Freshman Organic Chemistry II

CHEM 125b - Lecture 4 - Electronegativity, Bond Strength, Electrostatics, and Non-Bonded Interactions

Chapter 1. Generalization of Fractional-Order Rate Laws [00:00:00]

Professor Michael McBride: I want to start with a question that was asked by Ayesha after the last lecture regarding fractional-order of kinetics. You remember that in the reaction of methyllithium there was a pre-equilibrium between the tetramer and the monomer so that tetramer is dominant, but the monomer is reactive.

So remember what that meant was that in this pre-equilibrium the rate of association equals the rate of disassociation. So the concentration of monomer to the fourth power is proportional to that of the tetramer. Or if you take the fourth roots on each side it says that the concentration of the reactive species is one-fourth power in the amount of material you actually add, which is the dominant tetramer.

What Ayesha pointed out is this seemed to be similar to the case of a radical-chain initiator, which also had a fractional order. Remember the idea was that the initiator breaks to give two radicals, then the radical abstracts a hydrogen to give the radical, which is part of the cyclic mechanism in the radical-chain, the propagation steps of the free-radical reaction. But ultimately the machines are destroyed by two radicals getting together in the termination process, either forming X-X or X with R or R with R.

So at first glance this doesn’t look very similar to the equilibrium on the top. This isn’t an equilibrium at all. You’re not going back to the starting material; you’re not getting back to the peroxide. So this one is an example of a steady state instead of an equilibrium.

So when the initiation is constantly going on, the machine gets into a steady state of how fast it’s going around and disappearing at the same time, then you have the amount of termination equals the amount of initiation. So you have a steady amount of this, which is going to determine the rate of the reaction of the machine. And since the termination is proportional to the concentration of radical squared, the radical could be in the form either of X atom or in the form of the R•. But whatever it is, when two of them get together that does it. So the radical’s concentration squared, how much machine there is squared, is proportional to the initiation, to the concentration of this.

So this isn’t an equilibrium, as it was in the top. But really it’s the same deal. The number of machines then is proportional to the square root of the amount of the initiator. And notice that this is the same–association equals disassociation implies this relationship, termination equals initiation implies this relationship. 

So really they are the same thing, even though when she asked me the question I thought they were completely different. So I’ve been teaching this for 45 years or something like that, and it never occurred to me that these fundamentally were the same thing, the reason you have this fractional-order kinetics.

So this was a simplification and generalization that Ayesha came up with, and that’s what constitutes progress in science is when instead of having to memorize a zillion things, you find out they’re all the same and you just know that one. So congratulations to Ayesha, and keep up the good work all you people. That’s what makes it fun to teach you. After x years of teaching this I’m still learning.

Chapter 2. Electron-Pair Repulsion and Bond Dissociation Energy [00:03:44]

And I also ask questions. So on the 2009 exam I asked consider this chlorination reaction–the one we talked about at the end of last lecture–and certain bond dissociation energies, and then you’re supposed to figure out various things. But part G of this many-part question was why should the bond disassociation energies of N-Cl and R-Cl be so different, when those of N-H and R-H are so similar? So N-Cl is 46, N-H is 92, twice as big. R-Cl and R-H are about the same energy. Why is that?

Well, the answer I expected was based on this—lone-pair repulsion. So here on the left we have peroxide and ethane. They have the same number of electrons. And on the right we have N-Cl and methyl-Cl, the same number of electrons on the right.

Now if you look at the bond disassociation energies, this one is twice as strong as that one, and this one is nearly twice as strong as that one. So what makes the ones on the top so weak when really it’s the same number of electrons involved up and down?

Well, what you notice is that on top you have lone pairs on the atoms that are being bonded, which will repel one another. So the idea I expected was that that would weaken the bond.

Now is this plausible? Could the electron-pair repulsion explain the difference in bond disassociation energies between the top and bottom? Well, we can look at how the electrons actually are arranged, as we calculate it by quantum mechanics. And, of course, what it looks like depends on what contour we draw. This contour at 0.002 electrons per cubic Bohr radius is not very informative.

