Lecture 03. Reactions of Organometallic Reagents.
Nowick, James
UCI Chem 51C Organic Chemistry (Spring 2012) Lec 03. Organic Chemistry - Reactions of Organometallic Reagents - Instructor: James S. Nowick, Ph.D. This is the third quarter course in the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. Organic Chemistry 51C is part of OpenChem. 
 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 minus 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 in our group the use of primes and double primes and so forth is just my way of representing that there are 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. In 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 the 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. Now, 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 work up 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 an adduct with a new carbon-carbon bond. 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 the pheromone will lure the insect to the trap and then they'll fall in and die. 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 a partial negative charge in carbon and a partial positive charge on oxygen, on magnesium. Because 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 ion 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 carbon ion. 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 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. The sP2 has 33% s character, sp3 has 25% s character as you hold electrons more tightly the CH bond is more willing to give up H plus and give you a carbon ion. 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% 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 I wrote let's 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. All right well, Gouignard 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 phenolithium 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 IUPAC 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. 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. 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 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? Why doesn't the magnesium bond to the OH but instead bond to the O2? What's that? Oh, the question was why doesn't the magnesium bond to the OH? It doesn't, well there's no O8, 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. What was that? Did that look the same thing happened with water? 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. In that second equation are we supposed to have a term with you? Ah, 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 two equivalents or two lithium. And again this is very much part of the shorthand of writing organic reactions particularly when focused on synthesis. Good questions. So will organolethium reagents not be created if we don't have two lithiums? Ah, great question. Organolethium reagents not be created if you had only one lithium. Well now imagine you would what would happen would be you would have one mole of butyl lithium, one mole of butyl bromide, one mole of lithium you would get half a mole of butyl lithium. Now, here is 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 organolethium 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 ionic in the bond so I generally think of these as ionic. So I have to do the NH2 or the NH4O. 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 are thereabouts. 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. 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. Your chapter reminds you that you've already seen certain carbon electrophiles. For example, you've already seen epoxides. There are lots and lots of types of electrophiles you can generate your compounds with that you can react your acetylides and other carbon nucleophiles with. I'll just point out one example here that brings out a couple of additional points. 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. Butyl lithium is a great reagent for pulling off moderately acidic protons. For example, protons from alkynes and also protons from amines like diisopropylamine, 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 alkynolithium 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 N-buty. 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 N-butyl lithium treating our phenyl acetylene with it. That's going to give us our lithiated phenyl acetylene, our organolithium 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 attacked, 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 onto oxygen and opening up our ring. All right, so what we've done at this point is we've really over viewed a basic, a 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 the 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. So 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. I was wondering does that last reaction even come up where those lithium carbon bond for electron with the acid the apocyt. Does it have the same amount of acid in the bottom? Ah, great question. The question was when the lithium attacks the apoxide 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. The 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 so much? 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. So does chemistry rely on fossil fuel? Does the chemistry rely on, oh I love that question. 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 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. But yet you can't get a gallon of anything. You can't get eight pounds or seven pounds of anything for four bucks. I can't get a gallon of beer for four bucks. 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. Under 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 make big mouthfuls into bite size 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 methylbenzoate, 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 is in 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. Electrophils, on the other hand, have less going for them. We have more nucleophile and you can't just stop at this point. It's going to further get reduced. 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. Electrophils flow from our hydride onto the oxygen nailed 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 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. All right, 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. 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 benzoate. 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. That'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's going to go just like on the other one. We went via benzaldehyde. Here the reaction's 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 start with an abstraction. And 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 in check, so I'll write a balance loan here. All right, 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. 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 to elimination reaction, a less basic leaving group is OK. 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 methylithium plus methylbenzoate reaction was to substitute the methoxy group for a methyl group and get acetophenone. And then of course we couldn't stop at acetophenone because it's even more reactive than methylbenzoate, so another equivalent of methylithium adds. 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 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. What does it mean by a leaving group has less than 5 pKa? OK. 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. Now 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 retro-synthetic analysis and we can call this the process of thinking backwards about how to synthesize something. And the reason retro-synthetic 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-carbons 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 butylithium as a base, do I use methylithium. So retro-synthetic 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. Alright, 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 grignard 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 grignard reagent or an organolithium compound to an ester, it won't stop at the ketone. Is 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 retrosynthetic arrow, it means a thinking backwards arrow, organic chemists love arrows, curved arrows, equilibrium arrows, resonance arrows, this is a retrosynthetic 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, two 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 could 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 propyl metal reagent and propanol 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 at. So I would start with propanol, I've completed the requirement with each of my three components of four carbons or fewer, I've started with propanol, propanol it's commercially available, it fits the requirement, I'll add in just for the fun of it I'll use propyl magnesium chloride, I could use propyl lithium, I could use propyl magnesium bromide, propyl magnesium 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 and question. Absolutely, we could use chromium trioxide as an oxidizing agent, lots to choose from, nice in our retro synthetic 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 low and behold I have proposed a rational and selective synthesis of our target molecule for ethyl for 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. I will see you on Thursday, we'll start off with a ten minute quiz.
