 All right, give yourselves a big round of applause. So much of what happens in learning organic chemistry is not about me or about Johnny or about Kim. It's about you and actually working problems. And you're one-sixth of the way there with the quizzes and you're a good fraction of the way there with the homework. You're on board and you're keeping up. Now this quiz may have been a little easy. Chapter 19 wasn't that meaty but this gives you the flavor. The next one I'll probably ask you some questions about mechanism or about synthesis and curved arrows. We'll figure things that develop a good set of skills. But the quizzes are going to build on this problem-solving aspect of the course. And the skills that you've cultivated and the work that you put in in preparing for this and doing the homework is exactly what's going to set you up to excel in the class. So again, big round of applause for yourself. So today I want to move into the last section of chapter 20. Chapter 20 is kind of interesting because it has a ton of different stuff and it sort of is all over the map in terms of reactions. And I thought about it and what I've tried to do is really to group things into conceptual themes. And the last conceptual theme that I see from this introduction to chapter 20 and as I said, this is only a part of carbonyl chemistry because we're going to be continuing through four chapters, well five chapters, six chapters technically, 19 to I think 24 on carbonyl chemistry. It's a huge swath of the course in seven weeks. So in breaking out the last part of chapter 20, what I'd like to do today is to talk about selective reactions of carbonyl groups and the carbonyl family and also about the use of protecting groups in synthesis. So I'm going to start with sort of a conundrum I raised at in yesterday's class and that is that we talked about the addition of various types of nucleophiles to esters. We talked about addition of powerful hydride nucleophiles like lithium aluminum hydride and powerful carbon based nucleophiles like organolithium reagents or grignard reagents. And we said in those reactions you would typically take the ester, remember the ester is plus three oxidation state of carbon and not move one step down, one notch, one pair of steps down but move two steps down to the primary alcohol oxidation state. In other words, if we take an ester like methyl benzoate, it's the one I wrote in the previous class and we treat it with lithium aluminum hydride and then we carry out an aqueous workup with water as your textbook writes or H3O plus aqueous acid like HCl in water. The product of the reaction is benzyl alcohol and organic chemists are often really bad about writing small molecules and byproducts of reaction so I'll just remind us of course the other product of this reaction is methanol. And if I were thinking about a synthesis, maybe I wouldn't write it. If we go ahead and do the same chemistry, the same substrate methyl benzoate and now instead of treating it with lithium aluminum hydride, we treat it with say methyl lithium or methyl magnesium bromide and I'll specify that we're talking about two equivalents here although we really can't get away from problems happening from the lack of stopping happening and then again we do some type of aqueous workup. Now we've gone down and added two equivalents of the methyl group and gotten this tertiary alcohol. And the main point of this example is that we cannot stop these reactions even though they go through the intermediacy of benzaldehyde for the former and for benzophenone on the latter or for acetophenone on the latter we can't stop the first one at benzaldehyde or the second one at acetophenone. And so organic chemists are all about control. We're all about being able to choose and to develop tools to make the chemical compounds that we want to make and we have a conundrum here in the lack of selectivity, the lack of ability to select for the aldehyde or the ketone. I saw a question over there from the gentleman in the third row. For the exam are we going to have to put the byproducts? Great question, very reasonable question. When I'm thinking about a synthesis and I'm thinking about taking my methyl benzoate on through several steps to a product that contains that benzene and the carbon and the methyl is just sort of a throw away, I probably wouldn't write it. But if I'm thinking about writing an equation or a balanced equation in which I show all products of reaction or a mechanism I certainly would. Good question, I know and I know all these little variations. End up being of tremendous concern to you. Why do I sometimes draw lone pairs? Why do I sometimes draw curved arrows to show mechanisms in how I'm thinking? And other times we just zip from reactants to products. It's expressing different things. One is expressing a level of detail how the reaction is