Lecture 02. Reactivity of Carbonyl Compounds.
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Nowick, James
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. 
 So today we're going to really start from launching to the main video course and we'll start to talk about reactivity of carbonyl clouds. And this is going to encompass capture 20, but it's going to not just run through 20, but really all the way 24, captures 24, the next six weeks. This is the main piece of the course and this is also going to really be some of the intellectual height of organic chemistry. And we're going to see general themes of reactivity of carbonyl compounds and really it has to do with sort of two properties of carbonyl compounds. One is the reactivity with nucleophile, which is what we'll be focusing on today. And the other property that we're going to be learning about particularly as we move through the course into I think chapters 22, 23 and 24 is that there, well maybe I should say there are reactions with nucleophiles let me. Oh we'll keep it as reactivity. All right and the other will be there reactivity as nucleophiles. What do I mean by that? Well the first one is a little bit easier to see and is where we're going to be starting today and indeed in the next few lectures. And that is that the carbonyl group is electrophilic. The carbon of the carbonyl group is electrophilic and we're going to see how nucleophiles can come in and right now I'm just going to draw an arrow not really yet representing flow of electrons to say that nucleophiles are interested in reacting with the carbonyl carbon. The second concept reactivity as nucleophiles we're going to see as we develop the idea of enols and enolates, isomers and related species to carbonyl compounds. And at that point when we start to invoke that reactivity, what we're going to learn is that the alpha carbon, alpha carbon means the one over can react as a nucleophile and again that's not meant to represent the flow of electrons yet just showing the position and maybe I'll go ahead and emphasize that by saying alpha carbon. So in general as you go along from a carbonyl group you'd say the carbon next to it is the alpha carbon, the carbon one beyond that is the beta carbon, one beyond that is the gamma carbon. And here we'll refer to this as the carbonyl carbon. So in its own right this diagram doesn't really say very much but it sort of lays a framework. So let's start to move from this level of abstraction to a level of bonds and structure. And let's start with carbonyl group. And this isn't going to be acetone or any particular carbonyl group. I'm just drawing this as a general carbonyl group. And we'll be good people. We'll write our own pairs of electrons and focus on carbonyl group. So oxygen is more electronegative than carbon. Oxygen has a greater nuclear charge. It pulls the electron forward. In other words the carbon oxygen double bond, the bonds that comprise a sigma bond and a pi bond are while shared electrons while the covalent bond between carbon and oxygen they're not sharing equally. Those electrons spend a little more time at the more electronegative oxygen, the atom with the greater nuclear charge. And so I can represent that in terms of a partial negative charge and a partial positive charge. In other words since the electrons while spending time between oxygen and carbon spend a little more time at oxygen, we shift our charge over to oxygen. There's a partial negative charge. Our carbon is a little bit deficient in charge. There's a partial positive charge. We can also represent this type of structure by resonance structures. And remember resonance structures aren't one or the other but both at the same time to varying degrees or in some cases more than one if you're more than two. And so we can think of the carbonyl group as having a major resonance structure. And also a minor resonance structure in which instead of having a double bond and having a complete octet for carbon and a complete octet for oxygen, we can represent a minor resonance structure as having a single bond with a full negative charge on carbon on oxygen and a full positive charge on carbon. Now I'll bracket these structures to represent that it's, the two of them together we're thinking about. We're not resonating or vibrating or oscillating between the two structures. We're mostly the one on the left and a little bit of the one on the right. This is also representing the type of electron density that you would see in the molecular orbitals for the carbonyl group. The other thing I'll point out is we're going to be focusing a lot on arrows in this course. We'll use them to think about and communicate ideas of electron flow of equilibria and of resonance. And there are different types of arrows that we use to communicate different ideas. The double arrow like this communicates resonance. It communicates that it's both structures. It's not an equilibrium arrow which is two arrows pointing in the opposite direction. And it's not a flow of electrons where we draw an arrow originating at a electron and flowing at something at a once. In other words, originating at the nucleophile and flowing at the electrophile. And that actually brings us to our next element of reactivity and that is the reaction with nucleophiles. Now if you have a carbonyl group and again I'll draw it just as some generic carbonyl group not meant to represent acetone, not meant to represent any particular group and it encounters some nucleophile. Again for now we'll just represent that nucleophile in the abstract. They can react with each other. And so I'll use a reaction arrow to show that we're going from reactants to product. The product of this reaction now can be represented like so where our nucleophile has formed a bond to the carbon. And we can think about how that reaction occurs by thinking about the flow of electrons from the nucleophile, from the thing that has electrons to the carbonyl carbon, to the carbon that wants electrons. In our mind's eye we have in mind a picture where the carbonyl group is polarized and so electrons want to attack as carbon. In the very back of our brain we have a picture like this is very minor resonance structure guiding us and making us think from where the electrons want to flow. And of course we can't stop at this point because if we just left our electrons here now we would have 10 electrons around the carbon. We would have more than an octet. That wouldn't be possible. And so electrons have to continue to flow as electrons are flowing from the nucleophile to the electrophile to the electrophilic carbon in terms electrons are flowing up onto oxygen. And so when all is said and done we formed a bond between the nucleophile now represented by this line or these two electrons and we move two electrons onto oxygen and in turn put our negative charge there. Now as I said at that point I'm dealing with abstractions. The nucleophile here is not meant to be any particular nucleophile. But now let's come maybe to something more specific and in turn this is not meant to be any particular carbonyl group. But let's take a look at acetone and a nucleophile and sort of halfway between reality and an abstraction at this point. And so the nucleophile that I'll draw is hydride. And I think I'm going to put it in quotes. The reason I'll put it in quotes is you essentially never have hydride as an entity. The very end of our discussion here I'll show you an example of an actual hydride with a hydride anion that actually doesn't react as H minus or at least doesn't react as a nucleophile. But for now we'll worry about what our hydride source is in just a moment. For now just think about the abstraction of hydride and we'll have hydride reacting with acetone. And so electrons are going to flow from the hydride from the lone pair to the carbonyl. And again we can't stop at that point. Our electrons have to flow up from the double bond onto the oxygen atom. And so when all is done. Now we have an anion. It's the anion that would have been derived from isopropanol so we call it the isopropoxide anion. And it is a member of the broader family of what we would call alkoxide anion. So if you have an anion that's derived from an alcohol that's equivalent to removing H plus from an alcohol, you end up with an alkoxide anion. So a lot of organic chemistry question. Oh yeah, so does the nucleophile attack when the carbonyl is at its minor resonance structure? Or does the nucleophile cause the minor resonance? All right. I'm going to toss this question which is a good one out to somebody in the class. Does the nucleophile attack when it's the minor resonance structure or does it attack? Or does it attack cause the minor resonance structure? It wants to take a stab at it. Does it attack when it's the minor resonance? Okay, so now we have doesn't it happen in both because it alternates? It says it's a hybrid. A hybrid, yes. So I didn't like the word alternate but you've got the right idea. It's not one or the other. It's both resonance structures at the same time. So it's not like it's waiting for the minor resonance structure. They're both there at the same time. The problem is chemistry is all about monos, about conceptual monos. And the beautiful thing about drawings like this is they're simple. We can wrap our head around. That's super, super powerful. The problem, so that's the plus. The disadvantage is that these models are always incomplete. So there are layers of meaning in the models. And another approach is to look at molecular orbitals and electron density but it's harder to immediately wrap our head around things. The beauty about these drawings is you can scratch them in a stick on the sand or under paper and have an immediate grasp of reactivity. So it's not one, not the other but both at the same time. Now the second half of your question is interesting because of course the bond forming process is one where the electrons do start to approach. And as those electrons are approaching, as the nucleophile is approaching the electron, we're moving the other electrons continuously up onto the oxygen. So in a way this is what you would call a reaction coordinate. We're going in the process from the nucleophile to the electrophile being very far apart and having almost completely a double bond character in the electrophile and no bond between the nucleophile and the electrophile. We're moving the nucleophile into the carbonyl group. The carbon oxygen double bond is becoming more and more like a single bond and getting weaker and weaker until finally as we've gone over that energy of the transition state and down to the products, we've now formed that single bond. So that's the reaction we can have. Other questions? Good question. Should we be concerned with the stereochemistry? Ah, great question. Hang in there. Question, what should we be concerned with the stereochemistry? So let's start at the beginning. So the answer is yes and let's get to that point. All right. So organic chemists love to make stuff. We live and die by making new medicines, making new drugs, having reagents that cause chemical reactions and inventing new reagents. And so this is a huge part of organic chemistry. It's fun because you have an idea in your mind's eye and then you test it out. So two reagents that are used as a source of this, what I would say here is abstract or hypothetical hydride and ion is lithium aluminum hydride, too very popular in long standing reagents. Lithium aluminum hydride and sodium borahydride. Both of these reagents react as hydride with carbonyl compounds, particularly with aldehydes and ketones, which is what we're going to focus on today. The structure of lithium aluminum hydride is an aluminum hydride anion or tetrahydride anion as a lithium