CHEM 125a: Freshman Organic Chemistry I

Lecture 35

 - Understanding Molecular Structure and Energy through Standard Bonds

Overview

Although molecular mechanics is imperfect, it is useful for discussing molecular structure and energy in terms of standard covalent bonds. Analysis of the Cambridge Structural Database shows that predicting bond distances to within 1% required detailed categorization of bond types. Early attempts to predict heats of combustion in terms of composition proved adequate for physiology, but not for chemistry. Group- or bond-additivity schemes are useful for understanding heats of formation, especially when corrected for strain. Heat of atomization is the natural target for bond energy schemes, but experimental measurement requires spectroscopic determination of the heat of atomization of elements in their standard states.

 
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Freshman Organic Chemistry I

CHEM 125a - Lecture 35 - Understanding Molecular Structure and Energy through Standard Bonds

Chapter 1. The Limits of Molecular Mechanics Programs [00:00:00]

Professor Michael McBride: At the end last time we asked the question whether molecular mechanics programs are useful? And the answer is yes; in fact, they’re really nowadays indispensable for evaluating the energy of molecules and their shapes. But are they true? Are they really fundamentally correct, the same way we believe quantum mechanics to be fundamentally correct? And the answer to that is certainly no. We already saw how they depend on so many arbitrary parameters that are just adjusted to try to fit data from some molecules. And then there’s the hope that they’ll fit other molecules, and sometimes they do and sometimes they don’t. They usually do pretty well, but there can be special cases that differ for some reason, and it’s hard to predict those ahead of time. So one has to look at them with a certain grain of salt. Okay?

A good example of where they’re demonstrably incorrect is in the question of van der Waals radii; for example, the contact between two bromine atoms that are not bonded to one another. So we’re going to make a plot showing a bromine attached to carbon, and we’re going to do this in the same spirit that Dunitz and his crew looked at amines attacking carbonyls. We’re going to look at many, many crystal structures, and we’re going to have — we’re going to rotate the various structures so that they have a C-Br bond that lines up right there. And then we’re going to see how far it is to the nearest neighboring bromine, from some other molecule. Okay? And now what would we expect? Well there’s a certain van der Waals radius for bromine. So the next neighbor should have its bromine in contact with that one. So we should go that radius again, to get to a neighbor. And there should be various directions, of course, where the neighbor could be, but always at that same distance, if they’re in contact with one another. Okay? So there should be an arc of a circle, like that. So if we make these — if we rotate these various crystal structures that have been determined, so the C and Br line up there, and the other Br lies in the plane — the neighboring Br from another molecule — then the closest they should get is along that dashed red line.

Now, of course, there could be others that are further away, that aren’t in contact. And in fact there could be some that are a little bit closer. You’d really expect some to be a little bit closer. Because there’s van der Waals attraction, out to infinite distance; at least to great — to substantial distance, much further than where they’re actually in contact. They’re attracting one another. So this bromine is part of a molecule that has many atoms in it, and the other bromine is part of a molecule that has many atoms. And all these other atoms are attracting one another, pulling toward one another. Now, of course, it would just collapse, except for the fact that the ones, the closest ones, run into one another. But that means that they should get slightly closer than just precise contact, because they’re being pushed together by all the other atoms in those molecules. So we would expect neighbors to be slightly closer; some of the neighbors, at least, to be slightly closer than this van der Waals radius, or double the van der Waals radius; whatever the van der Waals radius means. Right? The question is where you get that.

But, at any rate, we could — you can look up in a book and see what they say the van der Waals radius of bromine is, and make a plot like this. And then you’ll put points on it, for all the different crystal structures of where neighboring bromines are. And they look like this. Here’s what I just said: To balance attractive and repulsive forces between neighbors, the closest atoms must be a little bit too close. Okay? And here’s what you see. And indeed, some of the atoms, in some of the cases, are a little bit too close. And, of course, many are further because other atoms are running into one another, of these two molecules. Okay? Well what do you find particularly curious about this plot? Anything? Chenyu?

Student: It’s not really circular.

Professor Michael McBride: I can’t hear very well.

Student: It’s not really circular.

Professor Michael McBride: It’s not circular. It’s an ellipse. Right? There are many more inside the red dashed line at the top than out to the side. So what do you infer from that, about the shape of the bromine atom?

Student: It’s not spherical.

