CHEM 125b: Freshman Organic Chemistry II

Lecture 26

 - Aromatic Substitution in Synthesis: Friedel-Crafts and Moses Gomberg


The Friedel-Crafts reaction creates new alkyl- or acyl-aromatic bonds, with or without cation rearrangement. Designing reaction sequences, especially those involving diazonium intermediates, so as to access a wide variety of substituted benzenes provides a good introduction to the challenges of synthetic organic chemistry. Aromatic rings with strong electron withdrawal can undergo nucleophilic aromatic substitution, which plays an important role in biochemistry. The special properties of phenyl-substituted alkanes, especially benzylic reactivity and steric hindrance, played an important role in the development of organic chemistry a century ago.

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Freshman Organic Chemistry II

CHEM 125b - Lecture 26 - Aromatic Substitution in Synthesis: Friedel-Crafts and Moses Gomberg

Chapter 1. Discovery of Friedel-Crafts Alkylation and Acylation [00:00:00]

Professor J. Michael McBride: OK, so we’re finishing up on talking about aromatic compounds, about the chemistry of benzene. You know, this was a big deal in the 19th century, in part because you had reactions that were clean. The compounds were often robust and easy to purify, crystallize, and so on. So many of the early developments in organic chemistry, especially in the second half of the 19th century, had to do with aromatic chemistry. Nowadays that we have more versatile techniques for analyzing things like NMR and for separating things, various kinds of chromatography, it’s possible to work on a much wider variety of molecules. So much recent chemistry is on other subjects, but much of the classical work was on aromatic compounds, and they still are important.

So the Friedel-Crafts reaction we were just beginning with, and we’ll talk today about some ideas about how you design syntheses, and about benzylic stabilization, the reactivity at positions next to an aromatic system. And about triphenylmethyl, which was a particularly important compound for the development of chemistry in the U.S.

So a subtext of what we’re talking about today is how chemistry developed in the United States. Because, as we mentioned last time, James Mason Crafts was an American, but almost all – essentially all his important work was done in France and published in French– almost all of it. So although he was an American, many people thought he was French.

You remember he started working in Wurtz’s lab, essentially as a graduate student– although he wasn’t enrolled as a student, I don’t believe. And met his seven-year-older colleague, a student– the first student of Wurtz– in that lab. And they published a number of papers together. Then he published nothing at Cornell and MIT– that was a textbook. And then he came back to Paris and published this enormous number of papers, 100 papers in 17 years, almost all of them with Friedel. And in particular, that paper–or that set of papers–in 1877, which was the Friedel-Crafts reaction.

Now how did they get into that? They got into that because they had been students together with Wurtz. And when he came back to work in Friedel’s lab, it was natural that they should follow up on the kinds of things they were doing at that time, on the kind of thing that Wurtz was interested in. So seven years before that timeline started, Wurtz published a paper on a new class of radicals. Bear in mind this is 1855, so this is just before Couper and Kekulé got the idea of valence. So this was when types and radicals were competing. So Wurtz was in France, and he wrote this in his thing about a special reaction of iodides with sodium. And if we translate that into modern atomic weights, it’s ethyl iodide and butyl iodide reacting with sodium to give ethyl butyl. Those are his new class of radicals. So hexane plus two sodium iodide.

Now, this is the infamous Wurtz reaction. You hear about the Wurtz reaction. The main place you hear about Wurtz reaction is on exam papers of elementary organic students, who figure you can make any carbon-carbon bond by taking two halides and treating them with sodium. But this is a horrid reaction. It gives terrible yields. It was possible for him to do this, because the products were often gaseous, but it was not at all a high yield reaction, because you get a lot of crossover. You can get ethyl with ethyl, you can get butyl with butyl. You can get radical reactions, you can also get ionic reactions of the ethyl sodium and butyl iodide. It’s a mess.

So the question that occurred to these guys in 1877 was, might other metals work better? In particular, how about aluminum? So they tried reacting alkyl halide, alkyl chloride with aluminum, with the idea of getting R2 and aluminum chloride. That would be an analogous reaction with a different metal. The question is, does it work? And of course, they were chemists. They did chemistry. And chem is try, as they say. So they tried it. So what they found was, it didn’t give a very good yield of what they wanted. It gave a lot of HCl. And notice that that’s very different, it’s not the metal that’s reacting with the chloride, it’s the hydrogen. Where did the hydrogen come from?

