CHEM 125b: Freshman Organic Chemistry II

Lecture 21

 - Functional Groups and Fingerprints in IR Spectroscopy; Precession of Magnetic Nuclei


Infrared spectroscopy provides information for analyzing molecular structure and for understanding bonding and dynamics. Although the normal modes of alkanes involve complex coordinated vibration of many atoms, the unusual strengths of multiple bonds give alkenes and alkynes distinctive stretching frequencies.  The intensity of characteristic out-of-plane C-H bending peaks allows assignment of alkene configuration. Characteristic carbonyl stretching peaks in various functional groups demonstrate the importance of pi- and sigma-conjugation.  The complex fingerprint region of IR spectra differentiates the subtle isomerism of polymorphic crystalline pharmaceuticals. A 90° phase lag between force and velocity explains the precession of tops and of magnetic nuclei in a magnetic field. Nuclear precession in the combination of a stationary magnet and a pulsed radio-frequency field can be visualized by means of the “rotating frame.”

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Freshman Organic Chemistry II

CHEM 125b - Lecture 21 - Functional Groups and Fingerprints in IR Spectroscopy; Precession of Magnetic Nuclei

Chapter 1. IR Frequencies of Alkanes, Alkenes, and Alkynes [00:00:00]

Professor J. Michael McBride: OK, so you have the exams back. I’ll have a good answer key for you soon. Currently, the answer key just has which slides of the presentation were the subject of each question. Maybe good news is that we’re moving on to a new topic. So it’s spectroscopy. And last time we started talking about IR spectroscopy, where the general idea, you remember, is that molecules vibrate. If they change their dipole moment as they vibrate, they can interact with light. That is, the electric vector of light can cause them to vibrate at a higher amplitude, but always at the same frequency.

And it’s important, as in pushing a child on a swing, that you push at the right time. If you just go whang, whang, whang, which some little kids do, the person won’t go on the swing. You have to push just when they’re coming back. So the frequency of the light has to be exactly the same as the natural frequency with which the molecule vibrates. But the molecule can vibrate in many ways, and these are called normal modes, and we were beginning to illustrate those last time.

We illustrated it with butane, but showed the spectrum of octane, which has the same features. And last time we saw that there were lots of degrees of freedom, even in butane, many more in octane. There are 36 normal modes there, but 72 in octane. And we looked at what some of them were. That that one was the coordinated stretching of the C-H bonds, as indeed with the field pushing up and down. And the next one, that blue one there or one of those peaks, is the hydrogens moving coordinatedly in and out in a normal mode, where all vibrate at exactly the same frequency. And that’s the frequency of that light. And that one, notice, the electric vector goes in and out of the board rather than up and down. So those were C-H stretches.

There are other C-H stretches like this one, which are sort of a breathing of the hydrogens, which don’t change the dipole moment. Therefore light has no handle on it. And that doesn’t appear in the IR spectrum. There is another kind of spectroscopy, called Raman spectroscopy, which can see that kind of vibration. And in fact, of the 10 C-H stretches that combine to give 10 normal modes involving C-H stretching, half of them have no handle, and you can’t see them in the IR.

But we can go on to other peaks. Now, those were C-H stretching at very high frequency. Now we’re at high low frequency, the high end of the low frequency peaks. And that particular peak is this motion of the hydrogens. What’s involved here? If you had to name that kind of a motion of a CH2 group, that you see in this one, what would you call it?

Student: Scissors.

Professor J. Michael McBride: Exactly, it scissors. Then this one– whoops, I went too far. Whoops, sorry. There we go. Notice the motion is almost entirely in the methyl groups at the end, not within the chain. It’s a bending. It’s, in fact, similar to the scissors motion of the CH2 groups, but this one is called something else. Could you give it a fanciful name, of what one methyl group is doing? It’s called the umbrella mode.

This little peak here turns out to be this motion. See how they’re wagging back and forth and any given CH2 is going like this? Sorry, there we go. That’s the wag.

And this last peak down here is this one. So the CH2’s are going like this, and that’s the rocking mode. So bear in mind that we’re illustrating with butane but showing octane, but they have these same motions. Obviously, octane has those same motions as butane does. And they’re all coordinated together at the same frequency, so it’s a normal mode.

And obviously, the ones that appear in the spectrum are IR active. So now, how about functional groups? Here we were looking just at a saturated alkane. What IR is especially good for is identifying functional groups. So that was octane, and here’s octyne. Incidentally, what does -yne mean as a suffix?

Student: A triple bond.

Professor J. Michael McBride: It means a triple bond. And methylacetylene is methyl on a triple bond, not methyl on a double bond. That gave some people problems on the exam. OK, so most of the peaks are the same. In fact, there’s one peak here, that one, which is almost precisely the same. Do you remember what that one was in octane? Well, we’ll see in just a second.