But if we go in further to higher densities to 0.01 to 0.05 to 0.25, now we see something different in these two things. You notice that in the bottom, the protons that are out here, the proton on the carbon, as opposed to where there is no proton on the oxygen, draws these electrons out. And the same thing here. The hydrogen that’s pulled out of the nitrogen nucleus and put out here to become CH₃, that draws these electrons out so that they don’t repel one another as much.

So that seems a plausible explanation, although it’s not quantitative. So we would say then that the repulsion between the electrons, which is less in the case where they’ve been drawn out, might be an important factor in stabilizing the ones where the electrons are drawn out. That was the answer I expected on the exam because we had talked about lone-pair repulsion. So that seems a plausible explanation for weakening O-O versus C-C or N-Cl versus C-Cl. 

But that wasn’t the answer I got on a lot of the–some people gave that answer, but a lot of them appealed to the difference of electronegativity, so that C-Cl has a big difference in the electronegativity, according to the Pauling scale of electronegativity. So carbon has 2.55, chlorine 3.16–they’re quite different in their electronegativity, whereas nitrogen at 3.04 is rather similar to the 3.16.

So what people had in mind then was there’d be an important resonance structure for C-Cl that would be C⁺Cl^–, and that would be–we associate the existence of resonance structures with stabilization.

Chapter 3. Heterolysis and Homolysis - Pauling’s Electronegativity and Bond Dissociation Energy [00:07:48]

Now we’ve seen that we can understand that thing in terms of the energies of orbitals, the concept that was involved there. We start with a bond either with nitrogen to chlorine, which would have its electrons there or carbon to chlorine, which would have its electrons down there. And then if we’re going to break the bond, those electrons, if we’re going to go to ions, like N⁺ or C⁺ and Cl^–, these electron pairs would have to go onto the chlorine. And they have to go farther when they’re on the nitrogen. 

So that mismatch in energies aids breaking them heterolytically for the carbon case. Where the carbon is mismatched, you don’t have to go up as much, so it’s easier to ionize. But that’s not what we’re talking about in bond dissociation energy. In bond dissociation energy we’re talking about breaking to two free radicals where one goes on one atom, one electron goes on the other from the pair. And now, as we talked about early last semester in terms of bond disassociation energy or homolysis, it’s the carbon-chlorine bond is stronger. So this is putting slightly more quantitative ideas to bear on the idea of why you have that difference in bond strength.

But the idea of electronegativity being associated with bond strength is an interesting one that’s worth tracing down a little bit. So I’m going to talk just a little bit today about electronegativity and bond strength.

The first use–I mentioned this last semester in English–of the word electronegativity was in 1837 in System of Mineralogy by James Dwight Dana. And we mentioned that James Dwight Dana was a professor here at Yale in the nineteenth century. He lived in this house that you’ve walked by, which is now a national historical monument. And he lived in the house with his wife, Henrietta Dana. Actually, it was Henrietta Silliman Dana was her maiden name. And she was the daughter of Benjamin Silliman, whose statue is just down past Nate out there, as you see him on the way up. I think his snow necklace is gone this week. So Dana was well married. That’s just to give you a little local color.

But electronegativity in the sense we use it. What sense would Dana have been involved in electronegativity? In 1837 what idea was going around where the negativity or positivity of atoms would be important?

Student: Dualism.

Professor Michael McBride: Dualism, right. The idea of what held things together. He thought if you could understand what was negative and what was positive then you could understand minerals in that case.

But electronegativity in the sense of organic chemistry, or inorganic chemistry for that matter, was popularized by Linus Pauling, beginning with this paper in 1933–or 1932 it was published. So this is a key passage in that paper. And remember that Pauling was pushing this idea of resonance–that you get stabilization when you can draw different resonance structures.

So he’s talking about the additivity of the energies of normal covalent bonds. He says, “There’s empirical evidence in support of the postulate that the energies of normal covalent bonds are additive.” Now what does he mean by additive? That is, that if you have atoms A and B, you could have AA, you could have AB, BB, and you could have AB. And his idea is that the bond strength in AB should be the average of what you have in AA and what you have in BB.

So he was a little confusing about that, because in some papers he used this concept using the arithmetic mean, but in others he talked about the geometric mean where it was the square root of the product instead of half the sum. But at any rate, the idea was that normal covalent bonds should be the average of the homonuclear bonds.

