Reactivity of Carbonyl Compounds

All good to go. So last time we were talking about the reactivity of carbonyl compounds. Today we're going to dive into chemistry. We have talked about how the HOMO of most carbonyl compounds or the most reactive electrons are the carbonyl oxygen. We've talked about how the LUMO or the electrophilic spot or the place where extra electrons could go for carbonyl compounds is. the pi star. We've talked about how when you attack an anti-bonding orbital like the pi star, you do two things. Number one, you make a new bond that's between the nucleophile and the carbonyl carbon, and that new bond is indicated by this arrow. The second thing you do is you break the carbon-oxygen pi bond. You got to do both. If you don't do both, you end up with a Texas carbon. That is a carbon with five bonds to it, which... We love the great state of Texas, but even in the great state of Texas, carbons only have four bonds. So, this is the first step for most of the reactions we will talk about, and it is an addition step. We call it an addition step because the nucleophile gets added and you get a tetrahedral intermediate making one molecule out of two. That is the first step of most of the reactions that we will talk about this semester. One thing we did not get a chance to talk about last time, which I would like to address briefly, is this addition step and the effect of the nucleophile on it. Basically, when we are asking the question of whether or not this addition step is reasonable, what we're basically asking is how reactive is the nucleophile relative to the tetrahedral intermediate that it makes. Notice that in the tetrahedral intermediate, you have a negatively charged oxygen. And so in some ways, how effective this first step is, or what its equilibrium constant is, can be summarized by just comparing how reactive is your nucleophile in the starting material relative to the negatively charged oxygen in the tetrahedral intermediate. We will see that nucleophiles that are at least as reactive as a negatively charged oxygen are good enough to attack directly. So what does good enough to attack directly mean? That means your nucleophile is at least as reactive. Another way of saying reactive is unstable. at least as reactive or as unstable as the negatively charged oxygen in the tetrahedral intermediate. You want to make a comparison? Fine. Draw, let's suppose, for a specific example, let's use OH-as the nucleophile, and we'll just keep this same aldehyde. And I want to know, is it reasonable for OH-as a nucleophile to attack the negatively, to attack the... carbonyl carbon to give you the negatively charged tetrahedral intermediate. Look at the tetrahedral intermediate, then look at your nucleophile, compare their reactivities. Both of them are negatively charged oxygens. If it gets more complicated than that, simply draw the conjugate acid of the nucleophile and draw the conjugate acid of the negatively charged oxygen. Look up their pKa's that will tell you their relative reactivity. So you okay on that? So is this a reasonable step or not? Great, totally reasonable. What would be unreasonable? Do I expect water to be able to attack an aldehyde directly? What's wrong with that? Water is more stable than a negatively charged alkoxide, which is what your tetrahedral intermediate is. In such cases, then, we will often have to do something else first. So for water, if we want water to be able to attack a carbonyl, in many cases we will need some kind of catalysis. An example of how that could be for water would be as follows. Let's use some acid catalyst. To use the aldehyde as a nucleophile in the first step, we protonate the oxygen, lone pair electrons on the carbonyl oxygen attack a proton from the acid, giving us this oxonium ion that we've talked about before. The oxonium ion is more reactive than the aldehyde starting material, and you can tell that that's the case because attack of water on the oxonium ion is suddenly going to look more reasonable. Water, the lone pair electrons on the water, the nucleophile, lone pair electrons on oxygen attack the pi star at the carbonyl carbon. You still get a tetrahedral intermediate. Only notice that the thing that gets made, the tetrahedral intermediate, is now we're dealing with a neutral oxygen, right? Can... a neutral nucleophile attack a positively charged oxonium ion to give you a tetrahedral intermediate that's positively charged and has a neutral OH group on it. See, neutral OH for neutral OH, that's reasonable. So that's one way you can tell whether a nucleophile can attack directly or whether you're going to need some kind of catalysis like protonation to have it happen first. We'll revisit that in a little bit. Some of you are going to take biochemistry at some point and you will learn about a few terms called specific acid catalysis versus general acid catalysis. This is an example of something that will be called later, not in this class, but later, specific acid catalysis. And what that means is that the acid base step, the protonation of the carbonyl, has to happen first. And then the addition step is the slow step in the reaction. There are other forms of this we will encounter when we get to chapter 14. Suppose that your nucleophile is an amino group instead of water, a neutral amine instead of water. You might still be worried about a neutral amine generating a negatively charged. oxygen intermediate, right? Let's just sort of draw how that would go. Here's the product after a neutral amine attacks an aldehyde. And there would be reasons to feel uncomfortable about that because you're comparing the nucleophilicity or the reactivity of the negatively charged oxygen in your tetrahedral intermediate with how stable your neutral amine is as a nucleophile. And if that comparison isn't coming easy, simply draw the conjugate acid. Here is the alcohol conjugate acid of the negatively charged oxygen, and look up its pKa, which if you did would be 16. Draw the conjugate acid of your nucleophile, which is the ammonium ion, which has a pKa of 11. In order for us to be really comfortable with direct uncatalyzed attack, we want our nucleophile to be at least as reactive as that negatively charged oxygen. And it's not. It's close, but not quite. And that reflects the fact that when you do reactions of aldehydes with amines, often you need an acid catalyst. What actually, so I'm going to say, um, You know, the answer to this question, is this okay, is I have substantial discomfort. I'm telling you this now because some people gloss over this substantial discomfort. In fact, your textbook actually will when it shows you this reaction. But any organic chemist that actually has been trained will tell you, when you go to the lab and do this reaction, you need an acid catalyst. So let me show you what the mechanism most likely is for an amine to attack an aldehyde. It turns out here it's a little bit different than it is for water attacking an aldehyde. For water, we had to protonate the aldehyde first in a fast step to get the oxonium ion. It turns out, for reasons that you can understand, the amine is more nucleophilic than water. The amine itself is more reactive than water. That's because nitrogen is less electronegative than oxygen, and you learned that in 351. So it turns out the amine, experimentally we observed, the amine doesn't have to wait until the acid catalyst protonates the carbonyl. It can actually attack during. So for amines attacking aldehydes, and this is mostly a preview, what we observe is that the nitrogen attacks the carbonyl carbon. in the same way that we've seen before, but as that process is happening, lone pair electrons on the oxygen pick up a proton from the acid catalyst. All of that happens in the same concerted step. It's relatively slow, and that gets you to an intermediate with a neutral oxygen. And I feel a lot better. That relieves my mild substantial discomfort because now I'm more comfortable with a neutral amine generating an intermediate with a neutral oxygen. So you're going to see examples of this. I would not expect you. To be able to predict the differences that I've just shown you between specific acid catalysis for the water attacking an aldehyde versus what I just barely showed you, which is general acid catalysis, I would not expect you to be able to predict that, oh, for water, you protonate first and then attack occurs in two different steps, whereas for the amine, you attack and protonate in the same step. But I would expect you to. Have the idea that the amine and water, because they're both neutral, probably aren't good enough to attack directly to give you a negatively charged oxygen in a tetrahedral intermediate. Is that, got the idea? Okay, that's the addition step. After that, and we're talking in general terms because so much of