 Okay, treat or trick, there's some candy floating around. Take some candy and keep passing the bag around until it's gone. That'll give you some free calorie energy to keep your brains going. Huh? No calories are free, especially when you get older. Okay, so we are moving on in this class. We're going to talk about a fundamental reaction and a set of mechanisms that are common to introductory organic chemistry courses and specifically eliminations. And really it's a very, it's a common set of reactions in which some sort of filled orbital donates into some sort of unfilled orbital to produce a pi bond. And all the things we're talking about you could potentially apply to the reverse reactions and we'll do some of that later. But we're going to talk about three types of elimination mechanisms. We'll talk about E1s and why are E1 reactions fast? It's because you have empty P orbitals, carbocation. That's an E1 elimination mechanism. They always involve carbocation. And they're fast because it's so easy to donate into an empty P orbital. Those are the best acceptor orbitals that we talk about. Okay, the alternative here is if we reverse the roles and instead of having a really good acceptor orbital, we have sort of this lame sigma star orbital like if we have a chloride leaving group. And we want to eliminate that out. So if you have a not so good orbital you're donating into, you better have a good nucleophile. That's an E1CB elimination mechanism. So non-bonding lone pairs especially for carb anions are very nucleophilic. That's why that's fast. It's not because your acceptor orbital sigma star is good. It's because your filled orbital is very nucleophilic. And then last, in this sort of series here, what if you've got not such a good orbital and you've got that's empty and you've got not such a good orbital that's filled, a CH bond? Well these ought to be the worst of the three possible elimination mechanisms. If you see an elimination occurring and you're trying to hypothesize what's the elimination mechanism, do not jump to this as your first choice. And I'm saying that because I know you will have a tendency to jump to that as your first choice because it's the most obvious mechanism. Okay, so let's talk about these three types of elimination mechanisms. And I'll start off by talking about E1 elimination mechanisms. They all involve carbocations and so that means we better have good leaving groups. That's when it's going to tend to be, those will tend to be prominent mechanisms for elimination. And not just good leaving groups but you want to have stable carbocations. That's another clue that it probably proceeds through an E1 elimination mechanism. First you generate a carbocation, then in a second step you pull off a proton. And then finally, another characteristic of these types of reactions are you have weak nucleophiles. Right? Under those conditions, if you had a strong nucleophile, you'd end up with an SN1 reaction. So it's typical of these types of reactions that you have this combination of factors going on. So let's talk about an example here, I guess one that has sociological significance, maybe scientific significance as well. This is a designer, I'm not sure I would call it a designer analog, an analog of a narcotic called demarole. So it's an opioid, looks just like morphine when it binds to the mu-opioid receptor. And of course, demarole has a carbon here and an oxygen over there and so somebody didn't know how to synthesize that so they thought, well, I'll just make this through a Grignard addition to a ketone and then I'll isolate that. And who cares whether the oxygen is there or there? It turns out that binds really well to opioid receptors just like morphine. The problem is if you're a sloppy chemist and your conditions are acidic, you can see what's going to happen here is that you're totally set up for this to generate a carbocation if you're not really particularly good at chemistry and you throw in too much acid or your propionyl chloride is contaminated with propionic acid and you're heating this up. So the problem is that while they were cranking out batches of this demarole analog, this morphine analog, let me draw this in a flat form. Hey, Chris, could you round up an eraser from? Oh, wait, here it is back here. So if you're cranking out batches of this, what did happen is that the batches they were cranking out were contaminated with this E1 elimination byproduct. So you can see what happens here is you eliminate this to give a carbocation and then in a second step, not in the same step, you pull off a proton from next door. And so now they were giving out in silver. I guess they were selling batches that were contaminated with this. This is actually innocuous. The problem is that the human body metabolizes this. I'll just write metabolism here into a compound that who would have known this? Nobody would have known this that causes destruction of the substantia nigra area of the brain. And so you end up with Parkinson's disease. And it's not slow onset Parkinson's like age-related Parkinson's disease. You instantly get Parkinson's disease and you don't recover from that. And it means you're just, you're catatonic. You sort of freeze up and that area of your brain is destroyed. That's turned out to be one of the number one tools now in Parkinson's research because they can induce Parkinson's disease in animal models. And it all came from this