 Continue where we left off on Friday and which means that it's going to look like I just changed my clothes on the camera here. So I want to continue where we left off and that's talking about SN1 reactions where you ionize some leaving group to give a carbocation. And we use the Hammond postulate on a regular basis to justify how fast are SN1 reactions when leaving groups pop out to give carbocation? And if we invoke the Hammond postulate we can simply look at the stability of the carbocations. And so how fast is it to take something like this and have a leaving group leave versus take something like just an allyl halide and have the leaving group leave from there? That's kind of what we're really asking and that's not a well-drawn error. So what we need to do is we need to think about well how nucleophilic are these pi bonds? How able are they to donate into the MTP orbital? How nucleophilic are they? Or how able are they to donate into the sigma star anti-bonding orbital? And all we need to do is draw out the energies of these different pi bonds, these canonical orbitals for a C-C bond and ask ourself what happens if we replace a carbon atom and a pi bond with a more electronegative atom? What happens if I replace this atom at the end here with a more and it doesn't matter which atom? What happens if I replace one of these two atoms with a more electronegative atom? You should predict that the energies of all the orbitals associated with this atom should drop down if I put something more electronegative there. So a carbon nitrogen pi bond should be less nucleophilic. Should be lower in energy and less nucleophilic than a C-C pi bond. And similarly a carbonyl pi bond should be even less nucleophilic. So immediately you have a prediction and your prediction is that this should be in terms of stability, that this should be the most stable and the carbonyl substituted carbocation should be the least stable because the C-O pi bond is the least nucleophilic. And likewise the ionization of a leaving group alpha to a carbonyl should be slower than the comparable allyl chloride leaving group ionization. And so that's just a basic understanding of electronegativity. I want to draw a series of pi substituted carbocation and try to compare those. And I want to put some numbers on how much stabilization we expect. So I'm going to draw a series of carbocation and I'll compare them all to a methyl cation which is absolutely awful, awful, hard to think about even. So there's a pH we use as an abbreviation for phenyl so you can draw a benzene ring there if you want. I'm often going to draw pH instead of a benzene ring. Here's a simple allyl cation. Here's a propargil cation that's with a triple bond. We called it propargil. We'll compare all of these to just a simple, plain, unsubstituted, nothing to stabilize it methyl cation. And then last I'll replace this with another cation that's got pi bonds associated with it but unfortunately it's got this N plus there. So that's a nitro group, a nitro substituted carbocation. And it's very easy nowadays using ab initio calculations or semi empirical calculations to calculate and compare the stabilities of these carbocation. And so I'll give you some stabilization energies. And these were calculated at the PM3 level of theory doesn't matter to you. The point is this is a gas phase calculation. You draw this and you ask what's the energy of this in the gas phase and you compare those energies. And what I want to do is I'll give you relative energies. The absolute energies aren't important. What's important is the relative energies. I'll arbitrarily assign the methyl cation a relative energy of zero and so relative to just a simple methyl cation, a benzene cation is 56 kilocalories per mole stabilized so it would be negative energy, lower in energy. I'll just draw it as because I'm talking about stabilization I'll keep all these numbers positive. How many times more stable is a benzene cation than a methyl cation? Lots, yeah I can't divide that by one point. I'm not good enough with that but it's a lot, right? A benzene group, a benzene cation is very stable. An allocation is pretty close. These are all in k-cals per mole which means you can convert them into factors of 10 simply by dividing by 1.4. Each 1.4 k-cals per mole corresponds to a factor of 10 in stabilization. And then finally I'll get over here to perpetual. It's not quite as good. Let me make that triple bond pi bond look really short so we can remember that the pi bond is very short, very stable pi bond, not very nucleophilic. So all of these are massively more stable. That's the effect of having a pi system next door. But not all pi systems are created equal. It's obviously bad to have this electronegative nitrogen with a formal positive charge here. This is actually 31 k-cals per mole higher in energy. You don't get stabilization of a cation with a nitro group, you get destabilization. So again, because I'm calling these stabilization energies, instead of putting minuses here, I've inverted the signs. Okay, what happens if you put carbocations in solvent? You can mimic that effect with semi-empirical calculations. And it's not exactly the same as measuring a number in water, but there's ways to mimic the effects of water salvation on these cations. And the point is they're not quite as stabilized anymore when you put cations in water because water intrinsically has this stabilizing effect. You get 23 k-cals per mole of stabilization energy at this, calculated at this level of theory for a benzyl carbocation. And when you look at an allyl compared to benzyl, what you find is they're not that different. So