 Okay, we're going to continue where we left off on Friday. We were talking about addition to pi star, so addition to carbonyls. If you're not totally up with this, go back to your sophomore organic chemistry text and work every mechanism problem in the chapters on carbonyl addition, esters, ketones, aldehydes, amids, and that'll help you sort of get back into the swing of things. When we left off, we were talking about stereoselective additions. And the idea that stereogenic centers that are next to carbonyl groups affect which phase you prefer to add to in a kinetic and irreversible reaction. And this is the basis for modern stereoselective synthesis of polyketide natural products. So we talked about this Felkin-On analysis. And I'm just going to make a little point here that this is On's name. What's On's last name? It's Nguyen. So, but he, in his first paper, he wrote it the Vietnamese traditional way. His last name is Nguyen, not On, but we still call it the Felkin-On analysis. Okay, so whenever you do some sort of addition of a nucleophile, in this case, it's an allyl boronate. We'll talk more about these so you can understand the mechanism later. This stereogenic center here with this poorly rendered methyl group that's hashed influences which phase you add to that carbonyl. And I showed you this Felkin-On analysis for how to predict which phase is preferred. And what I said is rule number one, when there's not a polar group, there's a carbon atom here. This is very big, but it's because it's just a carbon atom, you don't consider a polar. As you want the nucleophile to attack the opposite, from the opposite phase of the big group. So I've drawn out two renderings, two Newman projections for what I would see if I stood right here and sort of viewed down this carbon-carbon bond. That's the Newman projection. And all I have to do is imagine as I stand and look down this carbon-carbon bond is imagine what would I see as I looked down that bond. So there's a hydrogen atom sticking here. And if I orient it so that the big group is going this way, I'll have a hydrogen here. And that other carbon is back and then I'll have a methyl group sticking up. And if I swing this big group all the way around so that it's on your left-hand side, then I'd have the hydrogen group going up and the medium-sized methyl group sticking down. So those are the two reactive confirmations that would fit with the nucleophile adding from either your left-hand side or the right-hand side. And so remember one of the conclusions or one of our analytical rules for the Falconon analysis is that the nucleophile attacks from a Berge-Dunnitz trajectory opposite the big group. And the Berge-Dunnitz trajectory over here opposite the big group would look like this. And we want to choose the depiction. We want to choose the trajectory that minimizes eclipsing interactions. And that's this one right here, H is small. As we're coming in, forming a bond, we don't want to be bumping into some big group here. We want to be bumping into this. So I'll just write better and worse. And of course when you look at this, this is a paper that was published about a week ago or something. I typed in, if you'd like to practice, type in quote Falconon into the ACS Journal's website and then start working problems in papers. There's paper, 500 hits came up. You have lots of chances to practice, to be good at it. Because I promise you on one of our exams, I will ask you this, I expect you to be so good at this that in seminar when somebody mentions the word Falconon, you know the prediction ahead of time. And you can see it gets very hard when you start to twist around the carbon backbone and translate things. That's where all the work is. Okay, so that's the Falconon analysis. In this paper, they claim the sole product was the one predicted by the Falconon rules. Okay, so let's come back and look at something that is not Falconon, it is a powerful source of stereo control. So this is, if it's not Falconon, it is usually this alternative. If it's not Falconon, it's giving you high selectivity. There are cases when you get exactly the opposite of the Falconon prediction and in almost all cases it's because of this, because there was chelation that was affecting the reaction. And Cram was the first one to predict that this was possible and his model still holds, we often call it Cram chelation control. So chelation control gives the opposite result that you would predict from the Falconon analysis. Let me give you an example of a substrate that is well designed to give high levels of chelation control and in this particular case there was oxygen at the alpha position, that's the polar group and according to the Falconon analysis, this oxygen atom will bias those two faces that the nucleophile wants to attack opposite from the opposite side of the polar group which is oxygen. The polar group is not hydrogen, the polar group is not this