 Okay, we're going to continue. We're just going to finish up some little last bits on lecture nine, topic nine, which is addition to sigma star in concerted displacement reactions. And we've been talking about SP2 at SP3 hybridized centers. And I just want to remind you not to draw SN2 reactions at SP2 hybridized centers. And I think you know that you shouldn't do this, but I'm going to find that you're doing this. Every once in a while, you'll make this mistake. And I'm simply going to remind you, it's never easier to attack at SP2 centers than to attack at SP3. And it's actually extremely disfavored to attack at SP2 centers with nucleophiles. So recall, if you ever have a choice, and almost always you have a choice, in a case like this, the nucleophile will displace preferentially here at an SP3 center. And so let's, I've drawn four molecules down here so that we can try to compare some different rates of attack at SP3 centers and SP2 centers. And I'll put some actual rate constants here on these. So I'm sorry, I didn't mean to write k relative. These are actual rate constants in per molar per second. So you can convert them into relative rates if you'd like. So the bottom line is of these four substrates that I've drawn here, this pro-pargil substrate is the fastest. It's got an SP3 center that's being attacked. There's no steric hindrance from the alkyne. And in a way, this alkyne is helping accelerate SN2 reactions because the pi system is donating into that P-like orbital in the transition state. There's almost like a, there's P character that's increasing at that center in the transition state. This is a little bit faster than just a simple allyl system. So not a lot, about 50% faster. And in the case of allyl, you have to worry about the rotational state of this double bond. As I twist around this bond right here, I'll just draw a little line here. As I twist around that bond, that pi system is only activating that carbon center when it's in a particular orientation. Whereas pro-pargil, it's cylindrically symmetrical. You can spin all around there and it's always activating attack at that center and it's less hindered. Okay, so now let's take a look at these two substrates at the other end over here on your right-hand side. These are sp2 hybridized centers. And let's see how much slower these are. So here we're looking at something on the order of a factor of 10,000 times slower to attack at sp2 hybridized centers and do SN2 reactions. It's not impossible, but you need to have a really good explanation before you invoke sp2, or sorry, SN2 reactions at these kinds of sp2 hybridized centers. And of course you're probably not going to do reactions with this kind of alenohalide. More typically you'll be dealing with a vinyl halide. And so now it starts to look really ugly as far as a choice for what kinds of reactions are good and what kinds of reactions are bad. So generally don't design reactions that try to rely on SN2 reactions at vinyl bromide or vinyl halide. Don't invoke those kinds of mechanisms where you do SN2 reactions because then I'll remind you of these rate differences and ask you to explain why you think that's plausible. Let me give you a sense for, and I'm going to try to draw this out, I'll try to give you a sense for the kind of penalty that you pay and maybe in a way overestimating this. I'm trying to draw out a planar version of methane. Obviously you can't make planar methane, but it's easy to draw out planar methane in a computer and say tell me the energy of this. What you find is that planar methane is 130 kcals per mole higher in energy than regular tetrahedral methane. And I'll just write Td is my abbreviation for tetrahedral methane. That's the kind of penalty that you have to pay. Not the full 130 kcals per mole obviously. That would be insane. But you're paying some of that price in the transition state for trying to displace this. The transition state's planar at carbon. And there's one orbital that's orthogonal to the others that doesn't get used in that picture. So you don't pay that full 130 kcals per mole, but some of that penalty is included in trying to do attacks at sp2 hybridized centers. Okay, so I don't want to say that it's impossible to do SN2 react, well, I don't want to say that it's impossible to do SN2 reactions at sp2 hybridized centers. This shows you it's impossible. But if you wanted to take advantage of that, there are reactions where it's very clearly going through SN, well, I don't want to call this an SN2 reaction. I'll call it a concerted displacement. So here's a very, very cleverly designed system in which you're doing a concerted displacement at an sp2 center. In this case, somebody had this nucleophile absolutely poised here, five atoms away, ready to form a five membered ring. There's almost no entropic penalty to this. Entropy can cost you up to 12 kcals per mole in the transition state. And you can see what's happening here is that this alkoxy group is poised to push out that chloride in an SN2 type. It's not SN2 because SN2 requires two substrates to collide with each other. This is intramolecular, so you can't actually call this an SN2 reaction. So you can see how much you have to gimmick