 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 minus 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 in our group the use of primes and  double primes and so forth is just my way of representing that there are 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.  In 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 the 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.  Now, 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 work  up 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 an adduct  with a new carbon-carbon bond.  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 the pheromone will  lure the insect to the trap and then they'll fall in and die.  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 a partial negative charge in  carbon and a partial positive charge on oxygen, on magnesium.  Because 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 ion  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 carbon ion.  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 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.  The sP2 has 33% s character, sp3 has 25% s character as you hold electrons more tightly  the CH bond is more willing to give up H plus and give you a carbon ion.  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% 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 I wrote let's 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.  All right well, Gouignard 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 phenolithium 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 IUPAC 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.  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.  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  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?  Why doesn't the magnesium bond to the OH but instead bond to the O2?  What's that?  Oh, the question was why doesn't the magnesium bond to the OH?  It doesn't, well there's no O8, 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.  What was that?  Did that look the same thing happened with water?  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.  In that second equation are we supposed to have a term with you?  Ah, 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 two equivalents or two lithium.  And again this is very much part of the shorthand of writing organic reactions particularly  when focused on synthesis.  Good questions.  So will organolethium reagents not be created if we don't have two lithiums?  Ah, great question.  Organolethium reagents not be created if you had only one lithium.  Well now imagine you would what would happen would be you would have one mole of butyl  lithium, one mole of butyl bromide, one mole of lithium you would get half a mole of butyl  lithium.  Now, here is 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 organolethium 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 ionic in the bond so I generally  think of these as ionic.  So I have to do the NH2 or the NH4O.  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 are thereabouts.  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.  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.  Your chapter reminds you that you've already seen certain carbon electrophiles.  For example, you've already seen epoxides.  There are lots and lots of types of electrophiles you can generate your compounds with that  you can react your acetylides and other carbon nucleophiles with.  I'll just point out one example here that brings out a couple of additional points.  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.  Butyl lithium is a great reagent for pulling off moderately acidic protons.  For example, protons from alkynes and also protons from amines like diisopropylamine,  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 alkynolithium 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 N-buty.  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 N-butyl lithium treating our phenyl acetylene  with it.  That's going to give us our lithiated phenyl acetylene, our organolithium 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 attacked,  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 onto oxygen and opening up our ring.  All right, so what we've done at this point is we've really over viewed a basic, a 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 the 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.  So 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.  I was wondering does that last reaction even come up where those lithium carbon bond for  electron with the acid the apocyt.  Does it have the same amount of acid in the bottom?  Ah, great question.  The question was when the lithium attacks the apoxide 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.  The 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 so much?  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.  So does chemistry rely on fossil fuel?  Does the chemistry rely on, oh I love that question.  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 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.  But yet you can't get a gallon of anything.  You can't get eight pounds or seven pounds of anything for four bucks.  I can't get a gallon of beer for four bucks.  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.  Under 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 make big mouthfuls into bite  size 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 methylbenzoate, 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 is in 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.  Electrophils, on the other hand, have less going for them.  We have more nucleophile and you can't just stop at this point.  It's going to further get reduced.  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.  Electrophils flow from our hydride onto the oxygen nailed 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 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.  All right, 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.  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 benzoate.  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.  That'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's going to go just like on the other one.  We went via benzaldehyde.  Here the reaction's 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 start with an abstraction.  And 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 in check, so I'll write a balance  loan here.  All right, 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.  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 to elimination reaction, a less basic leaving  group is OK.  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 methylithium plus methylbenzoate reaction was to substitute  the methoxy group for a methyl group and get acetophenone.  And then of course we couldn't stop at acetophenone because it's even more reactive than methylbenzoate,  so another equivalent of methylithium adds.  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 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.  What does it mean by a leaving group has less than 5 pKa?  OK.  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.  Now 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 retro-synthetic  analysis  and we can call this the process of thinking backwards about how to synthesize something.  And the reason retro-synthetic 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-carbons 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 butylithium as a base, do I use methylithium.  So retro-synthetic 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.  Alright, 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 grignard 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 grignard reagent or an organolithium  compound to an ester, it won't stop at the ketone.  Is 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 retrosynthetic arrow, it means a thinking backwards arrow, organic chemists love arrows,  curved arrows, equilibrium arrows, resonance arrows, this is a retrosynthetic 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, two 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  could 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 propyl metal reagent and propanol 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 at.  So I would start with propanol, I've completed the requirement with each of my three components  of four carbons or fewer, I've started with propanol, propanol it's commercially available,  it fits the requirement, I'll add in just for the fun of it I'll use propyl magnesium  chloride, I could use propyl lithium, I could use propyl magnesium bromide, propyl magnesium  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 and question. Absolutely, we could use chromium trioxide as an oxidizing  agent, lots to choose from, nice in our retro synthetic 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  low and behold I have proposed a rational and selective synthesis of our target molecule  for ethyl for 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. I will see you on Thursday, we'll  start off with a ten minute quiz.