alkyl group, the polar group is this oxygen. But if you tried to use the Falconon analysis to predict which face you would get addition from if you added lithium aluminum hydride and so in this case the nucleophile is a nucleophilic aluminum hydride bond, aluminate hydride bond. If you tried to predict using the Falconon analysis, you'd get the incorrect prediction. So let me go ahead and draw out the products here and I'll give you some ratios. This group right here is called a benzyloxymethyl, some often called bomb, benzyloxymethyl, so I'll just write bomb here. And so the hydride that comes in can either come in from the top face and I'll write R1 over here for this other group and this is supposed to be an R2 down here and here's the other possible product where the hydrido group adds in from the back face and those are the two possibilities. And so in this case, the one that's favored significantly and this gives you a sense for what levels of stereoselectivity you can expect to see is you get about a 9 to 95 ratio favoring addition from the back face. And so why is that? That's the opposite of what you would predict if you had gone through all this trouble to draw out these two Falconon Newman projections. What's happening is that this happens to be a good, whenever you have oxygen at the alpha position, this turns out to be very good for chelation control. In other words, lithium wants eight electrons and so if you have a lithium cation floating around in there it loves to have oxygen lone pairs poking into that helping it to satisfy the octet rule. And in this particular case, you would expect just any alkoxy group at the alpha position to be good. This is even better because that extra oxygen can actually reach around and bind to the lithium through a three point chelation model. What that does is it forces this oxygen not to be 90 degrees to the carbonyl but it forces this oxygen to be in the same plane as the carbonyl. And so that now creates a face and this R2 group is now on the top face of that ring system, of the chelation ring system. So the nucleophile comes in and attacks from the back face. Okay, so top face is blocked and so the nucleophile comes which is the hydride group. All right, back face of this chelation ring here. So the top face is blocked. That's chelation control and it's, you find this occurring whenever you have, I'll show you some of the types of systems that give good results. So generally what you find, so what are the good chelating groups? In 99% of all cases it's some, well let me say 95% of all cases it's something like this. Where you have a simple alkyl, let me write alkyl here. Benzol would be one of the most common because it's easy to remove a benzol group. That's a common protecting group or perimethoxybenzol. So this would be good for chelation control because it allows some sort of a metal to grip those two, to be gripped by those two oxygen atoms. Here's what's not good, any kind of silo group and the more hindered the worse this will be. So I don't even want to draw anything. I'll just write not good here. There was recently a paper that was published that showed that there was a very specific case for diethyl zinc additions or alkyl zinc additions where you could have siloxy groups and it was so extraordinary and unusual that it was a JAX communication, just some special system where somebody finally got good chelation control with siloxy groups. You should as a general rule assume you are not going to get good chelation control if you have siloxy protecting groups at that position. You'll get Falcon on mode of addition. Okay, so what are some good metals for chelation control? And it's generally going to correlate with Lewis acidity. So titanium tetrachloride or any kind of titanium triflate. These are powerful oxophilic Lewis acids. I'll talk a little bit more about this in just a minute. So titanium, fantastic at being gripped by two metal atoms. Zinc, zinc chloride potentially. I would give you, you often see that used for chelation control zinc triflate, even better. Magnesium, magnesium perchlorate, magnesium bromide. So magnesium cations. They have a lot of partial positive charge. Probably, you can still see chelation control with lithium salts. So when you're using alkyl lithiums to add, oftentimes you'll, not every time, lithium, in this list lithium is the least good. So maybe I'll just put a pair of parentheses around that. But you can sometimes see a chelation control with lithium. Not always, but I would say more than 50% of the time. Okay, so how do you screw up chelation control? You compete with it. So for instance, if I were to add DMSO, you're not going to get any chelation control in DMSO. If you're using DMSO as your solvent. Why would any self-respecting metal want to coordinate with this lame system when it can coordinate to that ox? No metal will do that. Every metal will coordinate with the DMSO. If you