Freshman Organic Chemistry I

CHEM 125a - Lecture 24 - Determining Chemical Structure by Isomer Counting (1869)

Chapter 1. How Did Kekule Know Benzene Is Hexagonal? [00:00:00]

Professor Michael McBride: We’ve been going through the generations of organic chemistry, from Lavoisier to Berzelius and Gay-Lussac, to Dumas, Liebig and Woehler, and last time we got to Couper, and began with Kekulé, talking about their discovery, or invention really, of valence; in particular the tetravalence and the self-linking of carbon. But the other thing that Kekulé is most noted for is the structure of benzene, and we started looking at that a little bit last time with his sausage diagrams for drawing a structure. And this time we’ll go on in more detail with that idea in the determination of actual position in molecules, the positions of atoms in molecules. And this is largely due to students of Kekulé, so the next generation: Wilhelm Koerner, who was both in Germany and in Italy, as you’ll see, and Jacobus Henricus van’t Hoff, from the Netherlands, and also Albert Ladenburg, from Germany, and we’ll even have a little message of your own — [technical adjustments].

Okay, so we — Koerner, van’t Hoff, Ladenburg and E.P. Kohler, who is your great-great-grandfather academically. Okay, so benzene and molecular structure. This is the simplest of the so-called aromatic compounds. Obviously the name comes from the smell, that it has an aroma, but it was generalized from benzene which — benzene was discovered in 1825 by Faraday who isolated what he called bicarburet of hydrogen, which was C₂H, according to his formula, but actually it’s C₆H₆, because he had the carbon doubled and it’s six times what he thought it was. And he got it from gas oil, which is a byproduct. Quite a bit, you get a gallon — or they did at least then; I don’t know what happens nowadays — for making 1000 cubic feet of the oil, that was an illuminating oil; illuminating gas, pardon me.

So in 1833 Mitscherlich named the compound benzine, instead of bicarburet of hydrogen. And the origin of the name is interesting. It came from gum benzoin, via benzoic acid. So you knocked the CO₂, somehow, off of benzoic acid, and you get what Mitscherlich called benzine. Now gum benzoin itself has an interesting etymology, because it came originally from luban jawi, which I’m told is Arabic, meaning frankincense — that’s luban — and jawi is from java. So it came from the southeast Asia. But when the French saw this name, they assumed that lu was like le, la, les, and so on, the article. So they knocked it off. So it became ban jawi, which doesn’t mean anything in Arabic, but that then became benzoin and then benzine. Okay? The French, in 1841, named the alcohol version acide phénique, because it’s acidic, and we’ll see why shortly. We call it phenol, and we call C₆H₆ phenyl, and abbreviate it Ph for phenyl, or Φ, the Greek letter, for phenyl. And it comes from phainein, which means to bring light, because of its association with illuminating gas.

Okay, so we know that Kekulé’s structure for benzene is hexagonal, even though he wrote it with that sausage thing originally. Now why? On what basis did he think it was hexagonal? Well this is the paper where he proposed it’s hexagonal, and I’ll show you all the evidence he gave in his arguments for its being a hexagon. First, “All aromatic compounds” (so, for example, benzaldehyde as well as benzene and so on) “even the simplest are significantly richer in carbon than analogous compounds from the class of fatty substances.” Okay, so it’s rich in carbon; that’s the first evidence that it’s a hexagon. The second is that “Aromatic compounds, as for fatty compounds, there are numerous homologous substances; that is, those who composition differ by an integral number of CH₂ groups.” So if you replace a hydrogen by a methyl group, you add CH₂, right? Or if you put a CH₂ into a C-C bond, it becomes a little bit longer, a homolog. So there are homologous substances in the aromatic series. The third line of evidence was, “The simplest aromatic compounds contain at least six carbon atoms.” Okay? And the fourth line of evidence is, “All reaction products from aromatic substances show a certain family similarity, constituting the group of ‘aromatic’ compounds. More vigorous reaction can remove part of the carbon, but the major product contains at least six carbon atoms, and decomposition stops at this point, unless there is complete destruction of the organic group.” For example, burning it.

