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>> Good morning.
Let's try to get started on time today
because as you might remember, we have an exam Friday.
So today we're going to talk about NMR applications
and just some stuff that's hopefully interesting
and maybe ties some things together
that we've been looking at.
And then at the end we're going to talk a little bit
about the kinds of problems that might be on the exam.
But I'm happy to answer general sorts of questions
about it right now if anyone has them.
Yes, in the back.
>> Do you have a summary of your, like your lectures
and how they match with which sections in the book?
>> Mm-hmm?
>> I mean, do you suggest doing all the problems in the book,
too, in those sections?
Because some of them seem kind of irrelevant
to the lecture slides.
>> Well, so are you talking about the problems I assigned
or the rest of the problems in the book?
>> Not the problems you assigned, the ones in the book.
Like the example problems.
>> Okay, yeah.
There are bunch of example problems in the book.
There are things in the book that we did not go over at all.
And that's just fine.
So, I mean, for instance in Chapter 11 there's a whole bunch
of stuff about how lasers work.
And it's neat, and I hope you read it.
It's, you know, it's nice for your information,
but you're not going to be tested on it
because we didn't cover it at all.
We don't have time to do everything.
So if you want extra practice problems from the book,
I would say look for ones that are similar to ones I assigned
or similar to things that we did in class.
And then that might give you a good idea
of which ones are good to do.
There are definitely some that aren't going to help you
with specifically the exam material.
Anything else?
Mm-hmm.
>> Will you give us the [inaudible].
>> The Hermite polynomials, if you need to know them?
Yeah, I will.
For the most part, I'm not going to give you equations,
but something like that where you need a giant table
of wave functions, you're not expected to know that.
I will say if, you know, if you need to pull out your Chem.
1 notes and go over molecular orbitals
of diatomic molecules and, you know,
remember which way the sigma and pi orbitals go in energy
for a different diatomic molecules.
This is a good time to review that.
There's another question over here.
>> We don't need to draw, like,
molecular orbital diagrams, right?
>> Where did you get that idea?
>> Excuse me?
>> Where did you get the idea that you don't need
to draw molecular orbital diagrams?
>> I'm asking you, like, do we need to know how
to do that on the exam?
>> Yeah. I just said, if you need to review that,
make sure you go review it before the exam.
So I mean, the exam question is not going
to be draw the molecular orbital diagram of, you know, fluorine.
But in order to come up with term symbols
for diatomic molecules, you're going to need to know things
like the symmetries of the different molecular orbitals.
And so, I'm just recommending that's a good thing to review
if you don't remember.
Yes?
>> That's one of the [inaudible] symbols?
>> Mm-hmm.
>> For, you know, oxygen, so oxygen's paramagnetic, right?
>> Yup.
>> In one of them, it's the same electrons configuration,
but the term symbol is different.
>> Yup.
>> I was wondering why one of them is a negative and one
of them is a positive.
And I know it has to do with the symmetry of the orbitals,
whether they -- yeah, the parity.
But for the positive, I'm not sure why?
>> Well, so plus and minus -- so I think a lot of this I'm going
to -- the really specific questions I'm going to defer
to office hours and the review session tonight.
>> Okay.
>> But the plus and minus refers to, you know,
that I can answer quickly.
It refers to the symmetry about a reflection plane that's
through the internuclear axis.
So remember G and U is whether it's even and odd
with respect to inversion.
And then plus and minus is like you have two nuclei
and then there's a plane going, you know,
containing the internuclear axis,
and you reflect it top to bottom there.
And if you get the same thing it's plus,
and if you get a negative sign it's minus.
So that's how you tell that.
And so, you know, if you can remember all this stuff
and do it in your head and you get the right answer,
that's fantastic.
But if you don't remember, you know,
whether these molecular orbitals are G and U off the top
of your head, you might need to draw the picture and look at it.
And so you might want to make sure that you can do
that relatively quickly.
You know, you don't have to get credit, but you know,
that's one really good way to figure it out and also if you --
the more of your work you show,
the more partial credit you'll get
if you happen to make a mistake.
Other things I should mention.
The rules are the same as last time.