anion. The structure of sodium borahydride is a borahydride anion, BH4 minus, like the other is ALH4 minus, and a sodium cation. And you'll notice these structures are related. Boron is above aluminum in the periodic table, so in many ways they have similar properties. They differ in that aluminum is more electropositive than boron. The electronegativity of hydrogen is about 2.2. The electronegativity of aluminum is about 1.6. The electronegativity of boron is about 2.0. In other words, in all of these species, you have the bonds polarized to having more negative charge on hydrogen than on the metal. But in the case of aluminum hydride, they're even more polarized. In other words, aluminum hydride is a more reactive anion. Lithium aluminum hydride is more reactive. And you can see this in the laboratory immediately. If I take a spoonful of lithium aluminum hydride, a great powder and throw it into water, I better get back because it's going to release hydrogen with a big fizz and maybe even catch fire. And if I eat it too much, it'll explode. Zodium boron hydride is a white powder that will sort of bubble in water, but won't react nearly as fast releasing hydrogen gas. And there was a question here. Would NAH also release a hydride? NAH, we're going to come to in a moment. Great question. Let's talk about some reagents and their reactivity first. So in a way, that falls into the category of the exception and we're going to see its reactivity in a moment. So let me take an organic compound. We'll take a simple compound. It's a ketone. It's a ketone with a benzene ring on one side of the carbonyl group and a CH3 on the other side. You might see people write the CH3. You might see them leave it just as a line. Both are correct ways of writing it. The IAPAC name for this compound is one phenylethanone, but nobody is going to call it that. Everyone's going to call it aceto phenone because it's one of these compounds where the common name just really dominates. So I'll write out the systematic name just as a reminder. But again, the common name is going to predominate. And we're going to imagine treating this compound with lithium aluminum hydride. And now we're going to see how it reacts as hydride anion. As I said, aluminum is more electropositive than hydrogen. The bond between them is covalent, but it's a polar covalent bond. And it's one in which we can think of it almost as H minus or in other ways we can think of the electrons as flowing from the bond between aluminum and hydrogen onto the carbonyl carbon, toward the carbonyl carbon. And remember, I use an arrow from the electrons to the thing that wants electrons. I put two little arrowheads on here. It's representing that we have two electrons. And for now at least, I will be good at drawing my lone pairs of electrons and reminding us that we have two lone pairs and we have one more pair of electrons flow up there. And so when all is said and done, we end up with the alkoxide anion. And for now at least, I will be good at drawing in my hydrogen. And in addition to this, we have lithium cation. And in addition to this, we have alH3. And that's where I'm going to stop it at this point, is at the alkoxide anion. And that's how you can think of it, but I'll put a little bug in the back of your mind. In the back of your mind, you can think about the fact that aluminum doesn't have a complete octet. And in reality, this lone pair can now attack one of the lone pairs, can now attack the aluminum and fulfill its octet. So technically you'll end up with an aluminum alkoxide salt. All right, so this is kind of a mechanistic way of thinking about this reaction. So let's go and think about things from a synthetic point of view. So redemic chemists will often write equations as little recipes. Mix this with your compound and then mix in something else and then you get that. And organic chemists are very bad at writing balanced equations. Organic chemists often will just write reactants and products. Question. Is there a conservation of charge? Oops, good point. Thank you. So yes, so our lithium aluminum hydride, we have a lithium cation, an aluminum hydride anion, that's our reagent, the aluminum hydride transfers its hydride. At the end you have your lithium cation, your alkoxide anion, and alh3, which as I said will typically bond to oxygen. Also typically that undergoes further reaction with additional ketones. So often when you're doing the reaction in the laboratory you don't mix one mole of a ketone and one mole of lithium aluminum hydride but you might use two moles of a ketone or even three moles of a ketone and one mole of lithium aluminum hydride. So let me write a reaction as a synthetic organic chemist would. So now I'll be a little bit sloppier in the sense that I'm not going to write in my lone pairs of electrons. We're thinking more like a synthetic organic chemist. Not even going to write in my methyl group. It's implicit over here. And so step one is treatment with lithium aluminum hydride. I'll write it as LiAlH4, but sometimes you'll see it abbreviated as LaH, lithium aluminum hydride. And then I'll write in a second step in this recipe. And I'll write in the step H3O plus or your textbook writes H2O and I'll tell you about that in just a second. And then I'll probably also indicate the solvent that I would use to carry out this reaction. And typically I would carry out the lithium aluminum hydride reduction reaction in THF or in diethro ether. So this constitutes a little recipe. It says take some benzophenone, put it in a flask with some THF or some ether, throw in some lithium aluminum hydride, and then after a while after the reaction is done, throw in some acid or some water. Of course you can't go through one acid, you can't go to the stock room and say, give me a bottle of H3O plus. Look at you funding and say, oh, we don't have any bottles of H3O plus. We have aqueous hydrochloric acid, which of course is H3O plus and CO minus. We have aqueous sulfuric or concentrated sulfuric acid, which you can add to the water to make H3O plus and bisulfate N9 HSO4 minus. So that's a little bit of a short-term for adding an acid that you would get a strong mineral acid like H2O4 H2O4. Now, I think we call this step here the workup. I'll just write solvents and I'll just write workup here. And honestly, this particular one you could do with either aqueous acid or with water. Aluminum salts in water under the warm condition can form tremendous gelatinous things. You've all had lab, you've all shaken the septic funnel. You may have even encountered an emulsion. Who's encountered emulsion? I hate emulsions, don't you? The advantage of throwing in a ton of acid is it dissolves all the aluminum salts and gets rid of the emulsion. You can also add water in just the right way, a little bit of water, a little bit of sodium, a little bit of sodium sulfate, which makes your aluminum instead of gelatinous, makes it granular. So, either of those work for a workup and you'll see things written both ways. The main thing, of course, is that until we add something that provides a proton, whether it's acid, H3O plus aqueous HCl, or whether it's water, we have an alkoxide anion. But after the workup, you isolate your product, the alcohol. And as I said, organic chemists are typically very bad about balancing equations. So, typically this becomes a little recipe. And as I said, I'll just write it explicitly, our workup ends up taking our alkoxide anion, and we have the alkoxide anion plus H3O plus, plus hydroamine, or plus water under the right circumstances, and electrons flow from the alkoxide to the source of protons, and electrons flow from the hydrogen oxygen bond, onto the oxygen. And as I said, we've called this step. Questions or thoughts at this point? Ah, does it always make a secondary alcohol? Good question. If we had an aldehyde, we would get a primary alcohol. Of course, you couldn't have, you couldn't make a tertiary alcohol with hydride nucleophile, because by definition, hydride is going to add in a hydride. Next class, we'll learn about Brignard reagents and Organolithium reagents, instead of adding quote, unquote, H minus, can add an alkover to light, quote, unquote, metal minus, and you'll see how they pervade for, for example, tertiary alcohol. All right, there's a concept that really is useful in permeating your thinking, and that concept is oxidation state. I introduced it last time when I was talking about the carboxylic acid family, and I said in the carboxylic acid family, the oxidation state of an alcohol is, or the oxidation state of the carbon is plus three. In a ketone family, the oxidation state is plus two. If you remember in calculating oxidation state from general chemistry, you play a game where you divide up the electrons, giving them to the more electronegative atom, and then saying, okay, what's the net charge on the atom? If you have a carbon bound to another carbon, you divide them equally. So in a ketone, the oxidation state is plus two. I'll write ketone here. In our product, in our particular alcohol that we've, we've generated, secondary alcohol, our oxidation state is zero. So if you notice, by going from an oxidation state of plus two to an oxidation state of zero, we have undergone a reduction reaction. And the chemical that's done it then, our lithium aluminum hydride, is a reducing agent. It's the chemical or the reagent that carries out the reduction reaction. And there are many different reducing agents. I drew another one. I drew sodium borohydride. And let's talk about that. And I'll give you an example. Since the very astute gentleman asked the question about aldehydes, let's take, and there's primary alcohols. Let's go ahead and take an example of reduction of an aldehyde to a primary alcohol. And so we'll take benzaldehyde. Benzaldehyde is a very nice smelling compound that kind of smells of cherries. It's sort of the flavor of cherry coater, artificial cherry flavor. If you take, for example, sodium borohydride, the reaction would work with lithium aluminum hydride, but I'd like to introduce you to other common reducing agents. And in this case, you can actually run the reaction in a cell that's like methanol or ethanol. I said sodium borohydride reacts only slowly with water. If I tried to run a lithium aluminum hydride reaction in methanol, just like water, it's a source of hydroxylic protons. If I tried to run a lithium aluminum hydride reaction in methanol, I'd be rewarded by the whole thing as best spatter in the face and worst tattoo in the fire. So our product of reaction here is benzalazole. And we can think of this reaction occurring very much like the other reaction has occurred. So here's our aldehyde. Here's our borohydride anion. And the boron hydrogen bond is polarized. It's a polar covalent bond with more electron density on the hydrogen. And so borohydride anion can transfer a hydride adding it as a nucleophile to the carbonyl group. And again, we pick up our electrons onto oxygen. So now we get our alkoxide anion and BH3. And again, I'm skirting a little bit loosely because BH3 doesn't have a complete octet. And so typically it's going to form a bond with oxygen reversibly. Or we can think about it as forming a bond with oxygen. Now the nice thing about running this reaction in a hydroxyl exolvent is you've got this huge, huge excess of methanol. And so methanol can protonate the alkoxide anion. So here we have our alkoxide anion. We're in a ton of methanol. And so electrons can flow from the oxygen of the alkoxide anion to the proton of methanol. We put electrons back onto the oxygen. And after all is said and done in this equilibrium reaction now, driven by the mass action of having so much methanol in it. Now we have our alcohol in metoxide anion. I'm