Professor Michael McBride: It’s not spherical. Right? Okay. So the bromine atoms seem not to be spherical. Right? But every molecular mechanics program calculates van der Waals interactions between atoms as if they are spherical. The interaction depends only on the distance, not on the direction. Right? So clearly molecular mechanics programs are not perfect. Right? But still they’re pretty good, and you can get interesting information out of them — from them. Plus, they can tell you something about why something might be the way it is — right? — torsional energy and things like that. Okay, and in fact if you’re trying to deal with a big molecule, it’s hopeless to do quantum mechanics on it.

So here’s angiostatin, which is an anti-cancer drug. And this is the largest — at least as of a couple of years ago; I suspect it’s still the case — this is the largest molecule whose shape has been optimized by quantum mechanics. Right? And not a really high level of quantum mechanics. Here’s a description of it from Mike Frisch, from Gaussian Incorporated in 2003. He said:

“We optimized Kringle 1” (which is this thing; I don’t know why you call it Kringle 1) “with the AM1 method” (which is a very low-level of quantum mechanics) “using Gaussian 03. Plasminogen Kringle 1 contains 1200 atoms, which are made up of 642 non-hydrogen atoms and 578 hydrogen atoms. The job takes about 650 optimization steps, starting from the molecular mechanics geometry.”

So first they used molecular mechanics to get the approximate structure, and then they started adjusting it with quantum mechanics, step by step, until after 650 steps they got to a minimum of energy, and that’s this energy. But obviously this is not an everyday occurrence, to do something like this. So if you’re dealing with a big biological molecule, you have to use molecular mechanics. Right? But you know that fundamentally it’s not right. It may get a lot of things right, but don’t believe that it came down from Sinai on the tablets. Right? Yes Lucas?

Student: How long did it take to do just one calculation?

Professor Michael McBride: I don’t know. We could ask him sometime. He’s around here from time to time. I haven’t set him up to come in.

Okay. So quantum mechanics is pretty close, as we believe, to the truth. Molecular mechanics isn’t close to the truth but for practical purposes it’s indispensable.

Chapter 2. The Cambridge Structural Database and the Demands of Predicting Bond Characteristics [00:07:32]

Now, we wonder whether the standard structural model of molecules that we have is realistic in its geometric detail; things about bonds, bond lengths, bond angles and so on. And how are we going to know whether it is? We’re going to use X-ray diffraction to see just how regular bond distances are. Okay, now we’re helped in this by what’s called the Cambridge Structural Database. In Cambridge, in England, there’s this building where people sit and collect X-ray structure data from all over the world, and make it computer-searchable and so on, so that you can look at many, many structures. And you see that many structures are available. This is how many structures there are, as of any particular year, in the Cambridge Structural Database. And you see it’s up to about half a million X-ray structures have been determined, and put in the Cambridge Database, as of now. You can click on there and go to the website to see it, if you want; it’s interesting. Okay, and they predict that there’ll be half a million by 2010.

Okay, now how many atoms are there — this is how many structures — how many atoms are there? How many bonds in each structure? Well, as time goes on, X-ray is able to do more and more complicated molecules. So that at the beginning there were only about an average of twenty-seven atoms in a structure, but by 1982 there were forty-four, and by 1990 there were fifty-four, and after 2002, or something like that, there were seventy-three. So the molecules are getting bigger and bigger. And, as of now, as of at least this last January, the last issue of the database, there were thirty and a half million atom positions, x,y,z coordinates, available. Right? And, of course, there are many more bonds than there are atoms, because most atoms form several bonds. So there are greater than forty million bond distances available. Now the question is, how regular are they? Do carbon-carbon bond distances tend to be the same?

Well in 1987, when the Cambridge Database was much smaller than it is now, it was already large enough to do pretty good statistics on this thing. And here’s a paper published in that year, that you could look at if you want to, by people from the Crystallographic Data Center in Cambridge, that talked about this subject. Now here’s a table that shows how many different kinds of bonds they divided the experimental data into, so that within any given type the bonds would all be about the same length. Right? It’s not surprising that different types of bonds have different lengths. But if you subdivide — like double bonds and single bonds will be different lengths. Right? So those would be at least two types. But notice how many types they used. So, for example, there are 119 different types of carbon-oxygen bonds, that they did statistics on; not bonds but types of bonds. Okay?