take DMF, you're not going to get any chelation control. If you want to understand why the lone pairs on that oxygen or sodium nucleophilic draw the resonance structure of that using the nitrogen lone pair, and you'll see why that oxygen is so good at coordinating to metals. And undoubtedly, better than your substrate will coordinate to any metal. Alcoholic solvents seems pretty obvious to me why you wouldn't want to use that as a solvent for chelation control because your Lewis acid's going to coordinate to the alcohol lone pairs. And then the last, well, let me give you one more that would also be bad. This is dimethoxyethane, DME. And of course, that's got a built-in chelating group built into it, yeah. So let's say you do use one of these to see felkin on. Yes, would be the prediction as the default. Yep. Okay, so, and then last, I'm going to put, I hate THF for this because it's so hard to predict. Sometimes you'll get good chelation control and sometimes you won't. But typically, if I wanted to dissuade chelation or, sorry, if I wanted to prefer chelation control, I'd use diethyl ether, just stay away from this. Sometimes you get good chelation control but those lone pairs are actually pretty good at coordinating to things. So if you have a choice, don't use THF if you want good chelation control. Okay, I want to give you a ranking series for Lewis acidity. Like if you have a choice to throw in some sort of a Lewis acid and next time I'll have to remember to bring in a racer or a used Kleenex which might have more effective chelation of ink. Thank you very much. Okay, I'm going to give you a series of numbers that came from this study. Somebody just used a simple NMR analysis to gauge the capacity for different Lewis acids to coordinate to oxygen and act as a Lewis acid. And what's the goal of having a Lewis acid coordinate to some sort of an oxygen atom? It's to make the oxygen more electropositive. It puts a formal charge on the oxygen. And if I draw the Lewis, sorry, the resonance structure for this, I can see why when I draw the Lewis structure, sorry, the resonance structure for this, I think I can see why nucleophiles would be more likely to add to this system. So that puts a formal positive here. There's another resonance structure that puts a formal positive charge where the carbonyl is. So I'm not going to draw that particular resonance structure. So you can see why when you coordinate a Lewis acid, it becomes faster for nucleophiles to attack either here or here. I'll just write or. So Lewis acids increase the reactivity of carbonyl compounds. That's just a general principle. Okay, somebody was very clever and you may have learned this in spectroscopy class already, but these protons at the beta position of enones are really shifted downfield in your NMR. They get shifted way over into the aromatic region because there's so much partial positive charge here in an enone. And it doesn't matter whether you have the Lewis acid there or not. They get shifted way far to downfield. And the better the Lewis acid, the more that proton gets shifted. It turns out you can simply look at the chemical shift of this proton and that tells you about Lewis acid strength. So somebody was very clever to do that simple study. So I'm going to give you a list of Lewis acids and the results that you get from that. What I'll give you is the thermodynamics of that process. And then from the NMR, I'll tell you about how it made the enone, the pi star system, more reactive. That's what I mean by lumo lowering. It's pi, I'll just call that pi star reactivity. The lowest unoccupied orbital in this is pi star. So let's take a, I'll draw out a series of Lewis acids for which there's data. Boron trichloride, of course, boron has an empty orbital. There's chlorines on there that are electronegative. Compare that to aluminum trichloride. And to boron trichloride, this is much more common than boron trichloride. Triethyl aluminum, you could say the same for trimethyl aluminum. And then the last Lewis acid I have numbers for is stanic tetrachloride. Okay, so if you look at the, I'm going to tell you about the enthalpies of binding here. And I'll ask you to note something that's different. One of these is not like the others. Well, I guess maybe two of these are not exactly like the others. So most of these are similar. And there's two kind of outliers here. And I can't believe, I'm going to put that, because I can't, sorry, I can't believe that that's positive. So let me just put this here. I think that's negative. And I wrote in my notes positive. And I'll check on that for you. Here's what's unusual about this. What they all predict for you is that these powerful Lewis acids enthalpically prefer to be bound to oxygen in a carbonyl, except for one, boron trifluoride. So there's an entropy component that's very similar for all of these, but enthalpically you would predict that most of the boron