sorry, wouldn't metoxide be more happy without the H or without the H? Great question. Would the metoxide be more happy without the H or with the H? So in general, if you want to sort of keep one thing in mind, the basic rule would be all alcohols have pretty similar acidity. Water, I said to PK, was 15.7. That's sort of at one extreme methanol. It's about 16 ethanol, maybe pushing toward 17, 16 or 17, tert-butanol toward 18. Benzal alcohol is a little bit more acidic inductively. But the main point here is we've got a ton of methanol. Your methanol is 20 or so molar and you've got one molar of compound. So you've got this huge excess of methanol grinding this equilibrium to the right. And then you're going to do an aqueous workup in the end that's going to extract your compound into an organic salve. So you're right. It is an equilibrium. In this particular case, it's okay that it's an equilibrium. But that equilibrium, remember I was talking about PK last time and I said, the thing that we really want to be able to focus on in organic chemists, it does an equilibrium lie way to the left, way to the right, kind of in the middle or maybe a little more to the left or a little more to the right. That's how we think about things. And you're right. This equilibrium lies kind of sort of in the middle. Do you want to run a reaction with a liquid in your project and you react with methanol? Oh, ah. Well, if you've tried to run a lithium aluminum hydride reaction in methanol, the lithium aluminum hydride would probably first react with the methanol. But as I said, make hydrogen so vigorously and so rapidly that it would blow up in your face. So it would definitely react with the methanol. It might also react with the ketone, but I'm not going to try the experiment in the laboratory and see at the end when I wash my face off whether I had any reduced ketone. So in this case, in this case we started with benzaldehyde. And the oxidation state of benzaldehyde is plus 1. And by the time we've gone to benzal alcohol, again it's a reduction reaction. The oxidation state is minus 1. And that's not important in thinking about families because remember I said carboxylic acids and the whole carboxylic acid family, esters, amides, nitrioles, all that entire family is plus 3 oxidation state. They are more oxidized than aldehydes. And then aldehydes are more oxidized than primary alcohols. And you'll notice they go in twos and this is pretty common in organic chemistry. As you change oxidation states almost invariably you end up going into organic chemistry. The reason it's important to sort of keep this in the back of your mind is it helps you focus your thinking. When you change from a carboxylic acid to an ester through a chemical reaction, there's no change in oxidation state. You wouldn't think about using an oxidizing agent or reducing reagent to affect that change. You're changing the same oxidation state. You're changing from a ketone or an ester or an aldehyde to an alcohol. Then you're changing oxidation state and you think about using a reducing agent. If you're changing from an alcohol to a ketone or an aldehyde or an acid, you would think about using an oxidizing agent. There are a ton of high-grade reducing agents. Your textbook has given you some, I'll show you a few others. The general principle, the overarching principle of high-grade reducing agents is that you have a metal hydrogen bond, metal broadly defined, usually as a polar covalent bond where you have M with a partial positive charge and high-grade with a partial negative charge. By broadly defined, I mean that Goron is kind of a metal or a metalloid, aluminum is a metal, and there are many, many others. So the general principle, as I said, with lithium aluminum hydride technically after that first mole of hydride is transferred, technically what happens is the AlH3 reacts with the alkoxide and now this intermediate can transfer again hydride anion to another carbonyl group, which is why you may only use one mole of lithium aluminum hydride for several moles, for two moles or three moles, up to four moles, although typically you wouldn't use it of your electrophile, of your carbonyl compound. Your textbook gives an extremely example of this where you transform, you preform the reagent with three tert buttoxy groups on aluminum and this is a less reactive hydride source. There are two reasons for this. One reason why once you start to put alkoxide on the aluminum, it's less reactive is that the oxygen is electronegative and it's pulling electron density away so there's less of a partial minus charge on hydride. The bond is less polarized. Another reason is sterics, the tert-fuel group is very big and so that big set of tert-fuel groups gets around the way of this. So only the very most reactive carbonyl compounds react with a tetra-tritricerfutaxiome and hydride reagents. The oxygen metals that don't have a full octet can react as a hydride source. Your textbook mentions and we will talk more about in a moment or later lithium, a diasecudal aluminum hydride, abbreviated dival age, sometimes just dival, you will see it. And again, you have a polarized metal bond. Typically what happens in a dival reaction is the aluminum which doesn't have a complete octet first coordinates onto an oxygen or onto a nitrogen to get a complete octet and a formal negative charge on aluminum and then finish the hydride. I'll just show one other reaction. Sodium cyanoborohydride. It's another hydride reagent that's less reactive. Cyanogrups are electron withdrawing and so sodium cyanoborohydride is less reactive than sodium borohydride. There's a big alphabet soup of different reagents out there but you'll notice that the general theme in all of these reagents is that a metal is bonding the hydrogen. So the various student question was asked about sodium hydride. And that sort of is the oddball exception. Sodium hydride, NaH, actually reacts as a strong base not as a reducing agent. And there are a couple of reasons for this. All of the reagents I showed over there and the sodium borohydride and the lithium aluminum hydride have covalent bonds between metals and hydride. Sodium is even more electropositive than aluminum. Its electronegativity is about.9. And so the sodium hydride is actually an ionic compound and it's pretty darn insoluble in organic solvents. As a result you never really have a chance to have soluble hydride and it doesn't get into and react as a nucleophile. Plus because the bond is very ionic and because hydrogen is relatively electropositive compared to say chloride anion, sodium hydride is a very strong base. And so in general if you take sodium hydride with almost any source of protons, water or say methanol, it reacts. And so if I throw sodium hydride into say methanol, it reacts to form sodium methoxide and hydrogen gas. And so I was talking about acidities before and saying well you don't typically calculate things but you'll keep in mind where an equilibrium lies based on the difference in pKa's. Hydrogen has a pKa so this is an acid base reaction. Methoxide anion reacts with methanol to give methoxide anion. I've just shown it as NaOCH3 but of course that's Na plus OCH3 minus anionic compound plus hydrogen gas. So this is an acid base reaction. The methanol is an acid pKa about 16 or 17. The hydrogen is an acid on this side of the equation. Its pKa is about 35. It's a very weak acid whereas methanol is just a weak acid so that reaction, I won't even write it as an equilibrium. The reaction lies way, way, way to the right. pKa of 35, pKa of 16, difference of 19. If I were to write an equilibrium constant it would be 10 to the 19th just massively on the right. So that reaction ends up reacting as a base. Good point for some questions. The sodium hydride reacts bilimply in water. Absolutely. If you get water on it you see a beautiful bright orange flame. The big danger, if you were to take a scoop of it and throw it in water you'd get a big flash of fire. The big danger in the laboratory is when you're weighing it out you spill a teeny little bit and later on you're cleaning up and splattering acetone and water around. At that moment when the water hits an acetone is splattered around all the acetone and all the solvent goes up in fire. So when I work with sodium hydride what I'll generally do is get a wad of wet paper towels and wipe down my work area and I might see a little flash of orange. And that flash of orange under the big wet wad of paper towels doesn't do any harm and it prevents my feet from having to fire over and lighting lots of acid. How do you find oxidation state numbers? Okay great question. Alright oxidation state. State two examples here. We'll start with an extreme example. Carbon dioxide and methane. Carbon has four nuclear charges. We have eight electrons but we're going to play a game here. The game is that we give all those electrons, I'll even be good and write my Lewis acid structures, or write my Lewis electron dot structures. We play a game that we give all those electrons to oxygen and leave none for carbon because oxygen is more electronegative. Because we have a net four nuclear charge the oxidation state is plus four. Methane, carbon is more electronegative than hydrogen so we play the same game. But now we give our electrons to carbon and so we have plus four the nuclear charge minus eight the four electrons and our net oxidation state is negative. Now final, last case. Would the oxygen also have a negative charge? Yes the oxygen, I wasn't counting on the oxygen. The oxidation state of oxygen is negative two here because you have six nuclear charge minus eight electrons and the oxidation state of hydrogen here is plus one. I'm only counting for carbon. Why does carbon go from plus four to minus eight? You take the nuclear charge minus the number of electrons. So you take your nuclear charge and calculating oxidation state plus for the outer valence plus four. You have the one S which you don't count and two of the nucleons, two of the carbon ions. You count carbon spore across in the periodic table so carbon atom is carbon nucleus plus four electrons but now in calculating oxidation state in carbon dioxide we're taking all those electrons away. In methane we're giving it four more electrons. And in ethanol, let's take our last example ethanol and we play the same game here. We'll divide up our electrons in ethanol and for the metals we share and share a light. And so for ethanol our carbon is, this is our primary alcohol, our carbon brings four nuclear charges. We take away five electrons and we're at negative one oxidation state. Now let me tell you honestly, I am not generally calculating oxidation states in my head but it definitely is a romantic penis permeates my head. In fact I think in broad terms I think acid oxidation state, aldehyde oxidation state, primary alcohol oxidation state. Then I realize well ketones are shifted by one, secondary alcohols are shifted by one but they go in care. So when I'm thinking about stuff it's definitely not this calculation. This is just the underpinnings of my thinking. When I think about a reaction of say an alkene and a hydrogen I think carbon carbon jubbubon now gets replaced by carbon carbon single bond plus two carbons bound to hydrogen. That's taking my oxidation state of each carbon down by one or taking me down by next to a bond oxidation state. When I think about say forming a general diol from an alcohol I think oh I'm going up. And when I think about hydrating an alcohol I realize one carbon is going down, hydrating an alkene to an alcohol, one carbon is going down, the other is going up. They balance out. So for the soft side one why does the CO have a high electron? Why does the CO here? Sorry the carbon, how come the carbon