Now here are those 119 different types. There are sixty types that are made up of an sp3 carbon bonding to an O of type two. And there are seventeen types of bonds, which have sp2 carbon bonded to that type of carbon. O2 means the oxygen has bonded to two different atoms; O1 it’s bonded to just one atom. Right? There are four types where it’s an aromatic carbon that’s bonded to a divalent oxygen, and there are thirty-eight where an sp2carbon is bonded to an oxygen that makes bonds to only one carbon; that is, a C-O double bond. So there are thirty-eight different types of C-O double bonds. Okay? Or, in the case of C-N, there are ninety-seven different types of C-N bonds in their tabulation. And for carbon there are 175 different types of carbon-carbon bonds. And notice that there are twenty-seven different types of carbon-carbon bonds where both the carbons are sp3 hybridized. Right?

So they’re dividing this into lots and lots of different categories, in order to get regularity within each category. Now let’s look here at those twenty-seven different kinds of bonds that are carbon sp3 to carbon sp3. Okay, so for each type is given the mean and the median — two kinds of average of the bond distance; what the standard deviation is; what the distance is that splits off the shortest one-fourth, and the longest one-fourth; and then how many different examples there were to do this, to make the statistics on. Okay so now, in this table, C with a pound sign means any sp3carbon. So at the top there we see a CH3 bonded to CH2. So that’s the bond distance we’re interested in; methyl to CH2. But that CH2 has some other sp3carbon on it. Right? So that’s one type of bond. And you notice the mean distance is 1.513. And if there are two sp3carbons on the second carbon, then it’s 1.524 average. And if there are three of them, it’s 1.534. Now notice the standard deviation is about 1% of the bond distance. So within that class they’re all within about + or - 1%. Right? So a very small deviation. So they wanted to get them — they wanted to make the classes restricted enough that you’d have a predictable bond distance — right? — within 1%.

Okay, notice that crowding seems to stretch the bond. As you put more carbons on that second carbon, which is bonded to methyl, the bond to methyl gets longer and longer; longer by a little less than 1%, as you put more crowding on the second carbon. Here’s a graph of it, of that, the standard deviation and the mean. Right? So you see it gets, as you go from one R to two R’s to three R’s on the second carbon, the bond gets a little longer and a little longer; although they can overlap. Okay? Now then we go down to here, where both carbons are substituted. So the first carbon has two or three carbons on it, and the second carbon also has two or three carbons on it. And now you see there’s still more crowding. They get longer and longer. And, in fact, when you have three other carbons on each of the carbons whose bond you’re interested in, then those others repel one another, by van der Waals repulsion, and it gets actually pretty long. Right? It gets almost 10% — well 7% longer, or something like that. So anyhow, you can get these kinds of regularities, based on the statistics.

And if you wanted just an overall average carbon-carbon bond distance, it’s about 1.53 angstroms. Okay? However, if you have cyclopropane, or cyclobutane, then the bonds are, in the case of cyclopropane, unusually short, 1.51; and in the case of cyclobutane, unusually long, 1.55. Okay? Now if you have sp3- sp2, as a carbon-carbon bond, rather than sp3- sp3, then what would you expect? Would you expect a shorter bond or a longer bond, if one of them is sp2? Just guess.

Student: Repeat the question.

Professor Michael McBride: sp2means you’ll get better overlap, stronger bond; probably a shorter bond. Right? Just a little review for you here. Okay? And indeed, 1.50 is the average. So there are hundreds and hundreds of these, as you can see, in this particular paper. But if you want just order-of-magnitude things — actually it’s better than order-of-magnitude. But for carbon-carbon single bonds, it’s about 1.53 angstroms. For double bonds, 1.32. For triple bonds, 1.18. And for aromatic bonds, say within a benzene ring, these are one-and-a-half bonds. So not surprisingly they come between single and double. Okay? Notice they’re a little closer to double than they are to single; 1.38 to 1.32, as opposed to 1.38 versus 1.53. Does it surprise you that they’re a little closer to double than single? Notice, even if they were single bonds, but between carbons that were part of aromatic rings, they would be sp2- sp2carbon-carbon bonds. And we’ve already seen that those would be unusually short. Right?

So you don’t expect it to be one-and-a-half, the average between a normal single and a normal double, if you think about it with a little more sophistication. Okay, and then as we see here, a single, if it’s sp3- sp2, is 1.5; and a single, if it’s sp2- sp2, is 1.46. So the 1.38 is closer to being halfway between 1.46 and 1.32, than it is between 1.53 and 1.32. So depending on how accurate your model needs to be, you can get pretty good guesses, based on values you would find in a table. Now you can see some interesting departures from regularity. Here I’ll show it in aromatic carbon-to-nitrogen bonds. So there’s an aniline derivative. Aniline is aminobenzene. Okay, so there’s — that red bond is the bond we’re talking about, and there’s the distribution, a histogram, of distances. What’s funny about the histogram? Dana?

