Lecture 17. Enols, Enamines, and Enolates
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Vranken, David Van
 Advanced treatment of basic mechanistic principles of modern organic chemistry. Topics include molecular orbital theory, orbital symmetry control of organic reactions, aromaticity, carbonium ion chemistry, free radical chemistry, the chemistry of carbenes and carbanions, photochemistry, electrophilic substitutions, aromatic chemistry.
 Okay, so we've been talking about enols, enol ethers, enamines and enolates. And when we left off we were talking about enolates, but I want to bring you back just to sort of frame our discussion to this one example we had when we were talking about allyl silanes and allylboranes and allylboronates. I showed you allylborane additions as the prototype for many, many reactions in organic chemistry. And there was this pattern that if you had a carbonyl group that could first coordinate to an allylboron compound. That complex, even if it's only present in tiny, tiny amounts in solution can be super reactive because when you coordinate like this you end up with a more reactive electrophile at this oxo-carbenium ion, this oxonium ion species, and a more reactive nucleophile. The boronate becomes more reactive. This boron carbon bond becomes more nucleophilic. I've done another possible mechanism. If you take 202 you'll learn about 3, 3 sigmatropic rearrangements. This is a 3, 3 sigmatropic rearrangement. I would accept that if you drew that. But if you do this you'll end up with a boron oxygen double bond in your product, which is a resonance structure of what you get if you use the arrow pushing I showed over here. So before we had talked about this idea that when two things coordinate to make a complex the complex can be super reactive even if it's present in small quantities. So keep that in mind. Okay, so let's talk about this simple idea of deprotonating ketones in order to generate enolates. And if you have a substituent at the alpha position or two substituents you'll create stereochemistry either E or Z, either cis or trans geometry. And so you've got two different choices here for enolates that you can make. So with common bases, maybe I shouldn't say common, but common to a sophomore organic chemistry triethylamine. It generates a lot of enolate, but it generates a mixture of these enolates. And you can trap those out so that you don't, so it funnels off towards the enolate type products. With sodium hydride, with potassium t-butoxide, these are the kind of enolates that are really the common stuff in lower and undergraduate organic chemistry courses. These tend to favor this Z enolate. So again, there's an issue here with con and gold pralog. So in other words, O minus has the higher priority here and methyl has a higher priority than H. So that's the Z enolate right there. And that's thermodynamically favored. You're getting the thermodynamic product. The more stable enolate is the one that forms and you avoid this sort of bumping interaction. What's going on here is that this is an equilibrium going back and forth. If you have even the tiniest amount of enolate, or sorry, of ketone present, it will help to accatalyze the isomerization of these enolates. So let me show you how that works. Let's just suppose that we generate a small amount of this less favorable E enolate like this and you have any ketone left over in solution with alpha protons on there. That means you have a potential acid in your solution. So if I acknowledge the basicity of that alpha carbon and then use that to pull off that proton and don't worry about we're violating the three arrow rule, I'm just going to, we'll suspend that momentarily. The point is that you can see how this readily equilibrates your ketone and as the product of this you get either the E or the Z enolate out of this. As you can see how just a little bit of ketone can catalyze this isomerization and this isomerization, this transfer of protons back and forth is fast. If you want to make enolate and you want to end up with a thermodynamically more stable one, then it's okay to have a little bit of ketone floating around because it can accatalyze the isomerization so that you end up with a thermodynamically more favorable Z enolate. Okay, but usually we don't do that. That's actually the least common way to make enolates. So lithium diisopropyl amide LDA gives you kinetic enolates and you want to avoid this situation that I just painted up above. Let's go ahead and draw out a mechanism and let me start off by warning you, don't use this generic symbol B for a base when you draw the mechanism for LDA deprotonation. And I'll show you why. The mechanism for LDA deprotonation, remember all of these pictures that I showed you of aggregates, aggregates of lithium alkyl lithium compounds, lithium enolates, even LDA. When you put LDA in solution, even in THF, that lithium wants eight electrons and in this particular Lewis structure it only has two electrons and that sucks. So it's going to be coordinating things, THF molecules, lone pairs on there will be popping on and popping off. Eventually this oxygen atom on your substrate will coordinate to that lithium. And when it does, even if it's only a tiny, tiny amount of this species at equilibrium, you now have a situation analogous to that boron complex that I showed you before. So you now have a situation where this electrophile, this carbonyl is more electrophilic. That makes it easier to deprotonate here at this CH bond because those electrons will donate into the carbonyl. And that makes this, because I pick up a minus charge on lithium, it makes that nitrogen and the lone pairs on that nitrogen more nucleophilic and more basic. So by complexing this complex, it doesn't matter if it's 99% ketone and only a tiny amount of this complex, the tiny amount of this complex that you have floating around in solution is super reactive. And so now I have this six-membered ring transition state that shows why it's going to be so facile to deprotonate. So you have a really nice six-membered ring transition state for deprotonation. And the end product of this will initially still have all of this stuff coordinated. You don't initially immediately release the lithium. Lithium loves to coordinate to nitrogen. So if you want that to pop off in your mechanism, you need to pop that off as a second step. If you want to show how you generate a naked lithium enolate without an amine on there. Okay, so that's the mechanism for deprotonation by LDA. The important point about this is this goes through a chair transition state. So if you take Chem 205, organic synthesis, you're going to go into a great deal of discussion about the impact of this. I'll just sort of sketch that out for you now. And I'll draw some of these bonds in here, and then I'll fill that stuff in in just a moment. Let me try to make this a dash bond in front, and then I'll give you a chance to sketch that in in just a moment. So I'm going to fill in the blanks here as to what is what. I'm going to have a substituent, not equatorial or not axial, because this is going to be our carbonyl group right here. As we deprotonate here, this starts to lose its double bond character. And as we deprotonate here, as we take this proton off, we're going to start to break this bond. And I'm going to draw dashed arrows to show where we're breaking bonds. So because it's a chair, this substituent has an axial or an equatorial preference. I'm going to sketch in those axial and equatorial groups. And then back here, I'll go ahead and draw that nitrogen atom that is now so basic. So there's a lithium atom that's joining the oxygen and the nitrogen. And here's the isopropyl group. We went through a lot of work trying, okay, let's try lithium diethyl amide. Let's try lithium di-T-butyl amide. They've tried every single combination and it turns out that isopropyl is just perfect. It's very hindering but not too hindered that you can't adopt these kinds of transition states. And so when you adopt this transition state, these substituents here have a choice. If you had a choice between having your methyl axial or equatorial, you'd rather have it equatorial. Because the H, the least hindered species that wants to be axial. And the more hindered thing, methyl or whatever, is going to be equatorial. And that's why you prefer this particular enolate. And so you get, in this case, this would be the Z-enolate oxygen is opposite to methyl. So you get the kinetically less stable enolate when you use LDA because deprotonation goes through this chair transition state. So we call this the Ireland transition state. Robert Ireland formulated this, I think back in the late 1970s. Okay, so let's go ahead and take a look. I'll give you one sort of warning about this EZ. Awful, awful, confusing nightmare about E and Z. So first, and then I'll give you a warning about how sterics can override that. So let me just first draw out a picture for these reactions. So based on what I'm telling you, if you want to isolate the less stable enolate, you never, ever, ever want to have excess ketone floating around. Excess ketone catalyzes isomerization between cis and trans enolates. What you always do is you put an excess of the LDA in your pot. You make that first using butylithium and disopropylamine. And then you slowly, yep. Isn't that the E enolate that you drew? Did I draw the, oh sorry. Okay, maybe it's not confusing to you, but it's confusing to me. So yes, it is. Thank you for asking about that. And as I'm telling you now, LDA gives you kinetically the E enolate. Okay, so typically what you do is you add in your ketone or your ester drop wise to a solution of excess LDA. When you do this, there is no point in time. At no point is there ever excess ketone floating around. That would be a disaster. So really you don't add LDA to the ketone, you add the ketone to the LDA. We just refer to that as