 Good morning. Well, I am really happy today. So this is the first time that I've used electronic homework in one of the classes. And I was so happy when I looked this morning at 9 o'clock and found virtually everybody had engaged with the homework assignments. And so much of learning organic chemistry is actively working problems and actively trying to understand things. That I was really, really excited to see people are coming along with me on this process of learning. And it makes me very, very happy. So today we're going to continue our discussion of chapter 20. And I'd like to introduce organometallic reagents and talk about their reactions with carbonyl compounds and their reactions in general. And then we're going to talk about reactions of members of the carboxylic acid family with hydride nucleophiles and with organometallic reagents. And if there's time, we'll conclude by talking about use of these reagents and these reactions in synthesis. So when we were talking last time, we talked about hydride reagents with ketones and aldehydes. And we said sort of generically, if we have some ketone or aldehyde, now I'll write it generically just showing our groups that could be alcohol, that could be aromatic, that could be hydrogen. So generically a ketone or aldehyde. And we envision its reaction with some hydride nucleophile. Now remember when I'm writing something like this, H- in quotes, of course, there's no reagent that is itself a hydride nucleophile. But these are things like lithium aluminum hydride and sodium borohydride that serve as sources of hydride. And then if we carry out a workup with aqueous acid H3O plus or in some cases you can use water. And again, I'll put this in quotes because you can't go and buy a bottle of H3O plus. You could take sulfuric acid and pour it into water to make hydronium ion and bisulfate ion. You could take HCl in water and make H3O plus and chloride ion. And so you'd go to the stock room and ask for one of those. When you do that, you get a reduction of the ketone or aldehyde and protonation like so to give an alcohol. Now what I'd like to do at this point is to consider an analogous reaction, again, written at this point as an abstraction. So again, we'll take our ketone or our aldehyde and we'll imagine instead of adding our hypothetical hydride anion, we're going to add something that I'll say is R minus with a lone pair of electrons, meaning a carbon based anion. We'll talk more about it, not necessarily something you can get as a free anion and certainly not something that you could put in a bottle on its own. It would be part of an organometallic reagent. But again, right now, we're getting our view from 30,000 feet. And again, if you imagine some type of aqueous workup with H3O plus, now instead of adding in hydride to our carbonyl compound, we've added in an R group the use of primes and double primes and so forth is just my way of representing that they're various groups different or the same. Anyway, but again, in the view from 30,000 feet, the profound thing about this reaction as I've written it is that we formed a carbon-carbon bond. So much of organic chemistry, both the practical aspects of synthesizing useful molecules and the real intellectual beauty of the discipline is the fact that we can create great degrees of complexity, valuable complexity, medicines, analogs of natural products, natural products that are too scarce to get otherwise, probes to probe reactions and probe biology. We can do all of this through chemical synthesis in which we take little molecules that might be able to be purchased and build them up into big and complex and useful molecules that can cure cancer or fight disease or teach us things. So now we come down to the issue of what is our R minus? What is our source of our carbon nucleophile? And the first really, really valuable carbon nucleophiles that were developed were Grignard reagents developed by Victor Grignard. He received the Nobel Prize in 1912 in chemistry for this. And the basic idea is that you take some halide, a bromide, a chloride, or an iodide. Fluoride is sort of the odd ball among halogens. And if you go down your periodic table to astatine, it's not stable, it's radioactive. So organic chemists would never work with it. You can't really isolate it. Anyway, if we take an alkyl halide such as butyl bromide and we treat it with magnesium metal, if you've done one of these reactions in the laboratory, you'll have seen your magnesium metal is kind of bright and shiny and lightweight. It comes as turnings that have been worked off a big block of magnesium with a lathe. And you'll put them with your alkyl halide in a solvent. The solvent will be an ether solvent, either ether or THF typically, although sometimes other ethers can be used. THF is tetrahydrofuran. It's a cyclic ether. So it's an ether with a five-membered ring and it has lone pairs. It's kind of like diethyl ether with its ears pinned back. And either of these work well. The result is that you get a grignard reagent. I started with butyl bromide here and so I'll write the grignard reagent from butyl bromide. We call this butyl magnesium bromide and the ether solvent coordinates to the magnesium. The magnesium doesn't have a complete octet here. We have only four electrons around it, two from the alkyl group and two from the bromide. And so the ether solvent, the diethyl ether or tetrahydrofuran will coordinate to the magnesium and help give it the feeling of having a complete octet. Anyway, as I said, the broader category of this is a grignard reagent. And you can make these from anything from bromo or chlorobenzene to alkyl bromides to methyl iodide. And so this typically one makes a grignard reagent as part of a synthetic process. They're actually stable reagents. You can put them in a bottle. You can buy them. But in the laboratory, because they react with air and because they react with moisture, you would typically go ahead and immediately add a carbonyl compound. So let me show you