Student: There’s one that’s suddenly low.

Professor Michael McBride: There’s a hole in the middle. Right? It’s bimodal. Right? So it looks like there are two different kinds of bonds between nitrogen and aromatic carbon. Now let’s split them out, the ones that are in one and the ones that are in the other, and see why they have this difference. It turns out that if you do ones where the nitrogen and the benzene are co-planar, as shown above, then you have one nice histogram. And if the nitrogen is pyramidal, rather than being planar, then it’s another histogram. So it depends on whether the nitrogen is planar or pyramidal. But when the nitrogen is planar, it’s co-planar with the benzene. And for good reason. It’s so that the unshared pair on the nitrogen can be delocalized into vacant orbitals, π orbitals of the benzene ring. So that makes this carbon-to-nitrogen bond a little bit double bond-like, and shortens it. Whereas if the nitrogen is twisted, compared to the benzene, then there’s no reason for it to be flat, to have a p orbital that will overlap with the benzene orbitals. So it becomes pyramidal. But more importantly, there’s no double-bond character anymore, to the carbon-nitrogen bond. So the bonds are longer than the ones that are partial double bonds. So that’s why it’s bimodal. So you learn something from these statistical analyses. You have poor π overlap when the nitrogen-carbon bond is twisted.

Now, how complex must a model be in order to predict useful structures? So from the point of view of the Cambridge compilers, they wanted to get standard deviations and bond distance of about 1%, as you saw from the table. And in order to do this, as you saw, they needed 175 different kinds of C-C bonds, differing in these various ways, as we’ve illustrated; and 97 different types of C-N and 119 different types of C-O bonds. And 682 different kinds of bonds altogether, in their tabulation. Okay, now we have ambitious goals. We want to understand all stuff; chemicals that is. Okay? We wonder how things behave — their properties — and their chemical transformations as well. Right? And the key to doing this, the trick we’re going to use, is to understand the basis of structure — how the atoms are arranged — and energy. And we want to do this in terms of bonds. You could do it entirely without bonds; just use quantum mechanics to minimize energies of atoms, of arrangements of atoms, and calculate those energies. And you’d have that same information, structure and energy, but you wouldn’t understand anything, you couldn’t predict anything in your head. With bonds, you have a tool that will allow you to predict things.

Chapter 3. Calculating Chemically Useful Heats of Formation [00:21:37]

Okay, so let’s just see how well bonds do. Obviously there are going to be corrections, having to do with conformation and strain, as we’ve been talking about. But is there an underlying energy for a given composition, or a given constitution, a given set of bonds? Can you predict energy as well as structure? In other words, how standard are bond energies? If we’re going to do our understanding in terms of bonds, can you add up energies for all the bonds and get the overall energy for the molecule? Knowing, of course, that there are going to be corrections for strain and so on. Okay, so this started in 1868 already. In the very first issue, the second paper of the first issue of the German chemical journal, Chemische Berichte, which was the journal of the Chemical Society of Berlin, the second paper ever in that journal, was “On the Regularity and Calculation of Heat of Combustion of Organic Compounds.” So Hermann was interested in this because he was a physiologist and he wanted to know how much energy you get out of the food you burn up in your body. Right? So could you predict, on the basis of what substances you eat, how much energy could be available? Okay? So physiology led him to study this.

And there was another paper, very shortly after that, in the same journal, by Oppenheim, “On the Relationship of Heat of Combustion with the Constitution of Substances.” Right? So could you look at the constitution? He didn’t mean it yet, constitution. Notice this is 1868; still pretty early days. He didn’t mean constitution to mean nature and sequence of bonds. He just meant what was in there. Okay, so how did they try to do this? Could we do it, the way they did it, by predicting the heat of combustion, how much heat you get out of burning a thing, say just by how much carbon and hydrogen? We know you can burn carbon, coal, and get a certain amount of heat out. We know you can burn hydrogen gas and get a certain amount of heat out. Could you take a substance, know how much carbon and how much hydrogen is in it, and predict how much heat you’ll get out of burning it? Right? It would be great if that were true. Okay? So let’s see how well it works.