E1 elimination mechanism. Okay, so that's an example where you're totally set up to have a stable carbocation that's not just tertiary but benzylic. Let's take a look at an alternative elimination mechanism and probably the most common elimination mechanism that you will be exposed to if you do organic synthesis. It's a more common than E1 mechanisms. So the E1 conjugate base refers to the fact that you're going to have some conjugate base in there. You're going to pull a proton off and make a carbanion. So under basic conditions where you've got strong bases and that's pretty typical of organic chemistry nowadays. We use powerful nucleophiles like enolates. I'll draw this particular resonance structure of an enolate. We do lots of enolate chemistry now in organic synthesis. And so if you do an aldol reaction these would be typical conditions, aqueous sodium hydroxide. You can do these at room temperature. You end up getting a beta hydroxy carbonyl compound like this and this undergoes an elimination to give you an enone under the conditions of the reaction. It's reversible and thermodynamically it favors this, the enone under the conditions of this reaction. And so what's prominent about E1CB elimination mechanisms is that you have to have something that's capable of forming an acidic or something that's capable of forming a carbanion. So you have acidic CHs in there. And if you were okay with making the carbanion here in the first step, the enolate, well, you ought to be okay with making it here on the product. The CHs have about the same thermodynamic acidity. I mean, if anything, maybe it's got better thermodynamic acidity. So you can imagine deprotonating here to give the same kind of enolate that you had initially in the starting material before the aldol step. I'll draw the arrow pushing for the aldol. That's called an aldol reaction. We'll talk more about those later. Okay, and so now you've got this great nucleophilic lone pair here which is perfectly capable of pushing out an alkoxide leaving group in order to give that enone. And this is something that's very particular about carbanions and about anionic lone pairs. They are perfectly capable of pushing out hydroxides. If you somehow had some rule that hydroxides and alkoxides could not act as leaving groups, look at this. This reaction occurs at room temperature. So alkoxides are perfectly capable leaving groups. We'll see more of that later. So let me give you an example just to remind you that it's very common for alkoxides to act as leaving groups. You may have thought only halogens could act as leaving groups. Very frequently you'll run into intermediates that look like this in your reaction mechanisms. This is nothing more than an E1CB elimination intermediate, right? You already know that alkoxides can act as leaving groups when there's anionic lone pairs nearby because you've done this over and over again when you've drawn out mechanisms for hydrolysis of esters under basic conditions. Okay, so that's E1 reactions, E1 conjugate base. There's the conjugate base and the conjugate acid is before you pull the proton off. So whenever you have highly basic conditions and you can form stabilized carbanions, you will always go through this sort of two-step mechanism. And I'm going to urge you to not to do this, never, not ever, don't ever, never. And let me just emphasize this, ever, not ever. I hate even, I don't even want to draw regular arrows. I want to draw dashed arrows. Don't ever try to do this. That's an E2 elimination mechanism. If you do this, you're implying that it's better to add to this sigma star here. I'll just, I'll label that. There's an orbital here for a sigma star for a carbon oxygen. You're saying that it's better to donate into that antibonding orbital sigma star than it is to donate into pi star for a CO. And you know that's not true. We talked about this in the first week of class. It's easier to donate into pi star orbitals than it is to donate into sigma star. You make the enolate first and then you kick out the hydroxide in a second step. So resist the temptation to draw that as a one-step E2 elimination mechanism. Okay, last, the last mechanism is that E2. If you can't form stable carbocation and you don't have strongly basic conditions and the possibility for stabilized carb anions, then you can start thinking about one-step E2 elimination mechanisms. It's after you've depleted all of that stuff. Okay, so we'll go ahead and take an example. So if you can't form stable carbocation or if you can't deprotonate in order to make some sort of a carb anion, you don't really have too many choices left. So here would be an example of something that is definitely going through an E2 mechanism. Here's a benzene sulfonate leaving group. Sulfonates are good leaving groups like tosylate and mesolate. And if you throw in a decent base here, you could probably get away with triethylamine. In this particular case, they used alkoxide, potassium tbtoxide. It's a slightly hindered base so it helps slow down a normal and otherwise also slow. You can see how tricked out this substrate. This is a neopental center which really dissuades any SN2 reactions at this position. I don't see that this really has a whole lot of choice here except to give you the elimination product. And so in this case, the base, and let me draw this proton going downward because it'll just make me happier. There we go. And you can