up your system in order to make these reactions work. Notice also how cleverly designed this substrate is that you've got a methyl group here so that you can't do E2 eliminations across there. That would be a problem if you didn't have that methyl group. It would be a significant side reaction. And just to confirm that that's going through a concerted displacement, if you make the cis substrate, the other stereochemistry, there's no reaction with this. You don't get any cyclization. So I'll write no reaction, but what I really mean is no cyclization with the other geometry. So you really need this chloride to be anti and sort of poised here for this alkoxy group to push that out. Okay, so that's an example of, if you have something that's less tricked out than that, don't be proposing concerted displacements at SP2 centers. That's very challenging and not generally plausible. Let's take a look at another type of substitution that's possible but very rare. And that you'll have a tendency to overuse and that you will see overused in the literature when there's no mechanistic basis for those kinds of proposals. So we're going to talk about SN2 prime reactions. And I'll explain what that is in just one moment. This would be a typical case where you have to wonder, what's the mechanism for substitution on this system? If it's tertiary, if this is a quaternary center, if there's a tertiary center, if you substitute it, it gets very hard and very slow to add to that carbon atom. And so what you'll find is in rare cases that you can observe patterns of reactivity that look like this. Now how do you know that that is not an SN1 reaction? How do you know that the leaving group is not popping off to give an allocation and then in a second step? Well, you can check for things like solvent polarity dependence. Does it become much faster when you have polar solvents? That suggests an SN1 type reaction. I'll just give you some raw numbers here. If you compare rates of attack of a nucleophile on just a simple allillic chloride with nothing else on there, and you compare the rates of addition for here on this primary center, that's got no, this would be the SN2 pathway or, sorry, SN2 prime, and this direct pathway, let me make my lone pair bigger, the direct pathway where I attack directly here, we'll call it the SN2 pathway. The SN2 prime pathway is intrinsically over a thousand times slower. And here, in this case here, you're not worried just about, and there's a huge entropic penalty that's temperature, it's a temperature dependent entropic penalty. It's not just that you're worried about having the correct alignment between the nucleophile and the carbon chlorine bond. Now you have to worry in when these two molecules collide with each other is the pi bond perfectly aligned with the antibonding orbital of the carbon chlorine bond. Even more of an entropic requirement in the transition state. So there are cases where reactions are known to go through SN2 prime substitution. Let me give you an example. And in this case, they wanted to put on a leaving group that was pretty good but not too good, so they didn't use a tosylate because they had to be really careful to not get allillic cations. They didn't want to put too good of a leaving group here because they didn't want to form allillic cations. They wanted this to go through some sort of a substitution process. And so what they did is they heated this in neat pippardine. I'll just write, I'll just draw the structure for pippardine. So that was their solvent for the reaction. So they have a crummy leaving group here. This is not a fast reaction so they're using a very high concentration of their nucleophile to try to get something to go and heating the heck out of it for three days. So in this particular case, they got the substitution product. And here's their pippardine nucleophile and you can see it is added SN2. Now, again, this is a very heavily tricked out substrate. They took great pains to design something that wouldn't go through SN1 reactions. Notice this methyl group on here. In a way, it's kind of cool because it allows you to see that the nucleophile, and we'll talk about this in just one moment here, that the nucleophile added from the same face as the leaving group. In SN2 prime reactions, the nucleophile adds from the same face as the leaving group. And moreover, you can see that this methyl group allows you to tell that the nucleophile added from the same face, but that also makes it very hard for the nucleophile to attack here because the methyl group is sterically getting in the way. That means you're not directly attacking here. That's SN2 if you're attacking here. It's SN2 prime if you're attacking the double bond and that shifts over. Okay, so let me go ahead and draw out the orbital interactions here in an allylic system where there's some sort of a leaving group. And so I'll put, actually, I want to draw that leaving group pointing up. Let me make a little extra room here before you copy that. Okay, so let me go ahead and draw my leaving group here. And we'll draw a picture of the orbital interactions that are involved when you do this SN2 type process. So we've got some sort of a leaving group