 Okay, so let's pick up where we left off. We've been talking about alkali organometallics where you have organic compounds bonded to sodium. Thank you, Chris. Lithium, potassium, even magnesium, different column, but same idea. And so we talked about some various methods for accessing things like alkali-lithiums or grignard reagents. The first method that we mentioned is just direct reaction with metal, sort of like a reductive lithiation for organometallic chemistry. And really, you don't use this in the lab. This is not typical. This is the way they prepare commercial reagents, the cheap ones that you buy on huge scale, butylithium, methylithium, phenylithium, stuff like that. So that's not the kind of thing you would do in the lab. Traditionally, in organic synthesis, if you're making very complex organolithium reagents, you're usually most often using something like this. You do a lithium halogen exchange. You take cheap butylithium that you buy and then instantaneously, essentially, you react that with aryl iodides, aryl bromides, vinyl iodides, you get exchange to make the vinyl lithium or aryl lithium that you want, starting with cheap butylithium. That's more commonly the way that you will make an alkylithium in the lab. There's another exchange reaction, tin lithium exchange, where you can take any sort of a standing compound, for example, this one. You couldn't have a halide here because the lone pairs on oxygen would push that out. But you can make these kinds of alpha-stannyl ethers and do lithium tin exchange if you want to make this kind of alkoxy alkyl lithium. So those are three common ways to get access to alkylithiums. And so the fourth way that we'll talk about that is very common is reductive lithiation. This is not an simple exchange reaction. And so here's the way this typically works. The basic idea is that if you take lithium wire, that's the way you buy lithium. You buy these coils of wire stored in oil so it's not in contact with air or oxygen. And then you simply mix that with naphthalene, very cheap. This is what they make moth balls out of. You get something called lithium naphthalenide. Lithium naphthalenide. And so how do you think of what would you get if lithium dumped its electron into this system? How do you even depict that? We don't have a good depiction for that. Let me give you one way of thinking about that. I'll simply erase this bond here and I'll draw that pi bond as two electrons. And so now if we imagine dumping one extra electron, we can at least see where we might put that extra electron. And the problem is we have so many resonance structures we can draw that it just becomes that nobody draws it. So here's one resonance structure. Let's suppose I put that extra electron over here. Now you just imagine all the other resonance structures you can draw, maybe that would be a good exercise. And then you can get a sense for what lithium naphthalenide looks like. Okay, so it's very easy to make lithium naphthalenide just simply by mixing lithium wire with cheap naphthalene. And it is the most beautiful and intense blue you have ever seen in your life. And the beauty of this reagent is that it's super easy to make just by stirring these two things together. And then any place on the planet where you have a thiophenyl group, like here, for example, you simply drip in this reagent and instantaneously you get conversion from thiophenyl to lithium. So here would be an example. You take this lithium naphthalenide reagent and for whatever reason I'm going to draw out this weird structure that kind of looks uncomfortable. Minus 78 as you drip this in and there you go. You've got an alkyl lithium that you couldn't make from a halide that you could or easily you couldn't make that from a halide or you don't have to go through some toxic tin species. Very easy to put, try to put two of these electrons closer on one carbon. I hope you did a better job. At least I want to be clear or whatever. There's an extra electron in there. You can draw the resonance structures. Now one of the problems with this is at the end of the reaction you've got just a ton of naphthalene left over. There's just a ton of naphthalene in there after this reaction is over. So in order to make it easier to remove that a lot of people start with this naphthalene derivative and then they can just wash it away with an acid wash at the end. You know it slightly changes the reduction potential, your ability to put a dump of electron in but the main point here is you just want an easy way to do an acid wash and get rid of all this greasy naphthalene that's going to junk up your column. Okay so here's what's cool. So if you want to think about what's going on I'll give you a little snippet like react, a piece of reactivity. I'm not going to draw a full mechanism. We're not really talking a lot about radical mechanisms. You know as soon as I start talking about this naphthalene we have to start talking about donating an electron somewhere. So I'll give you one way to think about the initial structure. If you have a thiophenyl group attached to anything like T-butyl here's a way to think about what's going on. There's lithium cations floating around. They will coordinate to this sulfur and so if that dumps an