bond? Bookkeeping, when you have a carbon carbon bond you divide equally. You give one to each carbon in bookkeeping for oxidation state. Alright so that's what permeates my thinking on oxidation state. And as I said the biggies become the acid family, the aldehyde family, the alcohol family and then the ketone and secondary alcohol. Alright so I want to talk a little bit about reactivity toward nucleophile, toward hydride and other nucleophile. In a very good generalization, again there are always exceptions to generalization, but a very good generalization is that aldehydes are more reactive than ketones. And ketones are more reactive toward nucleophiles than carboxylic acids and esters. So this means for example if I were to mix a mole of sodium borohydride with a mole of an aldehyde, a mole of a ketone and a mole of an ester, I would expect the aldehyde to react first. In other words if I monitored the reaction by thin layer chromatography, the first species I would expect to disappear would be the aldehyde. So if I wanted to give a particular example just so I don't put this in the abstract, let's compare acetaldehyde, acetone and let's just say methylacetate CH3CO methoxy. So why can we think of this this way? Well think about our resonance structures. So you think about the carbonyl resonance structure and you'd say okay we have two resonance structures that we need to worry about in general for a carbonyl group. One resonance structure, the major resonance structure in which the carbonyl has a complete octet and the minor resonance structure in which we have separation of charge and the carbon doesn't have a complete octet. Now in the case of acetone we have two methyl groups here. Those methyl groups are stabilizing by electron donation. When you learned about carbocations you learned about hyperconjugation. You learned that a tert-butyl carbocation is more stable than an isopropyl carbocation which is more stable than an ethyl carbocation. So the carbon group is a tertiary carbocation which has three alcohol groups donating electrons by hyperconjugation feels less pain from that positive charge than a secondary carbocation and primary carbocation. The same principle over here. When you have alcohol groups on both sides of the carbonyl, that carbonyl is less electrophilic. It is more stabilized by electron donation. When you have an aldehyde like acid aldehyde it's more reactive because you have only one alcohol group stabilizing and if you have the one aldehyde with no alcohol groups from aldehyde it is extremely reactive because there is absolutely no stabilization of the partial positive charge. In the case of an ester you get a little bit of extra stabilization here because you can write a second resonance structure and I'll just go ahead and complete my lone pairs here. You can write a second resonance structure like so that provides an extra special stabilization of your positive charge of the carbons. So you get one additional resonance structure. I saw a question back. The question was if you have an aldehyde with a secondary carbonyl. So the question was very clever. If you have an aldehyde could you have a high drive ship? And the answer is that the partial positive charge is not enough on its own to introduce a high drive ship. However there are certain reactions. For example the pinnacle rearrangement where you can go ahead, actually not the pinnacle but there is another rearrangement where you can go ahead and push a high drive in once you've protonated the carbonyl. So the short answer is not for an aldehyde per se but your thinking is actually self. Another question. Can you explain how it's stabilized by electron donation? So you actually write a double bond between the carbons and then you write H plus. It's that exact same thinking. Oh I wouldn't go so far as to write such a structure. So if basically think of it as each of these CH bonds. Remember the hydrogen is more electropositive than carbon so each of these CH bonds puts more electron density on the carbon which inductively donates it in. So I'm going to take one last question and then I want to wrap up a couple of concepts that came up. Okay of course you asked there would it be better if the positive charge is on the carbon instead of the oxygen? All different resonance structures. Ah your question. Is this a better resonance structure than this? And the answer is no. And the reason the answer is no in terms of contribution is we have two competing ideas. We have the idea that atoms want to have complete octets and the idea that in general when there's a choice put the electrons on the, put the positive charge on the more electropositive the less electronegative atom. But the one that wins out in terms of being more important is octets. And so in an ester there are three different contributing resonance structures. That resonance structure is a little bit more important than the other resonance structure but they all contribute. They all stay on. Alright I want to bring up a concept that I asked about at the beginning and I promised to get to it at the end. And that was the concept of stereochemistry. So when we carry out a reduction reaction, when we for example take our benzo phenome and we treat it with lithium aluminum hydride and then we do a workup with water or aqueous acid as I said it's okay to use either. And I write this alcohol as our product. Even though I have written that molecule flat there is a stereogenic center in it. The carbon is a stereogenic center. By that I mean there are four different substituents attached to that carbon. It doesn't matter whether I draw it or don't draw it, it is. Just like it doesn't matter whether I tell you my middle name or I don't tell you my middle name, that S stands for something. So whether I write it or not we can have one enantiomer, we can have the other enantiomer and unless I do something special to only get one, unless I bring to bear some source of chirality I get both enantiomers. It takes chirality to make chirality. If I take an achiral reducing agent like lithium aluminum hydride or sodium borohydride and I take an achiral ketone I will get both enantiomers in equal amounts as the racemac. All of the proteins in your body, all of the amino acids in your body with the exception of glycine, all of the sugars in your body, all of the nucleic acids in your body. All of the lipids in your body are chiral. And in general, chiral molecules of one-handedness interact in a different fashion than the other. In other words, if I go to shake your hand and I instead present my other hand they don't fit as well together. And for that reason there is a tremendous interest in synthesizing chiral molecules of just one chirality as drugs and for many other purposes. So often the biological activity of one enantiomer is different than that of the other. In medicine one may have a desired effect, the other may be toxic. That's one of the reasons why many of the form babies were born due to polyvase where the enantiomer, which is given to combat nausea and pregnancy in the 60s, the enantiomer induced both effects. So in this reaction if we take, and your textbook focuses on one particular reagent called the CBS reagent. In part because the author of your textbook got her PhD with Nobel Prize winner E.J. Corey who invented this reagent. But in general there have been many sources of chiral hydride reagents that have been invented. And the general goal is instead of making one enantiomer and the other in equal amounts, you get just one enantiomer. In your textbook develops the idea of which one you get at this point in the thinking. It's not so important. But realize that if your man were a ketone and I came forward with this hand bearing a hydride, I would have transferred that hydride to the pump. And if I had come forward with this hand bearing a hydride, I would have transferred it to the back of your hand. And just in the same way we can bring in a hydride from one face of the carbonyl, from the front face of the carbonyl, to end up with the hydrogen pointing outward. To end up with the R stereochemistry or with the hydrogen coming from the back face of the carbonyl. In this case, end up with the S stereochemistry. Now I've written just one enantiomer. That's not quite true. It's very, very hard to get just one enantiomer. The molecules in your body, the enzymes, are so big and so good at engulfing substrate molecules that they're very good at generating just one enantiomer. In general, chemical reagents generate mostly one enantiomer. 99% is very good. 99.5% is very good. There's a magic number that Johnny knows for kilocalories per mole. And what is that number? 1.36 takeouts per mole at ambient temperature, about 1.4 kilocalories per mole that corresponds to a 10 to 1 difference in rate constant or a 10 to 1 difference in equilibrium constant. A parallel reagent that fits with 2.5 kilocalories or 2.7 kilocalories better on one face than another can select 99 times as good. Now I want to finish up with one last concept here. And what your textbook is talking about is really the big ideas here. And this is one of the problems I have with your textbook is sometimes, sometimes in trying to present the big ideas, you end up getting lost in the little pieces and never seeing the bars. So I'm just going to present one sort of more big idea that ties into the same thing. Imagine for a moment now, and this is in a way a little easier to wrap your head around than enantiomers and it doesn't really involve moving hands to demonstrate it. Imagine for a moment that we have tert-butyl cyclohexanone and the tert-butyl group is on one side. I've drawn it as a wedge. I've drawn it coming out. And we treat this with sodium borohydride in ethanol. Now the hydride can add either from the front face or from the back face. If I add from the front face, I get the OH pointing down and the hydrogen pointing up. If I add from the back face, I end up with the OH pointing up and the hydrogen pointing down. And we call the relationship, the stereochemical relationship between those two molecules. Diastereomers. And now even in the absence of any sort of chiral reducing agent, we don't get equal amounts because these two molecules aren't equal. We happen to get the 80 to 20 mixture. The mixture ratio isn't so important, but it's a little bit easier to see if we go ahead and we look at drawings. So the tert-butyl group is very big. I'm going to draw a chiracyclohexane. Remember, cyclohexane likes to sit with a large group equatorial. So here's our cyclohexanode. And now we have two ways that hydride can come in. Hydride can come in from the top or hydride can come in from the bottom. If hydride comes in from the top, we end up with this diastereomer, the transdiastereomer. And if hydride comes in from the bottom, now we end up with the axial LOA. We end up with the sixth diastereomer, the minor diastereomer. And the last thing I'll point out is organic chemists are control frees. When we synthesize molecules, we want to have absolute control of what we make. So naturally, for the same reason I talked about enantiomers, you want to make the right enantiomer of the drug. You want to make the right diastereomer of the one that you need. And so naturally, if nature did an 80-20 mixture of cis and trans, of trans and cis diastereomers, organic chemists have come up with ways for flipping to heat or death all of the cis. I'll see you next month and have a good weekend.