inverse addition. Okay, so here's the results that you get. If you take disopropyl, lithium disopropyl amide, and you add three pentanone to that, you add that drop wise to your solution. So you start off with a slight excess of LDA. What you'll get is not very good selectivity if you take a ketone like this. Of course you wouldn't really commonly take a ketone like this for modern synthesis. So you'll get a slight excess of this isomer here, this trans isomer. And that's not really very good. You see great selectivity. You see great selectivity in case of esters. If this is a little bit smaller, then it's a lot better to form this E isomer. Let me not say E plus C, let me say E slash C. And it's about 2 to 1, 70, 30 favoring the E isomer that I've drawn there. So the O lithium is trans. This is what you get based on that Ireland transition state. If you have an oxygen on the other side here, oxygen is smaller than ethyl. You get much better selectivity. And so now it's 94 to 6 ratio favoring this enolate. And then finally, if you really bulk this up, you could screw up that Ireland transition state and make it not really all that great. Like let's just imagine you put a T butyl or a phenyl group here. So now you favor this enolate about 2 to 98. So what's going on with the E and the Z issue? So what's the higher priority group here, ethyl or oxygen according to Konning-Goldpraylog? Oxygen is the higher priority. So when I want to judge what's opposite to the methyl, I look at the higher priority group and that's E. But now when I come over here, it's not the same anymore. It depends on what you have here, what your substituents here on this position. So in this case with the oxygen, it's 94% of the isomer that I'm showing you there. In other words, it will flip back and forth between what you call E and what you call Z depending on the Konning-Goldpraylog priorities of these substituents. And finally, we come over here to this O-lithium species. I didn't draw lithium very well there. And now maybe it would be more helpful if I called this, well, I would still call this, this is the Z isomer here because the higher priority group is O-lithium and the higher priority group here is methyl. But here the higher priority group is the ethoxy group. So you have to worry about this nomenclature issue. What is E and what is Z depends on whether you're working with ester enolates or ketone enolates. So the main idea is that you prefer these types of geometries due to the Ireland transition state, but you can mess that up if this gets too big and can spin it around the other way. Okay, and you'll talk about that a lot more in Chem 205. So the synthesis class. So you look at LDA in solution or lithium-amide beta. Or lithium-amide bases. They exist mostly as dimer. This is something that we've talked about before. And so typically you make LDA in THF. The ether oxygens of THF are coordinating. So this is what LDA looks like in solution. And so fortunately there's a small amount of this sort of a monomeric species that's floating around in solution. And that's what does all the deprotonation. And I don't know what the ratios are, but the equilibrium massively favors this dimer. Don't worry about drawing the dimer. I don't want to see the dimer, but just remember that it's aggregated mostly in solution like a lot of lithium enolates. It's the small amount of this stuff that's present that coordinates to your carbonyl and then goes through this Ireland transition state for deprotonation. So you can mess up all of this Ireland transition state deprotonation business. Let me come back and draw out. You don't always have to draw a chair in this class. So for example, if I asked you to deprotonate, draw a mechanism for deprotonation of this ester you would end up drawing out something that looks like this. You don't have to draw it in a chair shape in this class. I won't ask you questions related to that. But it's the small amount of the monomer that then coordinates to the carbonyl of an ester or a ketone and that's what generates these sort of intermediate species that then deprotonate through that six-membered ring transition state. So I take those protons off and I just attach that to the wrong position so I'm freaking out here. There we go. Ah, there we go. Okay. Okay, so we can mess this up. It's very easy to mess this up. What I showed you before is if you really add something bulky here, like instead of oxygen, if the t-butyl was directly connected, you can mess up this Ireland transition state, the six-membered ring transition state. But you can also mess it up by putting additives in your solution. Things that coordinate and complex very powerfully to lithium will make it harder and harder for your LDA to coordinate to your substrate. So for example, here's another deprotonation. And in this case, they're using hexamethyl disilazid. So instead of iso, so what's the difference here? Maybe I didn't do a good job of drawing this. So the big difference is this is commercially available and you can buy solutions to this. LDA, you make yourself. So if you've got the money, you can go ahead and buy this. Obviously, trimethylsilyl has, occupies more volume than an isopobyl group but the bonds are longer. So the silicon is farther away from the nitrogen. And overall, it's about the same effect as an isopropyl group. So again, you can buy this but LDA, you can't buy, you have to make that. So if you put an ester in with this in order to deprotonate the ester, whoops, the favorite isomer that you get now is not the one where the methyl group is on the same side as the ethoxy. Now the methyl group is on the same side as the olithium. And it's the selectivity is not perfect. So the e to z ratio here is typically somewhere, I'm going to say above 80 but I'll say 80 greater than 90 to less than 10 for this particular example in this case. And the secret here to getting this is they add HMPA to their solution. You'll see this used a lot less. There's other alternatives. But HMPA is a very powerful coordinator of lithium. Here's the structure for HMPA. If you'd like, you can draw a double bond O here but I'm going to draw this in its fitter ionic resonance form or the illid resonance form. Just to emphasize the fact that there's a lot of negative charge on that oxygen. And because of that, it loves to coordinate lithium. And it keeps the LDA from coordinating to the substrate. So you disrupt that six-member transition state. And ultimately you end up getting the opposite isomer is predicted from that, from that Ireland transition state. Okay, so HMPA, that's it over there, hexamethyl phosphoric triamide, that disrupts. So oftentimes if you see people who want to reverse their enolate selectivity, they'll add some powerful coordinator like HMPA into the solution. One gripe, legitimate gripe against this is this is a carcinogen. It's a powerful carcinogen. So people tend not to like to use those as co-solving. It's usually high-polygon like so it's kind of like purified. Oh, is it? Yeah. I've never used it before. I don't know. Let me show you one other way to make enolates. This one is very specialized and totally not obvious. Oftentimes I want to draw the simplest enolate and that's the one derived from acid aldehyde. And you don't make the enolate of acid aldehyde by deprotonating acid aldehyde. You make the enolate, so this is the enolate from acid aldehyde. So you don't make it like this. You don't make it by deprotonating acid aldehyde using LDA. It turns out that acid aldehyde is so reactive that as you're making the enolate, it starts doing aldol reactions with itself. So if you want to make LDA, you never want to have more acid aldehyde present in solution. What you do is you just put butylithium in THF. So the converse of this is if you don't want to make acid aldehyde enolate, don't leave your butylithium sitting around in THF at room temperature because they react with each other. So the reaction looks something like this. Let me draw the, well, let me redraw this with a bond here. That bond is basic and if you leave this sitting around long enough, it will do this. And so you'll medilate alpha to that oxygen atom to generate this medilated THF. This will happen at room temperature. So let me just put in here that's the half life for that. It's 107 minutes at room temperature. So don't put LDA or don't put butylithium in THF. It's like, oh, you guys are going to lunch? Oh, let me go with you. I'll just add the rest of my substrate later. No, that's a problem. So if you have more butylithium floating around, this can coordinate to that, right? Lithium loves to have oxygens coordinated to it. Yeah, it could be coordinated before if you're wondering, but I'm going to draw this in this sequence. And so now this can fragment. This super-duper nucleophilic, this super-duper nucleophilic carbon lithium bond there can donate into the antibonding orbital for this CC bond. Who would have thought that, right? You wouldn't have ever guessed that. And if you donate electron density into this carbon-carbon bond, if you draw out what that orbital looks like, there's electron density on the backside that starts to appear. And it's that electron density that comes in and donates into sigma star. But I'm going to draw my arrow pushing like this. Ultimately, you fragment to give ethylene. Did I just do that right? Yes, you fragment to give ethylene, and then you get your acid aldehyde enolate as a byproduct. And so then you can pop off the lithium and the R group, et cetera. And so that's what gives you the lithium enolate. Okay, so be careful when you put butylithium in THF. Don't let it sit around. You'll end up with this problem. Don't let it sit around at room temperature. You'll probably note from procedures that they're generally doing this at minus 78 reactions of butylithium in THF. So I want to ponder on this. You almost never will work with naked enolates. There's