in the gas phase, if you ever happen to be running reactions in the gas phase or in hexanes, which is very nonpolar environment, like then you should expect benzyl to be better at stabilizing carbocations than allyl. But if you run reactions in solvent like normal organic chemists or inorganic chemists or bioorganic chemists, then these are pretty much the same in terms of stabilization, allyl and benzyl. Okay, we get down to propadryl. That one's worse than both. That's what we expect. And of course, these are all relative to a methyl cation. And again, nothing much to say for the nitro substituted carbocation. That's pathetic. Don't have leaving groups leave when they're next to nitro groups because if you want to calculate how much less stable those carbocations are, they're way unstable. Okay, I want to make a point to you because oftentimes you're going to see numbers that were calculated through gas phase calculations. It's not trivial actually to do good solvated calculations. So in the gas phase, and so this has to do with calculations. Let me try to rationalize, why should this be true? This is even when all else is equal. Larger ions are more stable. So what I'm telling you is if you compare this carbocation, just a propyl cation like this, with a longer cation, what you'll find is that the larger cation is more stable. So let me draw this. Can we find some way to rationalize that? Why is it that if you did some calculation for the stabilization energy of this primary carbocation versus this one, that this one would look more stable? And the more carbon atoms you add to this chain, the more stable it would seem. And this is one of the things about calculations that you don't see a similar effect in solution. The way to think about this, or at least one way to think about this, is that both of these have an anti-bonding orbital associated with it, with that empty, with that carbocation. There's an empty p orbital there. And how do we think about what's stabilizing that? Well, there's CH bonds that are stabilizing that. I'm going to look at this CC bond. It can overlap with that empty orbital. And it can donate into that empty orbital. Sigma CC can donate into that empty p orbital on carbon. That's a stabilizing effect. It is always stabilizing to have filled orbitals interact with unfilled orbitals. That always leads to net stabilization. And so if I look over here, I've got a CC bond that can also overlap with that empty p orbital. But there's another CC bond right here that's anti-paraplanar to this one. And so I can highlight that. Let me draw out this CC bonding orbital here. And if I think about this bond right here, this donating into sigma star, that should raise the energy of this. Having this bond do a nucleophilic attack will weaken this bond and make it a better nucleophile. And you can continue this on through the entire chain. Every single interaction between filled orbitals and unfilled orbitals will help to stabilize this. So when you look at gas fakes calculations, be careful because anything bigger will seem more stable because there's more bonding orbitals interacting with anti-bonding orbitals. Even if when you put it into solution, this actually isn't really different in stability. So be careful of those, of the effects of those calculations. Okay, so let's go ahead and make one final comparison. I've been talking about lone pairs in our last lecture donating into carbocation ions. So I've just finished talking about pi systems, double bonds, triple bonds donating into and stabilizing carbocation ions. So let's come back and look at sigma bonds. And we're not done with sigma bonds. We're going to have another kind of lecture on that. This is one of my favorite carbocation ions, not because it's really interesting, but because it's easy to draw. So you'll find that I often draw t-butyl carbocation and sophomore organic chemistry in this class, not because they're super great, but it's just very easy to draw that carbocation. And it's a realistic carbocation. It's not hard to make tertiary carbocation in solution. One point that I want to make is that t-butyl carbocation ions are more stable than allyl or benzyl cations. And so why is it? What's the nucleophile that's donating into that empty orbital? And so let me remind you what's donating to that empty orbital in a t-butyl carbocation. What's stabilizing this t-butyl carbocation is the donation of these filled orbitals, the CHs. There's a canonical frontier molecular orbital associated with this. There's a sigma orbital, a sigma bond. And there's electrons in that bond. And they are overlapping in space with this empty P orbital. And if this one isn't well aligned, one of these three hydrogens will be well aligned. There's no way you can spin this methyl group and not have at least one of those bonds aligned with this empty orbital. And there's three of them. Back here I've got bonds that can align. Back here I've got bonds. There's always going to be some group on this t-butyl or tertiary carbocation. There's always going to be something that's aligned to donate into that. And it is true that these bonds aren't very nucleophilic. But they're close in space. And it's hard for them not to overlap in space with that orbital. So that's why t-butyl carbocations are more stable than just a simple allyl or benzene. So if I compare that with this, this isn't as good. That allyl cation is not as stable. And you wouldn't have to predict. Now of course this is not a typical carbocation. Usually there's extra substituents on there. And at that point, as soon as you put one extra methyl group, the substituted allyl cation