 I want to start off by thinking about a very simple reaction, the substitution reaction here. And I'd like you to think about what kinds of mechanisms you'd want to draw for this reaction in order to make a boron-nitrogen bond. So maybe you can sketch something out on your piece of paper. I'll sketch something out on my piece of paper here. I'm going to try to draw a molecular orbital interaction diagram that corresponds with the mechanism that I've drawn. And I'm not showing what happens to the HCl here. There's a loss of HCl. That's not important. So I'm going to sketch out the orbitals in these two components. One of them is a nucleophile and one of them is an electrophile. And what my arrow pushing says, it directly in one step this arrow pushing brings me to the product, done, case closed. Here's the problem with that. According to my arrow pushing, I'm taking one of the lowest energy orbitals, a sigma bond down here in energy. And I'm using it to attack one of the highest energy orbitals, a sigma star orbital. We talked about the relative energies of these filled and empty orbitals. We said that sigma orbitals are lower in energy than lone pairs. Based on that, the right thing for me to have done would have been to draw the lone pairs here and use that to attack, use that as my nucleophile. And the right thing for me to do is to completely avoid breaking this boron chlorine bond because that requires me to dump energy electrons into this super high energy orbital. Who would do that? There's an empty P orbital right here. That's where I ought to be interacting. So if I correctly interact the lone pair with the empty P orbital, I'll get all this interaction energy out. And that'll be a highly favorable reaction. That's the reaction that I should have drawn. Let me use a different pen color to draw the right arrow pushing because that would look like this. And it implies that I'll get an intermediate, at least one intermediate in this reaction. That's the correct first step. And you could have predicted it by knowing nothing more than the energies of those six canonical frontier orbitals. There's no pi bonds in each structure. So I left out pi and pi star. So you already knew how to answer this question just by knowing something about the relative energies of canonical frontier orbitals. It would simply make an adduct, actually. You'd have to have a base here to pull off the HCl. So boranes form complex, just form adducts with the means. Yep. Yes, so what this reaction will generate in an initial intermediate. And then if you have a base present, then you can pull off the hydrogen and lose the chlorine. So to correctly draw the charges on this, you'd end up with this illid type structure. And then that could go on and do other stuff. So the point is, even if this is the final product, you don't have to get there in one step when you look at the relative energies of sigma bonds versus lone pairs. Lone pairs are the better nucleophile. Pi orbitals are the better electrophile. Okay, so let's go ahead and take a look. At some types of empty P orbitals and compare them. I'm trying to keep this as simple as possible by making the substituents H. That's not really practical. You'll be interested in much more complex types of substrates with empty P orbitals. But if you consider the interaction of some nucleophile with the empty P orbital on this, if we drew the arrow pushing, you'd want to do something like this. And what I want to do is I want to compare the reactivity of this empty P orbital with other empty P orbitals as I march across the periodic table. So I've got a carbonymion over here. So it's a carbocation, but it doesn't have an octet. So we call this type of carbocation a carbonymion. A nitrenium ion, an oxenium ion, and a fluorinium ion. And so let's try to imagine measuring the rates at which these things react with nucleophiles. If I keep the nucleophile the same in each case, I can imagine adding to some empty P orbital. And we don't always draw these things. When we do arrow pushing mechanisms, we're not always going to draw that empty P orbital. I'll just draw it on the carbon case here. And you can just imagine them being there for these other atoms. So if you measure the rates of reaction of these types of ions, what you'll find is all react at equal rates. And I'm not sure if that matches with your intuition, but if you measured the rate at which these things have water attack them, you'd find that they all have rate constants for the reaction of about 10 to the 9th per molar per second. How can it be that they all would react at the same rates? Here's my prediction. That if I were to sketch out the energies of sigma star, that's crummy, pi star, that's high in energy, there's no pi orbitals there. And then get down to these P orbitals and compare them. If I think about the P orbital on carbon and then replace carbon over there with something that's more electronegative like nitrogen versus something with oxygen, as I replace carbon with more electronegative atoms in the same row, I should expect that this orbital, the empty P orbital ought to drop in energy. You kind of know, you know, fluorine's