And now, look at– that was 4-octyne, and now if we look at 1-octyne– a triple bond in a different position at the last two carbons– then we see exactly that same peak again. So all these things, octane, 4-octyne, 1-octyne, all have that sharp peak that’s circled there. And it’s this motion. It’s that same umbrella motion. And why is it so special? Why does it occur at exactly the same place in all of these? Because it has its own frequency, a certain frequency, different from the frequencies of the neighboring CH2’s. So it doesn’t couple with them. It’s a bad energy-match for that frequency with the neighbors, so it doesn’t couple very much. So it stands alone, and is the same in octane, in octyne, also in octene, as you’ll see shortly. OK, so that’s that one.

Now, there are some other peaks here that are new. Like, there’s one at really high frequency, 3315. Now what’s that? Here’s the molecule, and watch its motion. What’s mostly happening? And why is it different from anything you saw in octane? At higher frequency. First, what’s moving mostly? Ayesha?

Student: The C-H bond.

Professor J. Michael McBride: The C-H bond is stretching. And why is the stretching of this C-H bond different from the stretching of the bonds we saw in octane? Why is it a different frequency, so that it doesn’t mix up with other things? Bad energy-match, it doesn’t mix. Ayesha?

Student: It’s in the plane…

Professor J. Michael McBride: Can’t hear very well.

Student: It’s in the plane of the molecule

Professor J. Michael McBride: It’s true that it’s in the plane. In fact, it’s in the line in this case. All the atoms except for three hydrogens are on a line. But what’s special about that C-H bond as compared to others? We’ve seen things special about an acetylene C-H bond before in its chemistry.

Student: It’s sp hybridized.

Professor J. Michael McBride: It’s sp hybridized. Remember, it’s acidic. It’s unusually acidic. 25 is its pKa, because it’s a C-H bond, so it’s much stronger than the others. So it’s at higher frequency, as you’d expect for a stronger spring.

OK, now, at the other end of the spectrum, there’s this really strong peak at 630, and that’s this motion. It’s bending the hydrogen up and down, bending that C-H bond. And that turns out to be much easier than bending bonds that are normally hybridized.

So that’s that bending. So that’s characteristic of having an acetylene in a molecule. If it’s a terminal acetylene at the end of the chain, you’ll have these. If it’s not at the end of the chain, then you don’t have this kind of C-H.

Now, what’s this peak here at 2120? Give me some thoughts about what that might be. What kind of peaks come at the far left of the spectrum, at very high frequency?

Student: C-H .

Professor J. Michael McBride: C-H’s, because the H is so light. And then when you get bendings of C-H’s or stretchings of C-C, those come in the right half of the spectrum. So this particular peak at 2120 is in what’s a window usually. Usually, there’s nothing… There must be something really special that vibrates at that frequency. Any idea of what could make something… It’s not a C-H bond. It must be a C-C bond, but it’s a very special C-C bond.

Student: Triple bond.

Professor J. Michael McBride: What is it that would vibrate at very high frequency? Derek?

Student: The triple bond

Professor J. Michael McBride: The triple bond is three times as strong as a single bond, so it comes at unusually high frequency for a CC bond, and here is its motion. Whoops. So it’s a CC bond stretching.

Now, notice that in the spectrum of 4-octyne, the blue spectrum, you don’t have that peak. Why don’t you see the triple bond stretching in 4-octyne? How many carbons are there in octyne?

Student: They cancel out.

Professor J. Michael McBride: Right. There’s no change in the dipole moment, because it’s symmetrical as you move like that. If it’s at the end of the chain, it’s unsymmetrical, and you can see it. But if it’s in the middle of the chain, it doesn’t change the dipole moment.

OK, so you can tell the difference in two ways here, between a triple bond that’s at the end of the chain and a triple bond that’s in the middle of the chain. OK, so there’s no handle for the light to interact with that one.

So we’ve seen octane. We’ve seen octynes, two octynes, the 4- and the 1-. Now let’s look at octene, a different functional group. So again, most of the peaks look exactly the same, because they have long chains of CH2. But it’s trans-4-octene, so there’s a carbon-carbon double bond.

Now, there’s this big peak at 967. Now what could that be? Now here’s cis. That was trans-4-octene, here’s cis-4-octene. And now you see a peak at 1655. What do you suppose that is? It’s not involved in an H. It’s very high frequency, but it doesn’t involve H. Those are up around 3000. Any ideas for this one?

Student: The double bond stretching.  