almost always a lithium bound oxygen or something. If you're working with transition metals, they tend to favor carbon. But you're never working with naked enolates. There's always a counteron. But I want to ponder on this. You can imagine reactions that occur on carbon, and you can imagine reactions that occur on oxygen. So let's talk about that. How would you predict whether your electrophile is going to react with the carbon atom or the oxygen atom? Just come back to this issue of C versus O alkylation. And I don't have numbers just for a plain enolate. I have numbers for this stabilized enolate where there's an extra carbonyl group. This would be a beta keto ester. And what I want to do is I want to imagine what will happen if we take various ethyl electrophiles. So we're going to do an SN2 reaction and displace the X. But we're interested in this competition between attack, let me pick a different pen color here, between attack of the oxygen atom versus, let me use a different pen color, attack of the carbon atom. And how does X influence whether you're attacking on carbon or attacking on oxygen? So C versus O alkylation. If you want a high degree of C alkylation, then here's what you should do. What you should do is you could, you should make this as covalent and non-ionic as possible. Anything you can do to reduce the ionic nature of that oxygen will make it less capable as a nucleophile. So if you want very high ratios of carbon alkylation and less alkylation on oxygen, then I would pick the most covalent bond possible here. So let me draw out this, what do I have for the, I'll just draw a metal here, metal plus. In other words, is this metal floating off in space like I drew or is it covalently bound? If you choose lithium, that lithium is bound here. Actually, it's chelated between the two oxygens. That lithium is tightly bound to the oxygen. If you want to make this more naked, like more, more negative charge of oxygen, then move down the periodic table to sodium, to potassium, to cesium, which in some cases you'll use. And then finally, about as naked as you can get that enolate is something that, sorry, I'm trying to draw tetraethylammonium. It's a tetraalculammonium cation which can't coordinate at all to that O minus. In that case, it's a naked O minus. So in that case, what you would get is mostly O alkylation. There's the ethyl group. If you use tetraethylammonium, whereas if you use the lithium enolate, you'll get mostly C alkylation. And there's the new ethyl group on the carbon atom. So the more covalent you make that oxygen metal bond, the more covalent the less negative charge here, and the more you're going to see reactions on carbon. Okay, so what can we do to change the electrophile? So if I want to attack, if I want to favor attack here on this carbon, let's just review what is it that makes this oxygen so nucleophilic? In that case, it's the negative charge that makes reactions favorable. They're both nucleophilic. This oxygen is nucleophilic because it has lots of negative charge. This carbon is nucleophilic because the HOMO is biggest at this position, at this carbon atom. If I look at the HOMO, let me just sort of sketch out my little MO diagram here, MO. And there's all these orbitals in there. But there's this one orbital that's highest in energy, that's the HOMO. And what does that HOMO look like? What do those two electrons, where do they spend their time? It turns out those two electrons in the highest occupied molecular orbital spend most of their time centered above this carbon. So the carbon reacts because the most reactive electrons spend most of their time there, the oxygen reacts because it's got more negative charge. And those are completely different reasons. So if you wanted to make some electrophile react here faster, I wouldn't want to have an electrophile that had a lot of negative charge. I would want to have an electrophile that has very little negative charge. Sorry, very little positive charge. So if I take something like an alkyl iodide, carbon has about the same electronegativity as iodine. There's very little partial positive charge here. This reacts quickly because the LUMO is very big on that carbon atom. That long bond makes that LUMO, the sigma star, very big on the back end. And as I make my electrophiles more and more with electronegative groups on there, bromine is starting to become more electronegative than carbon. Chloride. Now I start, and now as I make my leaving groups, be more and more electronegative, a phosphate. So now that's very electronegative. Now there's a lot of partial positive charge here on that carbon atom. This reacts for, is very reactive for a very different reason. And finally at the bottom of my list here, I've got the tosylate, the sulfonate leaving group. Now there's a great deal of partial positive charge here. The more partial positive charge here, the more I'm going to see reactions that are governed by charge, and that means more O-alkylation. So I've got a set of numbers here to put things into perspective. So when X is equal to tosylate, ethyl tosylate, you get a C to O ratio of 11 to 88. In other words, you get more O-alkylation and then you get C-alkylation with ethyl tosylate, because it's mainly driven by charge. When X is equal to iodide, the leaving group there is equal to iodide, that doesn't mean care of. It means carbon to oxygen-alkylation ratio. It looks like a male term. So when you use ethyl iodide, you get more C-alkylation. So in other words, you don't totally reverse it to 99 to 1, but you can flip that balance back and forth. Okay. So let me go ahead and just summarize what I'm telling you about which things will react where. And I'll try to summarize that in two different ways. So how does this matter in synthesis? So if you want things that are going to react here on oxygen, it would be things that have a lot of partial positive charge. So for example, if you take anhydrides, they react faster on oxygen. You don't get C-acylation. Let me make that attack over here more clearly. You get O-acylation if you add a c-decan hydride. If you take TMS chloride, there's a lot of partial positive on that silicon. You get O-silylation. You don't get a lot of C-silylation. You make sililenoethers. They react faster on oxygen. If you take things like alkyl halides, like that tosylate, those react faster on carbon. And of course, that's what you care about mostly in organic synthesis. You care about making carbon-carbon bonds. Organic synthesis is dominated by reactions like this. Take alkaliadides or aldehydes. That's an aldol reaction. Those react on carbon faster. So those reactions are driven by HOMO-LUMO interactions, not by the fact that there's some amount of negative charge on carbon. Okay, so that's the big idea. Let's just review what we're saying about enolates. I'll go ahead and sketch out this picture of the HOMO. So if I look at the most reactive pair of electrons and we're looking down on top of the HOMO, there's this sort of set of orbitals that have pi symmetry above and below the plane of the board. And so in that HOMO, that pair of electrons spends most of its time above that carbon. It's bigger on the carbon than it is on oxygen or here. And tiny differences in the size of that orbital on carbon versus oxygen translate to huge differences in selectivity. If I were an electrophile looking for electrons and I wanted to overlap with an atom, I'd be wanting to overlap over here on this end because the most reactive pair of electrons spend their time there. So if you want reaction, so if you want C-alculation, you want electrophiles that have small partial positive charge. You don't want really charged electrophiles. And you want electrophiles that don't have, and you want electrophiles that have low lumos because it's a HOMO-lumodriven interaction. In contrast, if you want to react with this, then you want things that have a lot of partial positive charge, right? Reaction on oxygen, O-alculation is a charge-driven reaction. And you don't want it to be driven by HOMO-lumodriven interactions. So you want something with a high lumo. If I were to pick something that would really react quickly on oxygen, it would be like this. There we go. That will react really quickly on oxygen, much faster than on carbon, right? Of course, that's not very useful synthetically, but that would be an example of something that's all charge-driven. Okay, so sometimes you need tricks around this, or you need to find the right reagent balance. What happens if you do want to see acylate on the carbon of an enolate? So remember what I just told you, that if you take anhydrides, those will tend to form enol acetates. They react faster on oxygen. That's the general principle. If you do want to add an acyl group here, then what you do in this particular case, well, it won't be exactly that group, you can use something called Manders reagent. This is a very tricky sort of electrophile. Instead of having an oxygen-leaving group here on this side or a chloride-leaving group, use a carbon-based leaving group that doesn't put a lot of positive charge on carbon. This is not super electronegative like chlorine or oxygen, yet it can still act as a leaving group. And so as a result, when you use this kind of, this is called Manders ester, you end up getting acylation on the carbon atom. And it's just a little trick of replacing your typical chloride-leaving group with a carbon atom that's not very electronegative. Okay, one last thing to finish up with this. Parking's back to this example that I showed you over here with this boron. This is the way aldol reactions work. So if you take some sort of a carbonyl compound and you convert that to the lithium enolate, that may not look like a boron, an aloe boron species to you, but it's the same idea. I'm now just sort of flip this around. This is very similar to that aloe boron species. Instead of boron, we've