here. And I'm going to phase this orbital so that I, oh, I want to phase this, I want to phase this sort of anti-bonding here. So what we want to do is we want to donate electron density into this, this is my picture of a sigma star orbital. Maybe I should make this lobe a little smaller on the inside. So in other words, if you want to somehow displace this leaving group, you have to add to sigma star. How does it help you if you have a pi system nearby? Let me go ahead and sketch out pi star. And we'll try to imagine what would happen if we filled pi star with electron density? Normally pi star is empty, but if we started dumping electron density into pi star, what you'll find is you can make that overlap and you have to have the phasing correct, otherwise you'll get no significant double bonding interaction. This phasing tells you that as we fill normally empty pi star with electron density, what happens is that electron density will appear on the opposite side. This is a general principle when you add to pi system. So we'll talk about this more later. That if I add electron density, if I come up with a nucleophile here, the electron occupancy will tend to appear on the anti-periplanar side. It's the reverse of an elimination in a way. Okay, so what, in order for this orbital symmetry to work out, the nucleophile has to come in and attack from the same face as the leaving group. As you add, as you donate electrons into this orbital here, they start to appear on this backside and help you displace that from the other side. If you don't see this stereospecificity, you have no idea that it was, you have no evidence that it was SN2 prime. So notice that this is a fundamental requirement as that the nucleophile has to add in from the same face. That's a stereospecific requirement of SN2 prime reactions. You'll see people invoking SN2 prime reactions all over the literature and in most cases, there's no evidence for that. They're just trying to, because they don't have rules for arrow pushing, they're just trying to show some changes in bonding. Okay, so that's it for SN2 type concerted displacement reactions and so we're slightly going to switch gears here and talk about nucleophiles for concerted displacements that are a lot closer than this. We're going to talk about nucleophiles that are right next door, intramolecular relative to the leaving groups. There are all kinds of named reactions in organic chemistry, pinnacle, rearrangement, wolf rearrangement, Beckman rearrangement, courteous rearrangement, the list goes on a wolf rearrangement, the list goes on and on. Many of them are all categories of this, something called a migratory displacement. It's basically the same idea that's been used over and over and over again and maybe they haven't covered every single possible variant on this. All you have to do is just switch one atom out and put a new one in and you've got your own name to rearrangement. So I'm going to take you back to one of the oldest examples of this. This stays back to 1860. It's a very urgent origin of organic chemistry. The pinnacle rearrangement. And it was first demonstrated with this diol called pinnacle. You make pinnacle, I'm trying to draw this in a saw horse representation so that this group is specifically anti to this. So that there's a bond that's anti to the OH group. This is pinnacle. You take acetone, you dimerize it with a reducing agent. And you get this alcohol, it's dirt cheap and very common, very easy to access in 1860. And when you expose this to sulfuric acid, which was one of the few reagents they had back in 1860, something happens. Suddenly you get some product that doesn't have the same carbon skeleton anymore. And what's going on is that this is rearranging ultimately initially to give this oxonium ion with a T-butyl group on it. And of course you lose the proton and this is now just a simple ketone. They call this pinnacleone because classically you make it through a pinnacle rearrangement of pinnacle. Pinnacleone, it's a ketone. Even though it doesn't have the same carbon backbone as pinnacle. Okay, so this is a pinnacle rearrangement. What's going on here? The way to think about this is that the actual leaving group in this case is protonated OH. It's water is the leaving group. And so what is it that's going to make that act as a leaving group? There's a sigma star orbital on the back end. If this is going to leave, something is going to have to donate electron density into that sigma star orbital. Normally for carbocation formation, it's the CH bonds that are next door. But look at this bond that I've drawn right here. When it's anti-pare planar, the electrons in that sigma bond are donating into the anti-bonding orbital. They're helping to push out this water leaving group. And it turns out that this bond is weak. That bond is weak because there's a lone pair and we know lone pairs are good donors that's donating into sigma star for this bond and weakening it, making it a better nucleophile. So you can see by having this lone pair here, we weaken this bond and make it a better nucleophile. Weaker bonds are more nucleophilic. Those electrons in those orbitals are higher in energy. Okay, so we're going to see