trifluoride would not be coordinated to your oxygen. BF3 spends only a fraction of the time, less than 0.1% of the time bound to the carbonyl. So if BF3 is going to be a good Lewis acid, in the small fraction of time that it is bound, it better be having a very powerful effect. So if you look at the LUMO lowering in these cases, and then the units are completely unimportant here, this is the most powerful Lewis acid on this list. And arbitrarily, this is measured in electron volts. Who cares? You just want to know what's big and what's small in terms of the effect. Aluminum trifluoride. It's not as powerful as BCL3. It's a longer bond is really what you're doing. You're dropping down to a lower row in the periodic table. Finally, we get down to BF3 even though it spends very little time bound to the carbonyl oxygen. When it is bound, it has a very powerful effect. And so, you know, it's somewhere in this range, even though it doesn't bind as tightly as a triethylaluminum, overall it has a very powerful Lewis acid effect. Triethylaluminum is down here. And then finally, that Stannic tetrachloride at 0.61. Now, there's one common Lewis acid that I don't have in this list and I don't know the numbers for that. My intuition is that Tickel 4 titanium tetrachloride fits somewhere around in this region and I don't know where. It's an extremely good Lewis acid and I don't know why they don't have numbers for that. So, I hate to word you say least powerful because Stannic tetrachloride is a very powerful Lewis acid. But of the things on this list, the tin is the least powerful based on those numbers that were measured. Okay, so Lewis acid massively increase the reactivity. They increase the amount of partial positive charge here. They lower the energy of the lumo. So, if you had some sort of a pi star lumo that you were looking to donate into with your nucleophile, here's my orbital energy diagram just as a way to look at that. So, you know, when you coordinate a Lewis acid, it's not just a charge effect, when you coordinate a Lewis acid, you lower the energy of the lumo by making the oxygen atom more electronegative. All the atoms in this pi system, every orbital associated with that oxygen atom gets lowered in energy. Okay, so Lewis acid, it's a cheap and easy way to increase the reactivity of your substrates and the rates of your side reactions. I'm going to draw a series of carbonyl groups, functional groups and we'll put some numbers on the rates of addition of typical nucleophiles. I want to start off by just drawing an important resonance structure for you so we can key in on the fact that donation by neighboring atoms slows the addition of external nucleophiles. So, here's what I mean. So, if you have a carbonyl group and you're interested in some nucleophile adding in, if you want to mess that up, then put another group here with lone pairs and you'll mess that up and you'll slow that down. Right, if you've already got a lone pair that is right next door, that is always donating in, how can this compete by colliding with that carbonyl carbon? I mean this is always there, it doesn't have to collide. Concentration doesn't matter, you always get effective donation. So, when you draw the resonance structure for this, you draw resonance structures when you believe that there's going to be effective donation. And there's another resonance structure that reminds you of why this is so reactive and that other resonance structure looks like this that reminds you of why carbonyls are so electrophilic on carbon. Okay, so there's an MO depiction for this. Any time you can draw donation of filled orbitals and done filled orbitals, there's an MO depiction. It's in the notes, but I'm not going to take the time to draw that out. What I want to do is I want to draw out relative rates of addition. So, we can see how this group, it's not always a lone pair by the way, how groups that are next door influence the rates of reaction in a quantitative way. And then I'll come back and draw that orbital depiction thing. I'm going to start off here with formaldehyde, extremely reactive, extremely reactive. There's no lone pairs on those hydrogen atoms. There's no pi bonds on the hydrogen atoms. There aren't even any sigma bonds sticking up on the hydrogen atoms that can donate over into the carbonyl group. There's nothing. This is about as reactive as you can get. So, that's the top of our reactivity list. And so, if I compare that to a ketone like acetaldehyde, where I've at least got a CH bond, three of them, but at any point in time, one, maybe depending on the alignments, two, but not so perfectly, but you've got these CH bonds that can donate into the antibonding orbital. And they help decrease the reactivity of this carbonyl group. And if I put two alkyl groups on there and make it something like acetone, I'm not going to draw the CH bonds, they're there. Acetone is less reactive. Ketones are less reactive