occurring, the other is what we're forming or in the case where we're emitting the methanol, simply what we're forming as the product. And here I guess I will to keep consistent in the first panel here I will again sort of parenthetically write methanol is the product. But good question and a very, very legitimate concern, particularly when you're starting out learning and trying to figure out what people are expecting of you. So chemists have developed a number of different reagents to achieve selectivity in these types of reactions. Another question, another good question. So your textbook sometimes shows things as covalent bonds and sometimes as ionic bonds. So we're going to see that today to some extent. In general, lithium and magnesium when bonded to carbon are primarily covalent in their bonding. Sodium is a little more electropositive. It's more ionic in its bonding. Sometimes we'll write an ionic resonance structure. I mentioned in the last lecture. We'll sometimes in our mind think of methyl lithium as methyl minus with a lone pair, lithium plus. That's a non-bond resonance structure. But the main resonance structure is the covalent one. All right, so on to selectivity and control. So your textbook presents one reagent to do this reaction selectively. That reagent is dibal, diisobutyl aluminum hydride. So first let me write this as a synthetic reaction. And so if you take the reagent dibal, this reagent has three different names that people call it depending on the source you go to. The reagent is diisobutyl aluminum hydride. That's two isobutyl groups and an aluminum on hydride and a hydride on a hydrogen on aluminum. And you will see it written sometimes as dibal with an H on the end to indicate there's a hydride. Sometimes as dibal without the H on the end. And sometimes as dibah just because people want to catch the notion of the H. Anyway, if you treat your ester with dibal and then do an aqueous workup, and again you could add water or I would probably add aqueous acid, the product of reaction is the aldehyde. You can stop the reaction at the aldehyde. And again the byproduct of the reaction would be methanol as a byproduct. So the gist here is just as we've dealt with other sorts of nucleophiles like methylithium adding to a carbonyl, here we have a hydrogen nucleophile, hydride nucleophile on aluminum. And the overall result or partway through you get a tetrahedral intermediate so your methyl benzoate reacts with diasebutyl aluminum hydride to form a tetrahedral intermediate with your aluminum and your two isobutyls on there. But the main thing is because the aluminum oxygen bond is very strong, stronger and more covalent. Again, it's this notion of degree of covalency, degree of ionicness because aluminum is a little bit less electropositive than say, lithium. You have more covalent bond character. You form a very strong bond to oxygen. And you have a tetrahedral intermediate that's stable enough to not break down. I'll write it as more stable tetrahedral intermediate. In other words, your tetrahedral intermediate is sufficiently stable that until you carry out the aqueous workup until you add water or you add aqueous acid it sticks around. And remember the mechanism for this double reaction with lithium aluminum hydride. Hydride adds, you get a tetrahedral intermediate. The tetrahedral intermediate breaks down kicking out a methoxide leaving group. That gives you the aldehyde, another mole of hydride adds giving you a new tetrahedral species, your final alkoxide that protonates on workup. And so at this point you can stop the reaction. Now in practice, sometimes a chemist would choose to use dibal as a reducing agent to stop at the aldehyde. And sometimes a chemist would say, all right, look I'll just use lithium aluminum hydride, reduce it all the way down to the primary alcohol and then oxidize the benzal alcohol back up to the aldehyde. Sometimes in practice it's a little more efficient one way or another. As in many cases, there are multiple ways, multiple right ways of doing things. And in the laboratory a chemist is often concerned about what's easiest or what gives the best overall chemical yield, the best amount of product in the end. All right, so that takes care of achieving selectivity in our reaction to add a hydride nucleophile. Let's now take a look at achieving selectivity in adding an alkyl nucleophile. Now as I said before, carboxylic acids are part of a big family, a family in which your carbon is in the plus three oxidation state. And all of these species are easily interconverted. You'll be learning more about these interconversion reactions in the subsequent chapters. Carboxylic acids, esters, acid chlorides, acid and hydrides, amides, and even nitriels end up being all in this plus three oxidation state family of carbon. So in order to achieve selective addition of an alkyl group