becomes more stable. But just a raw allyl cation with only three carbons, it's not as good as a t-butyl carbocation. Okay, I want to come back and deal with an issue that's related to this effect of electronegativity. So sigma bonds can donate into carbocations. Even though you typically don't draw CH groups attacking things. Let me go ahead and start off by thinking about this. And what I want to think about is the ionization of this leaving group and what's available to help push out that leaving group. How fast is it for the chloride to pop out in these two cases? And so the question that we want to resolve is if we draw this out, if I try to draw this out, what are the mechanistic possibilities here? So one mechanistic possibility that I have is that this bond, I just told you that bonds that are nearby, nearby bonds can donate into anti-bonding orbitals. And so let's draw this carbon-nitrogen bond. It's got a pair of electrons associated with that canonical frontier orbital. You can imagine that maybe this bond could bend over and push out that leaving group, right? That's not hard to imagine. If I did that, because of the way I drew this arrow pushing, I have to draw my products like this. What I'm saying here is that I'm making a pi bond. I'm ionizing off a leaving group and I'm leaving this nitrogen without one bond. If I use that bond to attack, that means I'm taking the electrons away from nitrogen and I have to draw those as the products. That's the same thing I would have to draw for as a resonance structure for the t-butyl carbocation. There's another way to think about this, maybe. And that is that maybe the lone pairs on nitrogen could donate in order to give us some sort of a bonding interaction here. And those are two very different types of interactions. So what is it that stabilizes that makes ionization faster or fast for this? This is called a mustard, this class of compounds where there's a nitrogen beta to a leaving group. Why is it that mustard's ionized so fast to the point where they're used in chemical warfare and to kill cancer cells? It's not this. It's not the sigma bond. It's the lone pairs. It's not the sigma bond that's donating. It's the lone pairs that are donating. In other words, you don't break the first carbon and nitrogen bond, what you do is you make a second carbon and nitrogen bond. That's the way to think about the reactivity. So it's not that all sigma bonds are equally nucleophilic. Sigma bonds attached to electronegative atoms aren't very nucleophilic, and that's what you'll find. Let me give you another set of ionization reactions and some rates, and we'll try to rationalize why some are faster and why some are slower. So we had talked about this reaction of methoxy-methyl chloride, mom chloride where R is a methyl group. This turns out to be very fast. It's much, much faster to ionize this leaving group out here, vastly faster than if you just have a primary alkyl chloride, and it is true that you'll have some electronegative atom attached to the carbocation, but that's not what's important here. The important point is that this lone pair can easily push out that leaving group, and when I look over here at what's available, there's no lone pair on this carbon atom here. It's just a CH2. The only thing we really have are some CH bonds and some CC bonds, and then finally, if we compare just a simple alkyl chloride with a related system where we've moved the oxygen atom one atom farther away, what we find is, I'm not going to use much greater than, I'll just change that to greater than, we find that this is less reactive. It's slower to ionize out the chloride when we have an oxygen beta, so it matters where we put that oxygen atom over there. The lone pairs can help, but here the oxygen lone pairs, unlike nitrogen, the oxygen lone pairs aren't nearly as nucleophilic. You don't get this effect of the lone pairs. All you get is this crummy electronegativity effect here where oxygen is electronegative, and you can imagine what that's going to do to the carbocation stability. Okay, so why is it that this bond is not very, neither the lone pairs are nucleophilic enough or the bond, bonds to electronegative atoms are less nucleophilic. So here this CC bond can do a little bit to help push out, to help donate into the antibonding orbital, but the CO bond is just nowhere near as nucleophilic, and so you see a much slower effect. Okay, so oxygen can help ionize out leaving groups, but it really depends. The oxygen atom has to be right next door to that leaving group so that the lone pairs can push it out. Okay, so that's the end of this initial lecture on carbocations, on ionization reactions to generate carbocations, and I want to come back to this idea and really focus on, you know, it sort of looks nice to think about this idea of a CC bond donating into the antibonding orbital or a CH bond donating into that antibonding orbital and pushing out the leaving group, and I want to talk about some really good sigma bonds that can really, really soup up the rates of ionization, and that's the entire next lecture. I want to remind you of this relationship. We have got a name for this. What do I mean when I say viscinal? This is a viscinal relationship between two bonds, between two groups, and this is referred to as a geminal relationship. So you can see there's, this is sort of like a three-bond sort of organization, and this is a two-bond. You commonly use that language when you talk about proton and amarcoupling constants. You can have three-bond viscinal coupling and two-bond geminal coupling. Okay, so let's talk about this arrangement of leaving groups and