And that’s it. Those are the four lines of evidence. So would you conclude from that that benzene is a hexagon? It’s quite a stretch. So was this just a wild guess, based on the idea that carbon should be tetravalent? And even in a sense that doesn’t help much because there’s a question of what you do with that last valence of carbon, right? So he did have some support for this, but he didn’t cite it in the paper. He didn’t publish it until later. It had to do with counting isomers. So, remember this was the formula he drew in 1865 of chlorobenzene. But in the next year, when he started trying actually to say that it was a hexagon, he drew a hexagon — he didn’t try to use his sausages at this point — and he contrasted it with what you get with a triangle of carbons, with a carbon in the middle of each edge. Okay, so he labels the vertices, the carbon atoms, a, b, c, d, e, f, and counts isomers.

And this is his table of that count. So monobromobenzene, there’s only one form. Right? Obviously if it were a chain of six carbons, you could have it on the end, or the carbon that was next to the end, or the carbon next to that. Right? So there’d be isomers. But, in fact, only one was known, monobromo. Now how about dibromobenzene? It turns out that there are three modifications, he says: ab, ac, and ad. How about ae? ae would be next nearest neighbors, like ac, and af would be like ab. Okay, so three modifications. That’s consistent with its being a hexagon. Now how about tribromobenzene? There are three; drei modificationen. Okay, abc. So all successive, right? Or skip one. How about if you skip two? How about abe; should we have that one? How about abe? Kate, what do you say? Should we count abe as another one? Steve, what do you say?

Student: No.

Professor Michael McBride: Why not? Because I asked?

Student: Well because on the thing it says abc, abd, and ace.

Professor Michael McBride: Yes, but I wonder if you agree. Maybe he’s wrong. There’s precedent for people being wrong. Sam?

Student: Isn’t — abd is the same as abe.

Professor Michael McBride: Right, because if you count the other way, bae is the same as abd. Right? You go two in a row and then skip one, just go the other way around. Okay? Because it shouldn’t make any difference which way you go around. Okay? Or there would be ace, right? Where you skip every time. But there’s only one way of doing that. Okay, so three modifications. And if you go to four, five, six, it’s the same thing, except hydrogens are what you count instead of bromines. Okay? So that was consistent with experimental observation of how many isomers there were of these mono-, di-, and tri-substituted benzenes. Yes?

Student: What was he using to observe these modifications? What technique —

Professor Michael McBride: So how would he know they were different, that there were three different substances? Suppose he had them. He had to prepare them of course. Maybe they were not preparable, right? In which case you’d have a count that was too low. But how would you know you had three different things? How could you tell things were different? Yes?

Student: Boiling point.

Professor Michael McBride: Boiling point; melting point; color maybe, although these aren’t colored; density. Okay? Some physical property you have to measure. Right? The analysis obviously is going to be the same, because they’re isomers. And of course, if you’re not very good at measuring things, you might get a melting point between two samples that differs by two degrees. You guys are experienced in lab. Does that mean they’re different? What do you say Wilson? Your experience is that it doesn’t tell they’re different? Okay. So it’s not trivial. You have to be a good experimentalist. You have to be able to prepare them and you have to be able to tell when things are the same and when they’re different. But if you do, then you can count isomers. But you have to bear this in the back of your mind that you might be making a mistake because of your experimental observations.

Okay, now how about if it were a triangle, right? Then if you had a monobromobenzene, you could have the bromo either at a or at b; although it’s conceivable you couldn’t prepare one of them, or one of them would be reactive for some reason. Okay? And if you had two, you could have ab, or ac, or bd, or ad. There’d be four modifications instead of three, assuming you could prepare them all. Right? And then you can go on to the tribromo and see there would be six. Okay, so by counting isomers, you could get information about the structure. Okay?

In fact, the paper that I xeroxed this from had written in, in old brown ink, the letters from the bottom u, k, e, k, e, l, going around the ring, for a, b, c, d, e, f. And written next to these things was names for the compounds; elukek and so on; ekeluk, ekulek, for the order of these things. Do you see what’s funny?

Student: They’re anagrams of Kekulé.

Professor Michael McBride: Right. No, ekulek, they said, was Arabic for hexagon; that’s what was written in the book. So you’ve got humorists in all ages.

Okay, now let’s try applying the idea of isomer count to see if we can distinguish, among Dewar’s models of benzene that I showed you. So first there’s the one top and center, which is what we call the Kekulé structure of benzene; okay, single bond, double bond, single bond, double bond. How many isomers? Okay, how many monosubstituted isomers, if you put just one bromine in? And if that were the structure, a faithful representation of the arrangement, how many different ones could you get? Just one. Okay all the hydrogens are, in that sense, equivalent. Okay, how about if they’re di, how many could you get? There’s one, two, three, four, five. Right? So are they all different? Kate, what do you say?