And there was lots of that. There were as many moles of HCl as there were of the alkyl halide to begin with. But very little aluminum chloride and very little of the R2. Now when they tried to distill this stuff so they didn’t get the gas, but when they tried to distill they found higher boiling stuff, and it was poor in hydrogen, when they analyzed it. That means that it had a lot of double bonds in it, or rings, or both. So they decided it was actually benzene with R groups attached to it. Where did the benzene come from? Where did the hydrogen come from? From the solvent. Benzene. And an interesting thing about the aluminum: they found that they reaction was initially very slow. But the rate depended on how much aluminum chloride there was. As the reaction went on, you got a little bit of aluminum chloride. And the more you got, the faster it went, until indeed, it was almost impossible to control the reaction. Although it started very fast [correction: slow]. So they began to wonder whether aluminum had anything to do with it, or whether it was just aluminum chloride. And they decided that was true, they didn’t need aluminum, just aluminum chloride.

So this is completely different from what they set out to study, it’s not a Wurtz-type reaction at all, it’s completely different. You have to rearrange things. It’s an aluminum chloride catalyzed reaction of RCl with an aromatic compound. It’s an aromatic substitution. A carbon radical goes on in place of hydrogen, onto the benzene. So it’s a completely different kind of substitution from what they had set out to study. But notice that it’s rather analogous to the idea of chlorine reacting with catalysis by trichloroaluminum that we looked at last time, where the trichloroaluminum helps make the Cl- a leaving group. So it looks like it could be the same deal. Of course, this was mystery to them. This is much later insight about what the mechanism is. But it’s the same idea with RCl instead of ClCl. And it’s even possible that you can get SN1 versions of this, where the AlCl3 pulls off chloride and you have R+, if the R+ is sufficiently stable.

OK, so they published this first paper in 1877. The title of it was “On a New General Method for Synthesizing Hydrocarbons, Ketones, etc.” So a new way to form a carbon-carbon bond. So you could have RCl and an aromatic compound react in the catalysis by AlCl3, and get the substituted benzene and HCl. Now, they say it’s general, so they tried a number of different things. They tried R radicals that had one, two, and five carbons. In this very first paper, just a couple pages, they say they took about five weeks to do all this work that I’m telling you about. They also used bromide and iodide, it worked as well as chloride. They used toluene as well as benzene. In fact, toluene was better because they could get if purer to begin with.

So they said they could do it – they had a system where they could keep the thing warm and methyl chloride bubbling through– that’s a gas– night and day. So they ran it for a long time, for several days– they don’t say exactly how many– at 80°. And of course, you could do a substitution of methyl for hydrogen. You could do a second, a third, a fourth, a fifth. So from this, they got pentachloro- and hexachlorobenzene. Both of these were new compounds. So a new synthetic method means a brand new compound. So no one had ever seen these before.

How did they know what they had? Well, in the case of the hexachlorobenzene, they were able to prepare several hundred grams this way. Its melting point was 164, nowadays we say it’s 165, so they had very good stuff. They measured the molecular weight and they got the correct value within 2%, by measuring the density of the vapor. But a strongly confirmatory observation was that they could oxidize it. They tried a number of methods of oxidation, but the one that gave a good yield was to treat it with cold potassium permanganate for two months. So that wasn’t done in the five weeks. But they were able to get the hexacarboxylic acid, which remarkably enough had been known since 1799. Of course, in 1799 they didn’t know its structure. They didn’t know from structure altogether until 1858. But Baeyer had recently shown that this compound, which was isolated pure in 1799, was hexacarboxylic acid of benzene. Incidentally, the source of it in nature is a natural mineral called honeystone, which is this stuff. It comes in nature, probably from oxidizing graphite or something like that, I don’t know where it comes from exactly. But once you make a known product, then you know what this new stuff is. But the important thing was that they were able to make completely new compounds, and make old compounds in new, more efficient ways.

So this was alkylation. It’s Friedel-Crafts alkylation, putting an alkyl group onto benzene. So there were other alkylations. They found out they could use an alkyl group that had a benzene on it already, like benzyl chloride here, could react with benzene to give diphenylmethane. That was already known that this was a good way to make it. Or they could use chloroform and put three benzenes on it, and get triphenylmethane. And that was also known. But what was new, was that they could use carbon tetrachloride and do one substitution, then another, then another, then another and get tetraphenylmethane, which was an unknown compound. In fact, it wasn’t even known then, because it turned out that wasn’t what they had, although it’s what they said they had. Because it’s hard to know these things. It wasn’t that, because it’s too crowded to try to get four phenyl groups on a single carbon.