make the arrow end on the carbon or you can make the arrow end in the middle of the double bond. And I'll accept either one of those. I just want you to be precise about where you're showing the arrow ending. So you can show that middle arrow forming a pi bond or you can show it attacking this antibonding orbital here for the CO bond. And I'll accept either of those two depictions. Okay, so if you can't make carbocation ions easily or you can't make carbene ions easily next to leaving groups, then you can start to talk about concerted E2 elimination reactions. In this case, there's nothing on this carbon atom to stabilize a carbene ion. So you'd have to have a carbonyl or some sort of a septic group, from this stuff up here. I mean, this is nothing conjugated to any, there's nothing that would sufficiently stabilize that carbene ion. Okay, so let's look at the E2 in detail. So even though it's not really that common of an elimination mechanism, we really need to look at this in detail because there's so many reactions that this helps us to think about. So let me go ahead and draw it a sort of a conformational depiction here. And I want to be very particular as I draw about this sort of substituted carbon-carbon system about the anti-relationship, the required anti-relationship between the proton that you pull off and the leaving group. This is the idealized orientation for an E2 elimination reaction mechanism where this proton carbon bond here is anti to the leaving group. And when I have it like this, there's an anti-bonding orbital here for this, and I'll just label it. So we're donating into sigma star for this carbon X bond. So there's this anti-bonding orbital that has a lot of, the orbital lobe is very big back there. It's very easy to overlap with that. And so the electrons in this bond are weakened when the base attacks, and that makes it easy to push that out. If you look at the rates for reactions when this proton is anti, so let me write this out, that's an anti-relationship versus sin, it turns out the anti is faster. And it's not infinitely faster, but you can put some numbers on that. And I will put some numbers on that. Okay, so here's two different orientations of the substrate, and we're talking about a system where there's no special steric encumbrance to one relationship or the other. And so usually when you look at the rates of this sin elimination versus the anti-elimination, it's at least a factor of 10 better to do the anti. At least, I don't know of any cases where it's not at least 10 times better to do this, to do this anti-elimination. So how do you think about, how would one think about this? Let me try to depict the anti-bonding orbital for this CX bond. So there's an anti-bonding orbital that's sticking back here, and I'll try to draw some sort of phasing in here. That's sigma star for the CX bond. And why is it that this can donate? If you look at a canonical orbital for a sigma bond like this one, there is electron density on this back side. And it's smaller usually. Most of the electrons are spending most of their time in the middle between H and C. But there is a lobe on the back end here. Any time you draw a lone pair, there's a little lobe on the back end. Any time you draw a canonical sort of representative sigma bonding orbital, there's a lobe on the back end. So as you're putting electron density into here, this starts to fill up with electrons, and there is overlap. It's not zero. So, I mean, it's not impossible to do sin eliminations. They're just never as easy as anti-eliminations. And so because of that, I want you to be experts at seeing anti-relationships. This is critical, not just for the E2 elimination reaction, but for a vast amount of organic chemistry. And that's why I'm so picky, picky, picky, picky about how you draw chair cyclohexanes. Let's go ahead and take a look at two different substrates for an E2 elimination reaction. So this is a cyclohexane ring. And it's got a t-butyl group that very, very, very much wants to be equatorial, that bulky t-butyl group, by at least a factor of 1,000, does not want to be axial. And here's two different isomers of cis and trans isomers. I guess you'd call these diastereomers of this cyclohexal bromide. So if you do an elimination with potassium t-butoxide, which is a strong base, so this will be E2 elimination, that can't make a stable carbocation. So there's no E1. There's nothing there to stabilize a carbannine, so it's not E1CB. This is going to eliminate to an E2 elimination mechanism. And if you look at the relative rates for E2 elimination, this substrate is 500 times faster. And the reason why that's 500 times faster may not be obvious until we draw the protons that get eliminated. Here's one proton that's not interesting. But there's a proton down here that's axial, that's perfectly anti-periplanar at all times. That's perfectly anti-periplanar to that carbon-bromine bond that you break. And we come over here and we look at the protons that are beta to the bromide here, and neither of these protons is anti to the carbon-bromine bond. Let me emphasize for you the anti-relationships here. So here's an anti-relationship between this proton-carbon bond and the carbon-leaving group bond. That looks great. That's perfect. But we come over here to this carbon-bromine bond and we look at what's anti to that. It's these carbon-carbon bonds. And you're not going to do E2 elimination by