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>> I want to spend the next 3 lectures, 3 classes,
talking really closely about first order coupling
and the reason is that there's so much to be gained
by deeply understanding NMR spectra.
As I said, a lot of what one is going
to be doing is asking specific questions about stereochemistry
and being able to ask those questions is intimately linked
to understanding what's going on.
Also just in general for solving structures,
being able to read spectra, really read them at a level
that goes beyond the level of sophomore organic chemistry,
involves intimately understanding [inaudible].
So we're going to take a relatively slow path
through this.
In fact, we're going through the midterm exam only have 1-d
spectra on our exam so that we really focus
on understanding things.
So I want to start by kind of making the bridge
between last time's lecture where we talked
about magnetic equivalents and we talked
about non-first order systems and so last time was sort
of the bad and today is going to be the good.
So the bad is that I said a lot of the rules that you learned
in simple sophomore organic chemistry really
were oversimplifications.
There are very few systems that truly behave in the way
that you learned they should behave.
These are the first order systems.
So first order systems are anything
like AX systems, AMX systems, A2MX.
In other words, anything where coupled protons,
protons within a spin system are far apart in chemical shift
and if you do have 2 protons that are chemically equivalent
like we have in A2MX system
that those protons are both chemically equivalent
and magnetically equivalent.
We divided this and separated it from non-first order systems.
[ Writing on board ]
And these are systems in which you either have magnetically
inequivalent protons that are chemically equivalent
or you have protons that are similar in chemical shifts.
For example, a non-magnetically equivalent protons we saw,
for example, A, A prime, X, X prime systems and we talked
about just how ugly those systems could be.
Those were like the phthalate [phonetic] system
where I said no matter how far apart,
no matter how high a magnetic field you look
at dioctyl phthalate or ortho dichlorobenzene is never going
to get better than this complex pattern of lines
and then I said we have other systems like AB systems
where the protons are similar in chemical shift and ones
that are related to this, for example, ABX systems.
The good news about many of these types of systems is
that many of these non-first order systems behave very much
like first order and that you can start to apply some type
of simple, rational understanding to them,
which is more than I can say for an AA prime system,
XX prime system or an AA prime BB prime system.
Now sometimes these systems will look like first order,
which is great because sometimes you can analyze these types
of systems as first order and many times you can,
but what I tried to show you last time was how there are ones
that simply defy simple reduction.
>> So what do you use X or Bs based on the distance
or the separation of chemical shift not the actual distance
between them?
>> So let me show you exactly and let's take the AB system
because I think this is a great starting point
and what's nice is the AB system is going to be an archetype
for many sorts of systems
that although they're not first order we can apply first order
analysis to and we can start to see the distortions that occur.
So, a pure AX system is one in which you have a doublet
so it's 2 hydrogens that are J coupled.
Again, that's going to be the whole spin system
so I'll just put on XX and YY to represent some other nuclei
that aren't going to couple and not, of course,
something with a hydrogen on it where it's J coupling.
So you would have a doublet and then a big, big span between it
and then another doublet.
This little squiggly is just a break, break in the spectra.
If those 2 doublets are far apart in chemical shift,
then you're going to see them each as a simple 1 to 1 doublet.
Now as the distance between them becomes smaller.
In other words, either you have different substitutents
that instead of having them be very far apart they're closer
together in PPM or you simply went
to a lower field spectrometer, now you start
to see a distortion that we would call an AB pattern
where the inner line, and so now instead
of saying these are effectively very,
very far apart now I'm saying they are far apart like so.
In other words, this means, you know,
1 here and 1 way over there.
Okay, now the typical way
in which one characterizes this is the distance
between these line sis the J value, the distance
between these doublets
and technically one takes not the dead center of the doublet
but the weighted average because technically
with a multiplet the position of the multiplet is not
at its average but at its weighted average.
In other words, since this line is a little bit bigger we take
the center as just a hair over.
It's the weighted average.
In other words, if this line is 4 times as, if these lines are
in a 4 to 3 ratio and they're separated by .07 PPM, we'd say,
all right, you're .4 of the way over there; just a little hair.