Professor J. Michael McBride: The double bond stretching. Now, why do you see it in cis-4-octene, but you don’t see it in trans-4-octene? So that stretching can be symmetrical, right? And if it’s a trans double bond, like this, then it truly is symmetrical. There’s a center of symmetry there. But if it’s cis-4-octene, then if you stretch the double bond, and change the bonds from carbon to carbon, you change the dipole moment in this direction, not in this direction.

So if you have it cis like that, then the dipole change is actually in a different direction. Of course, normally you measure these spectra on solutions, so there’ll be some molecules that are oriented properly. But if you had molecules that were fixed, like in a crystal, you could tell which way the double bonds were oriented, by seeing which direction the electric vector could interact with it.

But notice that down at the low-frequency end, that has a big peak at 710, whereas the trans isomer had 967. Now, what was the really low-frequency peak in acetylene, remember, in the 1-octyne? Remember the one was really down low?

Student: The bending. 

Professor J. Michael McBride: That was the bending of the C-H. So these are down low, they could be bending of the C-H.

And let’s, just to compare, look at a double bond that has three carbons on it, so just one H that could bend out of plane. And it has a peak at 828. Now, of course, the others that don’t have that extra methyl group are cis and trans, but have two H’s, two H’s cis, or two H’s trans. And they’ll couple as they vibrate, because they’re near one another. So there’s perfect energy-match of these two.

So if you have any overlap, any coupling mechanism, mechanical coupling, if one can feel the other vibrating, then you should get two vibrations, coupled vibrations. That is, if we look at it this way, you have the 828, which is a single H, but if you had two of them at 828, and they could interact, you’d get a higher peak and a lower peak, which looks like what we see, in a way. But the higher peak comes in one compound, and the lower peak comes in another one. Where is the second of those two peaks that got mixed together? We could turn it that way to see the analogue to orbital mixing.

So let’s look at pictures of these vibrations. At the top, you see, is the molecule that has just one hydrogen. At the bottom, on the left, are two hydrogens on the double bond trans, and on the right, they’re cis to one another. Now, this is the vibration we’re looking at, out-of-plane vibration of the C-H. Everybody see what I’m talking about? It’s out-of-plane vibration of the C-H at the top.

Now, at the bottom, there are two CH’s. And notice that in this vibration, those two H’s move in the same direction, both up and both down at the same time. Why don’t I show the one where one moves up and the other moves down? Why isn’t that relevant for our discussion? Because there’s no dipole moment for that one. It cancels. One went up, the other went down. So to be IR active, to interact with the light, they have to move the same direction.

But notice, this is interesting, so they move in the same direction to be active. That one is at 828. This one, notice that when they both move up, it twists the bond. See how it twists the center carbon bond when the two hydrogens move up. But on the right… so that reduces the π overlap. It weakens the π bond when those p orbitals don’t overlap as well… But on the right, when they both move the same direction, the p orbitals still point so that they overlap with one another.

So the one on the left is harder to do because it weakens the π bond. It distorts the π bond, whereas the one on the right doesn’t distort the π bond. So the one on the left is harder to do and it’s higher frequency. So folding preserves the π overlap. The harder one and the easier one. The harder one is high frequency, the easier one is low frequency. So you can understand why you see only– when these two vibrations mix– why you see only one. You only see the one where they move in parallel. And it’s easier if they’re like this, than if they’re like this, where you’re twisting the bond as well.

So not only are the high- and low-frequency peak handy, because they diagnose being a cis or a trans double bond, you can tell which one you have, but also you can understand what it tells you about the bonding, why you have one that’s high and the other low.

Chapter 2. IR Frequencies of Carbonyl Groups: the Influence of Conjugation [00:16:43]

OK, but the real jewel in the crown of infrared spectroscopy is the carbonyl group. Why should it be such a great group? Why should it be so good in infrared spectroscopy? Why should it stand out and be something that you can really identify what kind of carbonyl group you have?

First, it absorbs very strongly. It interacts very strongly with light. What determines how strongly something interacts with light? How much you can push the atoms with the electric field of light? Luke, you got an idea?

Student: It’s the number of electrons?

Professor J. Michael McBride: Not just the number of electrons, but how much the dipole moment changes when you stretch. If the electrons change a lot when you stretch it, and it becomes much more plus-minus, if the plus-minus get separated more, then that’s a strong interaction with light. And the C-O is a polar bond, so it changes much more than a C-H bond does as it vibrates. So it’s going to be a very strong peak. It’s not going to be one of these wimpy little peaks.

Now, why does it stand out in its frequency? From one molecule to the next, you don’t have to worry so much about the coupling in interpreting it. A certain ketone, a ketone will always be very near where another ketone is, at the same frequency. Why? Because that bond is a double bond and nearby bonds aren’t double bonds. So the nearby bonds that it could couple its vibration with have very different frequency. So you have very little overlap, so you don’t change the frequency very much when it interacts with its neighbors. So it stands alone and is characteristic of a particular kind of carbonyl, and it’s very strong so that you can see it in the IR spectrum easily.