You get a cheat sheet.
You can use your calculator.
Please don't store a bunch of text in your calculator.
That's cheating.
But, using it as a calculator is just fine.
What else?
We need to post the seating chart,
but that'll be up today or tomorrow.
Everybody did a pretty good job last time about getting
in and getting seated.
Please try to do it even more efficiently this time.
The exam is kind of long, again.
So --
>> Noooo.
>> No, this is good, actually.
I swear. Because if I don't -- I do it on purpose.
I'd rather have it be a little long so that everyone runs
out of time rather than have to be more selective
about what to put on it.
Because it's a more effective measure of what everybody knows.
So if I -- for instance, if I only put NMR spectroscopy on it,
but it's really short, if that's the one thing you didn't
understand, then you're going to do really badly.
And it's not very fair because it test all the other things
that we went over.
So I'd rather put more different things on it
and measure what everyone knows more effectively.
And yeah, that means everyone's going to run out of time,
and it'll be a little stressful.
But it's a better measure of what people can do.
So, you know, just make sure that we all try to get here
on time and get organized quickly.
You guys did a pretty good job of that last time.
Yes?
>> Did you say that there's a review session?
>> There is a review session.
John, Mark and Jerry are doing that.
And it's tonight.
And it starts at 8:00 o'clock.
And what room is it in, you guys?
I don't remember.
>> 104
>> 104 Rowland?
Okay. Good.
All right, so any other general kinds
of questions about the exam?
It's going to cover rotational and vibrational spectroscopy,
things like interpreting spectra because we didn't do all
of that before we cut off for the end
of the material for mid-term 1.
There will be selection rules.
How do you know
which transitions happen and which ones don't?
You'll have to calculate or at least write
down Franck Condon factors for electronic transitions.
There will be some things like looking at a spectrum
and inferring what the energy level diagram
of the system looks like and vice versa.
Stuff like that.
You need to know about NMR both in terms of looking at matrices
for the angle of momentum operators and also, you know,
so the quantum mechanical representation of NMR on a,
you know, on a pretty basic level.
We haven't gotten that far into it,
but there is going to be a little bit.
And also being able to draw NMR spectra that are taken
under various conditions.
Question?
>> When do we use the transition dipole
and when do we use the Franck Condon factors?
>> Okay, so that's a good question.
So I wrote an unclear question on this.
So the transition -- when we were using the transition
dipole, that's just telling us,
okay is this transition allowed or not?
It's a yes or no question.
Just, are we going to see a signal?
Yes or no.
That's it.
The Franck Condon factor tells us the --
it tells us something about the amplitude of the signal
that we're going to see.
And so, you know, you could also get
from the Franck Condon factor that the amplitude
of that signal is going to be zero
because the states have zero overlap.
But it will also -- if it's not zero,
that will give us an actual number
for the intensity of the spectrum.
So, I will try to be extremely clear about what I'm asking
for in those kinds of questions.
Again, for this exam,
I recommend reading the directions really carefully,
because there are a lot of things where I have set things
up a certain way to try to save you time.
So it is long, but there are things
that could take even more time
if you don't follow the directions.
So please do.
If you don't understand them, feel free to ask us.
You know, the worst case scenario is we're going
to say sorry, I'm not going to answer that question right now.
But, you can ask.
I'd rather have everybody understand what's going
on than not.
Okay, so is that it?
Should we -- all right.
One more.
>> How do we ask, in the process of taking the exam?
Do we just raise our hands and wait until --
>> Yeah, just raise your hand.
I mean, we're pretty good at looking around
and catching people's questions.
We can't have everybody getting up and running around, so,
you know, please just raise your hand and stay in your seat
and we'll come around and talk to you.
All right.
Let's talk about NMR.
So last time at the end,
I talked about two-dimensional NMR.
And as by extension, multi-dimensional NMR
and how we use this to go about solving structures of molecules.
And we saw a simple example of an amino acid
and a complex example
of a two-dimensional spectrum of a protein.
I want to show you a little bit more
about how we actually solve structures
of molecules using this.
So, when we go through this process,
we start with two-dimensional or three-dimensional
or even higher-dimensional spectra.