So, if we call this distance Delta nu, typically if Delta nu
over J is much, much greater than 10.
[ Writing on board ]
We're in the situation like this and if Delta nu over J is less
than or equal to 10 and those are approximations,
then we're sort of into this AB situation.
By Delta nu I mean the difference in position in hertz.
So in other words, let's say the center of this line was
at 7.30 PPM and the center of this line was at 7.10 PPM
and let's just say here that our J value is let's say R,
what will work out?
What will work out well?
Let's say that our J value equals 17 hertz.
Now, imagine for a moment you're
on a very low field spectrometer.
Imagine you're on a 100 megahertz spectrometer what's
Delta nu at that point?
[ Pause ]
730 hertz.
Everyone agree?
>> Delta [inaudible] 20 hertz.
>> Delta, 20 hertz.
So at 20 hertz these guys would be hugely close together.
In fact, we'd have a situation that looked.
[ Pause ]
Like this.
At this point Delta nu actually will be just a hair further
apart because it's the weighted average.
I'm going to shift it over just a hair.
I'll make the outer lines just a little bit bigger.
This would be a situation where Delta nu over J is very small
where Delta nu is about 20 hertz and J is about 17 hertz.
If we had the same system at 500 megahertz,
what would the difference in,
what would Delta nu be for 500 megahertz?
[ Pause ]
A hundred hertz, right?
So at 500 hertz, 500 megahertz, Delta nu is equal to 100 hertz.
So you'll look at this situation
and at 500 megahertz you'd be more like this
and 100 megahertz you'd be like this.
So this is your AB pattern
and if they were ever closer they'd be like what I sketched
out before where the inner line would be huge
and the outer line would be very tiny.
What? That might, it would be like a 60 megahertz spectrometer
like one of the freshmen or sophomore and we actually have
like a 100 or maybe it's 60 in the sophomore lab it would be
like this or imagine the situation that instead
of having substituents that put these apart at .2 PPM,
imagine they were separated by .1 PPM, but the main thing
to keep in mind is for any given doublet no matter what the
center of this peak whether I looked at it
at a 500 megahertz spectrometer or at a 100 megahertz the center
of this peak is going to be 7.30 and the center of this peak,
again, weighted average center is 7.10.
Now 17 hertz is more characteristic
of a trans alteen [phonetic],
which was actually what I was doing when I was drawing this.
For something like this we'd be more
like about 7 hertz for a J value.
Thoughts or questions at this point?
[ Inaudible question ]
Okay, will the center move, so, if you improve the equipment?
So here we've gone from this is our 100 megahertz,
this is our 500 megahertz and the point is the center
of this peak for this whatever hypothetical compound this is,
the center of this peak is always at 7.3 PPM whether I'm
at 500 megahertz or 100 megahertz, but the distance
between the peaks because the number
of hertz per PPM is much smaller at 100 than at 500, the distance
between the peaks here is very far, it's 100 hertz apart
or relatively far, and over here it's only 20 hertz apart.
[ Inaudible question ]
The inner one?
The closer they are together the more they tent into each other
and that really is the difference between the AX.
[ Inaudible question ]
The center is related to the ratio of the bigger.
[ Inaudible question ]
Absolutely.
Absolutely.
Well, here the bigger one is at 7.29 PPM or 7.29 PPM
and the smaller one is at 7.31 PPM
and by here we've got these 2 lines and one of them is
at 7.2 PPM and the other is at 7.
whatever the number is.
Now what's valuable about looking at AB pattern
and understand it is it really becomes an archetype
for all sorts of systems that behave very near to first order.
So we were talking before about phenylalanine
and I guess the example I gave when we were talking
about spin systems was acetyl phenylalanine methyl amide.
[ Pause ]
And I pointed out that we had 1, so this is like a spectrum
in chloroform solution so I'll say NCDCL3, and we decided
that we had 1 spin system over here and the multiplicity
of this proton of the NH is a doublet because it's split
by 1 coupling partner.
Each of these protons they're non-chemically equivalent
so they split each other but they're going to be similar
in chemical shift, they're similar in the environment
so they'll both be at about 2 and a half, 3 parts per million.