attacking those carbon-carbon bonds. Right? It's essential that you see these types of anti-relationships because those are what set you up for elimination. Okay. So unless you've got some other special reason why not to, you should assume that E2 elimination reactions are occurring anti with an anti-relationship. So you may or may not be familiar with this term stereospecificity. Stereoselective reactions are reactions that prefer one isomer. They prefer to give you E alkenes versus Z alkenes or they prefer to give you a rethrodiastereomers over 3-0 diastereomers. Stereospecific means that one stereochemistry of starting material gives you one stereochemistry of product and the other stereochemistry of starting material gives you a different stereoisomer as the product. You can't know whether something is stereospecific until you test both stereoisomers of the starting material. That's the only way that you can make a claim that something is stereospecific. So let's go ahead and draw out a stereos, the result of stereospecific elimination of anti-protons, protons that are anti to leaving groups. And I'm trying to draw this in a way that you can clearly see with crystal clarity that this CH bond can adopt a conformation that's anti to that otosyl. There's only one proton on the beta carbon here. And so when you do an E2 elimination of this and in this case they did it with a thoxide anion, then you're going to preserve this relationship. Here's the relationship that you need to see to make sure you don't mess up when you draw out the product. We're going to have a double bond in the product. And let me draw out the way it is in my notes here. So, and we have to decide there's another, there's a methyl group in a hydrogen here and which one ends up where? And what you need to see is that see this phenyl right here? It is anti to this carbon methyl. It's anti in the starting material and it will be anti in the product. The carbon, this methyl is anti to phenyl. Let me draw the methyl down. And then I look over here and I see that this carbon methyl bond is anti to this carbon H bond. It will still be anti to each other in the product. And so it's important for you to see those anti relationships. Otherwise you might make a mistake drawing out one isomer versus the other. If you take the other diastere, here's how we know this is stereospecific. This group took the time to try the other diastereomer. This is a diastereomer now. And now the H on this tosyloxy carbon part right here, the H and the methyl are now switched. The H is coming out towards you. Let me make that more obvious. I'll make it clear what's coming out towards you here. There we go. And again, this is the same reaction also E2. And so now when we draw out the product here, what are the relationships that get preserved? See the carbon methyl here that's pointing forward? Anti to that is the other carbon methyl. They better be anti in the product. I mean if I'm eliminating out this proton tosyloxy pair of groups then I want those two methyl groups to stay anti. And I look at my carbon phenyl bond that better be anti to carbon H in the product. So that's a stereospecific reaction. One stereoisomer of the stereomaterial gives you one stereoisomer product and the other stereoisomer gives you a different stereoisomer product. You can't know whether something is stereospecific until you test both of the stereoisomers in the reaction. Okay, so let's take a look at a stereoselective reaction. In other words, I could also call this a stereoselective reaction because it prefers one stereoisomer. Each of those reactions is stereoselective. Any reaction that favors one stereoisomer is called stereoselective. So whenever you take just long alkyl chains, so you can imagine R1 and R2 here to be just big long alkyl chains and you have some sort of a leaving group and you subject this to E2 elimination conditions. So a base like triethylamine which is not so basic so you typically heat those. Or potassium tbutoxide at room temperature that's a reasonably strong base. You can't really get a, if this is just an alkyl group, secondary carbocation aren't stable. So there's no E1 elimination mechanism here. If you don't have a carbannine stabilizing group, you're not going to, on R2, you're not going to make a carbannine there. So now this can really only be an E2 mechanism. These prefer to give trans double bonds. And the way to see why that is true is to look at a Newman projection. If I stand in front of this, if I stand over here and I look down this carbon-carbon bond, what would I see in my starting materials? Maybe the confirmations in the starting materials will tell me about the kinds of interactions in the transition state for elimination and that is true. So if I stare down that bond and I draw a Newman projection, what would I see in front of me? Here's the preferred conformation. I would want some sort of a conformation in which, and I hope I'm getting this right here, that R1 and R2 look like this. This would be ideal. I don't want these things bumping into each other. I'm set up now so I can, as I pull this proton off, the electrons in that bond will help push out that carbon-x group. And of course, there is another proton here I could eliminate. Let me spin things around so that other proton is now aligned anti, and there we go. And I forgot to draw my last H here. Let me draw that. Okay, so these are just two different