Student: The purple and the red are the same —

Professor Michael McBride: You say the purple and the red are the same. Anything else?

Student: The two blue ones.

Professor Michael McBride: The two blues are the same. Now does everybody agree with that? Sherwin?

Student: With the purple you’ve got the red ones across a double bond.

Professor Michael McBride: Ah ha! the red one is across a square, and the purple one is across just a single bond. Right? So those are different, if this model is faithful in that respect. Okay? So, but we should take that one out, because that one goes across both a square and a single bond. [computer tone sounds] Ah, that’s somebody else; okay. Okay, now so there’d be four disubstituted isomers if it were that, and Kekulé found three. Right? Now how about if it were this structure? How many monos? If you replace just one hydrogen, how many different hydrogens? Elizabeth?

Student: Six.

Professor Michael McBride: Six. They’re all different. Right? So there’ll be six monos. How many di’s? Well let’s count them. So we’ll start from the top left there. There would be those sets of di’s, the ones connected by those arrows. And then we could start from the top — pardon me, five of those — then we could start from that one and there’d be four, and three, and two additional ones, and one, from that, that hasn’t been counted before. So we’d have fifteen disubstituted isomers. Right? Now how about these? Notice for these two that in both cases it’s AA, BB, AA. In the bottom one it’s turned on its side, but the same relationship, the symmetry of it. Right? Everybody see that? Okay, so these will have the same number of isomers, even though their structures are obviously different, but the same number of isomers. So there’ll be two monosubstituted isomers, either an A or a B. And now we’ve got the di’s to figure out, and it turns out there’ll be six. I’ll let you do that yourself. Okay, so two six there. Now how about if it were this, which is AA, BB, CC, because it doesn’t have as much symmetry. The red ones don’t have as much symmetry as the blue ones, although they’re the same as one another. So it turns out there are three monos, A or B or C, and there are nine di’s; and you can work those out if you want to. Okay, so you can’t distinguish among the red ones, or between the blue ones, but you can distinguish a black one from an aqua one, from a blue, from a red. Okay, so again you can get information about structure by counting isomers, if you’re good enough, and you can prepare them, and you can tell them apart.

Okay, now here’s a problem for you to do for Monday, and you can work in groups if you want to. But take — this is planar and two-dimensional, right? But in truth, because carbons are tetrahedral, when they form four single bonds, the others are not planar. Right? And I’ve drawn here how each of them would look in three dimensions, with some things, the wedges, coming out toward you, and other things, the dotted ones, going back into the page. Right? So if you consider these in three dimensions, how many isomers can you get? Right? That’s the question. Mono and disubstituted isomers. Will you be able to distinguish these, if they’re three-dimensional? And — that one, remember, is called Dewar benzene — and the one that’s not included was prismane or what’s called “Ladenburg” benzene. Remember I said we’d talk about Ladenburg; he’s coming up. So that’s another possibility, a three-dimensional symmetrical prism. Right? Which obviously Dewar couldn’t make with his models, because they would — unless he distorted them and bent them, right? Okay, so the question is, how many mono- and disubstituted isomers can you get from these, if they’re three-dimensional? So you guys can get together and talk that over and tell me on Monday. Okay?

Now there’s another way to get — to infer structure, and that’s from synthesis. So Kekulé, in 1867, reported that three acetone molecules, shown on the left there — you see the C double bond O and the two methyls attached to each central carbon. So three of those come together at a, at b, at c, losing water, and they form the double bonds on the right. So if that’s the way the stuff forms, then the three carbons in this compound called mesitylene should be on every other carbon of the ring — right? — because of the way it was put together by losing waters between them. So that’s another way of figuring out where something is, by synthesis.

Notice incidentally that there are little dots in the middle of some of these bonds. Why do you think those dots are there? I think it’s because this is a picture of a real model, that was made with wire coming out of balls, and that was the thing that screwed them together. Dewar had sent to Kekulé, when he wanted to come work with him, a set of these brass-strip models. So this I think is Kekulé’s version, from brass-strip models. Notice the funny bonds between the carbon and the oxygen, the double bonds, those triangles. Those are made of stiff wire that comes straight out and then were fastened together at an angle. Okay? Here’s a picture actually of the model that Kekulé used in his lectures to show benzene. And you can see that the carbon is tetrahedral. Okay? Yes?

