CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 34 - Acids and Acid Derivatives
Chapter 1. Reducing Carboxylic Acids to Carbonyl Groups [00:00:00]
Professor J. Michael McBride: OK. So more today about the lore of acids and acid derivatives, which you’ll find in chapter 17, 18. We’ll get on, maybe at the end, to chapter 19, but probably that’s for Friday.
At the end last time, we were talking about having an R-, that is an organometallic molecule, react with an acid, and showed that, of course, the first thing it does is act as a base, not as a nucleophile, and pull off the proton. But even at that stage, it’s possible for another one to come in and add to the carbon oxygen double bond to give this di-salt. And then when this is hydrolyzed at the end of the day, you get a ketone, so you could convert an acid to a ketone that way.
But it’s difficult. Because sometimes, it’s not possible to do it in two stages. Here you stopped at the dilithium salt. Then you got rid of the R- when you added acid and got the ketone.
But in the case of using hydride, rather than R-, as the base/nucleophile, again you first make the salt. But the second time around, you can get the aluminum, also, as the salt of the oxygen. And when you have that, that’s not a bad leaving group. It’s better than the O-lithium, the O-aluminum. So that it’s possible to do an elimination reaction where the electrons of the lithium-oxygen bond, or the oxygen anion, help push out the aluminum-oxygen anion, so that in situ, while you’re doing the reaction, it forms the aldehyde.
Bear in mind in the first step, in the first reaction, when we used the alkyl lithium, we got the carbonyl group back only after adding acid. Here it forms again while the reaction is going on. And of course, aldehydes are reactive with lithium aluminum hydride. So you don’t stop at the carbonyl. You go all the way to the alcohol adding two Hs. We’ll see a little bit later ways to try to control that.
Chapter 2. Decarboxylation Reactions [00:02:21]
Now, decarboxylation is another interesting reaction of carboxylic acids, and in particular, of diacids, like malonic acid, which has two acid groups attached to the same carbon.
This is this scheme that Mitsunobu used in his paper to talk about making the compound at the bottom right that we discussed last time or the time before that. Now remember, first he got this compound by forming that carbon-carbon bond. And it has two esters and a CN triple bond, but we said that it was possible to convert the ester into an acid. We’ll talk about this a little bit more. And the other ester into an acid, and the cyanide, also, the nitrile group, into an acid.
Now, I’ve drawn that acid in an interesting conformation here. Because you see a possibility– this doesn’t happen when you have the anion, but when you have the protonated carboxylate, and you have two of them, it’s possible to do this reaction. So you use these electrons to form a bond to that hydrogen. These go here, these go down here.
What’s the product at the top from this cycle? Can you see what it is? What did the carbon at the top become?
Student: Carbon dioxide.
Professor J. Michael McBride: Carbon dioxide, right. So you have CO2. That’ll fly off. It’s a decarboxylation. And you’re left with this, which is an enol. And of course, that can undergo an allylic rearrangement to the ketone form of the enol, and the H go on here. And then you have the product.
So when you have two carboxyl groups on the same carbon, and it’s protonated so that you can do this cyclic mechanism to transfer the proton, then you can easily lose CO2 and do a decarboxylation.
But there are other ways to do decarboxylations from acid. The CO2 could be a leaving group also from XCO2-, where X is something that can tolerate a negative charge, or from the RCO2 radical.
So for example, carbonic acid, the hydrate of CO2. It’s an acid; it’s also a base. So if you get this form, then you can see how it can undergo decarboxylation. So that’s just the reverse of the hydration of CO2. So carbonic acid and CO2 go back and forth. So that’s a case where the X can leave with an unshared pair.
Or carbamic acid, where it’s nitrogen that does the trick, and you can see the same kind of thing to get a decarboxylation.
Or we actually mentioned last semester already Kolbe electrolysis, where it’s a free radical that does it. So you start with a carboxylate anion, but you do an oxidation to remove an electron from it, which then makes this free radical.
But that free radical is fabulously unstable when R is an alkyl group. And it undergoes this shift of the electrons– single electrons, it’s a radical process– which is fast. Even at 4 Kelvin you can’t see an ESR spectrum of this thing. Even at that very low temperature it immediately loses CO2 and gives the free radical, even if it’s not a particularly stable free radical.