And then we have to go through these collections
of spectra and assign them.
So we determine specifically which amino acid
in the sequence do these particular peaks belong to.
And again, you need highly multi-dimensional spectra
to do that.
You have to walk through the backbone.
And you can recognize particular spin systems and say, okay,
that belongs to a valine.
And then, if you go across the backbone looking at, you know,
HCN kind of experiments and going
through the protein backbone, then you see, all right,
now I have a valine next to an isoleucine.
And you can start to eliminate like how many sites
in the protein contain that sequence.
And you made the protein, so you know that.
And so we go through and walk through the whole sequence
and come up with an address
for every proton, carbon and nitrogen.
Then we use some additional more long rayed experiments
to measure some restraints on the three-dimensional structure.
So here what we're interested in doing is measuring things
like dipolar couplings if it's a solid or an oriented sample
or NOEs, nuclear Overhauser effect cross-relaxation terms.
These are things that tell us about the distance
between one part of the protein and another.
So now we'll have, say, a dipolar coupling
between a valine that's on one side of the protein
in the amino acid chain and a tryptophan that's somewhere else
in the primary sequence.
And the only way that they can be close together is
because the protein is folded up.
And so these restrains tell us more about the distance.
Of course, since we have a whole lot of them,
we can't measure the distances very precisely.
If we had a specifically labeled sample with one C-13
and one N-15, we could measure the dipolar coupling
between those two nuclei and get the distance between them
to many, many decimal places
and we would have a very precise number.
People do that.
It's a useful experiment.
However, for protein structure determination
or anything that's moderately complicated, it's not so good.
Because in that case,
if I'm doing a proton carbon correlation, if I put in --
if I transfer magnetization from protons to carbon,
they don't just transfer magnetization
to the carbon they're directly attached to.
That magnetization also leaks all around and gets diffused
to different points in the protein because we have a lot
of protons in C-13 and N-15 all around.
And so that means that we get more information,
which is a good thing.
But it dilutes our magnetization
and we don't have these really precise distances.
So what do we do?
We basically take these distance --
instead of having these distance measurements very precisely,
we kind of bend them into small, medium and large, essentially,
so we can say, all right,
these pairs of residues are very close together.
Here are some that are in the middle.
And here are some other ones that are really apart.
And then we feed all of this
to molecular dynamics type simulation
that uses those measurements as restraints on the structure.
And so one of them on its own is not very precise,
but because we have thousands of them, you can put all
that in an MV simulation, minimize the energy
and hopefully get a reasonable structure out of it.
And then, of course, you know, you get this family
of structures and we have to go back through and check.
Some of these things end up being wrong,
the assignments are wrong, things like that.
And we iterate through a bunch of times,
and then eventually get a structure that looks
like it's starting to converge.
So this is how you solve an NMR structure.
We've talked about crystal structures before, you know,
where we're diffracting x-rays and then, you know,
solving an electron density map.
There are crystal structures and NMR structures
in the protein data bank.
And you know, they're all in there.
You know, you can kind of put anything you want in the PDB,
but they're different kinds of structures.
They give you different information
because they're measured quite differently
and they're really complementary.
So a crystal structure tells you, like it's a snapshot
of a protein that's more or less immobilized because it's
in this rigid crystal lattice.
And it gives you really high resolution, hopefully,
if the data is good, picture of where all the atoms are.
But it doesn't tell you about other things like mobility.
So in an NMR structure, so notice that, you know,
this particular protein, which is gamma-S crystallin.
This was solved in my lab.
It has this tail on the bottom.
That's the N-terminal end.
And you can see that in the family structures it doesn't
quite converge.
It's flopping around all over the place.
You can say, you know, well, that's not as nice
as a crystal structure where everything's locked down.
But that's kind of missing the point.
This is telling you information.
It's telling us that in fact,
that N-terminal tail is flopping around.
It's mobile.
As opposed to the C-terminal tail which you can see
over one the right, which is kind of tucked
up into the protein and it has a really well
defined conformation.
So, these are two different ways to get structures.
And they're really complementary.
So, going back to the process of how we get this,
if we look at some correlations.