So you’re going to compare, to graphite — that’s carbon — and hydrogen. Right? So we’d have a certain number of atoms per mole, and then we know what the heat of combustion is, per mole. You get 94r kilocalories out of burning a mole of graphite, a mole of carbon in graphite. You get 57.8 out of burning hydrogen. Okay? That’s per — I’m not sure, we’ll have to see, I think that’s per atom, not per molecule; we’ll see. Yeah, no, it’s per H2. Okay. So it’s 94.05 per carbon, and 57.8 per H2. Okay? Now you take some other substance, like ethene, C2H4. Right? So you should get two times 94.05, for the carbon, and two times 57.8; because it’s 57.8 for every two hydrogens. You’ve got four hydrogens. So two times that. So you should get 303.7 out of burning a mole of ethene. Right? In fact, you get 316. Right? Which isn’t bad. It’s an error of 12 kilocalories/mol, but that’s only 4%. So you’re pretty close already in predicting that. Now so we can be optimistic. Well let’s try something else. Notice, we could even understand why you get a little more heat out than you expected, because one of the bonds in C2H4 is not such a strong bond. Right? The second bond of the double bond isn’t so strong. So you expect the starting material not to be as stable as it might have been. It’s a little higher in energy than it would’ve been if that second bond was really good. So when you burn it, you get more heat out, because there was more in it to begin with. Right? It was less stable to begin with. So we can understand that.

Now let’s try to extend it to other things. How about cyclohexane? There we have 6 carbons and 12 hydrogens. You’d predict 911. It turns out to be 881. It’s off by 29. But that’s only a 3% error. Right? Not bad, we’re doing well. Now how about cyclohexanol. Right? There again it has — since it has the same amount — you don’t expect to get anything out of oxygen, because you don’t oxidize oxygen — right? — in burning. It’s only the carbon and hydrogen that you oxidize. So you predict exactly the same, 911; it’s the same amount of hydrogen and carbon. But you only get out 842, instead of 881. So now you’re off by 68 kilocalories/mol, or 8%. And if you take sugar, which is again C6H12, but O6 now, now you get out only 670. You’re off by 240. So you’re making a 36% error. Now it’s getting pretty bad. Right? Does that surprise you, that when there’s oxygen in the molecule you get less out by burning; less heat out by burning, when there’s already oxygen there? What is burning?

Student: Oxidation.

Professor Michael McBride: What do you do?

Student: Oxidation.

Professor Michael McBride: You’re oxidizing it. These are already partly oxidized, if they have oxygen in them. So you’ve already gone partway. You don’t expect to get so much out anymore. Right? Okay, so that makes sense. They’re partially pre-oxidized. So you make bigger — you make errors in this. So for fuel purposes, for physiology, this isn’t so bad; especially if you could include some kind of correction for every oxygen that’s already there, partway to the product. Then you’d get numbers that — you know, within 5, maybe 10% — are going to be right. So you actually could add up the — just from the composition — add up what atoms are there and predict how much heat you can get out by burning. Right?

Now, but might you need a more complex model, if what your goal is, is to predict chemically useful energies, rather than just ones that say how much energy you’re going to get by burning your chocolate bar or whatever? Okay? For physiology purposes, you might be content with plus or minus 5%. But for predicting the equilibrium constant between cyclohexane and cyclohexanol — suppose there was some reaction, oxidation reaction, where you can convert one to the other, and you’re going to be off by 1% say, which would be 9 kilocalories/mol. So you’re off by 1%. How off would you be in the equilibrium constant? How big a factor in equilibrium would nine kilocalories be? Do you remember how to do that? How do you go from energy to equilibrium constant? I’m going to keep boring in on this until you remember it, because it’s certainly going to be on the exam. Katherine?

Student: Ten to the — you do the three quarters.

Professor Michael McBride: 103/4ths of that. Okay, so 3/4ths of nine is about seven. Right? So by what factor would you be off in predicting an equilibrium constant, if you were off by 9 kilocalories/mol?

Student: 10-7th.

Professor Michael McBride: 107th. Right? Is that good enough, to be off by ten-million-fold? Not very good. Right? So you’re off by 107th. So a useful model clearly has to go, to get better numbers than you can get just from the composition, just from what atoms are there. You’re going to have to consider bonds — and that’s what I said we were going to do — to see how good it can be if you just have certain energies for certain bonds. So we need to go to constitution. So now when we’re going to talk about energy, there’s going to be a big, a really fundamental question. So here’s cyclohexane. It has a certain energy. What number should I put on it? Should I say it’s 13, 28, 562 million? Right? The question is one of our fundamental questions. What’s that?

Students: Compared to what?

Professor Michael McBride: Compared to what? What are we going to call zero? Okay? Now for practical purposes of doing an experiment, there’s a very good reference. and it’s the one we’ve just been using, without saying so — which is CO2 and H2O; the same amount of carbon in CO2 and the same amount of hydrogen in H2O. Right? Why is that a convenient reference point to take, from an experimental point of view? Lucas?