than aldehydes. That's just a general principle. Ketones are less reactive than aldehydes. And then finally, not finally, but majorly, I can crush the reactivity of this by putting something with lone pairs on there. Let's have an oxygen with lone pairs. And if I really want to slow down the rates of addition of nucleophiles, let me put good lone pairs that are nucleophilic next door, like nitrogen, lone pairs, amids, that's why your proteins are made out of amid bonds. It keeps them from hydrolysing spontaneously. Water doesn't add spontaneously to the backbones of proteins because of that donation. So I'm going to compare all of these functional groups to an amid. I'll assign that a relative rate of one. And we're interested in rates of addition of hydroxide. So if that's a relative rate of one, turns out that formaldehyde has a relative rate of addition of hydroxide of 100 billion, 100 billion times faster for formaldehyde to react with hydroxide nucleophile, anion nucleophiles. And compared to this, formaldehyde, adding that one extra methyl group cost you a factor of 100 in rate. So acid aldehyde is 100 times slower than formaldehyde. That's the effect of one methyl group. That's the effect of the CH bonds on this methyl donating into the carbonyl. There's a steric effect. That's a smaller part of that. It's mainly the donation of the CH bonds into the carbonyl. I come down to acid. The second methyl group is pretty similar to the first. It costs another factor of 100. And then when we get down to esters, there's a bigger jump, not just a factor of 100. Now there's a 10 to the third. So acetone, ketones are about 1,000 times more reactive than an ester. If you have some complex substrate with 10 million atoms and a ketone and an ester, you should be able to react faster with the ketone. You should be able to add butyl grignard reagents or other things faster to ketones than esters. And amides, of course, you should expect them not to react, and that's why it's very common to protect amines with carbamate protecting groups. They put a carbonyl next to the lone pair. That resonance donation is very important. So it simultaneously reduces the electrophilicity of the carbonyl, and it also reduces the nucleophilicity of the nitrogen lone pair. What I don't have a number for but is also very common is alkoxide, anions, I mean there's always something coordinated to that, sodium or lithium. Those are less reactive than amides, and I don't have a number. I'll simply say less than 1 on this reactivity list. Okay, so again, let me just come back to this. Like why is an ester so unreactive? What is this picture telling us, this resonance picture? If I have a lone pair donating into a carbonyl group like an ester, so let me dissect out this ester here, and I'll draw out these pieces of that puzzle. So let me start off by drawing the donor group here. If I want to think about what's the effect on the reactivity on this pi star system? What's the effect of this lone pair nearby? So let me draw out an orbital energy diagram to think about the interaction of this lone pair with this pi star system. So we can try to think about using orbitals and some sort of orbital energy diagram, what's the effect? So unfilled orbitals are higher in energy than filled orbitals. Here's my pi star for a carbonyl group. So here's my two pieces. Here's the carbonyl pi star. Here's the nucleophilic, I've sort of flipped it around there. So if I wanted to think about the interaction of these two pieces, if I mix two orbitals, I get two orbitals out. I get one orbital that's lower in energy and one orbital that's higher in energy. And what's the symmetry of these two new orbitals that I get out? They're pi-like, right? That's what I drew here. I drew pi. So let me label them pi something. So I get two new orbitals out. One's an antibonding orbital. One is filled. And so here's the little interaction diagram. And this is filled, the bottom orbital. What it tells me, right? Look at the orbitals that I get out. It's when I have donation like this. If I have donation where this lone pair donates into pi star, the system actually will now have a higher pi star. My MO diagram is telling me that it is now harder to add to an ester pi star because of that donation. Now, I don't expect you to draw these diagrams, right? I just expect you to look at this and say, oh yeah, there's lone pairs donating in. And that's why esters are unreactive. But if you ever need to, you can draw diagrams to back up your thinking or you sort of dissect this into a set of orbital interactions each with an associated energy. So four dimethylamino pyridine is a very common catalyst for organic reactions. So here's the structure of DMAP. There's a dimethylamino group at the four position of this pyridine ring. And Wolfgang Steglich was studying something called the Dock and West Reaction. Thirty years ago, this has nothing to do with the Dock