sigma bonds, and what are some surprisingly stable carbocation that can result from that arrangement? And in order to get us ready to think about that, I want us to become very good at drawing a specific type of bicyclic ring system. A nor-bornal ring system looks like this. Draw cyclohexane and draw one carbon bridge attached to it, and there's a very common classical way to draw this that looks like a coat hanger, and I just want to tell you don't draw it like this. I'll just write no, don't do that. Because when you do that, we're missing a very special relationship, and that's the point of this lecture, is special relationships between bonds. Instead, I want you to use your best chair drawing technique to draw nor-bornal ring systems. So let's practice drawing chairs. You know, when I was an undergraduate, I used to spend, count, I don't know how much time I used to spend, it's ridiculous. Just practicing drawing chairs on paper over and over and over again in the bathroom. Make some mist on the window or in the shower while I was eating. There's some special relationships that we're trying to gauge when we draw chairs. What I'm trying to do every single time I draw a chair is I'm trying to emphasize parallel bonds. This bond is exactly parallel to this one and exactly the same length. If I've drawn the chair correctly, that parallel relationship should come through. Here's another pair of parallel bonds that I'll point out to you. These bonds in front are parallel to the ones in back, the bond there. And it's not just this relationship, it's the fact that if I correctly draw an equatorial substituent that it is anti-paraplanar to those two bonds in the ring, perfectly parallel, perfectly the same length, and perfectly anti-paraplanar to those bonds in the same ring. And these relationships have to come out when you correctly draw a chair. So what I'd like you to do when you draw a norbornal ring system is draw it in a way that allows us to take advantage of your ability to represent anti-paraplanar relationships. So let's go ahead and start off with this for how to draw, how I would like you to draw norbornal ring systems so that we can see these anti-paraplanar relationships between bonds. So let's go ahead and start off. You'll notice that a chair is a six-membered ring but the norbornal ring system is composed of five-membered rings. So what I'd like you to do is I'd like you to start by drawing a chair and then remove one of the carbon atoms. Specifically, let's draw a bond across here like this, whoops. There we go. And now I have a five-membered ring. Maybe you could have just drawn this in very lightly. Let's practice again. And now I won't be so upset when I ignore the carbon atom on the side. Now the important point is I want to be able to draw perfectly axial bonds. When you draw axial substituents and I don't have another, here's a green pen. When I draw axial substituents on a chair, they should be perfectly straight up and down, not angled to the side, not leaning over perfectly straight up and down and all three of those axial bonds have to be parallel. Or I'll miss anti-paraplanar relationships. Let me draw the axial bonds on the bottom. Straight down, straight down, straight down. And if I've drawn this correctly, that means this bond here is anti-paraplanar to that one. And what that means is that this bond can perfectly donate into the anti-bonding orbital for the axial bond next to it. And those are the relationships that we want to see if we care about the interactions of filled orbitals with unfilled orbitals. I'm going to come over here and draw the axial bonds over here on my five-membered ring. Here's one. And now over here, I'm going to draw the other axial bond. And notice how clever I was not to draw this carbon atom directly below this one. Because if you do that, you're screwed, right? If you drew this carbon atom directly below that one, now you're going to have these overlapping lines and no human on the planet can tell what you're doing there. So you have to practice that. And now that I've drawn these two axial bonds with the same length, all I have to do is join those together, and there's my norbornal ring system, ready for me to emphasize anti-pair planar relationships. So again, why should you care so much about drawing things using chair-based confirmations? And I'm just going to give this to you as an example. This is actually a tricyclic compound. There's a bridge on this system right here. Let me draw that bridge in. I'm making this in bold to indicate that it's coming out towards you. And if we were to try to assess the ability of this oxygen, specifically the lone pairs on oxygen to come in and attack and displace that bromide, it does not look very good. It just does not seem very good. But let's try to reassess this and draw this using chair-based confirmations. So why should I be so particular about drawing a five-membered ring that's based on, and I didn't draw this end carbon here, that's based on chair-based cyclohexanes? If I draw it like this, it's very easy to fuse other chairs together. If you've correctly drawn your first chair, it's very easy to fuse chair rings together if you've correctly drawn them. And so the advantage of me drawing norbornal based on a chair is that it's very easy for me to fuse other chairs onto here. And there we go. So that's the 6-5 ring system. I can see there's an axial methyl group at this ring junction. There it is right there sticking up. Let me draw that axial methyl group. There we go. And I can see there's a norbornal ring system over here. Let me draw these bonds here in my norbornane ring system. And