 So today we're going to really start from launching  to the main video course and we'll start to talk  about reactivity of carbonyl clouds.  And this is going to encompass capture 20,  but it's going to not just run through 20,  but really all the way 24, captures 24, the next six weeks.  This is the main piece of the course and this is also going  to really be some  of the intellectual height of organic chemistry.  And we're going to see general themes of reactivity  of carbonyl compounds and really it has to do  with sort of two properties of carbonyl compounds.  One is the reactivity with nucleophile,  which is what we'll be focusing on today.  And the other property that we're going to be learning  about particularly as we move through the course  into I think chapters 22, 23 and 24 is that there,  well maybe I should say there are reactions  with nucleophiles let me.  Oh we'll keep it as reactivity.  All right and the other will be there reactivity  as nucleophiles.  What do I mean by that?  Well the first one is a little bit easier to see  and is where we're going to be starting today  and indeed in the next few lectures.  And that is that the carbonyl group is electrophilic.  The carbon of the carbonyl group is electrophilic  and we're going to see how nucleophiles can come in  and right now I'm just going to draw an arrow not really  yet representing flow of electrons to say  that nucleophiles are interested in reacting  with the carbonyl carbon.  The second concept reactivity as nucleophiles we're going  to see as we develop the idea of enols and enolates,  isomers and related species to carbonyl compounds.  And at that point when we start to invoke that reactivity,  what we're going to learn is that the alpha carbon,  alpha carbon means the one over can react as a nucleophile  and again that's not meant to represent the flow  of electrons yet just showing the position  and maybe I'll go ahead and emphasize that by saying alpha  carbon.  So in general as you go along from a carbonyl group you'd say  the carbon next to it is the alpha carbon,  the carbon one beyond that is the beta carbon,  one beyond that is the gamma carbon.  And here we'll refer to this as the carbonyl carbon.  So in its own right this diagram doesn't really say very much  but it sort of lays a framework.  So let's start to move from this level of abstraction  to a level of bonds and structure.  And let's start with carbonyl group.  And this isn't going to be acetone  or any particular carbonyl group.  I'm just drawing this as a general carbonyl group.  And we'll be good people.  We'll write our own pairs of electrons and focus  on carbonyl group.  So oxygen is more electronegative than carbon.  Oxygen has a greater nuclear charge.  It pulls the electron forward.  In other words the carbon oxygen double bond,  the bonds that comprise a sigma bond and a pi bond are  while shared electrons while the covalent bond  between carbon and oxygen they're not sharing equally.  Those electrons spend a little more time  at the more electronegative oxygen, the atom  with the greater nuclear charge.  And so I can represent that in terms  of a partial negative charge and a partial positive charge.  In other words since the electrons while spending time  between oxygen and carbon spend a little more time at oxygen,  we shift our charge over to oxygen.  There's a partial negative charge.  Our carbon is a little bit deficient in charge.  There's a partial positive charge.  We can also represent this type  of structure by resonance structures.  And remember resonance structures aren't one  or the other but both at the same time to varying degrees  or in some cases more than one if you're more than two.  And so we can think of the carbonyl group  as having a major resonance structure.  And also a minor resonance structure in which instead  of having a double bond and having a complete octet  for carbon and a complete octet for oxygen,  we can represent a minor resonance structure  as having a single bond with a full negative charge  on carbon on oxygen and a full positive charge on carbon.  Now I'll bracket these structures to represent that it's,  the two of them together we're thinking about.  We're not resonating or vibrating or oscillating  between the two structures.  We're mostly the one on the left and a little bit  of the one on the right.  This is also representing the type of electron density  that you would see in the molecular orbitals  for the carbonyl group.  The other thing I'll point out is we're going  to be focusing a lot on arrows in this course.  We'll use them to think about and communicate ideas  of electron flow of equilibria and of resonance.  And there are different types of arrows  that we use to communicate different ideas.  The double arrow like this communicates resonance.  It communicates that it's both structures.  It's not an equilibrium arrow  which is two arrows pointing in the opposite direction.  And it's not a flow of electrons  where we draw an arrow originating at a electron  and flowing at something at a once.  In other words, originating at the nucleophile  and flowing at the electrophile.  And that actually brings us to our next element  of reactivity and that is the reaction with nucleophiles.  Now if you have a carbonyl group and again I'll draw it  just as some generic carbonyl group not meant  to represent acetone, not meant to represent any particular group  and it encounters some nucleophile.  Again for now we'll just represent that nucleophile  in the abstract.  They can react with each other.  And so I'll use a reaction arrow to show  that we're going from reactants to product.  The product of this reaction now can be represented like so  where our nucleophile has formed a bond to the carbon.  And we can think about how that reaction occurs by thinking  about the flow of electrons from the nucleophile,  from the thing that has electrons to the carbonyl carbon,  to the carbon that wants electrons.  In our mind's eye we have in mind a picture  where the carbonyl group is polarized  and so electrons want to attack as carbon.  In the very back of our brain we have a picture  like this is very minor resonance structure guiding us  and making us think from where the electrons want to flow.  And of course we can't stop at this point  because if we just left our electrons here now we would have  10 electrons around the carbon.  We would have more than an octet.  That wouldn't be possible.  And so electrons have to continue to flow as electrons are flowing  from the nucleophile to the electrophile  to the electrophilic carbon in terms electrons are flowing  up onto oxygen.  And so when all is said and done we formed a bond  between the nucleophile now represented by this line  or these two electrons and we move two electrons onto oxygen  and in turn put our negative charge there.  Now as I said at that point I'm dealing with abstractions.  The nucleophile here is not meant  to be any particular nucleophile.  But now let's come maybe to something more specific  and in turn this is not meant  to be any particular carbonyl group.  But let's take a look at acetone and a nucleophile  and sort of halfway between reality  and an abstraction at this point.  And so the nucleophile that I'll draw is hydride.  And I think I'm going to put it in quotes.  The reason I'll put it in quotes is you essentially never  have hydride as an entity.  The very end of our discussion here I'll show you an example  of an actual hydride with a hydride anion  that actually doesn't react as H minus  or at least doesn't react as a nucleophile.  But for now we'll worry  about what our hydride source is in just a moment.  For now just think about the abstraction of hydride  and we'll have hydride reacting with acetone.  And so electrons are going to flow from the hydride  from the lone pair to the carbonyl.  And again we can't stop at that point.  Our electrons have to flow up from the double bond  onto the oxygen atom.  And so when all is done.  Now we have an anion.  It's the anion that would have been derived  from isopropanol so we call it the isopropoxide anion.  And it is a member of the broader family  of what we would call alkoxide anion.  So if you have an anion that's derived from an alcohol  that's equivalent to removing H plus from an alcohol,  you end up with an alkoxide anion.  So a lot of organic chemistry question.  Oh yeah, so does the nucleophile attack when the carbonyl is  at its minor resonance structure?  Or does the nucleophile cause the minor resonance?  All right.  I'm going to toss this question which is a good one  out to somebody in the class.  Does the nucleophile attack when it's the minor resonance  structure or does it attack?  Or does it attack cause the minor resonance structure?  It wants to take a stab at it.  Does it attack when it's the minor resonance?  Okay, so now we have doesn't it happen in both  because it alternates?  It says it's a hybrid.  A hybrid, yes.  So I didn't like the word alternate but you've got the right idea.  It's not one or the other.  It's both resonance structures at the same time.  So it's not like it's waiting for the minor resonance structure.  They're both there at the same time.  The problem is chemistry is all about monos, about conceptual monos.  And the beautiful thing about drawings like this is they're  simple.  We can wrap our head around.  That's super, super powerful.  The problem, so that's the plus.  The disadvantage is that these models are always incomplete.  So there are layers of meaning in the models.  And another approach is to look at molecular orbitals  and electron density but it's harder to immediately wrap  our head around things.  The beauty about these drawings is you can scratch them in a stick  on the sand or under paper and have an immediate grasp  of reactivity.  So it's not one, not the other but both at the same time.  Now the second half of your question is interesting  because of course the bond forming process is one where the  electrons do start to approach.  And as those electrons are approaching, as the nucleophile  is approaching the electron, we're moving the other electrons  continuously up onto the oxygen.  So in a way this is what you would call a reaction coordinate.  We're going in the process from the nucleophile to the  electrophile being very far apart and having almost completely  a double bond character in the electrophile and no bond  between the nucleophile and the electrophile.  We're moving the nucleophile into the carbonyl group.  The carbon oxygen double bond is becoming more and more  like a single bond and getting weaker and weaker until finally  as we've gone over that energy of the transition state  and down to the products, we've now formed that single bond.  So that's the reaction we can have.  Other questions?  Good question.  Should we be concerned with the stereochemistry?  Ah, great question.  Hang in there.  Question, what should we be concerned with the stereochemistry?  So let's start at the beginning.  So the answer is yes and let's get to that point.  All right.  So organic chemists love to make stuff.  We live and die by making new medicines, making new drugs,  having reagents that cause chemical reactions and inventing  new reagents.  And so this is a huge part of organic chemistry.  It's fun because you have an idea in your mind's eye  and then you test it out.  So two reagents that are used as a source of this,  what I would say here is abstract or hypothetical hydride  and ion is lithium aluminum hydride, too very popular  in long standing reagents.  Lithium aluminum hydride and sodium borahydride.  Both of these reagents react as hydride with carbonyl compounds,  particularly with aldehydes and ketones,  which is what we're going to focus on today.  The structure of lithium aluminum hydride is an aluminum hydride  anion or tetrahydride anion as a lithium anion.  The structure of sodium borahydride is a borahydride anion,  BH4 minus, like the other is ALH4 minus, and a sodium cation.  And you'll notice these structures are related.  