So it only did three substitutions, not four. And it was the triphenylmethyl chloride, but it didn’t have chlorine anymore, because they worked it up with water. And the water changed the chloride into an OH, an SN1 reaction. But that was a new compound for them. But they found that if you used phenyl chloride– chlorobenzene– if the chlorine was on the benzene ring, then it didn’t react. So it wasn’t perfectly general. But it could use alkyl or aralkyl, that is phenyl on a carbon. So you could have benzyl, but you couldn’t have aryl as the group.

But then notice that they also said it could be used to make ketones, how to make a ketone that way. So you start with a chloride that already has a C=O in it, an acid chloride. And when they reacted benzoyl chloride with benzene, they got diphenyl ketone, which was a known compound. But this was a better way to make it. With acetyl chloride, they could get a phenylmethyl ketone. That was a known compound. Here was an interesting one. With a diacid chloride, the phthaloyl chloride, they could get a compound which, I think, was unknown at that time with two phenyl groups put on. So it could react twice. Or it could react twice with the same benzene molecule to give that. And that was a compound that was known, anthraquinone. So they could get ketones, as well alkyl.

So this is a very, very general reaction. Very important for synthesis. So you could use acyl as well as alkyl. So you have not only Friedel-Crafts alkylation, but Friedel-Crafts acylation to made ketones.

Chapter 2. Avoiding Friedel-Crafts Rearrangements [00:13:26]

What is the question mark, what compound do you need to do a Friedel-Crafts reaction to get that compound? Name the question mark. Liang, what would you use?

Leon: Propane with a chlorine.

Professor J. Michael McBride: 1-chloropropane or n-propyl chloride. That compound. So does it work? We’ll try. A warning. Remember that once you put one alkyl group on– we talked about last time– it becomes more reactive. So there’s going to be a problem. If the product is 25 times as reactive as the starting material, then you’re going to get higher substitution instead of being able to get a good yield of what you want. But the good news was that if it was done for five hours with aluminum chloride– this was done by Ipatieff in 1940– he was able to get about a 45% yield up monosubstitution.

Now, when you read “yield,” you often wonder what it means. Naïvely, you say you get 45% as many moles of the product as you have of starting material. That’s what you’d read that, naïvely. But that’s not what they meant. Because if you used up half the starting material, so at the latter stages you have nearly as much of this as you do of the starting benzene, and this stuff is reacting 25 times faster-, there’s no way you’re going to get a 45% yield. So what they mean is 45% of the product is mono- as opposed to di-, tri-, tetrasubstituted. So in fact, probably only about 15% of the benzene was consumed. But still, they could get half of the product was monosubstitution at that point. OK, that’s great.

But the bad news is that of the monosubstitution, it’s not just the compound that Liang wanted to make. It’s 60% of that and 40% of the one we wanted. Now that was done… So that you could have rearrangement during the Friedel-Crafts alkylation. Now that was done at a slightly warm temperature. If they did it at -6°, then it turned around and you got more of the compound you wanted. But still, there’s rearrangement somehow.

How does the rearrangement occur? Well, remember the aluminum chloride comes up and starts pulling off the chloride and makes a complex like this with the incipient cation. You wouldn’t expect the primary cation to form easily, right? But you’ve the aluminum attached there, so it’s beginning to develop a charge. But that thing that you see there, that complex, can react with the benzene. That is, benzene comes in and displaces the AlCl4. So you get the intermediate with both H and the n-propyl attached to the same carbon, that cation. And then you lose the proton and you’ve got the product– one of the products. But the other product comes, because– so that gives the n-propyl benzene. But there’s a hydrogen here. And so what else can happen when you start pulling the chloride away?

Student: Hydride shift.

Professor J. Michael McBride: You can get a hydride shift. Bingo. So the hydride goes there, the cation moves to the middle, and now you get the other product. But notice if you do it cold, if you do at -6°, you get more of the initial reaction before the hydride shift occurs. So it’s possible to get the product you want, at least a substantial amount of it. But you have to be worried about rearrangement if the cation that you’re trying to put on is one that could rearrange to a more stable cation.

How about if you wanted to make acylation? There, you don’t get rearrangement. Why not? Because when you’re making this cation, it’s stabilized by an unshared pair on the oxygen. So it doesn’t rearrange. It’s already the most stable cation you can get from that. So with acylation, you don’t get rearrangement. But you need one equivalent, not just a catalytic amount, of aluminum trichloride, because it gets eaten up by the product, by associating with the carbonyl group– and by the starting material, as well actually. But it complexes that way, as well. But even though you’ve got a lot of aluminum chloride in there, you can easily get rid of it just by adding water, once you have the product. So now you have this unrearranged ketone.