electron into this benzene ring so let me just draw one of the double bonds in benzene as just two electrons. You can imagine that I could have drawn that as a pi bond and so what would happen if I put an extra electron in there? I put the lithium cation here. I put the extra electron over at one of these places and here's where I'll draw it like that. And now you can imagine this thing fragmenting. You can imagine this thing fragmenting in such a way that you would end up initially with a radical here on your carbon atom. And then fragmenting so I'd give one electron to here and I'm not going to show you how I get to that resonance structure but the resonance structure that I get to for this thiophenyl species is lithium thiophenoxide. So when you fragment this you get a carbon radical that desperately wants another electron and lithium is happy. The lithium naphthalenide is happy to dump an electron into a T-butyl radical and then I get this byproduct and that byproduct is floating around in your reaction mixture. It's not hurting you. Won't be more reactive than the alkyl lithium that you make. As a consequence of this, because the whole idea behind this is it involves formation of a radical intermediate that then gets converted into an anion, into an alkyl lithium, as you see this set of relative rates, do you get the fastest rates for reductive lithiation for tertiary alkyls? Remember tertiary alkyl lithiums tend to be much more reactive, screaming hot reactive than primary alkyl lithiums. But if you start with a thiophenyl, they form fastest. Alkyl lithiums form fastest when you have tertiary alkyl. That's faster than just primary. A thiophenyl is getting converted into primary alkyl lithiums and that's faster than, and in fact very hard to do this because it's very hard to make just a phenyl radical where there's a single electron in an orbital sticking off the side of a benzene ring. So this would be very, very slow to do just a thiophenyl ether conversion. Okay, so that's method number four, reductive lithiation. If you want to make an alkyl lithium, just put a thiophenyl somewhere, anywhere, hexane, t-butyl group, you'll be able to convert that into an alkyl lithium. Okay, so I'm going to come back to this idea of lithium halogen exchange. And you can make the same arguments for tin lithium exchange, but let me just remind you of the mechanism for this reaction. The important point behind this reaction right here, this exchange reaction and what makes it work is the fact that there's an equilibrium that favors this side and I want to talk about that equilibrium. Again, if you want to draw the mechanism for this reaction, it would look like this. You would have this initial eight formation of an eight complex, two bonds to the halogen. You'd have this iodate and now this bond, both of these bonds become nucleophilic and so you can imagine this attacking lithium and that's how you get the product. So, but let's forget the arrow pushing mechanism for now. I mean, I'd like you to know how to do that, but let's forget the arrow pushing mechanism and just focus on the thermodynamics here. What should we expect if we take a series of alkyl lithiums and we allow them to exchange? So somebody measured an entire series of compounds to figure out in which direction should this equilibrium be most powerfully directed? So everything here is compared to phenyl lithium formation. Let's go ahead and make a table where we look at the equilibrium constants for this reaction and I'm trying to write KEQ up here. I'm going to give you equilibrium constants for this process and so we're going to start off here with formation of vinyl lithium. And what we would find is that the equilibrium constant for this is very poor. Then finally phenyl lithium and I'm, let me try to be a little bit more space conscious here because I've got several compounds to show you. Cyclopropyl lithium, finally butyl lithium, not finally, but butyl lithium and then a secondary alkyl lithium. You're never going to make or work with cyclopentyl lithium but you might work with sec butyl lithium. That's a secondary alkyl lithium that's very common. So imagine taking away one of these carbons and just having a butyl, a secondary butyl group. So now this is a very powerful donor for lithium. So if you really want to drive lithium halogen exchange, go to a secondary alkyl lithium. If you want to drive lithium halogen exchange, you wouldn't just put in vinyl lithium, the equilibrium would lie on this side. Obviously if you put in phenyl lithium and phenyl iodide, you'd have a 1 to 1 equilibrium constant and if you really want to make a phenyl lithium, you start to use alkyl lithium. Butyl lithium will very, will, will drive this equilibrium toward phenyl lithium formation and finally the best would be a tertiary alkyl lithium. If you really want, and I don't have an actual number here, all I can say is it's greater than 10 million. It's greater than a secondary alkyl lithium. When you really see people want to absolutely drive alkyl lithium formation from a halide, they throw in T-butyl lithium as their reagent. So it's a pretty amazing, pretty amazing reaction and what's amazing is that this tin halogen exchange, the attack on iodide that I showed