So you remember Kolbe used that, unbeknownst to him, to make R2. R2 was ethane, remember? He thought he had made the methyl, but once molecular weights were determinable, it turned out that he actually had the dimer, ethane.
And in fact that reaction can work fairly well synthetically, to take two acids, knock out two CO2s, and have the Rs come together. That was done in 1848. But in 1973, they got a 50% yield by doing electrolysis that generates this free radical by losing CO2 from it, and the dimer was formed.
OK. So there are examples of decarboxylation.
But, also there’s another one that’s interesting. Which is to treat an acid with silver oxide, which gives a silver salt of the acid. Then if you treat that with bromine, you get R-Br. So you can convert a carboxylic acid to the alkyl bromide with one less carbon atom. A decarboxylation has been involved.
And it’s pretty easy to see how that would work. If you react the salt with Br2, the O- can do a nucleophilic substitution on bromine with bromide leaving to give this brominated intermediate. But now, if you have a free radical in there, it can attack the bromine to form the R-Br bond and generate this radical, which, as we’ve already seen, immediately loses CO2 and gives– so we have a free-radical chain that allows this brominated acid to become RBr.
Now, this reaction– to convert an acid to an alkyl bromide– is called the Hunsdiecker reaction, for a sort of curious reason, it seems to me. Between 1936 and 1944, the Hunsdieckers, husband and wife, published, I think, something like six patents for doing this, and one paper in the open literature.
But in fact, the same reaction had been done in 1861 and published in a leading German journal, Annalen, by Alexander Porfiryevich Borodin, who was obviously much earlier than the Hunsdieckers. This is his tomb in Saint Petersburg. And notice that it doesn’t have anything about doing this reaction on it, because he was not only a chemist, he was also a musician. So can you– I know some of you are good musicians. Can you see what this line is, up at the top? What is it?
Student: The Polovtsian Dances.
Professor J. Michael McBride: Polovtsian Dances by Borodin, from Prince Igor. And in fact, although he didn’t get any recognition in terms of the reaction being named for him– these people that came along 75 years later got the credit– he did get a posthumous Tony award. Because that tune, you know– [SINGS] dah dah dah dah dah dah dah dah dah dah dah– is the theme of the musical Kismet. So he came out OK, anyhow.
OK. But now, when we think about– so here’s a way to take RCO2 and make it into RBr. Right? When you’re thinking about possibilities of synthesis, you should think about, is there a way to go the other direction? To start with RBr and go to the acid with one additional carbon? So can you add a carbon to do that? Can anybody think of a way of doing that? To take RBr and substitute the Br with an acid group, CO2, put another carbon on the chain?
We’ll actually talk about this some more later. But we had shown an example that if you have cyanide– which is a nucleophile, so you could do an SN2 reaction, put cyanide there– but cyanide has the same oxidation state of the carbon as the acid does. So just treating it with water, and either acid or base– and we’ll talk about this little later on– and heat can do that.
So you can go either direction. You can take the acid to the bromide, or the bromide to the acid.
And that’s not the only way. Notice that this involves the R being R+, the thing with the LUMO, the σ* that gets attacked by the cyanide. But there’s another way to start with RBr where R is R-. It’s a HOMO that does the attacking.
So how would you convert RBr into something that’s like R-, something that has a high HOMO? Under what circumstances does R behave like R-, is it associated with a high HOMO?
Student: With a halogen?
Professor J. Michael McBride: When it’s associated with a halogen, it’s a low LUMO, right? And it’s the σ*. But how can you get the σ to be high energy?
Student: Use a Grignard reagent.
Professor J. Michael McBride: A Grignard reagent! So if you want R-, you make a Grignard reagent. And now you have to react that in order to get an acid. What will you react it with? What will it attack in order to put the carbon and oxygens on? Ruoyi?
Student: Carbon dioxide.
Professor J. Michael McBride: Carbon dioxide. It’s exactly the reverse of the reaction we’re talking about here. So you bring carbon dioxide in, you get the salt, treat it with acid, and you do it.
So there are many ways to skin these cats.
Chapter 3. Acid Derivatives and their IR Spectra [00:12:02]
OK. So we’re talking about acid derivatives. We’ve just been talking about acids. But acid derivatives are things where you have some other group there, X, with an unshared pair, not OH.
So one example that’s called an acid derivative is a nitrile, that we just talked about. Another one is a ketene.