Student: Doesn’t the story go that Kekulé also had a dream about a snake eating itself?

Professor Michael McBride: Yeah, that he made up in 1890. This is long after this. It’s conceivable that he had such a dream, but it was a story told for amusement.

Chapter 2. Koerner’s System of Isomer Counting and Chemical Transformation [00:20:10]

Okay, so here he is at the time he did this, in 1865, in Ghent. And he was surrounded with some hotshots. Koerner, for example, who was a young man. He acted as Kekulé’s secretary at this time, as well as a lab worker, and suffered from rheumatism. He came from Central Germany, which is a very sort of damp place a lot of the time, and cloudy, and so it wasn’t good for his rheumatism; nor was Ghent in Belgium. So, on his doctor’s advice, he went to sunny Palermo in Sicily. So Palermo is down there on the northwest coast of Sicily. And Sicily is a very interesting place in history. It was liberated in 1860, just before this time, after 2700 years of occupation by successive cultures. Okay? And the most recent one was Garibaldi, who came down with his Red Shirts in 1860 to free Sicily from the Bourbons, from the Spanish crown. Okay? And Cannizzaro came along and established this laboratory in Palermo. In 1860, he helped both Garibaldi, and the late Avogadro, and Gay-Lussac. Both of these were revolutionaries, although in different senses. Garibaldi was revolutionary politically, and Avogadro was revolutionary about knowing the atomic weight and molecular weight of compounds through gas density.

But in his way, Cannizzaro was a cautious person. This is what he said when he spoke in London, in 1872, on chemistry teaching. He said: “Above all we should take care that the pupils do not form to themselves any mechanical or geometrical conception of the cause and effects of the quantivalence” (that is, the valence) “of atoms. They must frequently be reminded that chemical facts neither prove not disprove anything relating to size, form, continuity, distance, relative disposition, etcetera, etcetera, of the atoms.” That is, don’t use these arguments for this purpose of finding where things — where atoms are in molecules. This is just a fancy that will appeal to students but isn’t anything real. This was a very cautious, conservative point of view, but was shared by all the leaders in chemistry.

“If we are sometimes obliged to speak of the relative position of atoms in the molecules, and even to give graphic representation of these positions, we must hasten to remark that these figures are nothing but artifices of the mind, intended to represent to the sight that which in reality we perceive only through the veil of transformations.” (So it’s only through reactions that you try to infer something about this; it’s not something real.) “But that we do not really know what it is that corresponds to that which we call position, either in space, or in the mutual action of different portions of matter.”

But he was one of the leaders, right? The followers, the students were more — didn’t take this, right? So Koerner, who had just come down, for his health, and joined Cannizzaro — in 1869 he came down, or he came down in 1866, I think, or seven — he wrote a paper called Facts Serving to Determine Chemical Position in Aromatic Substances. Right? So a completely different point of view from what Cannizzaro advocated three years later, right? To determine position. Cannizzaro, to be fair to him, was saying that you have to be different in what you teach and what you do research on. Teaching should be things you really know for sure, and research should be things where you’re trying to push back the frontier. But anyhow, Koerner wrote in this paper, in 1869:

“The dogma of the impossibility of determining the atomic constitution of substances, which until recently was advocated with such fervor by the most able chemists, is beginning to be abandoned and forgotten; and one can predict that the day is not far in the future when a sufficient collection of facts will permit determination of the internal architecture of molecules. A series of experiments directed toward such a goal is the object of this paper.”

Right? So that’s what he was trying to do. If he came along fifty years later, what would he have used, to find the position of atoms in molecules? X-ray, right? But he’s using only reactions and counting isomers. Right? So only the properties of things and elemental composition. Okay, “facts will permit determination of the internal architecture.” So this was a really new idea. And the paper, when it was published, was introduced and warmly endorsed by Cannizzaro, who went on to say, later, that you shouldn’t be teaching this to students, but it is a good thing to study. Okay? So Koerner had — there’s a thing called the Koerner Proof, which indeed he proved. But it’s not what he did in 1869. But this is very interesting. So you have, among disubstituted isomers, you have these possibilities. Okay? Now, how could you tell which is which? Suppose you had three bottles. One is black, one red, one blue, but they’re not colored, they’re just white crystals; or liquids maybe in this case. How could you tell which one is which? How could you put a label on the bottle? Well you could label them just with names, call them something like ortho, meta and para, which doesn’t mean anything. Right? But how do you know what structure they have? Cory?