rotomers of the same starting material. I'm just trying to get you to imagine what would happen if I spun this around so that green proton here, right? It's not distinguishable, but I'm coloring it green. What would happen if I spun that around so that green proton were anti to the leaving group? Now there's an obvious problem with any transition state that resembles this starting material conformation. If your transition state resembles this, that's going to start to hurt in the transition state. And that's why you favor these trans olefins in the products when you do an E2 elimination on just a linear, on just a straight chain alkane. Okay, one of the hardest, I think one of the hardest areas, the most confusing areas of all organic chemistry is the issue of competitive reactions between elimination and substitution. It's like, what a mind job. It's like, why do they start off with this in organic chemistry? It is so complex, E2, SN2, you have to worry about all kinds of, that's the last thing they should cover in any organic chemistry course. I'm going to try to talk about this sticky issue of substitution versus elimination. Any time you have a secondary alkohalide, let me write RX. And those are the interesting ones because they're typically have some stereochemistry, have the possibility for a stereocenter. Those are the interesting cases. If you try to add some strong nucleophile, strong nucleophiles tend to be strong bases, unfortunately. And so you usually see mixtures, mixtures of elimination and substitution reactions. You get mixtures of products, and that's awful. Let's take a look at an example of some rates. And I'll make a, we'll draw a very simple conclusion about the effects of temperature. So I can't imagine anything simpler than this. This is about as simple as you can get for a secondary alkyl bromide, isopropyl bromide. If we treat this with sodium ethoxide, we have two choices. The ethoxide can do an SN2 reaction to displace the bromide, but that's kind of crowded. And there's all these beta protons, three protons here, three protons there that you can pick from. And of course, this reaction does both. It gives you E2 elimination and it gives you SN2 elimination. Or sorry, SN2 substitution reactions. And I want to look at the rates for formation of these as we vary the temperature. So the rate constants that somebody took great pains to measure absolute rate constants. And don't worry about the units. They're per molar per second. Just look at the relative numbers. I'm looking at that number immediately that I'm thinking, that's slow, 10 to the minus 6 per molar per second is a slow reaction. And when I draw this out, the important point here is those are pretty similar, right? That's about the same. It means you're going to get about a one to one mixture of elimination and substitution reactions. So if you've designed your total synthesis of your natural product to rely on this as your final step, you're going to get screwed unless you want mixtures of two different things. I can't think of why that would be good. Let's look at what happens as we raise the temperature from about room temperature to about 120 whatever, about 100 degrees. So maybe you go to refluxing toluene. And there's other consequences to that. But if we just look at the raw effect of temperature, now immediately what you can see is that reactions are faster. You should expect that. Of course reactions are faster. As you raise the temperature, molecules have more energy when they collide with each other. But what you can see is that the SN2 reaction didn't gain in speed as much as the E2. This is a general truth in organic chemistry and it is not just E2 or E1. Generally heat favors elimination over substitution. Heat favors elimination over substitution. So maybe you want your reaction to be faster and you're trying to do a substitution, be careful. They're both going to get faster but the elimination will get even faster as you raise the temperature. So now you're looking at like a 60, 30, 70, 30 ratio of these two. So if you're trying to make the alkene, that's great. You can increase the selectivity. So let me just write this down. So heat generally favors elimination over substitution. If I come back to this especially for the E2 reaction, there's an entropic requirement. This H carbon bond can only push this out when it's anti or at least is best able to push this out when it's anti to that carbon chlorine bond. And this base will be most apt to attack that proton when it comes in exactly from the back side. There's all of this requirement for order in the transition state. Everything has to be aligned just perfectly. And the chances of you having this confirmation are more likely when you increase the number of collisions. That's the effect of temperature. So you don't have as much of a conformational entropy requirement for a substitution reaction. Those requirements are more stringent. The entropy requirements are more stringent for elimination and that's why it benefits more from temperature, from raising the temperature. Okay, I want to compare the two most common leaving groups because they're not equal. And you have a choice in organic chemistry what you make. I'm going to show you a reaction here. So this is just a long straight chain alkane with 18 carbons. And at the end we have some sort of a leaving group. So if you have an alcohol here, you can choose to convert it