you, this attack of an alkyl lithium attacking some sort of a halide and I'll just write arrow here. Well, let me draw another alkyl group. This attack like this on iodine to make this iodate is faster than SN2. That's what may be strange to you. It's vastly faster than an SN2 reaction. So in other words, you don't get this happening. I don't even want to put a regular arrow. I'll put a dashed arrow there to try to give you an idea. You don't get SN2 in these types of processes. You keep the temperatures low. You'll just get straight up exchange of lithium per halogen. Let me try to switch pens here because I feel like I'm running out. Okay, so very powerful and very easy to simply add butylithium to your alkyl iodides and convert those into alkylithiums. Okay, let me show you what I feel is the most obvious way to make something reminiscent of a carbanion and that's deprotonation. Can we just take alkanes or organic compounds and deprotonate them? And what are the limitations? Mechanistically, that seems like the simplest mechanism to me is simply deprotonating CHs. And the term that we're going to use for that is medallation. And so let me go ahead and come back and remind you. Like if you wanted to simply deprotonate an alkane, you want to obey thermodynamics. If you simply wanted to make an alkylithium like this by deprotonating ethane, you have to remember, you need something more basic than ethane. If you really wanted thermodynamically to have a plausible process. And it's hard to think of things that are that much more basic than ethane. But it's not hard to think of things that are more basic than ethane. You could just take an alkylithium and it ought to be able just, butylithium ought to be able to thermodynamically deprotonate these protons, these sp2 hybridized, these protons on sp2 hybridized carbons. And that ought to be way more, it ought to be even more facile to deprotonate the end of an alkyne. Just using something that's more basic than an alkyne. And so what kinds of things can we use? So typically, I don't have room here. I misallocated my space here, but, you know, you could use an amide anion to deprotonate. So let me remind you of the pKa's for this. So, and really this is a pKa prime. pKa prime means what if I had a proton there instead of lithium, right? I can't ask about the pKa of this because this isn't an acid. But if I stuck the H back there and you consider them acid, now I can talk about the pKa and we call it the pKa prime. So the pKa for simple alkanes is about 50. We talked about these pKa's. pKa's for an sp2 hybridized vinyl group is about 41 to 42. So for a simple terminal alkyne, it's about 24. And then I want you to remember this later. The pKa for, and we talked about this, for just a simple amine. To deprotonate an amine, like diisopropyl amine to make LDA, that has a pKa of about 36. Okay, so this is the thermodynamics. Let's talk about a reaction that is just completely impractical. Basically what I'm telling you is that thermodynamically you can do this. And I'll draw the H there. What I've just told you is that it's thermodynamically favorable to deprotonate benzene rings with butylythium, right? By nine orders of magnitude, the equilibrium constant favors formation of the phenylithium. But you have never seen anybody deprotonate benzene with butylythium. You've never seen that. And the problem is not thermodynamics. The problem is kinetics. It is just too slow. So there's no problem. Of course you could draw the arrow pushing mechanism where you take these super nucleophilic electrons in this bond here and attack the proton. It's just too slow. Okay, so how acidic do things have to be before you can do just simple metallation? Just simple deprotonation reactions of alkane type compounds. And I'll show you an example of where it can be acidic enough. And it's right here. Probably if you took a sophomore organic chemistry course in the United States, one of the first alkyl anion reactions you learn about is deprotonating acetylides. Just direct deprotonation. And you don't have to be very basic to have good enough rates for deprotonation. You can take butylythium and deprotonate in alkyne. You could take even a grignard, which is traditionally very nucleophilic but not very basic. And even this occurs at a fast enough rate. That's what's astounding. Grignards are not super basic. They're very nucleophilic. But it's surprising that even a grignard reagent is basic enough to pull off that proton at convenient rates. And I don't want to draw that as a lone pair. I'll draw it out like this. And so the byproduct of this would be ethane. Okay, so of course the equilibrium is favorable. You can look at the pKa's over there and see that this equilibrium favors this side by 50 minus 24 by 26 orders of magnitude. But what's amazing is that the rates are good. So it's very easy to deprotonate alkynes. Surprisingly easy. That's the basis of a sonogashira coupling. So in a sonogashira coupling, you take alkynes and palladium and triethylamine and copper. And under those mildly basic conditions, you're deprotonating that alkyne. I'll show you a surprising sort of factoid. So if you take sodium ethoxide and ethanol as the solvent for this and you do this substitution reaction, you