 Okay, so great job on the exam. It was very hard, I get that. Chris suggested to me it might be a little bit hard, but I wouldn't with it anyways, so I want to blame Chris for everything, but I'll take the heat on that. So you guys did great on a hard exam. We're continuing on on lecture 9, Addition to Sigma Star, and really there's two reactions that we're thinking about when we talk about addition to Sigma Star orbitals. One is proton transfers, where you attack a proton. The other is SN2 reactions where we attack carbon, and in each case the leaving group leaves simultaneously. It's a concerted displacement reaction. And we've spent the first part of lecture 9 talking about proton transfers. We're almost done, and I want to finish up our discussion of proton transfers by talking about equilibria. Whenever you draw out some long drawn out mechanism for an esterification reaction or an immune hydrolysis, anything, you tend to have lots of proton transfers in those reactions, and I want to encourage you every single time there's a proton transfer step in your mechanism to try to estimate the equilibrium constant for that reaction. And I'm going to give you one example. There's nothing special about this reaction. It's acid-catalyzed removal of a THP protecting group. If you were to draw out the mechanism for this reaction, really all of the business here is occurring at this pyrann, and I'll draw out just one of the many proton transfer steps in this mechanism. This PY stands for pyridinium. That's PPTS is pyridinium peritoluene sulfonate. It's really just toluene sulfonic acid mixed with pyridine so that pyridinium is the real acid in this reaction. And so let's draw out this reaction where pyridinium is acting as an acid and somehow reacting with this tetrahydro pyranol. This is a THP protecting group. And let's think about what's the first step in this process. And how much interesting thing, how much interesting, how interesting can we get with this step? So if I draw out the arrow pushing for this step, I'll end up with a transfer of a proton to give an oxonium ion that's oxygen with three bonds and a positive charge on there. We can say a lot about this. How likely is that? Is that a good mechanism? Why didn't I protonate this oxygen? Is that occurring under the conditions of the reaction? So let's go ahead and try to estimate the equilibrium constant for this step and every step where there's some kind of a proton transfer. And what power do we have? The power that we have is pKa's. We don't know. There aren't tables full of rate constants there. You just can't find that kind of information out there. But what we have at our disposal are lots and lots of tables of pKa's. It's very easy to do a titration in water. And so let's use that information. The pKa for a pyridinium ion, we talked about the pKa for an aminium ion before where you have a protein on an sp2 hybridized nitrogen atom. That's about 100,000 times more acidic than an ammonium ion for something like triethylamine. And the pKa for this is about 5. That's for any sp2 hybridized nitrogen with a proton on it. If we come over here to this protonated oxonium ion, that's kind of like a protonated ether, protonated alcohol. The pKa for this particular example is pretty close to the pKa for a protonated ether. So I'm going to use a pKa for minus 4. So this is not exactly a regular ether, but close enough for our purposes. As long as I know the pKa's of the two acids on each side of this equation, I can accurately estimate the equilibrium constant of this reaction. I told you that proton transfer reactions are fast. Probably under the conditions of the reaction, no matter what this does, this is probably an equilibrium because proton transfers are much faster than the reactions that break carbon-oxygen bonds, carbon-carbon bonds and things like that. So let's assume that this is an equilibrium and our picture for this reaction is that we're generating some amount of this and then this sits around and slowly does some dissociation step. Okay, so what's this super secret equation that we're going to use here? Our super secret equation, and I'm going to draw it in another pen color, is this. Keq for this step is equal to 10 to the, and it's going to look kind of sloppy and confusing here, but I'm going to write pKa for product minus pKa for reactant. Now I'm pretty bad with math, but even I can handle this equation, right? So let's go through and do the math here. So the equilibrium constant for the forward-looking process here is like this. Keq is equal to 10 to the products minus 4 minus the pK for the reactants, which is 5, which is plus 5, which is equal to 10 to the minus 9. I immediately have some deep insight into what's going on in this reaction. When you mix peridinium, peritoline sulfonate with some sort of an ether, what you can expect is that one out of