there it is. And now when I draw this bromide, the bromide is sticking up here. That means it's sticking out towards us. That means it's sticking on the same side as this bridge that's coming out. And now the important part is I now draw the hydroxyl group going down. There's an H that's axial pointing up. There's an OH pointing down. Here's the H bond. Here's the O. And now it's totally obvious to me that this ought to be the easiest bond forming reaction in the world. You could not stop this oxygen from attacking that bromide. If this got a whiff of base, there's a bond. There you go. You may be even without base. That will end up cyclizing, unstoppably. And you need to be able to do this without some fancy computer program that finds the minimum energy for you. You ought to be able to draw structures based on diamond lattices and based on cyclohexane chairs. I expect you to practice cyclohexane chairs. And I expect you to draw norbornane not like this. I don't want to see you draw it like this. I want to see you draw it like this over here so you can fuse on other six-membered ring chairs. And I'm going to give you problems where you need to see those relationships. Okay, so let's talk about why the norbornane ring system is special. It's going to help us to see the importance of bond alignment and proximity. Whenever I use the word solvolysis, that means I'm doing SN1 reactions and the solvent is the nucleophile. So if I say solvolysis, you might think about doing an SN1 reaction as ethanol. And then ethanol is the nucleophile. More typically, there were zillions of studies in the 1960s of solvolysis reactions. Most of those were done in acetic acid where acetate is the nucleophile. I'm going to start off by comparing two different norbornal ring systems or norbornane starting materials. In one of these, I'm going to have the leaving group sticking down here. And in this particular study, it was a sulfonate. It wasn't a tosylate where this is toluene. It was some sort of bromobenzene sulfonate, if I remember correctly. So they compared two different isomers. This particular position is called the two position of the norbornane ring system. So I guess you could have called this the two position or this the but by IUPAC, this is called the number two position. Okay, so let's go ahead and compare this with a diastereomer in which I've angled this instead of pointing downward upward. So these are two diastereomers of each other. There's an H atom there. Let me draw the H so we can see the relationship there. Here's two different diastereomers of a two norbornal ares sulfonate. And they are not equally reactive. Sometimes I like to darken bonds that are closer to me if I want to really emphasize in space what's close and what's far away. Maybe that's a little too long. Okay, so here's my two areing, here's my two norbornal. We have a name for this particular isomer because norbornane ring systems are so common. We call this the endoisomer. If you say, oh, that's that endo two norbornane sulfonate derivative, anybody, any chemist will know what you're talking about, we call this particular isomer the exo diastereomer, the exo isomer. So endo and exo and you simply need to remember that. Okay, let's think about ionizing this in acetic acid as the solvent. If we do solvolysis where the solvent acts as the nucleophile, actually in this case it was buffered, they threw in potassium acetate. So it's probably acetate anion that's attacking the carbocation. So the idea is you've got a leaving group that leaves to generate a carbocation and acetate comes back into attack. If we look at the reactivity of the endoisomer, I don't think it should be any surprise to you that you get this. You get the exo acetate as the product. But you don't really know the mechanism for this reaction just from the fact that you got inversion of configuration. You get a 100% clean inversion of configuration when you do this endoisomer. In other words, you start with the leaving group down, you end up with the nucleophile up, there's been an inversion at that center. If you take the exoisomer, you get exactly the same product. You get a 100% clean retention of configuration. So what's going on there? So first of all, it's very easy to conclude that this is not an SN2 reaction. All you have to do is run it under conditions where you decrease the concentration of the nucleophile and you find that the rate's not changing. 10 times less nucleophile, if it's SN2, ought to be 10 times faster. But if you change the conditions in a way that allows you to reduce the concentration of nucleophile, there's no difference in the rate. It's definitely not an SN2 reaction. What's more, and this is what's odd, is if I start with one enantiomer of this, the product is racemic. If I start with a single enantiomer of this, suddenly you have a racemic mixture for your product. When I come over and I look at this exoisomer, moreover, we find that it undergoes this substitution 350 times faster. Okay, so let's try to rationalize these facts. This was completely inexplicable in the early 1960s and there were fierce arguments about why these things should be true. I'm going to note something that's important with norbornain that makes it so special right off the bat and this is why I want you to be careful about drawing things. You have perfect bond alignment in the exoisomer. Let me draw out this exonorbornal type derivative again and I want to emphasize this relationship between the leaving group and in particular this axial bond. So let me highlight the leaving group. There it is and here's this bond here that's perfectly axial and it is perfectly