more electronegative. I want to add to that faster. Why is it that they all add at the same rates? They're all adding at diffusion controlled rates. You know, no reaction can work faster than the two reactants diffusing towards each other in solution. It's not that the fluorinem hind isn't more reactive, but right, this is already so reactive. Every single collision results in a successful reaction. You can't get more reactive than that or you can't end up at a faster rate. We're not going to get anywhere by making these comparisons is really the point that I want to make. We're really going to spend our time here thinking about carbocations, not these cations. And let me just make a more definite point for you. We're going to spend a lot of time thinking about carbocations that arise through SN1 ionization reactions. In other words, where some leading group leaves in order to generate a carbocation and then in the second step, some nucleophile will add to that empty P orbital. So that's a classical SN1 reaction. And what we are not going to think about is the ionization if I compare the rates to give nitrenium ions. We're not going to think about ionization to give oxenium ions. And we're not going to think about, there's no way, it just seems horrible for me to think about ionization to give a fluorinem ion. In fact, none of these processes where these things simply leave to give empty P orbitals, those aren't known in organic chemistry. That's not to say you can't have leading groups on here, but usually something will migrate over and push that out. You never have an empty P orbital on there. It's almost like an internal SN2 type reaction. Okay, so we're only going to spend our time talking about these kinds of empty P orbitals on carbon. And that's going to be the rest of our lecture for today and some of our lecture on Monday. So let's talk about this idea of making carbocat ions. P orbitals on carbon are important. Because we're interested in making carbon-carbon bonds, actually addition to P orbitals on boron are also important. So let's look in more detail at a reaction that I expect you to be familiar with. Let me go ahead and draw out a reaction coordinate diagram to remind ourselves of some characteristics of an SN1 reaction. If you have some sort of a nucleophile in the presence of some sort of carbon atom with a leaving group, you can think about this ionizing to form a carbocation that is definitely high in energy that no longer satisfies the octet rule. Carbocations are angry. They want more electrons. And then usually there's some sort of a nucleophile that attacks to make a product that's more stable. And if the product's not more stable, you wouldn't be running the reaction. So it wouldn't make any sense. And I'm going to draw that sort of a nucleophile attacking here. And we would still have our X minus and our nucleophile minus floating around. And so that's a classical reaction coordinate diagram. Let me make the point about this carbocation. So this is a two-step reaction that goes through a high energy carbocation and it's planar. So what are the implications of that? Let me just remind you that there's two types of carbocations. And here's one type of carbocation that I mentioned to you before. You can have too many bonds to carbon. And as long as I don't violate the octet rule, that is structurally okay. There's nothing intrinsically wrong with this species. This is one type of carbocation. We call that a carbonium ion. And I can't think of any case where we're going to talk about carbonium ions this quarter. So when I say carbocation, here's the other type of carbocation. Here's what I mean. I mean a carbon atom with an empty P orbital. So when I say carbocation, what I really mean is this type of carbocation, this is called a carbonium ion. Carbini, sorry, I left that off, a carbonium ion. So when I say carbocation, you can just assume that I'm talking about carbonium ions for the rest of the quarter, even though there's this other type called a carbonium ion. And let's not confuse those. Okay, so because the carbocation is planar, there's implications to that. Let's suppose I go through a lot of work to do an asymmetric reduction to make this alcohol. And then I want to substitute that alcohol was some sort of other group. Maybe I can do this in acetic acid. AC stands for an acetyl group. And so what I'm really interested in is the stereochemistry of that carbon bromine bond. Maybe I want to do a Grignard reaction. Or maybe I want to do some metal catalyzed nickel reaction. So what kind of stereochemistry do you expect for this substitution process? It has a lot to do with what you think the mechanism is. It could be an SN2 reaction. In fact, SN2 reactions are faster when there are pi systems nearby. But so are SN1 reactions. So really, if you say that this reaction is going to be stereospecific and give you the bromine going back, then you must have decided it's SN1, SN2. Whereas if you think you're going to get a mixture, a racemic mixture, then you've decided that it's SN1. And the reason it gives a racemic mixture is because this is planar. And because the nucleophile can add