So I mentioned that we can see individual spin systems
and identify what kind of amino acid it is,
here's a multi-dimensional sequence called a TOCSY
that tells us, you know, here we've got proton chemical shifts
and carbon chemical shifts correlated.
If you're used to looking at these things, you know,
there's only 20 amino acids that come up in normal proteins,
you get really good at seeing the patterns for the spin system
and identifying the amino acids.
So one of the first things that you do is go through something
like a TOCSY and identify the proton-carbon correlations
and see, you know, just label things
as to types of amino acids.
So, you know, here are just some, you know, some examples
of stuff that we can see where it is
because of the chemical shift.
So for instance, these -- the beta protons, the CH2's show
up at, you know, between 2 and 3 ppm in the proton dimension.
Same thing with these gamma protons.
And then we can walk through the backbone.
So now each of these slices is
from a three-dimensional spectra.
So we've got proton and carbon chemical shift again.
And we're going through the backbone.
And each of these lines that I'm drawing is a correlation
between amino acids that are next to each other in sequence.
And so we know exactly what's next to what.
And remember, we made the protein.
So we know the primary sequence.
If you took a protein of unknown sequence and gave it
to an NMR spectroscopist and said here, solve this structure,
we would have a really hard time.
It's not something you can typically do.
All right, so here's an example of a structure
that was solved using this kind of technique, you know,
with some representative data.
This protein is called Mistic.
It's a membrane protein
that helps other membrane proteins fold up.
So it's very popular in NMR labs.
This one was not solved in my group.
This is from Roland Riek's group.
And I want to point this out because we can do these kind
of correlations between nuclei not only among nuclei that are
in the same molecule, but we can do it across interfaces
between two molecules.
So one of the things that Roland Riek's group did
in this structure is they looked at correlations
between the protein
and particular parts of the membrane.
And so they were able to learn a lot
of about exactly how it's sitting in the membrane
from looking at the NMR structure.
Okay, so that's one application of NMR, you know,
looking at where we can go
with it beyond identifying simple molecules.
I want to talk about another one, again.
So we've seen this slide before,
looking at the relative sizes of interactions.
And we're going to come back to the quadrupole interaction.
So we just kind of brushed it off as, you know, hey,
you don't really see it in liquids except
that it makes adjacent nuclei relax.
But if we're talking about solid state NMR,
and particularly materials,
quadrupolar nuclei are really important.
There's a whole field of looking
at complex solid materials that's extremely important
in chemical engineering in particular,
and also physical chemistry where people develop techniques
to look at quadrupolar nuclei.
Okay, so here's a periodic table showing the NMR nuclei.
And everything that is blue has only spin 1/2 nuclei.
And everything that's white has no NMR active nuclei or it's
so radioactive that it doesn't stick
around long enough to tell.
And everything that's in pink is something
that has quadrupolar nucleus.
And so, you know,
when I different elements can have different kinds of nuclei
that could be spin 1/2 or quadrupole,
notice that hydrogen kind of has both.
That's because it has different isotopes.
So it has protons, which are spin 1/2
and then it also has deuterium, which are spin 1.
So some of these things do have both.
But even given that, if you look at the periodic table,
a whole bunch of the nuclei that are there are quadrupolar
or they spin greater than 1/2.
And if we weren't able to study that and deal with those,
we'd be missing out on a lot of what's going on in chemistry,
particularly in inorganic chemistry
and materials kinds of applications.
Okay, so what does it mean that something is quadrupolar?
A quadrupolar nucleus is something
that doesn't have a spherical distribution
of charge in the nucleus.
And, you know, what exactly does that mean?
What's the structure of the nucleus look like?
That's beyond the scope of what we're doing and you have to get
into some serious nuclear physics
to understand exactly what's going on.
But we can understand it on a conceptual level.
So, the quadrupole moment of the nucleus is interacting
with the electric field gradient.
And that's what produces some interesting line shapes
that can give us information about the structure
of the material or, you know, make the spectra really messy,
depending on what's going on.
This interaction is anisotropic.
It depends on the orientation with respect
to the magnetic field.
You know, we've talked about things like dipolar couplings
and chemical shift anisotropy.