aligned with the antibonding orbital with the sigma star antibonding orbital. If I draw out the antibonding orbital here, here's the antibonding orbital, that's sigma star for this carbon leaving group bond. That's the empty orbital there. If you put electrons into an antibonding orbital, you break the bond and this bond right here, this carbon-carbon bond right here is aligned to donate into that. It is overlapping in space. You cannot not have it overlap. It can't help but overlap with this antibonding orbital at all times. There's no option here in terms of flexibility of the structure. It is 100% of the time donating into that antibonding orbital and weakening this bond. And so that explains why this is so much faster. When you look over here at this endoisomer, you don't have this and it may look to you because of my crummy drawing. It may look to you like that bond is perfectly overlapping but it's not. And I wish I had a three-dimensional model to show you that that's not overlapping in space. Okay, so you get this perfect bond alignment that really weakens this exoisomer. And as a consequence, it's very fast to ionize out this leaving group. And so when I draw this carbocation intermediate, let me draw that empty orbital here. I almost like to draw the atoms that are attached on my carbocation, it helps me readily see what's stabilizing and what's not. Here's the carbocation intermediate. So now when my nucleophile comes in, it can add to the more accessible face. So in this particular example I gave you, acetate is the nucleophile and it can come in and attack this empty P orbital like that. But there's an antibonding orbital that is overlapping in space with the bottom lobe of this P orbital, that bond right there. It's the fact that this is donating from this side that makes it harder for the acetate to attack from the bottom. The transition state for attack on the top where you've got a partial bond here is the same type of transition state you drew for the leaving group leaving from the top. You know, if the transition state for leaving of the leaving group is lower in energy when it's exo, then you should conclude the same for nucleophilic attack as well as leaving group leaving. The important point is if I want to draw some sort of an interaction, some sort of a resonance structure if you will for this, where I have this bond donating into that empty P orbital, why don't I just shift it over? Why not just shift this over and have it bonded to that? And the way I'm going to depict that is I'm going to draw, start with the other type of chair. There's always two types of chairs that you can draw. And I, let me draw this so it's at the same level. That's going to drive me insane that it's not drawn at the same level. Well, that's not much better. Whatever, that's good enough. Okay, so there's my five-membered ring. And so now I'm moving, let me draw a dot so we can keep our eye on that carbon atom right there. And now I'm going to draw the dot on that carbon atom right here so I can really see it. And now I've moved this bond over so it's now attached to the dotted carbon. Now the carbocation is over on this side. And let me draw the empty orbital now. If I take the electrons away from this carbon atom, I should leave an empty P orbital there. And there's my plus charge. That's why you get racemic product. This is the structure of the norbornal cation. These are resonance structures. There's 50% of a bond right here and there's 50% of a bond to dotted carbon. The true structure of the norbornal cation has this sort of bridging interaction. There's no happy way for us to draw this with this two-bonding interaction. I'll struggle to draw this. I don't really know a good way to draw this. Nobody's ever figured out a good way to draw the two norbornal cation. You've got this carbon atom sticking down here. It's in back. Here's the carbon atoms in front. I'll put the dot on that one. And somehow or another you have to draw partial bonds to these two carbon atoms. Here's a partial bond there. Here's a partial. And it just looks ugly. Right? Moreover, you can't use these dashed bonds for arrow pushing. So I would urge you not to draw the norbornal cation like this even though that's a pretty good representation of the structure. So we're going to draw norbornal cations like this because with regular bonds we can do arrow pushing. Now what I want to urge you not to do is I want to urge you not to think of this as a stepwise process where this bond, well maybe, let me change that. Let me hold off on admonishing you not to do that. But it's not that this is very quickly migrating between the two. It's not just this is very fast because if you think of this as a chemical reaction where this bond comes over and forms a bond and then comes back like a windshield wiper, that means there's an energy barrier on a transition state. There is no transition state to this. It's not a chemical reaction. This carbon atom down here is simultaneously bonded to those two. It's not just quickly moving back and forth. Okay. So let's talk about this idea of, well, is it just doing a 1-2 shift? I don't know. Are 1-2 shifts fast or slow? We've heard of carbocation 1-2 shifts. This is basic stuff in sophomore organic chemistry. So let's talk about 1-2 shifts so that we can try to distinguish this resonance interaction from a stepwise 1-2 alkyl shift that has a transition state. So there's a toluene sulfonate on this cyclohexane derivative and if you draw this out with this tosylate leaving, you'll get a pretty crummy carbocation, a secondary carbocation. There's my secondary carbocation. That's awful. But fortunately, this thing can go downhill in energy if it simply moves this carbon bond over. And that's called a 1-2 shift. So there was an H here. Let me draw the H. I always like to see the things that are attached to my carbocations. And so the process of migrating over gives you a tertiary carbocation. That's thermodynamically favorable. So both methyl groups and hydrido groups can migrate and give you 1-2 shifts. Alkyl migrations tend to be faster. Let me just say this and get it right out there. 