to either the top face or the bottom face. And so you already knew that. I hope you already knew that carbocation's are planar and that the nucleophile can add from either face. That's the implication of that planarity. So generally, because the nucleophiles, I'm sorry I'm covering this up, but we're done considering this example. Generally, we use nucleophiles that are more nucleophilic than the leaving groups. If we didn't, then it probably wouldn't be some dynamically favorable. We use leaving groups that aren't very nucleophilic. And because of that, this barrier going to the products is lower in energy than the barrier going back to the starting materials. Because we use leaving groups that aren't very good nucleophiles. So this is generally true that this is lower in energy than this initial transition state. It's just a general property of typical SN1 reactions that are synthetically useful. You know, once you go through the trouble of making that carbocation there, just about any nucleophile that you put in is going to attack very, very quickly. The rate of the reaction isn't dependent on how fast things can add to a carbocation. Anything can add to a carbocation quickly. The rate of the reaction is dependent on how fast you form the carbocation. So, there's this very bizarre aspect to this reaction that the nucleophile structure usually doesn't affect the overall rate of the reaction. It doesn't matter whether I'm attacking the carbocation with methanol or ethanol or t-butanol. They all react fast. The slow step here is making the carbocation, not attacking with some sort of an alcohol. There's another interesting factoid about an SN1 reaction. It allows you mechanistically to provide evidence that it is going through an SN1 reaction. And that is if I add extra leaving group here, I can slow down the rate of the reaction. In other words, if I add, if I have a bromide leaving group here and I have 10 times more bromide, it makes it 10 times more likely that this will go backwards. I'll be shifting the shape of this free energy diagram by adding more X minus. The diagram that I drew out for you here assumes that everything is at one molar. But I can change the shape of this by going, by crossing out this standard state diagram and having things not at one molar concentrations. So if you add extra X minus, you can slow down the rate if this is going through an SN2 reaction. And if it's an SN, sorry, if it's an SN1 reaction, if it's an SN2 reaction, adding extra leaving group won't slow down the reaction at all. We call this the common ion effect, this idea that you can test to see whether it's really SN1 by throwing in some extra leaving group as an ion. Okay, so let's take a look at the kinds of leaving groups I'm going to show you throughout the rest of the quarter when we talk about SN1 reactions. There's all kinds. And I'm going to start off by talking about charged leaving groups that are common in SN1. And to help you out, I'm going to rank these and quantify them in terms of how good they are as leaving groups. We're going to start off with one of the best leaving groups. Maybe nitrogen gas is one of the best leaving groups. But triflate is second to that. And as far as stable things that you can put into a bottle potentially, triflate is about the best thing we can imagine. That's a trifluoromethane sulfonate. So instead of a mesylate with a methyl, triflate has a trifluoromethane. And that really stabilizes that anion. If you've ever worked with triflic acid, you know it's super-duper acidic. This is one of the best leaving groups that we've got. And it is substantially better than either tosylate or mesylate, which are also sulfonates. Tosylate has a toluene instead of trifluoromethane. Mesylate has a methyl instead of trifluoromethane. And triflate is about 10 to the 4th better as a leaving group. If you really want your reactions to be 10,000 times faster, wouldn't that be great if you could get your PhD 10,000 times faster? Here's a recipe for you to do that. Just a note for you, your side reactions will also be 10,000 times faster. That's the problem. So these are better than halides. Tosylate and mesylate are better than halide by a factor of about 1,000. And there's, and I think you learned about this kind of stuff as in your introductory organic chemistry courses, and that is that iodide is better than bromide is better than chloride by somewhat. And these are much better than ester leaving groups. And I'll write perinitrobenzoate. I've got this weird abbreviation. Here's what I mean. I mean a benzoate, so carboxylate leaving group. And this particular one, they tried to make it a little better by putting an electron with drawing group, but it's still not very good. You typically don't try to displace esters through SN2 reactions. It's not that you can't, but the halide is a much better leaving group, vastly better. And I'll put by 10 to, by a million fold. So if you really want to make a good leaving group of an alcohol, you don't make an ester. You make a halide typically. Or just as easily you can make a tosylate or amuselate. So these are charged anionic leaving groups. And you can