a typical synthetic carbon-carbon bond forming sequence that one might do in the laboratory. So we might take our butyl bromide and treat it first with magnesium and ether. You'll often see people write a slash in an equation. Often that slash is a way of saying the solvent is below the slash or if we're just writing a simple line of equation, we might put our solvent below the arrow. Then let's add a carbonyl compound for the purposes of teaching, for the purposes of this example, I'll take acetone as our carbonyl compound. And finally, let's do an aqueous workup. And I'll give us, I'll choose an acid here. Let's say aqueous HCl as the acid I would choose. And the product of this reaction now is a carbon-carbon bond forming adduct or an adduct with a new carbon-carbon bond. If you wanted to name the compound, we've now formed a six-carbon chain. So it's two methyl, two hexanol as our product. So this is very powerful because now we've taken some small compounds that you can buy and we've made a more complex molecule that you might not be able to buy. Many molecules with this type of structure, these types of structures like alcohols with long chains on them are insect pheromones, for example. So some of these types of synthetic products are used to make traps for insects like Japanese beetles where that will, the pheromone will lure the insect to the trap and then they'll fall in and die. All right, so I want to talk a little bit about the properties of Grignard reagents and organometallic reagents in general. So metals of course are much more electropositive than carbon. So any bond between a metal and carbon is either going to be a polar covalent bond or an ionic bond. If you want an index value or chemists like to keep electronegativity values in their head, it's a useful way of assessing the degree of polarity of a bond, magnesium has an electronegativity of 1.3, carbon about 2.5 or 2.55, so you can think of the bond between magnesium and carbon as a polar covalent bond. In other words, you can think of this as having a delta minus of partial negative charge in carbon and a partial positive charge on oxygen, on magnesium. Sometimes if you're writing a mechanism and you want to be quick about it, you could say, well, I'll write a non-bonded resonance structure. Even though I know it's primarily a covalent bond, I could write an ionic bond resonance structure so we could think of it as this. Maybe I'll put this in quotes here just to remind us. This is sort of our thinking. And so if I'm thinking about this from a mechanistic point of view, one way we could think about this, I'll take our acetone and our Grignard reagent written generically. One way we could think about this is that our organometallic reagent serves as a nucleophile and of course we draw a curved arrow mechanism. We help think about the flow of electrons and bonds by starting an arrow at the available electrons either as a lone pair or in a bond and then moving to the thing that wants electrophiles. In other words, from the nucleophile to the electrophile. As we did in thinking about hydride mechanisms, we can't go ahead and now have five pairs of electrons around this carbon atom. So concurrently as the nucleophile moves in, electrons move up onto oxygen. And so we continue our grammar, if you will, of writing our curved arrow mechanism now, writing a product in which we have our negative charge. And if I want to be good, organic chemists are rarely good. We love to throw away things that aren't necessary in our thinking, but if I want to be good, I'll draw the magnesium counter-eye on there. We can also think of this just as we did with lithium aluminum hydride, maybe a little more correctly or certainly a little more sophisticatedly. We can think of our covalent bond and keep in the back of our mind that it's a polar covalent bond and write a curved arrow mechanism in which we simply go ahead and now take the electrons from that bond, move it in like so, and again come out to the self-same product. So these are good ways of thinking, sound ways of thinking about the mechanism of the reaction. So one thing to keep in mind about Grignard reagents and indeed about most organometallic reagents, particularly ones in which there's a large difference in electronegativity between the carbon and the metal, is that organometallic reagents, like Grignard reagents, in general act as very strong bases. We can think about basicity in terms of the pKa of the conjugate anion. So if you have butyl magnesium bromide, the conjugate anion corresponds to an alkane, in other words where in the conjugate acid you have a CSP3 carbon bound to a hydrogen. The pKa for such a hydrogen is about 50. That's about at the very end of the basicity scale. That's about as basic as you can get for a carb anion. If you have an alkene, now you have a CSP2 carbon bound to a hydrogen, the electrons are held a little more tightly in this type of structure. The pKa now is about 44. The way of thinking of this is since the electrons are a little more stabilized, held a little more tightly in an orbital that has more S character, right? SP2 has 33 percent S character, SP3 has 25 percent S character. As you hold electrons more tightly, the CH bond is more willing to give up H plus and give you a carb anion. So an alkene is a little bit more acidic than an alkane. These are all compounds I call very weak acids and I usually put quotes around the acid because you would not get any evidence of acidity from the compound, say dissolving it in water and testing it with a pH meter. And yet in a Lewis acid, Lewis base reaction, you can think of an alkene say as a proton donor under certain circumstances. Now by the time you get to an alkyne, now you've got 50 percent S character in your CH bond and now these are reasonably acidic. They're still very weak acids. Your pKa is only 25. That's really, really, really weak still. It's not even like water which you think about as an acid forming hydroxide anion. It's still a very weak acid or