Student: Differences in —

Professor Michael McBride: Can’t hear very well.

Student: Differences in melting point.

Professor Michael McBride: Oh, differences in melting point will show they’re different. But how do you know which one will have the lowest melting point? You have to have some theory about melting point as a function of structure; which didn’t exist. Or do you have such a theory?

Student: Polarity.

Professor Michael McBride: Polarity. But notice, to be — yeah, one of them would be non-polar. But people weren’t measuring polarity at this time. So that was out of the question. Okay, here’s how Koerner did it. He added a third identical substituent and counted the isomers. So suppose you start with a black one, you add a third one, down to the right, or down at the bottom, or — and that’s all, because if you added down to the left it would be the same as down at the bottom, and if you added over to the left it would be the same as down to the — over to the left would be the same as down to the right. Everybody see that? You can only get two isomers, starting from there. If you start from the red one, you can get either of those. Can you get anything else, from the red one? Andrew?

Student: You get another one on the left.

Professor Michael McBride: Yeah, down to the left will be a different one. So you’ll get three from that one. And how about if you start with the blue one and put a third one in? Maria, what do you say? If you start with the blue one and put a third blue substituent in?

Student: You have the one in the middle.

Professor Michael McBride: Speak up a little.

Student: You have the one in the middle.

Professor Michael McBride: Right, you can get the one in the middle.

Student: And everything else is just —

Professor Michael McBride: Nothing else. Right? Only one possibility there. Okay? So one will give one, one will give two, one will give three; supposing that all the reactions work. Right? So it establishes the identity, not only of the disubstituted, but only also of the trisubstituted isomers. One comes from one of the original bottles, one comes from two, and one comes from three. So you know it. Pretty clever. And these were called — these are called now ortho, meta and para. But those aren’t the names that Koerner used. Or more properly, they are the names that he used, but not to name the way we do it now. If anybody, he had the right to name them, but he didn’t get it because he was off in left-field down in Palermo, and the people up in Germany, who were influential, were the ones that finally named them. But notice that this argument holds if benzene is a hexagon.

But if it were a pentagonal pyramid, or some other geometric structure, then this argument wouldn’t work at all. Right? So you have first to know that all the positions are equivalent in benzene; that it’s not like some of those Dewar structures we were just looking at. So how can you do that? This was proven, in this paper in 1869, by Koerner. The paper was published in Palermo, in the Giornale di Scienze Naturale ed Economiche di Palermo, which is the Journal of Natural Science and Economics of Palermo. It’s a beautiful big journal with big wide margins on the page and so on. But it had no circulation whatever. People in Germany, years later, were still trying to prove this, what he proved.

Chapter 3. Koerner’s Assumptions on Replacement and Experimental Distinguishability [00:29:53]

Okay, now here’s a question. Are the four valences of carbon equivalent; that is, are they all the same or might one of them be a thick bond, say, different from the other three? Right? Or, might it even be that two of the bonds are different; one’s long, or one of them’s curly. Might all four be different? Can you think of any evidence to say that the bonds in methane are equivalent? How could you prove this? Alison, you got any idea? What technique is Koerner using? Pardon me?

Student: To see if the isomers are —

Professor Michael McBride: Counting isomers. So what do you expect if you have four different bonds in carbon?

Student: It matters where you substitute.

Professor Michael McBride: Right. How many isomers would you get if there were four different?

Student: One substitute.

Professor Michael McBride: You could have a chlorine on the fat bond, the long bond, the curly bond or the normal bond; there’d be four isomers, if you could make them all, and they didn’t interconvert. Right? So there’s only one methyl chloride. Nobody’s found a second methyl chloride. So does that prove that they’re all equivalent? Yeah? Pardon me? Yeah?

Student: Not on its own. But they can attach more than one chlorine to all the bonds. So they know that they all sort of — they will work.

Professor Michael McBride: Okay. That doesn’t prove it alone, because maybe there’s only one that can be substituted. Right? That is, of the one — if it had the structure on the right, with four different bonds, maybe you could make only one monosubstituted one. How many disubstituted ones could you make then?

