 to really start the launch into the main need of the course and we'll start to talk about reactivity of the carbonyl bonds. And this is going to encompass chapter 20, but it's going to not just run through 20, but really all the way 24, chapter 44, the next six weeks. This is the main need 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. 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, I'll just look at it as reactivity. All right and the other will be 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 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 enolinks, 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 carbon yield 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 a carbonyl group to be acetone or any particular carbonyl group. I'm just drawing this as a general carbonyl group. You guys people write our lone 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 and 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 more than two. And so we can think of the carbonyl group as having a major 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. These structures to represent 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's different types of arrows that we use to communicate different ideas. The double arrow like this communicates resonance and 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 one section. In other words originating at the nucleophile and flowing to the electron. And that actually brings us to our next element of reactivity and that is the reaction with nucleophiles. So 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 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 for a 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 turn 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. 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 that's sort of halfway between reality and abstraction at this point and so nucleophile that I'll draw is hydride and I think I'm going to put it in quotes and 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 we have an anion, it's the anion that would have been derived from isopropanol so we call it the isopropoxide anion 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. And a chemistry question. Oh yeah so does a 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? Does it attack cause the minor resonance? Or does it attack cause the minor resonance structure? Does it attack when it's the minor resonance? Okay so now we have doesn't it happen in both because it alternates. It's 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 models, about conceptual models. 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 and electrophile being very far apart and having almost completely a double bond character in the electrophile and no bond between the nucleophile and 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 product we've now formed a single bond so that's the reaction. Other questions? The questions? 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 anhyde is lithium aluminum hydride, two very popular and long-standing reagents, lithium aluminum hydride, sodium borahydride. And 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, the H4 minus, LH4 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 just a moment. 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 was sort of bubbled in water but won't react nearly as fast releasing hydrogen gas. And there was a question here. Yeah, I'm sorry. Would NAH also release a hydride? NAH, we're going to come to it a moment. Great question. Let's talk about some reagents and their reactivities first. So in a way, we're going to, that falls into the category of the exception and we're going to see its reactivity in a moment. All right, 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 and the common name is going to predominate. And we're going to imagine treating this compound with lithium aluminum hydride. 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 another way is 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 a double, I put two little arrow heads on here as 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 with the alkoxide and at least I will be good at drawing in my hydrogen. And in addition to this, we have a 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 and I 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 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. Let's go and think about things from a synthetic point of view. The 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, 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 to the hydride, at the end you'll 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 ketone and one mole of lithium aluminum hydride, but you might use two moles of ketone or even three moles of ketone and one mole of lithium. 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 metal 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 and abbreviate it 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 diethyl ether. So this constitutes a little recipe. It says take some benzo phenol, 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. And 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 hand for adding an acid that you would get a strong mineral acid like HSO4. Now I think we call this step here the workup. So I'll just write solvent and I'll just write workup here. 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 a sept funnel. You may have even encountered an emulsion. Who's encountered an emulsion? I hate emulsions, don't you? The advantage of throwing in a ton of acid is it dissolves all of 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 hydroxide, 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 almolyl. 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 of the alkoxide anion plus H3O plus, plus hydroxide anion or plus water under the right circumstances. And electrons flow from the alkoxide to the source of protons and electrons flow from the source, from the hydrogen oxygen bond, onto the oxygen. And we call this step the workup. What are the 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 of alcohol with hydride nucleophile because by definition, hydride is going to add in a hydride. Next class, we'll learn about Brignard reagents and Herbanolythium reagents that instead of adding a quote-unquote H-minus can add an alkyl group like quote-unquote metal-minus, and you'll see how to make a tertiary of alcohol. All right, there's a concept that really is useful in permeating your thinking. And that concept is oxidation state. And 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. All right, ketone here in our product, in our particular alcohol that we've generated, our secondary alcohol 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 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. It's the very astute gentleman asked the question about aldehydes, let's take, or primary alcohols. Let's go ahead and take an example of reduction of an aldehyde to a primary alcohol. And so we'll take Benz aldehyde. Benz aldehyde 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. As I said, sodium borohydride reacts only slowly with water if I threw, 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 that best spatter would be in the face and worse had to be fired. A reaction here, Benzalethylhole. 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. Hydrogen bond is polarized. It's a polar covalent bond with more electron density on the hydrogen. And so for a hydride anion I can transfer a hydride and it has a nucleophile to the carbonyl group. And again, we pick up our electrons onto oxygen, oxide anion, inserting 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, but we can think about it. Now the nice thing about running this reaction in a hydroxylic solvent 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 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. And our alcohol in methoxide anion. The methoxide would be more happy without the H or without the H. Great question. Would the methoxide 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 towards 17, 16 or 17 terbutanol toward 18. Benzyl 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 cell. 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 need equilibrium like 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. Oh, ah. Well, if you tried to run a lithium-aluminum hydride reaction in methanol, the lithium-aluminum hydride would probably first react with the methanol. And as I said, they hydrate it 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 of these ketones. So in this case, in this case we started with benzaldehyde and the oxidation state of benzaldehyde is plus one. By the time we've gone to benzal alcohol, again, it's a reduction reaction. The oxidation state is minus one. Now, this is important in thinking about families because remember I said carboxylic acids and the whole carboxylic acid families, esters, amides, nitriels, all that entire family is plus one. 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 some organic chemistry. 