going to be on Wednesday during the regular class time. My office hours tomorrow have to be moved because I have to do these kind of make-up lectures with Sean tomorrow. And there's no group meeting for my group this week in case you were interested in showing. Thank you for your interest. I wanted to say something about this mechanism. I think it was Beth, Beth, who asked about this on yesterday during our discussion section. I looked up the mechanism for this O-silation in the presence of pyridine and they refer back to some earlier mechanistic papers. And I think the mechanism does involve, as somebody was, thank you for asking about that. Yeah, I didn't realize this, that even in the present, I mean the mechanistic studies were for DMAC, but from what I can see it looks very similar. These go through an en-silal intermediate. So the leaving group is not chloride in the catalyzed reactions, it's pyridine, which doesn't change our energy diagrams. But I learned something. I'd never known that those go through N-emidazoyl-silal intermediates before. I mean that's the more common conditions. Okay, let's go ahead and continue where we left off. You would never have known that. There's no way for you to have known that. Yeah, I wasn't asking you. I said write a plausible mechanism. There's no conceivable way you could have known that without kinetic studies. No, no, you couldn't have known that. For this problem, well from now on, if I ask you about this mechanism in the presence of pyridine or emidazole, I want you to know that silatia with TBS groups goes through an en-silal species. But there's no way you could have known that ahead of time. Okay, so let me come back to this. We've been talking about carbocations and I want to talk about a little bit more about bonding and cyclopropanes. So we're continuing on with lecture six. And we left off talking about this idea that there's a lot of p-character. In order to adopt these very strained angles, if I look at a cyclopropane, there's a significant amount of p-character. This isn't sp3 hybridized. If it were sp3 hybridized, it would imply that there's 75% sp character in this bond. But there's more than 75% p-character. So as a result of this very weird hybridization in the ring strain, these bonds inside of a cyclopropane ring have more p-character. And are therefore more nucleophilic than your average run-of-the-mill carbon-carbon bond. As a consequence, if this carbon atom is using more p-character to make this bond, it has less p-character to make the bonds outside of the ring. So as a consequence of that, these bonds out here have less p-character than a normal CH bond. And as a consequence, they're less nucleophilic. So that's what's weird about the hybridization of that carbon atom. And we can put some numbers on that. For example, if I compare a cyclopropyl carbanion with just a regular run-of-the-mill alkyl carbanion, I'll try to put something in an unstrained ring here so we can compare. If I compare the basicity of this carbanion with the basicity of a carbanion on a cyclohexane ring, which is normal sp3, what you'll find is that this lone pair is 10 to the 5th less basic, right? The bonds on the outside of the ring have less p-character. These bonds or lone pairs are things that are sticking outside the ring have less p-character. Those bonds or lone pairs are less nucleophilic and less basic than a regular bond. By a lot, 100,000 is a lot. Okay, so let's put some more numbers on this idea that bonding in cyclopropane can change nucleophilicity. And so I want to introduce you to a new term called cyclopropyl carbynil. That just means a carbon atom attached to a cyclopropane ring. And specifically, we're interested in carbocation that have cyclopropanes dangling off of them. And so let's see what happens in reactions that would make those kinds of carbocations. I want to consider SN1 reactions of some tosylates where tosylates is the leaving group. And if we just think about the relative rates here, I'll give you some relative rates relative to just ethyl tosylate, which you might expect as a pretty bad ionization reaction to give a primary carbocation. Of course, you can accelerate that by having a pi system next door. We have pi bonds next door. Why is this fast? It's fast because the pi system is donating into the anti-bonding orbital and helping to push out that leaving group. Or if you wanted to use the Hammond postulate and talk about carbocation stability, you could say, well, the carbocation is more stable and it looks like the transition state. But the bottom line is that pi systems next door, you should know that benzylic carbocations are stabilized and that benzylic leaving groups leave faster. And what I want to do is I want to compare this with a system where there are these strained bonds next door, not a pi system, but just a strained bond in that cyclopropane ring. So here's a bond that's sort of can potentially overlap in the right conformation with the anti-bonding orbital. This is 120,000 times faster than just a primary carbocation. All three of these are primary carbocations in theory. You get a sense that you may have thought that benzene rings and simple double bonds were all that for making alloy cations and benzene cations. But look at this, this is kind of pathetic looking in comparison. Cyclopropanes, this