Student: The one.

Professor Michael McBride: If the first one went in one place, the second one could go any of three. So you’d still say get three disubstituted. Unless you could only make one of those. Then there would only be two. So it doesn’t really prove it, the way a mathematician would like things proven, but it’s a plausibility, at least. Okay? But it turns out that — so the evidence is there’s only one isomer known, but it’s not really proof. Now Koerner, in order to make a real proof, had to make some assumptions, and these he stated in 1867, two years before the paper we’re talking about. And the first is direct replacement. And he wrote: “If one grants that in simple transformation the new substituent assumes the position of the element displaced….” So you can take a substituent off and put a new one there, but it will go in the same place that the old one came off. Everything doesn’t rearrange. You need that in order to make arguments like he wants to make. Right? And in fact that’s often true. It’s certainly logically parsimonious. It’s the simplest assumption you can make, and it usually is true. But there are rearrangements that do sometimes occur, and we’ll talk about those later on. But anyhow, for his argument you have to assume direct replacement.

And the second assumption is experimental distinguishability, which we’ve been talking about. “Most of the demonstrations are based on establishing the identity or difference of several substances of the same composition, but obtained by different reactions.” So you have to be confident that you can tell when things are the same and when they’re different. We just were talking about that. But Koerner was in an especially good position to do this, because he was really skilled in the lab. In fact, he got a Silver Medal at a Paris exposition, one of these international expositions, for his collection of beautiful crystals that he had lovingly prepared, and they had different shapes and so on. That’s one way you can tell things apart. And, in fact, this bottle here, says up at the top — I don’t think you can read it. But it says, Laboratorio di Chimica Organica, and it says Acido, Ac., Salicilico. So what is it? Salicylic Acid. This is Koerner’s sample, right? Of salicylic acid. And here are his beautiful crystals of 2,6-dinitro-3-bromotoluene. You can’t see them so well here. Well you can see that they’re still beautiful crystals, right? So those are some of his prize crystals. And here you can — so crystals are the best way to tell. And there are pictures of those things that were Koerner’s.

Chapter 4. The Koerner Equivalence Proof [00:34:23]

Okay, now here’s his proof, based on those assumptions. Okay, so there are three known isomers of hydroxycarboxylbenzene. So you don’t know anything about the structure. All you know is you is you have C₆H₄. You originally had C₆H₆, but one of the H’s has been replaced by OH, and another H by COOH. And the names of these things are salicylic acid, which I just showed you, hydroxybenzoic acid, and parahydroxybenzoic acid. So the names don’t mean anything, but those are just what the bottles are called. Right? Now, you want to use the existence of at least three. You don’t — there might be eighteen, for all you know, disubstituted ones. Right? But you know there are at least these three, and you can do experiments with them. So these are the experiments. First you treat it with HCl and heat it up — this was done by Graebe in 1866 — and it loses the COOH. Remember, we talked about how you could take benzoic acid, take the COOH off and have benzene. Here it’s done, take the COOH off, but you’re left with still the OH. And that compound, as we mentioned before, is called phenol. Right? But you can do the same trick with parahydroxybenzoic acid, and you get the same phenol, not a different one. What does that tell you? It tells you that the difference between salicylic acid and parahydroxybenzoic acid is not where the OH is. Right? They have to have the same position of the OH because — or at least equivalent ones — because you get the same phenol out.

And it was found also by Graebe that you could convert hydroxybenzoic acid to methyl anisate — that’s the OH becomes OCH₃; and so same for the acid. And then you can convert it to anisic acid, taking the CH₃ off on the acid group; from the ester, that is to say. And you can convert phenol by way of bromoanisole — Koerner, I believe, did this — and then with sodium and CO₂, to get the same anisic acid. Right? So the OH position has to be the same in hydroxybenzoic acid as it was in salicylic and parahydroxy. Right? So the conclusion is the difference among the three isomers that are known — salicylic, hydroxybenzoic acid, parahydroxybenzoic — is not where the OH is, since the O survives in identical compounds. So, for example, it couldn’t be like these three isomers of Dewar’s structure here. Suppose that were one of the acids, and that were another one, and that were the third. Then when you took the COOH off, you’d have three different OH compounds. Everybody with me? You see that? So the difference among these three, whatever it is that makes them different, can’t be where the OH is, because you can remove the COOH and they’re all the same.

