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21. Functional Groups and Fingerprints in IR Spectroscopy. Precession of Magnetic Nuclei

21. Functional Groups and Fingerprints in IR Spectroscopy. Precession of Magnetic Nuclei


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: 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. 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: 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. 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: 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: Can’t hear very well. STUDENT: It’s in the plane of
the molecule. PROFESSOR: 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: 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: 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: What is it that would
vibrate at a very high frequency? Derek? STUDENT: The triple bond PROFESSOR: 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: 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 bonding
stretching. PROFESSOR: 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’s 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: 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 pi overlap. It weakens the pi 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 bond. It distorts the bond, whereas
the one on the right doesn’t distort the bond. So the one on the left is harder
to do and it’s higher frequency. So folding preserves the
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. 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: 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 pi*– PROFESSOR: And pi*. 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 pi*, and it gets to 1681. So the C=O is weakened by
resonance, by occupancy of the pi*. 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: There’ll still be
resonance. It’s an unshared pair on oxygen. Should it be as strong as that
involving the nitrogen? STUDENT: No. PROFESSOR: Why? Why should mixing of the electrons into the pi* be weaker
for oxygen than it is for nitrogen? Here are the electrons on the
atom we’re interested in, on oxygen or nitrogen. STUDENT: Bad energy-match? PROFESSOR: 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 sigma*. So if you mix that with sigma*,
then you strengthen the C=O bond. You get a triple bond between
oxygen and carbon, and occupy sigma star. 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 sigma*, 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 pi
interaction, putting electrons in pi*. In this one, it’s putting
electrons in sigma*, 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
sigma*, so we’re not going to get the effect we got in the
acid, and the acid chloride and the ester. And we have pi 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
synperiplanar. 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. 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+-. 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. 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 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 forces 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: 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: 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: 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: 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 degrees
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 degree 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 degree 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 10-5th 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. 10-5th slower. You can also talk about it in K
degrees. 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. 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 degrees, 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 degree 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 degrees, 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.

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