So he uses, then, this plot where he’s measuring the energies of the bonds, bond disassociation energies, in unusual units for us, electron volts. You multiply by 23 to get kilocalories per mole. 

So here are H-X bonds that are normal, that is, calculated if they were normal. They’re the average of the H-H and the X-X for the different halogens, as you see, fluorine, chlorine, bromine and iodine. Now incidentally, there’s a little problem here because he was using the wrong value for the fluorine-fluorine bond energy in all of this work. He thought it was 65 kilocalories per mole; actually it’s 38. It’s a rather weak bond because of all these electron-pair repulsions that we talked about.

So his numbers are a little bit skewed, but at any rate, that was what he was doing. So this is theoretical. To get these red numbers he used H-H and X-X and just took their average.

Now observed values are this; the bonds are actually stronger than you expect. But they’re not all stronger by the same amount. And the reason is in Pauling’s theory, this resonance theory, that you could have the polar resonance structure, which makes it more stable.

What this suggests is that there’s this difference, ∆, between what you would expect if it were normal, the average, and what you observe, which is what you have when you have this polar resonance structure. And what he says here, the greater the ionic character of the bond the greater will be the value of ∆.

So the neat idea now is we could use ∆, which is based on experimental numbers, as a way to measure how important resonance is. So you could measure resonance stabilization by ∆, if you adopt his assumptions that normal bonds are the average, and now you measure these and the difference tells you how important resonance stabilization is. And then you try to see if there’s something you can correlate to predict how important resonance stabilization should be.

So he has his rather complicated Table 2 in the paper, which has observed bond energies, normal bond energies, ∆, and ∆^1/2. Now let’s show how that works by taking one particular block here.

So we’re comparing oxygen and fluorine. So here on the table we look at oxygen-oxygen, and its bond dissociation energy is 1.49 electron volts. Or you can look over here and fluorine-fluorine is 2.80. So if the bond were normal the oxygen-fluorine bond would be the average of those. Although remember, actually this is 1.65, it’s not 2.80. But put that aside for now.

Then if you look at oxygen with fluorine, the bond disassociation energy observed is 2.48. Now we could compare that. Notice it’s not the average of 1.49 and 2.80. It’s too high. So we could calculate what the normal would be, what the average is of O-O and F-F. And then we can take the difference. The difference is 0.33–that’s ∆. That’s how much, in his view, the O-F bond has been stabilized because of the resonance structure.

Now is there a way to predict how big that ∆ should be? Well, here’s the same deal with carbon and fluorine. And notice that the ∆ for carbon and fluorine is 2.20–much bigger than between oxygen and fluorine, which was only 0.33. Or you could look in a different block and see what it is for carbon and oxygen, it’s 1.0. 

So we know carbon to oxygen how much it is, we know carbon to fluorine how much it is, and we know oxygen to fluorine how big it is. Is there some relationship among these that we could–knowing two of them we could have predicted the third. Well, not really, because the difference between C and F and minus difference between C and O is not the difference between O and F. It doesn’t work. You can’t add up the polarity differences that way.

But then he tried something else. He tried taking the square root of ∆. That’s the brown numbers here. So this is a square root of those stabilizations. And now you see something interesting. That the difference between C-O and C-F is the difference between O and F, pretty close. So the square roots are about additive. Now I don’t think he had any good reason theoretically to say it should be the square root. But he fiddled with the numbers and he noticed that worked.

So the question is does it work for all the elements? If so, then from some you can predict the others. So here he has the ∆^1/2 for carbon versus other things. That is, how important is the resonance structure, the square root of the stabilization of the bond. How important is that for C-H, for C-N, for C-O, for C-F, for C-Cl, for C-Br? And can we use that to establish a scale of how different the polarities are among those different atoms?

So here he has carbon, and here’s hydrogen on one side, then bromine, chlorine and nitrogen, oxygen and fluorine on the other side. And the distances there are those numbers we just saw on the previous slide. How important after taking the square root is that energy difference, that stabilization. So we did it with carbon. So you get some values.

But if those values are fundamental, then you should get the same scale no matter what atom you used. You don’t have to do carbon-X, you could do nitrogen-X or oxygen-X or anything else-X, and you should always get the same scale, if this system is going to work. So that’s these other horizontal lines.