a billion molecules is protonated. By choosing to use PPTS, you've set up conditions where the ratio of deprotonated to protonated is 1 to 10 to the 9, you know, that's enough zero. So the ratio of these two species, and again, it's ratios that are intuitive to you when you do reaction mechanisms. For every billion molecules, only one of them at any point in time is protonated. And it's this that's sitting around waiting to slowly dissociate that. If you want this reaction to be faster, use a stronger acid. You'll have a higher fraction of these molecules protonated at any point in time. Yes? You need to know some. I'll give you some later that you have to memorize. I'll just give you a set and I'll say memorize these. So we're getting, that'll be in a later lecture. Okay. So, okay, what's the caveat? There's lots and lots of PKA's that have been measured out there because it's so easy to do a titration. They're almost all aqueous PKA's. So there are going to be some differences when you take into account, well, this is not exactly water, methanol is not exactly water, and especially in organic solvents. Feel free to use your aqueous PKA's anyways. It's better than nothing. It'll still give you insight. And unless I, I can't imagine asking you some question where it'll make a difference that you did it in an organic solvent. I'd rather you use an aqueous PKA than not to try to think in these terms. I'd rather you thought with, with these kinds of numerical terms using aqueous PKA's. How does it change? Depends. In this case I'm going from neutral to charged but you can imagine other cases where you go from negatively charged to neutral and that's what has an impact. In other words are you going from neutral things to charged things or charged things to neutral things? That's what affects these things. Okay. So let me step back. We're done with proton transfers. I said in this lecture we're going to talk about two different types of addition to sigma star. In the case of proton transfers we're talking about addition to sigma star for some sort of hetero-atom proton bond. And so I would draw it like this and I could take exactly the same nucleophile, exactly the same arrow pushing and attack a carbon atom and now it's not the same orbital anymore. Now I give it a different name. I call this an SN2 reaction as opposed to a proton transfer. Okay. So how are these different? What I mentioned to you before is it is fundamentally different to displace, to do a displacement reaction on proton because that's a first row atom. So remember our periodic table, H is over here in one corner of the periodic table in the first row. And then over here in the second row we've got carbon. So carbon is more massive. A methyl group is more massive than a proton by a factor of 15. If it's 15 times more massive it vibrates more slowly back and forth, you can look in an IR and confirm that. So immediately you should suspect that additions to carbon are going to be, that kind of displacement at carbon is going to be slower, significantly slower. But there's a difference here in what's driving these reactions. In proton transfers remember the actual mechanism involves formation of a hydrogen bond. So even though both of these types of reactions are governed by this fundamental equation that we talked about on day one, it turns out that proton transfers are mainly governed by charge. This is mainly a charge interaction. That's why proton transfers are so fast. There's very little of this filled MO, unfilled MO business powering this and making it fast. It's the fact that bases tend to have negative charge. Protons have a lot of positive charge and things that are acidic. Now why are SN2 reactions going to be fast? It turns out that it's not the charge that's making this, it's contributing for the reason why hydroxide can attack here. But I could take neutral thioethers and they'll still attack fast here. In most SN2 reactions, this is the dominant species. It's controlling that. So the arrow pushing looks very similar but different features for proton transfers versus SN2. Okay, so let's talk more about SN2 reactions. And those are the things that will form carbon-carbon bonds. So we have to be interested in SN2. But as you can tell from discussion section and maybe just generally talking with me, I hate SN2. I hate it. Because after one year of sophomore organic chemistry, it's the first reaction you learn. You learn it throughout the entire year. You use it throughout the entire year and you overuse it. You use it in every single case where you shouldn't ought to be using that. And so I'm going to do everything I can to dissuade you from overusing SN2 reactions. Okay, I want to start off by looking at the transition state for an SN2 reaction just in a very generic sense here. So if I draw the three things attached to the carbon that's undergoing