alcohol forming an acid giving up a proton and forming an alkoxide anion. And yet our pKa is about 25. And I'll show you an implication of that in just a moment when we start to talk a little more about alkanes. But the point of this comes back to what I was saying before about carbon, about Grignard reagents being very reactive toward water. A Grignard reagent acts as a base with water. And so if you expose a Grignard reagent like butyl magnesium bromide to water, the water acts like a Lewis, acts like a Bronsted acid and you get butane and bromomagnetium hydroxide, a mixed salt here. So this is a Bronsted acid, Bronsted base reaction if you think about it. We have water acting as an acid on the left half side of the equation. We have our alkane as an acid on the right side of the equation. And because the pKa difference is so humongous, the pKa of water is 15.7, the pKa of butane is about 50. I don't even bother to write an equilibrium arrow. The equilibrium constant is 10 to the 34th, right? It's 10 to the difference in pKa's. That reaction lies so far to the right that there's just no component to the left on it. And that's true with whether it's with water or an alcohol. And so just by comparison, imagine for a moment that I wrote, let me pick a particular alcohol. We'll pick ethanol as an alcohol. And so now we would get butane and a ethoxy magnesium bromide. And of course this would be the same with a carboxylic acid or just about anything that you would normally think about as even mildly acidic. Well, grignard reagents are one of a broader family of organometallic reagents. Another member of the family that reacts very, very similarly are organolithium reagents. And so if we wrote grignard reagents sort of generically as RMGX, that would be a generic way of writing a grignard reagent. We can write organolithium reagents generically as RLI. Organolithium reagents are formed by reacting alcohol or aryl halides. Again, we're talking iodide, bromide, chloride with lithium metal. If I write a balanced equation, say for bromobenzene and lithium, it takes two lithiums. And that makes sense if you think about it, right? We're carrying out a two electron process here and lithium has one electron. So magnesium has two electrons. And so a balanced equation becomes that we get phenyl lithium plus lithium bromide. And again, organic chemists are awfully, awfully bad about writing products of reaction. So I'll put the lithium bromide in parentheses because I might not write it. So typically if I were writing this as, say, a synthetic reaction and I imagine generating an organolithium compound and say reacting it with an organic compound, I'll give you an example of what I might write. So I might take bromobenzene or I might take chlorobenzene if I wanted. Treat it with lithium metal. Now organolithium reagents can be generated in ethers and ether can serve to coordinate but they also form clusters and so they actually can be generated in other solvents including hydrocarbons. I'm just going to skip the solvent here because it's much less important than in a Grignard reaction. And let's say as my partner, since I gave you a ketone before I'll take an aldehyde, the aldehyde I've chosen is pivaldehyde, that's the trivial name, or 2-2 dimethylpropanol would be the IAPAC name. And again, I'll imagine doing some type of aqueous workup. I'll just write H3O plus here to indicate I haven't specified the acid, it could be aqueous HCl, it could be aqueous sulfuric acid. One that I personally like to use in my own laboratory is aqueous ammonium chloride which is a very mild acid and very good for workups of reactions like this. Anyway, after our workup, the product now has a new carbon, carbon bond like so. And of course, because we've generated a stereocenter in the molecule, we've generated it as a mixture of two enantiomers. We've generated two different enantiomers in equal amounts. We've generated the racemate. All right, in part because your textbook mentions various different organometallic reagents at this point, I want to follow along and in part because I want to remind you of what I think are really, really useful items and part because I want to tie into this concept of pKa. I'd like to at this point talk about acetylide ions. Can I see? I have a question. For the second equation, what magnesium bromide is attacking the like of ethanol? Why doesn't the magnesium bromide bond to the OH but instead bonds to the ozone? What's that? Oh, the question was why doesn't the magnesium bond to the OH? It doesn't, well, there's no OH, oh, you're saying in that reaction because the ethanol has now given up its proton to react with the butyl part. So the proton has come off of the ethanol onto the carbon of the butane leaving an ethoxide anion which combines with the MgBr plus component. And it did. So we had in the case of ethanol, we had EtomgBr. In the case of water, we had HomgBr. So we had a very analogous reaction. Another question. Great question. In that second equation, are we supposed to have two lithiums? This is very typical of how organic chemists will write a reaction. So typically you would see that one might not, particularly as you became more used to writing reactions, it might be implicit. But indeed you would have two lithiums. So you could easily envision writing lithium parenthesis to equivalents or to lithium. And again, this is very much part of the shorthand of writing organic reactions, particularly when focused on synthesis. Good questions. Great question. Would organolithium reagents not be created if you had only one lithium? Well now imagine, what would happen would be you'd have one mole of butyl bromide, one mole of lithium, you'd get half a mole of butyl lithium. Now here's where the fun comes in. If I let that sit at a low temperature and used it quickly, I would have a reaction of one mole of butyl lithium, of one half mole of butyl lithium. But if I let it sit or tried to put it in a bottle and I now had that organolithium reagent sitting