and West Reaction. And when he was doing those very strange reactions of amino acids, he realized, hey, this type of catalyst has an extraordinary effect. Maybe I can use it for other reactions of carbonyl groups. And so what he showed is that if you take a typical acylation reaction and it's not fast enough for you, you can simply add 10 mole percent of this amine catalyst. Not triethylamine, not just pyridine, but DMAP. And when you add that, you'll isolate that alcohol. You'll perform acyl transfer reactions about 10,000 times faster. Just that 10 mole percent, of course, if you had 20 mole percent, it'll be 2 times 10 to the fourth faster. If you had 10 times more DMAP, you should get 10 to the fifth faster. So just that small amount of DMAP can give you phenomenal rate accelerations. And what we now know is that these reactions occur because DMAP is nucleophilic, and I'm not going to show all the arrow pushing for this. It attacks the carbonyl group faster than the crummy alcohol substrate. But what you could not have predicted ahead of time is that this intermediate, this acyl pyridinium intermediate, is more reactive than the original anhydride. And this is also true for acycloides. It's now faster to attack that carbonyl now that it's an acyl pyridinium. There's now this charged leaving group that's right next door that gives it more calomic positive potential, more of a positive partial positive charge on the carbonyl carbonyl. So this is the mechanism for DMAP catalysis. It gives you these, you know, this is a very simple and easy way for you to make your reactions faster. So it's a general principle of catalysis. So we call this nucleophilic catalysis because the DMAP had to act as a nucleophile and attack the carbonyl first. So if you look inside the ribosome, it's a massive structure, it's got a molecular weight on the order of 40 million, thousands of residues in there. And there is a single base residue, a single nucleotide in the ribosome. So in the mammalian ribosome, it's A246. Right, there's thousands of adenines in there. That's adenine 2486, 2486. All the ribosome for all that size boils down to this. The whole ribosome is there to position this one base, this one nucleotide, this one base, this one atom and that one lone pair next to two key functional groups, an ester and an amine nucleophile. So inside the active site of the ribosome, you've got an RNA ester with an amino acid hanging off the end so it's just an ester, it's nothing special about. And then you have your growing protein chain which is just an amine and the whole ribosome is there to catalyze this acylation reaction where you transfer this acyl group of a new amino acid onto the end terminus of a growing protein chain. And so in the end, you get a new peptide bond. But I'm not going to show you the whole amino acid, it doesn't matter, the point is that this is just, the whole ribosome boils down to a single lone pair. And so what's the mechanism for this in the active? You know, I draw this up here and then you look down here and it's, it kind of looks similar. You know, just like for DMAP you might think, well, maybe this lone pair here is donating and I can draw a resonance structure where I enhance the nucleophilicity of that nitrogen. Maybe that's what the ribosome is really doing. Maybe that's the whole purpose of this particular adenine base in the ribosome. It turns out that's not the role and you could never have predicted this ahead of time. It turns out that that one single lone pair in the ribosome is just there for this. It is just there to help the ribosome deprotonate this intermediate. After the protein amine attacks that ester, you generate this zwitterionic intermediate. And as soon as you pull this proton off, there's now a lone pair that can help push out the TRNA leaving group. And you can't push out that TRNA leaving group until you generate a lone pair on nitrogen. It's too slow. So there's two different very similar things here. There's some, there's just plain DMAP that goes through nucleophilic catalysis, right? You could imagine this adding to the acyl group and generating an acyl periodinium salt. But then there's also base catalysis. And so both of these are well-precedented in the literature. How do you tell the difference when you run a reaction? Whether something is going through nucleophilic catalysis or going through this kind of base catalysis. No, the adenine 2486 inside the ribosome does not attack the carbonyl group. It's just after this nitrogen attacks, it generates this intermediate. And then the key step in peptide bond formation is for the, this lone pair to pull this proton off this tetrahedral intermediate. That's the step that the ribosome is catalyzing that leads to rate acceleration. When you pull this, this proton off you end up with a lone pair that can assist pushing out the leaving group and not until you pull that proton off. Okay, so let's