displacement, so I'm trying to draw this edge on very three-dimensionally. And so you can imagine me displacing something like bromide over there or iodide. And in that transition state for an SN2 reaction, we have the nucleophile coming in and simultaneously displacing the leaving group. As this bond is getting shorter or as this distance is getting shorter, this distance is getting longer. Now in that transition state, if you look at the distribution of charges, it looks something like this, the important point is the carbon is starting to pick up positive charge in the transition state. As this carbon, as this X group is leaving with those electrons, even as the nucleophiles coming in, you start to build up positive charge. There is a transfer of charge from these R groups to this partially positive charge carbon in the transition state. And we'll see the impact of that in a second. So I'll just write this down. Negative charge, if these R groups are doing their job, negative charge is being transferred from the alkyl groups on the periphery. We would call these equatorial. We would call these up and down ligands on a trigonal bipyramidal structure. We would call those axial. Okay, let's, the implication of this idealized transition state is that electroneg, and this is true of any trigonal bipyramidal structure. It can be an organometallic species. It can be silicon. It can be phosphorus. Or it can be carbon. And it can be a transition state. Anything that is trigonal bipyramidal, electronegative substituents, prefer to be equatorial. Whoa, what did I say? That doesn't sound right. Axial. I'm immediately starting to freak out here. Prefer to be axial. Alkoxy groups, they prefer to be axial. Of course, that makes sense. Alkoxy tends to be something like a nucleophile. Electropositive groups, I don't have an abbreviation for this while I'll write out electropositive. So if you want to have fast SN2 reactions, of course, you want to have electronegative leading groups. Chloride, bromide, iodide, acetate, oxonium ions. You want to have electronegative nucleophiles, hydroxide, thiolate. And if you really want it to be fast, you want to have R groups attached to that center that's being substituted, excuse me while I tie my shoe. You want to have R groups attached to that carbon atom that's being substituted to be electropositive in nature. Let's see the impact of that on substitution rates. So if you look at the rates of SN2 substitution on various chloromethanes, what would you find is you increase the number of chlorines attached to that carbon that's undergoing substitution if I compare these. These all have just one carbon atom. And so what would I see if I compared the rates of these SN2 substitution reactions? What you'd find is that the fewer carbon substituents or the fewer chlorine substituents you have on there, the faster it is. This is faster than two chlorines. Dichloromethane is slower to do SN2 reactions on. That's convenient because you often use it as a solvent. When you do some SN2 reactions, I wouldn't but it's at least slower than chloromethane and that's not supposed to be arrow. I apologize for that. It's supposed to be greater than. We're talking about rates. Yeah. So and this is faster to substitute than chloroform. Chloroform is a better solvent for SN2 than, well, I mean it's also a little bit acidic on that proton but in general it's slower to do SN2 reactions on chloroform. That would be very rare for you to see somebody doing an SN2 reaction on chloroform. And carbon tetrachloride, very slow to substitute. And it all has to do with this preference for you not to have electronegative substituents attached to the carbon. And you might argue, oh, that's a steric effect. You don't know what you're talking about. That's just sterics. So let's take a look at a different example. I'll just take a look at propyl, well, butyl chloride. That's what I have numbers for versus this ethoxy. Now here, it's hard to argue that sterically, that this is more sterically encumbered. Oxygen has less sterics on it than the CH2 over there. And yet this one is five times slower. It's five times slower than a comparable SN2 reaction of just plain butyl chloride. That's not sterics. This is the effect of the fact that you've got this oxygen. And the oxygen has this polar effect. You could call it an inductive effect on that carbon that's undergoing substitution. There's no conjugation going on here. The oxygen is not directly attached to the carbon that's involved. But it's an important effect to keep in mind. Now it's a factor of five. Maybe that's not very big, but it's still an effect. Okay, so let me keep talking about this trigonal bipyramidal structure. Because this feature of trigonal bipyramidal structures is not just SN2 transitions, it's intermediates, anything with that trigonal bipyramidal structure. Let's take