 Okay, so we have spent so far the first part of this class talking about addition to empty P orbitals. Carbocations and reactions that involve carbocations that have empty P orbitals on carbon. Carbon wants to satisfy the octet rule. That's why it's so fast to add into carbocations. It's not the charge that makes it fast. It's the octet rule that makes it fast. Now we're going to skip pie star just momentarily. If I were to look at things that I want to attack, if I were some nucleophilic pair of electrons, what would I want to attack? I would want to attack pie star next, but we're going to skip that, we'll come back to that. And let's talk about sigma star, adding to sigma star orbitals. Just looking at these energies, you ought to predict that these ought to be the toughest reactions in organic chemistry. Addition to carbocations ought to be the easiest. Everything can do that. Everything can add to a carbocation. But now let's talk about things that are the hardest reactions. And I'll qualify that by saying it depends on what kind of sigma star orbital we're talking about. I'm going to sketch out a periodic table just roughly like this. This is my view of the periodic table. I'm going to try to put this second row a little higher here. And there's one atom, well there's two that are, there's only one that can form bonds. And that's hydrogen. Helium is not interesting. It's over there. So hydrogen is in a totally different row. And I want to consider that first. Because when you look at sigma star orbitals, they're not all created equal. So you can imagine some sort of an atom and I'll, let me just start off right here. I'll draw a picture of a nucleophile attacking a sigma star orbital. So if I have some nucleophile that displaces iodide in a concerted reaction, the reason why this works is because we're adding to the sigma star orbital for a carbon iodine bond. So let me sketch out that orbital. I'll give you my sort of rough approximation for what that orbital looks like. I'll add some phasing in here. Notice these two inner lobes of that orbital have opposite phasing. That's why there's no bond in the middle. If you add to this, you break the bond. So that's a typical SN2 reaction. But there's another type of sigma star orbital that's not slow but that's fast. And so let's consider that. The alternative would be if I had a pair of electrons attacking a proton. And now I'm going to reuse a different symbol here. I'll use the symbol base to represent that sigma star orbital. And this is a totally different row of the periodic table. Carbon is down here in the second row. Protons are in the first row. And whereas these reactions tend to be very slow, SN2 reactions tend to be slow, additions to this kind of sigma star orbital tend to be very fast. So we're going to spend some time talking about these sigma star orbitals to protons because they're fast. And then we'll talk about making carbon-carbon bonds or bonds to carbon here. So again, these are two different, I'm not going to claim that this is a good representation for an actual, I'll just draw it the same way though. Just so you can see the phasing. The actual sigma star is a slightly different shape for HCl. But it's the same idea. Sigma star orbital, so here the sigma star orbital that I'm adding to is sigma star for an HCl bond. And here it's sigma star for a carbon iodine bond. Okay, so let's talk about addition to proton and proton transfers. And I hate this. I hate talking about this because it just makes me feel awful. Because I told you on the first day the goal of an arrow-pushing mechanism was to break mechanisms down into elementary reaction steps. And I'm going to step aside from that just momentarily. Manfred Eigen was a German biophysicist who was interested in proton transfer steps. If you look at any enzymatic reactions or biological reactions, even most organic reactions, proton transfers play a major role in that. And what Manfred Eigen won the Nobel Prize for in 1967, what he learned how to do is he learned how to do ultra-fast reaction kinetics. He learned how to measure reactions that were so fast that nobody had previously been able to measure them before. And so let me go ahead and show you the mechanism that he discovered for proton transfer reactions. I'm going to draw a simple proton transfer reaction. And I'm not really intending for this to be dimethylamine. I just wanted to have two different atoms here. But let's consider momentarily the mechanism for this proton transfer reaction where I transfer a proton tonight from nitrogen to oxygen. This is a three-step mechanism to transfer the proton. That's what Manfred Eigen showed. Really disturbing stuff. In other words, what he showed is that the first step in any proton transfer reaction is formation of a hydrogen bond. That's the first step. There's an energy barrier and then you form a stable hydrogen bonded product that looks like this. And then there's a very fast bond vibration. There's an equilibration. Imagine this H atom. I didn't do just to this. I apologize for this. Let me, what I want to do is I want to move this over so you can tell the