a bunch of reactions that look like this. I'm just going to kind of change nitrogen, carbon, switch oxygen, every possible variant. And you just see these reactions done over and over. Okay, so that's the pinnacle rearrangement. Now how do you know that this is not just going to a carbocation? So I'm going to draw a slightly modified substrate here. Or I've got phenyl groups on each carbon in the dial. And then on one side I have a methyl group and on the other side I have a proton. Now if this were going through a carbocation, you'd be totally correct in believing that, well, it ought to be easier to form a carbocation at this center, right? If this is just carbocation formation, then what's going to happen is this OH is going to leave as water. You'll form a carbocation here and one of these two groups will migrate. But you don't lose this water. It's not the one over here at the tertiary center. You lose this water, this OH group as water. So when you look at the product for this reaction, there's some sulfonated sulfonic acid resin. It doesn't matter. It's just a solid phase version of sulfuric acid for this called napion. So the end product of this, what you can see is that the methyl group stayed bonded to the same carbon atom that had OH on it. What's happened is this phenyl group over here has migrated over. So if we draw this out with some, and I'll put back in this red, you can see what's happening here. Is that these lone pairs are donating, weakening that bond, and that's migrating over. Now there were, I guess there were four independent possibilities here. It could have been the methyl that migrated over, the phenyl here, the phenyl on this side, the hydride on that side, all of those were possible. In this case, it shows you that in the pinnacle rearrangement, the migratory aptitude, and you can run through a series of these in order to gauge which groups migrate faster. And if you measure that, what you'll find is that aryl migrates the fastest. That's greater than CC double bonds. That's faster than hydridol groups. And then those migrate faster than tertiary, which is faster than secondary, which is faster than primary. Another thing, you're not going to use sterics to argue what migrates faster here. You might look at the fact that there's electron deficiency in this group here, and the better it is able to accommodate that, the better you'll be at migrating over. So why is it that, if you wanted some sort of a reasoning for why is it that phenyl groups migrate faster? When you look at 1, 2 carbocation migrations, it turns out that carbocation migrations are faster. And again, this is not a carbocation with phenyl rather than alkyl because the double bonds here can donate in at the same time that the sigma bond is donating in. And you might be saying, wait a second, you're saying that a sigma bond can donate into an orbital at the same time that a pi, yes, yes, yes. If I have an empty orbital in some acceptor system, more than one component can donate into that orbital at the same time. That's, MO calculations take all that stuff into account. If you drew some sort of a resonance structure where this pi bond donates in, it'll look kind of like cyclopropyl carbinyl. That's why phenyl and vinyl migrate so quickly. In the transition state for migration, it looks kind of like cyclopropyl carbinyl because not only is this sigma bond donating in there, the pi bond is also donating in. Okay, so pinnacle, everything I'm going to show you is basically the pinnacle rearrangement with some sort of a variation. So let's take a look at some common variants. So this first one we'll go through is called the Typhono-Demianoff reaction and it's very easy to make visceral hydroxia means. So now they have these osmium catalyzed reactions to make those or you can add nitromethane into a carbonyl and then reduce the nitro group. That's how you make methamphetamine. Of course, the trick here is you need to convert that amine into a leaving group and the best way to do that is with something called nitrous acid. Typically, usually in chemistry, we write HONO. That's not, it looks like an acronym. It's not an acronym, it's a molecular formula. This is nitrous acid and so that's what they're using in order to convert the amine group into what's called a diazonium leaving group and initially you get an intermediate that has an extraordinary leaving group on there. I can't think of any better leaving group. It converts nitrous acid, converts amines into diazonium groups, right? Nitrogen gas is now the leaving group. How do you get better than that? It's very hard and you can see the system is perfectly set up, I've got bonds here depending on how I tilt this dinitrogen. I can, just depending on how I rock this back and forth at any point in time, one of these two bonds here will be aligned anti to this bond and can just shift right on over. And so let's draw that out. So now instead of having water as a leaving group, I've got a vastly better leaving group diazonium. And so you can imagine it's just migrating over. And what is it that makes this bond good at migrating? Well, it's being weakened. What's weakening this bond? Well, there's lone pairs here on this OH