just gives you a sense for how awesome instability cyclopropyl carbinyl cations are. So let's go ahead and just draw out those cations. Here would be one cyclopropyl carbinyl cation. And I'll compare that to something that you might think is a great cation. I'll put a benzene ring on a carbocation. Then I'll add another benzene ring on a carbocation. And then I'll add a third benzene ring on the carbocation. This is called a trityl cation, triphenyl methyl. This is still more stable. This primary carbocation is still more stable than a trityl cation. Just to give you a sense for, yeah, yes, you can. I don't know, these are easier to attack. So there's other sites to attack that are kinetically not reversible whereas trityl cation addition is reversible. Okay, so cyclopropyl carbinyl cations are super duper stabilized. Look for reactions that make cyclopropyl carbinyl cations because those will be fast reactions. So I want to talk about the importance of the orientation of those orbitals. And I'm going to struggle here to try to draw out what kinds of orientations are good for stabilizing. So let me try to draw out a carbocation with an empty P orbital here. And the single best orientation for that cyclopropane ring would have each of these bonds sort of aligned so that they could both simultaneously donate into one lobe of the P orbital. That's the optimal conformation of a cyclopropyl carbinyl cation. Or sometimes you can tilt it so just one of them is perfectly aligned. That's also a good orientation. But in contrast, if you spin this around the wrong way, you'll get completely opposite effects. So here's my empty P orbital. There's my carbocation. If I spin this around so that this CH bond up here, maybe that's not a good three-dimensional drawing. So that this CH bond up here is aligned, that's lousy. That's a bad orientation for a cyclopropane ring. So you really want to have these, either one of these bonds ideally perfectly aligned, perfectly parallel to that empty P orbital. That's the ideal alignment. Let me, that should say, what was I, it looks like I was going to write something bad, but I meant to write bad, bad. If you want to write BA, never mind. If you want to write something else bad, it's bad. Okay, so let me compare two different systems. I want to look at a constrained system because I think maybe it'll be easier, maybe it'll be easier to see these effects. You can start off, I'm going to draw an adamantine ring system and you can start off by drawing a chair conformation. And I just want to be able to see exactly where that anti-bonding orbital is. There's an empty orbital there. And so why is it, what are the things that make this possible to ionize, this tosylate leaving group? There's an empty sigma star orbital here and there are things in this adamantle structure that are donating into that empty orbital. So specifically this axial bond over here and then one in back are aligned with that empty orbital and are overlapping in space with that empty orbital. They're donating electrons. So it's the donation from this sigma bond into this anti-periplanar, into the sigma star orbital of this anti-periplanar carbon oxygen bond. That's what makes it possible to ionize out this tosylate group. If you put, if you construct a compound that has a cyclopropane ring right here, that becomes way faster. It strains the bonds. It strains this bond so that this carbon atom has to use more p-character and when there's more p-character in this bond, when you have that cyclopropane ring there, it's 10 to the 9 faster. So a billion times faster with the cyclopropane there because that's a bond that's perfectly aligned. You're also straining this bond down here but that doesn't do you any good. It's not overlapping with the anti-bonding orbital. So let's take a look at a system that's not so well designed. Let me draw another atom, another adamantel system. I'll also start with a chair but I'm going to put the tosylate leaving group at a different position so I'll draw this with my axial bonds going down like this. And so now I have the capacity potentially to make a tertiary carbocation with that. Let's take a look at the bonds that are oriented with that anti-bonding orbital. Here's the anti-bonding orbital, it's on the backside of that tosylate leaving group, that's my sigma star orbital. And these bonds over here are aligned. Here's a bond that's aligned with that anti-bonding orbital. Back here is a bond that's aligned. Here's a bond that's aligned. They're anti-periplanar to that o-tosylate leaving group. If I make the mistake of putting the cyclopropane over here, that doesn't strain this bond. This is now like one of the bonds that's on the outside of the cyclopropane ring. This is now 1,000 times slower and I don't have room here but you would write with, I'll just write with cyclopropane, with CP. So with the cyclopropane, that slows it down. So you have to be careful how the cyclopropanes are oriented relative to that leaving group. It's not simply sufficient to be next to the leaving group. You have to worry about the orientation of those bonds in the strained ring. Okay, so I want to compare. Let me try to give you a resonance picture for what's going on and it's going to be very strange looking when we try to think about how do I draw this intermediate where there's