see this huge range of reactivity, 13 orders of magnitude just among this list here from the best to the worst. So now let's look at some common neutral leaving groups that start off charged with a positive charge, but then when they leave, they leave as neutral molecules that are very common in SN1 reactions. And we're going to see them over and over. So this would be one of the most common leaving groups that would be either water or an alcohol. So if I protonate an ether, this can walk away with the electrons and generate a carbocation. And that would give you a neutral alcohol methanol as a leaving group in this case. I'm not going to give you relative orders here. If I have an ester, I can make an ester act as a leaving group. This is the basis for a large amount of peptide synthesis chemistry. The idea is that you protonate this carbonyl and now that very easily allows this to walk away as a carboxylic acid leaving group that's neutral. And you always protonate the carbonyl. You don't protonate this carboxyl oxygen. You protonate the carbonyl oxygen to make that a better leaving group. So another thing that we're going to frequently see throughout this quarter to make carboxylates better leaving group is to convert this into an oxonium ion, not by adding a proton but by adding trimethyl silo. Silicon loves oxygen. Silicon oxygen bonds are strong. So it's very easy to use sililating agents like TMS triplet in order to o-sililate carbonyl oxygens. And so then this can leave to generate a silo ester. And then an extremely important class of leaving group, mechanistically it arrives in a different way is this phosphonium ion. If you can somehow transfer a phosphorous atom onto here and have this phosphonium ion, this turns out to be a fantastic leaving group. We mentioned the stability, the extraordinary stability thermodynamically of a phosphorous oxygen bond. It's never going to leave this way and give the electrons away, take the electrons away from phosphorus. The electrons are always going to go here. And I can draw this making a double bond or I can just leave the electrons on oxygen. Those are both equally acceptable. And in this case, I'm going to make it go to oxygen so you have P plus O minus. Okay, so those are common leaving groups that we're going to see throughout this quarter. Look for them. Look for reactions that will generate these kinds of intermediates because that means you've got a leaving group that's ready to take off. Okay, so let me bring you back to a classical introductory concept that really messed you up in your introductory chemistry course, in organic chemistry. So I'm going to state something. There's two variations on the Hammond postulate and I'm going to state that for you now. So the first one has to do with, and then I'll sketch it out so you can see what I'm talking about. So that's one statement of the Hammond postulate. Highly endothermic, and we're talking about elementary reactions. Highly endothermic elementary reactions involve late product like transition states. Highly exothermic elementary reactions where you're going way downhill in an elementary reaction step with one transition state. They involve early transition states that look very much like the starting material. So let me try to sketch this out for you so you can see graphically, what does this mean? What is all this stuff about early and late transition state mean? I'm going to start off by drawing a reaction coordinate diagram again, so I keep wanting to put delta there. This is just free energy. That little superscript means standard state. We're just pretending everything is at one molar concentration. That's not an important distinction for this. And so what you have to imagine is if you start off with some substrate here, this is called mom chloride that's used to make methoxymethyl ether protecting groups. So you throw in an alcohol, the alcohol substitutes the chlorine. And the reaction doesn't involve the alcohol doing an SN2. The reaction involves this lone pair attacking and making that come out. And so this chlorine is going to move farther and farther and farther and farther away. But by the time you've reached the transition state, the energy peak, that chlorine is almost all the way gone. You're almost at the point where the chlorine is so far away you wouldn't even call it a partial bond. In other words, the transition state for this reaction, the high energy point, looks a lot like the product of this elementary reaction, the cationic product. In the transition state, the chlorine is so far away. So here's the transition state. I'll try to sketch that out for you. And I'll exaggerate it. There's a partial negative charge, partial positive charge. So that's what we mean by product like transition state. It's like you can't even tell that there's any bonding interaction there. And so we call that a late transition state. Like as this chlorine gets farther and farther and farther and farther and farther and farther away. There's this reaction coordinate. You could make the reaction coordinate, the carbon-chlorine bond or distance between those two. So that's an example of a late transition