All of these things behave as a second rank tensor with respect
to the magnetic field.
So that means that we can spin
about the magic angle and average them out.
The magic angle.
I don't know if we talked about this before.
Did I tell you about magic-angle spinning or no?
No? Okay. There's a little section about it in your book.
So, essentially what it is is we have these interactions
that behave as a second rank tensor, dipolar coupling,
chemical shift anisotropy,
magnetic susceptibility differences, things like that.
And in a liquid, these all average
out because the molecules are moving around isotropically
on the timescale of the experiment.
In the solid, that's not true.
They're stuck in some rigid place, you know,
whether it's a crystal or a glass or something like that.
And they're not averaged out by the environment.
So, if we just took a spectrum of a static solid,
a lot of times we would see a big mess
because we have these very complicated interactions
that are all overlapping.
If we want to simplify that spectrum,
then often what we do is we'll spin at the magic angle.
And all that is is the angle
where the second order Legendre polynomial happens
to go to zero.
And you can think about it in a simple geometric way like this.
If you imagine a cube, so you have you x, y,
z coordinate system and there's a cube,
draw a line between vertices of the cube that are farthest apart
and think about that angle.
That's the magic angle.
So if we spin about that, we're averaging in the x,
y and z direction all equally.
And so that helps us get rid of a lot of these interactions
that behave as second rank tensors in space.
So the quadrupolar interaction is more complicated than that.
It behaves as a fourth rank tensor,
at least the second order quadrupole term does.
And so spinning at the magic angle doesn't make it go away
completely because there's not just --
there's not just one degree of freedom.
You would need two degrees of freedom
to average it out completely.
So one way you can do that is by spinning
at two different angles during the experiment.
And people have done that.
We actually do it in my lab for different reasons.
Or, you can use techniques where one
of the averaging processes is done in actual space
and the other one is done in spin space.
All of that is a little more advanced
than we're going to get for now.
But let's just look at what the interactions look like.
Okay, so here's what the quadrupolar Hamiltonian
looks like.
So there's a [inaudible] on the electric quadrupole moment,
which again, has to do with this field gradient.
And it also depends on this spin quantum number for this nucleus.
So that could be -- it goes in increments of a half,
so it could be 3/2, it could be 1, it could be 7/2.
There are quadrupolar nuclei with all kinds
of different spin quantum numbers.
It also depends on this electric field gradient tensor.
And so here's what we get for something like a spin 1 nucleus.
So instead of just having two possible values
for the spin angular momentum, you know, we can have --
for a spin 1/2 we have plus or minus 1/2 and we interpret
that as up or down relative to the magnetic field.
If we have a spin 1, that has these three states
with M sub L values of 1, 0, and minus 1.
And so that means that when we look at its NMR spectrum,
if we just have one nucleus, and it's a spin 1,
it's going to give us a doublet
because it has these two possible transitions.
And so what would it be nice for you to know about this
at this point as far as quadrupolar nuclei?
You should know that they exist.
You should know what their spin states look like.
And you should be able to do simple things
like write matrix elements for operators like IZ and I plus
and I minus, stuff, you know, stuff like that that we went
over in class, you know, the simple ones
in these kind of bases.
And that is about it.
Oh yeah, I guess the other thing is you should also be able
to draw NMR spectra for them as we did in some
of the homework problems where you said, all right,
you have a spin 1/2 coupled to a spin 3/2,
what do the spectra look like?
You should know how to do those.
But that's about it.
So let me tell you a little bit about an application of this
that we work on in my group.
So again, the main application of this in, you know,
out in the field is solids, materials that, you know,
aluminosilicate glasses are popular ones.
But all kinds of materials were the state that you want to look
at is a solid and, you know, not necessarily a crystal
so you can't do crystallography.
Zee lights, all kinds of things like that.
Here's an example from my group.
So, we don't work on solid materials except
for proteins occasionally.
But one thing that we use quadrupolar nuclei
for is looking at orientation in biomolecules
in the context of membranes.
So we want to look at membrane proteins stuck in a membrane
in as close to native-like conditions as we can.
Membrane proteins are really important.