cyclopropanes donating into an empty P orbital? Oh, you mean the shape, in other words, when I draw a carbocation, why don't I draw it like ChemDraw? But the actual shape of an empty P orbital, it looks like a sphere and I guess I never know how to draw this. It kind of bothers me to draw them that way. So if you look at the shape of an empty P orbital and draw a sphere around this carbon atom, it actually looks like half a sphere up here and half a sphere up there. That's the actual shape of a P orbital. It doesn't look anything whatsoever like this. Yeah, so whoever invented ChemDraw in the mid-1980s drew P orbitals like this and it's too bad they didn't draw them more like a half egg shape. This is just one empty P orbital. It's phased like this. I sometimes don't draw the phasing in just because. Yeah, so if I draw the extra parts of this empty sigma star orbital, it looks like this and if I include the phasing, it's hashed here, not hashed here so there's no interaction then hashed here again. So usually I just draw the big part. Yes, yeah, sometimes I'll draw like this but when I'm, I prefer to draw like that actually. Okay, so empty, so let's draw a resonance picture for a cyclopropyl carbynol cation. If I draw a cyclopropyl carbynol cation and I want to think about how do I represent this? There's really no perfect representation. If I really believe that this bond is donating into that empty P orbital like this, well why don't I draw a resonance depiction that looks like that? Here's what your resonance depiction will look like. It'll look like this. If I really believe this bond is donating into that then there ought to be pi character here in that cyclopropyl carbynol cation. There's another resonance structure I can draw is, well if I really believe that this carbon bond is donating into that empty P orbital, maybe I should think about this carbon bond as shifting over. Here's a different resonance structure I could draw where I lift this bond away from this dotted carbon here and then I leave a carbocation at that center. These are all valid resonance structures for a cyclopropyl carbynol cation. As hard as it is to think about three-dimensionally. What you find just to convince you that these resonance structures are valid ways to think about a cyclopropyl carbynol cation. If you make cyclopropyl carbynol cations, so if you take an amino methyl cyclopropane and you treat it with nitrous acid, you generate diazonium leaving groups, those leave in water, what you get is a distribution of products that contains this substitution product, you get this substitution product and you get this substitution product. You get all three of those substitution products. They're not all in equal ratios, that doesn't matter that much. You get the same distribution of products, you get the same products if you start with this leaving group on a cyclobutane ring. So you're sort of entering the same kind of carbocation if you start with the cyclobutane. Okay, so whenever you see the possibility to ionize things off of cyclobutyl, they're actually a little bit slower but they end up giving you this very similar kind of carbocation. They give you the same type of carbocation that looks like a cyclopropyl carbynol. Okay, so cyclopropane bonds are magnificently strained. They're super nucleophilic. Yes? Yeah. No, that was supposed to be, yeah, so when you have this cyclopropane here, it's now a thousand times slower. And if you make it without the cyclopropane, that's relative to the one that has no cyclopropane here. Because this bond has less P character. Note that this bond is outside the cyclopropane ring. Those bonds that are out sticking off the edges of the cyclopropane, those bonds have less P character and are less nucleophilic. It's the bonds within the ring that have more P character. And these aren't overlapping in space with that empty orbital. Is that? Yeah, in this case, this is within the ring. This CC bond is within the ring. So when you synthesize this derivative that has a CC bond here, this CC bond is within the ring and so super nucleophilic, screaming hot nucleophilic bond there. By having this carbon-carbon bond be part of a three-membered ring. Okay, so let's talk about a different type of nucleophilic bond. So everywhere, look for cyclopropanes and leaving groups next to them. Better than a t-butyl, better than a tritle cation, just to have one cyclopropane bond nearby. I want to draw, I want to remind you of the effects of making bonds longer. What happens when I take a bond and I make it longer? And I just want to compare the canonical molecular orbitals for bonds where, for example, I simply make things longer. So let's go ahead and start off by thinking about a carbon-carbon bond. And this will be a t-butyl group here. I want to do that so I can keep things similar. So I want to think about the orbital for that bond. There's a sigma CC orbital for this that's filled. And there's a sigma star orbital that's equally higher in energy for that CC bond. And so what happens if I drop from carbon down to the next row of the periodic table from carbon to silicon? What happens is the bond gets longer. So let me try to emphasize that in some ridiculous way like this, just so you can see. So you can't have SIH3 easily. So the values are reported for trimethyl. But the prediction is