state. And if I wanted to sketch out a reaction with an early transition state, I can take exactly that same cation. Here's an exothermic reaction. And then if I have some nucleophile in here that can attack this like an alcohol, I can imagine this coming in and attacking that. The expectation is, boy, you don't have to get very close. You know, you can just sort of somewhere scooch over here and be kind of close to this and you've already got some bonding interactions. That anti-bonding orbital is so hungry for energy or for electrons that anywhere in the vicinity, well, I'll just make this my partial bond, if that alcohol gets anywhere in the vicinity of that empty p-orbital, you've already moved over the barrier and you're ready to form a bond. So this is an example of an early transition state. The transition state looks almost like the carbocation. And then finally, when I get to the product here, at least of this step, there's a bond here ready to be deprotonated by some sort of a base that you have in the reaction mixture. Okay, so the power of the Hammond postulate is that we really want to say something about rates. We're always going to be worried, which is fast, which is slow, which arrow pushing mechanism is likely, which arrow pushing mechanism is unlikely. When I talk about whether your arrow pushing mechanism looks correct or not correct, I am usually talking about whether I think it's faster or slower than some alternative. And that means we are interested in transition states. Transition states determine rates for reactions, not product stability, not starting material stability, but transition states. And it is very hard to think about transition states. But the Hammond postulate allows us to do something that really simplifies our lives. It's so hard to think about structures where there's partial bonding and dashed lines and partial charges. And so the Hammond postulate allows us to, instead of focusing on these transition states, allows us to focus on the cation intermediate. We can predict the rates of SN1 reactions, ignoring the transition states and only looking at the carbocation stability. That's the power of the Hammond postulate. We're very good at looking at carbocation and deciding whether they're very stable, and we'll practice at that. And the reason why this works, and you have to always keep this in mind, it's because the carbocation looks like the transition state. And that's why the Hammond postulate works. The big mistake that you will make is starting to apply this to other reactions. Oh, I see what he's saying. They're telling me that stable products always mean faster reactions. No. Stable products correlate with faster reactions only when you're going uphill in energy. The other half of the time, the carbocation stability has nothing to do or is not, it's the other half of the time, the product stability, it has nothing to do with the rate of reaction. The other half of the time, it's the starting material that matters, and in every single case, the general rule that always applies is that it's the transition state energies that matter. And so we're just trying to find a way to dodge the fact that transition states are hard to draw and hard to think about. Here's what's going to happen ultimately to you. I want to draw out three sorts of species here. I'm going to draw this reaction out in more detail. Let's draw out the starting materials here for this ionization reaction. Let's draw out the transition state where I have a partial bond. And then I'll draw out the carbocation product of that first step. So the Hammond-Poslut tells us that this ought to be a fast SN1 reaction because these lone pairs are donating into the P orbital and making that a more stable carbocation. This is a fast SN1 reaction. And I can tell that because it makes a stable carbocation. Now I could just have easily applied all of that thinking. Oh, it's the lone pairs that are making this a fast SN1 reaction. I could just have easily applied that to the transition state where I've got this sort of, this isn't quite a sigma star orbital. It's not quite an empty P orbital. It's somewhere in between. Like what is this? Is that a P orbital? But there's something interacting. It looks sort of like a carbon chlorine sigma star or an empty P. We don't know. There's some sort of donation. What you'll find is as we go through this course, you will find yourself using these kinds of donation arguments more and more, not on the carbocation intermediates. You're going to find yourself applying these types of orbital interaction arguments to the starting materials. Everything we said about the carbocation, we could just have easily applied to the transition state and we could just have easily applied that thinking. To explain why this is fast, we could just have easily applied that thinking to the starting materials. We could just as easily have said, this is a fast SN1 reaction because the lone pairs are donating into sigma star, the anti-bonding orbital. And when you donate electrons into an anti-bonding orbital, you break the bond. That's why it's called an anti-bonding orbital. This SN1 reaction is fast and