a look at something else that has trigonal bipyramidal structure. So let's take a look at some sort of a molecule like ammonia. And of course, this is not interesting with protons on there, but imagine having three different alkyl groups. And now this is a chiral center. If I have three different alkyl groups on here. But you'll have a hard time isolating one and antimirror versus the other because it's very, very fast for a means to invert. And if I draw out what the transition state for that inversion looks like, it looks trigonal bipyramidal like this. And in that transition state where everything is planar, as it flips up like an umbrella, in this transition state, you're going to have the electrons in this p orbital. If the H's are planar, there must be a leftover p orbital here. I'm not going to phase the orbital. And then finally, this thing inverts so that the lone pair is now at the bottom. So very, very fast for a means, this process. And if instead of H you have alkyl groups, you, those are not, those centers are not considered stereogenic because this inversion is so fast. Okay, what happens if I replace nitrogen with other atoms? And I'll give you the inversion barriers for these as I replace nitrogen with other atoms in the periodic table. So what am I giving you? I'm giving you inversion barriers. And you can compare this by substituting nitrogen and ammonia for these other types of ligands. So what happens if instead of nitrogen, I have a carbanion? Let's just create ourselves a mini periodic table here. So if I have a carbanion with a lone pair here, how fast do carbanions invert like ammonia? If you look at, and I'm giving you some calculated numbers, so what's important is kind of not the exact number here but the sort of, the relative numbers are important. So if I look at carbon minus, 13 k-cal per mole inversion barrier is actually very fast. That means it's fast at room temperature. And indeed, if you try to make chiral carbanions where that's a C minus there, you will have a very hard time trying to work with and isolate those before they invert. Good luck. Put nitrogen there. It's a little slower but still fast at room temperature. Just slower than carbon, I guess is all I can say. And finally, if it's an alkoxide, well, that's not too meaningful because there's two lone pairs but at least we have a number for that and it's 1.7. Okay, so the important point here is not the differences between all of these. What's important is we drop to the row below this and go to third row atoms. Watch what happens to the barriers. That's supposed to be oxygen plus, by the way. Phosphorus, and I don't know why I'm putting 0.7. If you wrote 35 and 38, you get the big picture here. I'm regretting having done that. Okay, there's a huge jump going from second row atoms to third row atoms. This inversion type process that we saw for ammonia, silicon, anions, they just don't invert. If you somehow had a way to make a silicon anion with three substituents on there, you should expect that to retain its stereochemistry. Even at elevated temperature, you'd have to heat that very hot to get a silicon anion to invert like this and lose stereochemistry and to racemize. Phosphines, you can buy chiral phosphines where the phosphorus atom is the chiral center. They don't do this. This kind of an inversion barrier means you have to heat it to get it to invert and you'd have to heat it very hot. Sulfur plus, again, that's not really that relevant as something you'd have to worry about inverting, but you can see that everything in this third row is now slow to invert. So it's a general principle. The carb anions are fast to invert. Well, these numbers came from calculations. They calculated the energy of the transition state and subtracted the energy of the grad state. K-cals per mole, sorry. Yeah, everything, every number I give you in this, I should have written that down, but every number I give you is K-cals per mole. So there are cases where you will have seen that people have made and worked with configurationally stable alkyl lithiums. For all intents and purposes, for the purposes of inversion, you can think about the transition states for inversion of this, alkyl lithium is being similar to the carb anion. Your prediction should be that if you put electronegative substituents on here, on this carbon atom, it should invert more slowly. If you put electronegative substituents on that carbon atom, it should be harder to gain access to that trigonal bipyramidal transition state. That's why whenever you see people work with configurationally stable, and let me make this configurationally interesting by having three different groups on there, there always will be some sort of a heteroatom attached to the carbon, an electronegative heteroatom. That's when you can start to slow down this inversion process. Let me draw the back end