But suppose what you really wanted in your heart of hearts, Liang, was not the ketone. You wanted propylbenzene. Ah, that’s what you want. How can you get there? By removing the O and replacing it with H2. And there are reagents for doing that. Like the Clemmensen reduction, treat with the zinc and hydrochloric acid. Now that’s a very strong acidic medium, so maybe other things in your compound wouldn’t stand up to it. Propylbenzene is one thing, a more complicated molecule is something else. Maybe the thing you want won’t tolerate acid, but it will tolerate base. Then you could use the Wolff-Kishner reduction, with hydrazine and potassium hydroxide. But notice it’s done at 200°, so that’s a very vigorous basic thing. And, if neither of these work– both of them are cheap and give good yields– but if neither of them work, there are other reagents nowadays that you could use. Those are the classical reagents for changing the acyl to an alkyl. But the important thing to see here is that you’ve now been able to make that compound without worrying about the rearrangement, by going by way of the ketone.

Chapter 3. Synthetic Accessibility via Aromatic Substitution; Diazonium Salts [00:19:48]

So that’s a synthetic strategy. So let’s think about what things you can get from benzene, what groups you can put on using aromatic substitutions that we talked about last time, and also the alkyl and acyl Friedel-Crafts reactions. We said you could put deuterium on with D2SO4. Chlorine on with chlorine and AlCl3, or bromine with bromine and FeBr3. Or other metal salts could work as well, these so-called Lewis acids. Or you could use the Friedel-Crafts acylation and make a ketone, or alkylation and make an alkyl compound– as long as it doesn’t rearrange. Or you could even put CO on with HCl. HCl will protonate CO to make the cation that then puts on HC=O as a substitution reaction. So that one would work too. Or you could put a nitro group on, we said, with nitric acid, sometimes with sulfuric acid, if you need it to be more reactive. Or you could put on SO3H, that is SO3 can get protonated by sulfuric acid and then the HSO3 can go on and give these sulfonic acid products. So there are lots of different things you can put on benzene this way.

And in fact, the sulfonic acid can come off again, the SO3 can come off again if you treat it very hot with steam. And you might wonder, why would you do that if you went to all the work to put it on? And you’ll see shortly why that is.

You can also get other compounds by changing these groups into something else. For example, the nitro group. You can reduce it, either catalytically with palladium or by using a reagent and a metal, like tin and HCl and get aniline, the aminobenzene. So NO2 can become NH2. And then you could treat NH2 with nitrous acid. It’s sodium nitrite and HCl. The HCl protonates it, you get nitrous acid. And that changes NH2 into N2+, which is called a diazonium compound, benzenediazonium chloride, that’s called. Now why in the world would you make that?

Well, it has a resonance structure here that has plus on the nitrogen. And that can be the cation that does substitution, electrophilic aromatic substitution. And notice that you’re making it in the presence of something that’s a very active receptor for that. So you could make this compound, which is important because it’s yellow. It’s a yellow dye. It’s called aniline yellow, and it was the first such dye that was made; the first of hundreds of these so-called azo dyes that are all different colors. I looked it up on the web to see if I could get a picture of something dyed with it, and find that it’s very popular for skateboards nowadays, to use aniline yellow.

But hundreds of these were made. This was where activity was in organic chemistry. This was before synthetic drugs. This was before plastics. This was before the use of petroleum the way it’s used nowadays. So there weren’t these commercial driving forces for developing organic chemistry. But dyes were important. So the last half of the 19th century there was a lot of work on things like this. So hundreds of these were developed. And they have fancy names. Like Bismarck brown, named after the Iron Chancellor of Germany. It was a brown color. And in fact, notice that there were two of these reactions that went on to give this Bismarck brown. It made two of these azo groups in there. I looked up Bismarck brown on the web, and find out that it’s used as a histological stain. There it is.

So this was a big deal.

But that’s not the only reason that these diazonium chloride was important. N2 is a fabulous leaving group. The N-N triple bond is the strongest diatomic bond. So it very easily leaves as N2. And notice that that’s a very clean reaction, too. When it leaves, what does it leave behind if you lose N2 from benzene diazonium cation? It’s the phenyl cation. A cation that’s very hard to make. Because, we talked about that before, it’s a very strong bond and not a good place to have a vacancy, in an sp2 orbital. So there are lots of synthetic uses for the diazonium chloride besides making dyes.