Now let’s think about how these are regarded as acid derivatives. There, as we just mentioned in the last slide, the carbon has a +3 oxidation level, because X is electronegative, something with the unshared pair there. And in nitrile, it’s also +3. So you can go between these things and acid just by treating them like water. Although you may need catalysts and a high temperature in order to do it. Nitriles are not so easy to do. Remember he refluxed it for 24 hours or something like that, Mitsunobu did, in order to do this.
OK. That’s fine. But over here, the carbon is +2 in its oxidation level. It doesn’t look like you’re dealing with the same thing. However, this other carbon is -2. And if you go to the acid, it’s -3, +3. So it’s, again, no change in overall oxidation level. So you don’t need an oxidizing/reducing agent, and you can just do that with water. So ketene will react with water to give a carboxylic acid. So all these things are called acid derivatives.
First, just a word about the spectroscopy. In fact, I’ll go in about three seconds through the first slide, because we’ve already done it in some detail. Remember, we looked at carbonyl stretching frequencies in the IR, and saw that aldehydes and ketones were similar, but the amide was much lower frequency, because of the unshared pair on the nitrogen being involved, stabilized by the carbon oxygen π*, and therefore weakening it to make it a weak bond.
But in the case of the ester the bond was, in fact, strengthened a bit. And the case of acid chloride, it was strengthened quite a bit. And that the reason for that was that you could get this other curious resonance structure. So we already did this, and I’m not spending any time just reminding you of it.
The one thing I’ll talk about that’s new is the anhydride, which is of course an acid derivative where the X group that’s attached to the carbonyl is another acid. So here we see an IR spectrum, and here is the percent transmission. So this is very little light getting through. A strong absorption means a small number here.
And if we look in this region, here’s the carbonyl region, we see that there are two peaks, which isn’t surprising, because there are two carbonyls. There’s one at 1819, which is very strong, and another one at 1750, which is considerably weaker.
Now two carbonyls, two IR frequencies, two IR peaks. That seems to make sense. Except that it’s symmetrical. Wouldn’t you expect them to be the same?
Student: There’s an asymmetrical frequency.
Professor J. Michael McBride: Pardon me?
Student: There’s an asymmetrical frequency.
Professor J. Michael McBride: OK. So they couple. They could either both be stretching at the same time, or one can be stretching while the other’s shrinking. And if they’re close enough together, separated only by one oxygen, they can interact mechanically one another, and one of those is easier than the other. So there are two normal modes, both involve both of the C=O bonds. Right? And we can understand how it would be. Because if they’re in phase, both stretching at the same time, then the light will interact very strongly with it. Because both of them were being stretched by the same direction of the electric field of the light. And it turns out that that’s difficult to do. It’s difficult to stretch them both at the same time. It’s better to have one stretch when the other shrinks, in terms of energy.
So this one is hard to do. That means it’s at high frequency. But it’s quite intense, because they’re both going the same direction.
The other normal mode would have one going up while the other’s going down. Now, if they were strictly parallel, there would be no net interaction with the light for that. But if they’re twisted a little bit, or splayed out a little bit, then there will be some interaction with light. They won’t exactly cancel one another. And that’s what happens here. So you have a strong peak where they’re both stretching at the same time, and a weaker peak where one is stretching and the other is shrinking.
Now, if you compare that with this cyclic imide, which is another acid derivative where nitrogen is attached to the acyl group, and now we look at the peaks here, and you see a strong peak at low frequency and a weak peak at high frequency. So it’s exactly the opposite. Can you see why it would be the opposite?
Because now, when they both stretch at the same time… when one stretches and the other shrinks, the oxygens are moving in the same direction. So they’re both being pushed the way light wants them to go. So that one, which is now the easy one to do, where you stretch one and shrink the other, is now very strong. It interacts strongly with light. But the one where they’re in-phase, here, is now very weak. They nearly cancel one another.
That’s just an interesting additional example of IR coupling. But it helps you see how you use the IR to interpret stuff about structure, as to whether two carbonyls are parallel or antiparallel most of the time.
Chapter 4. Interconversion of Acid Derivatives, Saponification [00:18:18]
OK, now, the interconversion, the chemistry of the acid derivatives. So the idea is that some nucleophile comes in and forms the bond, and then the other one leaves. So that’s a substitution. But of course, it involves a tetrahedral intermediate. It’s not like SN2. You actually have an association to give it a stable intermediate, followed by dissociation. So it’s not a pentavalent transition state. It’s a tetrahedral, tetravalent carbon intermediate.