know, so what's the mechanism for this substitution to replace this chloride with an ethoxy group? What's completely amazing is that if you measure the rates, the rate is first order. There's a first order dependence on the concentration of sodium ethoxide. Why is it that if you add 10 times more sodium ethoxide, the rate gets 10 times faster? What that tells you for certain, it is not a simple SN1 reaction, right? For SN1 reactions, the rate has nothing to do with the nucleophile. It's how fast you form the carbocation. So it can't be an SN1 reaction, but it's also pretty unlikely that this is SN2 to attack at the tertiary center. It turns out that what's happening here is that the mechanism involves deprotonation initially to make this terminal alkyne. Who would have thought that? Nobody would have thought that, right? You wouldn't have thought that. I wouldn't have thought that. And now the loss of chloride is very fast. It's slow here to make a carbocation, but now way over there that a satellite anion makes it now fast to remove the chloride. And I'm going to go straight to a weird looking resonance structure. In other words, let me now use this with that negative charge and now this pi bond becomes very nucleophilic and it helps to push out that leaving group. And what you get is a carbene. That's now a carbene. This is weird. There's two, four, six electrons on this system. It's like a carbocation and a carb anion all on one carbon. And that reacts very quickly with nucleophile in solution. So you can draw it as the sodium ethoxide that's attacking or I'm not going to show all the steps but or ethanol attacking but that is now very quickly attacked. Not on this end but attacked back over here where the leaving group left. And so then you can redraw this and again I'm not going to show all the steps going back to this final product. So how do they know that this carbene is there? Well what they did is they ran this reaction in the presence of compounds that have double bonds and what they showed is they were able to get cyclopropanes instead, right? You can't explain this kind of reaction formation of these products without invoking a carbene. Unless you invoke a carbene you're not going to get to this. So that's how some other major piece of evidence that there's a carbene involved. Okay so very difficult to pull protons. There's not a problem thermodynamically with pulling protons off of sp2 centers. The problem is one of kinetics. So thermodynamically butylithium ought to be able to make things like this simply by deprotonating. With alkynes it is fast enough. You've got a huge thermodynamic favorability for pulling protons off the ends of alkynes and it's kinetically facile. So let me show you a trick, a simple trick for deprotonating, directly deprotonating sp2 hybridized, protons on sp2 hybridized carbons. And it's all about entropy. Well maybe not all about entropy. It's partly about entropy. And the basic idea is this. It has to do with prior coordination. And we're going to see this idea over and over and over again in aldol reactions and enol boronate chemistry. So let me start off by simply laying out a problem, a synthetic problem. And there used to not be a good solution to this problem. So if you wanted to do electrophilic aromatic substitution of phenol for example with a simple electrophile like methyl iodide, how would you do this? So first of all it's not easy to make methyl cations. You can under some conditions do Friedl crafts with methyl halides, but it's not easy to do this. And if you did some Friedl crafts thing, how do you stop from getting the para substituted product? So an electrophilic aromatic substitution, you get mixtures of ortho and para. And so here's how you get around this. The way you do this is you use deprotonation. And you use this oxygen atom in an indirect way to help you deprotonate that CH. So what you do is you first convert your phenol into some sort of an, not an ester, but an ester like derivative. You need something that's stable. So in this case they make a carbamate because the lone pairs on nitrogen are donating into the carbonyl. It makes it hard for things to attack that carbonyl. When you have that nitrogen next door, let's call it a carbamate. And this is one of the better derivatives for helping bases come in. This is one of the better functional groups for helping bases come in and deprotonate that. So when you treat this with butylithium, which is a good nucleophile and a good base, the fastest thing that happens is not for the carbon-lithium bond to attack the carbonyl. That's not the fastest thing that happens. The fastest thing that happens is that this lone pair comes in and attacks the lithium atom. In other words, these things start to associate with each other. And watch what happens when we do that. So there's my butylithium. It's now coordinated to oxygen. And I have to put the charges right here if I want this Lewis structure to be correct. If I put a third bond to oxygen, that's now an oxonium species with a positive charge there. But more importantly, the lithium is now a lithiate. And as soon as I draw that negative charge there, I realize, wow, in contrast to the butylithium I started with, now every single bond to this lithium