got a lithium atom. And lithium loves to coordinate to oxygen. And so let's draw that coordinated species. Even if it's present in tiny, tiny amounts, this coordinated species now has a more reactive electrophile and it has a more nucleophilic nucleophile. Now that that's a lithiate, that whole enolate part becomes more reactive. So the important one is even if there's a tiny amount of this in solution, even if it's mostly dissociated like this, even if there's a tiny amount of this, all of your C-C bond-forming reactions go through this mechanism, six-member transition. And it is also a chair. And it's that chair, like nature of the transition state that allows you to predict stereoselectivity in aldol reactions. If you take synthesis, chem 205, you'll learn 204. I think you'll learn a lot more about that. Okay, I'm going to finish up by showing you, I'm going to switch over to a sort of slide here. Actually, let me, so we just gave a handout in class. Everybody got this handout that shows something called the Meyer reactivity tables. And I want to talk about this, and maybe I can pose it as a question. I want to talk about the power of these tables. I've shown you a bunch of enol ethers and enamines and enolates and which ones are most reactive and which ones are least reactive. And I've shown you a bunch of electrophiles, carbocation, things like that. And so at some point, you're going to get down to wondering, gee, is that a good reaction or is that a bad reaction? So we've talked about enol ethers, enols, enamines. Here's an enol silo ether. It's a very common nucleophilic species. Let's suppose I mix that in with this aminium ion. We know that aminium ions are reactive. Well, that reaction occur. Right? We've learned all this stuff about organic chemistry. And we still can't answer the simple question of whether this reaction of a good nucleophile, that's supposed to be a bond. I just made it a little bit too thick. We still can't answer the simple question of whether a nucleophile will react with an electrophile. And that seems kind of pathetic. How can it be that nobody's ever measured that? And the point is that Herb Meyer measured this stuff. He's a hero in the field of organic chemistry. So the question is, does this have enough oomph? And does this have enough tug in order to form a carbon-carbon bond? And so here's how you address this. Let me try to turn on this. In other words, to answer that question, you need somebody to take their group and tell everybody in the group, okay, you're going to measure kinetics for the next two years. Right? Who was able to get anybody to do that heroic work? And Herb Meyer did it. Okay, pause. Sorry, I got to let this warm up here. Okay, this is our course website. You guys probably recognize this. I'm going over to links here. There's a links page. And on the link page, I added a link to Herb Meyer's logarithmic reactivity scales. And he had to deal with this. He wanted to address this simple issue of why can't we just look this up in some table to figure out which reactions work well and which reactions don't work well. And so when you go to that link to Meyer's reactivity site, he's got all kinds of tables. I'm going to click on the first one. Just to show you, I wish there was some way to go to full screen without the, whatever. I don't know how to do that. Okay, so here's the Herb Meyer reactivity scales. Let me try to explain what you're looking at. He's got this whole set of numbers. Over here on this side, he's got all the nucleophiles. Let's just take a look to make sure we can identify the nucleophiles. Oh, there's just a simple double bond attacking things. And it's way down, well, it's inverted here. The worst electrophiles are on top, sorry, nucleophiles are on top, and the best nucleophiles are on the bottom. So as we get down to the bottom, look, there's an enol silo ether. That's more nucleophilic than a regular alkene. Oh, look, here's an allyl stannane and an allyl silo, an allyl silane. Those are better nucleophiles than simple alkene. So as you walk downward here, nucleophiles get better and better and better and better and better. And this is logarithmic. These are factors of 10 differences as you go from minus 6 to plus 6 all the way down. Okay, so let's just try to gauge. What's the difference in nucleophilicity between just a simple terminal olefin and an allyl siloane? How many times more reactive is an allyl siloane than just a simple terminal olefin? So the way to answer this is I'll come over here and I'll look at a terminal olefin, and that has a nucleophilicity parameter of minus 2.25. Forget the second number there for the moment. Now let's find an allyl siloane. Let's just find a simple one like allyl trimethyl silo. That's 1.79. So what's the difference between 1.79 and minus 2.2? I can't do that math so quickly. So it's about four. Four orders of magnitude. Immediately these tables are useful. Allyl silanes are about four orders of magnitude more nucleophilic than simple C-C double bonds. Well, it matters what electrophiles you're looking at. And so what Meyer did that made these tables useful to organic chemists is he looked at reactions that organic chemists care about. He said, I'm going to look at reactions that form C-C bonds. So he started off by taking Ben's hydro cations and he started looking at all kinds of nucleophiles adding to that. So most of these reactions that he looked at with nucleophiles, he was looking at the attack to make a new C-C bond. And it's not that big a stretch to say, well, that's the same as if I look at this, right? It's the same idea. It's a stabilized carbocation. You could redraw that as a stabilized carbocation. So his tables are valid for people who care about C-C bond formation. And in general, that's you. So you can gauge relative reactivity if you walk up and down this table, relative nucleophilicity. So let's look at the most reactive nucleophiles that he measured. He couldn't measure everything, right? He didn't measure butylythium. That's off the charts. But some of the more reactive things he has here are stabilized enolates. He doesn't have any unstabilized enolates. That would be even lower than his measurement techniques allowed. He's got simple amines on here. He's got enamine somewhere over on this side. So you can gauge reactivity. And it's very powerful oftentimes to say, hey, this is 10 orders of magnitude more reactive than that, more nucleophilic. Okay, conversely, he also made measurements for electrophiles. He took enol-cyl-lethers and allosilanes and looked at their reactivity with all kinds of electrophiles. And so let's come back up to the top of his charts here for the most reactive species. The most reactive species he looked at were simple, well, these aren't simple carbocadions. We're just resonance stabilized carbocadines. So here's a diphenylmethylcarbocadion, carbenium ion. And if you want to gauge, well, how much does a methoxy group stabilize that and tone down the reactivity? You can come down and look. And if you walk across here, it says 2, it says 3, so plus 3 versus a diphenyl, which is about 5.9. So that's about 3 orders of magnitude less reactive when there's a methoxy in place of a benzene ring. So you can gauge the relative reactivities of electrophiles and specifically the kinds of electrophiles that form carbon-carbon bonds. What's powerful here is that you can predict the rates of reactions at room temperature. You can always extrapolate to higher or lower temperature. Things are faster at higher temperature or slower at lower temperature. But at room temperature, without having going into the lab, you can predict which reactions are faster or slower by plugging in these numbers. You've got an electrophilicity parameter. You've got a nucleophilicity parameter here that's these first numbers. The second number here is sort of like a fudge factor that varies just a little bit. They're usually close to 1, but they're not always 1. So look over here, there's this fudge factor that's exactly 1, which means it has no influence. But over here, there's other fudge factors. For example, furan has a fudge factor of 1.31. What that says is that furan is kind of picky about the electrophile. The sensitivity of that nucleophilicity depends on which electrophile you're reacting. So that's kind of like a sensitivity parameter. I usually just forget that because I'm too stupid to take that into account and do this simple arithmetic. There's a very powerful rule of thumb that you can use associated with these tables. And here's the rule of thumb. And that doesn't look like that pin is working too well. So the rule of thumb is that you can expect reaction at room temperature at reasonable rates like within a day. That would be a reasonable. That's what we mean by react. If the sum of these two numbers is greater than 5. Now very rarely in organic chemistry do you have this kind of simple predictive power. Right? This is what you want to get out of this class. You want to be able to predict which reactions are fast, which reactions are slow. And if you're not sure, if you didn't get an intuition by sitting and listening to the lectures, you can go to these tables and look these kinds of things up. If things are really electrophilic or super nucleophilic or both, then those numbers, this number will get higher and higher and higher. Better electrophiles will make this number higher and better nucleophiles will make this number higher. And so if the number is greater than 5, if that sum is greater than 5, then you can expect things to react at room temperature. Okay, so that's it for CC double bonds as nucleophiles. And you've got lots of problems on the problem set and help you practice that stuff.