hydrogen bond is about two angstroms long. A typical C-H-O-H or NH bond is about one angstrom long. So you can tell that this is not fully bonded to the oxygen because it's closer to nitrogen. There's a second step in this mechanism where this proton vibrates back and forth. You could measure the frequency of that in an IR spectrum. How fast? It'll have a stretch around 3,500 wave numbers. That's the frequency. So when I redraw this, there's another extreme form where this is now an oxonium ion and this is now hydrogen bonded to the N minus. And these are not resonance structures. The hydrogen atom has moved. If the atoms move, it's not a resonance structure. And there's an energy barrier for that. Very small, but there's an energy barrier. And then in the last step, you break that hydrogen bond and now the hydrogen atom is transferred over to the oxygen. Now we already said do not use arrow pushing for data bonds or hydrogen bonds. We can't use this for arrow pushing. If you try to take electrons out of this bond, there isn't really, it doesn't represent an orbital. So we're stuck now. How are we going to represent this using arrow pushing? And I'm going to make this very simple. Here's our rule for arrow pushers and that means you. Draw H bond, draw hydrogen atom transfers as one mechanistic step. And let me change that to proton transfers, not hydrogen atom. I don't, when I say hydrogen atom, I mean radical. In other words, when you draw the arrow pushing for this, I want you to do what you've been doing from day one. Just take this lone pair, attack the H N bond or the H and then break the H N bond. That's how I want you to represent it. But every time you do that, I want you to remember that it is a mechanism that is dependent on hydrogen bond formation. And as soon as you forget that, as soon as you forget that hydrogen bond formation is one part of this three-step mechanism. So I want you to draw it like this but every time you do that I want you to remember it involves H bond formation. Because if you forget that, you're going to suddenly run into cases where reactions don't make sense anymore. And it's when you remember that this is a three-step mechanism. That's when you have something to fall back on. Okay, so let's take a look at an important fact about proton transfers that has to do with reaction geometry. So when protons transfer back and forth between atoms, they prefer linear geometries. And I feel like I need to justify this because you will have a tendency to do it through nonlinear geometries. And I want to tell you why you shouldn't do that. So there's good evidence for the idea that proton transfers involve ideally and involve linear transition states. If you look at crystal structures involving hydrogen bonding, remember hydrogen bonding is essential for proton transfers. If you look at hydrogen bonds in crystal structures, they tend to be linear. Take a look at any protein crystal structure or neutron diffraction structure where you can actually see the positions of the hydrogen atoms and you look at hydrogen bond geometries, they tend to be linear. 180 degrees. In other words, the hydrogen isn't like way over here. It's right between the two atoms. So you have 180 degree bond angles between oxygen, proton, and nitrogen. Or alcohol to oxygen and hydrogen bonds doesn't matter. Zillions of crystal structures and zillions of hydrogen bonds. You can look at all of them and this is what you're going to see. Second piece of information. And I don't know how much, this may not make much sense to you yet, but when you take chemistry 202, you're going to learn about kinetic isotope effects. What happens if I replace this proton with a deuterium? Well, deuterium is more massive. It's got twice the mass of a proton. You expect things to slow down. Everything should get slower. It's twice as massive. It turns out that how much slower depends on the geometry. It's not twice as slow. It's more than that for a typical kinetic isotope effect for protons. What you find is if you look at proton transfers for compounds that have deuterium instead of proton, and I've got B to represent base and A to represent acid, yes, it slows down when you change a proton for a deuteron, but it slows most when the transition states are linear. When you design molecules where this has to be bent or you design molecules where this has to be transferred in a linear fashion, suppose maybe it's like a five-membered ring. It slows most when proton transfers involve linear, let me just write, with linear transition states. And that's an evidence that, well, indeed the preferred transition state is a linear proton transfer, those kinetic isotope effects. So don't worry about the meaning of this or how to interpret that yet. You'll learn about that in Chem 202. Okay, so everything we know about proton transfers suggests that proton transfers want to be linear. And here's why this matters for you as someone who's going to push arrows in mechanisms. Very frequently you're going to run into cases like this where you have an intermediate that looks like this and you're going to want to somehow explain how