OK, so that’s why these are so great, the carbonyls. So here’s a ketone– or pardon me, an aldehyde, acetaldehyde. Notice how very strong a peak it is. It’s strong and it’s independent. It doesn’t mix much with its neighbors. Now, there’s a ketone, right? Acetone. Notice it’s about 12 cm-1 lower frequency, but that’s reproducible. The ketones would come at a little higher frequency, aldehydes come at a little lower frequency– or pardon me, vice versa. So you can tell which is which.

Now, here’s an amide, where we put a nitrogen on the carbon-oxygen double bond. Now, that went to a substantially lower frequency, 1681. Why should the C=O bond, double bond of an amide, be weak and vibrate at lower frequency, a weaker spring? Anybody got an idea of how the nitrogen would do that to the C-O double bond? Yigit how about you?

Student: The interaction between the lone pair of nitrogen and the π*–

Professor J. Michael McBride: And π*. So you put electrons from the nitrogen into the antibonding orbital of the C=O, weakening the C=O bond and lowering its vibration frequency. So there’s the unshared pair, we mix them with π*, and it gets to 1681. So the C=O is weakened by resonance, by occupancy of the π*.

OK, now, suppose we look at an ester, where we have an oxygen next to the carbonyl. Can I get someone to predict that for me? Po-Yi, what do you think? Where is the ester going to come, compared to, say, a ketone and the amide, the one with nitrogen?

Student: Well, there will still be resonance.

Professor J. Michael McBride: There’ll still be resonance. It’s an unshared pair on oxygen. Should it be a strong as that involving the nitrogen?

Student: No.

Professor J. Michael McBride: Why? Why should mixing of the electrons into the π* be weaker for oxygen than it is for nitrogen? π*; here are the electrons on the atom we’re interested in, on oxygen or nitrogen.

Student: Bad energy-match?

Professor J. Michael McBride: But oxygen, as you say, has a higher nuclear charge. Lower, it doesn’t mix as much. So it shouldn’t lower the frequency as much as 1681. So I think you would predict that it’d be someplace between 1715 and 1681. Make sense? Yep.

There it is, it’s higher frequency, 1746! So there’s something different. Not only is it not as good as nitrogen, it goes the opposite direction. Now how about if we put a halogen on there, like chlorine? That’s higher still, 1806. It’s true its electron pair is lower, so it won’t mix as much. It won’t weaken it as much as nitrogen did. But why does it strengthen it to have a halogen there?

So here’s a different kind of resonance. You have an unshared pair– on the on the right, we used an unshared pair of the adjacent atom to interact with the carbonyl. Now we’re going to use an unshared pair on the oxygen of the carbonyl to mix with an orbital of the C-X bond. What orbital could that mix with? It could mix with σ*.

So if you mix that with σ*, then you strengthen the C=O bond. You get a triple bond between oxygen and carbon, and occupy σ*. So the appropriate resonance structure would be one with no bond to that X group. Of course, that’s not the dominant structure, but it’s one that causes the bond to get, the C=O bond, to get stronger. So it moves to high frequency. So we can see that that interaction is important.

And in this case, the mixing with the σ*, the C-O is strengthened by resonance. So in the case of amide, it was weakened by resonance. In this case, it’s strengthened by resonance. In that case, it was a π interaction, putting electrons in π*. In this one, it’s putting electrons in σ*, and making a triple bond.

Look at this case here, where we have a C=C double bond. Now we don’t have the low σ*, so we’re not going to get the effect we got in the acid, and the acid chloride and the ester. And we have π electrons that could do the mixing. So that looks like a case where this resonance structure could be significant, and it would weaken the C=O double bond. But the peak is, in fact, at higher frequency than a ketone, 1720. Now, why is this so? Notice that in this case, you have two double bonds separated only by one single bond, so they both have a better energy match, these double bonds, than when you had a double bond and only single bonds in the vicinity.

So now you can mix them. You can make them in-phase and out-of-phase. And here, notice that one is stretching while the other is shrinking. It’s out of phase, and it’s mostly a vibration of the C=O. And that’s the one that’s at high frequency. And notice it’s a double peak, because you have the other one when they’re in phase as well. And here’s the one where it’s in phase and it’s mostly a vibration of the C=C, rather than the C=O. But notice, they’re both stretching at the same time.

So we have this situation. We mix them, and we get 1720 and 1683? Notice there’s this peak at 1618. This is a really neat case, because the doubling that you see in that strong peak isn’t what you think it is. It’s not the in-phase and the out-of-phase. What is the 1618? And what’s the source of the doubling?