So you can treat it with HBr and copper bromide, cuprous bromide, CuBr, to put a bromine on in place of N2. Or chlorine on in place, or cyanide on in place of N2. Now all these reactions, called Sandmeyer reactions after the guy who developed the idea of using the copper bromide, probably go first by an electron transfer. So you have CuBr2-, which is formed from Br- plus the copper bromide. So that gives an electron to one of the antibonding orbitals in nitrogen. So you generate this radical instead of the cation, and a radical from the copper reagent. But that’s very unstable, and it loses nitrogen instantly, N2. So you have that radical. But now these radicals react with one another to put the bromine on. You have copper bromide, which then adds bromide- again, and recycles. So the copper is catalytic.

So you could do it also with chlorine, also with cyanide. You could do it with iodide. You don’t even need the copper there, because I- itself can give the electron to the diazonium compound. You could do it with BF4-, which is probably different. That’s probably not donation of electron. That one probably does involve losing N2 from the cation to form the phenyl cation, which then pulls off a fluoride from BF4-. So that’s probably an SN1 reaction, where N2 leaves. It’s a great leaving group. And probably the same for the reaction with water, to put OH on.

You could also react it with NaNO2 to put NO2 on. But what’s the utility of that? Why would you want to do that? Is that going to be, you think, a way you’re going to make your fortune? To sell nitrobenzene making it this way? What’s the problem? Yeah?

Student: The NO2 is a cation?

Professor J. Michael McBride: No, that’s not the problem. The NO2 here is an anion, NaNO2, is an anion. We talked about NO2+ before, but this is NO2-. So this would do the trick. But commercially, this is bound to fail. Matt?

Matt: You used that to get the diazonium.

Professor J. Michael McBride: Yeah, remember, where did we get the diazonium salt? We got it from the amine. Where did we get the amine? We got it from the nitro group. So we could run our factory day and night and turn starting material into product at pretty good yield. But you’re not going to make a fortune that way.

But there is utility for that, but it’s not obvious at first. And the same thing is true here. You can react with hypophosphorous acid, H3PO2, not phosphoric acid, and that puts H on, instead of the N2. But that’s going even further back, to benzene. Why in the world would you want to do that? Well, we’ll show you what’s the utility of that one?

Now suppose you want to prepare a certain compound, like para-nitrochlorobenzene. How are you going to make it? Well, you can start with nitrobenzene. And we know how to put a chlorine on. How do you put chlorine on? What reagent do you use to put chlorine on? What, in general, to put any X on in place of H, what form of X do you need? What’s the thing that does the attacking? X what? X radical? X-? X+? It’s electrophilic aromatic substitution. You need X+. So if we want to put chlorine in, we need Cl+. Where do we get Cl+? Cl2 plus AlCl3. AlCl3 pulls the chloride off, we have the equivalent of Cl+, and that’ll do the trick.

Except it won’t do the trick. Why? Why won’t this work? We talked about it last time. Because nitro is deactivating. It’ll make the reaction slow, but still it’ll happen. But not only is it deactivating, what else is it? Meta-directing. It’s not going to give this product, it’s going to give the meta product. So are we out of luck? Is there a way of making the para-chloronitrobenzene? How would you do it? Sebastian?

Sebastian: Start with the Cl.

Professor J. Michael McBride: Start– so that one’s not going to work. But start with the– because you’d get the wrong isomer. But if you start with the Cl and put the nitro on, now the reaction is more reactive. Nitro is a very strong reactive reagent. And chloro is ortho/para directing, so you’re going to get the compound you want. Great.

Now suppose what you want is meta-chlorophenol, hydroxybenzene? That’s the one you want to make. So how are we going to do that? Now because both O and Cl are ortho/para-directing, so we can’t decide which one we’re going to put on. Furthermore, we don’t have the O+ reagent to put O on. We have the Cl+ reagent, but we don’t have the O+ reagent. So how are we going to get these things in the meta position?

Well, we could put the chlorine on. Remember, at least we got meta things by that mistake we made before. And now if we could convert nitro to OH, then we’ve got the product we want. Any way of getting OH? Those of you with rapid photographic memories? You convert nitro to amine. I think you can reduce it with this reagent. I’m not sure. But if not that one, there’ll be another one that allows you to reduce it. Then HONO converts it to the diazonium, and water converts that to OH. So we were able to do it that way by way of the nitro group. Converting the diazonium compound to OH, the diazonium salt. Or you could also make it a halogen there, or a nitrile, or NO2, or H. But of course, if it were NO2, you could do it a different way.