we could have predicted that simply by looking at the starting material and not the carbocation and talked about filled orbitals interacting with unfilled orbitals. Okay, so fast SN1 reactions, that's going to be the general role when we have good donor orbitals nearby. So let's take some examples of some very fast SN1 reactions. But they don't call them SN1 reactions in your introductory organic course. Lone pairs are fantastic donors. Here's a great one. Nitrogen lone pairs are very nucleophilic. If I put that next to something with a leaving group, this is about the best SN1 reaction I can imagine. And you might not have called that an SN1 reaction. But when we first introduced you to SN1 reactions in sophomore organic chemistry, this is what we should have shown. That's fast. Lone pairs are nitrogen pushing things out. It makes a carbocation. You might have drawn a different resonance structure for that. So what's another example of what you should expect to be a fast SN1 reaction? Acetals are intrinsically unstable. Any time I see two heteroatoms with lone pairs attached to an SB3 center, what I'm thinking is they're pushing each other out. Each one is trying to push the other one out. And the second I tickle this other one with a proton, it's gone by. So whenever you see acetals, aminals, you need to start to think about kicking out a leaving group, one or the other. Now it doesn't always have to be a lone pair nearby. You probably know that allilic substrates are very fast to push out leaving groups. And that's not, you could argue, oh, that's because it gives a stable allilic carbocation. No, that's not exactly why. It's because this pi bond is donating into sigma star and weakening the bond. That's why that's fast. I didn't have to talk about the cation. I didn't have to talk about the transition states. It's the same sort of thinking. And then finally we get to this. And now it's not quite so obvious what's donating into sigma star. We know that this is leaving. Something is weakening this bond. And in this case, it's all of these CHs that are nearby that are donating. And they're not good donors. Sigma CH, that bond is not very nucleophilic. But there are electrons in that bond. And they are overlapping in space with this empty sigma star orbital. If I try to sketch that down here, there's sigma star for carbon oxygen. That's why T-butyl carbocations form so quickly. It's because these CH bonds are donating into that anti-bonding orbital. And it's not as quick as allil. And it's not as, well, depending on the substitution. And it's not as quick as either of these ways to form a mini-amiens or oxycarbenium ions. Okay, so SN1 reactions. The good ones have donor orbitals right nearby. That's when SN1 starts to look really good. So let's take a look at shapes of carbocation and how they relate in a dramatic way to stability. Okay, let me go ahead and redraw a hybridization diagram for you. And what I'm going to do is I'm going to try to sketch out the energies of empty molecular orbitals. And I'll remind you that 2S orbitals are really low in energy and 2P orbitals are very high in energy. And just at a conceptual level, let's suppose these were filled. I'm just trying to think conceptually here. If these were both filled, which one of these orbitals is it easier to take away the electrons from and leave empty? The P. It's easier to take those away. These are really happy. It's very hard to grab electrons to make an empty orbital that has lots of S character. You do not want to leave S character orbitals empty because electrons will want to go into those orbitals. They don't want to be empty. They want to be filled. It's the high energy orbitals that want to be empty. So let me go ahead and sketch out some empty orbitals here for you. Let me go ahead and draw out a 2P orbital. Typically, they're empty. If you have some P orbital, that's a carbocation. And so what would happen if we compared this type of an empty orbital to an empty SP3 hybridized orbital or to an empty SP2 hybridized orbital? The more S character that we mix in, the harder it is for that to exist without electrons in it. So if I think about all the stable carbocations that I'm used to seeing, they usually look like this. That's a typical carbocation, a t-butyl carbocation. It's got an empty P orbital. What you're not used to seeing is carbocations that are bent. Now, when have you seen that? We just start off by saying that carbocations tend to be planar. That's because of this. And what you've not seen is stuff that looks like this. You may have seen this, but far less frequently, this carbocation sticking off the side of a double bond. And I've never heard of or seen anything that looks like this ever. I don't care how good of a leaving group you put on there. It didn't go through that. You might have substituted the leaving group at the end of an alkyne, but it didn't go through this carbocation. You do not want to have an empty orbital with all that S character. Okay, so let's take a look at some examples. And I'll start off with this SP2 hybridized carbocation. So we can be clear about what I'm saying. Here's what I'm telling you. There's