Student: Well that’s what you get in burning.

Professor Michael McBride: Ah, so all you have to do is burn it, see how much heat you get out, and you know the energy relative to CO2 and H2O. So that’s called the heat of combustion. And it turns out to be 881.6 kilocalories/mol, for cyclohexane. We were just talking about that. Okay? Now so it’s easily measured. That’s a good comparison. But if I hadn’t told you that we wanted an experimental comparison, that we wanted sort of a simple-to-think-about comparison — maybe if you couldn’t even measure it — what might you think of as a point of comparison? If you’re going to talk about — you got a whole bunch of different compounds, you want to know how stable is, compared to what? What are we going to use as the what?

Student: Unbonded atoms.

Professor Michael McBride: One — it could be unbonded atoms. That would be a really good one, because then when we start with atoms, we put them together and make bonds, and if there’s a certain energy for each bond, then we would be comparing our molecule, or our model of a molecule, with all these bonds, to the atoms. That’d be a very good one. But that’s going to be a very hard one to deal with experimentally — right? — to go to atoms. There’s another thing that you could actually have in a bottle, that might make sense to compare to, and that would be the elements, in their standard state. So graphite; how stable is it compared to graphite being the carbon and H2 gas being the hydrogen? Right?

So that, if you’re comparing to the elements in their standard states, that difference — notice that cyclohexane is just slightly more stable than the elements bonded to one another. Right? And that’s called the “heat of formation”. That’s just a definition; that’s what the heat of formation is. It’s the energy compared to the elements in their standard state. Okay? Now, but how could you measure that experimentally? How would you get heat of formation? You know the energy of cyclohexane relative to CO2 and H2O, by doing combustion. How would you know it relative to graphite and hydrogen? You can’t mix hydrogen and graphite and have it suddenly magically become cyclohexane and measure the heat involved in that transformation. But how could you get it? Do you see?

Student: [inaudible].

Professor Michael McBride: There’s a neat trick to use.

Student: Burning graphite and hydrogen.

Professor Michael McBride: Ah, you could burn the graphite and the hydrogen, see how much heat they give off, and then take the difference. Right? So that’s what you do. You burn that and get 911.1, and the difference then, 29.5, is the heat of formation of cyclohexane, from the elements. Well done. Okay, but Andrew says it would be great, for purposes of — oh and in fact these heats of formation are available in tables for lots and lots of compounds. And this brings up a question. This table is taken — and there are several others here for alkanes, cycloalkanes, alkenes, alkynes, aromatic hydrocarbons and everything. You can look these up. You can find them on the Web too, all these heats of formation. So it’s how stable a particular substance is. And it’s not just molecules. Notice they also have atoms and radicals, inorganic compounds, cations, and anions; all of them relative to the elements. So you can find tables like this. These particular tables I took from a book by Streitwieser, Heathcock and Kosower, which we’ve often used as a textbook in the course. I think it might be out of print now, though the publisher may still have some copies. But would you like to have a textbook for second semester, so we don’t have to continue with the Wiki stuff?

Students: Yes.

Professor Michael McBride: Now, so if we’re going to do that, we have to make the decision. Are we going to get a slick, new book, that has a lot of nice color pictures in it and so on, and the very latest thing? Or are we going to get something like Streitwieser and Heathcock? Which I think is a wonderful book, but you’re going to have find secondhand copies, or maybe the publisher still has a few and so on. These new books, they charge an absolute arm and a leg for them, as you probably have found out. So I want to do it however you want to do it. Actually, why don’t you think about it a little bit? I like this book, but there are plenty of other books that are just fine for our purposes. And we won’t be 100% in the book, you won’t be surprised to hear. But there’ll be a lot of stuff that could be useful in having a textbook. Chenyu?

Student: Could you tell us the level of clarity or quality in instruction, clarity [inaudible].

Professor Michael McBride: It’s not going to make a lot of difference. In fact, we could probably get one of these, you know, outlines of organic chemistry, little paperback books and so on, that would probably have the stuff we wanted. But then you probably couldn’t hold your head up against your roommates, and so on, if you have a little dinky book like that. [Laughter] When I took organic chemistry, the professor was Professor Fieser, who wrote lots and lots of textbooks, and he had just written one for nurses, for a one-semester course. So we took two semesters and used a book that was about that thick. It was — I wasn’t too embarrassed, I didn’t know enough to be embarrassed at that time. And it was a good book. I don’t have it anymore. Somebody borrowed it from me and didn’t give it back. It was a neat book. But anyhow, we can talk about this a little more next time. But think over whether you want to fork out for a shiny new book, or would rather have a book, which I think is equal in quality, maybe better even, that you’d have to go buy secondhand copies or something, but it’d save you some money. So think about it, and we’ll talk about it next time.