what we're going to do in the next two or three years, chapters is all variations on this theme. So I know some of you are anxious to get to what about lithium aluminum hydride and what about NADH and we will try to do that but if you understand the general picture the details are going to be a lot easier. Yes? Yeah, I haven't showed you the end of the reaction. The acid does, in most cases, get regenerated. And we'll do that later on in Chapter 14 and beyond. Yep. Okay, others? That was the addition step. Let's talk about what happens after the addition step. And we're going to be general again for a moment. Take a step back. Here's your general carbonyl, and here is your nucleophile. And let's just suppose now that the nucleophile is... good enough to attack directly? We've already answered that question. That's the addition step. That is an ugly arrow. I will pause at various points during the lecture to make the drawings a little bit prettier, and I'm sorry about that. This is the addition step, which we have talked about. Here is your tetrahedral intermediate. So let's just take as a given that that occurs. The question of what happens next is really depends on a couple of things. It depends on the identity of the nucleophile. It depends on the identity of the Y group, which we're going to think about, is it a leaving group or not? And it's all going to be, whether or not it's a good leaving group, is going to be relative to that negatively charged oxygen in the tetrahedral intermediate. So basically, we're going to ask this. same question is your nucleophile a good enough leaving group relative to the negatively charged oxygen where good enough here means the same thing as it meant earlier that is it must be at least well good enough leaving group means as good or better at as a leaving group as a negatively charged oxygen. Okay? Is the nucleophile a good enough leaving group relative to a negatively charged oxygen? Is the Y group a good enough leaving group relative to a negatively charged oxygen? So let's answer the first question. Is the nucleophile a good enough leaving group relative to a negatively charged oxygen? If that answer to question one is yes, then what happens will essentially be the reverse of the addition step. Meaning, if the nucleophile is a decent leaving group, which let's just color it blue for yes, it is a decent leaving group. Then we can expect the reverse of the addition step. of the addition reaction to happen. What that looks like is that electrons from the negatively charged oxygen will kick back down to form, to reform the oxygen carbon pi bond and the nucleophile leaving group will leave. That's the microscopic reverse of the addition step. We call that elimination. So one thing that can happen is elimination of the nucleophile to get you Back to starting materials. This happens frequently. It is often reasonable and this happens in reactions when they are reversible and can go both ways. Students in general are uncomfortable with that situation. They want it to be decided, but Often reality is more complicated than that. Frankly, a lot of biochemical reactions are reversible. You should be super grateful that they are because it's what's keeping you alive. So you've got to get used to the discomfort. So an example of a reaction that we have done today where the backwards reaction might be reversible would be this one. Notice your nucleophile, the OH group, which I'm now going to highlight in blue as our nucleophile, came in. attack the carbonyl carbon to give a negatively charged oxygen, is the reverse step, elimination of the OH group, reasonable here? Totally, right? A negatively charged oxygen should be able to kick off another negatively charged oxygen. In this language, as we discuss these reactions, I'm going to call the negatively charged oxygen the kicker-offer, though that is not... I'm sure grammatically correct. And we could talk about what's going on with the electrons in the orbitals here, but it's actually maybe not necessary for our purposes, so I think I'll gloss over that. Go ahead. Is that because gold and gas are 16? Yes, exactly. Is that because the conjugate acid in the example I just showed you, if you want to know whether it's reasonable for the kicker offer to kick off a group, Go ahead and kick off the group, look at it as a leaving group, draw its conjugate acid, find the pKa, which you noted is 16, do the same for the kicker off, or if they're about the same, then great, that's a reasonable step. Often when reactions are in equilibrium, like this, you'll need to do something downstream later on to push the reaction in a direction that you want. Okay? So if yes, then the reversible step, then the step of nucleophile elimination is fine. Let's think about another option. What if the answer to question two is yes? That is, is the Y group, which I guess since we need more colors, let's make it orange. If the Y group is, I'm sorry, every time I say the word Y group, I'm back in my freshman year. Did they have Y groups for you guys when you started at BYU? No? Some of you skipped freshman orientation because you were too cool. I was not too cool for freshman orientation. And yes, I can still remember. the cheer that we came up with in our Y group. And no, I will not, under any circumstance, perform it for you. All right, so if the Y group is a decent leaving group relative to the kicker offer, then the following elimination of the Y group is possible. That looks like this. Electrons on the negatively charged oxygen kick down to form a pi bond and the Y group leaving group leaves, giving you this product. Notice that if the nucleophile is eliminated, you go back to... starting materials and it's like there wasn't a reaction. If the Y groups eliminated, you've done two things. You did an addition reaction as step one, and then you did an elimination reaction as step two. The overall product of an addition reaction followed by an elimination reaction is substitution. This is an example of an overall substitution reaction if you compare starting materials to products that proceeds not via a concerted mechanism like you saw with the SN2 reaction, but rather addition first, then elimination. And we will talk about this substitution reaction. We will often call it nucleophilic acyl substitution. The substitution is now clear where that comes from. Nucleophilic means the nucleophile is doing the substitution, swapping out the Y group. Where does acyl come from? I'm going to highlight in gray an acyl group. This is a carbonyl group. Attached to something else that's carbon the stuff in gray is an acl group so nucleophilic acl Substitution is we swap out the leaving group on the acl group for something else this happens all the time in biology it happens happens all the time in chemistry. It's how soap is made. It's also how the peptide or amide bonds in your proteins that are being made by the ribosome right now inside you are being made. This reaction happens all over. Interfering with it is how beta-lactam antibiotics work, but it all happens via this kind of mechanism. Okay, questions so far? Yes? Okay, so the question is, if the nucleophile is a leaveable nucleophile relative to negatively charged oxygen, sort of, and the Y group is different, how do you determine whether you're going to go back or go forward? Sometimes there's a difference. a clear answer, sometimes there's not. Let me take you through that, but go ahead. The same question or different question? Alright, can we pause and we'll answer and come back to you. Alright, so classic example of that is a really highly efficient nucleophilic a cell substitution reaction starting with an acid chloride and let's say the methoxide nucleophile hopefully you can tolerate me abbreviating methoxide as this if you can't I don't care enough to change There's a New York Times article from about 10 years ago by someone who compared her OCHEM class to a bad boyfriend that was sort of soaking up her whole life and time and was not giving her very much in return. And what I just said, I don't care enough to change. is bad boyfriend language. So there you go. I'll send that article out to you if you want to see it. It actually has some good points about the value of OCHEM. But in any case, yes, negatively charged. charged oxygen can attack a carbonyl directly to give a tetrahedral intermediate. Then we compare at this stage the goodness of leaving groups and the best leaving group leaves. In this case, we compare Cl minus, the Y group, with OME minus, the original nucleophile, and we ask at this stage which is better. for the kicker offer to kick off the Cl-or the OMe-. And to make that decision, you simply draw what the product would be if Cl-got kicked off. Then you compare the Cl-leaving group with the OMe-nucleophile. You see which is the best