lots of reactions that can substitute vinyl halides. You can have conjugate addition elimination reactions. You can have metal catalyzed reactions. There's all kinds of reactions for substituting bromine with some sort of a nucleophile. And none of them went through this. There are no reactions where you substitute vinyl halides that involve the bromide just popping off to lead this cation. The important point is I'm not telling you you can't make vinyl cations. I'm just telling you you can't make them like this. So I'll use a dash line. Don't do that. So there are some reactions that go through vinyl cations. I'll give you sort of an example of a known intermediate. So this is an example of a vinyl cation. So what I'm telling you is if this reaction went through this intermediate it didn't involve loss of some sort of a leading group. How do you generate cations like that? Here's how this one was generated. It took advantage of entropy and the fact that it was very hard for this aminium ion to get away from the alkyne. It's so likely as this thing is flopping around, it's bouncing into the alkyne over and over, that increased the chances that the empty orbital, the empty pi star orbital on this would interact and it increased the chances that that would attack. And then once you attack this can now be attacked by some nucleophile. So it's not, there's all kinds of examples where you generate vinyl cations. You just don't generate them through SN1. You generate them by adding things to alkynes. That's how you generate vinyl cations. Okay, so let's examine the importance of shape here because what you can see is I'm kind of torquing the angles here. I start from, I guess, 120 degree angles to, I mean I'm changing the hybridization in the shape. So the ideal carbocation wants to be planar. And so what happens if I somehow constrain a system so it can't adopt this ideal planar shape? I can make it very, very hard to ionize in an SN1 process. So I'm going to start off by drawing an adamantel system. I want you guys to practice drawing chairs. It's going to be important in this class and there's nothing that's going to help you other than just drawing lots of chairs. So I'm going to draw an adamantel system simply by adding some line. You can add some axial groups to a cyclohexane chair. And so I can imagine that if I have a bromine atom on here, you know, maybe I can ionize and still make a planar carbocation. But these groups in the carbocation, they don't have infinite flexibility in how much, in how much room they can leave when this thing starts to go planar. And so let's go ahead and think about this process. And what I want to do is compare this with some other polycyclic carbon compounds where maybe it's not so easy to become planar. And I don't think that this, I'm going to draw a 222. This is called a bicyclo 222 octane. There's two bridge heads here. And it's not, the name is not important but I may refer to this later. So in IUPAC nomenclature, we call this a bicyclo 222 octane ring system, two carbons, two carbons, two carbons. There's three carbon bridges. That's where that nomenclature comes from. So maybe it's not totally obvious to you but it's harder for this carbon to go planar when it's constrained like this. It's harder for this carbon to move down right in the center of these three carbon atoms here so that that can adopt a planar structure. And I can make this even harder by drawing out an orbornil ring system. So now I've got only a one carbon bridge here. That makes it really hard for this carbon to move up and be planar as that leaving group is leaving. And if you look at the relative rates of SN1 reactions here, that's a poorly drawn arrow. If you look at the relative rates, I don't know what the absolute rate constants are here but I can tell you in a relative sense how fast do these bromides leave. I'll give this a relative rate of one. This one is a thousand times slower. And then when I have just a one carbon bridge here and it's really hard for that carbon to move up so that that can be planar, now it's, I don't hold on. I have to count how many zeros here. One, two, nine. There's nine zeros. One, two, three. Does it matter? It doesn't matter really how many zeros. It's just really hard. It's really hard. When I keep something from adopting a planar structure, it is really hard for this resulting carbocation to be planar. And that just shows you how expensive it is to empty out a p orbital. If you look up there, if you're trying to make an empty orbital but it's got S character in it and it's not planar, that becomes really, really expensive. Very expensive. Okay, so let's talk about some of the donors. Loan pairs are fantastic. I wish I could have a lone pair donor next to these to donate into the antibonding orbital and help push out that leaving group. So we're going to use carbocation. It's kind of as a framework for thinking about donors, different types of donor orbitals. And so what we'll do is we'll end up spending some time thinking about if I draw some sort of MO diagram here, we'll think about the ability of carbocation like that. And I'll sketch this out. That's an