1-2 alkyl migrations to carbocations are fast, very fast. Now let me give you a calculated energy barrier. If you take just this very simple T-butyl substituted carbocation like this and you think about the rate for this methyl group to migrate over, they've calculated the transition state energy for that. It looks like this. There's a methyl group in the transition state. There's a methyl group that's simultaneously bonded to those two carbons. The two carbons have equal charge. This is a transition state for that 1-2 migration. It is very low in energy. The transition state energy here is 5 kilocalories per mole. In Chem 202, when you take mechanisms 2, you'll learn more about what does that mean? 5 kcals per mole, transition state energy. That means this is very fast. The half-life for this is on the order of about, it's less than 10 seconds. I don't know. It's on the second long time scale. Very quickly moving back and forth. If you're trying to do some reaction where you want to trap one carbocation before alkyl groups shift, that's sometimes a problem. That's very often a problem. You probably learned for fetal crafts reactions that it's hard to trap the tertiary carbocation before those shifts start taking place. So 1-2 migrations are fast. Even though, in theory, this would be a very fast 1-2 migration, it's now very easy nowadays to do transition state calculations using electronic structure calculations. And they tell you that this is not a transition state. This is actually a stable entity with partial bonding. It's not a transition state. Transition state or energy barriers where stable molecules are energy wells. And this is the true structure of the normal Boronkelon. It's not a fast migration. It's a slow one. Okay, so let's talk about other types of shifts because you will frequently in the older literature see alkyl shifts flung all around in mechanisms and overused. Can you have one three shifts or one four shifts? Yes, you can, but they are very, very rare. Let me give you an example of, and I'm going to try to draw a stop sign here, an eight-membered ring. I don't even know. Let me see if I can do this. Let me just draw a stop sign the way I know how to draw a stop sign. Eight-membered ring. There's my eight-membered ring. It's not so good. See, I never practiced over and over again drawing eight-membered rings. I only practiced drawing chairs so you can see how sucky that is. Okay, so let's draw two substituents over here. I'm going to draw a methyl group on one side of this eight-membered ring and then way over on the other side of the ring, I'm going to have a leaving group. And when you allow this, when you heat this up enough for this, it takes about 45 degrees in a polar solvent. That's the kind of temperature it takes for, you know, on a reasonable time scale for this tosylate to pop out. The product that you get out of this reaction is still an eight-membered ring. It still has the methyl group on there. But the tosylate is gone. And there's now a double bond right next to where that methyl group was. And initially it doesn't look so weird until you realize, whoa, what that means is you had to lose this H atom on this side. You can't get to that other strut, well, I guess it's possible the methyl group somehow moved to the other side, but you'd still be stuck trying to explain what happened to the atoms, the other Hs that were next to there. What's happening here is eight-membered rings are very crowded. Eight-membered rings are the hardest rings to make in organic synthesis because of something called transannular interactions. You cannot stop bumping of atoms across an eight-membered ring. These are called transannular interactions. When you're trying to make them, they kill you and your reaction rates for cyclization. And for the particular carbocation that you generate when you ionize out that tosyl group, that tosylate group, you get a carbocation that has this empty P orbital sticking towards the other side of the ring. Here, let me try to emphasize which bonds are closer to us here by sometimes I like to darken bonds that are closer to us to help us see spatial relationships. So you have this P orbital and here's this H atom right there with its electrons in that sigma bond. And that is not a very nucleophilic bond. We all know that. But it's right there over. It is physically overlapping in space. I didn't do this justice. I mean there's a steric interaction in the starting material where this H is already close. That CH bond is already close to the anti-bonding orbital for the tosylate. So it is not hard for this H atom to simply jump across, it's overlapping in space with that empty orbital to jump across. So in other words, you can get longer range alkyl or hydride shifts but they are very rare. So let me go ahead and redraw this product here. And here's the tertiary carbocation that you get as a result and then there's an H atom over here that gets plucked off on the other side by the acetate. This is done in acetic acid that gives you this resulting elimination reaction. So you can get longer range shifts. They are very rare and it requires super special molecular shapes