one. If you can't remember, draw the conjugate acid and look up the pKa. I'm probably getting that number wrong, but it's no contest. The PKA, the CL is multiple billions of times more better as a leaving group than OME minus. So that's how you predict that that reaction is good. There might be cases where it would be difficult to predict, and we'll see examples of that. In cases where maybe the leaving groups are similar, probably this is an equilibrium reaction, and which way it goes is going to depend on some other factors that you introduce. Go ahead. So the difference between whether or not you can do acid is that it's straight with the... The nucleophile, yeah. The difference in whether or not you need an acid catalyst is going to be all about the strength of the nucleophile. But you end up with the same thing. You're substituting for Y group. But right. Whether or not it's acid catalyzed, you're going to end up with the Y group being substituted out by the nucleophile. Yep. Now this will illustrate something. This whole thing that we just went through with the acid chloride going to the ester. and it was simply an issue of leaving group versus leaving group. That will explain something I told you last time, which is that if we rank the reactivity of the carbonyl derivatives, I told you it was easy to make a less reactive carbonyl derivative from a more reactive one, but not easy to go the other way. What we just did was we took an acid chloride in category one, super reactive, and we turned it into an ester. No problem. Why wasn't it a problem? Because we replaced a better leaving group with a worse leaving group. That's downhill in reactivity. That's favorable. So this ranking reflects those leaving group differences that we talked about. All right. So with that, oh, we do need to consider one final third case, which is... What if the answers to question one and two are both no. Then you get to the tetrahedral intermediate and the Y group is not a good enough leaving group to leave and the nucleophile is similarly also not good enough of a leaving group to leave. Relative to the kicker offer. And so there we stop. There the addition reaction has happened. Starting materials there abbreviated. The addition step can happen, but there will be no subsequent elimination step. There we stop. I can't draw an octagon fast. So there we stop. And yes, I am distracted enough to color that red. Until the reaction is done and we do what is called aqueous workup. Aqueous workup is a step we do at the end of the reaction where we add water to protonate any of the negatively charged oxygens that are there. Normally, we don't want to... In OCHEM, by the way, we're notoriously bad at showing you what the counter ion is, but it's always there. Generally, when we're doing reactions in the lab, we don't want to isolate the sodium salt of something. It's trickier to deal with from a practical point of view than just the neutral compound. So we will do an aqueous workup step where we add water, and that will simply do the acid-base reaction to protonate. the negatively charged oxygen and then we're going to end up with the following product where both the R group, I'm sorry, both the nucleophile and the Y group and the protonated kicker-offer are all still in the molecule and this is going to be called an addition product. You will be tempted to think about nucleophilic addition to carbonyls as distinct and not even the same as nucleophilic acyl substitution, you should not. That's an unnecessary distinction. These two molecules should be in the same drawer, in the same room, in the same building in your mind palace. Because there, if you have one, I don't know whether that actually works or not. I've been watching Sherlock on the BBC and it's fun, but I don't know whether it works. Any case, they're the same thing. It's just all about leaving group ability. Let me give you an example of a reaction where neither the Y group nor the nucleophile are good leaving groups. And this will get into some of the stuff that we will do in chapter 13. So imagine that you're, for fun, we'll stick with the aldehyde. The Y group for the aldehyde is H. Y minus would be H minus. which is a terrible, horrible, no good, very bad leaving group. You can tell if you draw its conjugate acid and go look up its pKa, which is 35. Not a great leaving group. Certainly not relative to negatively charged oxygen. Enter a reagent that we will call sodium borohydride because that is its name. This is an example of what I will call a hydride delivery reagent. As I said, H-is a terrible thing on its own. You can buy sodium hydride, which is NAH. It's a sodium salt. It's not very soluble at all in organic solvent. And so... The hydrogen never dissociates from the sodium, so the hydrogen can't be a nucleophile. When we want to use H-as a nucleophile, we need a different strategy. And the one that works is to use a molecule that can carry H-without too much trouble, but is also not that bad at giving it away. Sodium borohydride, if I were to draw out the Lewis structure, is just boron, an sp3 hybridized boron bonded to four hydrogens. The negative formal charge is on the boron, but if you actually look up electronegativities, boron is less electronegative than hydrogen. So even though the negative formal charge is on the boron, most of it is out on the hydrogens. Now this is different, because normally with reagents that you've seen before, the hydrogens have a... Delta plus on them, they're electron poor. Not so with sodium borohydride. The hydrogens are electron rich. And what can happen is that sodium borohydride can hand off its hydride, H minus. The H minus leaves the boron. It's never floating off on its own. It, in fact, is handed off directly to the carbonyl carbon and attacks the pi star, breaks the pi bond. And that... gives us, well that's not the color we wanted, that gives us the addition product. Here the nucleophile H-is delivered by the borohydride nucleophile delivery or hydride delivery reagent and we end up with this tetrahedral intermediate. I will highlight, except we lost the oxygen, I will highlight in green the kicker offer group, highlighter not pen, Two hydrides, they're both equally terrible at being a leaving group, and so there we stop. And there we stop until aqueous workup. It turns out, actually, that sodium borohydride can be used in methanol, which is a protic solvent. This is a solvent that has protons that are available for the taking if you have a strong enough base. And so what likely happens at this stage... is that the negatively charged oxygen picks up a proton from methanol, and then you get your alcohol as a product. Now, in another life, you may have been tempted to memorize putting on a flash card and trying to stuff it in a drawer in your mind palace that the aldehyde reacts with sodium borohydride to give you a primary alcohol. That is a boneheaded use of your mind palace. Mine Palace real estate is very expensive. Do not memorize things you don't have to, okay? You don't need to memorize this reaction because you can figure it out if you know a couple of things. If you know about leaving groups, if you know about carbonyl chemistry, and then I suppose you might have to memorize the borohydride is a hydride delivery reagent. Do you see how the consequences of this reaction follow automatically from your understanding of mechanism and reactivity? You don't need to memorize that sodium borohydride reacts with aldehydes to give primary alcohols and chant it as a mantra because that's just the natural consequence. Aldehydes are reactive at the carbonyl carbon. Nucleophiles attack the carbonyl carbon to give a tetrahedral intermediate. What happens next depends on whether you got a good leaving group. Okay? Good. So that's actually this chapter in a nutshell. What we're going to do now is talk about some variations on hydride delivery reagents. And we're going to talk about some variations on the actual electrophile. So far, what I showed you was hydride delivery to an aldehyde where you don't have a good leaving group on there. But we need to expand our view. So let's just, sure, fine. Trying to figure out what's best to do next. Let me introduce the other players in terms of hydride delivery reagents, and then we'll see how they interface with the various carbonyl starting materials that you could use. So you've already met sodium borohydride, and this is... a reagent that's good enough to react with aldehydes and ketones because aldehydes and ketones are top level reactivity carbonyl molecules. Here is a version of sodium borohydride that is even more reactive. It is called lithium aluminum hydride. Every semester somebody wants to know why did we switch the counter ion from sodium to lithium and is there such a thing as sodium aluminum hydride? The answer is