And it's very useful to have their structures.
So you see numbers thrown around like membrane proteins make
up a third of the proteome and something
like 60 percent of all drug targets.
And the reason for that is that, you know,
these are really what are allowing things to pass in
and out of our cells, including information.
So all of our senses, you know, the antennas
for those are membrane proteins, for instance.
And all kinds of other things involving transporting food
into the cell, waste out of the cell.
All of that.
It all goes through membrane proteins.
So it's really useful to be able to solve their structures and,
you know, figure out how to design drugs
to interact with them.
However, it's difficult to get their structures
because they don't crystallize very easily,
they need interaction with membranes and also,
looking at them by solution state NMR is really difficult
because they have to be stuck in membranes.
So we're interested
in developing NMR techniques to do this.
So one of the things that we do, and other groups do this
as well, it's a pretty large field, we're interested
in putting these membrane proteins in membrane mimetics
that are tractable to NMR but they're realistic enough.
And so the system that we often use for that, that's again,
pretty common, is this mixture of short chain
and long chain lipids that self-assemble
into oriented membrane domains.
And here, you know, I'm showing them as little cartoons,
little disks that align in the magnetic field.
They don't look exactly like this.
But one of the tricks that we use with deuterated molecules is
that we put D2O in with our bicelle systems.
And, you know again, deuterium is a spin 1,
so if it's in an anisotropic environment
where everything is not getting averaged
out by the molecular motion, then we'll see a doublet.
And so when we want to map out what kinds
of phase transitions the bicelles are undergoing,
you know, we're interesting in having a system that is aligned
so that we can stick our membrane protein
in it and do NMR on it.
If you look at this cartoon, you know, again,
they don't look exactly like that, but the important part is
if you have these mixtures at low concentration,
we get tiny little micelles that are no good
because the protein won't fit in there
in its native conformation.
And if we get all the way up to high temperature,
we have large vesicles which are, again,
no good because they're heterogeneous.
You know, we're not going to see a good alignment of our sample.
Each protein isn't going to be in the same environment.
And in between we get these oriented systems that are useful
for looking at the protein structure.
And so the way we measure that is we throw in D2O
and we take spectra of the deuterium.
So you know again, a lot of times you hear about, you know,
you take your NMR spectra of organic molecules
in deuterated solvents because you don't want to see a bunch
of protons from the solvent.
And so you think that deuterium is silent to NMR.
It's not. It's a good NMR nucleus.
It's just that it's a spin 1, and so if we want to look at it,
we have to tune the probe to a different frequency
and actually investigate the deuterium.
And so here we're just using the D2O as a probe to spy
on the bicelles and tell us whether they're oriented or not.
You could also put deuterium on the lipid molecules themselves.
And people do that.
But it's really expensive as opposed to throwing
in D2O, which is cheap.
So it makes a good starting point for this kind of study.
It's cheap and it's easy.
We can tell whether things are oriented or not.
So that enables us to do things like look at, you know,
here's a protein from a thermophilic organism that,
if you see here in our particular bicelle recipe
that we made, you know, the details are too much
to go into for right now.
But, if we look at this at 55 degrees, the distance
between the two peaks of a doublet is really far apart.
That means we have a lot of orientation in the sample.
And that enables us to look at this thermophilic protein
at a temperature that's close
to its operative temperature, which is important.
You know, we want to be able to study these things
at the temperature where they're actually active.
Okay, so I pronounce us done with NMR.
And you know, so basically that's everything that we need
to know for the second mid-term.
Give me a second to get the document camera thing set up
and let's talk a little bit about some of the kinds
of problems that there might be.
[ Background Sounds ]
Okay. Can you see that?
Is that good?
All right.
So obviously we don't have time in the 15 minutes
or whatever that's left to go over, you know, all the kinds
of problems that could be on the exam.
But I just want to start sort of close to the end
of what we talked about and what we've had maybe a little bit
less time to practice.
And go over a few of these problems.
So sort of the last thing that we've talked about is NMR.
So let's look at that.
Okay, so the idea here is -- let's look at Part B here.
>> Professor?
>> Yeah? Will this go up on the Website?