for a long time with more butyl bromide, I would get E2 elimination or because the reagent is strongly basic to give butene and butane or I would end up having SN2 displacement to give octane or both. Great question. So yes, I would definitely use two equivalents. All right, well at this point I want to talk about acetylide anions and sort of follow along with our textbook but also because thematically it fits in. So as I said acetylene alkynes in general are especially acidic while there's still very, very weak acids, pKa of about 25, there's strong enough acids that very, very strong bases can pull off their proton. So for example, if I treat an alkyne with sodiumid, I get the sodium acetylide anion. Now sodium's a little more electropositive than lithium. Lithium has an electronegativity of 1, sodium of 0.9. By the point you get to organosodium reagents, they're pretty, pretty ionic in the bond. So I generally think of these as ionic. Ah, great, NH2. NANH2 is sodiumid and if I'm going to balance my equation and as I said organic chemists are usually very bad about this, ammonia, thank you, thanks very much. Ammonia is the other byproduct of reaction and now we see ourselves very much in the situation of a Bronsted acid, Bronsted base reaction. So we have pKa of about 25 and pKa of about 38 and so a difference of pKa means that equation lies way, way, way to the right, 10 to the 13th equilibrium constant you're there about. So basically I throw one mole of sodium amide, one mole of an alkyne and I get essentially all acetylide anion and I'll write a balanced equation or I'll write a synthetic equation here in which I say take propyne. I treat it with sodiumid, NANH2 which you can make by dissolving sodium metal in ammonia with a little bit of iron and it reacts to give sodium amide and for the heck of it, again I'm just trying to give us a range of carbonyl compounds, for the heck of it I'll take 2-butanone and then since I mentioned that I like aqueous ammonium chloride as a source of acid for a workup I'll just demonstrate what I would do in my own laboratory which is to use aqueous ammonium chloride and the product of this reaction is this, it is the alkyne with the alcohol. This is of course racemic, in other words your textbook points out very nicely that you're going to add your nucleophile from both the front face and the back face of the carbonyl and so we have one enantiomer in which the OH is pointing back and the alkyne is pointing out, we have another enantiomer in which it's added from the back face and now the OH is pointing out and the alkyne is pointing back and we will get equal amounts of these, both of the R and of the S. Although this chapter is focusing primarily on carbonyl compounds it's also beginning to introduce ideas of organic synthesis and carbon-carbon bond forming reactions and so your chapter reminds you that you've already seen certain carbon electrophiles so for example you've already seen epoxides so there are lots and lots of types of electrophiles you can generate your compounds with that you can react to your acetylides and other carbon nucleophiles with and I'll just point out one example here that brings out a couple of additional points. So butyl lithium is commercially available, it's a common source of a highly basic organometallic reagent that can be used not only as a nucleophile but also as a very strong base and so butyl lithium is a great reagent for pulling off moderately acidic protons for example protons from alkynes and also protons from amines like diisopropyl amine which you'll see later. Anyway butyl lithium would react with our alkyne, remember our pKa of the alkyne is about 25, our pKa of butane is about 50 and so it would react again in an acid base reaction to give now an alkynal lithium reagent and so we can use this as well as a way of making anions and I'll just give you one example or this as well as a way of making carbon-carbon bonds. So just to give you some diversity in your chemicals in the molecules that you see I'll take say phenyl acetylene and we could envision treating it with butyl lithium. You'll often see butyl lithium written as in buley. N means normal, normal is just a fancy way of saying it's the regular, it's the one butyl lithium. Rather than say the lithium being at the two position of butane which is called sec butyl lithium or being on a tertiary butyl group which is called tert butyl lithium and is a stronger base. Anyway let's envision using the common reagent in butyl lithium treating our phenyl acetylene with it. That's going to give us our lithiated phenyl acetylene, our organolythium compound and just to demonstrate the point of other reactivity we can picture the reaction say with an epoxide and then again an aqueous workup H3O plus and the product of this reaction is an alcohol just as we've been generating in all of the reactions I've shown thus far but what's interesting about this is now the alcohol instead of being connected directly to the carbon that had been or directly to the carbon where the nucleophile attack it's one over you can think of this as R and the alkyne lithium reacting with the epoxide and so we're going to draw electrons flowing from the carbon lithium bond into the carbon oxygen bond pushing electrons on to oxygen and opening up our ring. So what we've done at this point is we've really overviewed a basic of fundamental reaction of carbonyl groups and we've introduced these compounds these reagents that make for very very strong nucleophiles hydride nucleophiles like lithium aluminum hydride in particular to a lesser extent less reactive sodium borohydride and then various organometallic reagents like Grignard reagents and organolithium reagents. At this point I want to sort of broaden out our thinking and start to talk not just about ketones and aldehydes but more broadly about the reactivity of the carboxylic acid family and to just put this into context I mean carboxylic acids essentially all compounds in which we have carbon in the