Well, here’s a spectrum of methyl vinyl ketone, that same compound with a double bond adjacent to the carbonyl, measured in argon at 13 K. And you can see it’s got those two peaks, 1718, 1696, but then it’s got that 1623. These aren’t exactly the same frequencies we saw in the previous slide, because that was measured in solution, this is measured in an argon solid. So there’s a little influence of the neighbors on the frequency.

But if you photolyze that at 308 nanometers for two and a half hours, and then look at the difference between the spectrum you have now and the spectrum you had before, see that some peaks have gotten stronger, they point up, and those that have gotten weaker, if you take the difference, go down.

So you see that peak at 1696 went down, not up. So what does this mean? It means that there are two different compounds there. One is becoming the other, so the one that’s being formed is getting stronger, the one that’s going away is getting weaker and appears negative when you do the difference. And that’s 1696. So that original doublet we looked at is not from the same compound, those two peaks, they’re from two different compounds. But the 1696, if photolyzed, can become the 1718.

So if you calculate the positions for that compound shown at the top, you see the C=O, and the C=C, so those are the 1718 and 1623. We got that. That’s that compound. So the in-phase and the out-of-phase vibrations. What’s the other one? It’s this. What’s the difference? It’s the conformation around that central bond. They’re both planar, but one is like eclipsed and the other is like anti.

And if you calculate the spectra for the one where the two double bonds are trans to one another, you see a very strong peak at 1696, and hardly anything at the low frequency. Why is there hardly anything at the low frequency? Because that one, which is mostly a C=C vibration and a little bit of coupled C=O, has them so that one tends to cancel the other. The C=O doesn’t move very much but it’s a big dipole, so it can cancel it. And at the top, they reinforce one another, so the mostly C=C vibration is rather strong, because the C=O is helping it out.

So one of these just called syn periplanar. It’s flat, periplanar, but adjacent to one another. And the other is anti periplanar. OK, there’s one more peak there, and I don’t know what that is that went away. It might be a combination of some other freak transitions. That’s something we don’t want to talk about now, it’s too complicated. But anyhow, here’s a picture of these vibrations. We’re in-phase, where that’s the low frequency, both stretching at the same time. But mostly the C=C, the C=C is stretching nine times as much as the C=O.

But here’s the actual amplitude. The actual amplitudes are very subtle for these things. And in fact, we talked about this a little bit last semester, about how, when we were doing quantum mechanics and vibration, how much things actually move. And here’s the out-of-phase normal mode, mostly the C=O vibration, a little bit of C=C. And as one stretches the other shrinks. So the C=O is moving six times as much as the C=C, and the actual amplitude is like that. And if you look at Lecture 8 last semester, you can see where we talked about how vibrational amplitudes are.

Here we were talking about two things: identifying functional groups, characteristic peaks, strong peaks that you can identify with particular functional groups, and you can look them up in tables, as I suspect you did in lab already, to see what functional group would come where. And that’s very valuable. We used it here more to understand how strong bonds are, and resonance and so on.

Chapter 3. IR Fingerprints in Pharmaceutical Characterization [00:29:54]

In the real world, it can enter in a lot of cases. One interesting case is a multibillion dollar pharmaceutical, which crystallizes in different crystal forms, and those forms are separately patented. The drug is called Paxil. These are taken from the patent for that compound, paroxetine hydrochloride. This is the anhydrate, in called Form A, or Polymorph A. Polymorph means different shapes, different crystals. Now, what do we see in this spectrum? We see C-H stretching peaks over at the left, and we see fingerprint regions on the right. All these complicated combinations of things. But is there anything we can identify? Well, here, this sort of big shoulder in here, turns out to be characteristic of ammonium ions, NH+ plus. So we know it’s an ammonium salt from that.

Now, what are these peaks at really high frequency? They’re much higher. Those that are hydrogens on a double bond, sp2, are indeed higher frequency than hydrogens on a single bond, sp3, but they’re not as high as this. And what those turn out to be due to is water. And notice than when water is hydrogen bonded like this, you have OH’s that are at the end of the chain, that have a very high frequency, and OH’s that are involved in hydrogen bonding, which are lower frequency, because they don’t just go one way. It’s not so bad to move toward the other oxygen.

OK, so we see two peaks here, which is interesting because this molecule is called an “anhydrate,” but, in fact, it has water in it. So that was sort of a weird thing about it. OK, so here’s the second form, Form B. And again, it has ammonium. There’s ammonium functionality in there, C-H and fingerprints. And here’s the third form, same deal. Now, notice incidentally, that they had a lot of sample in this one, so a lot of it had zero transmittance. Those C-H peaks didn’t go all the way, weren’t sharp at the bottom, because they just got cut off. No light got through the sample.