group that are donating into the anti-bonding orbital. It's like the OH lone pairs are almost doing an SN2 attack breaking this CC bond, which weakens it and it migrates over better. Okay, so that's what happens. You can see what's going to happen here is you'll end up with a ring expansion. And I'm not sure how good of a job I'll do here with drawing this out, but you end up with a seven-membered ring. And that's traditionally what Tiffano-Demminoff is used for, it's for doing ring expansions on ketones. And you end up with an extra methylene group there whenever you need that. Yes, I would do it like this. The end product, or the end of this, so I'm leaving out a step, sorry. We'll have a proton still in the carbonyl. And then you can deprotonate. So again, you don't buy, you can't buy nitrous acid. There's no bottle of nitrous acid in the lab. What you do is you take sodium nitrite and you add hydrochloric acid to that. And I'm not going to go through the mechanism. It's like five or six steps of aero pushing, but basically you can protonate nitric acid in order, sorry, this nitrous acid and get watered as a leaving group as part of the diazonium formation mechanism. So you make nitrous acid by protonating sodium nitrite and then that reacts in an acid-catalyzed reaction with amines. Okay, so again, you can see the analogy. They just took OH2 and replaced it with nitrogen gas as a leaving group, same idea. Okay, let's extend this idea from a sigma bond here. What would happen if I replace this single bond here with a double bond? It'll still work, yeah. They're both getting protonated rapidly. Proton transfers are fast. Stick a proton on, pull it off, stick a proton on, pull it off, stick a proton on, pull it off. So it'll, there's rapid and reversible protonation that's occurring in this reaction. So it's not that you're not protonating this, it's just that you don't get migratory displacement until you protonate this one. It's faster. That's the limiting step. Okay, so this is an alpha-diazo carbonyl compound and this is kind of a worthless structure to draw. It doesn't give us any insight into reactivity. Actually, let me change this. I'm going to draw it as a more popular resonance form and this is still worthless for arrow pushing as far as thinking about reactivity. Let me draw, I'll draw a different resonance structure for this that'll help you to more clearly see what kind of reactivity you should expect for this system. And until you draw this alternative resonance structure, I don't feel like it's entirely clear what you ought to be drawing. Let me start off by doing this. I'll take these electrons here and this nitrogen, donate them here, move those over and that'll help you to more clearly see what kind of reactivity to expect from this system. So that's really the best resonance structure to help you think about what's going to happen when you heat this or if you hit it with a photon of light and jack in 80 kcals per mole of energy. So under those kinds of conditions, heat or light, these will rearrange to give ketines and then the leaving group you can see is going to be nitrogen gas. I'll just sort of sketch that in there. Okay, so how do we think about the orbital arrangement here that allows this to rearrange to give ketine? Well, once again, I can see it's the R group that is migrated over. So this is the leaving group. That's pretty clear. Why is that leaving? It's leaving because the R group is pushing it out. It's like the R group is the nucleophile here and you might say, well, that's a crappy nucleophile. Why would that push anything out? Well, there's a lone pair here that is always aligned anti, that is always aligned to donate into the anti-bonding orbital here that weakens that R carbon bond. That R carbon bond is nucleophilic because of that lone pair which is donating into that and that allows that to migrate over and so you can heat or light or the common ways to do that reaction. You know, you never, I'm just looking at this example. You never know the impact of organic chemistry around you. You look at all the microchips that are in your computers and cell phones, et cetera, video projectors, they're made through organic chemistry. So it turns out that the photoresist resins that they use to pattern those microcircuits are organic compounds. It's organic photochemistry that's driving all of that and I'll show you one of the earliest versions of a photoresist that's used to photopattern integrated circuits. It's a process called photolithography. This is called Novolac DNQ resin. DNQ stands for diazo-naftoquinone and that's this wolf rearrangement substrate that goes through this mechanism here. Somebody sat in their organic chemistry mechanisms class and heard about this wolf rearrangement and I guess they thought they were going to apply it. There's a sulfone type group there that helps modulate the properties. In this case they use light to put in the energy that allows this rearrangement to occur. And the, I'm not going to draw through the entire mechanism here but in this case, when you do this rearrangement what happens is you go from this water insoluble organic goop that they spread on microchips and wherever you shine