Okay, that’s the first step. And the conclusion is the sites occupied by OH are either identical, or equivalent. Here’s what’s equivalent. So suppose they could be this. There’s OH, COOH. See did I? Yeah here. So you could have that OH, that OH and that — no, there’s; okay, sorry I’m messing this up. There’s one, one, one, two, three. The OHs in those are not the same position — one’s the top right, the other’s the top left — but they’re equivalent, right? Once you got it, you could flop if over and they’re the same as one another. Okay, so we know at least that much. So let’s call the OH position, which is either the same or exactly equivalent, call it ω, just for a name. So OH is in the position of Hω; that we know, in all three compounds. Can you guess what you do next? Next — yeah?

Student: Probably take the OH off and leave the COOH on.

Professor Michael McBride: Ah! Take the OH off and leave the COOH. Right? And that’s possible too. You can treat it with phosphorus pentachloride, which converts OH to Cl, and then with sodium and water, which takes the chlorine off and leaves hydrogen. So you get benzoic acid. And what do you do next? Do it to one of the other compounds, right? There, and you get the same benzoic acid. Or you do it here — and I think Koerner is the guy who did that, although I can’t — haven’t found a paper to that effect. But anyhow, he says that you get the same benzoic acid. So what does that prove? So the COOH positions are not fundamentally different in the three compounds. So the OH positions are equivalent, and also the COOH positions are equivalent. How can they then be different, the three compounds? Because they’re clearly different. It must be their relative positions that are different. Okay? The three isomers do not differ because of the absolute position of the COOH. So the three sites of COOH are equivalent, and we can name them x and y and z. None of them can be exactly the same, because if they were exactly the same then those hydroxybenzoic acids, two of them at least, would’ve been the same, if the COOHs — because we know the OHs were the same. Okay? So, for example, it could be this way in Dewar’s benzene. You could have that position for x, that for y, that for z. And they’re all different with respect to ω. Okay? The ω’s will all be the same. So that would be a possible structure, according to this. In this particular case, ω would be the same as x, y and z, as a site for a single substituent, if it were this structure.

So Arppe, a chemist, had made a compound which he called nitroaniline, which was benzene that had NH₂ and NO₂ in it, and it was known by transformations, that the NH₂ was ω, that position OHω, and the NO₂ was in y; that is, you could interconvert groups and make it into one of those known hydroxybenzoic acids. So you knew it was ω and y. Okay? Now, you do a transfer — and it’s related to that particular hydroxybenzoic acid. Okay? Now, you replace the NH₂ by Br — and we’ll talk about these next semester, what particular reactions did that — and then you change NO₂ to NH₂, and then you replace the NH₂ by chlorine. So now you’ve got benzene with bromine; that must be in the ω position, because it replaced the first NH₂. And then chlorine is in y because it substituted for the second NH₂, which came from the NO₂, which was in y. Okay? So it’s ω and y. Can you see what he’s going to do next? He replaces the NH₂ by Cl, and then changes NO₂ to NH₂, and then replaces it by Br. Why? So now he’s prepared a compound which clearly has chlorine in the position ω and bromine in position y. I see pursing of brows and so on. Is everybody clear? So he’s prepared two — from the same compound he’s prepared a bromide chloride. Right? One of them is ω bromide y chloride, and the other has them exchanged. And what do you think he observed? They’re the same. What does that show?

Student: [inaudible].

Professor Michael McBride: It shows that ω is the same as y, because otherwise these would be different when you put different atoms in them; assuming you have the experimental technique to tell when things are different. Okay? So now — so it couldn’t be, for example, a pentagonal pyramid, which one had bromine in y, and the other had chlorine — chlorine in y, the other had chlorine and ω and bromine in y. It couldn’t be that, those would be different. Okay? So there must be at least four equivalent substituted positions: ω, y, x and z. Okay? And that would be consistent with that structure we showed before, this one. So it could be that. It doesn’t have to be a hexagon.