OK, now if we look at the fingerprint region, there’s Form C, there’s Form A, and here’s Form B. Now, what you look for, if you’re going to dispute a patent, is the guy who’s violating your patent making your compound? So you try to find peaks in here that are characteristic of your compound, and see whether there’s some of that in what the other guy’s trying to sell. So you look around here and try to figure out where in this forest of things can we find a peak, even if we have no idea what the normal mode is, that we’ll be able to distinguish.

And it turns out that those peaks listed below there in green, blue, and red are ones that occur in the corresponding form and not in the others. And in particular, those peaks there, the 665 and the 675 are very good for distinguishing B and A, right? So there was a patent dispute that involved, can one detect 5% of the B, the 675, in the presence of 95% of the unprotected one? In which case, they would be violating the patent on B, if there were 5% of it there. So this is the legal world and IR being involved in it.

Chapter 4. The Precession of Magnetic Nuclei [00:33:49]

So we’ve seen here spectroscopy used both for structure but also for dynamics, how things are moving. And we saw it first in electronic spectra, where we saw the electrons sloshing up and down at a rate that had to do with the difference in energy of the two orbitals that are involved. We’ve just seen vibrational, and we’re going on to nuclear magnetic resonance, which is at a much lower frequency, radio frequencies, and involves magnetism which we haven’t discussed yet. And it involves precession, which we haven’t discussed yet. But there are problems on the Chem 125 web page that will help you understand what’s involved in precession.

Now, here’s a young Michael Faraday, we showed this picture last year, and he was a chemist, remember? And he discovered benzene in illuminating gas. We talked about that last semester. But that’s not all he discovered. He also discovered magnetic induction, and he invented the idea of magnetic and electric fields. He thought they were real, that there were real little filaments there for these lines of force. And the idea that you can get magnetism from electric current and electricity from changing magnetism. So that’s generators and so on. If you’re interested in seeing things about that, here’s a website at Florida State University Magnet Lab, that tells you things about Faraday and that. And 30 years later then, Maxwell, who knew math, which Faraday didn’t, built these into the comprehensive theory of light and electromagnetism.

Now, precession is involved, and you’ve seen precession before. Here’s a top, and remember the amazing thing about this is that, if you put it on here it doesn’t fall down. Whoops. It stays standing up, and that’s an amazing thing. If it’s not spinning– let me slow it down just a bit. And now, that’s precession, when it goes around like this. Remember how a top goes like that, the axis goes like this. And, of course, if it’s not spinning, gravity just makes it fall down.

So why doesn’t it fall down? Why does it fall around rather than falling down? Now, you can see it better if we use something bigger, so we’ll go to a bicycle wheel here. And I think I’ve got that on the next slide. OK, so we have a bicycle wheel here. I hang it up on this thing. There we go. Whoa, get it hooked. There we go. Now, here I’m holding it up. The strings pulling that way, but its weight, if I let go, is going to make it fall. There’s no big deal there. And that’s shown. The force down of gravity at the red arrow and up on the string puts a twist on it, which makes it move right at the top and left on the bottom. No big deal there.

But it’s different, as you know, if it’s spinning. So let me start spinning here. [spins bike wheel] And now when I let go– obviously, when I’m holding it there’s no torque on it– but if I let go, it falls around rather than falling down. Now, people talk about vectors and right-hand rules… but why not a left-hand rule, I wonder… and so on. But it’s something else to really understand why this is doing– why there’s precession in a situation like this.

So normally it would just fall down, but if it’s spinning it falls around. And Feynman, in his lectures, said something interesting about this. He said, “many simple things can be deduced mathematically more rapidly. They can be really understood in a fundamental or simple sense. The precession of a top looks like some kind of miracle involving right angles and circles, and twists and right-hand screws. What we should do is understand it in a more physical way.” Why does it move around instead of falling down? So that’s what we’re going to do here.

And it turns out to have to do with phase again, and the difference between force and velocity. So let’s think about this. Let’s take a point on the rim of this. And say… Notice that there’s going to be a torque from this wanting to go down here and being held up here. And that’s transmitted through the spokes to the rim. So everything above the center feels a force to your right, and everything below the center feels a force to your left when I let go. Everybody with me on that?

So let’s take a point on the rim. Here it feels no force at all, because it’s moving neither right nor left. As it goes up it gets more and more force to the right. Everybody with me? Still more force to the right, more force to the right. Now less force to the right, less zero, and now it feels a force to the left as I let go. So we can make a plot of the force as a function of position on the rim. So if it’s in the picture on the screen here, when it’s above, the force is to the right; below, it’s to the left. And this is what the force should look like. When it’s in the front or in the back, there’s no force. When it’s above, there’s a force to the right, and when it’s below, there’s a force to the left.