light it does this photochemical version of the wolf rearrangement and then you get a ketene and so then when they expose this to aqueous sodium hydroxide this gets converted into a water soluble sodium carboxylate. I'm not going to show you all the steps for attack of hydroxide on a ketene but it's like a three step thing going on there. And so now wherever you shine light those little parts of your goopy film become water soluble and can be washed away to leave these patterns, these microcircuit patterns. They have much more sophisticated photochemical resist resins nowadays that are based on catalytic proton generation. This is one of the earliest versions of photoresists, these DNQ type resins. Novolac is just a polymer. You don't need to know about that, not that it's not cool but you don't need to know about it. Okay, so Hoffman and Curdius rearrangement. Let's switch from carbon to nitrogen. I'm just going to do a quick pen comparison to see if this other black pen is better. I think that's a little better. Okay, so a classical Hoffman rearrangement you can see the analogy here to this, to this Wolf rearrangement. I'm simply going to replace this atom right here with a nitrogen atom. And the leaving group is not that important in this particular case with the Hoffman. It's a bromide. And in order to make this work better you typically have to have this with a negative charge on there. And so let me go ahead and draw out a resonance. So imagine taking an n-bromoamid and deprotonating the proton from an n-bromoamid. It's easy to make n-bromoamids. I'll give you the conditions in just a moment. And what I'm going to do is I'm going to draw a resonance structure once again. Not this one. This is not a really good resonance structure. But as soon as I draw this I can see how much it resembles that Wolf rearrangement thing over there. And look at that. Looks just like that, well, not exactly, but very close to that Wolf rearrangement. So now you can imagine, okay, bromide is the leaving group. But why is that going to leave? Something is going to push that out. This R group here is going to push that group out. And why is that R group so nucleophilic? It's because this lone pair is donating into the antibonding orbital. It can't not be donating. The lone pair is overlapping in space with the antibonding orbital for this R carbon bond. And so the products of Hoffman rearrangements initially are isocyanates. And usually you do something with those. Most uninterestingly what you do is you hydrolyze them down to a means. So there's another version of this reaction you can do with azides. And so let me go ahead and draw out, allow you to imagine, if you will, an azide that's an N3 group. Here's the more common way people draw azides. N3, but if you draw it. N3, then you can't really see that it looks a lot like that wolf rearrangement. That's an azide group right there. It's, I'm just drawing a good resonance structure for that. Okay, so same idea whether it's nitrogen gas as a leaving group or an enbromo amide. Typically how do you use this to make a billion dollars? Here's one way. You take a, so this is the way you do a Hoffman rearrangement. It used to be called a Hoffman degradation because the end result here is you're going to remove this carbon so that this carbon is now directly connected to the nitrogen atom there. So initially what happens is that in the presence of base, you're forming a nitrogen bromine bond. You deprotonate here, you attack bromine, and so you make this weak, displacable nitrogen bromine bond, your enbromo amide. So that's an initial intermediate. And then the sodium hydroxide that you're stirring around and deprotonates this, and that's what initiates this whole, this whole Hoffman business. So the end result in this case, I won't draw the whole species here, I'll just draw the isocyanate. And so in the presence of sodium hydroxide, you can't stop, you can't isolate this if you're stirring it around in sodium hydroxide. This ends up hydrolyzing to give you something called fentermine. So back in the 1990s, there was this diet pill called fenfen, it was a prescription diet pill. So half of it was fentermine, the other half was some of course as it always is some derivative of speed, methamphetamine, so, and Wyeth-Irst made billions of dollars off of that until people started getting heart attacks and then they had their asses sued off for billions of dollars. So that's why most companies are very scared of making diet drugs now a days because they're worried about that. Okay, let's talk about the courteous rearrangement. It's this azide business up here, this azide thing where it's the same idea, but it's, I think conceptually it's easier to think about making an azide. So in a classical courteous rearrangement, you start with an acid chloride or some sort of activated ester and then you have an azide anion floating around. You can literally add in sodium azide to acid chlorides. And again, this is the common resonance structure you draw for an azide, so let me draw the more useful resonance structure. So I'm just going to start by drawing a resonance structure for this that allows me to see the relationship between this and that