Now, which group will leave from that intermediate, X or Y? Well, there’s a hierarchy of these things. The easiest thing to leave is chloride, then carboxylic acid, then an alcohol anion, then a nitrogen anion. And you can see why this would be. Why so? Because that’s the order of stability of the anions, right? As exemplified in the pKa.
So if you could make the most energetic one, the top one, the acid chloride, then by adding the others successively, you could make it into any of the other three; or the next one, the acid anhydride, into either of the other two; or the ester into the amide, and so on.
So the important thing is to get up at the top of the ladder at the beginning. And you can do that. The same way you converted alcohol to RCl, you can convert an acid to the acid chloride by using SOCl2.
And, as I was saying, out of order before, this order of the hierarchy is clear because of the pKas. They’re in the order of the stability of the anions.
But it’s not just the stability of the anions that makes them have that order. Because as we saw before, you get resonance stabilization of the electrons that are on Y. That is, whatever this group is here, chlorine, oxygen, oxygen, nitrogen. Right? Those electrons are stabilized by the vacant orbital here, as denoted by these resonance structures.
So even if you didn’t have the question of the stability of the anions, the stability of the derivatives would still be the same. It likes most to be with nitrogen, because it can get the special stability from stabilizing the high-energy electrons of the nitrogen. It’s lower if it’s an oxygen. It’s still lower if it’s a carboxylate group, because although it’s an oxygen’s unshared pair, as in the case of alcohol, that electron pair on that oxygen is already stabilized by its own carbonyl group. So it’s not as available for being stabilized by the one here.
So this one is not as stable as that, this one is not as stable as that. And what that means is, you don’t have to do this with base in order to get these anions, in order to have this reactivity scale. You could do the same thing if you had acid-catalyzed. And we’ve already seen it in the case of Fischer esterification, where now it’s a question of alcohol and water. That is, are you going to put on OR, or are you going to put on OH? And we’ve gone through the acid catalyzed mechanism of Fischer esterification, which converts OH to OR, or OR to OH.
And you can also do transesterification. You can take one ester and a different alcohol and have the OR’ replace OR”, or something like that. You can go back and forth from one ester to the other.
But the equilibrium constant for this acid-catalyzed process is one. There’s not much difference among the different Rs. But you make it go the way you want in a Fischer esterification by using the law of mass action. You put in a whole bunch of alcohol, then you get that ester. Or you put in a whole bunch of water, and then you get the carboxylic acid. So going back and forth between ester and acid also is very much like these transesterifications.
But if you use base, it’s a different story. So you could have hydroxide come in and OR- go out. That’s fine. And it looks like it should be an equilibrium constant of about one. But in fact, this reaction has a special name, saponification, and it has an equilibrium constant much greater than one. Now, why should there be a difference between OR and OH, an enormous difference? It’s because this isn’t the end of the line. That OR- that disappeared can get back into the act. In what way? Sebastian?
Student: Lose the hydrogen
Professor J. Michael McBride: Right. This is not called an acid for nothing. So it can transfer the proton to the OR-.
And now that’s very favorable. Because the pKa of the acid is about 5, the pKa of a typical alcohol is about 16. So they differ by 11 orders of magnitude. That means the equilibrium constant is about 1011 , when you factor in that you don’t stop at the acid, but go to the anion of the acid and to the alcohol.
Now, why does this thing have a special name, saponification, this particular substitution reaction? Because you have these things called triglycerides, which are long chain acids attached as esters to a three carbon unit.
You know what else you call those things besides triglycerides?
Professor J. Michael McBride: Fats. And if you treat fat with lye, you’re doing something that’s been known through millennia. Right? Because lye is what you get when you take the ashes from a fire and filter water through it, and then evaporate the water. You know what you get? What lye is? It’s potassium hydroxide. So treat it with KOH, and you have this reaction take place. So you get carboxylic acids, long-chain carboxylic acids, and you get alcohols on all those three.
So these long-chain fatty acids, the salts of them, the potassium salts, are called what? Why is it called saponification? What’s savon or sapon?
Professor J. Michael McBride: It’s soap. So that’s how soap was made. And still is made. And that stuff is called glycerin, that trialcohol.