Why do I say about 3 parts per million?
Well, they're off of a phenyl group
so if we were methyl group off
of a phenyl I'd say 2 parts per million.
It's an ethylene so that pushes it to like 2
and a half parts per million.
They're beta to a couple of electron withdrawing groups,
they're beta to a nitrogen, they're beta to a carbonyl,
so then it's going to shift them downfield
by about another half a PPM.
So we'd expect them to both be at about 3 parts per million
but probably not to be on top of each other.
So each of these is going to show up at as a DD
and that DD is going to be part of what looks
like an ABX pattern because this is part of an ABMX system.
M is something that's far apart from either A and B and C
and so forth and X. So we have 1 proton that's going
to be way downfield and nitrogen protons are typically
at about 7 parts per million.
One proton that's going to be moderately downfield
because it's next to an electron withdrawing group and it's alpha
to a carbonyl and beta to a phenyl group so this is going
to be about 4 and a half parts per million.
Then these guys that are both going to be close
to 3 parts per million.
So we have far apart from this and H far apart from alpha
and the alpha is far apart from the beta.
So this guy here is going to be split by 3 different protons.
So he's going to be a DDD if all of the Js are different or a TD,
and we'll talk more about these or DT,
if 2 of the Js are the same.
Or a quartet if all 3 Js are the same within the limits
of experimental error.
[ Pause ]
So the one I really want to draw our attention
to then is these 2 hydrogens here because now this type
of AB pattern really can serve as an archetype
for more complex patterns that are non-first order
but are close to first order.
So an ABX pattern is something where you have the AB pattern
in which each line is further split.
So imagine this type of pattern here with some level
of separation, but now with each
of these 2 lines split into a doublet.
So what you see is line line, line line, and then line line,
line line for these 2 protons.
[ Pause ]
This is for the ABX system.
So this is what we're seeing right at about 3 PPM.
[ Writing on board ]
Which of them is which?
[ Inaudible question ]
Okay, so stereochemically one of these protons is pro R
and the other proton is pro S. They're diastereotopic protons.
If we can go ahead and so, for example, if I replace this
with a deuterium in the thought experiment, then the ranking
of the carbons, the ranking of the 4 constituents
on here becomes highest rank, next rank, next ranked and so
that would become an S center.
So this is a pro S proton and this proton is pro R.
If I knew the geometry here, for example,
if I knew the phenyl group preferred to point in one way
or another or I could by Nuclear Overhauser Effect experiments
and the like detect certain proximities,
then I would be able to get an experimental correlation
or predicted correlation based on say proximity
to anisotropic groups of 1 proton and 1 peak.
So right now I don't know which is which,
but with additional experiments --
this is obviously just a sketch --
but with additional experiments in context, yes, you can figure
out with diastereotopic proton is which.
Other thoughts and questions?
[ Inaudible question ]
Do they ever look the same height?
Great question.
So, right now I've made a sketch for a situation
in which these are relatively near to each other.
In other words, maybe they're separated by a tenth of a PPM.
So if they're separated by a tenth of a PPM,
the JAB here is going to be about 14 hertz.
At 500 megahertz that would be a separation if they're separated
by a tenth of a PPM about 50 hertz.
So Delta nu over J would be about 50 to 14 about 3,
but if they were very, very far apart, if something held these
in very different magnetic environments,
then you would see the outer lines getting bigger.
[ Writing on board ]
And the peaks becoming more like a regular doublet of doublets.
So if they were further apart, it would look more like this,
these guys would be bigger.
They would be tenting into each other less.
Other questions?
These are really important and this is one
of the reasons I'm going really slow over this.
[ Inaudible question ]
Coupling, coupling is always going to be mutual.
So if we call this, let's name our peaks 1, 2, 3 and 4
and let's name our peaks 1 prime,
2 prime, 3 prime and 4 prime.
So the JAB is going to be 1 minus 3 and 2 minus 4.
They will within the limits of experimental error be the same
and JAB here will be the same within the limits
of experimental error.