and the N-bromo amide up above. And I'm not going to draw the whole nitrogen leaving group. But again, you can see the arrangement of species here. Lone pairs on this O minus, look how nucleophilic that is, donating into the anti-bonding orbital, weakening this R-carbon bond so that R-carbon bond is weak and nucleophilic. And look at that, it's anti-paraplanar to the leaving group. You can't get better than that. And that just pushes out the leaving group. So the R-group sort of shifts over so that it's now bonded to the nitrogen atom. And so that's called the courteous rearrangement. And it's just like that, a Hoffman degradation, Hoffman rearrangement. Okay, so very easy to make azides. I'll give you one more practical example of this reaction, a one-step reagent if you don't like making acid chlorides. That's very common for doing this courteous rearrangement. And I'm just going to put an R-group here. It doesn't matter what it was. Right, you couldn't take this and make an acid chloride very easily. You'd end up with polymer or some sort of cyclic sulfite ester there, which wouldn't be very good. In this particular case, and I've got to get these pieces right here, just one moment here. So you can see they're replacing this carbon-carbon bond to that carboxylic acid with a carbon-nitrogen bond. So there was an isocyanate intermediate involved here. And the reagent that you use for this is diphenyl phosphoryl azide. Instead of making an acid chloride, diphenyl phosphoryl azide, sometimes it's written DPPA, makes an O-phosphate leaving group here. And then the azide leaving group from this comes back in and displaces the phosphate leaving group. So there's an intermediate that's involved in this reaction that looks like this. And then the phosphate anion gets displaced. And again, I'm not going to draw the ester exchange mechanism here. So that's the leaving group. And then this undergoes that courteous rearrangement. And then the hydroxy group cycleizes onto the isocyanate, just a very common one-step way to make isocyanates from carboxylic acids using a one-step do-it-all reagent. Okay, how else do you make a billion dollars with concerted rearrangements? I'm going to use this other board so I have room. So you can see the lesson here. It's like, gee, I can take a second row atom in one of these rearrangements and replace it with a different second row atom. And it works pretty much the same. And that's true. I'm going to start off by drawing out a very big natural product. And you're not going to have room to draw out this whole natural product. I'm going to draw out a derivative that is, so this in and of itself is not a billion-dollar drug. This is called an oxime. It's made by simply treating a ketone with hydroxylamine, oxime. They're usually crystalline. In the old days, people used to make melting point derivatives of everything and get melting points. But we don't do that anymore. We have NMR. So this is the oxime that's derived from a billion-dollar pharmaceutical. And again, you don't have to draw all of this stuff down here. Below these maybe top three atoms, it's not really all that interesting, at least from the standpoint of chemical mechanism. What I'm drawing out is erythromycin. It's an antibiotic. You may have taken it. The old, the original version of this was called biaxin, sold by Baer. It's expressed in bacteria. And they make this on the kilotone scale. They can express this. That's why everybody wants to use polyketide biosynthesis because this is the example that everybody uses to show that once you get the genes for natural product biosynthesis into an organism, you can grow up kilotone quantities at the cost of growing up feces. It's very, very sound technology. So of course, Baer sold this as antibiotic baxin. They wanted to protect their market share. So what did they do? They did what every company does. They patented every single conceivable derivative of this starting material. Anything you could possibly think of, they patented. Except for this. Let me try to get this in here correctly. What they didn't patent was the ring expanded version of this. And I'm just drawing the top end here from, let me sort of draw this over, so from here on over. They missed something. You know, you'd think you cover everything, that you're ingenious enough to patent everything. But they didn't cover this. And of course, Pfizer thought of this. And so what they realized is, gee, whenever I have a ketone, maybe there's some way to insert a nitrogen atom there using a migratory displacement reaction. And so that's what they do. Well, basically the whole idea is you take the ketone, you make the hydroxylamine, and then you make the otosyl derivative of that ketone and it does something called the Beckman rearrangement. And it makes this intermediate. And again, I'm not going to draw the whole rigmarole thing in here. You don't need to see everything that's involved. What you can see has happened is that somehow or another this carbon atom that used to be right here has now shifted up and formed a bond with nitrogen. And see this hydroxyl group over here? It's cyclized around onto that. I'll draw the mechanism a little more detail in just a