plus three oxidation state. I will get to these several questions in just a moment. Esters being another member of this family acid chlorides so I'm going to paint with a very broad brush and later on we're going to get to a more specific understanding of the reactivity of this broad family but right now I'm going to paint with a very broad brush I'll include in the family acid and hydrides and maybe to wrap up in main members of the family I'll talk about amides. But before we discuss their reactions with organometallic reagents I saw several questions. I think there was one here, one there. Yes, great question. The question was when the lithium attacks the epoxide does it attack from the top or the bottom? Does it occur with inversion of stereochemistry? And indeed it does although there is no stereochemistry at the center that we formed here you could imagine let's say having two different substituents here like a hydrogen and a deuterium and we would get inversion of stereochemistry. Typically epoxides are attacked by nucleophiles by basic nucleophiles at the less sterically hindered carbon. So for example if I had the epoxide with a methyl group called propylene oxide, this one is trivially called ethylene oxide. If I had the epoxide with a methyl group on one side propolythium the alkyne would attack the carbon that didn't have the methyl group on it. Does that make sense to you? Another question I saw one young lady, same question and why do we like to use butyl stuff? Great question. So all of our hydrocarbons ultimately come from petroleum and so one of the ways in which butyl lithium is made is by first cracking, heating petroleum very hot to get smaller fragments and one of the fragments that's easily isolated is butene and the butene can then be taken on to various types of products including butyl bromide for example. So this is one of the reasons that butyl is used. You can buy propolythium but you can buy great big bottles of butyl lithium, methyl is another common one. Does the chemistry rely, oh I love that question. Yeah, almost all of organic chemistry ultimately goes back to petroleum, some of the chemicals that we use as simpler building blocks come from other sources like carbohydrates and you know modern not ancient plant sources. But yeah, almost all of organic chemistry comes back to petroleum so the seeds that you're sitting in have a plastic that may be poly, I don't know, it's polypropylene or something and a covering that's a synthetic fabric like nylon. All of those have come from petroleum and so I look at petroleum as much as I hate to pay $4.30 at the pump when you think about the amount of stuff you're getting, you're getting a gallon of gasoline. This is too valuable to be burning up because there are so many things that you can make from it. You can't get a gallon of anything. You can't get 8 pounds or 7 pounds of anything for $4.00. I can't get a gallon of beer for $4.00 and yet we go ahead and we burn it so yes, petroleum is incredibly valuable to organic chemists and one of my dear colleagues' favorite questions, one that I won't be asking you on an exam because it's too open-ended and far too ridiculously complex for you at this point in your sophistication is to go ahead and write a synthesis of a steroid, let's say cholesterol or testosterone you've seen in your current chapter some steroids starting with petroleum so that would be one of her favorite questions. All right but I want to go now to the reactions of esters and of various members of the carboxylic acid family and I thought your textbooks presentation particularly in the reduction section may have been a little bit confusing because there are a lot of subtleties and when I think about a subject I like to think about it in terms of sort of the general rule and then exceptions to it and there are a ton of little exceptions and your textbook has picked one and we're going to, they're talking about lithium aluminum hydride with amines and we'll talk about that later but right now I want to paint with a very broad brush a sort of general reaction of lithium aluminum hydride. They're these strong nucleophiles, these potent nucleophiles. Hydride sources particularly lithium aluminum hydride just reduces the crap out of everything. Alkyl lithium reagents like methyl lithium just add to everything as much as they can. Ditto for grignard reagents with certain exceptions so again a very broad brush view from 30,000 feet but we're going to take a specific reaction. We're going to take methyl benzoate here to exemplify our point. We're going to imagine treating it with lithium aluminum hydride. I'm deliberately writing this as a synthetic reaction. We'll talk about what's happening in a moment and then an aqueous workup with, I'll just write generically H3O plus and you might do this reaction say in THF and I'll be a good person and write a balanced equation or at least actually I won't write a balanced equation but I will at least write my two organic products of the reaction. Your organic chemist is typically focused on the big stuff but I'm going to write this product, benzal alcohol and the other organic product, methyl alcohol, methanol and collectively then these two constitute the organic products of reaction and what this ends up illustrating is a new property that we haven't yet seen called an addition elimination reaction and we're going to see that this addition elimination reaction goes through an intermediate of benzaldehyde and to a certain extent maybe with the exception of amides here if I took any of the compounds that I'm erasing and treated them with lithium aluminum hydride you would get a similar reaction of them. All right so as I said this is a big mouthful and I like to bake big mouthfuls into bite sized pieces so let's think in sort of broad mechanistic terms here. All right I'm going to think about, I'll write out our benzoate component, I'll write out our methyl benzoate and I will at least for the moment try to be a good person and write