 Okay, so we've been talking about enols, enol ethers, enamines and enolates.  And when we left off we were talking about enolates, but I want to bring you back just  to sort of frame our discussion to this one example we had when we were talking about allyl  silanes and allylboranes and allylboronates.  I showed you allylborane additions as the prototype for many, many reactions in organic  chemistry.  And there was this pattern that if you had a carbonyl group that could first coordinate  to an allylboron compound.  That complex, even if it's only present in tiny, tiny amounts in solution can be super  reactive because when you coordinate like this you end up with a more reactive electrophile  at this oxo-carbenium ion, this oxonium ion species, and a more reactive nucleophile.  The boronate becomes more reactive.  This boron carbon bond becomes more nucleophilic.  I've done another possible mechanism.  If you take 202 you'll learn about 3, 3 sigmatropic rearrangements.  This is a 3, 3 sigmatropic rearrangement.  I would accept that if you drew that.  But if you do this you'll end up with a boron oxygen double bond in your product, which  is a resonance structure of what you get if you use the arrow pushing I showed over here.  So before we had talked about this idea that when two things coordinate to make a complex  the complex can be super reactive even if it's present in small quantities.  So keep that in mind.  Okay, so let's talk about this simple idea of deprotonating ketones in order to generate  enolates.  And if you have a substituent at the alpha position or two substituents you'll create  stereochemistry either E or Z, either cis or trans geometry.  And so you've got two different choices here for enolates that you can make.  So with common bases, maybe I shouldn't say common, but common to a sophomore organic  chemistry triethylamine.  It generates a lot of enolate, but it generates a mixture of these enolates.  And you can trap those out so that you don't, so it funnels off towards the enolate type  products.  With sodium hydride, with potassium t-butoxide, these are the kind of enolates that are really  the common stuff in lower and undergraduate organic chemistry courses.  These tend to favor this Z enolate.  So again, there's an issue here with con and gold pralog.  So in other words, O minus has the higher priority here and methyl has a higher priority  than H. So that's the Z enolate right there.  And that's thermodynamically favored.  You're getting the thermodynamic product.  The more stable enolate is the one that forms and you avoid this sort of bumping interaction.  What's going on here is that this is an equilibrium going back and forth.  If you have even the tiniest amount of enolate, or sorry, of ketone present, it will help  to accatalyze the isomerization of these enolates.  So let me show you how that works.  Let's just suppose that we generate a small amount of this less favorable E enolate like  this and you have any ketone left over in solution with alpha protons on there.  That means you have a potential acid in your solution.  So if I acknowledge the basicity of that alpha carbon and then use that to pull off that  proton and don't worry about we're violating the three arrow rule, I'm just going to, we'll  suspend that momentarily.  The point is that you can see how this readily equilibrates your ketone and as the product  of this you get either the E or the Z enolate out of this.  As you can see how just a little bit of ketone can catalyze this isomerization and this isomerization,  this transfer of protons back and forth is fast.  If you want to make enolate and you want to end up with a thermodynamically more stable  one, then it's okay to have a little bit of ketone floating around because it can accatalyze  the isomerization so that you end up with a thermodynamically more favorable Z enolate.  Okay, but usually we don't do that.  That's actually the least common way to make enolates.  So lithium diisopropyl amide LDA gives you kinetic enolates and you want to avoid this  situation that I just painted up above.  Let's go ahead and draw out a mechanism and let me start off by warning you, don't use  this generic symbol B for a base when you draw the mechanism for LDA deprotonation.  And I'll show you why.  The mechanism for LDA deprotonation, remember all of these pictures that I showed you of  aggregates, aggregates of lithium alkyl lithium compounds, lithium enolates, even LDA.  When you put LDA in solution, even in THF, that lithium wants eight electrons and in  this particular Lewis structure it only has two electrons and that sucks.  So it's going to be coordinating things, THF molecules, lone pairs on there will be popping  on and popping off.  Eventually this oxygen atom on your substrate will coordinate to that lithium.  And when it does, even if it's only a tiny, tiny amount of this species at equilibrium,  you now have a situation analogous to that boron complex that I showed you before.  So you now have a situation where this electrophile, this carbonyl is more electrophilic.  That makes it easier to deprotonate here at this CH bond because those electrons will  donate into the carbonyl.  And that makes this, because I pick up a minus charge on lithium, it makes that nitrogen  and the lone pairs on that nitrogen more nucleophilic and more basic.  So by complexing this complex, it doesn't matter if it's 99% ketone and only a tiny  amount of this complex, the tiny amount of this complex that you have floating around  in solution is super reactive.  And so now I have this six-membered ring transition state that shows why it's going  to be so facile to deprotonate.  So you have a really nice six-membered ring transition state for deprotonation.  And the end product of this will initially still have all of this stuff coordinated.  You don't initially immediately release the lithium.  Lithium loves to coordinate to nitrogen.  So if you want that to pop off in your mechanism, you need to pop that off as a second step.  If you want to show how you generate a naked lithium enolate without an amine on there.  Okay, so that's the mechanism for deprotonation by LDA.  The important point about this is this goes through a chair transition state.  So if you take Chem 205, organic synthesis, you're going to go into a great deal of discussion  about the impact of this.  I'll just sort of sketch that out for you now.  And I'll draw some of these bonds in here, and then I'll fill that stuff in in just a  moment.  Let me try to make this a dash bond in front, and then I'll give you a chance to sketch  that in in just a moment.  So I'm going to fill in the blanks here as to what is what.  I'm going to have a substituent, not equatorial or not axial, because this is going to be  our carbonyl group right here.  As we deprotonate here, this starts to lose its double bond character.  And as we deprotonate here, as we take this proton off, we're going to start to break  this bond.  And I'm going to draw dashed arrows to show where we're breaking bonds.  So because it's a chair, this substituent has an axial or an equatorial preference.  I'm going to sketch in those axial and equatorial groups.  And then back here, I'll go ahead and draw that nitrogen atom that is now so basic.  So there's a lithium atom that's joining the oxygen and the nitrogen.  And here's the isopropyl group.  We went through a lot of work trying, okay, let's try lithium diethyl amide.  Let's try lithium di-T-butyl amide.  They've tried every single combination and it turns out that isopropyl is just perfect.  It's very hindering but not too hindered that you can't adopt these kinds of transition  states.  And so when you adopt this transition state, these substituents here have a choice.  If you had a choice between having your methyl axial or equatorial, you'd rather have it  equatorial.  Because the H, the least hindered species that wants to be axial.  And the more hindered thing, methyl or whatever, is going to be equatorial.  And that's why you prefer this particular enolate.  And so you get, in this case, this would be the Z-enolate oxygen is opposite to methyl.  So you get the kinetically less stable enolate when you use LDA because deprotonation goes  through this chair transition state.  So we call this the Ireland transition state.  Robert Ireland formulated this, I think back in the late 1970s.  Okay, so let's go ahead and take a look.  I'll give you one sort of warning about this EZ.  Awful, awful, confusing nightmare about E and Z.  So first, and then I'll give you a warning about how sterics can override that.  So let me just first draw out a picture for these reactions.  So based on what I'm telling you, if you want to isolate the less stable enolate, you never,  ever, ever want to have excess ketone floating around.  Excess ketone catalyzes isomerization between cis and trans enolates.  What you always do is you put an excess of the LDA in your pot.  You make that first using butylithium and disopropylamine.  And then you slowly, yep.  Isn't that the E enolate that you drew?  Did I draw the, oh sorry.  Okay, maybe it's not confusing to you, but it's confusing to me.  So yes, it is.  Thank you for asking about that.  And as I'm telling you now, LDA gives you kinetically the E enolate.  Okay, so typically what you do is you add in your ketone or your ester drop wise to  a solution of excess LDA.  When you do this, there is no point in time.  At no point is there ever excess ketone floating around.  That would be a disaster.  So really you don't add LDA to the ketone, you add the ketone to the LDA.  We just refer to that as inverse addition.  Okay, so here's the results that you get.  If you take disopropyl, lithium disopropyl amide, and you add three pentanone to that,  you add that drop wise to your solution.  So you start off with a slight excess of LDA.  What you'll get is not very good selectivity if you take a ketone like this.  Of course you wouldn't really commonly take a ketone like this for modern synthesis.  So you'll get a slight excess of this isomer here, this trans isomer.  And that's not really very good.  