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. But when 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 hydride reducing agents. Your textbook has given you some. I'll show you a few others. The general principle, the overarching principle of hydride reducing agents is that you have a metal hydrogen bond, metal broadly defined, polar covalent bond where you have M with a partial positive charge and hydride with a partial negative charge. By broadly defined, I mean that boron is kind of a metal or a metalloid, aluminum is a metal and there are 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 group. Your textbook gives an extreme example of this where you transform or you preform the reagent with four, with three tert butoxy 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 tritrefuse-toxial aluminum hydride reagent. Even 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 diasecule aluminum hydride, abbreviated dival H or sometimes just dival, you will see it. And again, you have a polarized metal bond. Now, 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 transfers hydride. I'll just show one other reagent, sodium cyanoborohydride. It's another hydride reagent that's less cyanobroops or 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 students' 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. All of the 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. Remember 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. Hydride anion reacts with methanol to give methoxide anion. I've just shown it as NaOCH3 but of course that's Na plus OCH3 minus an ionic 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. That 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. For some questions, the sodium hydride react violently at water, absolutely. If you get water on it you see a beautiful bright orange flame and 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. But 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 splattering water around and 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 then flash of orange under the big wet wad of paper towels doesn't do any harm and it prevents must be from having a fire over and lighting lots of acetone over fire. How do you find oxidation state numbers? Okay, great question. Oxidation state, state two examples here will start with an extreme example, carbon dioxide and methane start with carbon dioxide. 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 with the electron to oxygen, we play with all those electrons to oxygen and we've none for carbon because oxygen's more electronegative. Because we have a net four nuclear charge, the oxidation state is plus four. There's more electronegative than hydrogen so we play the same game but now we give our electrons to carbon so we have plus four in the nuclear charge minus eight before electrons and our net oxidation state is negative. Less case. Sorry Professor, wouldn't 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. So why does carbon go from plus four to minus eight? I don't get it. Why this, no, no, no, 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, right, you have the one S which you don't count in two of the nucleons, two of the carbon ions, right. So 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 to oxide we're taking all those electrons away in methane we're giving it four more electrons. Okay. And in ethanol, let's say our last example, ethanol to play the same game here will divide up our electrons in ethanol and for the metals we share and share alike and so for ethanol our carbon is, so this is our primary alcohol, our carbon brings for 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 penance permeates my thinking. In fact, I think in broad terms, I think acid oxidation state, aldehyde oxidation state, primary alcohol oxidation state and 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 shovel bond now gets replaced by carbon carbon single bond plus two carbons bound to hydrogen that's taking my oxidation state to be carbon down by one or taking me down by next to a 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 alcohol I realize one carbon's going down, hydrating an alkene to an alcohol, one carbon is going down, the other is going up, they balance out. Sorry, question so for the soft side one why does the CO have why do you like carbon? Why does the CO, here? Sorry the carbon, how come the carbon? Bookkeeping, when you have a carbon carbon bond you divide equally you give one to each carbon in bookkeeping for oxidation state. All right so that's what permeates my thinking on oxidation state. The biggies become the acids family, the aldehyde family, the alcohol family and then the ketone and secondary alcohol. So I want to talk a little bit about reactivity toward nucleophile, toward hydride and other nucleophiles in a very good generalization. Again there are always exceptions to generalization but a very good generalization is the aldehyde's active in ketones and ketones are more reactive toward nucleophiles than carboxylic acids and esters. One way to think about this, 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 CH3, 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 carbonyl 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. About carbocations you learned about hyperconjugation. You learned that a tert-butyl carbocation is more stable than a isopropyl carbocation which is more stable than an ethyl carbocation. In other words a tertiary carbocation which has three alkyl groups donating electrons by hyperconjugation feels less pain from that positive charge than a secondary carbocation and primary carbocation and it's the same principle over here. When you have alkyl 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 alkyl group stabilizing and if you have the one aldehyde with no alkyl groups from aldehyde it is extremely reactive because there is absolutely no stabilization of the work. 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. You get one additional resonance structure. I saw a question. The question was if you have an aldehyde with, 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 ship 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. Well, I always like to think of it just, so the question is can I explain how it's stabilized by electron donation? I like to think of the methyl groups as just electron donating. In the case of a carbocation I really would write a non, remember the non bond resonance structures where you actually write a double bond between the carbons and then you write H plus. It's that exact same thinking. Oh, I lose those so far as the right sense. It 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. I'm going to take one last question and then I want to wrap up a couple of concepts that came up. Okay, well of course the S 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 octet. And so in an S 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 home. All right, I want to bring up a concept that was asked about at the beginning and I promised to get to it at the end and that was the concept of stereochemistry about a reduction reaction. When we for example take our benzo phenome treated with lithium aluminum hydride and then we do a work up 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. There are four different substituents attached to that carbon. 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. We can have the other enabler and unless I do something special to only get one, unless I bring to bear some source of chirality I get both enantiomers. It's 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 racinates. 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 because 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 polyamide 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 in your textbook, your textbook focuses on one particular reagent called the CBS reagent. In part because you're 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. And the general goal is instead of making one enantiomer, and the other in equal amounts, you get just one enantiomer. And 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 you're a man with 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 the 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 to 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 or 99.5% is very good. There's a magic number that Johnny knows for kilocalories per mole. And what is that number? 1.36 take-outs 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. They're getting, 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 on the trees. 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 an anti-immers 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. 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. 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. And so here's our cyclohexanone. 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, end up with this diastereomer, the transdiastereomer. And if hydride comes in from the bottom, now we end up with the axioloate, we end up with the six diastereomers, 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 than 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 to flip it and either get all of the cis or all of the trans. I'll see you next one and have a good weekend.