Now why would you want to do H to make chlorobenzene? That seems stupid.

So the key reaction here is HONO, nitrous acid, for converting aniline, the aminobenzene into any of many different things. Here’s a picture of the Yale chemistry class, Sheffield Scientific School, in 1898. So a lot of distinguished looking guys sitting here. And they’re showing the tools of their trade, as we mentioned before. But there’s a little box at the feet of the guy who’s playing a flask like a banjo, here. And do you see what it says on it? Hot or cold, OH NO! So this is a little pun here. It’s HONO. Because this is the chemistry they were doing at that time. Making diazonium things and converting one thing to another.

So suppose what you want to do is convert toluene into 1,3,5-trimethylbenzene. What’s the problem? Just do Friedel-Crafts and put methyl groups on, right? What’s wrong with that? The alkyl groups are ortho/para directing, you won’t get this isomer. If you try that, it’s going to fail. But you could put nitro group on in the para position. We already talked about that kind of thing with chlorine instead of methyl as an ortho/para director.

And now what are we going to do? We’ve got not only the wrong group, nitro instead of methyl, but we’ve got it in the wrong place. But this is an indirect synthesis. Notice what you can do is reduce it, the nitro to amine. And now you could react that. Or if it’s too reactive– remember that the rocket fuel was a nitration of something like that– you could acetylate it, if that’s necessary. I don’t know whether it would be. And now you do the Friedel-Crafts reaction. Now this is ortho/para directing, methyl, so it would tend to put it here or here. But this is more strongly ortho/para directing because of the unshared pair. And it would put it here or here. So it’s a stronger ortho/para director than the methyl is, that unshared pair.

So it goes to there. So now you have that NH2 group there. So we’ve got the methyls where we want them, but we’ve got an NH2. What are you going to do about that? What do we convert NH to? What did the people in 1898S do? They treated it with HONO. Make the diazonium salt, and then react that with hypophosphorous acid to convert it to hydrogen. So the nitro, then the amino group was put on in order to direct where other things are going. And then you have to take it off again. So there was a utility to removing it with hydrogen.

Now here’s a problem for you to work on. How to make para-dinitrobenzene with the reactions that we know about here. So you can work on that on your own.

Chapter 4. Nucleophilic Aromatic Substitution and NADH Reduction [00:35:15]

Now here’s an interesting compound. 2,4-dinitrofluorobenzene. And the interesting thing about that one is it reacts within an amine to replace fluorine by the amine. So this is an aromatic substitution. But it’s different from the ones we’ve been talking about. Can you see why? Why it’s not like the electrophilic aromatic substitutions we’ve been talking about? What makes the nitrogen reactive? Arvind? What makes the nitrogen reactive of the amine?

Student: It has a partially negative charge, right?

Professor J. Michael McBride: Well it’s a little negatively charged, what’s its high HOMO?

Student: The unshared pair.

Professor J. Michael McBride: Unshared pair. So that’s what’s going to make it reactive. Now we have nucleophilic aromatic substitution. The things we’ve been talking about before are +reagents. Cl+, carbon+, NO2+. This is high HOMO instead of a low LUMO. Now that product actually is important, to make it yellow. Because it was used when you do chromatography, if you spray it with this stuff, where the amines are will turn yellow. So it’s a good visualization reagent. And it identifies the amino acid at the end of a chain, because if you put this on the end of the chain, where there’s NH2, and then cleave it apart and see what the pieces are, the one that’s yellow was the one that was on the end of the chain. So this stuff was called Sanger’s reagent, after Frederick Sanger, who used it to determine the amino acid sequence of insulin. In fact, he invented this way of doing it, putting this color label on and he sequenced insulin in 1955. And three years later, he got the Nobel Prize for this, Sanger did.

Incidentally, how do people know the sequence of proteins nowadays? Do they do it this way? Do you know? How do people know what the sequence of a protein is going to be? They do it by sequencing the nucleic acid, right? And in fact, Sanger got the 1980 Nobel Prize for sequencing DNA. So he got two Nobel Prizes for sequencing.

But this clever thing, that he, as a good chemist, figured out how to do, is nucleophilic aromatic substitution. Now how is it well designed, this reagent, to do that? Well for one thing, it’s yellow. The product is yellow. But another one is that this anion that you get is very good because it’s got a nitro group where the charge is. Or there. So it’s activated for forming that intermediate by two nitro groups. Then you lose the proton, and then you lose fluoride.