It will be 1 prime minus 3 prime and 2 prime minus 4 prime.
In this case since it's an ABX pattern, the coupling
with the other proton so the coupling
with the remote partner, we'll call it AX, JAX equals 1 minus 2
and 3 minus 4 and, again, those will be the same
within experimental error.
So let's say for a moment this is 14 hertz that's also
14 hertz.
Let's say for a moment that this is 6 hertz,
actually it looks the way I've draw it it looks more
like about 9 hertz so let's say this is 9 hertz, that's going
to be 9 hertz within the limits of experimental error.
Here this distance will also be 14 hertz as will that distance.
This distance the way I've drawn it looks like it's
about 12 hertz and that looks like it's about 12 hertz.
[ Inaudible question ]
Ah, where?
Oh, no, no, no.
One minus 2 will not equal to 2 minus 3.
So over here JBX equals 1 prime minus 2 prime
and 3 prime minus 4 prime and, yeah,
if you look at this pattern
and you draw a splitting tree diagram, we split into a doublet
so that's our big J and then each
of those lines is further split with a small j
and so you get this pattern of line line, line line,
and if I call my lines 1, 2, 3 and 4, 1 minus 2 is the small j,
3 minus 4 is the small j, 1 minus 3 is the big J,
2 minus 4 is the big J.
[ Inaudible question ]
From the AB coupling, the geminal coupling
in this particular case, and the source of the small j is
from the coupling to, in this case, this nucleus over here.
[ Inaudible question ]
That's going -- beautiful question --
that's going to depend on the geometrical relationship
on the Karplus curve
so typically you will not have exactly the same coupling
between 1 of the 2 protons, let's say the pro S
and this proton versus the other proton and this proton.
So rather than my, let me put up some real data.
You'll see exactly the same thing
but at least it will be a nice, nice chance for you
to have a real spectrum and I know I passed this out before,
but we didn't look as deeply at this spectrum.
Now let's look at it with a fresh pair of eyes.
Let's look at it more deeply.
So I didn't use this exactly compound.
I just grabbed this right off of the Aldridge [phonetic] webpage
and remember you can go to sial.com, www.sial.com,
to get yourself lots of spectra.
It's a great way to check out your ideas
and your understanding of things.
So here we see a real compound.
I've shown you this before.
This is phenylalanine in D20.
[ Pause ]
So unlike the example that I, the hypothetical example I gave
in chloroform, this one doesn't have an amide here it has an
amine here and it has a carboxylic acid.
In D20, those exchange and so this becomes ND2
and you essentially see no coupling and don't see it.
Actually there's DCL here so this really becomes ND3 plus
and you don't see coupling.
This becomes D. So this system here what remains
of the 2 methylenes and the methine,
so this is an ABX system and so you see coupling here
and this is your phenyls and this is your HOD
and this is your alpha proton and these are your beta protons.
So this is a very real example of what I've sketched out
and you'll notice the distance between these 2 lines does,
indeed, match the distance between those 2 lines.
In other words, the 1 to 3
or 2 to 4 distance does match the 1 prime to 3 prime
or 2 prime to 4 prime distance, and you'll also notice
that the 2 couplings with the alpha proton are a little bit
different, a little bit different from each other.
So it looks like if I had to eyeball it here
that our coupling here, this distance between 1 and 2 or 3
and 4 is about 6 hertz and the distance here, the distance
between 1 prime and 2 prime or 3 prime
and 4 prime is about 8 hertz.
So you'll notice now the alpha proton is split into a doublet
of doublets so each of these is a DD ABX pattern,
DD ABX pattern, and then you'll notice
that our alpha proton is a DD
and the doublet reflects the 2 different couplings.
In other words, the distance between lines 1 and 2 or 3
and 4 corresponds to this coupling,
to this 6 hertz coupling, and the distance between lines 1
and 3 and 2 and 4 corresponds to this coupling
to the 8 hertz coupling.
[ Pause ]
Thoughts or questions?
[ Inaudible question ]
Well, great question.
So the question is if the alpha proton is a doublet
of doublets shouldn't it be leaning a lot more?