Boron is above aluminum in the periodic table,  so in many ways they have similar properties.  They differ in that aluminum is more electropositive  than boron.  The electronegativity of hydrogen is about 2.2.  The electronegativity of aluminum is about 1.6.  The electronegativity of boron is about 2.0.  In other words, in all of these species,  you have the bonds polarized to having more negative charge  on hydrogen than on the metal.  But in the case of aluminum hydride,  they're even more polarized.  In other words, aluminum hydride is a more reactive anion.  Lithium aluminum hydride is more reactive.  And you can see this in the laboratory immediately.  If I take a spoonful of lithium aluminum hydride,  a great powder and throw it into water,  I better get back because it's going to release hydrogen  with a big fizz and maybe even catch fire.  And if I eat it too much, it'll explode.  Zodium boron hydride is a white powder  that will sort of bubble in water,  but won't react nearly as fast releasing hydrogen gas.  And there was a question here.  Would NAH also release a hydride?  NAH, we're going to come to in a moment.  Great question.  Let's talk about some reagents and their reactivity first.  So in a way, that falls into the category of the exception  and we're going to see its reactivity in a moment.  So let me take an organic compound.  We'll take a simple compound.  It's a ketone.  It's a ketone with a benzene ring on one side  of the carbonyl group and a CH3 on the other side.  You might see people write the CH3.  You might see them leave it just as a line.  Both are correct ways of writing it.  The IAPAC name for this compound is one phenylethanone,  but nobody is going to call it that.  Everyone's going to call it aceto phenone  because it's one of these compounds  where the common name just really dominates.  So I'll write out the systematic name just as a reminder.  But again, the common name is going to predominate.  And we're going to imagine treating this compound  with lithium aluminum hydride.  And now we're going to see how it reacts as hydride anion.  As I said, aluminum is more electropositive than hydrogen.  The bond between them is covalent,  but it's a polar covalent bond.  And it's one in which we can think of it almost as H minus  or in other ways we can think of the electrons  as flowing from the bond between aluminum and hydrogen  onto the carbonyl carbon, toward the carbonyl carbon.  And remember, I use an arrow from the electrons  to the thing that wants electrons.  I put two little arrowheads on here.  It's representing that we have two electrons.  And for now at least, I will be good at drawing my lone pairs  of electrons and reminding us that we have two lone pairs  and we have one more pair of electrons flow up there.  And so when all is said and done, we end up with the alkoxide anion.  And for now at least, I will be good at drawing in my hydrogen.  And in addition to this, we have lithium cation.  And in addition to this, we have alH3.  And that's where I'm going to stop it at this point,  is at the alkoxide anion.  And that's how you can think of it,  but I'll put a little bug in the back of your mind.  In the back of your mind, you can think about the fact  that aluminum doesn't have a complete octet.  And in reality, this lone pair can now attack one of the lone pairs,  can now attack the aluminum and fulfill its octet.  So technically you'll end up with an aluminum alkoxide salt.  All right, so this is kind of a mechanistic way  of thinking about this reaction.  So let's go and think about things from a synthetic point of view.  So redemic chemists will often write equations as little recipes.  Mix this with your compound and then mix in something else  and then you get that.  And organic chemists are very bad at writing balanced equations.  Organic chemists often will just write reactants and products.  Question.  Is there a conservation of charge?  Oops, good point.  Thank you.  So yes, so our lithium aluminum hydride,  we have a lithium cation, an aluminum hydride anion,  that's our reagent, the aluminum hydride transfers its hydride.  At the end you have your lithium cation, your alkoxide anion,  and alh3, which as I said will typically bond to oxygen.  Also typically that undergoes further reaction with additional ketones.  So often when you're doing the reaction in the laboratory  you don't mix one mole of a ketone and one mole of lithium aluminum hydride  but you might use two moles of a ketone or even three moles of a ketone  and one mole of lithium aluminum hydride.  So let me write a reaction as a synthetic organic chemist would.  So now I'll be a little bit sloppier in the sense that I'm not going to write in  my lone pairs of electrons.  We're thinking more like a synthetic organic chemist.  Not even going to write in my methyl group.  It's implicit over here.  And so step one is treatment with lithium aluminum hydride.  I'll write it as LiAlH4, but sometimes you'll see it abbreviated as LaH,  lithium aluminum hydride.  And then I'll write in a second step in this recipe.  And I'll write in the step H3O plus or your textbook writes H2O  and I'll tell you about that in just a second.  And then I'll probably also indicate the solvent that I would use to carry out this reaction.  And typically I would carry out the lithium aluminum hydride reduction reaction in THF  or in diethro ether.  So this constitutes a little recipe.  It says take some benzophenone, put it in a flask with some THF or some ether,  throw in some lithium aluminum hydride, and then after a while after the reaction is done,  throw in some acid or some water.  Of course you can't go through one acid, you can't go to the stock room and say,  give me a bottle of H3O plus.  Look at you funding and say, oh, we don't have any bottles of H3O plus.  We have aqueous hydrochloric acid, which of course is H3O plus and CO minus.  We have aqueous sulfuric or concentrated sulfuric acid, which you can add to the water to make H3O plus  and bisulfate N9 HSO4 minus.  So that's a little bit of a short-term for adding an acid that you would get a strong mineral acid like H2O4 H2O4.  Now, I think we call this step here the workup.  I'll just write solvents and I'll just write workup here.  And honestly, this particular one you could do with either aqueous acid or with water.  Aluminum salts in water under the warm condition can form tremendous gelatinous things.  You've all had lab, you've all shaken the septic funnel.  You may have even encountered an emulsion.  Who's encountered emulsion?  I hate emulsions, don't you?  The advantage of throwing in a ton of acid is it dissolves all the aluminum salts and gets rid of the emulsion.  You can also add water in just the right way, a little bit of water, a little bit of sodium,  a little bit of sodium sulfate, which makes your aluminum instead of gelatinous, makes it granular.  So, either of those work for a workup and you'll see things written both ways.  The main thing, of course, is that until we add something that provides a proton,  whether it's acid, H3O plus aqueous HCl, or whether it's water, we have an alkoxide anion.  But after the workup, you isolate your product, the alcohol.  And as I said, organic chemists are typically very bad about balancing equations.  So, typically this becomes a little recipe.  And as I said, I'll just write it explicitly, our workup ends up taking our alkoxide anion,  and we have the alkoxide anion plus H3O plus, plus hydroamine, or plus water under the right circumstances,  and electrons flow from the alkoxide to the source of protons, and electrons flow from the hydrogen oxygen bond,  onto the oxygen.  And as I said, we've called this step.  Questions or thoughts at this point?  Ah, does it always make a secondary alcohol?  Good question.  If we had an aldehyde, we would get a primary alcohol.  Of course, you couldn't have, you couldn't make a tertiary alcohol with hydride nucleophile,  because by definition, hydride is going to add in a hydride.  Next class, we'll learn about Brignard reagents and Organolithium reagents,  instead of adding quote, unquote, H minus, can add an alkover to light, quote, unquote, metal minus,  and you'll see how they pervade for, for example, tertiary alcohol.  All right, there's a concept that really is useful in permeating your thinking,  and that concept is oxidation state.  I introduced it last time when I was talking about the carboxylic acid family,  and I said in the carboxylic acid family, the oxidation state of an alcohol is,  or the oxidation state of the carbon is plus three.  In a ketone family, the oxidation state is plus two.  If you remember in calculating oxidation state from general chemistry,  you play a game where you divide up the electrons,  giving them to the more electronegative atom,  and then saying, okay, what's the net charge on the atom?  If you have a carbon bound to another carbon, you divide them equally.  So in a ketone, the oxidation state is plus two.  I'll write ketone here.  In our product, in our particular alcohol that we've, we've generated,  secondary alcohol, our oxidation state is zero.  So if you notice, by going from an oxidation state of plus two to an oxidation state of zero,  we have undergone a reduction reaction.  And the chemical that's done it then, our lithium aluminum hydride,  is a reducing agent.  It's the chemical or the reagent that carries out the reduction reaction.  And there are many different reducing agents.  I drew another one.  I drew sodium borohydride.  And let's talk about that.  And I'll give you an example.  Since the very astute gentleman asked the question about aldehydes,  let's take, and there's primary alcohols.  Let's go ahead and take an example of reduction of an aldehyde to a primary alcohol.  And so we'll take benzaldehyde.  Benzaldehyde is a very nice smelling compound that kind of smells of cherries.  It's sort of the flavor of cherry coater, artificial cherry flavor.  If you take, for example, sodium borohydride,  the reaction would work with lithium aluminum hydride,  but I'd like to introduce you to other common reducing agents.  And in this case, you can actually run the reaction in a cell that's like methanol or ethanol.  I said sodium borohydride reacts only slowly with water.  If I tried to run a lithium aluminum hydride reaction in methanol,  just like water, it's a source of hydroxylic protons.  If I tried to run a lithium aluminum hydride reaction in methanol,  I'd be rewarded by the whole thing as best spatter in the face  and worst tattoo in the fire.  So our product of reaction here is benzalazole.  And we can think of this reaction occurring very much like the other reaction has occurred.  So here's our aldehyde. Here's our borohydride anion.  And the boron hydrogen bond is polarized.  It's a polar covalent bond with more electron density on the hydrogen.  And so borohydride anion can transfer a hydride adding it as a nucleophile to the carbonyl group.  And again, we pick up our electrons onto oxygen.  So now we get our alkoxide anion and BH3.  And again, I'm skirting a little bit loosely  because BH3 doesn't have a complete octet.  And so typically it's going to form a bond with oxygen reversibly.  Or we can think about it as forming a bond with oxygen.  Now the nice thing about running this reaction in a hydroxyl exolvent is you've got this huge, huge excess of methanol.  And so methanol can protonate the alkoxide anion.  So here we have our alkoxide anion.  We're in a ton of methanol.  And so electrons can flow from the oxygen of the alkoxide anion to the proton of methanol.  We put electrons back onto the oxygen.  And after all is said and done in this equilibrium reaction now,  driven by the mass action of having so much methanol in it.  Now we have our alcohol in metoxide anion.  