You see great selectivity.  You see great selectivity in case of esters.  If this is a little bit smaller, then it's a lot better to form this E isomer.  Let me not say E plus C, let me say E slash C.  And it's about 2 to 1, 70, 30 favoring the E isomer that I've drawn there.  So the O lithium is trans.  This is what you get based on that Ireland transition state.  If you have an oxygen on the other side here, oxygen is smaller than ethyl.  You get much better selectivity.  And so now it's 94 to 6 ratio favoring this enolate.  And then finally, if you really bulk this up, you could screw up that Ireland transition state  and make it not really all that great.  Like let's just imagine you put a T butyl or a phenyl group here.  So now you favor this enolate about 2 to 98.  So what's going on with the E and the Z issue?  So what's the higher priority group here, ethyl or oxygen according to Konning-Goldpraylog?  Oxygen is the higher priority.  So when I want to judge what's opposite to the methyl, I look at the higher priority group  and that's E. But now when I come over here, it's not the same anymore.  It depends on what you have here, what your substituents here on this position.  So in this case with the oxygen, it's 94% of the isomer that I'm showing you there.  In other words, it will flip back and forth between what you call E and what you call Z  depending on the Konning-Goldpraylog priorities of these substituents.  And finally, we come over here to this O-lithium species.  I didn't draw lithium very well there.  And now maybe it would be more helpful if I called this, well, I would still call this,  this is the Z isomer here because the higher priority group is O-lithium  and the higher priority group here is methyl.  But here the higher priority group is the ethoxy group.  So you have to worry about this nomenclature issue.  What is E and what is Z depends on whether you're working with ester enolates  or ketone enolates.  So the main idea is that you prefer these types of geometries due  to the Ireland transition state, but you can mess that up if this gets too big  and can spin it around the other way.  Okay, and you'll talk about that a lot more in Chem 205.  So the synthesis class.  So you look at LDA in solution or lithium-amide beta.  Or lithium-amide bases.  They exist mostly as dimer.  This is something that we've talked about before.  And so typically you make LDA in THF.  The ether oxygens of THF are coordinating.  So this is what LDA looks like in solution.  And so fortunately there's a small amount of this sort of a monomeric species  that's floating around in solution.  And that's what does all the deprotonation.  And I don't know what the ratios are, but the equilibrium massively favors this dimer.  Don't worry about drawing the dimer.  I don't want to see the dimer, but just remember  that it's aggregated mostly in solution like a lot of lithium enolates.  It's the small amount of this stuff that's present that coordinates to your carbonyl  and then goes through this Ireland transition state for deprotonation.  So you can mess up all of this Ireland transition state deprotonation business.  Let me come back and draw out.  You don't always have to draw a chair in this class.  So for example, if I asked you to deprotonate, draw a mechanism for deprotonation  of this ester you would end up drawing out something that looks like this.  You don't have to draw it in a chair shape in this class.  I won't ask you questions related to that.  But it's the small amount of the monomer that then coordinates to the carbonyl of an ester  or a ketone and that's what generates these sort of intermediate species  that then deprotonate through that six-membered ring transition state.  So I take those protons off and I just attach that to the wrong position  so I'm freaking out here.  There we go.  Ah, there we go.  Okay. Okay, so we can mess this up.  It's very easy to mess this up.  What I showed you before is if you really add something bulky here, like instead of oxygen,  if the t-butyl was directly connected, you can mess up this Ireland transition state,  the six-membered ring transition state.  But you can also mess it up by putting additives in your solution.  Things that coordinate and complex very powerfully to lithium will make it harder  and harder for your LDA to coordinate to your substrate.  So for example, here's another deprotonation.  And in this case, they're using hexamethyl disilazid.  So instead of iso, so what's the difference here?  Maybe I didn't do a good job of drawing this.  So the big difference is this is commercially available  and you can buy solutions to this.  LDA, you make yourself.  So if you've got the money, you can go ahead and buy this.  Obviously, trimethylsilyl has, occupies more volume than an isopobyl group  but the bonds are longer.  So the silicon is farther away from the nitrogen.  And overall, it's about the same effect as an isopropyl group.  So again, you can buy this but LDA, you can't buy, you have to make that.  So if you put an ester in with this in order to deprotonate the ester,  whoops, the favorite isomer that you get now is not the one  where the methyl group is on the same side as the ethoxy.  Now the methyl group is on the same side as the olithium.  And it's the selectivity is not perfect.  So the e to z ratio here is typically somewhere, I'm going to say above 80  but I'll say 80 greater than 90 to less than 10  for this particular example in this case.  And the secret here to getting this is they add HMPA to their solution.  You'll see this used a lot less.  There's other alternatives.  But HMPA is a very powerful coordinator of lithium.  Here's the structure for HMPA.  If you'd like, you can draw a double bond O here but I'm going to draw this  in its fitter ionic resonance form or the illid resonance form.  Just to emphasize the fact that there's a lot of negative charge on that oxygen.  And because of that, it loves to coordinate lithium.  And it keeps the LDA from coordinating to the substrate.  So you disrupt that six-member transition state.  And ultimately you end up getting the opposite isomer is predicted from that,  from that Ireland transition state.  Okay, so HMPA, that's it over there, hexamethyl phosphoric triamide, that disrupts.  So oftentimes if you see people who want to reverse their enolate selectivity,  they'll add some powerful coordinator like HMPA into the solution.  One gripe, legitimate gripe against this is this is a carcinogen.  It's a powerful carcinogen.  So people tend not to like to use those as co-solving.  It's usually high-polygon like so it's kind of like purified.  Oh, is it?  Yeah.  I've never used it before.  I don't know.  Let me show you one other way to make enolates.  This one is very specialized and totally not obvious.  Oftentimes I want to draw the simplest enolate and that's the one derived from acid aldehyde.  And you don't make the enolate of acid aldehyde by deprotonating acid aldehyde.  You make the enolate, so this is the enolate from acid aldehyde.  So you don't make it like this.  You don't make it by deprotonating acid aldehyde using LDA.  It turns out that acid aldehyde is so reactive that as you're making the enolate,  it starts doing aldol reactions with itself.  So if you want to make LDA, you never want to have more acid aldehyde present in solution.  What you do is you just put butylithium in THF.  So the converse of this is if you don't want to make acid aldehyde enolate,  don't leave your butylithium sitting around in THF at room temperature  because they react with each other.  So the reaction looks something like this.  Let me draw the, well, let me redraw this with a bond here.  That bond is basic and if you leave this sitting around long enough, it will do this.  And so you'll medilate alpha to that oxygen atom to generate this medilated THF.  This will happen at room temperature.  So let me just put in here that's the half life for that.  It's 107 minutes at room temperature.  So don't put LDA or don't put butylithium in THF.  It's like, oh, you guys are going to lunch?  Oh, let me go with you.  I'll just add the rest of my substrate later.  No, that's a problem.  So if you have more butylithium floating around, this can coordinate to that, right?  Lithium loves to have oxygens coordinated to it.  Yeah, it could be coordinated before if you're wondering,  but I'm going to draw this in this sequence.  And so now this can fragment.  This super-duper nucleophilic, this super-duper nucleophilic carbon lithium bond there can donate  into the antibonding orbital for this CC bond.  Who would have thought that, right?  You wouldn't have ever guessed that.  And if you donate electron density into this carbon-carbon bond,  if you draw out what that orbital looks like,  there's electron density on the backside that starts to appear.  And it's that electron density that comes in and donates into sigma star.  But I'm going to draw my arrow pushing like this.  Ultimately, you fragment to give ethylene.  Did I just do that right?  Yes, you fragment to give ethylene, and then you get your acid aldehyde enolate as a byproduct.  And so then you can pop off the lithium and the R group, et cetera.  And so that's what gives you the lithium enolate.  Okay, so be careful when you put butylithium in THF.  Don't let it sit around.  You'll end up with this problem.  Don't let it sit around at room temperature.  You'll probably note from procedures that they're generally doing this  at minus 78 reactions of butylithium in THF.  So I want to ponder on this.  You almost never will work with naked enolates.  There's almost always a lithium bound oxygen or something.  If you're working with transition metals, they tend to favor carbon.  But you're never working with naked enolates.  There's always a counteron.  But I want to ponder on this.  You can imagine reactions that occur on carbon,  and you can imagine reactions that occur on oxygen.  