yes there is such a thing as sodium aluminum hydride. I don't know what difference it makes. It probably makes a difference to use lithium for reasons that we don't need to worry about but that are purely practical and I again don't care enough to find out so that's sort of what you're stuck with. Lithium aluminum hydride is more reactive than sodium borohydride. The reason for that, if you look at the periodic table, and again, I will draw these reagents out in their Lewis structures, boron... Boron and aluminum are from the same column of the periodic table. Only aluminum is down one row. What happens to electronegativity as you go down a row of the table? Anybody remember? It's less. So remember, the whole reason boron could deliver hydride is because boron is less electronegative than hydrogen, such that the hydrogens are all electron-rich. Aluminum is even less electronegative than boron and so the hydrogens on aluminum hydride are even more electron rich. The delta pluses are even bigger. So much so that lithium aluminum hydride cannot use protic solvent. You cannot use protic solvent with lithium aluminum hydride. If you do... The hydride acts as a base to remove a proton from whatever thing that has a pKa of less than 35. You make hydrogen gas, and it's exothermic, and in the lab, heat and hydrogen and oxygen often cause fires. So this is a reagent that is a matter of practicality. You can't use protic solvents. We won't worry too much about the solvent issue in this class, though you may see some things in the text. where your textbook author describes what solvents are being used. A couple of interesting things about how these reagents interface with carbonyls. Remember, the most reactive carbonyls that we're going to talk about in this chapter are the aldehydes and the ketones. Esters, sort of one row down. Sodium borohydride is less reactive than lithium aluminum hydride. Sodium borohydride can react with aldehydes and ketones, but not esters. Why? Because sodium borohydride is not reactive enough, and esters are less reactive than aldehydes or ketones. In contrast, lithium aluminum hydride is reactive enough. that it can not only react with the most reactive carbonyl derivatives, the aldehydes and the ketones, but also with the less reactive ones like esters. Okay, so let's show you some examples and some consequences. I already showed you a reaction of sodium borohydride with an aldehyde to give you a primary alcohol. Let's now talk about the ester. shall we? And let's react an ester with lithium aluminum hydride. Again, we've got to use lithium aluminum hydride because sodium borohydride is not reactive enough. So I'm going to draw a lithium aluminum hydride like this. Lithium is the counter ion. I will often forget to draw it. And again, I do not care enough to change on that. But hydride gets delivered to the carbonyl oxygen. I do care enough to change about a lot of things. Repentance is very important. But repenting for not showing counter ions is something I'm willing to delay for a while, I guess is what I'm saying. Aluminum hydride delivers the hydride to the carbonyl carbon to get a tetrahedral intermediate. There's the negatively charged oxygen. There's the methoxy group. And there is the hydride that was just delivered. Now... What's going to happen next? This is the Y group. This is the nucleophile. This is the kicker-offer. Do we have a good leaving group relative to a negatively charged oxygen? Yeah, methoxy could leave relative to that negatively charged oxygen, and indeed that is what happens. So we have an addition step followed by an elimination step to give us from the ester, an aldehyde, and then we have that methoxide leaving group, which goes away and compare with the lithium counter ion, I suppose. That's an addition and elimination reaction. We could call that overall a substitution reaction, right? In going from starting materials to products, overall that is substitution. You should note that this is a special kind of substitution reaction. This is called a reduction reaction because in organic chemistry, reduction means increasing the number of carbon-hydrogen bonds and... or decreasing the number of carbon heteroatom bonds. We went from an ester to an aldehyde. That is a reduction. So there we go. Now we kind of have a problem, and that is that through this reduction reaction, we converted the ester into an aldehyde, and the aldehyde is now more reactive than the ester. We're going uphill in reactivity. The reason we can do that is because lithium aluminum hydride is such a reactive reagent. You can pay for uphill transformations as long as you're pairing them with downhill transformations, and that reduction reaction is favorable. Carbon-hydrogen bonds are strong. But you now have your aldehyde, and we know that lithium aluminum hydride can react with aldehydes too. So the next thing that happens here is another molecule of lithium aluminum hydride. I'm oversimplifying a little bit here in ways that should you find out what I'm doing, you will thank me for not telling you. Another molecule of lithium aluminum hydride will deliver a second equivalent to the aldehyde, getting another addition step to get another... tetrahedral intermediate. Again, negatively charged oxygen. Now the two things attached, both are nucleophile. Now the, I'm sorry, the color labeling is just going to totally break down at this point. Hopefully you can forgive me for not showing. The oxygen cannot kick off H minus here or H minus there. There we are stuck with our stop sign. I should just have that on copy and paste. Until the end of the reaction, whereupon we add water, aqueous workup. Sometimes people will add this as a aqueous water. Isn't that funny? Aqueous workup. Yeah, when someone tries to sell you all brand new extra special aqueous water, don't give them your money, please. All right. So. That's the product. When we use lithium aluminum hydride, you have one substitution reaction and then the following addition reaction, which is, again, itself a reduction. Two rounds of reduction to take you from the ester to the aldehyde and then all the way to the primary alcohol. The reason we don't stop at the... aldehyde is because the aldehyde is more reactive than the ester and it's going to compete with the ester for lithium aluminum hydride in practice we add enough lithium aluminum hydride so that we know we're going to go all the way to the alcohol yes can you use uh Can you get a hydrogen from something else besides aluminum hydride instead of adding water? No. What's left after the hydride gets delivered is actually ALH3, which is neutral. Aluminum is still less electronegative than hydrogen. Those hydrogens are really not acidic, so they're not a good proton source. With excess? With excess, no. Okay. So what do you need to memorize about what I just showed you versus what don't you need to memorize? memorize. You probably are going to need to memorize that lithium aluminum hydride can react with esters and aldehydes. You're probably going to need to memorize that lithium aluminum hydride will not stop at the aldehyde. Other than that though, the consequence of it being a primary alcohol you can figure out based on the details of the reaction. Now we should introduce the question, what if we wanted to invent a reagent? Suppose that we have a molecule that's an ester, and we want to stop at the aldehyde because we really, really want that aldehyde. We don't want to go all the way to the primary alcohol. What if we could invent a reagent that would be like lithium aluminum hydride, reactive enough to react with the ester, but could somehow not react with the aldehyde? That's a tall order. It's really hard because aldehydes are more reactive than esters. Your text presents two solutions. It asks you to memorize without any explanation. Actually, there are some explanations that are somewhat hand-wavy. I'm going to show you one of them. The other one you can still use and memorize. But one of them I think you can understand how it works, and it's actually pretty interesting. I'll set up the problem, and then I think we're going to have to be done for the week. The reagent is like lithium aluminum hydride, only it's not negatively charged. This is called di- isobutyl aluminum hydride and because we don't like writing out all of those words we call this dibal h and some organic chemists just drop the h and just call it dibal and we're all fine with that um interestingly this reagent is used to reduce esters to the aldehyde but notice you How it's different from lithium aluminum hydride. It's not even negatively charged on the aluminum. Your text doesn't explain this, but what happens is really, really cool, and I want to show it. But we are out of time. So with that cliffhanger, I will see you next week on Monday in the Marb.