You notice these guys are really tenting into each other
and this one is just barely tenting in.
So now look this is a 300 megahertz spectrum.
So the distance between these 2 looks like it's
about a tenth of a PPM.
So they're separated by about maybe it's two-tenths.
So, the distance between these 2 is about 60 hertz
because it's 300 megahertz so it's 300 hertz per PPM.
So they're separated by about 60 hertz and the J value is
about 14 hertz so that's the case where Delta nu over 12 is
about 4 or 5 whereas here the difference
between the alpha proton and the beta protons is about 1 hertz,
about 1 PPM, about 300 hertz, and the J value is 6 and 8.
So this is a case remember I said the big difference
between the AB and AX type of situation this is a case
where Delta nu over J is very big, 300 versus 6
or 8 is a factor of well over 10.
So they're effectively far apart
and you get very little tenting inward.
Other thoughts?
[ Inaudible question ]
So great question.
So this is in D20 the most hydrogens
on heteroatoms exchange and so most hydrogens on heteroatoms.
As a matter of fact, I will say I can give you exceptions
but I will say hydrogens on nitrogen, oxygen are replaced
with D. So they get replaced with deuterium.
Deuterium shows up in completely,
shows up not 500 megahertz but at about 80 hertz, 80 megahertz.
So they don't show up in the same spectrum
and the J values are so small that for intents
and purposes you don't see coupling plus they're exchanging
very quickly.
[ Inaudible question ]
Even without DCL they will exchange.
Because of DCL the amine is protenated
so the amine is an ammonium group and because
of DCL it dissolves whereas phenylalanine
in just pure water wouldn't be nearly as soluble.
All right so I've started to hint that different types
of coupling relationships have different coupling constants
and what I'd like to do at this point is to talk
about typical coupling constants and see how we can use them
to enhance our understanding.
So, this coupling constants generally
if you needed just one number to keep
in your head you could keep 7 hertz or 6, 6 to 8 hertz.
Let's say we're talking SP3 to SP3 right now
and if I needed a number, actually, I'm going to put this
as a general CH to CH.
I'll show you double bonds in a second but if you need 1 number
to keep in your head, 6-8 hertz or 7 hertz is a great number
to keep in your head without confirmational preference.
I'll say without a confirmational biases.
What do I mean by a confirmational bias?
Well, if there's a strong, strongly held relationship,
for example, if 2 hydrogens are locked
in an antiperiplanar relationship
so here we have 180-degree dihedral angle,
now our J value is going to be bigger.
It's going to be about 8 to 10 hertz.
So, for example, if you have 2 axial protons.
So I'll put this as JAX, AX, that would be a typical example
for 8-10 hertz where you're locked
into an axial relationship.
If you have something locked into an equatorial relationship
where now you have axial equatorial or equatorial,
now you're talking about a 60-degree dihedral angle
and so a typical JAX equatorial
or J equatorial equatorial is on the order of 2.
I'll put little tildas, little squiggles here just to indicate
that that's approximate 2 to 3 hertz.
[ Inaudible question ]
This is based exactly off of the Karplus curve.
So a general way of thinking about coupling is
that coupling comes from interaction of the nuclei
with electrons in the bond that polarize the next bond,
that polarize the next bond.
If at one extreme you have 180,
a 90-degree dihedral relationship
between those 2 bonds, you get no overlap of orbitals.
If at the other extreme you have 180, you get very good overlap
and antiperiplanar relationship
and at 0 you also get a good overlap
and so you see very large coupling constants at 180
or at 0 and very small at 90 or 60.
So, Karplus relationship is basically a relationship
between theta, your dihedral angle, and J,
and as sort of a general relationship would be if we go
from 0 to 90 to 180 that we go at 0 it's about 8 hertz.
We have kind of a cosine wave going down at 90 to a minimum
and up to about 10 at 180 degrees.
[ Pause ]
Now this is sort of for general kind of plain vanilla,
it's modulated, the coupling constants are modulated
by electronegativity and hybridization.
In general, electronegative substituents lead
to a smaller coupling constant.