talk about a simple test here. To distinguish these two, I heard John Browman who was one of the three editors for Science Magazine referred to this as the most brilliant experiment he had ever seen in his life. And maybe he was being overly, it's an experiment to distinguish these two possible mechanistic pathways. And what I want to do is I want to draw a picture for you of solid phase synthesis resin. These are just tiny, tiny polystyrene beads. These are the same types of solid phase resins you use for ion exchange, for peptide synthesis, for some types of DNA synthesis. And so what I want you to imagine is taking two pools of resin, two piles of resin here of beads. And in one pile of beads, I've got attached to that an a-seal derivative. And so in this particular case, Jules Rebeck was clever enough to put on not an acid chloride, not an anhydride, but he had a nitrofenol leaving group. So that turns out to be a good leaving group, this nitrofenolate. And he found some way to hook this up to the plastic beads. And so what he wanted to test, what he wanted to test would is how quickly, if I have another pile of beads and I mix these beads all together, if I have one pile of beads, how quickly would this a-seal group get transferred to make a new amide bond? And what he showed is that if you simply mix the beads in solution, there's no transfer. The a-seal groups never get transferred over here. If you physically take the beads and mix them together and then grind them, you still don't get any amide bond formation. But then what he showed is if you take simple catalysts like DMAP or emidazole, which is the side chain of histidine, which has a very similar effect to DMAP, what he showed is that as soon as he added emidazole to solutions that had these two sets of beads floating around, that the emidazole would come in and pick up this group, fly around to an a-seal emidazoleum intermediate and transfer the a-seal groups to the other beads. In other words, you've got two different phases here, two solid phases that don't react. But as soon as you have a soluble species that's capable of nucleophilic catalysis in there, that that leads to a chemical reaction. So here's the end product of the reaction. And this reaction proceeds through an a-seal, emidazoleum intermediate. It works because the emidazole attacks and then pops out the leaving group and then it floats around carrying this a-seal group. And those a-seal, emidazoleum intermediates can float around between the beads and then come and be attacked by the nitrogen lone pairs. So it's called the three-phrase test and it is proof that it's not just the basicity of the medium. It's not just the basicity of the solvent that matters. You have to have some group that's capable of nucleophilic catalysis that can literally pick up groups and move them to other beads. And so it's an easy, I say it's easy to test. You actually have to synthesize the two types of polymer supports. But this is what allows this transfer to occur. Okay, so that's a three-phase, the three-phase test can distinguish nucleophilic catalysis from base catalysis. Okay, so I want to look at electronic effects and strain effects on the reactivity of carbonyl groups. And so what I have numbers for is this equilibrium where water is acting as a nucleophile. So yeah, that's not a very interesting nucleophile but that's what I have numbers for. I'm going to draw my carbonyl group in this resonance depiction form. Because I want, you can think of carbonyl in your mind. But that's a carbonyl group. I'm just drawing the charge-separated resonance structure. And because it reminds us that we're interested in reactions in which a nucleophile attacks a carbonyl. So if you look at water as the nucleophile, it turns out it's easy to get numbers for this where water acts as a nucleophile. And it's more than one step in the mechanism. That doesn't matter. But we're interested in the equilibrium constant for formation, this is called a hydrate for formation of this nucleophilic adduct called a hydrate. Okay, so let's look at the equilibrium constant as we change the R groups on this system. And I'm going to compare this to a molecule that you ought to have an intuition for and that's acetone. It's like the prototype ketone, the simplest ketone you can have. So if we look at the equilibrium constant for addition to water, it's not, I mean it can't be zero, but it's not zero. I mean there's some of the hydrate when you just get a bottle of acetone that's wet. There's hydrate in there. If you just take acetone right out of the bottle, if it's not dried in some way, you might expect to see a little smidgen of OH stretches in your IR spectrum. Okay, so what happens as we replace this with groups that are not as good at donating? So here's what I have numbers for. What would happen if I replace one of the methyl groups of acetone with a carbonyl group? This isn't as nucleophilic