We've talked about how when you attack an anti-bonding orbital like the pi star, you do two things. Number one, you make a new bond that's between the nucleophile and the carbonyl carbon, and that new bond is indicated by this arrow. The second thing you do is you break the carbon-oxygen pi bond. You got to do both.

If you don't do both, you end up with a Texas carbon. That is a carbon with five bonds to it, which... We love the great state of Texas, but even in the great state of Texas, carbons only have four bonds. So, this is the first step for most of the reactions we will talk about, and it is an addition step.

We call it an addition step because the nucleophile gets added and you get a tetrahedral intermediate making one molecule out of two. That is the first step of most of the reactions that we will talk about this semester. One thing we did not get a chance to talk about last time, which I would like to address briefly, is this addition step and the effect of the nucleophile on it. Basically, when we are asking the question of whether or not this addition step is reasonable, what we're basically asking is how reactive is the nucleophile relative to the tetrahedral intermediate that it makes. Notice that in the tetrahedral intermediate, you have a negatively charged oxygen.

And so in some ways, how effective this first step is, or what its equilibrium constant is, can be summarized by just comparing how reactive is your nucleophile in the starting material relative to the negatively charged oxygen in the tetrahedral intermediate. We will see that nucleophiles that are at least as reactive as a negatively charged oxygen are good enough to attack directly. So what does good enough to attack directly mean?

That means your nucleophile is at least as reactive. Another way of saying reactive is unstable. at least as reactive or as unstable as the negatively charged oxygen in the tetrahedral intermediate.

You want to make a comparison? Fine. Draw, let's suppose, for a specific example, let's use OH-as the nucleophile, and we'll just keep this same aldehyde.

And I want to know, is it reasonable for OH-as a nucleophile to attack the negatively, to attack the... carbonyl carbon to give you the negatively charged tetrahedral intermediate. Look at the tetrahedral intermediate, then look at your nucleophile, compare their reactivities. Both of them are negatively charged oxygens. If it gets more complicated than that, simply draw the conjugate acid of the nucleophile and draw the conjugate acid of the negatively charged oxygen.

Look up their pKa's that will tell you their relative reactivity. So you okay on that? So is this a reasonable step or not?

Great, totally reasonable. What would be unreasonable? Do I expect water to be able to attack an aldehyde directly? What's wrong with that? Water is more stable than a negatively charged alkoxide, which is what your tetrahedral intermediate is.

In such cases, then, we will often have to do something else first. So for water, if we want water to be able to attack a carbonyl, in many cases we will need some kind of catalysis. An example of how that could be for water would be as follows.

Let's use some acid catalyst. To use the aldehyde as a nucleophile in the first step, we protonate the oxygen, lone pair electrons on the carbonyl oxygen attack a proton from the acid, giving us this oxonium ion that we've talked about before. The oxonium ion is more reactive than the aldehyde starting material, and you can tell that that's the case because attack of water on the oxonium ion is suddenly going to look more reasonable.

Water, the lone pair electrons on the water, the nucleophile, lone pair electrons on oxygen attack the pi star at the carbonyl carbon. You still get a tetrahedral intermediate. Only notice that the thing that gets made, the tetrahedral intermediate, is now we're dealing with a neutral oxygen, right?

Can... a neutral nucleophile attack a positively charged oxonium ion to give you a tetrahedral intermediate that's positively charged and has a neutral OH group on it. See, neutral OH for neutral OH, that's reasonable.

So that's one way you can tell whether a nucleophile can attack directly or whether you're going to need some kind of catalysis like protonation to have it happen first. We'll revisit that in a little bit. Some of you are going to take biochemistry at some point and you will learn about a few terms called specific acid catalysis versus general acid catalysis. This is an example of something that will be called later, not in this class, but later, specific acid catalysis.

And what that means is that the acid base step, the protonation of the carbonyl, has to happen first. And then the addition step is the slow step in the reaction. There are other forms of this we will encounter when we get to chapter 14. Suppose that your nucleophile is an amino group instead of water, a neutral amine instead of water.

You might still be worried about a neutral amine generating a negatively charged. oxygen intermediate, right? Let's just sort of draw how that would go. Here's the product after a neutral amine attacks an aldehyde.

And there would be reasons to feel uncomfortable about that because you're comparing the nucleophilicity or the reactivity of the negatively charged oxygen in your tetrahedral intermediate with how stable your neutral amine is as a nucleophile. And if that comparison isn't coming easy, simply draw the conjugate acid. Here is the alcohol conjugate acid of the negatively charged oxygen, and look up its pKa, which if you did would be 16. Draw the conjugate acid of your nucleophile, which is the ammonium ion, which has a pKa of 11. In order for us to be really comfortable with direct uncatalyzed attack, we want our nucleophile to be at least as reactive as that negatively charged oxygen.

And it's not. It's close, but not quite. And that reflects the fact that when you do reactions of aldehydes with amines, often you need an acid catalyst.

What actually, so I'm going to say, um, You know, the answer to this question, is this okay, is I have substantial discomfort. I'm telling you this now because some people gloss over this substantial discomfort. In fact, your textbook actually will when it shows you this reaction.

But any organic chemist that actually has been trained will tell you, when you go to the lab and do this reaction, you need an acid catalyst. So let me show you what the mechanism most likely is for an amine to attack an aldehyde. It turns out here it's a little bit different than it is for water attacking an aldehyde. For water, we had to protonate the aldehyde first in a fast step to get the oxonium ion.

It turns out, for reasons that you can understand, the amine is more nucleophilic than water. The amine itself is more reactive than water. That's because nitrogen is less electronegative than oxygen, and you learned that in 351. So it turns out the amine, experimentally we observed, the amine doesn't have to wait until the acid catalyst protonates the carbonyl. It can actually attack during.

So for amines attacking aldehydes, and this is mostly a preview, what we observe is that the nitrogen attacks the carbonyl carbon. in the same way that we've seen before, but as that process is happening, lone pair electrons on the oxygen pick up a proton from the acid catalyst. All of that happens in the same concerted step.

It's relatively slow, and that gets you to an intermediate with a neutral oxygen. And I feel a lot better. That relieves my mild substantial discomfort because now I'm more comfortable with a neutral amine generating an intermediate with a neutral oxygen.

So you're going to see examples of this. I would not expect you. To be able to predict the differences that I've just shown you between specific acid catalysis for the water attacking an aldehyde versus what I just barely showed you, which is general acid catalysis, I would not expect you to be able to predict that, oh, for water, you protonate first and then attack occurs in two different steps, whereas for the amine, you attack and protonate in the same step.

But I would expect you to. Have the idea that the amine and water, because they're both neutral, probably aren't good enough to attack directly to give you a negatively charged oxygen in a tetrahedral intermediate. Is that, got the idea?

Okay, that's the addition step. After that, and we're talking in general terms because so much of what we're going to do in the next two or three years, chapters is all variations on this theme. So I know some of you are anxious to get to what about lithium aluminum hydride and what about NADH and we will try to do that but if you understand the general picture the details are going to be a lot easier. Yes? Yeah, I haven't showed you the end of the reaction.