Notice that in the case of carbon the difference between carbon and oxygen is 1, the difference between carbon and fluorine is 1.48, and then the difference between oxygen and fluorine is 0.48. And experimentally, if you measure relative to oxygen, oxygen to fluorine is 0.58, not bad. So here are the others compared to oxygen for the O-X bonds, and here they are relative to hydrogen.

And notice that as you come down with oxygen, fluorine, chlorine, nitrogen, bromine, carbon, they’re sort of the same. It’s the same order and they’re about the same. So this isn’t the last word in theory. But it looks like something pretty fundamental.

And then here he used relative to both H on one end and F on the other. So he called these things, this scale, electronegativity. And here he gives the increasing electronegativity, which is just those numbers, the square root of the stabilization of the bond from the previous slide, and here are the values for hydrogen, phosphorous, iodine, sulfur, carbon and so on. And these were the days when they still did hand-written figures in the Journal of the American Chemical Society, which was a long time ago. 

And then below that, he looks at–I couldn’t figure out for a long time what the heck this was, but finally I figured out–he doesn’t say in the paper–but it’s a picture of the periodic table, which has been distorted so that it shows every atom at its proper electronegativity. 

So here are the halogens, here are the chalcogenides, nitrogen, phosphorous, carbon and so on. In other words, he’s taken the periodic table and he’s distorted it like that so that the numbers go from red to yellow, as you go from left to right.

So that’s where electronegativity in the current sense–the Pauling scale of electronegativity, that’s where it came from. And notice that what it really depends on, is what the periodic table depends on, is that as you go across here, you have the nuclear charge is increasing. The screened nuclear charge–we talked about that last semester. And as you go down, the number of nodes in the valence shell is increasing. That’s what gives the periodic table its character, that’s what explains electronegativities. 

So the idea of this was to push the concept of resonance. Now is it surprising that we find that bond strength correlates with differences in Pauling’s electronegativities? Remember, χ_P is Pauling’s electronegativity values. 

Remember how we got into this question? We said why are these polar bonds like N-Cl unusually weak compared to bonds that don’t have that difference in electronegativity? And we said it was repulsion of electron pairs, but some people said it’s electronegativity. 

Now I ask you, is it surprising here that bond strength correlates with Pauling’s electronegativity scale? Has anybody have an opinion on that after what I have been saying? I’ve gone through this pretty fast, so I don’t expect you to, but I was impressed by the question I got after last lecture, so I thought I’d give it a try here. Nate? 

Student: No, because you used the bond strengths to determine–

Professor Michael McBride: Yeah, no. Because that’s where the electronegativities came from in Pauling. Remember, it’s the square root of the bond strength stabilization that he used to get the electronegativity scale. So it’s not surprising at all. Nate, you’re exactly right. The Pauling electronegativity scale was defined by the differences in bond strengths from the averages.

Now in the first issue of Nature this year–this year has been proclaimed the International Year of Chemistry, so you’re sitting in the right class here. So the first issue of Nature was celebrating chemistry and they had a bunch of essays. And one of them on pages 26 to 28 was by Philip Ball, and it was called “Beyond The Bond.” Have we moved beyond bonds now–can we just use quantum mechanics to talk about things?

Remember, this was the question we asked at the end of last semester. If the history of chemistry hadn’t developed the way that it did in terms of bonds, and we just had quantum mechanics, would anybody ever have talked about bonds, if you could just calculate all these numbers? And the little squib about this article said, “More than ever before, new techniques show the bonds to be a convenient fiction, albeit one that holds the field of chemistry together, finds Philip Ball.” 

But one passage in that I thought was particularly timely. He said, “By the 1960s, for all Pauling’s salesmanship, MO theory was generally agreed to be more convenient than his resonance theory for most purposes.”

So when I was a student, resonance theory was a really big thing and molecular orbital theory was just coming on. But it’s really different now. Now molecular orbital theory is what people mostly focus on, and only certain communities tend to talk about resonance. But elementary teaching tends to lag behind where people are. So resonance is still a big thing in AP courses I suspect, and so on. 