this proton gets from here to over here. Maybe I'll draw a methyl group here so these oxygens look a little bit different. And somehow or another you have to explain how that proton's going to get to the other oxygen, otherwise you can't finish your mechanism. And you will be tempted beyond belief to do this. Oh, look at this. Look how easy it's right there next door. Look how close that is. It's like two angstroms away. That's not the mechanism. Based on everything we know about proton transfers and tetrahedral intermediates, the mechanism is two steps. In other words, if you're under basic conditions, draw B to represent base. If you're under acidic conditions like a fissure stratification, draw A minus to represent the conjugate base of an acid. And then pick up the proton first to generate an intermediate. In other words, if we're trying to explain how that proton moves over to here, don't do it intramolecularly. That's not the way it works. It involves some initial proton transfer that's very fast. Proton transfers like this are fast. Here's my B and here's my H. I need to draw that. And now look at that. I've drawn it right next to this O minus. I can take that right off. It's never faster to transfer that under realistic organic chemistry conditions. It is never faster to reach over and have this lone pair somehow magically end up with a linear trajectory. There's no way that that O minus spends time over here. The only way you're going to get that proton over there is if you draw a transition state where it's got a 90-degree angle between oxygen, proton, and oxygen. And I can't, I guess I could think of something less ideal but that is so non-ideal. It will be faster for some basic molecule to pluck off that proton and re-deliver it. So that's how I want you to draw proton transfers and tetrahedral intermediates through intermolecular two-step mechanisms. Okay, so why do I think this should be so fast? Why do I believe that that ought to be such a fast reaction? Faster than that four-membered ring transition state where I just reach over and grab that. I think that ought to be lousy. And in contrast, I think it ought to be fast to transfer protons intermolecularly. I want to draw out a table for you where we ponder on this simple equilibrium as I vary the acid. I want to look at the transfer of protons. This is data that was collected by Eigen. It was pretty insightful. This is meant to be a rate constant. Here's a rate constant for the backward reaction. So I start off with two things. I end up with two things. And I'm just transferring a proton back and forth between water and this A minus species. And what I'll remind you of is that the equilibrium constant for this reaction, when you know the rate constants is equal to the forward rate constant divided by the reverse rate constant. You might have interpreted equilibrium constants in terms of concentrations, but there's a kinetic view of equilibrium constants. And you should recognize this equation. This is the acid ionization equilibrium equation. This equation that I just drew up here is the basis for pKa's. So let me go ahead and draw out a table here where I consider some proton transfers back and forth to water for various different acidic species. So let me make a table here. And I'll start off by drawing the structure of the acid. And then we'll write down the pKa. And then we'll write down some rate constants that Eigen measured. Here's acetic acid, hydrogen sulfide. There's two protons attached to sulfur. I'll draw ethyl acetoacetate. You may have learned in your sophomore organic chemistry courses that those CHs between the two carbonyls are very acidic. I'll draw out an ammonium ion. I'll draw the bond there. And then finally I'll draw nitromethane. Nitro groups strongly stabilize carbannions next door. If you've ever made methamphetamine from nitroethane, you took advantage of that. I hope you didn't do that, but whatever. And I'm going to draw the pKa's here for these species. I could look these up in any pKa table. And I expect you to know these pKa's or approximately that. We'll talk about these more later. I'll expect you to memorize some pKa's. And what I want to do is I want to tabulate these forward and reverse rate constants for these different proton transfers. So in other words, up here in my equation, if I replace HA with hydrofluoric acid, and I think about the transfer of a proton from HF where water comes along and acts as a base and plucks off that proton, what I would find is the rate constant for water plucking off this proton from hydrofluoric acid is about 10 to the 8th. And these are in per molar per second. I'm not going to write the units in every case. 100,000 times per molar per second. If everything was at present at one molar concentration, it would happen at 100,000 times a second. That's how fast water would pluck a proton off of that HF. If I look at acetic acid, which of these two is more acidic? HF or acetic acid? HF is more acidic. Acetic acid is less acidic. Well, that's kind of comforting. Acetic acid is less acidic and it transfers its proton more slowly to water. If we look at