we're going to use a different member of the carboxylic acid family, an acid chloride. And you'll learn about how to form acid chlorides later on. Now if you just treated almost any member of the carboxylic acid family with an organometallic, oops, an organometallic, cool iPhone. With an organometallic nucleophile. So if we treated it with the say methylithium or methyl magnesium bromide we would have, and again I'll indicate that we're talking about two equivalents but the main point is that we can't really stop at one equivalent. If we, in other words if you added one equivalent it wouldn't stop at the ketone in the other case. We'd just get a mixture of the alcohol and unreacted ester in some ketone. It wouldn't be a controlled reaction. So if we imagine treating with methylithium and then in a second step, again our aqueous workup and I'll just write this as H3O plus because in my lab we'd certainly throw aqueous acid in here. The problem is we'd be in the exact same situation that we were with methylbenzoate. We would have two equivalents of methyl nucleophile adding here. We really don't have control. And this is essentially the same theme. There are little variations. Your textbook, this is kind of the same theme with lithium aluminum hydride. If we treated lithium aluminum hydride with an acid chloride or a carboxylic acid or an ester it would always reduce it down. There are little variations in reaction of grignards and organolithiums with carboxylic acid. There are little variations which your textbook gives you an example on in the reaction of lithium aluminum hydride with amides to give you a means. But for the overall general principle it's members of carboxylic acid family with these powerful nucleophiles, lithium aluminum hydride, grignard reagents and organolithium compounds all get addition of two equivalents and all kick out the leaving group. Question? No. And you could easily write excess. In my own laboratory people would probably use two and a half or three equivalents in order to do this to make sure that there was enough of the reagent to achieve the reaction. But yeah it would be a minimum of two equivalents. But as I said if you used less you wouldn't still wouldn't be able to stop and controllably have a ketone product. All right. The reagents that people have developed to make this reaction selective because what this is all about is selectivity. It's achieving the reaction that you want to occur. The reagent that people have developed to make this selective is organocuprate reagents. I'll write the reagent first and then I'll show you how you make it. So the reagent here is dimethyl, is lithium dimethylcuprate CH32CULI. It's a carbon nucleophile again. Now we have carbon bound to copper. Copper is even less electropositive than lithium or aluminum. And so it's even more of a covalent bond. The exact mechanism of the reaction goes beyond the scope of this class but in the very last lecture if we go on schedule and talk about chapter 26 I'll be giving you a little bit of a flavor of the very special reactivity of transition metals like copper and palladium. But that will be later on if we have time. Anyway the overall result of this reaction is that you can controllably add one equivalent of your methyl group to your member of the carboxylic acid family. In this case specifically your acid chloride for reasons of the mechanism. An organocuprate reagents can be made by the reaction of two equivalents of methylithium with copper iodide to give you CH32CULI which as I said is called lithium dimethylcuprate and you'll also if you want to write a balanced equation which as I said organic chemists tend to be very bad at you'll get lithium iodide as the other byproduct of the reaction. There are many subtleties in this chemistry. Many of which go beyond the scope of the course. Certain groups like ethyl groups due to what's called beta hydride elimination don't work as well in this reaction. Your textbook gives you examples of vinyl cuprate. Those work well. Certain cuprates like acetylienes on copper don't transfer. At this point I'm just giving you and your textbook is just giving you a teeny taste of this big wide world of selective reactions here. Another way one can actually achieve this reaction selectively is to add a little bit of copper to a grignard reagent but again we're not going to focus heavily on any of that right now. I just want to give you a little flavor as your textbook does of this notion of control in a reaction. Question. For the exam it is a very good idea to know arrow pushing for organolithium reagents and grignard reagents and lithium aluminum hydride and this overall principle of reaction of the carboxylic acid family. The addition elimination reaction of the first nucleophile followed by the addition of the second nucleophile. But for more specialized reagents like the