a good mechanism in which I write lone pairs of electrons and try to keep track of my charges and to keep things simple I'm going to write rather than writing out all of lithium aluminum hydride I'm going to write hydride just as this abstraction of a hydride anion and the first thing that hydride does is it's a good nucleophile, a potent nucleophile we've talked about the reactivity in general of carbonyl groups. The carbonyl group is an electrophile, electrons flow from the nucleophile to the electrophile and on to the oxygen and I'll try to be a good person and write all of my lone pairs all three lone pairs of electrons around the oxygen and then two on this other oxygen. Now this is not a stable species, this is not something that you can isolate, it's an intermediate and so I'm going to remind us of the fact that it's an intermediate by drawing it in a bracket. We have a special name for this intermediate because we've gone from a trigonal carbon, a carbon with three things around it to a tetrahedral carbon, a carbon with four things around it, I'm going to call it a tetrahedral intermediate and as I said the tetrahedral intermediate isn't stable, the tetrahedral intermediate can break down, electrons flow back down from the oxygen, they push out methoxide and now we get a new carbonyl at this point we've gotten benzaldehyde, the intermediate that I mentioned and methoxide anion with its three lone pairs of electrons on it. The reaction doesn't stop at this point. Esters are less electrophilic than aldehydes, in other words, aldehydes are more electrophilic than esters. We can write resonance structures for esters or a resonance structure in which the methoxy group donates electrons into the carbonyl and makes it less electrophilic. Aldehydes on the other hand have less going for them and so we have more nucleophile and you can't just stop at this point, it's going to further get reduced and so here's our aldehyde, here's our H minus again and again I'll try to be a good person and put it in quotes and electrons flow from our hydride onto the oxygen now to give rise to an alkoxide anion and at this point that's what will sit around in your flask. I haven't drawn the counter ions or anything until you do a workup and I'll just write this as workup, meaning adding some acid or adding some water, I'll put this again in our sort of 30,000 foot view of H plus and the product of this reaction is benzal alcohol and the other product of the reaction is we'll also protonate our methoxide so the other product of our reaction is methanol and I'll just put that in parenthesis here. So I want to show you some generalities, I want to show you some analogy in this and so at this point I'll write essentially the same reaction with just a slight difference on it. So before we could have been thinking about lithium aluminum hydride I said let's consider lithium aluminum hydride at this point let's consider methyl lithium and so I'll take our same ester, I'll take methyl benzawate but I'll treat it first with methyl lithium and I'll write in parenthesis two equivalents, I'll try to remind us that we're using a full amount of it and then secondly we'll treat this with some aqueous acid and the product of this reaction now instead of adding in two hydrides instead of adding in two hydrogens we've added in two methyls, it's essentially the exact same thing and so now we've gotten an alcohol as our product in which we've added in two methyl groups. The reaction is going to go just like on the other one we went via benzaldehyde here the reaction is going to go by way of the ketone called aceto-phenone but just as in the case of the other one we can't stop at the ketone with one equivalent of methyl lithium. The ketone is more reactive than the ester and so as it now is sitting around it immediately reacts as well and so that's our sort of view at 30,000 feet of the reaction of these very, very strong nucleophiles with members of the carboxylic acid family. As I said there are some exceptions, some differences but we can kind of catch this general spirit of this on the following equation and so I'm going to write this, I don't always like the way your textbook presents things particularly with a lot of abstractions because from my way of thinking it's easier to start at the concrete and then work to the abstraction for a computer I think it's very good to like start with an abstraction and you know it can spit out all the examples but we've just looked at two examples. So now I'm going to write the abstraction and I'll say many members of the carboxylic acid family and I'm going to write again this sort of abstraction of nu minus dot dot so some type of strongly basic nucleophile that encompasses all of the reagents that we've been talking about. So I'll say strongly basic nucleophile that includes for example lithium aluminum hydride, RLI, organolithium reagents, RMGX. In other words all of these species have in common that they have a bond between a metal relatively electropositive metal and a highly electronegative species hydride or I'm sorry a more electronegative species hydride or lithium or carbon. So basically the generalities of this type of reaction are that in general we get and I'll write parenthesis XS here just to avoid confusion. When you have XS in general you're going to observe addition of two equivalents of your nucleophile which I guess I've written as NU plus Z minus. So that would sort of be the biggest abstraction and I guess I'll try to be good and keep my electrons and check so I'll write a balance loan here. So what we've looked at here is an addition elimination reaction and we've seen this general principle that when you have something like a methoxy group on a member of the carboxylic acid family the reaction doesn't stop. Things go on. Now to many students the first time they see an addition elimination reaction they find it confusing and here's why they find it confusing. Here's our let's say our ester and we've just added a nucleophile to it