>> My lectures are always posted on the Website.
There's a video.
>> Okay.
>> Am I going to scan this piece of paper
and put it on the Website?
No. But there is a video.
Okay, so the idea is we want to look
at the carbon decoupled NMR spectrum of this molecule.
And so let's look at how we want to draw this.
So it's a proton spectrum, so the first thing you want
to do is label your axis.
So chemical shift, proton, ppm.
Zero goes over here and it's going to end around 12.
That's an important part of the problem.
Things that you should be worried
about when you're predicting the NMR spectrum of a compound.
One is how many protons are there that are not equivalent?
And, you know, make sure
that you read the directions really carefully.
So in this particular case, I remember when I gave this exam,
it was a couple years ago, I said, you know,
sketch the spectrum of this molecule.
And some people only wrote down the answer for the protons
that are labeled A and B, but you know,
they didn't read the directions.
In this case it was draw the whole thing.
So just, you know, make sure you're --
make sure you're actually following what the question is
asking for.
Okay, so how many protons do we have that are inequivalent?
What do you think?
How many different kinds of them do we have?
>> Six.
>> Yeah, so we've got these guys and this one.
And what do you think?
Are this one and that one equivalent?
>> No.
>> No, right?
Because of symmetry.
Okay, so the next thing that we want to worry
about is the chemical shift.
And one thing that I am going to give you as part
of this exam is a basic table of chemical shifts as far
as where different functional groups show up for protons
because I'm not really worried
about anyone memorizing those numbers.
I'm also just -- I'm going
to roughly stick them in at this point.
The main thing I'm worried about in that is getting stuff sort
of in the right order as far as where it falls on the spectrum.
You know, so I'm not going to argue with anyone
about whether something shows up at 7 ppm or 7.5 ppm.
It's not important.
The important thing is getting things
in roughly the right places.
And I will give you a table to use for that.
Okay, so first let's look at the methyl groups.
So those are going to be over here, you know, close to 0,
or between 0 and 2, say.
And so we've got two of them.
What do you think?
Which one is going to be closer to 0?
This one or this one?
>> The lower one.
>> The bottom one?
>> Yeah.
>> Yeah. How do you know?
[ Inaudible Audience Responses ]
Yeah, you guys are all on top of this.
I shouldn't have worried that we need
to spend a lot of time on it.
Okay, so the first one we're going
to draw is this bottom one.
So that's going to go somewhere, you know, we'll put it
at 1 point something ppm, say.
You know, maybe around 2, you know,
we'll guess down here somewhere.
We also have to worry about the splittings.
Okay, so what is the signal of that going to look like?
Is it going to be a singlet or a doublet or a triplet or what?
>> Triplet.
>> Yeah, it's a triplet.
So we have --
All right, so we called those A and B. I'm going to call this C,
D, E, F -- yes, that's arbitrary, but that's fine.
Okay. So this is E. Now,
how about our other methyl group here, B?
So we already said it shows up a little bit farther
to the left than the other one.
What does its signal look like?
>> Like a singlet.
>> Yup, it's a singlet, because it is not split by anything.
Okay, so then the next thing, we have some aliphatic proton here.
So that's the CH2.
And that's split by its neighbor, the methyl group.
So it's going to be a quartet.
And that's, you know, in here somewhere.
And again, this is sort of the level of, you know,
care that I expect with, you know, with exactly
where you put the chemical shift.
Questions?
>> How's the CH2 less [inaudible] shielded
that the OCH3?
Like it's marked [inaudible].
>> You are so right.
I was not thinking of those things altogether.
So yeah, these are issues you can have.
But again, I'm going to give you a table of that.
So it'll be hopefully easier to keep track of.
Yes?
[ Inaudible Audience Question ]
>> You know, at some point it just gets too hard to draw,
so yeah, ideally, if you had all the time
in the world you should draw them so that they're the same,
you know, so that the relative intensities indicate how many
there are.
But in reality here, like on this quality
of a drawing telling the difference between one proton
and three, I'm just not so concerned about it.
So yeah, ideally you should know that, but it's not necessary
to try to indicate it with the drawing.
Okay. So we've got our methyl and aliphatic protons.