organocuprate reagent as I said the mechanism goes a little beyond the scope of the course. All right so this brings us to a notion here of control and specifically the notion of chemo selectivity that is achieving control for the reaction of one group but not another and in a way we've already seen that notion in these two reactions and that we're controlling the reaction to get just to the aldehyde or ketone oxidation state. But really the best sort of the archetypal examples of chemo selectivity are when you have two functional groups and you want to carry out a reaction on one of those functional groups to the exclusion of the other functional group. So let us take and again like many of my examples in the class they'll be a little bit artificial in the sense that you might not immediately have a reason to do this let's take a derivative of a compound called pterothalic acid where we have half of it as the aldehyde and half of it as the ester. And if we imagine treating this bicarbonyl compound, this bicarbonyl compound with lithium aluminum hydride and then with aqueous acid, lithium aluminum hydride just blasts down all of the members of the carboxylic acid family. It reduces all of them. It's a super powerful source of hydride nucleophile. And so our ester group gets reduced and I'll just write it as CH2OH to tell us that we have the alcohol and our aldehyde gets reduced to the alcohol. In other words, this reaction is not chemoselective. It doesn't hit just one group. Now conversely, if we take a milder reducing agent like sodium borohydride, then we do get selectivity. So I'll just write ditto mark for the reactant here and then I'll write sodium borohydride. Typically people use alcohol salvants like methanol for a sodium borohydride reduction. If we carry out the reaction of this ester aldehyde with sodium borohydride, then we will selectively reduce the aldehyde group without reducing the ester group. We say that sodium borohydride is chemoselective. It reduces aldehydes and I might add ketones but not esters and so that principle of control is something that's very useful because it allows you to make a product that you desire and as I said, so much of organic chemistry is about making molecules that often are useful for example as drugs. I want to give you a couple of other examples of selectivity and I borrowed these directly from your textbook. There are many ways to do these types of reactions and I thought this example which is straight from your textbook would actually illustrate these points very nicely. So if we take a compound like cyclohexenone which has both a carbon-oxygen double bond, a carbonyl group and a carbon-carbon double bond, we have two different groups that can be reduced. The carbon-oxygen double bond and the carbon-carbon double bond. Reagents like sodium borohydride or lithium aluminum hydride and I'm not going to write all the conditions. I'll just write sodium borohydride or lithium aluminum hydride reduction. Reduce the carbonyl group. These reagents are nucleophilic toward carbonyls. They add to the carbonyl group. If we take the same cyclohexenone and instead perform catalytic hydrogenation on it, your textbook writes one equivalent of hydrogen and I will add or low pressure. In other words, the way you could do this reaction is either to go ahead and control the amount of hydrogen that you're adding or carry out the hydrogenation quickly with just a balloon of hydrogen at atmospheric pressure monitoring by TLC and you do this in the presence of a palladium on carbon catalyst or a platinum catalyst. Your textbook uses palladium on carbon as an example. You can selectively reduce the carbon-carbon double bond and so go from cyclohexenone to cyclohexanone. Hydrogenations of carbonyls are more difficult. It requires more effort to reduce the carbon-oxygen bond. In other words, hydrogenation is chemoselective for the carbon-carbon double bond. But if you beat on it, your textbook writes it as H2XS. That might be a big balloon of hydrogen and monitoring the reaction for a long time or sometimes people will carry out these reactions at higher pressure. I'll say high pressure. By high pressure, it can mean anything from bicycle, tire tube pressure of like 75 PSI in a special glass bottle to an apparatus that's steel with a tank that can hold thousands of PSI aptly called a bomb. If one hydrogenates under high pressure, again with a catalyst, it can be palladium on carbon or platinum oxide or any of that platinum family of metal, then you can reduce both the carbon-carbon double bond and the carbon-oxygen bond. And again, one is achieving control, one is achieving chemoselectivity in this type of reaction. There are entire books on the topic of catalytic hydrogenation, in fact, a book by that title. So there's a lot of different ways to do this reaction. And again, your textbook is