NU minus to form our tetrahedral intermediate and that tetrahedral intermediate isn't stable. It breaks down. It kicks out the OR group. In other words our electrons flow like so to push out our OR group that serves as a leaving group and the first time people see this having already been through 51B and learned an SN2 displace was it 51B or 51A you learned SN2 displacement. A 51A and seen an SN2 displacement reaction people start to think about this and say whoa what's going on here? We don't see an OR group like a methoxy group leave in an SN2 displacement reaction but this is a little bit different. Here a less basic leaving group is okay. See in an SN2 displacement reaction you've got to crowd five things around carbon. You don't have a stable intermediate. That leaving group was perfectly happy attached to the carbon and yet something's coming in and pushing it out and it's going over this high energy barrier, this transition state to make that atom leave and so that atom that leaves has to really, really want the electrons. In other words a good leaving group in an SN2 displacement reaction has to go ahead and have a very stabilized anion or conversely because stability of the anion means acidity of the conjugate base conversely a strongly acidic conjugate base. So in this case it's a little different because there's nothing bad about the tetrahedral intermediate. We haven't had to crowd anyone in here. It's easy to get to this point but now it can happily kick out the OR group and what it gets in return is a carbon oxygen pi bond which is very strong and so there's no problem in getting your tetrahedral intermediate together. It's not pentavalent. There's nothing bad about it but it's very good to go downhill and to break down and kick out the leaving group and get back your pi bond. So what I always like to think about here is in an SN2 displacement reaction you've got to have a good leaving group but in the case of this reaction in addition elimination reaction a less basic leaving group is okay and by comparison I'd say maybe less than 5 pKa is good for a leaving group in an SN2 displacement or maybe even an E2 elimination. Thoughts or questions at this point? Doesn't it work like a substitution? Indeed. This is a substitution reaction so the very first step of our methyl lithium plus methyl benzoate reaction was to substitute the methoxy group for a methyl group and get aceto phenone and then of course we could and stop at aceto phenone because it's even more reactive than methyl benzoate so another equivalent of methyl lithium adds but indeed it is a substitution reaction unlike an SN2 substitution reaction here you can have a leaving group that's a little bit less acidic like an alcohol or an alkoxide. If you added a water and acid guess I'm not exact. Exactly following you mean oh you mean in the second step would we? Yes. So in the workup with acid we would protonate indeed and one more question. Okay what do you mean great question what do you mean by a leaving group has less than 5 pKa? In an SN2 displacement reaction chloride bromide iodide are wonderful leaving groups pKa of the conjugate acid respectively about negative 6 whatever number you use about negative 6 for HCl for the conjugate acid about negative 8 or thereabouts for HBr about negative 10 for HI and you can go a little bit less good leaving group. I can write an equation I can give you an example of a case where instead of having a very strong acid counter you know for the conjugate acid in an SN2 displacement you could go slightly from negative you know negative 6 for chloride into the positive range I could give you an example as low as 5 for the pKa beyond that in an SN2 displacement it pretty much isn't going to occur unless like in the case of an epoxide you have ring strain pKa of the conjugate acid in an epoxide is 17 for an alcohol but you've got that roughly 30 kilocalories per mole ring strain associated with the oxirane ring making it pop open. So that's the rare case in an SN2 displacement like reaction where you can actually have what's essentially alkoxide as leaving group but it's spring loaded. However in the case of an addition elimination reaction absolutely no problem kicking out methoxy group or ethoxy group pKa of the conjugate acid 17. Great question. All right at this point I want to move on to the final point that I want to make in today's lecture and to bring us to some ideas of synthesis and showing you how powerful organometallic reagents are in carbon-carbon bond forming reactions but also the process by which organic chemists think about using these reagents and using reagents in general to build up molecules and so I'm going to give you a little bit of a contrived example here but it's very much like the type of thinking that people use as they become more sophisticated. So the example I'm going to give us is to synthesize a particular compound I've chosen it as an example to illustrate a point but indeed something very much like it could say be a useful insect pheromone that you might want to make for a trap for say Japanese beetles. So we're going to try to develop a synthesis of 4 ethyl 4 octanol from compounds containing 4 carbon atoms or fewer. There's nothing magical about 4 carbon atoms but many if you look at commercially available compounds in general most of them are small and you can buy a whole range of different compounds and more big and more complex compounds often are not as available. So you can buy many compounds containing 4 carbon atoms or 3 carbon atoms or 2 or 1. Of course you can buy plant-decontaining 5 but for purposes of this example we're going to say 4 or fewer. And what we're going to do with this is illustrate the way chemists think the process of retrosynthetic analysis and we can call this the process of thinking backwards about how to synthesize something. And the reason retrosynthetic analysis is so powerful, the reason that the process of thinking backwards is so powerful is it's so easy to get caught up in