is now more basic and more nucleophilic. That's what that negative charge tells me. It's now more reactive when it's coordinated than when it's just floating around. In other words, when it's close to these ortho positions, that carbon-lithium bond is now more basic. And so when it's coordinated, you have faster rates of deprotonation than you do when the butylithium is just floating around and colliding with stuff. So this is the intermediate that you get. And so ultimately, you can draw a lone pair hanging off that phenyl and then have that re-pick up a lithium ion. So you end up getting this, you know, the lithium that's dangling over here now re-coordinates, so I'm skipping that step. You can fill in that extra step on your own. And so now you get this intermediate that looks like this. And then there's a lithium. So in other words, in the final product, this oxygen is still coordinated to that lithium. And so now when you throw in your electrophile, like methyl iodide, this is very reactive with electrophiles, that carbon-lithium bond. It doesn't have to be methyl iodide, but you could use that as your electrophile if you wanted. So now you use this to attack whatever your electrophile is going to be, there's that nucleophilic carbon-lithium bond. There we go. And now at your leisure, you can hydrolyze that carbamate back down to an alcohol. So this is one common way to make methylation work. You take slow reactions and the simple trick of having a coordinating group that soups up the reactivity now speeds up reactions nearby. So we call that directed. And there's all kinds of groups you can have here. They generally involve, you know, five member, four, five member, six member type ring transition states to bring this basic bond close to the CH that you want to be programmed. Okay, so let's extend this idea of the importance of this type of pre-coordination a bit. What's the hybridization of these carbon atoms in ethane? Now SB3, what's the hybridization of these carbon atoms in an alkene? SB2, what's the hybridization of this carbon atom here in an alkyne? SB. So the more S character, right, 50% S character versus 25, the more S character, the easier it is to deprotonate. The more S character, the easier it is to deprotonate. The less P character, the easier it is to deprotonate. So let me just remind you of hybridization in a cyclopropane ring. When you look at the hybridization, in order to pinch down to make these 60 degree angles, you end up with these carbon, carbon bonds here being about, and SP5 seems weird, it's about 80% P character is the main, I don't even know if I'm doing the math right, but there's more P character than normal in this bond. What's important is if you're using extra P character to make these carbon, carbon bonds, you have less P character to make these bonds outside. The hybridization for these bonds outside is approximately SP2.2. It's closer to SP2 hybridized. Those bonds on the outside of a cyclopropane are more like this. It should be easier to deprotonate cyclopropanes than it is to deprotonate just a simple alkane. It looks SP3, but those aren't SP3. Okay, so keep that in mind. Okay, so we're going to talk about two reactions of oxyranes or epoxides, and let's go ahead and start off with this very traditional reaction. You take an epoxide and you treat this with a lithium amide base, and the question is what happens? So what happens with an epoxide and a lithium? So lithium amide, well, it's nucleophilic, it's basic, but you don't use lithium amides as nucleophiles. I can't think of any case you would purposely ever do that. When you use lithium amides, you're using them as bases, and here's the base-promoted reaction that you see in this case. It's an E2 elimination. Right, there's a pro here. Here's the leaving group. The carbon-oxygen bonds are leaving groups, and there's the proton that gets pulled off. So that's an E2 elimination, and lithium amide is a very powerful base. Here's what's going on in this reaction. In other words, it's not that this lithium amide comes along. What's the most basic part of this? It's the lone pair. It's not the nitrogen-lithium bond. It's the lone pair that's the most reactive part, but that's not the first thing that happens. The first thing that happens is that the lone pairs on this oxyrane come along and coordinate with that lithium. And when it does that, remember, let's keep track of our charge as I now have a positive charge on oxygen. What does that do, that positive charge on oxygen? It makes these carbon-oxygen bonds easier to break. It makes the oxygen a better leaving group. And at the same time that you're making an oxygen better, a better leaving group, let me finish the other part of this, you now make this nitrogen more basic. Every part of that nitrogen atom attached to that lithium now becomes more basic. So this bond is easier to cleave. This nitrogen atom is now more basic. And so now you have this accelerated deprotonation. The important part here about this is that you end up with a, there's a requirement for a syn elimination. And they've shown this by putting deuterium on one side and proton on the other. They've shown that in every case it's the proton on the same face as the epoxide that gets pulled