that this bond is longer. It's less stable. There's less orbital overlap, less effective orbital overlap. The bond is less stable. That means it's higher in energy. And that means the sigma star orbital is lower in energy. That's not so important. What's important is that this is now higher in energy. And if I drop yet one more, if I drop further below the periodic table to tin, let me make this even better. Even, I don't know if that looks longer. Then your expectation is whatever silicon does, tin can do better. Anything silicon is known to do, a carbon-tin bond is better at it. They're more nucleophilic. They stabilize carbocations more. They're more toxic, I guess. They stink more. They always do things sort of one-upping carbon-silicon bond. Your prediction is that carbon-silicon bonds should be super nucleophilic. Longer bonds are more nucleophilic. So let's see the implications of that for stabilizing adjacent carbocations. So I'm going to draw a simple comparison. So we've talked about before the effect of these CH bonds on a T-butyl carbocation. It's these CH bonds that are stabilizing a T-butyl carbocation. If you compare that to a T-butyl carbocation that has just one single carbon-silicon bond next door, so trimethyl silo, everything else the same. With that single carbon-silicon bond there, with that trimethyl silo group, this becomes more stable by 38 k-cals per mole. So how many times more stable is this when it has this carbon-silicon bond at the beta position? We call this a beta silo cation. Two words, beta silo cation. How many times more stable is that? I mean, whatever that is, divide by 1.4, it's 10 to that power. So that's something like 20, some 10 to the 20, something or something like that. It's a huge amount more stable, 10 to the 20s, something more stable. If you convert this into factors of 10, it's a massive increase in stability. So look for reactions that can generate beta-silocarbon cations. Now what I'll tell you is it is actually not very common to generate beta-silocarbon cations through SN1-type ionization reactions. But as we go through this course, what you're going to find is later on, you're going to see 100 million reactions that generate beta-silocarbon cations like this. And we'll get to this. This is the way you most commonly see beta-silocarbon cations formed in organic synthesis. Every time you see an allylsilane or a vinyl silane, you need to be thinking about how smoking hot, maybe I shouldn't say hot, it's how incredibly stabilized this carbocation will be. That's not just a regular double bond. That double bond can make a beta-silocarbon cation. So look for a little exilanes. Look for a little exnanines. Look for any chance you have to put this long carbon-silicon bond next to a carbocation. Okay, so this is not just the carbon group. It's not just carbon-silicon tin. Any longer bond is more nucleophilic. And so that means every single time you've seen a metal and it doesn't matter whether it's palladium or cobalt or rhodium, every single time you've seen a metal that has a chance to have a beta-carbocation, that's going to be a super stabilized carbocation. Let's all draw this there. So all beta-metal carbocations are stabilized. So carbon metal bonds are longer and they're very nucleophilic. Yeah, the bond is longer. So if I have a longer bond, it's higher and it's less stable, it's higher in energy and if it's higher in energy, it's more nucleophilic. Hyperconjugation means that this bond is donating into the empty orbital. That's just a different word for what we're talking about. I have to go look at that. Here's why it's complex. So if I put silicon here and carbon here, I still have longer bonds. The problem is that this is now farther away from the carbocation and I can't remember. I've looked that up before and now I can't remember. It's nothing special. I can't think of anybody who ever puts leaving groups alpha to silicon and gets a beneficial effect at it out of that. So my intuition is that no, it's not. The fact that this bond is longer doesn't make up for the fact that this bond is longer and it's farther. You have to donate way a longer distance over to the empty P orbital. Okay, so some bonds can be very nucleophilic and can very much stabilize carbocations. That's the whole point of that lecture. So now let's go on to a new topic. Or maybe it's the same topic. It's just a thousand different ways you can donate into an empty orbital. And, you know, it's not like every reaction goes through carbocations. My hope is that you'll be able to take these ideas of long bonds, strained bonds, those kinds of ideas and apply them to other kinds of empty orbitals like pi star and sigma star. When we get there. So I want to talk about some other types of groups that you can have next door to leaving groups. And that's the main idea here. And I want to start off by talking about a topic that we've already talked about. And that is what about heteroatoms? So what I want to do is I want to consider the ionization of this leaving group in order to give a carbocation. And so, yeah, there's an oxygen lone pair up on top. That's okay. Don't worry about that. The main point I want to make is what's the effect of having another heteroatom right here? And let's put this into the context of DNA and RNA. Or this could be NAD plus. It's an enzyme