And then I'm going to draw a break in the middle of this axis
because now I've got everything else spread out.
And so I want to indicate that there's some stuff
in the middle before we get to, you know, the aromatic protons.
And so those are all going to be sort of over here.
And as far as what order you get them in, you know, for again,
for this level of problem,
as long as you get something reasonable, it's fine.
They're all going to be pretty close together.
And you know, here two of them are going to be doublets
because they're split by, you know, one neighbor.
And the one in the middle is going
to be a double of doublets.
And here's something about intensities
that I do actually want to see.
So we have a quartet looks
like the middle ones are higher intensity.
And a doublet of doublets has all of them the same height.
And so, I would like to see that as, you know,
crude as the drawing is going to be, I'd like to see
that you understand that.
And so, I just wanted to go through this to show, you know,
I know these problems where you have
to draw something are kind of hard.
I want to show sort of what level of detail I expect and,
you know, what kinds of things are not necessarily going
to really count against you at this point.
Because I know you have a finite amount of time,
and you can't do everything in the time constraints.
So I'm worried about, you know, do you get it roughly
in the right place as far as chemical shift.
You're going to have a table.
And getting the splitting patterns right.
Question?
>> I was just wondering how did you know whether it's going
to be -- whether that proton would be a triplet
or a doublet of doublets?
>> Well, so a doublet of doublet s means it's split
by one neighbor over here and then another neighbor over here.
So it's split by two non-equivalent protons.
If it's a quartet, then it's split
by three equivalent protons.
Yeah, so it's a different effect.
Yes?
>> So this molecule is
like symmetrical [inaudible] those two protons would be
identical and they would be a triplet.
>> Yup. So, that is true, If it were identical,
then it would have two identical neighbors.
Okay, so now let's look at the next one.
Okay, so we said explain the splitting pattern
of the resonance of proton A. So proton A,
we said is a doublet of doublet.
And then the question is don't just describe it,
explain the physical reason for the effect.
So just describing it would be you write
down it's a doublet of doublets.
You know, okay, but that doesn't say
that you understand exactly what's going on.
So you would need to say something like it's split
by two non-equivalent neighbors.
And you know, even better if you mentioned that the reason
for the splitting is that its neighbors can be either up
or down and that adds to or subtracts
from the main magnetic field.
So this is definitely the kind of stuff
that you might see come up.
Let's see, what else do we have in relationship to this?
I might also ask you about things like --
here I mentioned that we are quickly running out of time.
I had a huge plan for all the examples that we could go over,
and of course, this stuff always takes longer than I thought.
So, we'll have to do some of that in office hours.
And you know, maybe I'll share some of it with the TAs to use
at their review session as well.
Okay, so we also have --
we don't have time to do the whole thing, but let's just talk
through it a little bit.
This last question.
Sketch the A and X NMR spectra for an AX2 spin system
where A is a spin 1, X is a spin 1/2
and the J-coupling is 25 hertz.
So, okay, so what this should tell you is that, you know,
they're called A and X.
That means they're two different types of nuclei
so they don't show up in the same spectrum.
So the A spectrum and the X spectrum and different things.
So you know, maybe the A is a deuterium or maybe it's an N14.
I guess that's a more realistic example.
And X could be a proton.
And so you're not going to see the N14
and the proton in the same spectrum.
So you have to draw two separate things.
And so you know, here's our axis.
I didn't tell you what nucleus this is,
so you can't get too excited about the actual values.
But you know that without the splitting, A is a spin 1,
and so it's going to have a doublet if it's just by itself.
If it had no interaction with anything,
it would have a doublet.
Then we know that it's split by two protons.
And so each of those protons are going to split it into,
you know, it's going to be split into a triplet,
but it had two peaks to begin with.
So here's what that looks like.
It's a doublet of triplets.
And the J-coupling is 25 hertz.
So that's this distance between them.
And I will let you figure out the second part of the problem,
how do you draw the proton spectrum
when it's split by that spin 1?
Okay, we are out of time.
If you want to do more examples,
come to office hours today and tomorrow.
Go to the review session. ------------------------------6515476ad6f9--