Okay, and anyhow, that’s where I got this table. Now, let’s look at the heat of formation, from this book, of cycloalkanes. And so cycloalkanes are CH2 groups in a ring. Right? Six CH2 groups is cyclohexane. Two CH2 groups is cycloethane, or ethylene. Right? So now here’s the heat of formation there; how big the ring is — two carbons, 3 carbons, 4 carbons and so on — and what the heat of formation is. Now let’s try to make sense out of this. We could look at the heat of formation, divided by how many carbon or CH2 units there are in there. It could be they’d all be the same. That would be the simplest thing, if all we had to do is take how many CH2 groups there were there, add that up, and we would get what the stability is. And it’s — they’re not too bad. But again, compared to what? Right? So you’re going to be — the biggest error is 6, that I see there, and a more typical error is about 4 say, or something like that. So 4 — that would be, if you were off by 4 kilocalories/mol — ah, but it’s per CH2 group. Ah, so it gets worse, right? If you have a lot of CH2 groups, there’s a lot of energy. So if it were only CH2 group, you’d off only by a factor of 1000, if you had that kind of error. Okay?

Well wait a second. This isn’t an error. This is just what it is per group. Right? So it’s — well let’s go one step further here. The minimum is 4.9. That’s for cyclohexane, our old favorite, which isn’t strained, remember, or at least not very strained, at least chair cyclohexane. Okay, so let’s suppose that that’s the standard, and everything else is a little bit higher in energy, more positive, because there’s some kind of strain there. Okay? How big is that strain? Well let’s compare what it would be. So we’ll call that — you’d get — well say you’ll get 4.9 reduced, compared to the elements, if you have an unstrained ring. And now we’re going to look at the others and see how strained they are. Now notice, down at the bottom here, 13, 14, 15, 16, these big rings have almost exactly the same energy. So they’re not strained. Right? We would get a very good prediction of what their heat of formation is if we just multiply from the value we got from cyclohexane. But the others here are a little bit off.

So strainless theory is n times -4.9 predicts our heat. So you see that we have a strain energy, which is the experiment, that column shown, versus theory; the n times 4.9. And here are the strain energies you see. Note, almost no strain energy down at the bottom, just a few kilocalories/mol. That’s not going to hurt us very much. But up at the top we have strain — the first one is bad, ethylene, we know, because the double bond isn’t so strong. Okay, that’s fine. But then we have all single bonds, but small ring strain, because you don’t have such good overlap. And we talked about that. Okay, so 6 to 27 kilocalories of strain there. We already understand why that is. But here’s an interesting one. These ones, which are bigger than 6 — right? — bigger than 6 members in the ring, but they have strain, and that strain is called transannular, across the ring. And you can easily see why it is. There’s a view of a model of cyclohexane, at the top, and below is cyclooctane.

Now of course you’d — Baeyer thought it would be strained because you’d have to bend the bonds out. But Sachse said you don’t have to bend the bonds out, all you have to do is pucker it. Right? But if you start puckering too much, opposite sides will run into one another. And you can really see this if you look at space-filling models of these. Here, the repulsion between those two that are across the ring, you can see the hydrogens really crunch into one another. So these intermediate sized rings suffer from transannular strain. So you can see this by comparing heats of formation. Which remember — how do you — what experiment do you do to measure heats of formation? What actually would you do when you went into lab? You have these substances. What do you do with them in lab, in order to measure their heat of formation?

Student: Calorimetry.

Professor Michael McBride: You’d burn them. Right? What you actually measure is the heat of combustion, right? But then if you also know — if somebody out there has for you measured the heat of combustion of graphite, and the heat of combustion of hydrogen, then you have the numbers that enable you to work up your data and find out what the heat of formation of your substance was. Right? So now this book, in Appendix III of this book, has average bond energies. And this is what Andrew wanted. Right? He wanted to be able to start with the atoms, add up all the bonds, and get the energy. And, in fact, here is a table that purports to do that; that says that these are energies you can add up to get what the heat of atomization would be. And you’ll notice that it makes a little distinction among carbon-carbon bonds, gives different values for double and triple bonds, than for single. That’s certainly sensible. And the same for nitrogen.