cofactor. So what would you predict about the ability to this, for this base in DNA or RNA, purine or pyrimidine, to act as a leaving group? How easy? This is a source of mutations in DNA. What's the effect of having a hydroxyl group here? Does that make it faster? Does that make it slower? What we said is that these lone pairs are never nucleophilic enough to bend over and push out the leaving group. The only effect of having an alkoxy or hydroxy group next door here is that this slows down. This is an electronegative atom. It makes it harder for you to form an oxocarbenium ion, a carbocation at this center. So this ends up being slower. And it's about 150 times slower. I mean, it's not a factor of 10 to the 9 or 10 to the 20, but that's still significant. It's slower to depurinate or lose the pyrimidine or the purine base from RNA than it is from DNA. When nature made nicotinamide, adenindynucleotide cofactors where there's this leaving group built in, this is the base that's attached to NAD plus. Nature made it with this extra hydroxyl group here. It didn't use a DNA fragment to make that. That slows down the loss of that group. Not ever do you want that group to act as a leaving group for cofactor type reactions. So that's the effect of having this hydroxyl group as it slows down the loss of this leaving group here. So oxygen is not what you want to have at this visceral position, visceral to your leaving group if you want to have a fast ionization reaction. But that's only for oxygen. Let's see what happens when we switch to nitrogen. It kills me. This is a clinically used anti-cancer drug sold as mustergen. Please get off of your ass and get out of here and go invent a better drug than this. It's very toxic. You can design something better than this. I'm positive of that. The whole idea behind this anti-cancer drug is this. This nitrogen is not oxygen. Nitrogen lone pairs are dramatically more nucleophilic than oxygen lone pairs. And if you have a nitrogen and a mean that's visceral to your leaving group, that can help push out. And it's not the nitrogen carbon bond. It's the lone pairs that help to push out that leaving group. And so what this does is it makes a strained aziridinium. So it's not an aziridine. It's got too many bonds. So you call it an aziridinium, kind of like ammonium. That's an aziridinium ion. And this now has strain. And that strain allows it to be opened up relatively quickly by nucleophiles like DNA. And so this leads to not just a single alkylation event. Those are very easily repaired by cells. But what it does is it cross-links because now after this opens up, you can imagine the same thing happening on the other side. And those types of cross-links are virtually, let me draw the DNA up here because I've run out of room. So now the same thing can happen on the other side and you can cross-link another strand. And those types of problems with DNA are virtually unreparable. Cells kill themselves. They undergo apoptosis if that's a problem. So nitrogen lone pairs, yes, they can participate in what we call neighboring group participation. Whereas oxygen isn't helping. The oxygen lone pairs are less nucleophilic. You couldn't have known there'd be that distinction. You couldn't have known that the difference in nucleophilicity for a nitrogen lone pair is enough to allow it to make those strained rings. So if you're looking for evidence that these things, more evidence that these kinds of aziridinium ions are important intermediates, if you synthesize this kind of chloroamine and there's nothing special about the chloride, it would be the same if you made the toslate. And you do substitution reactions with this where you have some sort of a nucleophile. This is one of the products that you get. Let me draw some stereochemistry here so you can see what's happening. It'll be completely inexplicable to you why you would get a six-membered ring product unless you know about this aziridinium ion formation. So you can tell what's happening here. What's happening is that this lone pair is helping push that out to make a three-membered ring and then the nucleophile opens up on the other side. So in fact, you get a mixture of both substitution products from this. And I'm not going to draw the intermediate. I'll leave that to you to draw the three-membered ring aziridinium ion in that case. Oxygen lone pairs are less nucleophilic and less basic. So just to put some, I'll bring this back to acidity and basicity. Let me take this off. I'll just, so if you look at the relative basicities of these lone pairs, you can go to a pKa table and look up a protonated oxygen versus a protonated amine. And the pKa's will tell you the relative basicity of those lone pairs. And so in terms of relative basicity, the pKa prime, pKa prime means if I put a proton on there, for this is on the order about minus 2 to minus 4, depending on whether this is R or H. For an ammonium ion, it's 10. That's the pKa for a lysine side chain or triethylammonium, those sorts of things. 