But any rate, it’s not surprising really. Certainly it’s not surprising as you correlate with Pauling’s scale, because that’s what he used to define bond strengths. But the idea of polar bonds having high bond dissociation energy in terms of what we talked about first, how far the electrons have to go up when you break the bond to get two free radicals, we expect that energy mismatch will strengthen bonds.

So accrued correlation of the Pauling electronegativity scale plotted vertically with other measures of how tightly electrons are held, the orbital energy. For example, how much energy it takes to pull an electron away from an atom. Or how much energy does it take to pull an electron away from a charged ion. Both of those things are measures of orbital energy. And that’s what’s called Mulliken’s electronegativity scale, which he published two years later in 1934, the average of ionization potential and electron affinity.

So this one, Mulliken was one of the chief originators of molecular orbital theory, which in the middle of the century was in real competition. It was a little bit like radical theory and type theory 100 years earlier. There was a lot of competition between them. But Mulliken’s theory, which has to do with orbital energy we can understand, and we see that it correlates more or less well with Pauling’s electronegativity scale. So fine. So electronegativity does have something to do with it, but electronegativity really is not sort of a fundamental property. It was measured in terms of these bond stabilizations.

OK, that’s just a little bit of a sidelight to what we’re actually talking about. The chemistry and physics of alkyl halides we’re going to talk about now.

Chapter 4. Alkyl Halides – Electrostatics, Non-bonded Interactions, and Solvent Properties [00:27:04]

Free-radical halogenation introduces a functional group into alkanes. Normal alkanes don’t have a functional group, unusually high HOMO or low LUMO. Once you’ve done a free-radical reaction on those, now you have this unusually low LUMO, σ*, so you could begin doing things with it. So what we’re going to talk about now, having talked about free-radical reactions, is the ionic chemistry involving that σ*. So we’re going to talk about the alkyl halides, which now we know how to prepare.

So these are nucleophilic substitution, and acidity is much the same thing. We mentioned this preliminarily last semester. So there are non-bonded interactions and solvation, which is key for ionic reactions.

And I just noticed that all the work I did since last night is not in here because I’ve re-done that slide. So the question is am I going to put that in here or not? It would take about two seconds–well, it’ll take about a minute to break here and start again with the most recent version, which I can’t sacrifice all the work I did since last night. I stayed up late. So I’m going to save this. Pardon me. Because by mistake this morning I loaded the version that was on the web. OK, I’ve got that saved now, nearly. Oh, it’s saving it over the web, that’s the problem. I hadn’t reckoned on that one.

Are there any questions while we’re pausing for this?

Just incidentally, how much did you learn about electronegativity in high school? Was this a big focus of your AP course? 

Student: Yeah.

Professor Michael McBride: So now you know where it came from–it came from Dana down the hill, and then Pauling. 

There we go now. I want the other version that’s here I think.

So I’m going to do that last slide again because that’s where I noticed things were bad.

[here is the end of the interruption]

So free-radical halogenation introduces a functional group. And we’re going to now be looking at that functional group, the chemical and physical properties of alkyl halides, which involve σ*, the functional group we’ve been able to put in. So we have S_N1 reactions at which those electrons go onto X in order to break the bond. And we have those in which the new electrons have gone into σ* in order to help the X leave, and that’s the S_N2 reaction. And if the atom in the middle is H instead of R, then this same thing is acidity. We introduced this concept last semester, but we’re going to talk about it in more detail in terms of reaction mechanisms now.

Now in the process, this makes ions. So we have to go beyond the molecules themselves in thinking about this reaction. Because non-bonded interactions and solvation turns out to be key for ionic reactions. You can’t just look at the reacting species; you have to look at the surroundings, which makes everything ever so much more complicated.

Now this Henri Victor Regnault, who lived, as you see, in the middle of nineteenth century. And you know the first letter of his last name because it said that that’s where they R comes from, it stands for Regnault because he did a lot of thermodynamic stuff. So PV=nRT is this man.

Now in 1850 he wrote a book, The First Elements of Chemistry, and there’s one passage about chemical affinity here that I think is worth looking at. He says, “For chemical affinity to act freely, substances must be dispersed.”