hydrogen sulfide, that's even less acidic than acetic acid. And it's kind of nice to see that it transfers its proton more slowly to water. It's kind of comforting to see that. I come back over here to this ethyl acetoacetate and it's 10 to the minus 3. That's way slower. So I can see ethyl acetoacetate, just looking at the pKa's, I can see that ethyl acetoacetate is less acidic and it's now much slower. Now I look at ammonia and it's actually 25 but I'll just write 10 to the 1 to keep everything in base 10. Well, that doesn't make so much sense when I'm something, right, now it's getting faster again. Now I'll come back over here and I'll look at nitromethane and now it's really slow, 10 to the minus 8. That is a very slow reaction. It's not easy for water to deprotonate nitromethane. We need stronger bases for that. So what I want you to note is it looks like for some of these, there's this kind of correlation between acidity and the rate at which it donates a proton to things. Now I want to look at the reverse rate constants. How quickly does a hydronium ion put a proton back on there? And what you see is the rate constant here is about 10 to the 11. This looks faster than diffusion control. Usually diffusion control reactions where things collide max out at about 10 to the 9 but these were run in water so there's an advantage to transferring protons to water. That's a very fast second order rate constant and it's fast because water is the solvent. I look at acetic acid, water plucks the proton, water, sorry, hydronium ion delivers a proton to acetate very quickly. I look over here at HS, it's very quick. These are all diffusion controlled. It can't get any faster than that. Every single collision between hydronium ion and fluoride minus results in bond formation. Every single collision between hydronium and acetate minus results in OH bond formation. So now I come back down to this ethyl acetoacetate and well that's not diffusion controlled. Something is weird here. Let's look at ammonium ion. Well that's about diffusion controlled. And then I'll come down to nitromethane and it's kind of troubling me. There are two outliers here. What are the outliers that don't seem to make sense? It's kind of obvious to see here. Hydronium re-delivers a proton back to the anion at diffusion control rates except for this case. And these were also the cases where I suddenly my monotonic decrease that seemed to match pKa didn't work so well. So this is kind of an exception. There's some weird thing going on here and there's some weird thing going on right here with these two. How are these two different from the other two cases? Or the rest of the cases in the table? Yes. We're transferring a proton from a carbon atom. The general rule is we can construct a general rule except for these two cases so let's construct a general rule. I don't think I have enough room there so let me come over here and use this white board. The general rule is the proton transfers to and from carbon are generally fast. Both two, sorry, two and from heteroatoms are generally fast. Oxygen, nitrogen, fluoride, sulfur, those types of proton transfers are fast. Heteroatoms, proton transfers two and from heteroatoms are fast. And in a lot of these cases they're diffusion controlled. That's why I never worry about using or drawing out proton transfers to and from oxygen, for example, up above. That's why I'm so confident that it's fast through a two-step mechanism because proton transfers tend to be fast. So don't worry if you end up drawing out some mechanism that has many, many steps because of proton transfers because those are fast. It's not the number of steps in your mechanism that makes a mechanism correct. You can have very long, completely valid mechanisms as long as you have lots of proton transfers in there. Don't worry about it. If you have a long mechanism that doesn't involve proton transfers maybe there was a shorter mechanism and you didn't see it. But never be afraid to draw out proton transfer steps. They're generally fast. The exception is proton transfers to and from carbon. Proton transfers to and from carbon are usually slow. So I come back over here and I look at the pKa's for ammonium and this carbon acid right here. They have about the same pKa. Same pKa but you look proton transfers to and from this carbon atom are about a 10,000 times slower than from nitrogen. So it's not the pKa that matters. It's the fact that it's bound to carbon. And why is it so slow to transfer protons to and from carbon? It's because those involve hydrogen bonds that suck. Hydrogen bonds like this are not really good. Hydrogen bonds are mainly calomic and there is not a lot of positive charge on the protons that are attached to carbon. That's what Manfred Eigen showed. He showed that hydrogen bonding was an essential step in proton transfers and you don't get good hydrogen bonds when the H is bound to carbon. Okay, we're going to stop right there and talk more about addition to sigma star orbitals attached to protons and then when we come we'll also be able to start talking about SN2 reactions where you form bonds to carbon.