12 orders of magnitude difference in basicity. This is 12 orders of magnitude more basic and approximately that much more nucleophilic. Yeah. Oh, because oxygen's more electronegative than nitrogen. So oxygen's electronegative. There's more protons in the nucleus. And because this is more electronegative, if you try to make a carbocation, when there's an electronegative atom here, it kills you. It's really bad if this is fluorine. It really destabilizes that carbocation to have an electronegative atom here. Yeah, like without it, it's faster with the hydrochloride. Yeah, it's electronegative. So if you have an electronegative atom here, the lone pairs aren't helping you. All you get is the bad electronegativity of oxygen that's destabilizing the carbocation. So, yes. So put it in the mean right there on that position. Yeah. If it's anti to the leaving group, it'll help push it out. Okay, so let me, where did this mustardion thing came from? It came from people trying to kill each other. So these are casualties due to chemical warfare agents in World War II. I mean, and here's one of the principal chemical agents that they were lobbing back and forth at each other during, sorry, World War I, during World War I. And this is not a nitrogen mustard. It was a sulfur mustard. So when you hear people talk about chemical warfare and mustards, they're referring to the sulfur compound. And it turns out that it's like, where does sulfur fit in? Is that like oxygen? And the lone pairs don't push out leaving groups? Or is it more like nitrogen? Well, sulfur's in the same row as oxygen. It's called a chalcogen. But because these lone pairs, because the sulfur is, the bond is long, the lone pairs are bigger. Everything is bigger. Bonds are longer. The lone pairs are bigger. These lone pairs are more nucleophilic on sulfur. And so it turns out in this case, you do get neighboring group participation for sulfur, for selenium. Once you drop below that second row, also, sulfur is not as electronegative as oxygen. So you suffer less of a penalty in ionization reactions. But when you pop this thing off, you end up generating this, what looks like a weird species here. This is not, it looks kind of like an epoxide, but it's not an epoxide. If this were neutral, we'd call this an episulfide. But it's not even an episulfide. There's a positive charge there. We call this an episulfonium ion. You don't call a protonated epoxide an episuponium, I don't know what you'd call it. But it's like an epi, it's the name comes from something like epoxide, but it's a sulfonium ion. Okay, and then in the presence of water in your lungs, unfortunately, this opens this up. And I won't show all the steps. But the important point is you start cross-linking proteins, mainly, it's too fast. This is too fast to happen and wait around for DNA. This starts to generate two moles of HCl gas for every one of these, for every one of these molecules that gets in your body, and the HCl causes edema and starts blistering your skin, very bad stuff. Okay, so sulfur is more like nitrogen. For a different reason, the sulfur lone pairs are more nucleophilic than oxygen lone pairs. Why is that slow? Yeah, sterics, which is very hard for nucleophiles to attack at tertiary centers. So keep that in mind. So let's put a thioether right next door to that o-tosylate. We can see those lone pairs are ready to help push out. Just like I told you, those lone pairs are ready to help push out this tosylate. If you look at the products you get out of this reaction, you get two different products. If you use ammonia as your nucleophile and you get those products in a 4 to 1 ratio. So let me go ahead and draw out that episulfonium ion intermediate that you generate. So the reaction starts off with this ionization reaction where the sulfur lone pairs are pushing out the leaving group. And so you get this episulfonium ion. And now, if you're using an amine as the nucleophile, in this case ammonia, that ammonia has two choices. The ammonia can attack over here at the primary position or it can attack at the more substituted position. And you can tell from the products that the dominant mode of attack is for the amine to attack at the more substituted position. That's the general rule. Nucleophiles attack at the more substituted position. When you have positively charged three-membered ring-onium ions. So if you learn some sort of a rule in sophomore organic chemistry that, oh, you always attack more quickly in SN2 reactions at primary or at less substituted positions, that's wrong. And it's wrong in this case of three-membered rings. So let's be clear about where this is true. So if you have an epoxide like this, it's faster to attack at the primary position. If you protonate it, it prefers to attack at the tertiary position. Yes, it would still attack at the, it doesn't go through a carbocation. So if you make an, a quaternary ammonium, notice that there's a positive charge here. Nucleophiles prefer to attack at the tertiary position. If you have the episulophonium ion, and again, it doesn't matter whether it's quaternary or, you know, whether it's secondary or tertiary, nucleophiles attack at the more substituted position. If you have a mercurinium ion, we don't cover oxymercuration reactions anymore. We used to cover those in, in sophomore organic chemistry, but mercury is toxic. These will prefer to attack. You always get attack, what we refer to as Markonikoff addition of the nucleophile to the tertiary position. It's opposite if you don't have the positive charge. The effect of this is that you're really, the, I'm, maybe I ought to be exaggerating these structures in a way that