off. It's not an anti-elimination like an E2, so it's not really an E2. E2's are anti. This is a syn elimination. Okay, so that's one reaction of epoxides with bases. And again, it looks kind of like an E2, so that doesn't really, that's not a mind bender. I didn't do a good job of drawing this proton here, so let me redraw that. There we go. Sorry for that crummy drawing. Oh, and I forgot to break this. There we go. Okay, so epoxides, you can do eliminations with them. You relieve ring strain. That's cool. So again, if you were able to resist the temptation to do an SN2 reaction on the epoxide, then that kind of made sense. Okay, let's talk about another reaction. And I'll start off by showing you the observation that led people to realize something cool is going on, something else cool goes on with epoxides and bases. Oh, here's another epoxide. It's not a five-membered ring. So when you treat this with lithium diethyl amide, you get the expected syn elimination product, this kind of thing that looks like an E2 elimination but is syn. Yield is only 16% for this. It's not very efficient. Turns out that the major product in this reaction is a bicyclo-octane system with two five-membered rings. So how do you explain that? There's something non-obvious that's going on. I hope you can immediately see that you're not going to explain that unless you can figure out how to make these CH2 groups on the other side reactive or something so reactive, maybe you're not making these reactive, but you're making something so reactive that it can react with those CH2 groups. That's the only way you're going to make a five, that kind of five-membered ring system. I'm going to redraw this epoxide in a different conformation to remind you that when you have eight and nine-membered rings, there's significant trans-annual interactions. It doesn't look like a stop sign. It looks more like this, and there's very bad bumping. If you have to cyclize an acyclic molecule to make an eight-membered ring, this is going to just screw you. It's going to kill those kinds of bumping interactions. But it also really enhances the possibility that things here will react on the other side. So you can imagine if you have some sort of an epoxide here, and you have an alkylithium, there's another competing reaction, but with things that are basic, amide bases in this case or potentially alkylithiums. So not only does the oxygen coordinate reversibly, but you can also get this deprotonation. Remember the CH bonds on the outside of strained three-membered rings are more acidic than normal, thermodynamically more acidic and kinetically more acidic than normal. And so you end up getting deprotonation of these CHs on epoxides. So you end up getting medallation on the epoxide ring, competing with that ring opening. This is a competitive process that competes with that ring opening. And so here's what I don't think you could have guessed, is that this pops open. I wouldn't have guessed that. I would never have guessed that. That this pops open. And when that pops open, it gives you what's called a carbenoid intermediate. It's not a carbene, because there's a lithium attached to where you would have had a carbanion. So eventually this coordinates with another lithium. Let me just leave that as an O minus for now, otherwise it'll be confusing. So you end up with this species. It's like, oh, it's a carbocation. That sucks. Well, but at least there's some, somewhere on here, there's a nucleophilic pair of electrons. So you relieve ring strain. You've got this very electron-rich carbon atom that relieves ring strain. So this is a carbenoid. It's sort of like this. It's not a carbene with an empty p orbital. A carbene would have an empty p orbital and a carbanion sticking out here, but there's no carbanion. There's a carbon-lithium bond. That's called a carbenoid intermediate. If you took away the lithium, that would be a carbene. So let me just label that carbenoid. And as you can tell, it's electron-deficient. That carbon only has six electrons. So these C-H bonds on the other side of the ring are poised to form a bond across that ring. So you add here and at the same time that you're donating into that empty p orbital, sorry, I'm on the wrong structure. Okay, here we go. There we go. So you've got this empty p orbital, very unhappy, very unhappy, take these electrons, transangular interact, and maybe I should readjust the shape of my ring to put these closer. I could reshape this so that the bond, this was, whatever, I don't even think I'm using the right carbon atoms. I'm not going to take the time to redraw this, but you could readjust the shape to where you could see that this is directly across from that. And then this nucleophilic bond at the same time can come and add to that and pull the proton off. You basically do a C-H insertion across the ring with this carbenoid. And who would have guessed that? You could never have guessed that in a million years. That's something where somebody has to show you that and then you go, holy cow, can it do that? Yeah. Okay, so, and again, that wasn't predicted, somebody stumbled on that reaction. Okay, I want to stop and try to debunk a fiction here by, in a disturbing way, by showing you some pictures. I hope I