into a bromide using PBR3 or the Appel reaction we'll talk about later with triphenylphosphate. Or you can convert the alcohol into a tosylate leaving group. And so which one should you choose? Well, it depends on what you want to happen in these reactions. So you have a choice between SN2 products. So SN2 substitution and I'm not going to draw the whole formula there. Just I'll just write C16 versus E2. Two competitive reaction pathways. So if you throw in potassium tbutoxide with this and X is equal to bromide, you don't get a whole lot of SN2 substitution here. It's mostly elimination. And of course that's volatile. It comes off, you never see it. It's like gee, my reaction looked so clean. Well, maybe it's not so volatile with C16, but this is a common issue. Okay, and then versus X equals tosylate. So what? Same nucleophile, same backbone, same amount of hindrance essentially. Now when you have a tosylate leaving group, you get much more substitution and much less elimination. So why is there this difference between halides and tosylates? Now by many measures, you might say that they have about the same leaving group ability. That's a pretty good generalization. Bromides and tosylates are good leaving groups for very different reasons. So at the very beginning of this class, we said that there are three kinds of factors that come into play as reactants come together. We said there's calomic effects, things related to charge. We said that there are steric effects, just electrons don't want to occupy the same points in space, and we said that there's interactions between filled and unfilled orbitals. So let's take a look at how that explains this contrast in the outcomes of these reactions. Let me go ahead and draw out a substrate where I have a bromide anti to some sort of a proton. And why should this kind of a system, why should an alkyl bromide be good for elimination reactions? The reason why alkyl bromides are good for elimination reactions is let me try to super overemphasize this in some freakish way. There we go. So there's sigma star. And I'll phase this with some sort of phasing here for this sigma star orbital for my carbon bromine bond. Carbon halogen bonds are long, they are weak, and sigma star is very big. It is very easy for a nucleophile looking for some empty orbital to come overlap with this. Now there is some partial positive charge here. Bromine is also electronegative. But that's a relatively minor contributor here. So I'll just draw my diminutive little partial minus. So with a carbon halogen bond you've got this big sigma star and a relatively small charge. If you have iodine here, iodine is about the same in electronegativity as carbon. In fact, if you look at the original Pauling scales, Pauling said that carbon is more electronegative than iodine. So you can't be claiming that there's some charge that's helping you attack here in the case of iodine. Now let's contrast this with a sulfonate. This could be a mesolate or a methyl or a tosolate. If I have toluene there, it doesn't matter. What matters is this sulfonate. Now we've got a decent sigma star here. This is supposed to be carbon oxygen. There we go. So there's my sigma star orbital. That's the empty orbital. Nucleophile is looking for some empty orbital to overlap with. And there, you can overlap with that orbital. That's pretty good. The big factor here is the significant amount of partial positive charge at that carbon. Oxygen, especially with a sulfonate there, is much more electronegative. So now the alkoxide will attack quickly here. The negatively charged nucleophile will attack quickly here because the charge, the partial positive charge, the colomic effect is very big. Helps the nucleo. So when you come over here and you look at tosolate, it's 99 to 1 favoring attack at this carbon. But when you have the bromide here, there's very little, that's supposed to be partial positive. Sorry about that. So the carbon is partial positive. That's, you know, these are supposed to be partial positive charges. That's the effect of an electronegative atom as it creates partial positive charge. So now there's not a big contributor for the alkoxide minus to be colomically attracted through electrostatics to that carbon atom. Yeah. Nitrile groups. You mean what would I predict for this elimination to eliminate out? I would predict this would be very poor because, you know, the bond is not particularly long. The, since the bond is not long, you don't expect a big sigma star orbital here. It's not particularly electronegative like a heteroatom. So I would expect it to be very difficult to pull a proton off here and simultaneously attack here or to even do SN2 reaction. Now there are many examples where you do this, but that's, you know, this is more like an E1CB elimination mechanism. Is this what you're, is this what you're getting at? That's very common. So cyanide can't act as a leaving group if you've got a good enough, right? This lone pairs there all the time, always overlapping. So there's a better chance to push that out. Okay. So let's switch over. I'm tired of protons and maybe you're tired of protons too. So let's go to something more interesting. Let's just imagine for a moment that we put a metal here and I wouldn't really call that a metal, but those are semi-metals officially. So whatever we were saying about protons, you can imagine a long nucleophilic bond. Long bonds are more