of that orbital down there at the bottom. That's when you start to slow down this inversion enough to where you can actually work with those kinds of carb anions before they racemize at that center. It's when you have an electronegative heteroatom. So in this particular case, the transition state energy for inversion of carb anions like this is about 23 kcals per mole. And that's right at the point where if you keep it cool, minus 78, you can work with those configurationally stable. Oxygen substituted alkyl lithiums before they invert. Okay, so those are all properties of trigonal bipyramidal transition states. Okay, so let's drop down to that third row, silicon and phosphorus, and let's talk about the substitution. So oftentimes you'll see stuff like this. We're going to see a lot more of this later on in the course, maybe Mukayama-Aldol reactions where you have o-silo groups, protection of molecules with TBS chloride, when you're going to want to substitute that silicon off of there. Whenever you draw substitution reactions at silicon, they are never SN2. It's never SN2. It's slow to achieve these trigonal bipyramidal transition states. Trigonal bipyramidal sucks for these third row atoms. So the transition state to get closer to that is just not good. It's better to have an intermediate. So if I draw out the correct mechanism for substitution of that silicon to pop off the carbonyl, what I want to do is do this and then stop drawing. Don't pop off the leaving group as we're attacking. This silicon, I mean it's not great, but it's perfectly happy here with these five things attached with silicon, and maybe that's going to bother you a lot. And you won't want to do it, even though I'm telling you, don't do SN2, you're going to have a tendency to want to pop that leaving group out, don't pop it out. You can have six bonds to silicon. There's no octet rule for silicon, that's the second row. You can make stable structures, silicon 8s. You'd have six bonds on silicon. There's nothing wrong with these five bonds here. Now, after you do this, now you can break that silicon oxygen bond to make the carbonyl leave. So again, even though I'm telling you this and it sounds like a simple rule, I'm going to catch you every time you try to do SN2 at silicon and say, hey, we said don't do SN2 concerted reactions at silicon. Okay, so the important point is it's the intermediate that really is trigonal bipyramidal. It's not the transition state. In the transition state, you're getting closer and closer to trigonal bipyramidal. Okay, so let's, anytime you see some sort of pentavalent silicon, let me just say something about the language I'm using. Let me pull this back down for a second. You might have heard me refer to this as a silicon 8. I had that ending 8 on there. Every time you hear me add 8, pallidate, cuprate, silicon 8, hydridoborate, that 8 tells you there's a negative charge on that atom. And whenever there's a negative charge on that atom, every single bond becomes nucleophilic. So I'm usually really particular about this kind of language because it's informing you of something. It's telling you something about the reactivity. So listen for that. When you hear me refer to something as a cuprate, pallidate, boronate, hydridoborate. Okay, because these intermediates in substitution processes resemble trigonal bipyramidal transition states, all those same rules apply. If I were to show you this silicon 8, this is the preferred geometry for this. So notice in this intermediate, I've got acetoxy displacing from the back side in a two-step mechanism. Notice how careful I was to draw a trigonal bipyramidal intermediate there in the silicon substitution. When you look at stable silicon molecules or intermediates in processes, the electronegative substituents want to be axial in that intermediate. The electropositive groups like R groups or H want to be equatorial. So in general, when you substituted silicon, nucleophile tends to attack opposite from the side of the electronegative leaving group. So when you substitute with TMS chloride and you've got some chloride leaving group, the alkoxy group is going to come in from the opposite side of the electronegative leaving group, that's the preferred geometry. If you tried to come in from the side, you're going to have to pay a price, a stereoelectronic price for that if you somehow had some reaction where you wanted to come in from the side and not come in from the backside. It's not impossible, but it costs you. It's not as favorable as coming in from exactly the opposite of this bond that you're breaking. That's favored coming from the backside. So just to put some numbers on that, if I compare the energies of this versus this, where I have one of the fluorine's equatorial now and not both of the maxial, this structure is 8.3 k-cals per mole higher in energy to have an electronegative