empty p orbital on carbon. To interact with various types of donor orbitals. And which ones would I expect to be more stabilizing? And so the best types of donor orbitals will be non-bonding lone pairs. Lone pairs on nitrogen, lone pairs on oxygen. Those are the things that are going to donate into carbocations when they're next door and make them more stable. So if I put a substituent right here and here's my substituent, if this is the, if hydroxyl with a lone pair is the dot, that's going to be a very high energy donor. Here's another substituent I could replace for the dot, maybe I could have some sort of a vinyl group here. So pi is not quite as good as a lone pair, but that's still pretty good. And then the least nucleophilic donor I could have next door to that carbocation would be if the dot is equal to some sort of a CH bond. Here's that CH bond right there. That bond can potentially be a nucleophile and can donate into carbocations. So you already, so we've already discussed this. Higher energy donors will donate better. And so our best carbocations will have donors next door. Let's put some numbers on how big of an effect that is, depending on the lone pair. All lone pairs are not created equal. Let's try to quantify how much relative stabilization we get by having nitrogen lone pairs versus oxygen lone pairs versus other types of lone pairs. Next to a carbocation. But when nitrogen lone pairs are really great. They're very nucleophilic. Amines are basic because of that. So what I want to do is I want to compare this carbocation. Really it's in a mini-amiion. I've just drawn it as a carbocation type resonance structure. Right, if I draw the structure with a pi bond, you'd call it in a mini-amiion. And this is really a protonated carbonyl. I'm just disguising it as a hydroxyl stabilized carbocation. So all of these have lone pairs on them. And then finally I'll compare these to something that looks kind of offensive to me. It's just a plain naked methyl carbocation that has nothing next door that can stabilize that. There's no lone pairs on hydrogen. There's no sigma bonds sticking off of hydrogen that can donate into that empty P orbital. It's totally unstabilized. When you look at the stabilization energy, we can compare all of these to a simple methyl cation that has nothing next door that can donate into this. This has no stabilization energy associated with it. What I'll find is that a fluorine substituted carbocation is six kilocalories per mole stabilized related to a methyl carbocation. Six k-cals per mole stabilized. Now you know as well as I do that fluorine is electronegative. So yes, fluorine is electronegative and it's trying to suck electrons out of this and getting nowhere. But the lone pairs on fluorine are donating into that carbocation and it's a net beneficial effect. Fluorine stabilizes carbocation. It's a second row atom and you can make decent pi bonds by doing that donation. Of course hydroxyl lone pairs are better substantially. How many times more stable is a fluorine carbocation than a methyl, a fluorine substituted? So 1.4 k-cals per mole would be 10 times more stable. So how many times does 1.4 go into that? Three, four, it's about four. It's about 10 to the fourth more stable. That's just using that 1.4 k-cals per mole. Just by having this fluorine lone pair nearby. And when I go to a hydroxyl lone pair, I'm not good enough at math to figure out how many 1.4s go into that. But it's a lot. That hydroxyl substituted carbocation is way more stable than a fluorine substituted carbocation. And then when we get over here to nitrogen, that's now 20, what is that, 29 k-cals per mole more stable. I can't convert that off the, I'm not good at dividing by 1.4. But it's probably like 10 to the 20th or something. So vastly more stable. So these are energies in k-cals per mole. And you can convert those into numerical ratios of stability. Okay, so one last sort of little snippet of information about lone pairs. Here's a type of carb, you could consider this to be a carbocation. I can draw a resonance structure for this that looks like this, that's called an acylium ion. You've seen these before in electrophilic aromatic substitution when you did Friedl-Kraftz isolations. And that might have looked really weird to you at the time when you were drawing those out. But that's not really weird at all. It's about as stable as a t-butyl carbocation. Even though it looks weird and it's got this weird triple bond in there and you might think, oh my God, positive charge on oxygen, I can't have that, sure you can. This has, this structure tells you that every single second row atom has filled octets of electrons, that's fantastic. You'll never get to that with the t-butyl carbocation. So these lone pairs are good donors and they can stabilize this. So it's about as stable as a t-butyl carbocation. Okay, we're going to continue on there. We've got a little bit of stuff to finish up with carbocations when we come back on Monday. But continue to push arrows, but somewhere underneath all that arrow pushing, there's all this cool ability to use molecular orbitals to explain why you're pushing the arrows.