Now, here’s the next step. So you start with a compound that’s known to be in the meta series. It’s got OH in ω, and nitro, NO₂ — he wrote AzO₂ — in x. And you have another one that has again OH in the position ω, but bromine in x, instead of nitro. And now you start doing substitutions. Remember, he used “meta” differently from the way we do. Okay, so first he put in a bromine someplace, and then a nitro someplace, and got bromonitrometanitrophenol. Okay, he added a bromo and a nitro group. So you know that’s the benzene core, with H₂, with a nitro someplace, with a bromo someplace. Those are the ones you just put in. But the others are still where they were. OH is in ω and nitro is in x position. And now you go to the other compound and put in two nitro groups. So you don’t know where those two nitro groups are, but you know ω ‘s — that OH is in ω and bromine is in x. Everybody see what you’ve done now? And you know what he observed? Those two compounds are identical. Can you see what conclusion you get from that? These are identical, right? But one of them has bromine in x, and the other has nitro in x. How can they be identical? There must be a second x position that’s — there must be two positions that are equivalent, right?

Now, could any of those positions be ones we already know? Could it be ω? We already know ω, x, y and z are the same, right? Could this new x be one of those? Could it be ω? No, because we already got ω, so it can’t be ω. Could it be y or z? No, because those would’ve started from a different compound in the first place, if it were y or z. Right? You know that this is in x. It’s not y, not z. Right? So there must be a second x, right? We’ll call it x’. There must be two positions, x and x’. They’re not equivalent to x, or y. They’re not the same as x or y or z or ω. It’s a new one. Okay? So that’s what they are. So there must be at least five equivalent positions. Now that would be consistent with that pentagonal pyramid, right? All the five around the base of it are at five equivalent positions. But you can’t get three different disubstituted ones among those positions. Right? So it couldn’t be that. So none of Dewar’s models work with this, although the perfect hexagon will work; a perfect one, not alternating single and double bonds. And the equilateral triangular prism, the Ladenburg benzene will also work. So as I said, it would work with either of those, except that the one on the right doesn’t give the right number of disubstituted isomers.

Now Koerner’s own argument, that the sixth position is equivalent, was faulty, because it hinged, in a subtle way, on assuming that the benzene was hexagonal before you proved it was. It wasn’t obvious, but it’s true if you look at it carefully. But it is possible, using group theory, to construct a logically rigorous argument on the basis of his evidence. His last step of his argument was wrong but his evidence was adequate to prove it. So he was right in his intuition that this proved it, and he was right in his conclusion that all the positions are equivalent. So he deserves enormous credit for formulating the logic of the first true structural proof; not just one of these plausibility things, like there’s only one monosubstituted methane — chloromethane, therefore all the positions are equivalent. Right? That’s just plausibility. But this is a real rigorous proof. Now how many isomers do you think there are of C₂H₄Br₂? Think about it. Ethane with two bromines, how many isomers?

Student: Two.

Professor Michael McBride: Pardon me? Somebody guess.

Student: Two.

Professor Michael McBride: John?

Student: Three.

Professor Michael McBride: And how would you get three?

Student: Because there’s a double bond on the —

Professor Michael McBride: No double bond. There’s six things on it. So forget that one.

Student: I was thinking of the wrong one.

Professor Michael McBride: Yeah. If it were a double bond you’d be right. But there are no double bonds. Corey?

Student: Two.

Professor Michael McBride: And what would you call them? How would you distinguish them, or describe them?

Student: One would have one bromine on each carbon and one would have two bromines on one carbon.

Professor Michael McBride: Good. So you could have two bromines on one carbon or one bromine on each carbon. So there should be two isomers. Now, in that same journal, the same volume of the same journal, in 1869, in Palermo, a guy named Paternó said the three isomers of dibromoethane, supposing they actually exist, can easily be explained without assuming any difference among the four valences of carbon. You don’t need curly bonds, long bonds and stuff like that. And here’s how he explained it, with this picture. Right? So both, on the leftmost they have two on the same atom. The other two have one on each atom. And how are they different — one on each carbon atom — but how do they differ? In the phase of rotation, around the carbon-carbon single bond. Right? And at the bottom he says — or actually notice there’s a little black thing in the middle. What do you think it is? Dana, what do you think?

Student: Some way to join the —

Professor Michael McBride: Right, it’s a case of rubber tubing. Because this is an artist’s picture of an actual model made out of sticks and balls. Right? And sure, those things aren’t super-imposable. Right? But he says at the bottom: “It is superfluous to say that this is only a way of representing the facts, and that all these ideas need to be tested experimentally.” So we’ll see next time what kind of reception this received.

[end of transcript]

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