leaving group on the side. How many times more stable is it when the electronegative groups are axial? So we can convert k-cals per mole into factors of 10, meaning you can, I'm not smart enough to divide that by 1.4. A millionish, I'll take that. So it's a millionish times more stable. What that means is if you draw some sort of reaction where this things comes in from the side rather than opposite to the leaving group, it's like a millionish times slower as a penalty. Now, of course, this can't reach around. That has an entropic advantage, but you should worry if you've got your alkoxy group attacking silo chloride from the side, because that's a million times slower than if it could have attacked from opposite to the leaving group. Okay, so let's translate this into biology. This is essentially a model for DNA. It can be alkoxy groups there. So if I had this sort of a, this sort of, this would be a phosphonate ester, and I displace this or hydrolyze this with alkoxide. And many, many enzymes do phosphoryl transfer. Polymerases, ligases, anything that transfers DNA, kinases, phosphatases, they all involve some sort of oxygen atom attacking a phosphate ester. I want to start off by drawing the resonance structure for this so that we can decide what's the preferred trajectory of attack. Hydroxide can add from four different types of directions here in order to displace that alkoxy group. But according to our rule, what's the most electronegative group that's attached to this phosphorus atom? I've got two methyls, those are equal. I've got O minus, and then I've got neutral oxygen. What's the most electronegative group attached to this? No, not O minus. Because O minus is negative. It's got plenty of electron density. The neutral O. It's kind of convenient. That's what we're going to displace. And if you look at the idealized transition state for this, the ideal transition state would be to have, I'll draw it as a solid arrow, will be to have the hydroxide attack anti to the ethoxy group. That's the preferred trajectory of attack. It's not to attack opposite the P double bond O. It's to attack opposite the leaving group. And this goes through an intermediate that we call a phosphorane. And then in a second step, you pop out the leaving group. It's not concerted. And if you wanted, you could have drawn out this P double bond O resonance structure the whole way through. Every single enzyme that has ever been studied, kinases, ligases, phosphatases, polymerases, every single one obeys this rule of stereoelectronics. No enzyme has ever been found in which the nucleophile does not attack anti to the bond that's leaving. The bond that's, the group that's leaving always has some sort of a hydrogen bond to, I won't draw it to poison the structure, but it always has some hydrogen bond to it that further increases the electronegativity of that group. So nature understands this simple rule and applies it every single time it designs an enzyme. Okay, so now I'm going to go back to something that I hope will be well known to you and very obvious, but it's not so obvious that I can skip it. I'm going to give you a set of four numbers here for, these are approximate relative rates for nucleophilic displacement of alkyl halides. I'll start off with alkyl, with ethyl halide. I'll compare that to this, which now has a second, an extra methyl group on there. I'll put a third methyl group on there. And then finally, I'm going to make a, draw out another primary alkyl halide here. And I'll give you the relative rates for SN2 reactions. So this should just be clear. We're talking about SN2 rates. This is the fastest one of the bunch, the primary alkyl halide. I hope you already knew that primary alkyl halides are faster for SN2 than substituted alkyl halides. This has one extra methyl group at what we refer to as the alpha position. So we refer to this as the alpha position relative to X, and we refer to that as the beta position. The more things you add to the alpha position, the slower it gets to do SN2. It becomes more slow. I'll qualify that in just a moment. And so it's six times slower when we add that one methyl group on there. And if we compare this to T-butyl, now it's on the order of 100 times slower than these other two. It's very slow to do SN2 substitutions at tertiary centers like this. But finally, when we get to this system, which is still primary, right? The first one is primary. This is primary. This has beta branching. The methyl groups are at the beta position. And unfortunately, there is no way that a nucleophile can come in and attack from the backside without bumping into one of those methyl groups. There is no possibility for that to avoid the bumping. And so now we're looking at an SN2 rate that's about 100 times slower than T-butyl. So we call this a neopental center. This is called neopentane or a neopental halide. So