moment. But this is the intermediate. And this is basically an amide. They reduced that down and methylated in order to make Z-throw max. I don't know, there's now a zillion formulations for this. I'll just write Z-throw. So maybe you've had any ear infection or some sort of bacterial infection. Z-throw max took over market share from bears by axon. Okay, so let's draw out the arrow pushing for this Beckman rearrangement. And it looks something like this. And the important point to remember about the Beckman rearrangement, I'll draw out the otoscel intermediate here, is that it's stereospecific. When you make an oxyne, you get E and Z isomers. And you don't easily control that unfortunately. Usually you get mixtures of E and Z. And they don't rapidly interchange. And the only migration that you get is the group anti to the otoscel leaving group. And so what happens is this group is oriented here to displace out that phenomenally good otosceloxy leaving group. And the intermediate that you get looks kind of disturbing. It looks kind of like an acyliomion from electrophilic aromatic substitution. We call that a nitriliomion. And it's not that bad. You may think that this looks bad to you, but at least every single second row atom has a filled octet of electrons. So that's no big deal to make this nitrilium. And usually when you expose these to water, water will come in an attack in order to make an amide group. I'm not going to draw all the steps in that mechanism. And let's see, I had R prime here. So I won't show all the steps involved in that attack. But in the case of Z-throw, in the case of a rethromycin when they did this rearrangement, it was an intramolecular hydroxyl group that attacked initially. Okay, so that's a Beckmann rearrangement. And it's the same kind of idea. You've got this orbital rearrangement here that allows a group that's anti to displace a leaving group. And these are always aligned with each other. This R carbon bond is always aligned with that O-tossil antibond, the nitrogen-oxygen antibonding orbital. Okay, the last example of a migratory displacement, that's not the last one that I'm going to cover. I'm sure there's many, many more examples of these. These are really kind of two related reactions. I guess I would call the rearrangement itself a crigé rearrangement. And here's the idea. The idea is that you've got some sort of a ketone or a carbonyl. And when you treat these with peroxides, we'll talk about this more later, peroxides turn out to be very, very nucleophilic functional groups, more so than a regular alcohol. So initially you'll get an adduct and I'll draw it in a very suggestive way here. Here's my peroxy hemiacetal, that's what I would call that. Where I have two oxygens bonded to a carbon, call that an acetal and this is a hemiacetal because it's OH. And I'm drawing it suggestively here so you can see this completely weak oxygen-oxygen bond. Right? That's a weak bond. It's easy to break. If you want to make explosives while you're on the plane next time, take some hydrogen peroxide and acetone, that's why you now have to check all your liquids now because of this reaction. So in this particular case, again, we've got this lone pair that's donating in, that's weakening that bond, makes it easier to act as a nucleophile. So this weak nucleophilic bond now just shifts right on over. It overlaps with the antibonding orbital and there you lose that leaving group. And oftentimes you'll run this under slightly acidic conditions where you can protonate this and make that an even better leaving group. So you can see what happens is you cleave a carbon-carbon bond in essence and you end up with an ester. So where you used to before have a carbon R attached directly to the carbonyl, now there's an ester. There's an oxygen atom that's been inserted in between. So it's a great way to do ring expansions. And here's something you couldn't have known with respect to migratory aptitude is that if you have two very similar carbon substituents, one of them is more substituted, it's the more substituted side that reacts in preference. So one version of a reagent that you can use for this type of Bayer-Villager reaction is a peroxy acid. So not a peroxide, but a peroxy acid. These are very common for this reaction. So it provides an acid and it provides a weak oxygen-oxygen bond. And the end product for this Bayer-Villager reaction now has an extra atom inserted and there's two different products but the major product involves oxygen being inserted on the more substituted side. And this is completely general. And the ratios, this will give you a sort of a typical ratio. These are the isolated yields, 87% and 5%. Okay, so weak oxygen-oxygen bonds. You can just replace carbon with oxygen. There must be a zillion other variations on this and I would think they've all been tried. Jesus, this stuff's been going on for 140 years or something like that, but maybe not. There's your proposal. Okay, so that's all we're going to cover for right now. I can't even remember what we're going to cover when we get back, but we're done talking about addition to sigma star orbitals either by nucleophiles that come in intermolecally or by neighboring bonds.