Now this was, of course, writing in 1850 this was before bonds or anything. But they knew you don’t you just mix solids together; you have to disperse them. And you can’t just grind them. 

“Since dispersion by mechanical pulverization is incomplete,” the molecules have to touch molecules, not just particles of stuff touch particles. “They must be taken into the liquid or gaseous state. Previous chemists expressed this fact by saying, ‘Corpora non agunt, nisi soluta.’” Substances do not react unless they’re dissolved. So this dissolution, the fact of what’s around the things is what’s going to be important in understanding reactions.

So the theory of organic chemistry became manageable in the first place when people started studying gases, as you remember. Because it’s often possible to focus on the simple unit with strong interactions, that’s molecules are the individual units, and bonds within them are the strong interactions with well-defined geometries and energies, as we discussed last semester. And you neglect the much more numerous and complex intermolecular interactions with the solvent. But actually we’re going to have to consider those, because we’re interested not just in the gas phase but in solution.

So it was essential to go into the gas phase in order to get a toehold on understanding chemistry in the first place. But if you want to understand real practical chemistry you have to understand solutions.

So these weak intermolecular interactions give organic materials many of their most valuable properties. Most of the properties of what’s important in your bodies are not things that involve making and breaking bonds, but changing non-bonded interactions between things. And that’s what we’re going to be looking at for a little while now. So non-bonded classical energies.

Now suppose you have just Coulomb’s law, charge-charge interaction–that’s where we started last semester. That’s, as you know, the only true source of chemical potential energy. There’s gravity and physics of magnetism, but those are tiny. They’re nothing compared to Coulomb’s law. There’s the strong force–that’s for physicists, that’s way above what we deal with in chemistry. The Coulomb’s law is where we live. And it’s proportional to 1/R as we know. And the constant for the proportionality is –300, 332 kilocalories per mole when you use angstrom units.

So this provides a very long-range attraction. Even when you’re 3 Å apart, the energy is still 100 kilocalories per mole. Actually, now where should we start with zero? We could start with zero when they’re infinitely far away, and then we use Coulomb’s law to get how the energy goes down. Or if we’re thinking of breaking the bond we could start at the bond distance, or the distance between two ions, and see how much it takes to pull them apart a certain distance.

So a bond is 1.5, let’s take bigger. Let’s suppose they’re 2 Å apart. That means they’re at 332 over 2. The energy is –165. If we go to 3 Å then it’s –100. So we’ve gone up in energy by about 100 kilocalories per mole in going from 2 to 3 Å. And we have a lot further to go on the way out. 

This is very different from covalent bonds where when you get to 3 Å that’s the end of the line. So covalent bonds are typically 100 kilocalories per mole, but you get there very quickly. So they’re very stiff. As you break the bond, you go up in energy really quickly.

Coulomb doesn’t go up as rapidly, but it goes up very, very far. So it’s long range. And that’s why solvents are so important, because another molecule can be rather far away and still influence ions where it wouldn’t influence free-radicals.

But this is attenuated by a solvent. If you have a dielectric constant, as it’s called, in the solvent, that reduces the magnitude of the Coulombic energy or the Coulombic force.

Now how big are these values of ε? Here are some values taken from the Jones book. Notice that hexane is a factor of 2. So that’ll take that constant from 300 down to 150–still quite big. But if you go to something like water at 78, that cuts it way down. This into this is something like 4. So it’s only 4 kilocalories per mole rather than 332 kilocalories per mole, if you’re operating in water, with water all around the ions. So that’s makes an enormous difference how these ions are going to interact, what solvent you’re in. It makes quite a difference what solvent.

And notice down to acetone, the solvents might be considered polar–that is, they have pretty big dielectric constants. So they really attenuate the energy of interaction between the ions. But chloroform, ether, and hexane are not very polar; they don’t cut it down very much. 

Now this has to do with things like the solubility or the melting point of salts like sodium chloride, or tetramethylammonium iodide. Because notice that these ions are rather small, so the charges are close together. But these ions are big, so they’re substantially further apart. And that’s going to make an enormous difference when you’re starting to pull them apart. If you go to twice the distance here, you really cut this way down.