neopental are the slowest SN2 reactions based on sterics. So that kind of substitution pattern where you have beta branching, three beta substituents next to a leaving group. Those are the worst of all possible SN2 reactions. Don't propose stuff like that. Don't design syntheses based on those. It'll kill you if you try to do that. Not as much as this. Well, I mean, there's still an orbital that's back here. It's, you know, if I came in from the side here, then maybe it would be worse. But we're coming in from the back end. My line drawing doesn't do it justice. Maybe that's why it's good for us to memorize the numbers. Yeah. If you have an electronegative substituent beta on any of these, it'll be slower. It'll be even more slow. But you're asking in terms of relative numbers. I'd have to go back and look at that alkoxy ethyl group. It was like a factor of five slower. So imagine if I put an alkoxy group here, maybe five times slower than that. I'm just estimated. Yeah. Wait, say that again? If I fuse this to something to have like a chain like this so that it was leaving, is this? It looks to me, I'd have to look at that. I would expect that to still be slow, still be a problem. But would it still, would it be this much slower? Maybe not. In other words, there could be ways you could torque it away to make a little extra room. So I wouldn't know the numbers for that. Okay. So let's talk about three-membered rings. Because this will turn this on its head. It's like you think you know everything once you know these rules I do. I'd be like, oh, I got sterics, slower, tertiary sucks, never substitute tertiary centers. Of course we know that. I hope everybody knows that. So what I want to do is I want to draw some generalizations for you. Whoops. And here's what I want to draw. So I want you to contrast these two species. Just a simple epoxide where we're opening with alkoxides. So if you take a sodium alkoxide and you open up an epoxide, you'll get addition, there's nothing surprising about this. You get preferred addition to the primary center. But if you do that kind of ring opening under acidic conditions, there's no alkoxide roaming around when you've got sulfuric acid in there. That's not what makes the reaction fast. If you do reactions with catalytic acid, it's the neutral alcohol that's attacking. And it's not the epoxide that's opening up, it's the oxonium ion, the protonated epoxide. And in these cases, preferred attack happens at the tertiary center every time. And this is a general principle. If you have some kind of rule in mind that attack at tertiary centers is always slow, that's wrong. Here's the exception. And it's general. So it doesn't matter whether you're looking at aziridinium ions, oxonium ions. It doesn't matter whether that's a H or whatever. Mercurinium ions, we don't teach oxymercuration reactions anymore, but we used to. It's, your prediction ought to be whether it's metal, oxonium, aziridinium, that the nucleophile always attacks at the more substituted center. Focke-roxidation, if you know that reaction, it's all the same. More substituted centers get attacked more quickly. The fallacy is that if you draw it where these bonds are equal in size, then it's not so obvious what's going on. This bond's already partially broken. It's already a longer bond in the intermediate. I just didn't draw it that way. So if you want to make it easier to understand, you could draw these kinds of things like this where you have a tertiary center. And here's this prebroken bond with this leaving group just ready to leave. I mean, maybe that'll help you understand why it's faster to add to the tertiary center. But this, you have to be very careful with these three-membered ring-onium ions. As far as I know, that's not true for four-membered rings, five-membered rings, six, just three-membered rings where this is true. So let me give you another series of compounds and we'll look at the relative rates for SN2 reactions. So here's another butyl chloride. In this case, it's a Finkelstein reaction where you exchange chloride for iodide. So you use these by sodium iodide and acetone. And these are always SN2. Well, for most typical substrates that you would use. So let me go ahead and change the structure of this alkyl chloride here and we'll look at the rates for this SN2 substitution process. And your tendency may be to want to take your SN1 thinking and apply it to this. And just remember, this is SN2 we're talking about here. So I'm going to assign this a relative rate of 1 and we'll compare these other substrates in terms of how fast iodide displaces chloride. If you look at this alkyl chloride here, sterically it's pretty similar to the butyl. I just didn't have the number for propyl. So sorry about that. Propyl and butyl will be pretty similar. It's 33 times faster. So in other words, when you have pi systems nearby, allyl, benzyl, not only are they faster for SN1, they are also faster for SN2. It's faster for everything. Substitute's faster than that. What's surprising and maybe not so obvious is when we go to this, that would definitely be slower for SN1, having an electronegative cyanogroup there. You don't want to make a cation next to this electron acceptor. But if you're doing SN2 reactions, that's a benefit. 