So let's talk about that.  How would you predict whether your electrophile is going to react  with the carbon atom or the oxygen atom?  Just come back to this issue of C versus O alkylation.  And I don't have numbers just for a plain enolate.  I have numbers for this stabilized enolate  where there's an extra carbonyl group.  This would be a beta keto ester.  And what I want to do is I want to imagine what will happen  if we take various ethyl electrophiles.  So we're going to do an SN2 reaction and displace the X.  But we're interested in this competition between attack,  let me pick a different pen color here,  between attack of the oxygen atom versus,  let me use a different pen color, attack of the carbon atom.  And how does X influence whether you're attacking  on carbon or attacking on oxygen?  So C versus O alkylation.  If you want a high degree of C alkylation,  then here's what you should do.  What you should do is you could, you should make this  as covalent and non-ionic as possible.  Anything you can do to reduce the ionic nature  of that oxygen will make it less capable as a nucleophile.  So if you want very high ratios of carbon alkylation  and less alkylation on oxygen,  then I would pick the most covalent bond possible here.  So let me draw out this, what do I have for the,  I'll just draw a metal here, metal plus.  In other words, is this metal floating off in space  like I drew or is it covalently bound?  If you choose lithium, that lithium is bound here.  Actually, it's chelated between the two oxygens.  That lithium is tightly bound to the oxygen.  If you want to make this more naked, like more,  more negative charge of oxygen,  then move down the periodic table to sodium, to potassium,  to cesium, which in some cases you'll use.  And then finally, about as naked as you can get that enolate is  something that, sorry, I'm trying to draw tetraethylammonium.  It's a tetraalculammonium cation which can't coordinate at all  to that O minus.  In that case, it's a naked O minus.  So in that case, what you would get is mostly O alkylation.  There's the ethyl group.  If you use tetraethylammonium, whereas if you use the lithium  enolate, you'll get mostly C alkylation.  And there's the new ethyl group on the carbon atom.  So the more covalent you make that oxygen metal bond,  the more covalent the less negative charge here,  and the more you're going to see reactions on carbon.  Okay, so what can we do to change the electrophile?  So if I want to attack, if I want to favor attack here  on this carbon, let's just review what is it  that makes this oxygen so nucleophilic?  In that case, it's the negative charge  that makes reactions favorable.  They're both nucleophilic.  This oxygen is nucleophilic because it has lots  of negative charge.  This carbon is nucleophilic because the HOMO is biggest  at this position, at this carbon atom.  If I look at the HOMO, let me just sort  of sketch out my little MO diagram here, MO.  And there's all these orbitals in there.  But there's this one orbital that's highest in energy,  that's the HOMO.  And what does that HOMO look like?  What do those two electrons, where do they spend their time?  It turns out those two electrons  in the highest occupied molecular orbital spend most  of their time centered above this carbon.  So the carbon reacts because the most reactive electrons spend  most of their time there, the oxygen reacts  because it's got more negative charge.  And those are completely different reasons.  So if you wanted to make some electrophile react here faster,  I wouldn't want to have an electrophile that had a lot  of negative charge.  I would want to have an electrophile  that has very little negative charge.  Sorry, very little positive charge.  So if I take something like an alkyl iodide,  carbon has about the same electronegativity as iodine.  There's very little partial positive charge here.  This reacts quickly because the LUMO is very big  on that carbon atom.  That long bond makes that LUMO, the sigma star,  very big on the back end.  And as I make my electrophiles more and more  with electronegative groups on there,  bromine is starting to become more electronegative  than carbon.  Chloride.  Now I start, and now as I make my leaving groups,  be more and more electronegative, a phosphate.  So now that's very electronegative.  Now there's a lot of partial positive charge here  on that carbon atom.  This reacts for, is very reactive  for a very different reason.  And finally at the bottom of my list here,  I've got the tosylate, the sulfonate leaving group.  Now there's a great deal of partial positive charge here.  The more partial positive charge here, the more I'm going  to see reactions that are governed by charge,  and that means more O-alkylation.  So I've got a set of numbers here  to put things into perspective.  So when X is equal to tosylate, ethyl tosylate,  you get a C to O ratio of 11 to 88.  In other words, you get more O-alkylation  and then you get C-alkylation with ethyl tosylate,  because it's mainly driven by charge.  When X is equal to iodide, the leaving group there is equal  to iodide, that doesn't mean care of.  It means carbon to oxygen-alkylation ratio.  It looks like a male term.  So when you use ethyl iodide, you get more C-alkylation.  So in other words, you don't totally reverse it to 99 to 1,  but you can flip that balance back and forth.  Okay. So let me go ahead and just summarize what I'm telling you  about which things will react where.  And I'll try to summarize that in two different ways.  So how does this matter in synthesis?  So if you want things that are going to react here on oxygen,  it would be things that have a lot  of partial positive charge.  So for example, if you take anhydrides,  they react faster on oxygen.  You don't get C-acylation.  Let me make that attack over here more clearly.  You get O-acylation if you add a c-decan hydride.  If you take TMS chloride, there's a lot of partial positive  on that silicon.  You get O-silylation.  You don't get a lot of C-silylation.  You make sililenoethers.  They react faster on oxygen.  If you take things like alkyl halides, like that tosylate,  those react faster on carbon.  And of course, that's what you care about mostly  in organic synthesis.  You care about making carbon-carbon bonds.  Organic synthesis is dominated by reactions like this.  Take alkaliadides or aldehydes.  That's an aldol reaction.  Those react on carbon faster.  So those reactions are driven by HOMO-LUMO interactions,  not by the fact that there's some amount  of negative charge on carbon.  Okay, so that's the big idea.  Let's just review what we're saying about enolates.  I'll go ahead and sketch out this picture of the HOMO.  So if I look at the most reactive pair of electrons  and we're looking down on top of the HOMO, there's this sort  of set of orbitals that have pi symmetry above  and below the plane of the board.  And so in that HOMO, that pair of electrons spends most  of its time above that carbon.  It's bigger on the carbon than it is on oxygen or here.  And tiny differences in the size of that orbital  on carbon versus oxygen translate  to huge differences in selectivity.  If I were an electrophile looking for electrons  and I wanted to overlap with an atom, I'd be wanting  to overlap over here on this end  because the most reactive pair of electrons spend their time there.  So if you want reaction, so if you want C-alculation,  you want electrophiles that have small partial positive charge.  You don't want really charged electrophiles.  And you want electrophiles that don't have,  and you want electrophiles that have low lumos  because it's a HOMO-lumodriven interaction.  In contrast, if you want to react with this,  then you want things that have a lot  of partial positive charge, right?  Reaction on oxygen, O-alculation is a charge-driven reaction.  And you don't want it to be driven by HOMO-lumodriven interactions.  So you want something with a high lumo.  If I were to pick something that would really react quickly  on oxygen, it would be like this.  There we go.  That will react really quickly on oxygen,  much faster than on carbon, right?  Of course, that's not very useful synthetically,  but that would be an example  of something that's all charge-driven.  Okay, so sometimes you need tricks around this,  or you need to find the right reagent balance.  What happens if you do want to see acylate on the carbon  of an enolate?  So remember what I just told you, that if you take anhydrides,  those will tend to form enol acetates.  They react faster on oxygen.  That's the general principle.  If you do want to add an acyl group here, then what you do  in this particular case, well,  it won't be exactly that group,  you can use something called Manders reagent.  This is a very tricky sort of electrophile.  Instead of having an oxygen-leaving group here  on this side or a chloride-leaving group,  use a carbon-based leaving group that doesn't put a lot  of positive charge on carbon.  This is not super electronegative like chlorine  or oxygen, yet it can still act as a leaving group.  And so as a result, when you use this kind of,  this is called Manders ester,  you end up getting acylation on the carbon atom.  And it's just a little trick  of replacing your typical chloride-leaving group  with a carbon atom that's not very electronegative.  Okay, one last thing to finish up with this.  Parking's back to this example that I showed you over here  with this boron.  This is the way aldol reactions work.  So if you take some sort of a carbonyl compound  and you convert that to the lithium enolate,  that may not look like a boron,  an aloe boron species to you, but it's the same idea.  I'm now just sort of flip this around.  This is very similar to that aloe boron species.  Instead of boron, we've got a lithium atom.  And lithium loves to coordinate to oxygen.  