So sodium fluoride has a melting point of something like 801 ºC, but cesium iodide, the same charges with bigger ions, is 200 degrees lower in its melting point. And the solubilities reflect the same thing. It’s easier to get the solids apart if you’re starting with bigger ions.

So the big interaction is charge-charge, but there are others that are important, too. Charge with dipole, so if you have a molecule that’s not charged but has a dissymmetry in the charge that’s within it, so it has a plus and a minus end, then those can interact, and that’s proportional to 1/R², so it dies away more rapidly.

And if the dipole can orient so that you have to average over different ways it could be oriented, it would be oriented more likely in the favorable than in the unfavorable way, but that depends on the temperature. And if you have that situation it turns out to be R^–4 over temperature, because of how much is favored one way or the other. 

Then if you go to a charge with something that’s non-polar, but could become polar if it’s put in the field of another charge, so you push the nucleus one way, the electrons the other way. And now you get a charge interacting with an induced dipole. That depends on R^–3–1/R³. Then you could have two dipoles interacting with one another–that’s also 1/R³. Or you could have two induced things that are acting with one another.

Have you ever seen that before? How the electrons are distributed in this one? When they’re over here–the ones here tend to be over there, and when these are here, these are here. Where did we encounter that before?

Student: Correlation.

Professor Michael McBride: Correlation energies, right.

So that turns out to be R to the [minus] sixth power. So notice that these others, other than the Coulomb, they fall off very, very rapidly. And they start very weak. So you might think you could neglect them, but there could be a lot of them. That’s the problem when there are many solvent molecules around.

Now let’s look at some trends in the halides. And whatever textbook you have you’ll find this–When we talk about different functional groups we’re talking about the alkyl halides now. Typically, chapters in all the books you look at will start with something about the physical properties of those compounds and then go on to their chemical properties. So this is a little bit about physical properties of the halides.

So here we look at the van der Walls radius of the halogen, X. And here we look at the bond distance between X and carbon. And you notice they’re almost exactly the same. And we learn something from that, which is that non-bonded distances are something like twice-bonded distances. Because the radius of an atom is about the same as the distance from the nucleus of the atom to the nucleus of the carbon. So that means that the covalent radius is about half of the van der Waals radius. 

Now here’s a property, the dipole moment. How polar is the molecule? The dipole moment shown here, what do you notice about that as you go fluorine, chlorine, bromine, iodine? What’s the first thing you notice about the trend?

Student: It doesn’t work for H.

Professor Michael McBride: Derek?

Student: There’s nothing with H?

Professor Michael McBride: Well, I didn’t put it. If you have H, it turns out there is no dipole moment because it’s with methyl. So H-CH₃ is a symmetrical molecule, and the center of the electrons is the same as the center of the nuclei.

But the red is a trend as you go–well, forget hydrogen because it’s not really a halogen. But here’s the radius, the bond distance. What’s special about this one, about the orange one? Lauren?

Student: It goes the opposite way.

Professor Michael McBride: Pardon me?

Student: It goes the opposite way. It goes down.

Professor Michael McBride: It goes up and down, it’s not monotonic. What do you learn when you see trends that are not monotonic? Matt? 

Student: It’s dependent on two factors.

Professor Michael McBride: There must be two things going on. But what’s the other thing? We’re interested in how much charge is separated. Here we have CH₃ and a halogen. So the dipole moment is how big the charges are and how much they’re separated.

Now we already see here how much they’re separated, that is, what the distance is. But how big is the charge? So if we look here–this is incidentally measured in debye units, which is 4.8 times the charge times the separation. Now we know the separation and we notice that it’s not monotonic. And here’s the charge, how many electrons have been transferred. And that’s quite monotonic.

But what we’ve seen here is that the dipole moment is the product of two properties–how much charge and how much separated. But they’re opposing trends. As one goes up the other goes down. And although they’re both monotonic, one is non-linear, more non-linear than the other. So they don’t exactly cancel one another in going up and down and you get weird behavior when you have conflicting non-linear trends. So there were two things going on to give the dipole moment.

Now you can see a similar kind of thing in another measure of the size of atoms, which is so-called A-value. That’s how much is axial compared to equatorial in cyclohexane. Axial is more crowded than equatorial is–we talked about this last semester. So if we know what the energy difference is between axial and equatorial, we have some measure of how big that group is. So it’s E_axial – E_equatorial in kilocalories per mole, so that measures substituent size.