helps you to remember this, but this bond is just a lot longer when you have these onium ions. It's already prebroken before the nucleophile comes in. So, yes, it is sterically encumbered to attack at that tertiary position, but the bond is already partially broken in all of these three types of structures. So, this is the general rule. So, it's only when it's a neutral three-membered ring that you attack at the less substituted position. What if the more, meaning that what if you had something like this, and it was maybe over here, but very bulky? And then, let me try to make it equal. Let me try to put another group on this side. So, now they're both secondary. I'd predict you'd attack here faster. Maybe you can ask me afterwards, and then I'll try to answer that. So, I'm not getting it, but you can ask me afterwards. Okay, let's look at some other, and I've just finished covering a bunch of examples where you make strained three-membered rings. And let me step away from that, because it's not like every neighboring group has to make a strained three-membered ring. It is kind of surprising, isn't it, that you can have neighboring groups that will readily form three-membered ring intermediates. One thing I want to urge you not to do, and I'm going to do this, and you just have to remind me to, hey, stop doing that, is I want you to stop calling these, or if you're about to, to not call this an SN2 reaction. So, if you do some substitution of this by some nucleophile, that's not SN2. By definition, SN2 means that the transition state involves both nucleophile and the electrophile. They have to collide 10 times higher concentration of nucleophile means 10 times higher reaction. That's the definition of SN2. Rate equals rate constant times concentration of Rx times concentration of nucleophile. By definition, this is an SN1 reaction. So, I don't have an alternative to this SN1, SN2 language and nomenclature that we use, but the whole SN1, SN2 system breaks down when you want to describe this. That's not SN2. That's actually an SN1 reaction, even though it involves a concerted displacement. Okay, let's take one more, one more example of neighboring group participation that involves lone pairs. You draw two different substrates that have toslates here, and in each case, I'll draw it. There's an acetate group that's nearby. And, of course, oxygen is electronegative, so maybe that's not so good, but you can get a beneficial effect out of this. And it matters stereochemically the way that neighboring group is disposed. So, if I look at the relative rates for ionization, for popping out of this toslate group, what you'd find if you measured the rates at which toslate leaves from these, you'd find that this other isomer over here, the anti-isomer, is 600 times faster. It's 600 times faster to have the anti-isomer. And you can kind of guess the reasons why. I'm going to start off by drawing this in a confirmation that can help you see what's going on. This chair can flip back and forth between different chair type confirmations, and it will eventually reach a confirmation in which this toslate is oriented where this carbonyl lone pair can help push that out. And unlike all those other examples that I showed you, if you think, oh, but this doesn't look so good, there's those axial bonds, but this ring has none of that strain of the three-membered rings. So maybe it has to generate a small amount of this diaxial conformation, but now there's nothing stopping this. There's no ring strain stopping this from cyclizing around to give, and now I'm going to draw this with a flat structure, to give an oxonium ion intermediate. So here's the intermediate that you would get in this reaction. You would now have the, I'm just now realizing that I'm, I think I'm drawing the enantiomer of this. So don't, if you want, you can make this dashed and dashed and bold and dashed. Sorry about that, I didn't mean to do that. So you get this intermediate. Now where essentially you can draw an equivalent of the resonance structure with the double bond here. So that's an oxonium ion with three bonds to oxygen, this particular resonance structure. And so now the nucleophile, in fact, can attack either side of this. You'll see 50% of the attack occurring here, and if you draw the other resonance structure, you can see why the other 50% of the attack occurs down here at the bottom. So you get both enantiomers out of this. So ultimately you'll end up getting, if you're doing the reaction in acetic acid, so let me write this down with the, I'll write acetalysis to indicate that my nucleophile here is acetate. So in the end, you'd end up with half of your acetate anion attacking here, half of your acetate anion attacking there. So you get a racemic mixture, even if you started with one in, with a single enantiomer of your substrate. Okay, so that's neighboring group participation by lone pairs. We're going to have to stop there. And when we come back, we're going to consider a little bit more what it's like to have double bonds near leaving groups, not directly adjacent where you can form a carbocation, but somewhere just a little bit farther away. Can you have double bonds poised somewhere else to help push out leaving groups? And of course the answer is yes. And we'll see how much that can be worth in terms of rate accelerations.