just giving you a little taste which is fine because it gives you a lot of the needed thought processes. Now, in these reactions on this blackboard, we've looked at taking the oxidation state down. In first case, we took the oxidation state of the carbon-oxygen double bond down. We reduced the oxidation state of the carbon. In the second reaction, we've reduced the oxidation state of these two carbons. And in the third reaction, we've reduced them both. Just as there are many ways to reduce the oxidation state of carbon, there are many ways to increase the oxidation state of carbon to oxidize carbon. And your textbook gives you a little taste. They give you three related sets of oxidation conditions to oxidize a primary alcohol up to a carboxylic acid. All the ones that they use, that they show you involve chromium. It's a classic potassium dichromate or sodium dichromate or chromium trioxide in sulfuric acid and water are all essentially equivalent. All of these oxidize primary alcohols up to carboxylic acids. There are newer, more modern reagents that don't use toxic carcinogenic chromium. Your textbook leaves those out and that's fine. As I said, you can write an entire book. Ditto, there are ways to stop at the aldehyde oxidation state and your textbook gives you one such set of reagents, the use of silver oxide in aqueous ammonia and that will stop you at the aldehyde oxidation state. And again, what this is all about is about control in chemo selectivity. All right, so I guess by conclusion I will just write, there are many other selective oxidizing agents. All right, at this point I would like to turn my attention to another problem. So we've looked at a problem of chemo selectivity and at this point I'd like to turn my attention to the problem of regioselectivity and we've kind of seen this issue already with the hydrogenation question and I'd like to give of us the same type of example and bring back our cuprate reagents. So here's our cyclohexenone again and if we treat our cyclohexenone with a powerful nucleophile, we already saw lithium aluminum hydride and I'll now give you an example of methyl lithium or a Grignard reagent. I'll choose methyl magnesium bromide just as an example and then we carry out an aqueous workup. We add, just like the lithium aluminum hydride did, we add to the carbonyl group. The powerful nucleophile attacks the carbonyl group. What's interesting with this alpha beta unsaturated carbonyl compound is if we treat it with an organocuprate reagent, so I'll just write a ditto mark and then I'll bring back our lithium dimethyl cuprate and again we'll carry out some type of aqueous workup. Typically you would use acid. Now we get a very different sort of addition. The addition is not to the carbon oxygen bond but rather to the carbon bond in what we call a 1,4 addition and I'm going to contrast that to the previous example which we call a 1,2 addition. And I'd like us to think, as I said, we're not going to really fuss about the mechanism of the cuprate reaction in detail and why, or to put it more specifically, why copper does what it does. The reasons for the details of the difference are rather subtle but we're going to focus on the house and the watts, the house and the watts that are happening. So in the 1,2 addition, here's our carbonyl compound, here's our alpha beta unsaturated carbonyl compound. I call it an alpha beta unsaturated carbonyl compound because the carbon that's next to the carbonyl group, we call the alpha carbon and the carbon that's 1 over, we call the beta carbon. So in the first example, our nucleophile which we could think of as methyl minus or we could think of it as a methyl bound to a lithium but I'll just write it as nu minus, our nucleophile attacks the carbonyl group and I'm going to draw in a lone pairs of electrons to help us think about this and I'll draw a curved arrow from the nucleophile, from the thing with the electrons to the electrophile to the thing that wants the electrons and I will push electrons up onto the carbonyl oxygen because of course if I stopped at this point I would have 10 electrons around the carbon and we would be in trouble. So at this point in the mechanism we have formed an alkoxide anion, the nucleophile has added and that's how your first reaction goes. And one of the things that was guiding our thinking in thinking about the reactivity of the carbonyl group is we recognize that although the carbon oxygen bond is a covalent bond it is a polar covalent bond. We embodied that last time or the other time when we talked about this we embodied this in two ways. One was writing a delta minus on the oxygen and a delta positive on the carbon. The other way of embodying or thinking about this is to write two different resonance structures and so I can write one resonance structure like so, the main resonance structure or major resonance