details when you try to look at things from a forward point of view that it's very hard to see how to get there from here from compounds containing 4 carbon or fewer. Oh well we're using organometallic reagents, do I use an organolithium, do I use a grignard, do I use an ester, do I use an acid chloride, do I use butylythium as a base, do I use methylithium? So retrosynthetic analysis is kind of like thinking your way through a chess game. We're going to start with big pictures and then work our way to the details. So let me show you how I would think about this particular example. And you'll encounter on this week's discussion problems several examples very similar to this that involve processes of thinking backward at various levels of sophistication. All right, so let me draw out my compound. So it is, let's see, 1, 2, 3, 4, 5, 6, 7, 8. So here's my target molecule. Now when I first think about this, I could look at this and say oh well, okay, he said 4 carbons are fewer and we just learned about addition of organometallic reagents to esters so I could envision adding, taking some ester would have to be a methyl ester of propanoic acid because that's 4 carbons, I couldn't use any more. And I could envision forming this bond by adding in butylythium and maybe again for the point of view of view from 5,000 feet or 30,000 feet, I'm just going to think in abstractions of some metal whether it's a green yard or a lithium doesn't matter. And again I could think well here we add in some metal and that kind of catches the strategy but we've got a problem here and the problem is how do we get selectivity? I just said you can't really add one equivalent of a green yard reagent or an organolithium compound to an ester, it won't stop at the ketone. Was that your question? All right, so let me take this same idea and see if we can think backward a little bit. So we know, see here I'm trying to do it all at once and I kind of got myself going here, it's a lot better than saying we're going to think forward but let me think a little bit backwards. We know that an alcohol can be formed by a ketone and an organometallic reagent. So we could imagine, so I'm going to use this arrow here, this arrow, this big open arrow means a retro synthetic arrow, it means a thinking backwards arrow. Organic chemists love arrows, curved arrows, equilibrium arrows, resonance arrows, this is a retro synthetic arrow. So I could envision going ahead and going backward to this ketone and I could envision, let's see did I do that right, nope, I could envision this ketone to hexanone and a butyl metal reagent and that would work, I could add butyl lithium or butyl magnesium bromide to two hexanone and that would be okay to make that alcohol but then I'd need a way to make this ketone and oh, well okay, we can make that ketone, we've only learned reactions that form carbon-carbon bonds to make alcohols at this point in this course. So I could imagine making that ketone by oxidation of the corresponding alcohol of three hexanol and again right now I'm not going to worry which reagent to use, whether I use chromium trioxide, whether I use potassium chromate, whatever but now I look and I say oh wait and I can think backwards again now to the point of a propymetal reagent and propanel and at the strategic level we've now broken this molecule apart. We've used the process of thinking backwards to figure out how we can put this molecule together and now having completed our retro synthetic analysis, now we're ready to go forward and worry from the strategy to the tactics of okay, what reagent do we choose, how do we do our details and so the last thing I'll do in solving this hypothetical problem is to show you the synthesis that I've worked out. So I would start with propanel, I've completed the requirement with each of my three components of four carbons or fewer, I've started with propanel, it's commercially available, it fits the requirement. I'll add in just for the fun of it, I'll use propomagnesium chloride, I could use propolythium, I could use propomagnesium bromide, propomagnesium iodide, I'll carry out my workup with aqueous HCl, I could use aqueous sulfuric acid, I could use aqueous ammonium chloride, I could probably be lazy and write H3O plus over the arrow but I'm going to go to the stock room and get some chemicals and I'm going to ask them for some HCl. The product of that reaction after workup is 3-hexanol, I now am ready to oxidize my 3-hexanol, you learn lots of reactions last quarter for oxidation, they taught you potassium dichromate in sulfuric acid and water sometimes called Jones reagent, there are many, many reagents based on chromium-6, sodium dichromate, chromium trioxide, there are lots of safer alternatives including alternatives based on bleach but we're going to use reactions that you know so we're going to use potassium dichromate. Then at that point we have our 2-hexanone end question. Absolutely, we could use chromium trioxide as an oxidizing agent, lots to choose from nice in our retrosynthetic analysis, not having to worry about getting that right yet and now having the leisure of going and choosing our reagents and choosing our tactics. Finally, to complete our synthesis I'm going to take butylythium and I'll do an aqueous workup and again I'll use HCl, I've written it below the arrow here, written it above the arrow there, doesn't really matter, chemist is going to read it the same way and lo and behold, I have proposed a rational and selective synthesis of our target molecule for ethyl-4-octanol and this art of recognizing something in a molecule and seeing where it comes from is going to grow and grow in the course. Right now we've seen an alcohol and we said oh, I know how to make an alcohol. I can make an alcohol by adding in two carbon nucleophiles, I can make alcohols by adding in carbon nucleophiles. Later on we're going to see all sorts of other families of carbonyl compounds. All right, I will see you on Thursday. We'll start off with a 10-minute quiz.