nucleophilic to tin or to silicon. So tin is better longer and more nucleophilic than silicon. So if you had either of these metals here, then what you'd find is that it's very easy if you have some sort of a leaving group. I'm using protonated hydroxyl as an example to overlap with this sigma star for a carbon oxygen bond and to push out the leaving group. These are particularly capable. Now these could be, these are drawn as one step mechan, or sorry, as two step E1 type mechanisms. We already talked about this. This is not anything new. We already talked about the fact that carbocation ions are vastly stabilized by beta silicon or beta tin. In fact, any long bonds will do this. Let's take a look at some numbers. I can't remember whether I gave you these numbers before. If you look at the relative rates for how easy it is to form carbocation ions, and then in a second step, or it can be two steps, you can use this long bond to donate over and form a pi bond like this. Sometimes you have to have a nucleophile add in to make that metal an eight complex, stan eight, silicon eight. The important point is that you can really favor E1 type reactions if you have a beta metal bond. Let's take a look at the relative rates for these types of elimination mechanisms. They've been measured. It turns out that tin, those long tin bonds are about the same as mercurinium, or, and I think of mercury is very similar to things like palladium. So tin bonds are very long. So metals, transition metals in the periodic table comparable to tin, and those are about a million times faster than long silicon bonds. So I'll just write 10 to the 6 here. Tin is about 10 to the 6 better than silicon. Just, I mentioned this before. Anything that silicon can do, tin is better, but unfortunately it's also more toxic. So, and that's again about 10 to the 5th better than having a proton for an elimination. So if you really want to eliminate, you want to put a beta silicon nearby. If you have a proton on one side and a silicon on the other of a carbocation, it's the silicon that's going to eliminate, that's, that's going to eliminate to give you these types of double bonds. If you really want to control an E1 type elimination, have silicon nearby. It'll help you make the carbocation and eliminate out. Okay, so let's take a look at some silicon reactions. So again, tin is great, silicon is less toxic, non-toxic, I guess I would say. So there is an elimination process called the Peterson olefination, and there's two variants of this reaction, and I'll show you those two variants. So let's just imagine that we've got some sort of beta silo alcohol, and in this case, I'll draw a stereochemistry here associated with the starting material. I'm not going to draw the H's at this position. So we have a trimethyl silo group. We have an alcohol beta to that, and there's a particular stereochemistry, the two methyls are coming forward instead of one forward and one back. And so there's two different ways for you to eliminate out the silicon and the oxygen. You have a choice. You can treat it with acid, or you can treat it with base. You can either convert this into a good leaving group, or you can convert that into a good, well, we'll talk about a different variant on that. So let's talk about making this into a good leaving group. You don't always have to use acid. In this particular case, they treated this with trifluoroacetic anhydride, and acylated this alcohol. So you can also do these by protonating that with under acidic conditions. But in this case, what I'm showing is they make a better leaving group by acylating that. And I wish I had room here because I want to draw that carbon-silicon bond as long. So this now turns this into a better leaving group, and it's not a great leaving group. Trifluoroacetate is not like a tosylate leaving group. But what happens is that makes this good enough as a leaving group that this carbon-silicon bond can stabilize formation of a carbocation. This will spontaneously pop out and give you a beta-silo carbocation, a highly substituted carbocation. And let me see if I, I'll just draw like this. That's supposed to be planar, but you get the idea. I'm just trying to draw it in the same shape there. So you get this super-stabilized beta-silo carbocation. The electrons in this long, I didn't draw it long, but this long nucleophilic carbon-silicon bond are donating into that empty P orbital. And so as soon as the nucleophile comes in and attacks, and I'm sorry I don't have room to draw the whole thing in here, but I'll draw it like this. A trifluoroacetate that got popped out is floating around. It's very easy for this to attack silicon. Silicon loves oxygen. It loves electronegative oxygen. So then this can come in and attack the silicon and don't do SN2. It'll initially form a siliconate with five bonds perfectly happy. And then that siliconate with negatively charged silicon and it's long and it's, and it's nucleophilic bond there will then push over. So I'm not going to draw both of those steps. And so that's what gives you this in which this, whoops, in which this cis relationship, you end up preserving this cis relationship between these two methyl groups here. Okay, so that's one version of the Peterson olefination. There's a different version of the Peterson olefination in which you treat this with a very strong base and you don't have covalently coordinating counter ions. You tend not