structure and then I can write a second resonance structure that reflects the polarization of the bond where the carbon doesn't have a complete octet. And it's not as important to resonance structure because a resonance structure in which the carbon doesn't have a complete octet is less of a contributor. But I can also write a third resonance structure and the third resonance structure is to continue to move the electrons of the double bond much as you do in an allylic cation which you learned about in the previous quarter and now you have your positive charge over at the beta position. And in order to help avoid confusion I want to draw in my key hydrogens here so that you can see that just like the resonance structure at which you have the positive charge on the carbon yield carbon, this last resonance structure, this other minor resonance structure has a positive charge on the beta carbon which has one other hydrogen. In other words it doesn't have a complete octet either. So collectively these three resonance structures make up a more complete picture that help us think about the reactivity and in the case of the 1,4 addition, again we're going to skip the whys of it at this point in your learning of organic chemistry but in order to understand the reactivity on the 1,4 addition reaction you can say okay now I can see why in some cases a nucleophile is going to react at the carbonyl carbon and at other cases because there's also a partial positive charge at the beta position at some cases the nucleophile is going to add to the beta carbon and now we just push our electrons like so to represent and think about the reactivity here. Now that kind of took us halfway through the conjugate addition through the 1,4 addition of the cuprate because we did a second step where we did an aqueous workup and the product of that reaction which I've just erased it was on the bottom of your right hand blackboard was a ketone and at this point we're halfway through we don't have a ketone and we're about to add a source of aqueous acid. Now the species that we've just generated here is called an enolate anion and this is your first exposure to the enolate anion you're going to be learning more about it in a moment more specifically you're going to be learning more about enolate anions in chapter 23. So here is your enolate anion and just like your ketone you can think about it having multiple resonance structures. Again this is going to come back more later on so we can think about one resonance structure indeed it's going to be a major resonance structure, the major resonance structure which I've written on the left and a second resonance structure which I've written on the right and you're going to as I said be seeing more about these in chapter 3 but for now what you can think about is that in the workup step of the organocuprate reagent you protonate the enolate anion. You can think of it in occurring in two different ways. I probably like to think of it more in the way that I'm writing over here. You can think about the proton going onto the alpha carbon to give rise to your ketone product and in terms of Occam's razor in terms of the greatest simplicity that's probably the way I like to think about it. Your textbook writes the mechanism slightly differently and that's okay too. Your textbook likes to think about the proton going on to the oxygen like so and this product is called an enol and the enol is in equilibrium or goes to form the ketone by a reaction which again you will be learning more about in chapter 23 called tautomerization. I'm going to write that again. We call the ketone and the enol forms tautomers and we call their conversion tautomerization. This issue of selectivity and achieving control is one that cannot always be done by picking the right reagent. In other words, sometimes you can say oh yeah I want a Grignard reagent here, I want a Cooperate reagent here, I want lithium aluminum hydride here, I want DiBal there but there are cases where one gets into a situation where no amount of choice of the right reagent alone is going to do something, allow you to do something that you want to do and let me show you an example of this and how chemists work around it. Imagine that I wanted to generate a Grignard reagent from 3-bromo-1-propanol and I eagerly envisioned saying okay I'm going to go ahead and treat my 3-bromo-1-propanol with magnesium and diethyl ether in hopes or in expectations of getting this Grignard reagent. We wouldn't get it. The problem with this Grignard reagent is it's inherently unstable, it can't be formed and I gave you an example in our previous lecture that sows the seeds for destruction here. If we take a Grignard reagent and treat it with an alcohol like ethanol, this is the example from the previous class, I pointed out that Grignard reagents are very strongly basic and alcohols are