Now, I’m going to ask you a different question. Where is the velocity to the right fastest? Where’s the force to the right fastest– largest force, at what position on the rim?

Student: Top.

Professor J. Michael McBride: At the top. So it’ll move the furthest, it goes the fastest when I let go [clarification: for non-rotating wheel]. It goes much slower here. OK, that’s fine.

Now I’m asking a different question. Where is the velocity fastest to the right? Well, let’s start here. It’s beginning to feel a little bit of force to the right. But if it’s rotating, it’s still feeling force to the right. More and more, and more, and more force to the right, more force to the right. Maximum force to the right. And now less, and less, and less, and less, and less, and less, and less, and less, until it gets to zero.

But all this time, during this half rotation, that point that I’m holding is feeling force to the right, and what does that mean? It’s accelerating to the right. Everybody with me on this? So where is its velocity maximum to the right?

Student: Where force is minimum?

Professor J. Michael McBride: Suppose I start pushing somebody on a bicycle, and I keep pushing and pushing. At the beginning I don’t push very hard, then I push real hard, and then I don’t push very hard, and finally I let go. Where are they moving fastest, assuming no friction and so on? Yigit?

Student: At the very end.

Professor J. Michael McBride : At the very end. All the time I’m pushing I’m accelerating it. So where is the velocity fastest on this to the right? Tell me when I get there?

Student: Now.

Professor J. Michael McBride: There. And that’s exactly what it does when it moves like this. And then it’s being slowed down, because you’re pushing to the left. And here it’s got zero velocity, you’ve cancelled it. And now it starts moving to the left and reaches its maximum velocity to the left here. Then it starts slowing down again and has zero velocity to the left or right here, and then it has its maximum. So it moves like this. Isn’t that neat? So it’s the difference between force and velocity. That velocity, you have to integrate force over the time that’s involved.

So if you look at where is the velocity maximum. It’s going to be 90° after the force is maximum. So it’s going to be the maximum velocity when it’s in front. Velocity to the right. And as it keeps going, it then slows down. And it has its maximum velocity to left at the bottom [correction: back], and then it comes back up. So there’s this 90° phase lag between the periodic force and the velocity. And that’s why it falls around. So the velocities look like that. So there’s a 90° phase lag, and it falls around rather than falling down.

Now, you’ve heard of nuclear spin. So these protons [correction: nuclei], ones that have an odd atomic weight, either an odd number or an odd weight, we know that they spin. And the reason we know that they spin is that when they’re in a magnetic field, they precess. And how do we know that they’re– when you put the force on them. Obviously, if there’s no force, this doesn’t precess, it’s when you try to twist it. So if this is a magnet that’s spinning, and you put it in a magnetic field that tries to twist it, then it goes around and precesses. So that’s how you know that these nuclei are magnetic. Hydrogen, fluorine, and phosphorous, those are the common isotopes, but for carbon and oxygen they’re uncommon isotopes.

And this is how fast they precess in a magnetic field. They precess like this, and you can measure the frequency of their going around and around, and we’ll mention that in just a second. And if a field happens to be 23.5 kilogauss, which is not an uncommon field, then protons go around the fastest, 100 million times a second. Fluorine is almost as fast, but phosphorous, carbon, oxygen are less fast. So that, just to give you– Connecticut Public Radio broadcasts at 90.5 MHz, and WCBS is at 0.88 MHz, 880 kHz. So these are radio frequencies.

Now, the electron also spins, but it’s way off scale up at the top, 66,000 MHz. It’s a much more powerful magnet than the nuclei. OK, so a megahertz multiplied by 3 and times 10–5 is wave numbers. Remember, in IR we talked about frequencies in wave numbers, how many waves in a centimeter. But the radio frequencies are very, very, very slow compared to that. 10–5 slower. You can also talk about it in °K. And 100 MHz then, where the proton resonates in this frequency, is equivalent to 0.01 K.

And it’s quantized, so that the direction of the magnetic field is either more or less with the applied field, or more or less against the applied field. But it’s not like a compass. It’s not like you can have any angle and different energies for each angle. It can either have this or that. That’s quantization of spin. Actually, quantum mechanics is much easier to do with spin, because you can only have two values, where for position or energy, you could, in principal, have any value.

The equilibrium ratio of the up to down is 1.0003. So that means that if you have 200,000 nuclei, there are three more pointing up in the favorable direction than pointing down. It’s really, really wimpy, the Boltzmann factor, the energy difference between those. And O-17 occurs at 6% in natural abundance and C-13 at 1%, but these others, like protons, are 99.98%, and only 0.02% deuterium.