to want to use lithium counter ions here. You want things like potassium that don't bond strongly to oxygen. And so when you do that, I'm going to flip around this central bond in my molecule. I'm going to flip around when I draw this so that I can see the oxygen can adopt a confirmation in which it's sinned to the silicon. And so when I do that, if I flip around this bond, if I twist around that bond, of course I have to, if I move that hydroxyl so it's pointing up, I have to move this methyl so it's pointing back. So it may look like there's a sleight of hand here. And this is just paying an homage to the fact that oxygen loves silicon. I just told you that. Oxygen loves silicon. And so you'll end up with this four-membered ring of siliconate, and notice my use of language there. Siliconate tells you there's a negative charge on silicon. Don't freak out over the five bonds to silicon. Again, silicon is not a second row atom. You can have six bonds. Don't worry about that. Silicon loves oxygen. So now all the bonds to silicon are nucleophilic. And so this whole contraption here can fall apart like this, and so that will pop out a siloxide anion and now give you the transoliphon. You get to choose with a Peterson olefination whether you get the cis isomer or the trans isomer as long as you have one beta-syl alcohol to choose from. And so how do you think about this? Well, this goes back to the idea that if I draw out this, this, let me not use the same pen color. If I draw out this sigma orbital here, there's also electron density on that backside here, and that can overlap with this sigma star anti-bonding orbital as sigma star for the carbon oxygen bond. And so this canonical sigma orbital for this nucleophilic carbon sigma bond does have electron density on the backside. It can help push that out. This is not a paricyclic 2 plus 2 reaction analogous to the Diels-Alder reaction. There are lots of concerted reactions that are not paricyclic. This would be an example of one of those. Okay, so these are two variants of the Peterson olefination. You may never do a Peterson olefination in your life, but you, I feel like it's very likely if you go into synthesis that you'll end up using this type of elimination reaction. And here's a more traditional example of an elimination that involves a siliconate. If you synthesize complex molecules, you're always looking for ways to use orthogonal protection strategies. You don't want it where every single one of your 12 protecting groups all comes off with acid because oftentimes you want to take off one protecting group and leave the others intact. So sometimes you'll use this thing called a Teoch protecting group, trimethylsiloethoxy carbonyl Teoch T stands for trimethylsilo. And so if you treat this with a naked fluoride anion, so a tetrabutylammonium fluoride, we usually call this T-baff, but I'll draw it out in this case for you. So that's a non-coordinating counterion. The tetrabutylammonium doesn't coordinate the fluoride. If anything coordinates the fluoride, it'll suck the electrons. It'll suck the nucleophilic acid. This is a source of naked fluoride anion. So when you add T-baff, that fluoride comes in and attacks that silicon. So yes, oxygen loves silicon. Fluoride loves silicon even more. We talked about the bond strength of a silicon fluorine bond. It's kind of like the strength of a carbon-carbon triple bond. Fluorine loves silicon. So you end up with this siliconate. Now all the bonds to silicon are nucleophilic. And in this case, they're nucleophilic enough to pop out this carbamate anion-leaving group. So by making that bond nucleophilic, you can push this out and just that allows. And then you lose carbon dioxide either on the workup or on the reaction conditions. And that's how you take off a TAAC protecting group without exposing your reaction to acidic conditions. You may have occasion to want to make Grignard reagents that look like this. You can't make these. And maybe it's obvious after talking about that TAAC why you can't make those. And the reason you can't make those is something called the board reaction. If you have a nucleophilic bond right here, look how nucleophilic that is. And you're hoping that that's going to attack some carbonyl. It's never going to get there. Because it's going to do this first. While you're making your Grignard reagent, it's going to be decomposing on you through this reaction to eliminate ethylene. So, and this is true for any nucleophilic carbon metal bond. So if you take something like this and you treat it with sodium, and we'll talk about why you don't make alkyl sodiums later on in the course. But in this case, you could do Grignard as well at magnesium. If you convert this into a nucleophilic bond, it will immediately snap open. If you've got a nucleophilic bond here, that will immediately open up. And maybe you want that to open up. But in this case, you wouldn't be able to make a Grignard reagent or some nucleophilic center here because it's simply going to pop that open. Okay, we've got just a little bit left to cover here for, excuse me, for elimination reactions. So when we come back, I'll finish up talking about these nucleophilic bonds that are baited oxygen. And then we'll talk about the importance of stereospecificity. Instead of talking about nucleophilic carbon, carbon metal bonds, we'll try to extend this to other types of bonds that can push out leaving groups.