The normal electrophilic aromatic substitution, X+ comes on, forming an intermediate with plus, plus plus, and then H+ leaves. Now something minus comes on, or a high HOMO, the amine, minus, minus, minus, and then F- goes away. Now why was F such a good thing to choose? Because F, generally, is not such a great leaving group. The other halogens are better as leaving groups. It’s good because the leaving is not the rate-determining step of this thing. The slow step is that initial addition. So the more electron withdrawing the substituent is, the better it will be. So fluoride is the one to use. So it’s activated by fluorine. So Sanger, a great chemist, figured out a good reagent and got rewarded with a Nobel Prize. In fact, two Nobel Prizes.

Now, another place this crops up is in NAD+, which is this compound. Have you heard of NAD+ and NADH? Some of you? You will when you take biochemistry, if you haven’t heard about it yet. These are oxidizing/reducing agents. So you have, say, an alcohol with, a secondary alcohol so you have an H there, and now you can do a nucleophilic aromatic substitution. Notice what’s going onto the benzene is H with the electrons, H-. It’s helped out by the unshared pair on oxygen coming in here. And this, then, is a good place to put the electrons, on the nitrogen. So you get a ketone that got oxidized, and this NADH.

Now the nice thing is that this is reversible. The energies of these two are closely enough balanced that you can run it backwards and forwards. That means this reagent, NAD+, can be used as an oxidizing reagent, NADH can be used as a reducing agent. So you can make an alcohol into a ketone or a ketone into an alcohol. And these are key molecules in biological oxidation and reduction.

And notice the way they do it is by transferring H-, accepting or giving back H-. So that’s that nucleophilic aromatic substitution.

Chapter 5. Benzylic Reactivity, Steric Hindrance, and Moses Gomberg [00:40:44]

So benzylic reactivity was related to pKa. Remember, what’s the pKa of normal alkanes, roughly? How acidic is RH? Just a regular old alkane, roughly? pKa?

Student: 50-something.

Professor J. Michael McBride: 50-something, 50 few. But if you have allylic H, it’s 43. So it’s something like seven powers of 10, or nine kilocalories better than a normal alkyl group, because of allylic stabilization. If you have phenyl of vinyl [correction: benzyl], instead of the carbon-carbon double bond, it’s even two powers of 10 better, or 12 kcal/mole. And if you have two phenyl groups on it, diphenylmethane, it’s much better. It’s another 11 kilocalories better. Not quite as good as the first one because there’s not as high a HOMO any more, once one has begun to lower it. So there’s a little bit less. Predict what it’s going to be if you put three phenyl groups on. Luke, you look ready to go on making a prediction here. How many kilocalories?

Luke: 10.

Professor J. Michael McBride: Another 10. It’s 3. The third one does hardly anything. Why? Because it’s so crowded that the phenyls have to twist, and don’t overlap very well. So steric hindrance in triphenylmethyl reduces the overlap with the anion you want to stabilize, so it reduces it by 25% from diphenylmethyl. OK, so the crowding is an important thing.

And this brings us to Moses Gomberg and the triphenylmethyl radical, which is 110 years old this academic year. It was announced in October, 1900. So October we were already underway in this class.

So Moses Gomberg was a speaker when this building was dedicated in 1923, one of the featured speakers. But he came from a very different background from Crafts. Remember Crafts was the son of an industrialist in Boston and could afford to go 17 years to Paris to do his research. But Gomberg was born in the Ukraine. He was Jewish and his father got in political trouble, and there were pogroms going on, so he and Moses left and came to Chicago, where Moses– he was born in ‘66, so here he’s 18 years old, and he’s working in the stockyards in Chicago for two years. But then, somehow, he gets to the University of Michigan two years later.

And here’s his transcript from the University of Michigan. He graduated in 1890. So let’s look at what kind of thing you would have taken if you were getting a BS in chemistry, as it says up here, at Michigan in 1890. There were distribution classes. So he took French, Rhetoric, Logic, and Psychology. And then these are science classes, physics and math. And here are chemistry-related classes. There are metallurgy, scientific freehand drawing, geology and minerology, and chemical literature in German and French. And the big one in the middle, not too surprisingly, is chemistry. So the distribution wasn’t quite what it is here nowadays.