3,000 times faster for SN2. If you have to do SN2 reactions, make sure there's a pi system next door. 35,000 times faster. I wish my research could go 35,000 times faster. And I'm telling you how to do that if you're doing SN2 reactions. Okay, I want you to contrast the transition states for SN2 and SN1 reactions. And let's start off by thinking about SN1. We've already had a lot of practice with that. And so I want you to think about this process where you simply ionize that. And what's happening in the transition state for this? I want to try to draw a transition state and I'm going to break these down into some really fundamental orbitals here. So in my transition state, I'm going to try to make this almost pyramidal, not yet. It's a very late transition state. This X is starting to get farther and farther away. It's not quite planar yet. It's trying to get planar. And here's this pi system next door, this double bond. So here, I'll try to draw that sort of bending backwards a little bit. And I'm going to try to sketch out some orbitals here. In the transition state, this is starting to look almost like a P orbital. It's not quite pure P, but the late transition state for an SN1 reaction, it almost looks like a carbocation. And meanwhile, this X group is leaving and it's kind of overlapping with its orbital with this empty P orbital as it's leaving and marching away. So the transition state in terms of if I were to decompose it into these orbitals, it would look something like this. So the bottom line is that with an SN1 reaction, with that empty P orbital there, and this is a general property of SN1 transition states, there's kind of not enough electrons. That's a sad, sad carbon atom that has an empty P orbital. Not enough electrons as we get closer and closer to this carbocation intermediate in the transition state. And so maybe you can tell why it might be happier. If right next door, we had a filled, I'm trying to make sure I phase this correctly. So here's that pi double bond. Look how great it is that we've got this filled orbital right next door to help donate electron density into that sad, sad, carbocation type species that's starting to develop. So that's why SN1 is faster when you have that little bond next door. Now I want to contrast this situation. I hope I have enough room here. With a carbonyl next door. And not an SN1, I want you to think about an SN2 reaction where a nucleophile is displacing the leaving group, pushing it out like this. And let's draw out that transition state. Now the transition state is total, not just close to planar, but is exactly planar. Pretty close to exactly. Depends on which of these is whatever. It's close to planar. Okay, so let me go ahead and draw this out in that transition state. It's different now. I've still got this sort of p orbital shape here. If it's planar, if this carbon is planar, there has to be a p orbital or an MO, a molecular orbital in the transition state that looks like a p orbital. I've got this leaving group leaving with its pair of electrons, that's a filled orbital. I've got this nucleophile with its pair of electrons, that's a filled orbital. We've got all these orbitals that are filled now. Leaving group, nucleophile. The problem here is too many electrons. There's too many electrons in this transition state. With the nucleophile coming in with all its electrons, there were already enough electrons in here to have covalent bonds and now you're coming in with your extra electrons, that's the problem. And so now we have a very different solution. Now having not just a pi bond here, but a carbonyl here is a better acceptor. Now we have pi star orbitals here that are very low in energy and good at donating and let me phase this and I'll phase this so we can see that as we start to build up too much electron density and we start to fill this orbital, it gives us somewhere to go. We've got somewhere to donate the extra electron density. So SN1 reactions, not enough electrons. SN2, we're looking for a donor to really help us. And that's why carbonyl is better than a simple C-C pi bond. Say again, yeah this, sorry I didn't label this, this is pi star now because the extra electrons, as the electrons fill this orbital, now that can donate into this anti-bonding orbital for CO. So you're giving the electrons somewhere to go. Okay, I'm going to have to stop there. We're almost done with SN2, thank God. And we'll pick up with that when we come back on Monday.