Chapter 4. Measuring Heats of Atomization with Bond Energies [00:42:18]

So the question is, does this work? Can one sum up bond energies to get useful heats of atomization? So how well can bond energies predict the heat of atomization, and where does the heat of atomization come from? So again you’re going to do these heats of combustion, but you’re going to need somebody to measure a new value for you; not just the heat of combustion of graphite and the heat of combustion of hydrogen gas. And you’ll see what you need here. Because we’re going to now compare, not with graphite and hydrogen — which is something you can have in a bottle — but we’re going to compare with the atoms, which you can’t have in a bottle. Now how are we going to measure the heat of atomization? We’re going to do it indirectly, by this line. Right? We know the heat of combustion of our substance, or we measure it. We know the heat of combustion of graphite and hydrogen. So all we need to know is the heat of atomization of graphite.

How hard is it to get a carbon atom out of graphite, or a hydrogen atom out of H2? If we knew that, then we could figure out that that’s going to be [1650.6] kilocalories/mol; that’s to get out six carbons from graphite and twelve hydrogens from H2 gas, in order to make the C6H12 of cyclohexane. Right? So if you knew that number, the [1650.6], the heat of atomization of graphite and hydrogen, then you could add all those together, with the proper sign, and get that it’s [1680.1], the heat of atomization of cyclohexane. And then you could try to parlay that into what the bond strengths are, in cyclohexane. So what you need is the heat of atomization of graphite and the heat of atomization of hydrogen. But how can you know the heat of formation for an atom? You do it by spectroscopy. So we’ve seen things like this before, a Morse-type curve. And we know that you have an X-Y diatomic [molecule] down in the bottom there. And if you — it takes a certain amount of energy to get to X plus Y, to get to the atoms. If you knew that energy — and that’s what we want to know; how hard is it to break these apart? H2, for example, into 2 H atoms?

But you notice that you could do this with light, and if you knew what the minimum amount of light is, to go up — of course, it’s going to be quantized, right? There are going to be given levels, until you get to the level where it dissociates. And then, as you’ll remember from Erwin Meets Goldilocks, it’s not quantized anymore. You can have any energies you want. So if by spectroscopy you can see when it changes from being individual lines, to being a continuum, where any energy goes, then you know what that energy is. So you’d know how much energy it takes to go from the molecule to the atoms. So you can do that by spectroscopy. And in the case of a hydrogen molecule, it turns out to 104.2. But from that you get two hydrogen atoms. So it’s 52.1, is the heat of formation from the gas of a hydrogen atom. So we know that; because the heat of formation of H2 is defined as zero. That’s the element in its standard state already. You can do the same thing with oxygen and get that it’s 59.6 to make an oxygen.

But how do you do carbon? Because the standard state of carbon is not C2. Right? And indeed you can’t buy C2 in a bottle to do the experiment either. But what you can get in a bottle is CO, carbon monoxide. So if you did the experiment on CO, you’d find out that that number is 257.3. And now you have what you need; because that’s the difference between CO and a C atom plus an O atom. But you know where CO is, compared to graphite and oxygen, by burning CO, and you know where oxygen is, relative to oxygen in its standard state, because we just did it with O2. Right? So there’s burning graphite is 26.4. Graphite to a single oxygen atom is that heat of formation of the oxygen atom. So all we need to have is that difference, and we know what the heat of formation of carbon is. It’s going to be 171.3. Okay? So we’re able to do it by doing spectroscopy on CO.

But this was done — people got the number 171.3, but then they began to worry. And there were two Nobel Laureates, plus Edward Teller — who was big in the atomic energy community in the United States, you may have heard of him — so big time physicists who absolutely disagreed with 171.3. Because what they said was that the spectroscopy was misleading, that actually it was giving one of — the light that was going in was giving one of the atoms, not in its normal state, its low energy state, but in a higher state. So the number that you predicted, that you measured is 171.3, was not to a normal carbon atom, but to an excited carbon atom. Right? So that the true energy of a carbon atom would be lower than that. And so they thought — one set of them thought it would be 141 kilocalories/mol, and another thought it would be 125. So these numbers are all very precise. You can measure these positions of spectral lines very accurately, but you don’t know how to interpret it. It could be way off, by twenty or almost fifty kilocalories/mol. So you need some other way of doing this. And for many years I’ve brought in a guest lecturer to explain how he did it. But unfortunately he died a year and a half ago. So I’ll have to tell you what he would have told you, but we’ll wait until next time for that.

[end of transcript]

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