And so let's draw that coordinated species.  Even if it's present in tiny, tiny amounts,  this coordinated species now has a more reactive electrophile  and it has a more nucleophilic nucleophile.  Now that that's a lithiate,  that whole enolate part becomes more reactive.  So the important one is even if there's a tiny amount  of this in solution, even if it's mostly dissociated  like this, even if there's a tiny amount of this,  all of your C-C bond-forming reactions go  through this mechanism, six-member transition.  And it is also a chair.  And it's that chair, like nature of the transition state  that allows you to predict stereoselectivity  in aldol reactions.  If you take synthesis, chem 205, you'll learn 204.  I think you'll learn a lot more about that.  Okay, I'm going to finish up by showing you,  I'm going to switch over to a sort of slide here.  Actually, let me, so we just gave a handout in class.  Everybody got this handout  that shows something called the Meyer reactivity tables.  And I want to talk about this,  and maybe I can pose it as a question.  I want to talk about the power of these tables.  I've shown you a bunch of enol ethers and enamines  and enolates and which ones are most reactive  and which ones are least reactive.  And I've shown you a bunch of electrophiles,  carbocation, things like that.  And so at some point, you're going to get down to wondering,  gee, is that a good reaction or is that a bad reaction?  So we've talked about enol ethers, enols, enamines.  Here's an enol silo ether.  It's a very common nucleophilic species.  Let's suppose I mix that in with this aminium ion.  We know that aminium ions are reactive.  Well, that reaction occur.  Right? We've learned all this stuff about organic chemistry.  And we still can't answer the simple question  of whether this reaction of a good nucleophile,  that's supposed to be a bond.  I just made it a little bit too thick.  We still can't answer the simple question  of whether a nucleophile will react with an electrophile.  And that seems kind of pathetic.  How can it be that nobody's ever measured that?  And the point is that Herb Meyer measured this stuff.  He's a hero in the field of organic chemistry.  So the question is, does this have enough oomph?  And does this have enough tug in order  to form a carbon-carbon bond?  And so here's how you address this.  Let me try to turn on this.  In other words, to answer that question,  you need somebody to take their group  and tell everybody in the group, okay,  you're going to measure kinetics for the next two years.  Right? Who was able to get anybody to do that heroic work?  And Herb Meyer did it.  Okay, pause.  Sorry, I got to let this warm up here.  Okay, this is our course website.  You guys probably recognize this.  I'm going over to links here.  There's a links page.  And on the link page, I added a link  to Herb Meyer's logarithmic reactivity scales.  And he had to deal with this.  He wanted to address this simple issue of why can't we just look  this up in some table to figure out which reactions work well  and which reactions don't work well.  And so when you go to that link to Meyer's reactivity site,  he's got all kinds of tables.  I'm going to click on the first one.  Just to show you, I wish there was some way to go  to full screen without the, whatever.  I don't know how to do that.  Okay, so here's the Herb Meyer reactivity scales.  Let me try to explain what you're looking at.  He's got this whole set of numbers.  Over here on this side, he's got all the nucleophiles.  Let's just take a look to make sure we can identify the nucleophiles.  Oh, there's just a simple double bond attacking things.  And it's way down, well, it's inverted here.  The worst electrophiles are on top, sorry,  nucleophiles are on top, and the best nucleophiles are on the bottom.  So as we get down to the bottom, look, there's an enol silo ether.  That's more nucleophilic than a regular alkene.  Oh, look, here's an allyl stannane and an allyl silo, an allyl silane.  Those are better nucleophiles than simple alkene.  So as you walk downward here, nucleophiles get better and better  and better and better and better.  And this is logarithmic.  These are factors of 10 differences as you go from minus 6 to plus 6 all the way down.  Okay, so let's just try to gauge.  What's the difference in nucleophilicity  between just a simple terminal olefin and an allyl siloane?  How many times more reactive is an allyl siloane  than just a simple terminal olefin?  So the way to answer this is I'll come over here  and I'll look at a terminal olefin,  and that has a nucleophilicity parameter of minus 2.25.  Forget the second number there for the moment.  Now let's find an allyl siloane.  Let's just find a simple one like allyl trimethyl silo.  That's 1.79.  So what's the difference between 1.79 and minus 2.2?  I can't do that math so quickly.  So it's about four.  Four orders of magnitude.  Immediately these tables are useful.  Allyl silanes are about four orders of magnitude more nucleophilic  than simple C-C double bonds.  Well, it matters what electrophiles you're looking at.  And so what Meyer did that made these tables useful  to organic chemists is he looked at reactions  that organic chemists care about.  He said, I'm going to look at reactions that form C-C bonds.  So he started off by taking Ben's hydro cations  and he started looking at all kinds  of nucleophiles adding to that.  So most of these reactions that he looked  at with nucleophiles, he was looking at the attack  to make a new C-C bond.  And it's not that big a stretch to say, well,  that's the same as if I look at this, right?  It's the same idea.  It's a stabilized carbocation.  You could redraw that as a stabilized carbocation.  So his tables are valid for people who care  about C-C bond formation.  And in general, that's you.  So you can gauge relative reactivity if you walk  up and down this table, relative nucleophilicity.  So let's look at the most reactive nucleophiles  that he measured.  He couldn't measure everything, right?  He didn't measure butylythium.  That's off the charts.  But some of the more reactive things he has here are  stabilized enolates.  He doesn't have any unstabilized enolates.  That would be even lower than his measurement techniques allowed.  He's got simple amines on here.  He's got enamine somewhere over on this side.  So you can gauge reactivity.  And it's very powerful oftentimes to say, hey,  this is 10 orders of magnitude more reactive  than that, more nucleophilic.  Okay, conversely, he also made measurements for electrophiles.  He took enol-cyl-lethers and allosilanes and looked  at their reactivity with all kinds of electrophiles.  And so let's come back up to the top of his charts here  for the most reactive species.  The most reactive species he looked at were simple,  well, these aren't simple carbocadions.  We're just resonance stabilized carbocadines.  So here's a diphenylmethylcarbocadion,  carbenium ion.  And if you want to gauge, well,  how much does a methoxy group stabilize that  and tone down the reactivity?  You can come down and look.  And if you walk across here, it says 2, it says 3,  so plus 3 versus a diphenyl, which is about 5.9.  So that's about 3 orders of magnitude less reactive  when there's a methoxy in place of a benzene ring.  So you can gauge the relative reactivities of electrophiles  and specifically the kinds of electrophiles  that form carbon-carbon bonds.  What's powerful here is that you can predict the rates  of reactions at room temperature.  You can always extrapolate to higher or lower temperature.  Things are faster at higher temperature  or slower at lower temperature.  But at room temperature, without having going into the lab,  you can predict which reactions are faster  or slower by plugging in these numbers.  You've got an electrophilicity parameter.  You've got a nucleophilicity parameter here  that's these first numbers.  The second number here is sort of like a fudge factor  that varies just a little bit.  They're usually close to 1, but they're not always 1.  So look over here, there's this fudge factor  that's exactly 1, which means it has no influence.  But over here, there's other fudge factors.  For example, furan has a fudge factor of 1.31.  What that says is that furan is kind of picky  about the electrophile.  The sensitivity of that nucleophilicity depends  on which electrophile you're reacting.  So that's kind of like a sensitivity parameter.  I usually just forget that because I'm too stupid to take  that into account and do this simple arithmetic.  There's a very powerful rule of thumb  that you can use associated with these tables.  And here's the rule of thumb.  And that doesn't look like that pin is working too well.  So the rule of thumb is that you can expect reaction  at room temperature at reasonable rates like within a day.  That would be a reasonable.  That's what we mean by react.  If the sum of these two numbers is greater than 5.  Now very rarely in organic chemistry do you have this kind  of simple predictive power.  Right? This is what you want to get out of this class.  You want to be able to predict which reactions are fast,  which reactions are slow.  And if you're not sure, if you didn't get an intuition by sitting  and listening to the lectures, you can go to these tables  and look these kinds of things up.  If things are really electrophilic or super nucleophilic  or both, then those numbers, this number will get higher  and higher and higher.  Better electrophiles will make this number higher  and better nucleophiles will make this number higher.  And so if the number is greater than 5,  if that sum is greater than 5, then you can expect things  to react at room temperature.  Okay, so that's it for CC double bonds as nucleophiles.  And you've got lots of problems on the problem set  and help you practice that stuff.
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