sufficiently acidic that they protonate them. If we take butyl magnesium bromide and treat it with ethanol we get butane plus ethoxymagnesium bromide, basically the magnesium salt, the magnesium alkoxide salt of ethanol. And so this hypothetical Grignard reagent that we would like to generate is unstable, it can't be formed, it can't be isolated, it can't be generated. It immediately reacts with itself with the alcohol quenching the Grignard reagent. And so to come around, to work around problems like this, organic chemists will use something called a protecting group. Protecting group is an idea that you're going to hide, you're going to mask one part, reactive part of the molecule, you're going to protect it from a reagent or from conditions that would otherwise destroy it. And at this point in your textbook they introduce one class of protecting groups, you'll learn about more later, protecting groups that they introduce are silo ethers. And so the general gist of protecting an alcohol as a silo ether, particularly a tert-butyl dimethyl silo ether, is to take your alcohol, treat it with a chlorosilane, tert-butyl dimethyl dichlorosilane is particularly popular and a base, the base that's widely used is called amidazole. It's a good base for this particular reaction. I guess I won't draw in lone pairs. And the product of this reaction is our silo ether. By the way, we call this reagent TB-DMS chloride and then we're going to call our ether the TB-DMS ether. I'll write it out here as R-O-T. Actually, I will write it first showing you the silicon SI dimethyl tert-butyl. And we also get, if you want to write a balanced equation, we'll also get protonated amidazole, the amidazole hydrochloride. But again, organic chemists typically don't pay attention to these byproducts of reaction. Now, the big idea behind protecting groups is that they are easy to put on, they hide the alcohol and then they're easy to take off. So before I show you a whole sequence of reactions, let me show you how to take the protecting group off. So we have our TB-DMS ether and I'll just write this as R-O-T-B-D-M-S. If you notice me slurring my words here, many organic chemists abbreviated further, just like we said, di-bal-H, they abbreviate further. Many abbreviated further as a TBS ether. Anyway, if you treat your TB-DMS ether with tert-butyl ammonium fluoride, again, now we get sort of into the land of acronyms, tetrabutyl ammonium fluoride is a big fat tetrabutyl ammonium cation with fluoride as an anion. It's a source of fluoride that dissolves in organic solvents like tetrahydrofuran and fluoride loves silicon. Oxygen likes silicon a lot, fluoride loves silicon. It forms a really, really, really strong bond to silicon. When we treat our TBS-DMS ether with T-Baf we get back our alcohol and again, organic chemists are terrible about writing byproducts of reactions but I will, for this one time, write tert-butyl dimethylsilofluoride as a product. So this brings us to an application of protecting groups because the reason that people want to put protecting groups onto alcohols are to solve synthetic problems and do something they couldn't otherwise. So let me show you an example of doing something we couldn't otherwise. We want to generate the equivalent of this green yard reagent that I just said you couldn't make. So imagine that we take the Bromo 1 propanol just as we did before but instead of treating it with magnesium and ether we treat it with TB-DMS chloride and emidazole. Now we would get the TB-DMS ether as I said Johnny would write it as the TBS ether to save room. Now if we treat our TB-DMS, see I can't even write all of those letters TB-DMS ether with magnesium and diethyl ether now we can generate a green yard reagent that's the equivalent of the green yard reagent that we couldn't do. And if we then let's say added that green yard reagent so I'll just continue my equation over here. If we imagine adding that green yard reagent to let's say a generic ketone or aldehyde I'll just write it as R carbonyl R prime to indicate that we're not dealing with any specific ketone or aldehyde. Now we could generate this otherwise impossible to achieve product and the only thing that would be different about this product than having generated that green yard reagent that we couldn't generate is we have the TB-DMS group on that. But we know what to do with that. We simply take this and we treat it with Tbaff and the silo group comes off and we've generated a product that we couldn't otherwise get. One of the tremendous intellectual triumphs of organic chemistry is the ability to think backwards about reactions. It's retro synthetic analysis and it's one of the reasons that my academic grandfather and the academic father of the textbook E.J. Corey received the Nobel Prize in chemistry and so I want to show you an example of this very important