Now, there’s a very interesting book about lore in the Appalachian mountains called The Foxfire Book. Has anybody seen that? There was a movie made from it a long time ago, actually. But a bunch of high school English students had the project of going out and talking to people who knew how to do things, and then writing the traditional lore.
And one of the people they talked to was the grandmother of one of the girls, whose name was Pearl Martin, and who made soap. And so she showed the girls– she had a big kettle, and put in lard, and put in lye, and stirred it up. And she says, there, now the soap’s making, and so on. And so she got the soap out, and she said, you know, “I could wash my hair in it.” And the girls were astounded that she would wash her hair in this stuff that came from lie. But she said, “I ain’t afraid to wash my hair in it.” She said, “That there lard eats the lye.” What did she mean by that? She, of course, hadn’t studied chemistry, at least I don’t believe, although she had very good practical knowledge of it. That she knew that she could wash her hair with this stuff, even though you put lye in it. The girls were dumbfounded at this.
But the reason is that the lard “ate” the lye. The lye was like RO-. pKa of about 16. But this last stage that makes the reaction go makes it a carboxylate with a pKa of only 5, so it doesn’t hurt your hair.
So there’s some tradition for you.
Chapter 5. Selective Reduction of Acid Derivatives [00:27:16]
Now, an interesting question is, what happens when the nucleophile is R-, or H-, rather than these other things? So I’m going to put a little bit of this on the board. And there’s a– see, I should put this up, put this thing over on the other screen. Actually, it’s supposed to be there. Yeah, there it is. So let me put this up– get some light here.
OK. So what if you have R- or H- as the nucleophile?
So suppose we start, and these are for reactions with acid derivatives. Suppose we use the acid fluoride. Then if we use hydride, for example, lithium aluminum hydride– so the AlH4-, high HOMO there– OK, so hydride can come in here. Put minus on the oxygen. Chloride can leave. And we get aldehyde.
But the reaction won’t stop there, because we already know that aldehydes are reduced by lithium aluminum hydride. So we’ll go on to put two hydrogens on, and make this, and ultimately when you add an acid, you’ll get all the way to the alcohol, not just to the aldehyde.
So it’s not just a substitution of chloride by hydride, but you go on all the way to the product. Or if you have an alkyl lithium, or an alkyl magnesium bromide.
You do the same kind of thing. You get the– now it would be a ketone first, because this is an R group. But again, you go on and get the alcohol, having put on two new R groups. So if that’s the kind of alcohol you want, with two R groups, there’s two R groups of the same kind that you can get from an alkyl halide, then that’s a good way to do it.
Now, if you use an ester, I’ll just write, ditto. If you react it with lithium aluminum hydride, you get all the way to the alcohol, replacing now the OR-, instead of the Cl. Or if you use alkyl lithium or alkyl magnesium bromide, you go all the way to the alcohol.
In fact, a particularly interesting one is to use two chlorines or two esters. Can you see what would happen in that case, if you reacted those with the hydride, or with the R-, an alkyl metal compound? What’s going to be different?
Well, if you went through, you replace the chloride by R-, same kind of trick we’re doing here, or H-. Then you replace the other chloride by H- or R-, you get an aldehyde or a ketone. And then you add to the carbonyl. So now we put three groups on. So we have a COH at the end with three R groups.
You could do the same thing with the hydride, but there wouldn’t be much point in making methanol that way. You’re using expensive reagents to make something cheap. But if you want something that has three of the same R groups on it, that could be a practical way of going about it.
It’s slightly different if you’re using an amide. So suppose you use lithium aluminum hydride. Now we’ll get– just call this R, here. So we have NH2 and H- that came in.
Now, what we’ve done above is then to lose this group that was originally there. Lose chloride, or lose O-, RO-. But it’s a little different here. Because with the aluminum salt here, the aluminum helps make this oxygen a better leaving group than it would otherwise be. So instead of coming like this and losing the nitrogen, what happens is the unshared pair helps you lose the oxygen. So the nitrogen is not lost. It’s still there, as N double-bond now, H, and two Hs on there.
But now that’s rather like a carbonyl group, right? So the H- can get… the lithium aluminum hydride can get in the act again. And the product from lithium aluminum hydride reduction of an amide is an amine.
OK. So now these are examples, then, of going all the way to the COH. The carbonyl is completely gone, and you added two, or in one case, three groups.