I'm sorry, wouldn't metoxide be more happy without the H or without the H?  Great question.  Would the metoxide be more happy without the H or with the H?  So in general, if you want to sort of keep one thing in mind,  the basic rule would be all alcohols have pretty similar acidity.  Water, I said to PK, was 15.7.  That's sort of at one extreme methanol.  It's about 16 ethanol, maybe pushing toward 17, 16 or 17,  tert-butanol toward 18.  Benzal alcohol is a little bit more acidic inductively.  But the main point here is we've got a ton of methanol.  Your methanol is 20 or so molar and you've got one molar of compound.  So you've got this huge excess of methanol grinding this equilibrium to the right.  And then you're going to do an aqueous workup in the end that's going to  extract your compound into an organic salve.  So you're right.  It is an equilibrium. In this particular case, it's okay that it's an equilibrium.  But that equilibrium, remember I was talking about PK last time and I said,  the thing that we really want to be able to focus on in organic chemists,  it does an equilibrium lie way to the left, way to the right,  kind of in the middle or maybe a little more to the left or a little more to the right.  That's how we think about things.  And you're right.  This equilibrium lies kind of sort of in the middle.  Do you want to run a reaction with a liquid in your project and you react with methanol?  Oh, ah.  Well, if you've tried to run a lithium aluminum hydride reaction in methanol,  the lithium aluminum hydride would probably first react with the methanol.  But as I said, make hydrogen so vigorously and so rapidly that it would blow up in your face.  So it would definitely react with the methanol.  It might also react with the ketone, but I'm not going to try the experiment in the laboratory  and see at the end when I wash my face off whether I had any reduced ketone.  So in this case, in this case we started with benzaldehyde.  And the oxidation state of benzaldehyde is plus 1.  And by the time we've gone to benzal alcohol, again it's a reduction reaction.  The oxidation state is minus 1.  And that's not important in thinking about families because remember I said carboxylic acids  and the whole carboxylic acid family, esters, amides, nitrioles,  all that entire family is plus 3 oxidation state.  They are more oxidized than aldehydes.  And then aldehydes are more oxidized than primary alcohols.  And you'll notice they go in twos and this is pretty common in organic chemistry.  As you change oxidation states almost invariably you end up going into organic chemistry.  The reason it's important to sort of keep this in the back of your mind is it helps you focus your thinking.  When you change from a carboxylic acid to an ester through a chemical reaction,  there's no change in oxidation state.  You wouldn't think about using an oxidizing agent or reducing reagent to affect that change.  You're changing the same oxidation state.  You're changing from a ketone or an ester or an aldehyde to an alcohol.  Then you're changing oxidation state and you think about using a reducing agent.  If you're changing from an alcohol to a ketone or an aldehyde or an acid,  you would think about using an oxidizing agent.  There are a ton of high-grade reducing agents.  Your textbook has given you some, I'll show you a few others.  The general principle, the overarching principle of high-grade reducing agents is that you have a metal hydrogen bond,  metal broadly defined, usually as a polar covalent bond where you have M with a partial positive charge  and high-grade with a partial negative charge.  By broadly defined, I mean that Goron is kind of a metal or a metalloid, aluminum is a metal, and there are many, many others.  So the general principle, as I said, with lithium aluminum hydride technically after that first mole of hydride is transferred,  technically what happens is the AlH3 reacts with the alkoxide and now this intermediate can transfer again hydride anion to another carbonyl group,  which is why you may only use one mole of lithium aluminum hydride for several moles,  for two moles or three moles, up to four moles, although typically you wouldn't use it of your electrophile, of your carbonyl compound.  Your textbook gives an extremely example of this where you transform, you preform the reagent with three tert buttoxy groups on aluminum  and this is a less reactive hydride source.  There are two reasons for this.  One reason why once you start to put alkoxide on the aluminum, it's less reactive is that the oxygen is electronegative  and it's pulling electron density away so there's less of a partial minus charge on hydride.  The bond is less polarized.  Another reason is sterics, the tert-fuel group is very big and so that big set of tert-fuel groups gets around the way of this.  So only the very most reactive carbonyl compounds react with a tetra-tritricerfutaxiome and hydride reagents.  The oxygen metals that don't have a full octet can react as a hydride source.  Your textbook mentions and we will talk more about in a moment or later lithium, a diasecudal aluminum hydride,  abbreviated dival age, sometimes just dival, you will see it.  And again, you have a polarized metal bond.  Typically what happens in a dival reaction is the aluminum which doesn't have a complete octet first coordinates onto an oxygen or onto a nitrogen  to get a complete octet and a formal negative charge on aluminum and then finish the hydride.  I'll just show one other reaction.  Sodium cyanoborohydride.  It's another hydride reagent that's less reactive.  Cyanogrups are electron withdrawing and so sodium cyanoborohydride is less reactive than sodium borohydride.  There's a big alphabet soup of different reagents out there but you'll notice that the general theme in all of these reagents is that a metal is bonding the hydrogen.  So the various student question was asked about sodium hydride.  And that sort of is the oddball exception. Sodium hydride, NaH, actually reacts as a strong base not as a reducing agent.  And there are a couple of reasons for this.  All of the reagents I showed over there and the sodium borohydride and the lithium aluminum hydride have covalent bonds between metals and hydride.  Sodium is even more electropositive than aluminum.  Its electronegativity is about.9.  And so the sodium hydride is actually an ionic compound and it's pretty darn insoluble in organic solvents.  As a result you never really have a chance to have soluble hydride and it doesn't get into and react as a nucleophile.  Plus because the bond is very ionic and because hydrogen is relatively electropositive compared to say chloride anion, sodium hydride is a very strong base.  And so in general if you take sodium hydride with almost any source of protons, water or say methanol, it reacts.  And so if I throw sodium hydride into say methanol, it reacts to form sodium methoxide and hydrogen gas.  And so I was talking about acidities before and saying well you don't typically calculate things but you'll keep in mind where an equilibrium lies based on the difference in pKa's.  Hydrogen has a pKa so this is an acid base reaction.  Methoxide anion reacts with methanol to give methoxide anion. I've just shown it as NaOCH3 but of course that's Na plus OCH3 minus anionic compound plus hydrogen gas.  So this is an acid base reaction.  The methanol is an acid pKa about 16 or 17. The hydrogen is an acid on this side of the equation. Its pKa is about 35.  It's a very weak acid whereas methanol is just a weak acid so that reaction, I won't even write it as an equilibrium.  The reaction lies way, way, way to the right. pKa of 35, pKa of 16, difference of 19.  If I were to write an equilibrium constant it would be 10 to the 19th just massively on the right.  So that reaction ends up reacting as a base.  Good point for some questions.  The sodium hydride reacts bilimply in water. Absolutely. If you get water on it you see a beautiful bright orange flame.  The big danger, if you were to take a scoop of it and throw it in water you'd get a big flash of fire.  The big danger in the laboratory is when you're weighing it out you spill a teeny little bit and later on you're cleaning up and splattering acetone and water around.  At that moment when the water hits an acetone is splattered around all the acetone and all the solvent goes up in fire.  So when I work with sodium hydride what I'll generally do is get a wad of wet paper towels and wipe down my work area and I might see a little flash of orange.  And that flash of orange under the big wet wad of paper towels doesn't do any harm and it prevents my feet from having to fire over and lighting lots of acid.  How do you find oxidation state numbers? Okay great question.  Alright oxidation state. State two examples here. We'll start with an extreme example.  Carbon dioxide and methane. Carbon has four nuclear charges. We have eight electrons but we're going to play a game here.  The game is that we give all those electrons, I'll even be good and write my Lewis acid structures, or write my Lewis electron dot structures.  We play a game that we give all those electrons to oxygen and leave none for carbon because oxygen is more electronegative.  Because we have a net four nuclear charge the oxidation state is plus four.  Methane, carbon is more electronegative than hydrogen so we play the same game.  But now we give our electrons to carbon and so we have plus four the nuclear charge minus eight the four electrons and our net oxidation state is negative.  Now final, last case. Would the oxygen also have a negative charge?  Yes the oxygen, I wasn't counting on the oxygen. The oxidation state of oxygen is negative two here because you have six nuclear charge minus eight electrons and the oxidation state of hydrogen here is plus one.  I'm only counting for carbon.  Why does carbon go from plus four to minus eight? You take the nuclear charge minus the number of electrons.  So you take your nuclear charge and calculating oxidation state plus for the outer valence plus four.  You have the one S which you don't count and two of the nucleons, two of the carbon ions.  You count carbon spore across in the periodic table so carbon atom is carbon nucleus plus four electrons but now in calculating oxidation state in carbon dioxide we're taking all those electrons away.  In methane we're giving it four more electrons.  And in ethanol, let's take our last example ethanol and we play the same game here.  We'll divide up our electrons in ethanol and for the metals we share and share a light.  And so for ethanol our carbon is, this is our primary alcohol, our carbon brings four nuclear charges. We take away five electrons and we're at negative one oxidation state.  Now let me tell you honestly, I am not generally calculating oxidation states in my head but it definitely is a romantic penis permeates my head.  In fact I think in broad terms I think acid oxidation state, aldehyde oxidation state, primary alcohol oxidation state.  Then I realize well ketones are shifted by one, secondary alcohols are shifted by one but they go in care.  So when I'm thinking about stuff it's definitely not this calculation.  This is just the underpinnings of my thinking.  When I think about a reaction of say an alkene and a hydrogen I think carbon carbon jubbubon now gets replaced by carbon carbon single bond plus two carbons bound to hydrogen.  That's taking my oxidation state of each carbon down by one or taking me down by next to a bond oxidation state.  When I think about say forming a general diol from an alcohol I think oh I'm going up.  And when I think about hydrating an alcohol I realize one carbon is going down, hydrating an alkene to an alcohol, one carbon is going down, the other is going up.  