Now here are the values for the halogens, and hydrogen incidentally, which is defined as zero because you’re comparing hydrogen with hydrogen. So here it is. And again, you see it’s not monotonic–it goes up and down. So there must be two things involved. What are the two things? 

One of the things is how big the van der Walls radius of the atom is, which clearly gets bigger as you go fluorine, chlorine, bromine, iodine. But why does the effect on axial versus equatorial go up and down. Why doesn’t it just go up? 

Student: The bond distance between the carbon and the substituent. 

Professor Michael McBride: Because the bond gets longer as the atom gets bigger. So it’s not monotonic, that suggests there’s a competition. So here’s fluorine–that’s what’s shown here. And now watch this model as I change from fluorine to iodine. What happened? The atom gets bigger, not shown to scale, but the bond gets longer. So you have a larger van der Waals radius, but it stands off further, so it’s not so dramatic, the difference.

Now compare that if the group there is CH₃. Because CH₃ has this effective van der Waals radius. So that’s how I decided where to put it on this scale. It’s got the right van der Waals radius. But the bond distance for carbon-carbon is much shorter than it is for bromine and iodine or chlorine for that–or it’s about the same as fluorine or a little bit bigger. And now when we put these two things together in order to get the A-value–we got this big group and it’s come in some, compared to the halogens.

Now you see there’s a really wild outlier in this thing because you now have a big group, the CH₃, and it’s short. So even though the radius of methyl is between bromine and iodine, it’s effect on this is much bigger because it sits in closer.

Now the boiling points of the alkyl halides. Here’s a table–we could have taken it from any book, this particular one is from the Jones text–that shows different alkyl groups and the hydrogen, fluorine, chlorine, bromine, iodine, what are the boiling points?

Now let’s just see if we can understand this. I’m going to go just a little bit further because we started late with people getting here late, and we had that little break in the class.

So let’s look at the boiling point. So methane, methyl with hydrogen, is very low boiling, and then it gets higher and higher and higher. That is, the liquid tends to hold together more, not to go into the gas phase–the molecules are holding together more.

Now the question is why. Why when we replace hydrogen by fluorine, by chlorine, by bromine, by iodine does the boiling point go up? What causes the molecules to hold together?

Student: Intermolecular forces.

Professor Michael McBride: It’s not covalent bonding. We’re talking about non-bonded interactions now. What could make them hold together? We know the fundamental forces–it’s Coulomb’s law as colored by dipoles and things like that.

Student: Hydrogen bond? 

Professor Michael McBride: Taka, pardon me?

Student: Hydrogen bond?

Professor Michael McBride: Hydrogen bond is an idea. Now you’re not going to get hydrogen bonds when you don’t have hydrogen. So the halogens that’s not going to work for. And the next thing we’re going to talk about is hydrogen bonds–that’ll come in the next class now.

Any other ideas of what holds them together? Nate?

Student: Polar interactions and dispersion forces?

Professor Michael McBride: Ah, you can have polar interactions. So you could have dipoles interacting, even though there are no charges.

So let’s look at the dipole moment, which was on that previous table. And you see that methane has zero, but fluorine has 1.85. So that’s going to allow the molecules to hold together. So raise the boiling point because you could have them lined up this way.

But it’s not just polarity. Because if we go across here, the polarity went up, but then it goes down again. But the boiling point keeps going up. So there must be something else other than these dipoles holding together.

And that other thing is polarizabilty. How easy it is to shift the electrons in the molecule, and how many electrons there are to shift. So you see if you look at fluorine, chlorine, bromine and iodine atoms, fluorine atom is very non-polarizable. When you bring a charge or a dipole near it, it doesn’t distort very much.

But chlorine distorts more, bromine, and iodine the most because it has the most electrons, the most weakly held. And CH₄ is not polar, and also it’s not very polarizable. The hydrocarbon is not very polarizable given that it’s five atoms and these are just one atom. 

Next time we’ll start at this point and go in and try to understand the boiling point of the alkanes without the halogen on them. Thanks for your patience with changing horses in midstream.

[end of transcript]

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