anymore. The CH's here donate into the carbonyl and make it harder for water to add. But a pi system here is not as nucleophilic. What you see, and there's also this sort of dipole effect which destabilizes this di carbonyl structure, it turns out that water, the hydrate form of that, is much more preferred. So now it's like a, I don't know, like a 6 to 10 ratio, what I don't even know what this is, 0.6 equilibrium constant, a 6 to 10 ratio of hydrate, 6 out of every 16 molecules now exists as the hydrate. So this is a better donor. This CH bond is a better donor in a way than this pi bond, and there's this dipole thing which is opposing. If I go to acid aldehyde, it's very hard to work with acid aldehyde. Of course, that's, if you're doing some methodology project and you need to pick acid aldehyde as one of your substrates, it's going to suck because it's hard to get dry acid aldehyde. It's very volatile, it's hard to get it dry, and if you don't get it dry, it exists as the hydrate in a 1 to 1 ratio if there's water present. If you don't have CH bonds here on this side to donate in, it turns out that carbonyl is very reactive even with water. Then we get formaldehyde. Remember the super reactivity of formaldehyde? There's very little formaldehyde in formaldehyde. I mean, you don't buy formaldehyde, you buy a polymer of formaldehyde. So here's what you buy when you buy formaldehyde, you buy this infimic polymer called paraformaldehyde in which one carbonyl group on one formaldehyde is attacked, another carbonyl group. And you have to heat the heck out of it, you have to crack it with heat in order to liberate free formaldehyde. Alternatively, you buy solutions in water or methanol. Methanolic solution is called formalin, and those exist as adducts. If you really want to soup up the reactivity of a ketone, put trifluoromethyl groups on there. Those are awful, awful donors. You can imagine that carbon fluorine bonds have no nucleophilicity to them. There's no nucleophilicity to these carbon nucleophones. Those aren't going to attack anything ever. So there's very little donation here, and at the same time that there's very little donation by this carbon fluorine bond, fluorine is also very electronegative. It's the most electronegative atom in the periodic table. And so that increases the partial positive charge there. So if you look at the effect here, this is all hydrate. If there's any water around, it'll sponge out the water and exist as this hydrate diol. There's a drug called lubeprostone. Some of them I don't know that I think you can guess what lubeprostone is, but it helps to lubricate your gastrointestinal tract. That's not the important point. But lubeprostone exists like this. Let me try to draw out the structure for lubeprostone. Lubeprostone is actually a hydroxy ketone. But the medicinal chemists who designed this were very clever. They put this, so this is not the whole structure for lubeprostone, but they were very clever to put a diphthoromethal group on one side of the carbonyl. And guess what that does to the equilibrium for cyclization? If you don't have those two fluorine groups on there, then you get more of this free open chain form. But once you put that diphthoromethal group in there, the equilibrium very heavily lies on the side. Of this closed chain form. And that's the active species that binds to whatever receptor, some ion channel, I think inside of cells. And so that's just somebody using this simple little trick of putting carbon fluorine bonds instead of carbon age bonds. Okay, so let's take a look at the effects of strain. Cyclopropanone, you can buy cyclopropanone, you can't buy cyclopropanone as cyclopropanone, but you can buy hydrates of cyclopropanone or ethoxyacetals, hemiacetals. Of cyclopropanone, there's nothing that exotic about that. It massively prefers to exist as the adduct and it massively does not want to exist as the ketone. And this is the effect of ring strain. If you look at the intrinsic preference for a hydrate for a carbon atom that has four substituents, this wants to be 109 degrees. A carbon with four substituents. If you look at a carbon atom that has three substituents like this, that carbonyl group wants to be 120 degrees. And in this case, because it's a cyclopropane ring, it doesn't get to be either. The fact is that when you've got three carbons bonded to each other, all those angles are 60 degrees and all these angles are 60 degrees. So it's basically a matter of which sucks less. You don't get to be 60. It sucks less to be like this. At least 109 is closer to 60 than 120. And so cyclopropanone, again, you can't buy that. You can only buy acetals of cyclopropanone. They're not happy. Okay, so I'm going to go ahead and stop right there. And when we come back, I'll talk about various ring sizes for ketones, cyclopentanone, cyclohexanone versus free acetone. And then we'll talk about some other types of pi star systems.