The acid does, in most cases, get regenerated. And we'll do that later on in Chapter 14 and beyond. Yep.

Okay, others? That was the addition step. Let's talk about what happens after the addition step.

And we're going to be general again for a moment. Take a step back. Here's your general carbonyl, and here is your nucleophile.

And let's just suppose now that the nucleophile is... good enough to attack directly? We've already answered that question. That's the addition step. That is an ugly arrow.

I will pause at various points during the lecture to make the drawings a little bit prettier, and I'm sorry about that. This is the addition step, which we have talked about. Here is your tetrahedral intermediate.

So let's just take as a given that that occurs. The question of what happens next is really depends on a couple of things. It depends on the identity of the nucleophile. It depends on the identity of the Y group, which we're going to think about, is it a leaving group or not?

And it's all going to be, whether or not it's a good leaving group, is going to be relative to that negatively charged oxygen in the tetrahedral intermediate. So basically, we're going to ask this. same question is your nucleophile a good enough leaving group relative to the negatively charged oxygen where good enough here means the same thing as it meant earlier that is it must be at least well good enough leaving group means as good or better at as a leaving group as a negatively charged oxygen.

Okay? Is the nucleophile a good enough leaving group relative to a negatively charged oxygen? Is the Y group a good enough leaving group relative to a negatively charged oxygen?

So let's answer the first question. Is the nucleophile a good enough leaving group relative to a negatively charged oxygen? If that answer to question one is yes, then what happens will essentially be the reverse of the addition step. Meaning, if the nucleophile is a decent leaving group, which let's just color it blue for yes, it is a decent leaving group.

Then we can expect the reverse of the addition step. of the addition reaction to happen. What that looks like is that electrons from the negatively charged oxygen will kick back down to form, to reform the oxygen carbon pi bond and the nucleophile leaving group will leave. That's the microscopic reverse of the addition step.

We call that elimination. So one thing that can happen is elimination of the nucleophile to get you Back to starting materials. This happens frequently.

It is often reasonable and this happens in reactions when they are reversible and can go both ways. Students in general are uncomfortable with that situation. They want it to be decided, but Often reality is more complicated than that.

Frankly, a lot of biochemical reactions are reversible. You should be super grateful that they are because it's what's keeping you alive. So you've got to get used to the discomfort.

So an example of a reaction that we have done today where the backwards reaction might be reversible would be this one. Notice your nucleophile, the OH group, which I'm now going to highlight in blue as our nucleophile, came in. attack the carbonyl carbon to give a negatively charged oxygen, is the reverse step, elimination of the OH group, reasonable here? Totally, right? A negatively charged oxygen should be able to kick off another negatively charged oxygen.

In this language, as we discuss these reactions, I'm going to call the negatively charged oxygen the kicker-offer, though that is not... I'm sure grammatically correct. And we could talk about what's going on with the electrons in the orbitals here, but it's actually maybe not necessary for our purposes, so I think I'll gloss over that.

Go ahead. Is that because gold and gas are 16? Yes, exactly.

Is that because the conjugate acid in the example I just showed you, if you want to know whether it's reasonable for the kicker offer to kick off a group, Go ahead and kick off the group, look at it as a leaving group, draw its conjugate acid, find the pKa, which you noted is 16, do the same for the kicker off, or if they're about the same, then great, that's a reasonable step. Often when reactions are in equilibrium, like this, you'll need to do something downstream later on to push the reaction in a direction that you want. Okay?

So if yes, then the reversible step, then the step of nucleophile elimination is fine. Let's think about another option. What if the answer to question two is yes?

That is, is the Y group, which I guess since we need more colors, let's make it orange. If the Y group is, I'm sorry, every time I say the word Y group, I'm back in my freshman year. Did they have Y groups for you guys when you started at BYU?

No? Some of you skipped freshman orientation because you were too cool. I was not too cool for freshman orientation.

And yes, I can still remember. the cheer that we came up with in our Y group. And no, I will not, under any circumstance, perform it for you.

All right, so if the Y group is a decent leaving group relative to the kicker offer, then the following elimination of the Y group is possible. That looks like this. Electrons on the negatively charged oxygen kick down to form a pi bond and the Y group leaving group leaves, giving you this product.

Notice that if the nucleophile is eliminated, you go back to... starting materials and it's like there wasn't a reaction. If the Y groups eliminated, you've done two things.

You did an addition reaction as step one, and then you did an elimination reaction as step two. The overall product of an addition reaction followed by an elimination reaction is substitution. This is an example of an overall substitution reaction if you compare starting materials to products that proceeds not via a concerted mechanism like you saw with the SN2 reaction, but rather addition first, then elimination.

And we will talk about this substitution reaction. We will often call it nucleophilic acyl substitution. The substitution is now clear where that comes from.

Nucleophilic means the nucleophile is doing the substitution, swapping out the Y group. Where does acyl come from? I'm going to highlight in gray an acyl group.

This is a carbonyl group. Attached to something else that's carbon the stuff in gray is an acl group so nucleophilic acl Substitution is we swap out the leaving group on the acl group for something else this happens all the time in biology it happens happens all the time in chemistry. It's how soap is made.

It's also how the peptide or amide bonds in your proteins that are being made by the ribosome right now inside you are being made. This reaction happens all over. Interfering with it is how beta-lactam antibiotics work, but it all happens via this kind of mechanism.

Okay, questions so far? Yes? Okay, so the question is, if the nucleophile is a leaveable nucleophile relative to negatively charged oxygen, sort of, and the Y group is different, how do you determine whether you're going to go back or go forward? Sometimes there's a difference. a clear answer, sometimes there's not.

Let me take you through that, but go ahead. The same question or different question? Alright, can we pause and we'll answer and come back to you.

Alright, so classic example of that is a really highly efficient nucleophilic a cell substitution reaction starting with an acid chloride and let's say the methoxide nucleophile hopefully you can tolerate me abbreviating methoxide as this if you can't I don't care enough to change There's a New York Times article from about 10 years ago by someone who compared her OCHEM class to a bad boyfriend that was sort of soaking up her whole life and time and was not giving her very much in return. And what I just said, I don't care enough to change. is bad boyfriend language. So there you go.

I'll send that article out to you if you want to see it. It actually has some good points about the value of OCHEM. But in any case, yes, negatively charged. charged oxygen can attack a carbonyl directly to give a tetrahedral intermediate.

Then we compare at this stage the goodness of leaving groups and the best leaving group leaves. In this case, we compare Cl minus, the Y group, with OME minus, the original nucleophile, and we ask at this stage which is better. for the kicker offer to kick off the Cl-or the OMe-.

And to make that decision, you simply draw what the product would be if Cl-got kicked off. Then you compare the Cl-leaving group with the OMe-nucleophile. You see which is the best one. If you can't remember, draw the conjugate acid and look up the pKa. I'm probably getting that number wrong, but it's no contest.

The PKA, the CL is multiple billions of times more better as a leaving group than OME minus. So that's how you predict that that reaction is good. There might be cases where it would be difficult to predict, and we'll see examples of that. In cases where maybe the leaving groups are similar, probably this is an equilibrium reaction, and which way it goes is going to depend on some other factors that you introduce. Go ahead.