can boot up the projector here. So we can talk about the structures of these things that I've been drawing for you and how it's all a lie and how we're going to deal with that. Oh, cool. Okay, so when I show you these alkyl lithiums and things of that nature, how should we think about those? So I'm pretty sure if you've taken an undergraduate organic chemistry class in the United States that you were introduced to alkoxide and intermediates in a way that looks like this, this is an excellent way for you to help you understand the basicity and reactivity of the alkoxide because you were trained to think, oh, when I see it minus charge, I know that this is basic. And so, what's the better depiction of these? Now, I would argue maybe that this is better or if you get to a case of an enolate, what does an enolate look like? How should we depict these? Let me show you some crystal structures of these compounds so we can get a better idea of how to think about these, not just in the solid state but in solution. Here's the structure of potassium tbutoxide. So over here, if you've been using this depiction, how many bonds are there to potassium here? Zero over here. At least here, there's one. How many over here in the real structure? Three. You know, to me, one bond looks closer to reality than zero bonds, right? There is no such thing as free tbutoxide floating around. There's no such thing, not even close. Let's come down here and look at an enolate, a lithium enolate. We're going to talk a lot more about those later. How many bonds to lithium here? There's none. You know, even here, when I draw this, that's at least better than this. But in reality, you don't have one bond to lithium. You've got three at a minimum here in these types of systems. Here's the oxygen. Each oxygen is connected to how many lithiums? There are three lithiums. And each lithium is connected to three oxygen. This, in reality, are the kinds of structures you have in solution when you stir things around. Every single time that you run reactions and things don't make sense, come back and remember these pictures here. Okay, you might make some argument. Oh, well, of course, there's lone pairs all over these oxygens. That's why it's coordinating so heavily. That's why you get all this aggregation. I've got potassium, but I've got lone pairs everywhere. And the only way to make them happy is to have the lone pairs coordinate to something. So let's go down and look at an alkylithium. There's no lone pairs on carbon, right? When I draw an alkylithium like this, I don't have any lone pairs to talk about. And so how would I think about the structure of alkylithium in solution? In solution, not a crystal structure where you could be biased by some packing forces or lack of solvent. If you put this in THF, this is the structure of butylithium in THF. And there are many studies that show that this is true. And it's actually an equilibrium. It's like, what is going, like, how do you even, let me try to unwrap this structure for you. So each, here's, each butyl chain right here. And each butyl lithium, carbon-lithium bond is part of this sort of cubic lattice with four carbons and four lithiums, right? It's nothing like this, a free carb anion and a free lithium. How many bonds to lithium in the real structure here? There's four. There's not zero, there's four. Now in THF, of course, when you have lots of THF present, there's an equilibrium. You'll see this other just dimer of butyl lithium. But again, lithium in this dimer has four. So over here, you can see each lithium is bonded to three carbons. And over here, each lithium is bonded only two. It's an extra THF molecule that is helping the lithium to satisfy the octet rule, right? The lithium wants to satisfy the octet rule, yeah. What's the charge on lithium and how many male octets? If you do the formal charges here, you'd have minus three charge on lithium. So these are dative bonds I'm drawing. They're not charge-balanced. These are not true Lewis structures. In a Lewis structure, I'd have to put minus three on lithium. Okay, so how do we, let me just help you deconstruct this butyl lithium thing because it's disturbing to see this kind of stuff. And I don't know if this is going to make it more disturbing or less disturbing, but let's try to understand the reactivity of butyl lithium. When you look at butyl lithium, here's what you ought to be thinking. Oh my God, that bond is nucleophilic. That's what you ought to be thinking. And at the same time, you ought to be thinking, what the, this lithium only has two electrons and it wants eight. Lithium wants an octet of electrons just like every other second row atom. And so how are you going to make that happy? What you're going to do is you're going to take this super duper duper nucleophilic carbon lithium bond and you're going to use it to donate into the empty orbitals on lithium. Right, so now that's at least a little better. Okay, that's a little better. Now lithium at least has four electrons, but it still wants eight. And this is what's going to drive other butyl lithiums to attack this. You'll notice in that other structure I showed you, I never filled in the carbon hydrogen bonds. Now that I do that, how many bonds are there to carbon