They balance out.  So for the soft side one why does the CO have a high electron?  Why does the CO here?  Sorry the carbon, how come the carbon bond?  Bookkeeping, when you have a carbon carbon bond you divide equally.  You give one to each carbon in bookkeeping for oxidation state.  Alright so that's what permeates my thinking on oxidation state.  And as I said the biggies become the acid family, the aldehyde family, the alcohol family and then the ketone and secondary alcohol.  Alright so I want to talk a little bit about reactivity toward nucleophile, toward hydride and other nucleophile.  In a very good generalization, again there are always exceptions to generalization, but a very good generalization is that aldehydes are more reactive than ketones.  And ketones are more reactive toward nucleophiles than carboxylic acids and esters.  So this means for example if I were to mix a mole of sodium borohydride with a mole of an aldehyde, a mole of a ketone and a mole of an ester,  I would expect the aldehyde to react first.  In other words if I monitored the reaction by thin layer chromatography, the first species I would expect to disappear would be the aldehyde.  So if I wanted to give a particular example just so I don't put this in the abstract, let's compare acetaldehyde, acetone and let's just say methylacetate CH3CO methoxy.  So why can we think of this this way?  Well think about our resonance structures.  So you think about the carbonyl resonance structure and you'd say okay we have two resonance structures that we need to worry about in general for a carbonyl group.  One resonance structure, the major resonance structure in which the carbonyl has a complete octet and the minor resonance structure in which we have separation of charge and the carbon doesn't have a complete octet.  Now in the case of acetone we have two methyl groups here. Those methyl groups are stabilizing by electron donation.  When you learned about carbocations you learned about hyperconjugation.  You learned that a tert-butyl carbocation is more stable than an isopropyl carbocation which is more stable than an ethyl carbocation.  So the carbon group is a tertiary carbocation which has three alcohol groups donating electrons by hyperconjugation feels less pain from that positive charge than a secondary carbocation and primary carbocation.  The same principle over here. When you have alcohol groups on both sides of the carbonyl, that carbonyl is less electrophilic. It is more stabilized by electron donation.  When you have an aldehyde like acid aldehyde it's more reactive because you have only one alcohol group stabilizing and if you have the one aldehyde with no alcohol groups from aldehyde it is extremely reactive because there is absolutely no stabilization of the partial positive charge.  In the case of an ester you get a little bit of extra stabilization here because you can write a second resonance structure and I'll just go ahead and complete my lone pairs here.  You can write a second resonance structure like so that provides an extra special stabilization of your positive charge of the carbons.  So you get one additional resonance structure. I saw a question back.  The question was if you have an aldehyde with a secondary carbonyl. So the question was very clever. If you have an aldehyde could you have a high drive ship?  And the answer is that the partial positive charge is not enough on its own to introduce a high drive ship. However there are certain reactions. For example the pinnacle rearrangement where you can go ahead, actually not the pinnacle but there is another rearrangement where you can go ahead and push a high drive in once you've protonated the carbonyl.  So the short answer is not for an aldehyde per se but your thinking is actually self.  Another question.  Can you explain how it's stabilized by electron donation?  So you actually write a double bond between the carbons and then you write H plus. It's that exact same thinking.  Oh I wouldn't go so far as to write such a structure.  So if basically think of it as each of these CH bonds. Remember the hydrogen is more electropositive than carbon so each of these CH bonds puts more electron density on the carbon which inductively donates it in.  So I'm going to take one last question and then I want to wrap up a couple of concepts that came up.  Okay of course you asked there would it be better if the positive charge is on the carbon instead of the oxygen?  All different resonance structures. Ah your question. Is this a better resonance structure than this?  And the answer is no. And the reason the answer is no in terms of contribution is we have two competing ideas.  We have the idea that atoms want to have complete octets and the idea that in general when there's a choice put the electrons on the, put the positive charge on the more electropositive the less electronegative atom.  But the one that wins out in terms of being more important is octets.  And so in an ester there are three different contributing resonance structures.  That resonance structure is a little bit more important than the other resonance structure but they all contribute. They all stay on.  Alright I want to bring up a concept that I asked about at the beginning and I promised to get to it at the end.  And that was the concept of stereochemistry.  So when we carry out a reduction reaction, when we for example take our benzo phenome  and we treat it with lithium aluminum hydride and then we do a workup with water or aqueous acid as I said it's okay to use either.  And I write this alcohol as our product.  Even though I have written that molecule flat there is a stereogenic center in it.  The carbon is a stereogenic center.  By that I mean there are four different substituents attached to that carbon.  It doesn't matter whether I draw it or don't draw it, it is. Just like it doesn't matter whether I tell you my middle name or I don't tell you my middle name, that S stands for something.  So whether I write it or not we can have one enantiomer, we can have the other enantiomer and unless I do something special to only get one, unless I bring to bear some source of chirality I get both enantiomers.  It takes chirality to make chirality.  If I take an achiral reducing agent like lithium aluminum hydride or sodium borohydride and I take an achiral ketone I will get both enantiomers in equal amounts as the racemac.  All of the proteins in your body, all of the amino acids in your body with the exception of glycine, all of the sugars in your body, all of the nucleic acids in your body.  All of the lipids in your body are chiral.  And in general, chiral molecules of one-handedness interact in a different fashion than the other.  In other words, if I go to shake your hand and I instead present my other hand they don't fit as well together.  And for that reason there is a tremendous interest in synthesizing chiral molecules of just one chirality as drugs and for many other purposes.  So often the biological activity of one enantiomer is different than that of the other.  In medicine one may have a desired effect, the other may be toxic.  That's one of the reasons why many of the form babies were born due to polyvase where the enantiomer, which is given to combat nausea and pregnancy in the 60s, the enantiomer induced both effects.  So in this reaction if we take, and your textbook focuses on one particular reagent called the CBS reagent.  In part because the author of your textbook got her PhD with Nobel Prize winner E.J. Corey who invented this reagent.  But in general there have been many sources of chiral hydride reagents that have been invented.  And the general goal is instead of making one enantiomer and the other in equal amounts, you get just one enantiomer.  In your textbook develops the idea of which one you get at this point in the thinking.  It's not so important.  But realize that if your man were a ketone and I came forward with this hand bearing a hydride, I would have transferred that hydride to the pump.  And if I had come forward with this hand bearing a hydride, I would have transferred it to the back of your hand.  And just in the same way we can bring in a hydride from one face of the carbonyl, from the front face of the carbonyl, to end up with the hydrogen pointing outward.  To end up with the R stereochemistry or with the hydrogen coming from the back face of the carbonyl.  In this case, end up with the S stereochemistry.  Now I've written just one enantiomer.  That's not quite true.  It's very, very hard to get just one enantiomer.  The molecules in your body, the enzymes, are so big and so good at engulfing substrate molecules that they're very good at generating just one enantiomer.  In general, chemical reagents generate mostly one enantiomer.  99% is very good.  99.5% is very good.  There's a magic number that Johnny knows for kilocalories per mole.  And what is that number?  1.36 takeouts per mole at ambient temperature, about 1.4 kilocalories per mole that corresponds to a 10 to 1 difference in rate constant or a 10 to 1 difference in equilibrium constant.  A parallel reagent that fits with 2.5 kilocalories or 2.7 kilocalories better on one face than another can select 99 times as good.  Now I want to finish up with one last concept here.  And what your textbook is talking about is really the big ideas here.  And this is one of the problems I have with your textbook is sometimes, sometimes in trying to present the big ideas, you end up getting lost in the little pieces and never seeing the bars.  So I'm just going to present one sort of more big idea that ties into the same thing.  Imagine for a moment now, and this is in a way a little easier to wrap your head around than enantiomers and it doesn't really involve moving hands to demonstrate it.  Imagine for a moment that we have tert-butyl cyclohexanone and the tert-butyl group is on one side.  I've drawn it as a wedge. I've drawn it coming out.  And we treat this with sodium borohydride in ethanol.  Now the hydride can add either from the front face or from the back face.  If I add from the front face, I get the OH pointing down and the hydrogen pointing up.  If I add from the back face, I end up with the OH pointing up and the hydrogen pointing down.  And we call the relationship, the stereochemical relationship between those two molecules.  Diastereomers.  And now even in the absence of any sort of chiral reducing agent, we don't get equal amounts because these two molecules aren't equal.  We happen to get the 80 to 20 mixture. The mixture ratio isn't so important,  but it's a little bit easier to see if we go ahead and we look at drawings.  So the tert-butyl group is very big. I'm going to draw a chiracyclohexane.  Remember, cyclohexane likes to sit with a large group equatorial.  So here's our cyclohexanode. And now we have two ways that hydride can come in.  Hydride can come in from the top or hydride can come in from the bottom.  If hydride comes in from the top, we end up with this diastereomer, the transdiastereomer.  And if hydride comes in from the bottom, now we end up with the axial LOA.  We end up with the sixth diastereomer, the minor diastereomer.  And the last thing I'll point out is organic chemists are control frees.  When we synthesize molecules, we want to have absolute control of what we make.  So naturally, for the same reason I talked about enantiomers,  you want to make the right enantiomer of the drug.  You want to make the right diastereomer of the one that you need.  And so naturally, if nature did an 80-20 mixture of cis and trans, of trans and cis diastereomers,  organic chemists have come up with ways for flipping to heat or death all of the cis.  I'll see you next month and have a good weekend.
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