But you might actually want the intermediate. You might want the aldehyde. So the question arises, is there a way to stop it, so you can get that intermediate product? And you can do that. You can stop by tuning the nucleophile, so that it’s able to react in the first reaction but not reactive enough to react in the second.
So for example, instead of using lithium aluminum hydride, use lithium aluminum hydride but that instead of having other hydrogens on it, has O-t-butyl taken three times. Now, there are two things about having that in place of hydrogen. The first is that the oxygen is electron-withdrawing. It stabilizes anions. So it makes the electrons here that are going to be doing the attacking– the sigma electrons– it makes them lower in energy and less reactive.
What else about having the t-butyl group here would help in mitigating the reactivity of that tri-t-butoxy lithium aluminum hydride? Rahul?
Student: Steric hindrance.
Professor J. Michael McBride: Right. It’s big, so it’s hard to approach. OK. So it’s reactive enough to take the acid chloride to the aldehyde, but not to take the aldehyde in the second step. But it’s used at very low temperature, at dry ice temperature, so that that second reaction won’t happen.
Or you can use, in a very similar situation, a reagent called DIBAL. Again, you use it at minus 78°. And it’s again an aluminum hydride, but now again, it doesn’t have the oxygens, but it has two isobutyl groups which, again, are big and help tame it. So that one can be used with the ester. And take it over to the aldehyde.
Or another way of getting from the acid chloride and stopping at the aldehyde is to do it with H2 and a catalyst. So the acid chloride, reduce it to put H2 on the ends of the bond. But of course, the aldehyde can also be reduced. So you tame the reactivity of the catalyst by adding a nitrogen-containing compound, quinoline. This is called the Rosenmund procedure.
We saw this before. The idea of a catalytic hydrogenation where you make the catalyst less reactive in order to stop it. Remember what that was, way back when?
Student: Acetylene to alkene.
Professor J. Michael McBride: Right. The Lindlar catalyst, where you wanted to take a triple bond and go just to a double bond, but not from the double bond to a single bond. It was, again, the case of adding something that would poison the catalyst.
And finally, just as a reminder, it’s possible to have an acid chloride and react it with a very, very weak carbon nucleophile that only will react if you soup up the chloride by using AlCl3 to help make that a good leaving group. Now you can use a really weak carbon nucleophile, high HOMO on carbon, that will react to give the ketone, but won’t react with the ketone, because it’s so weak. Here you did a special thing to help make the chloride reactive, but it doesn’t help make a ketone you get reactive.
Remember what the nucleophile is you used in this? An acyl chloride/AlCl3? What is it? Acylation. So you could use benzene as the nucleophile. We talked about it from the point of view of something attacking benzene, but now we’re looking at it the other way. Benzene attacking the something else. Right? So we make the aromatic ketone that way, by Friedel-Crafts acylation. But again, it’s a question where you could do the first one, but not the second one, by tuning the reactivity of the reagents.
Chapter 6. Nitriles and Ketenes [00:40:45]
OK. Now acid-, base-hydrolysis of RCN. Let me just use the other board here for a moment. So we talked before about converting the CN to a carboxylic acid. And we can we can go through acid- and base-catalyzed mechanisms of doing that. It’s very simple, and I’ve got a blackboard realization of that, just to trace through it quickly.
So you have the nitrile, the CN group. Protonate it. You have a resonance structure, that has + on carbon. React that with water, so now you’ve got the C-O bond. Lose a proton, so you have the alcohol here now. But then protonate the nitrogen. Now we have this. But if you take a base and pull off this proton, you have the amide.
In other words, this is an allylic rearrangement. It’s a hydrogen here that goes to the other end of this double bond.
So you can get partway to the acid. You get from this acid derivative to this acid derivative, the amide.
But now suppose you protonate this back again and get to this one, which I’ve redrawn here, and have that C+ attacked by water. So now a second water attacking the carbon. And take the proton off here, and put it on the nitrogen, and then have the NH3 leave with the electrons. You have this cation, lose a proton, and now you’re all the way to the carboxylic acid.
So there are two stages of substituting water here. First adding water to give the amide, and then changing the amide acid derivative into the acid itself. So that’s acid catalysis. And I won’t spend more time on it now, but it would be good for you to be able to do that.