to allow those to occur. You might see, see the advantage to this one is when the hydride migrates over, it leaves you with a tertiary carbocation. And let me state that in another way. The reason why this CH is so nucleophilic is that there's a lot of other bonds that are anti-pare planar to it weakening that bond. And if this were opposite a secondary, that CH might not be, it wouldn't be as nucleophilic. So will it give you the same hydride migration? I don't know but your prediction should be that it should be slower. So maybe if you heat it to high enough temperature you'd get it. So you might get a mixture in that case of double bond on this side versus double bond on the other side. Okay, so in the end the thing that I want to point out to you is that when you're, I'm going to give you a bunch of problems on the problems that have to deal with norbornal and I told you don't bother drawing it like this, we can't push arrows with this. What I'll tell you is go ahead and push arrows, go ahead and if you'd like to, to make it easier. I will allow you to pretend with your arrow pushing that norbornal systems are migrating in a stepwise process. So when you draw mechanisms involving norbornal cations like this I will allow you to do this like this. But I want you to know that that's not really stepwise, that these are actually resonance structures of each other. I will allow you to draw it as a stepwise process back and forth but I'd like you to remember that that's stepwise. Just agree that you're going to remember or sorry that that's, that this is, these are actually resonance structures of each other. Okay, so let's talk about a, I've got just enough time to mention another type of bond and I don't have time fully today to impress upon you how nucleophilic this bond is but I'll try. So we just finished talking about norbornal. It's an example of the effects of perfect bond alignment and proximity. So instead of talking about bonds that are perfectly aligned I want to talk about bonds that are very nucleophilic and I want to start off by making a comparison of hybridization, different hybridization, bonds with different hybridization and how that affects nucleophilicity. Something we've already talked about. What I want to do is I want to compare these types of molecular structures and see how bond angle correlates with hybridization. So what's the hybridization in an allene or in an alkyne? This carbon is forming sigma bonds on each, on opposite sides. What's the hybridization of the hybrid orbitals that the carbon atom is contributing? That's sp hybridized. How about over here if you have an alkene or a ketone or, yeah, it's actually this, I meant to draw an angle here. So sp, sp2 and then finally we get to tetrahedral carbon and we would refer to that as sp3 hybridized. If I look at the bond angles here what's the bond angle right here that's not an arrow pushing arrow. So 180, how about the bond angle for alkenes or ketones? So 120, how about in a perfect methane molecule? I can't remember the .5, I'm just going to leave 109. Okay, how much p character is there in this case? That's 50% p character. How much p character in this case? It's about 67, 66, doesn't make any difference whether you say 66 or 67 and finally we get to sp3 and it's 75% p character. What's happening to the nucleophilicity of these sigma bonds? The bonds are becoming more nucleophilic, not the pi bonds but the sigma bonds are becoming more nucleophilic as we increase the p character. So now I'm going to draw something maybe that's not quite so familiar. I'm going to draw cyclopropane. What's the bond angle in cyclopropane? Well it depends on what bond angle I'm talking about. I've got bonds in the center that the C, C bonds but then I've also got these C, H bonds. And if you look at a simple cyclopropane ring, what you would see is that the bonds in that ring have to be about 60 degrees. I'm sure there's some sort of geometry math class you could take that would teach you to back out. Yeah, in an equilateral triangle the bond angles are 60 degrees. But what's interesting is that in order to attain that bonding it has to use weird combinations of orbitals to achieve that 60 degree bond angle for the C, C bonds in the ring. So that's the C, C, C bond angle. But in contrast the H, C, H bond angles this is shorter than you're used to than anything you're used to. The H, C, H bond angles are larger than what you're used to for a tetrahedral type carbon. Those are 120 degrees. If you know, if you know the bond angles you know something about hybridization. Bond angle correlates with hybridization. And so what you would find is if you look at the percent P character for something that has a 60 degree bond angle it turns out that it's about 83 percent P character. Holy, that's a huge amount of P character in there. P character correlates with nucleophilicity. Those bonds ought to be very, very, very nucleophilic. And when we look at these C, H bonds the bonds that are sticking outside of the ring what you'd find is that they only have about 66 percent P character. It's like an SP2 hybridized bond. It's like a C, H bond stuck on an alkene. Those C, H bonds on a cyclopropane. So the expectation is if you have anything that's electron deficient next to these bonds that they can donate into those and stabilize them or push out leaving groups. Let me go ahead and show you the idealized arrangement for a cyclopropane next to, I'm going to go ahead and stop there. So when we come back on Wednesday we're going to talk about what it means for carbocations when they're next to cyclopropane rings. Those bonds are super nucleophilic.