So the difference between whether or not you can do acid is that it's straight with the... The nucleophile, yeah. The difference in whether or not you need an acid catalyst is going to be all about the strength of the nucleophile.

But you end up with the same thing. You're substituting for Y group. But right.

Whether or not it's acid catalyzed, you're going to end up with the Y group being substituted out by the nucleophile. Yep. Now this will illustrate something.

This whole thing that we just went through with the acid chloride going to the ester. and it was simply an issue of leaving group versus leaving group. That will explain something I told you last time, which is that if we rank the reactivity of the carbonyl derivatives, I told you it was easy to make a less reactive carbonyl derivative from a more reactive one, but not easy to go the other way. What we just did was we took an acid chloride in category one, super reactive, and we turned it into an ester. No problem.

Why wasn't it a problem? Because we replaced a better leaving group with a worse leaving group. That's downhill in reactivity. That's favorable. So this ranking reflects those leaving group differences that we talked about.

All right. So with that, oh, we do need to consider one final third case, which is... What if the answers to question one and two are both no. Then you get to the tetrahedral intermediate and the Y group is not a good enough leaving group to leave and the nucleophile is similarly also not good enough of a leaving group to leave.

Relative to the kicker offer. And so there we stop. There the addition reaction has happened. Starting materials there abbreviated.

The addition step can happen, but there will be no subsequent elimination step. There we stop. I can't draw an octagon fast. So there we stop. And yes, I am distracted enough to color that red.

Until the reaction is done and we do what is called aqueous workup. Aqueous workup is a step we do at the end of the reaction where we add water to protonate any of the negatively charged oxygens that are there. Normally, we don't want to...

In OCHEM, by the way, we're notoriously bad at showing you what the counter ion is, but it's always there. Generally, when we're doing reactions in the lab, we don't want to isolate the sodium salt of something. It's trickier to deal with from a practical point of view than just the neutral compound.

So we will do an aqueous workup step where we add water, and that will simply do the acid-base reaction to protonate. the negatively charged oxygen and then we're going to end up with the following product where both the R group, I'm sorry, both the nucleophile and the Y group and the protonated kicker-offer are all still in the molecule and this is going to be called an addition product. You will be tempted to think about nucleophilic addition to carbonyls as distinct and not even the same as nucleophilic acyl substitution, you should not.

That's an unnecessary distinction. These two molecules should be in the same drawer, in the same room, in the same building in your mind palace. Because there, if you have one, I don't know whether that actually works or not. I've been watching Sherlock on the BBC and it's fun, but I don't know whether it works.

Any case, they're the same thing. It's just all about leaving group ability. Let me give you an example of a reaction where neither the Y group nor the nucleophile are good leaving groups.

And this will get into some of the stuff that we will do in chapter 13. So imagine that you're, for fun, we'll stick with the aldehyde. The Y group for the aldehyde is H. Y minus would be H minus. which is a terrible, horrible, no good, very bad leaving group.

You can tell if you draw its conjugate acid and go look up its pKa, which is 35. Not a great leaving group. Certainly not relative to negatively charged oxygen. Enter a reagent that we will call sodium borohydride because that is its name.

This is an example of what I will call a hydride delivery reagent. As I said, H-is a terrible thing on its own. You can buy sodium hydride, which is NAH. It's a sodium salt.

It's not very soluble at all in organic solvent. And so... The hydrogen never dissociates from the sodium, so the hydrogen can't be a nucleophile. When we want to use H-as a nucleophile, we need a different strategy.

And the one that works is to use a molecule that can carry H-without too much trouble, but is also not that bad at giving it away. Sodium borohydride, if I were to draw out the Lewis structure, is just boron, an sp3 hybridized boron bonded to four hydrogens. The negative formal charge is on the boron, but if you actually look up electronegativities, boron is less electronegative than hydrogen.

So even though the negative formal charge is on the boron, most of it is out on the hydrogens. Now this is different, because normally with reagents that you've seen before, the hydrogens have a... Delta plus on them, they're electron poor.

Not so with sodium borohydride. The hydrogens are electron rich. And what can happen is that sodium borohydride can hand off its hydride, H minus. The H minus leaves the boron.

It's never floating off on its own. It, in fact, is handed off directly to the carbonyl carbon and attacks the pi star, breaks the pi bond. And that... gives us, well that's not the color we wanted, that gives us the addition product.

Here the nucleophile H-is delivered by the borohydride nucleophile delivery or hydride delivery reagent and we end up with this tetrahedral intermediate. I will highlight, except we lost the oxygen, I will highlight in green the kicker offer group, highlighter not pen, Two hydrides, they're both equally terrible at being a leaving group, and so there we stop. And there we stop until aqueous workup.

It turns out, actually, that sodium borohydride can be used in methanol, which is a protic solvent. This is a solvent that has protons that are available for the taking if you have a strong enough base. And so what likely happens at this stage...

is that the negatively charged oxygen picks up a proton from methanol, and then you get your alcohol as a product. Now, in another life, you may have been tempted to memorize putting on a flash card and trying to stuff it in a drawer in your mind palace that the aldehyde reacts with sodium borohydride to give you a primary alcohol. That is a boneheaded use of your mind palace. Mine Palace real estate is very expensive. Do not memorize things you don't have to, okay?

You don't need to memorize this reaction because you can figure it out if you know a couple of things. If you know about leaving groups, if you know about carbonyl chemistry, and then I suppose you might have to memorize the borohydride is a hydride delivery reagent. Do you see how the consequences of this reaction follow automatically from your understanding of mechanism and reactivity?

You don't need to memorize that sodium borohydride reacts with aldehydes to give primary alcohols and chant it as a mantra because that's just the natural consequence. Aldehydes are reactive at the carbonyl carbon. Nucleophiles attack the carbonyl carbon to give a tetrahedral intermediate.

What happens next depends on whether you got a good leaving group. Okay? Good. So that's actually this chapter in a nutshell. What we're going to do now is talk about some variations on hydride delivery reagents.

And we're going to talk about some variations on the actual electrophile. So far, what I showed you was hydride delivery to an aldehyde where you don't have a good leaving group on there. But we need to expand our view. So let's just, sure, fine. Trying to figure out what's best to do next.

Let me introduce the other players in terms of hydride delivery reagents, and then we'll see how they interface with the various carbonyl starting materials that you could use. So you've already met sodium borohydride, and this is... a reagent that's good enough to react with aldehydes and ketones because aldehydes and ketones are top level reactivity carbonyl molecules. Here is a version of sodium borohydride that is even more reactive. It is called lithium aluminum hydride.

Every semester somebody wants to know why did we switch the counter ion from sodium to lithium and is there such a thing as sodium aluminum hydride? The answer is yes there is such a thing as sodium aluminum hydride. I don't know what difference it makes. It probably makes a difference to use lithium for reasons that we don't need to worry about but that are purely practical and I again don't care enough to find out so that's sort of what you're stuck with. Lithium aluminum hydride is more reactive than sodium borohydride.