intellectual process of thinking backwards about making organic molecules. And so I'm going to give you again a little bit of a contrived problem that gives you the flavor of synthesis. Let us ask how would we synthesize the molecule 15 octane diol from compounds containing four carbon atoms or fewer? So I'll say compounds containing four carbon atoms or fewer. Again, just like in my previous example, I've done this to give you sort of a small catalog of organic compounds that you can think of without necessarily saying there aren't bigger compounds available. But since we've already seen molecules like imidazole and t-baff and so forth, our reagents that may have in TBDMS chloride are reagents that may have many carbon atoms. I will open up our catalog and any reagents that you require. And while to my knowledge, 15 octane diol isn't an insect pheromone and it isn't a drug, it's certainly representative of the types of molecules that synthetic organic chemists make. So of course, one of the things is knowing what 15 octane diol is and so I will draw out the structure of it. And of course, you have to think, all right, there's an octane chain, so there's an eight carbon chain with an alcohol at the one position. So one, two, three, four, five, six, seven, eight and an alcohol at the five position. Now, we've started to learn to recognize that different groups can come from different sorts of connections and often there are lots, lots and lots of correct ways to put together molecules. We're looking to build a molecule here. And so we know, for example, since we're looking to make it from smaller pieces, that we probably want to put it together by carbon-carbon bond forming reactions. Alcohols could come from ketones, alcohols can come from addition of grignard reagents to ketones or aldehydes. So we could imagine breaking this molecule apart and in this view that I mentioned from 30,000 feet last time, you can look at this molecule and say, okay, I could envision forming this carbon-carbon bond by some form of addition of an organometallic reagent to a carbonyl compound, in this case for maldehyde. I could envision forming this carbon-carbon bond from a molecule where we add a carbon nucleophile like a grignard reagent or an organolithium compound to an aldehyde. These are both perfectly good ways to put the molecule together. Both of them, however, involve a fragment in one case that's seven carbon atoms. In one case, that's five carbon atoms. But if I envision forming this carbon-carbon bond, now things look really, really attractive. We've split the molecule into two equal pieces. My hypothetical example gave you only up to four carbons to work with. And so that's really, really convenient here. And so we could think in terms of retro-synthetic analysis and say, yeah, a good retro-synthetic disconnection would be to break this molecule into this four-carbon grignard reagent, this four-carbon aldehyde. And remember, this is the view from 30,000 feet. And we already know that we can't make this four-carbon grignard reagent because we know that grignard reagents containing alcohols and grignard groups are inherently unstable. But now we have all of the tools that we need in our toolbox to solve this problem because we know that we're clever enough to work with compounds that are equivalent to this grignard reagent by use of protecting groups. And so we can approach this problem with confidence and now work our solution through in a forward direction. We start with four bromo, one butanol. We protect it with TB-DMS chloride and amidazole to get the corresponding TB-DMS ether. We take our TB-DMS ether, we treat it with magnesium and diethyl ether, and then we add to that grignard reagent. If you've generated one in the laboratory, you'll typically take metal turnings, magnesium turnings, put them in a dry flask, maybe flame dry the flask. You'll add your ether, you'll add your grignard reagent, you'll crush the magnesium turnings until the reaction starts and you generate your grignard reagent. Once you've generated your grignard reagent, you will immediately add to it butanol. You'll probably distill your butanol or your TA will distill your butanol so it's nicely pure. You'll carry out an aqueous workup with aqueous. Let me choose an acid here. I should commit. Let's say aqueous HCl. You literally could do this at the chemistry stock room. You would go to them and ask them for all of these reagents. A little short on space here. I've got to crunch in my TB-DMS group. And now in the final step of our synthesis, you simply treat this TB-DMS ether with T-Baff and low and behold, you've generated your 1,5-octane diol. All right. Well, this is the way we think about carbonyl chemistry. We've learned a lot of reactions of them. As we continue into the next chapter, we're going to learn new reactions of carbonyl groups. Thank you.