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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 minus 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 in our group the use of primes and double primes and so forth is just my way of representing that there are 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. In 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 the 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. Now, 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 work up 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 an adduct with a new carbon-carbon bond. 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 the pheromone will lure the insect to the trap and then they'll fall in and die. 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 a partial negative charge in carbon and a partial positive charge on oxygen, on magnesium. Because 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 ion 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 carbon ion. 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 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. The sP2 has 33% s character, sp3 has 25% s character as you hold electrons more tightly the CH bond is more willing to give up H plus and give you a carbon ion. 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% 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 I wrote let's 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. All right well, Gouignard 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 phenolithium 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 IUPAC 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. 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. 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 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? Why doesn't the magnesium bond to the OH but instead bond to the O2? What's that? Oh, the question was why doesn't the magnesium bond to the OH? It doesn't, well there's no O8, 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. What was that? Did that look the same thing happened with water? 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. In that second equation are we supposed to have a term with you? Ah, 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 two equivalents or two lithium. And again this is very much part of the shorthand of writing organic reactions particularly when focused on synthesis. Good questions. So will organolethium reagents not be created if we don't have two lithiums? Ah, great question. Organolethium reagents not be created if you had only one lithium. Well now imagine you would what would happen would be you would have one mole of butyl lithium, one mole of butyl bromide, one mole of lithium you would get half a mole of butyl lithium. Now, here is 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 organolethium 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 ionic in the bond so I generally think of these as ionic. So I have to do the NH2 or the NH4O. 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 are thereabouts. 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. 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. Your chapter reminds you that you've already seen certain carbon electrophiles. For example, you've already seen epoxides. There are lots and lots of types of electrophiles you can generate your compounds with that you can react your acetylides and other carbon nucleophiles with. I'll just point out one example here that brings out a couple of additional points. 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. Butyl lithium is a great reagent for pulling off moderately acidic protons. For example, protons from alkynes and also protons from amines like diisopropylamine, 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 alkynolithium 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 N-buty. 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 N-butyl lithium treating our phenyl acetylene with it. That's going to give us our lithiated phenyl acetylene, our organolithium 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 attacked, 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 onto oxygen and opening up our ring. All right, so what we've done at this point is we've really over viewed a basic, a 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 the 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. So 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. I was wondering does that last reaction even come up where those lithium carbon bond for electron with the acid the apocyt. Does it have the same amount of acid in the bottom? Ah, great question. The question was when the lithium attacks the apoxide 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. The 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 so much? 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. So does chemistry rely on fossil fuel? Does the chemistry rely on, oh I love that question. 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 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. But yet you can't get a gallon of anything. You can't get eight pounds or seven pounds of anything for four bucks. I can't get a gallon of beer for four bucks. 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. Under 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 make big mouthfuls into bite size 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 methylbenzoate, 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 is in 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. Electrophils, on the other hand, have less going for them. We have more nucleophile and you can't just stop at this point. It's going to further get reduced. 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. Electrophils flow from our hydride onto the oxygen nailed 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 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. All right, 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. 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 benzoate. 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. That'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's going to go just like on the other one. We went via benzaldehyde. Here the reaction's 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 start with an abstraction. And 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 in check, so I'll write a balance loan here. All right, 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. 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 to elimination reaction, a less basic leaving group is OK. 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 methylithium plus methylbenzoate reaction was to substitute the methoxy group for a methyl group and get acetophenone. And then of course we couldn't stop at acetophenone because it's even more reactive than methylbenzoate, so another equivalent of methylithium adds. 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 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. What does it mean by a leaving group has less than 5 pKa? OK. 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. Now 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 retro-synthetic analysis and we can call this the process of thinking backwards about how to synthesize something. And the reason retro-synthetic 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-carbons 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 butylithium as a base, do I use methylithium. So retro-synthetic 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. Alright, 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 grignard 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 grignard reagent or an organolithium compound to an ester, it won't stop at the ketone. Is 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 retrosynthetic arrow, it means a thinking backwards arrow, organic chemists love arrows, curved arrows, equilibrium arrows, resonance arrows, this is a retrosynthetic 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, two 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 could 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 propyl metal reagent and propanol 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 at. So I would start with propanol, I've completed the requirement with each of my three components of four carbons or fewer, I've started with propanol, propanol it's commercially available, it fits the requirement, I'll add in just for the fun of it I'll use propyl magnesium chloride, I could use propyl lithium, I could use propyl magnesium bromide, propyl magnesium 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 and question. Absolutely, we could use chromium trioxide as an oxidizing agent, lots to choose from, nice in our retro synthetic 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 low and behold I have proposed a rational and selective synthesis of our target molecule for ethyl for 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. I will see you on Thursday, we'll start off with a ten minute quiz.
organic chemistry is actively working problems this through chemical synthesis in which we take about the properties of Grignard reagents and very much part of the shorthand of writing reaction is this. It is the alkyne with the alcohol. This is of course racemic. In other compounds in which we have carbon in the plus taken on to various types of products including this product, benzal alcohol, and the other