But you can also do it with base. So in base, you attack the CN triple bond, as if it were a CO double bond. Generate this, put a proton on, and now do the allylic rearrangement, the same thing we did before, except that now instead of being acid-catalyzed, it’s base-catalyzed, so we’re halfway there. We’re to the amide, once we put the last proton on.
And now we can go from the amide, attack it, and then lose the NH2, probably in a protonated form. Then we have the carboxylic acid and drive it all the way to completion in base, as in the case of saponification, by making the anion, the stable carboxylate anion.
OK. So there, both acid and base could be used. But it’s a hard reaction to do. It has to be heated a lot. And sometimes people even do it using biological catalysts to try to get it down to a lower temperature.
Now, a ketene, we said, was another acid derivative. So if you have an acid chloride and react it with an amine, we’ve just been talking about reactions like this, where the product– you lose chloride, add nitrogen, and the product is an amide, the carbonyl with nitrogen next to it.
But this is a special case, because there are three alkyl groups on the nitrogen. Do you see how that makes it special? Ayesha? If you use the nitrogen as a nucleophile, and attack, and chloride leaves, then you have N+. Right? Because you’ve used the unshared pair to make that new bond. And normally, what would you do with that N plus? How would you get rid of the charge?
Student: Lose a proton.
Professor J. Michael McBride: Normally, you’d lose a proton from the amine. But this time, we used a trialkyl amine, so it can’t do that. So in fact, it doesn’t do it. It probably attacks sometimes. Chloride probably leaves. But chloride can come back and the nitrogen leave, because it couldn’t become stable by losing a proton. So that probably goes on.
But another reaction could happen instead, which is for the nitrogen base to attack a proton here. So you could do an elimination reaction, lose HCl, and you generate, then, ketene, this funny molecule, which is sort of an analog of CO2„ except instead of an oxygen, you have CH2 group here.
Now, that group then could be attacked by a nucleophile, the same way CO2 is attacked. And now you have the nucleophile attack, and that’s an enolate, so we could draw the resonance form of it and we could put a proton on there from the protonated nucleophile and get the product, plus the nucleophile anion, which is now able to do it to the next ketene.
OK. So what we’ve done is convert a ketene into an acid derivative, depending on what nucleophile we use. And you can use lots of different nucleophiles. Here’s, for example, taken from that Jones textbook, is sulfur as the nucleophile. This, they said, was done at dry ice temperature for three days, and gave a thioester in 93% yield.
Or instead of base-catalyzed like this, it could be aci- catalyzed. So reaction with t-butyl alcohol with acid. So the acid can protonate the oxygen, then this oxygen can attack, right, as the nucleophile, lose the proton, and you get t-butyl ester, 87% yield.
Or you can use a nitrogen. This is done in ethanol water, and it gives a 75% yield.
Of course, these other things could’ve reacted. The O-, or HO-, could have reacted if it were in base. But it’s probably, although it isn’t said so here, it’s probably in acid. So it’s probably an acid-catalyzed reaction.
Chapter 7. Insertion into the Acyl-R Bond: Baeyer-Villiger Oxidation [00:47:35]
Now, then you can have insertions into acyl-R bonds. Let me just start this, and we’ll finish it up next time. So we’ll talk about the first reaction. You remember this picture? Who’s interesting in this picture that we talked about before? This guy here. Remember who that is? What did he do, if you can’t remember his name?
Student: Made triphenylmethyl.
Professor J. Michael McBride: He made triphenylmethyl, right? That’s Gomberg. So this is the same picture. And there’s Baeyer, who was the boss of the lab. But this is Victor Villiger, a Swiss who was working in the Munich lab at the time. And just at 1900, so the same time that Gomberg was making triphenylmethyl, the Baeyer-Villiger reaction was invented, which inserts oxygen into an acyl bond.
So let me just do that one quickly on the board, and then we’ll quit.
So what we’re doing is taking R, O, R, and we’re going to insert an atom in here to make it R, O, OR. So change a ketone into an ester.
And the way it’s done is to react it with a peroxy acid. So now we have RO-, and it is acidic, so we’ll take the OO-,, O, O.
And now, normally what we think of is these electrons come in and lose this group. But we don’t have a leaving group here, right? So instead what happens is that these electrons come here. So we have double bond O at the top, R, and now what’s attached here is O, this O, and attached to that O is the other R, and this good anion has left.
So we ended up by inserting an oxygen into the C-R bond. And we’re going to talk more about that next time.
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