And here he is as a student in the analytical laboratory. After his freshman year, he averaged nine and a half chemistry labs a week. That’s labs, not hours in lab. Nine and a half different lab sessions a week. And two-thirds of them were analytical chemistry. That’s what he’s doing, he’s doing a titration here. So there was no spectroscopy then.

Then after he graduated and got his PhD– he was I think the second PhD in chemistry at the University of Michigan. And it’s interesting that he studied chemistry. You might think as an immigrant who had worked in the stockyards, he would be trying to do some practical work. So, in this three year period when he was an undergraduate, there were, I think, something like 400 graduates in pharmacy from Michigan and about five in chemistry. And he took chemistry. So that’s very interesting.

But anyhow, then he went to learn in Germany in Munich with Baeyer. So there’s Baeyer, 61, the head of the lab. And here’s his assistant, Thiele, who was 31. And here’s Richard Willstatter, who was 24 then, the fair-haired boy in the lab, who got the Nobel Prize in 1920 as an organic chemist. And here’s Moses Gomberg. He’s 30 years old, older because of his family history than these other people, almost as old as Thiele.

But here’s what Willstatter the Nobel laureate, later said about him. “Moses Gomberg was Thiele’s coworker in the student laboratory. He was very reserved and modest, kept entirely to himself, and never chatted in or out of laboratory. Some years later, the work he carried out in the United States on the triphenylmethyl radical, a case of trivalent carbon. became famous.”

And you know what I think the key phrase is here? “in the United States.” Because remember, Crafts’s work was done in France, nothing in the United States. But here, just two decades later, the key work is being done in the United States. So we’re going to see how this happened.

“This brilliant experiment, one of the most beautiful in organic chemistry and one which few people credited at first, gave great impetus to chemistry and would have been worthy of any distinction.” So he was nominated, Gomberg was, for the Nobel Prize a number of times, but he never got it.

Now how did he get into this? Well, just before he got to the lab, Thiele, the guy in the front row who ultimately drank himself to death in France, had a student named Heuser who made this compound called AIBN, azoisobutyronitrile. And from it, in 50% yield, prepared a new carbon-carbon bond. And this is a different way of making carbon-carbon bonds. The way you do it is to lose nitrogen. So N2 comes out and the two radicals come together.

So when Gomberg came along into the lab for just a semester, or maybe two semesters it was, he said, “Under the sponsorship of Professor Thiele, I have followed up on these reactions…” So he worked on this compound, this idea of knocking nitrogen out. And then he went for the third term to Heidelberg to work with Victor Meyer, who had introduced the idea of steric hindrance. This was a stressful time to be in Heidelberg, because Meyer committed suicide during this period.

But an interesting question in steric hindrance was this tetraphenylmethane. Remember, Friedel and Crafts thought they had made tetraphenylmethane, but in fact they had made triphenylmethyl chloride, because it’s so crowded. So Friedel-Crafts wouldn’t make it. They tried with phenyl anions, phenyl-metal compounds, to make tetraphenylmethane, and they couldn’t do it. So this was a real challenge – a grand challenge, was how to make tetraphenylmethane. Not that anybody thought it would be good for anything, but it was a symmetrical, beautiful compound and how do you make it?

 And Gomberg succeeded. “I have tried to solve this problem in a completely different way.” So the way he used was what he had learned the previous two terms in Munich. So in Heidelberg, he tried making this compound, which you get from benzene diazonium salt and triphenylmethyl chloride, the stuff that Friedel-Crafts could make.

So he tried it to see if it would work, and at 110° with copper there, from eight grams of the starting material he got 0.3 grams of this stuff. How does he know what it was? Well he discusses the solubility, but that’s not a very good key to what the compound is. But he did the analysis. And this is what he found. And this is the theoretical. So it agreed very well with that. And what else might you be able to do? The molecular weight would be very key. And how do you get the molecular weight? You can get it from the vapor density. But this has very low vapor pressure. But you can get it by freezing point lowering.

So that’s what he did, he tried freezing point lowering. He had 300 milligrams of this material and he used 100 milligrams for this purpose. And he found that he lowered the melting point by 0.289°. This is real, careful work. He’s being paid off for having spent nine and a half labs a week for three years at Michigan. Oh, it’s boiling point, not freezing. The solvent boiling point elevation, which implied 306 for the molecular weight, and he calculated 320. So very good. So then he went back to Michigan, to Ann Arbor. He stayed there the rest of his life working at the University of Michigan. And what did he do when he got back? I’ll tell you next time.

[end of transcript]

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