The reason for that, if you look at the periodic table, and again, I will draw these reagents out in their Lewis structures, boron... Boron and aluminum are from the same column of the periodic table. Only aluminum is down one row.

What happens to electronegativity as you go down a row of the table? Anybody remember? It's less.

So remember, the whole reason boron could deliver hydride is because boron is less electronegative than hydrogen, such that the hydrogens are all electron-rich. Aluminum is even less electronegative than boron and so the hydrogens on aluminum hydride are even more electron rich. The delta pluses are even bigger. So much so that lithium aluminum hydride cannot use protic solvent.

You cannot use protic solvent with lithium aluminum hydride. If you do... The hydride acts as a base to remove a proton from whatever thing that has a pKa of less than 35. You make hydrogen gas, and it's exothermic, and in the lab, heat and hydrogen and oxygen often cause fires. So this is a reagent that is a matter of practicality. You can't use protic solvents.

We won't worry too much about the solvent issue in this class, though you may see some things in the text. where your textbook author describes what solvents are being used. A couple of interesting things about how these reagents interface with carbonyls.

Remember, the most reactive carbonyls that we're going to talk about in this chapter are the aldehydes and the ketones. Esters, sort of one row down. Sodium borohydride is less reactive than lithium aluminum hydride. Sodium borohydride can react with aldehydes and ketones, but not esters. Why?

Because sodium borohydride is not reactive enough, and esters are less reactive than aldehydes or ketones. In contrast, lithium aluminum hydride is reactive enough. that it can not only react with the most reactive carbonyl derivatives, the aldehydes and the ketones, but also with the less reactive ones like esters.

Okay, so let's show you some examples and some consequences. I already showed you a reaction of sodium borohydride with an aldehyde to give you a primary alcohol. Let's now talk about the ester.

shall we? And let's react an ester with lithium aluminum hydride. Again, we've got to use lithium aluminum hydride because sodium borohydride is not reactive enough.

So I'm going to draw a lithium aluminum hydride like this. Lithium is the counter ion. I will often forget to draw it.

And again, I do not care enough to change on that. But hydride gets delivered to the carbonyl oxygen. I do care enough to change about a lot of things. Repentance is very important. But repenting for not showing counter ions is something I'm willing to delay for a while, I guess is what I'm saying.

Aluminum hydride delivers the hydride to the carbonyl carbon to get a tetrahedral intermediate. There's the negatively charged oxygen. There's the methoxy group. And there is the hydride that was just delivered.

Now... What's going to happen next? This is the Y group. This is the nucleophile. This is the kicker-offer.

Do we have a good leaving group relative to a negatively charged oxygen? Yeah, methoxy could leave relative to that negatively charged oxygen, and indeed that is what happens. So we have an addition step followed by an elimination step to give us from the ester, an aldehyde, and then we have that methoxide leaving group, which goes away and compare with the lithium counter ion, I suppose.

That's an addition and elimination reaction. We could call that overall a substitution reaction, right? In going from starting materials to products, overall that is substitution. You should note that this is a special kind of substitution reaction. This is called a reduction reaction because in organic chemistry, reduction means increasing the number of carbon-hydrogen bonds and...

or decreasing the number of carbon heteroatom bonds. We went from an ester to an aldehyde. That is a reduction. So there we go. Now we kind of have a problem, and that is that through this reduction reaction, we converted the ester into an aldehyde, and the aldehyde is now more reactive than the ester.

We're going uphill in reactivity. The reason we can do that is because lithium aluminum hydride is such a reactive reagent. You can pay for uphill transformations as long as you're pairing them with downhill transformations, and that reduction reaction is favorable. Carbon-hydrogen bonds are strong. But you now have your aldehyde, and we know that lithium aluminum hydride can react with aldehydes too.

So the next thing that happens here is another molecule of lithium aluminum hydride. I'm oversimplifying a little bit here in ways that should you find out what I'm doing, you will thank me for not telling you. Another molecule of lithium aluminum hydride will deliver a second equivalent to the aldehyde, getting another addition step to get another... tetrahedral intermediate.

Again, negatively charged oxygen. Now the two things attached, both are nucleophile. Now the, I'm sorry, the color labeling is just going to totally break down at this point.

Hopefully you can forgive me for not showing. The oxygen cannot kick off H minus here or H minus there. There we are stuck with our stop sign.

I should just have that on copy and paste. Until the end of the reaction, whereupon we add water, aqueous workup. Sometimes people will add this as a aqueous water.

Isn't that funny? Aqueous workup. Yeah, when someone tries to sell you all brand new extra special aqueous water, don't give them your money, please.

All right. So. That's the product.

When we use lithium aluminum hydride, you have one substitution reaction and then the following addition reaction, which is, again, itself a reduction. Two rounds of reduction to take you from the ester to the aldehyde and then all the way to the primary alcohol. The reason we don't stop at the...

aldehyde is because the aldehyde is more reactive than the ester and it's going to compete with the ester for lithium aluminum hydride in practice we add enough lithium aluminum hydride so that we know we're going to go all the way to the alcohol yes can you use uh Can you get a hydrogen from something else besides aluminum hydride instead of adding water? No. What's left after the hydride gets delivered is actually ALH3, which is neutral.

Aluminum is still less electronegative than hydrogen. Those hydrogens are really not acidic, so they're not a good proton source. With excess?

With excess, no. Okay. So what do you need to memorize about what I just showed you versus what don't you need to memorize? memorize. You probably are going to need to memorize that lithium aluminum hydride can react with esters and aldehydes.

You're probably going to need to memorize that lithium aluminum hydride will not stop at the aldehyde. Other than that though, the consequence of it being a primary alcohol you can figure out based on the details of the reaction. Now we should introduce the question, what if we wanted to invent a reagent? Suppose that we have a molecule that's an ester, and we want to stop at the aldehyde because we really, really want that aldehyde.

We don't want to go all the way to the primary alcohol. What if we could invent a reagent that would be like lithium aluminum hydride, reactive enough to react with the ester, but could somehow not react with the aldehyde? That's a tall order.

It's really hard because aldehydes are more reactive than esters. Your text presents two solutions. It asks you to memorize without any explanation. Actually, there are some explanations that are somewhat hand-wavy.

I'm going to show you one of them. The other one you can still use and memorize. But one of them I think you can understand how it works, and it's actually pretty interesting.

I'll set up the problem, and then I think we're going to have to be done for the week. The reagent is like lithium aluminum hydride, only it's not negatively charged. This is called di- isobutyl aluminum hydride and because we don't like writing out all of those words we call this dibal h and some organic chemists just drop the h and just call it dibal and we're all fine with that um interestingly this reagent is used to reduce esters to the aldehyde but notice you How it's different from lithium aluminum hydride. It's not even negatively charged on the aluminum.

Your text doesn't explain this, but what happens is really, really cool, and I want to show it. But we are out of time. So with that cliffhanger, I will see you next week on Monday in the Marb.

Transcript for:Reactivity of Carbonyl Compounds

Transcript for:
Reactivity of Carbonyl Compounds