Chapter 5. Radio Pulses and the Rotating Frame [00:49:35]

OK, now, this is a really neat concept, the rotating frame. And let me do this, and then I’ll let you go, because it’s a holiday coming up. So there’s the applied magnetic field, and remember we said that the nuclear field, the axis about which it’s spinning, can either be more or less up, or more or less down but at a particular angle. So suppose it’s like that. And it precesses because it’s spinning.

And the rate of precession in that field we just talked about is 100 MHz, that’s how fast it goes around. So notice that that precessing proton has a constant vertical field. That doesn’t change as it precesses, the vertical component. But the horizontal component goes around, like that. And that looks neat, because if that’s going around at 100 MHz, that’s going to be like an antenna, and it’ll generate light that’s coming out. It’s like the vibration of C=O, charge going back and forth. You see it going back and forth as it is like that. So it looks like we might be able to detect that broadcasting.

So will it generate a 100 MHz radio frequency signal? No. It would if you just had one proton. But of course you have a mole of protons, or at least a millimole or something of protons. And so some we’ll be here, some will be here, some will be here, some will be here, some will be here. And if you look at the horizontal component– although the vertical components all add– the horizontal components that are changing in time cancel. Everybody see that? So you don’t get any net signal coming out.

So the horizontal fields cancel, but there’s a trick that allows you to do it. And that is, suppose that instead of us sitting here and watching them precess, suppose we orbit around them at 100 MHz. We would certainly get dizzy if we went around them at a 100 MHz. But what would a proton look like if we were going around it at 100 MHz? And it’s going around at 100 MHz? It would look like it’s standing still. So that makes it much easier to solve the problems, to work in a rotating frame so that they’re not moving around from our point of view at 100 MHz. They’re just standing still. So we look at these and they’re just standing still, as if there were no applied field.

Now we take a very weak magnetic field that we generate, a radio frequency field, that’s horizontal. In our frame it’s horizontal, pointing out that way. Now, of course, in truth, it’s going around at 100 MHz. But from our point of view, it’s just sitting there. And from our point of view, there is no applied field, because our going around canceled that. So what will those protons do as we look at them? They;ll precess around that field. See what I’m saying? From our point of view they’ll just be this one weak little magnetic field, and it’ll appear in a constant position relative to us, because we’re whipping around so fast. And so these things will precess around that. But they now they make all different angles, because they’re quantized with respect to that one. They’re quantized with respect to the vertical field.

So they’ll precess like that. That particular one would go, and all the others would go precessing at the same rate. So that’s about 0.1 MHz rather than 100 MHz, so this is 1000 times weaker field. But how long do we do it? How long do we put this field on? Just long enough to make them go 90°, like that. So now they’re pointing out toward us like this. But, in truth, they’re not statically pointing out toward us. They’re going around at 100 MHz, because we’re going around at 100 MHz.

And now there’s a net field that’s oscillating in time, if we go back and take ourselves back to the laboratory, and not spin around like that or orbit like that. We see this whole set going around like that. So if we put in this thing called a 90° pulse, now we can hear these things broadcasting. And what will they tell us? They’ll tell us how strong the torque is.

I didn’t do that here, but let me do it. [spins bike wheel] Now, if this is a magnet, I can twist it by putting a magnet field perpendicular to it. Now, suppose I made the field that’s perpendicular stronger, to twist it harder. So that’s how fast it’s precessing. What happens if I put this wrench on? Is it going to go faster or slower, if I hang it on here? Faster or slower?

Let me speed it up again here. [spins bike wheel] Incidentally, that’s another question. If I go faster, if I make it spin faster, will it precess faster or slower? It precesses only because it’s spinning. If it weren’t spinning it wouldn’t precess. So suppose I spin faster, does it go faster or slower? Faster? It should go faster?

No, because the amount of time that’s involved going here, that we’re pushing is shorter if it’s spinning faster. So in fact, it goes slower. Isn’t that neat?

And now if I twist harder, it goes faster. See how it’s going faster now? And then it slows down, if I take that off? It’s not an ideal demonstration, I grant you. In fact, it goes slower if you– pardon me, it goes faster if you put more torque on and it’s spinning at the same rate. So that’s what these problems are, for you to think about.

So I’ll just finish in a second here. So we went into the rotating frame long enough to get into 90°, and now we go back, and now we see a signal coming out at 100 MHz. So that’s how you do the NMR experiment. They don’t like that, and that’s the signal you see. So it’s 100 MHz in the laboratory frame. Until they relax and go back to where they started, then the signal goes away. So that’s what we’re going to talk about some more. Have a good break.

[end of transcript]

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