there? Right, if you have some sort of bias against carbon with five bonds, forget that bias. Right, there's nothing wrong with having, we've talked about this before. There's nothing wrong with five bonds to carbon. Just don't violate the octet rule. We have not violated the octet rule. This carbon still has eight electrons. There's eight electrons in this carbon atom and if I take these electrons and I share them between these two lithiums, I'm not adding extra electrons. I'm just sharing those electrons. So we don't, you know, this may be disturbing to you, but we have to get over that. And I'll show you one last picture here before we go back to the chalkboard type lectures, but just to move down a little lower in the periodic table to cuprates. So we're going to draw some cuprates in just a moment, but I want to remind you it's like this alkyl lithium thing. Like if you draw an alfennel cuprate or a fennel copper species, we don't draw it like this when we draw things above the arrow or mechanisms. But this is the structure of a fennel copper compound. Look, I've got this benzene ring and I've got two bonds to copper for each of these carbons. And here's the cuprate. So over here with a cuprate where there's two alkyls on copper, now look at this. It's like each carbon on the benzene ring that's bound to copper is also bound to a lithium atom. Okay, so the structures that we draw, when you draw a butyl lithium, are not right. They're not the correct structures. And I want you to keep that in mind. And I'll tell you how we should deal with that as we go along here. Okay, let me go ahead and show you how I would like you to deal with this. So if you go look up the structure of sodium t-butoxide or sodium methoxide, it's aggregates. It doesn't look like this. And yet, when we do arrow pushing, I want you to draw it as the monomer. I want you to draw potassium t-butoxide as the monomer. I want you to draw sodium alkoxides as the monomer. But I never want you to forget that these are aggregated in solution. You're going to be dealing when you do synthesis with compounds that have other oxygens and chelatable groups all over the place. Don't forget that that sodium is not happy with just those two electrons there. So when you draw mechanisms, just to deal with how should we draw mechanisms for things like this? I don't want you to draw alkoxide minus like this. I don't want you to draw this. I don't even want to draw a regular arrow here because it hurts me to think about this. You're not going to simply have this dissociate into two components. If you calculated using Coulomb's Law what it costs you to pull a sodium away from an alkoxide, that's so energetically expensive it's not realistic to do this. There is nothing you can do with these lone pairs that you cannot also do with these lone pairs. So when I give you a sodium alkoxide or potassium tb-toxide leave it on there and do this, right? We don't have to dissociate the sodium or the potassium or the lithium first. It'll come off eventually. And when I ask you to do this stuff I'm fighting all of your previous training probably for drawing mechanisms but that's too bad. So you'll get an oxonium intermediate. Now you can dissociate the sodium. Now you're not pulling a minus away from a plus and depending on the concentration actually the bromide might come back in first, poke some electrons into sodium and then dissociate but we can simply draw it like this at least now we're not trying to separate a plus from a minus, right? That's what makes this so painful over here at least the most painful to me. Okay, so this is closer to the way you ought to be dealing with these types of reactions. This is the way I would like you to push arrows. Don't pop off the metal from the alkoxide. Just don't do that, okay? So here's what is also okay, what I would like you to be able to do for your arrow pushing. Frequently we'll come across things and it doesn't have to be butylythym, it could be a grignard reagent, right? I don't want you to draw this dissociating into some sort of ridiculous thing like this. This is nothing like what's going on in solution. I just want to use a different pen color to say don't do that. Don't make it dissociate to give an alkyl anion. That's not what grignards look like. That's not what alkyl lithiums look like. If you want to do some sort of nucleophilic attack there's nothing you can do with this lone pair and the minus that you can't do with this. There you go, right? Just use the sigma bond to attack things. Right here there's a lone pair. That's the more nucleophilic component but there's no lone pair on carbon in butyl magnesium bromide or things of that nature. And now you can pick up the magnesium. So this is the way I would like you to push these types of electrons. Okay we don't have time to quite finish up. I'm going to talk a little bit more about enolates, right? I'm not going to, we're in this class, we're not going to draw all these complex aggregates. We're just going to draw a monomeric species. And I'll make you feel a little bit more comfortable with drawing a monomeric species when you draw lithium enolates. So when we come back on day we'll finish up on this and then hopefully move on to alkenes as nucleophiles. Yes?