In general, if you've got an SP2,
SP2 bond between the 2 atoms like a double bond,
you have bigger J values.
So let me show you a couple of examples.
So as I said, first number to keep in your head is
about 7 hertz, but now if you want to think
about some oddball situations, you can think
of like an aldehyde where now you have an electronegative
oxygen and you have an SP2 carbon here
and aldehydes are very funny in that your coupling constant is
on the order of 2 to 3 hertz so that's sort of unusual.
Alkenes, it's good the keep.
These are numbers that really are worth keeping
at your fingertips.
For a CIS alkene we're talking typically on the order of say 7
to 12 hertz with let's say 10 hertz being typical.
For a trans alkene we're talking maybe 12
to 18 hertz or 14 to 18 hertz.
Let's say 14 to 18 hertz with maybe 17 hertz being typical.
So these are all examples of vicinal couplings.
One more example falling right in that sort
of general 7 hertz range.
Let's take on a benzene as an example.
On a benzene as an example, we're talking maybe
for ortho coupling maybe 6 to10 hertz
with maybe 8 hertz or 7 hertz typical.
[ Pause ]
Two bond couplings tend to show more variation
than 3 bond couplings.
So, for example, if you have 2 carbons on a methylene group,
on an SP3 hybridized carbon with different substituents
on the carbon you can see anything from say, oh,
5 to 20 hertz depending on the electronegativity
if it's just sort of carbons on here maybe 14 hertz is typical.
[ Pause ]
If you have an SP2 carbon on a double bond,
you're talking maybe 0 to 2 hertz.
Maybe 1 hertz being typical.
[ Pause ]
So all of these are examples of vicinal and geminal coupling.
In other words, 2-bond and 3-bond coupling.
These are J2 HH couplings, these are J3 HH.
If you have anything more than that, if you have greater than
or equal to 4 bond coupling,
we're talking long-range coupling.
So, for example, J4 HH would be an example
of long-range coupling.
Normally in saturated systems, you don't see coupling,
but if you have a system
where you have certain geometrical relationships,
then you may see it.
So, we're talking about say a carbon not with its neighbor
but with a carbon 1 over.
So where does this come up?
Usually if you have intervening double bonds.
So, for example, [inaudible] systems.
We've talked a little bit about this in discussion section.
Typically let's say 0 to 3 hertz depending
on the geometrical relationship.
Metacoupling on a benzene same type of thing.
Let's say 1 to 3 hertz.
[ Pause ]
The only real situation that you can actually see a visible
splitting where you have just SP3 carbons is
if you have a locked relationship what is sometimes
called W coupling.
So usually you need a locked relationship
where you have a series of antiperiplanar bonds.
This occurs, for example, in the norbornene ring system.
They call it W coupling because you have a W-like relationship.
So say these 2 hydrogens
on a norbornene you can see you have the series
of antiperiplanar relationships
that make a W. Norbornene actually is loaded
with W coupling so you have another W relationship
across the ring like this
and there's even a third geometrical W relationship
hiding in the molecule like so.
[ Pause ]
So I want to pass out, there's 1 table that's really useful
in your book, and I want to pass it out just
because the stuff that's in the back of your book is
so much better when it's passed in front of your eyes rather
than simply waiting there in the back
of the book undiscovered and uncared about.
So this is the Appendix F I mentioned before and I just want
to show you how many good things are hiding
in this 1 little appendix here.
So everything we've talked about
and more is hiding in this appendix.
So we have our [inaudible] coupling and this is all going
to come up on homeworks.
If you're wondering what your typical [inaudible] couplings
are, you'll find answers here.
If you're wondering what happens if you have a double bond next
to a double gone, you'll find answers here.
If you're wondering, we've already seen
in the homework we've already seen coupling across acetylenes
and so how many bond couplings is that?
[ Inaudible question ]
Four-bond coupling, right?
So 1, 2, 3, 4 bond coupling, but if you ask as many people do
when they come up with some of the homework assignments here,
well, can you couple further?
Can you get 5 bond coupling?
The answer is right here waiting,
waiting for you to read about it.
Right over here typically you see a little coupling
across there.
Coupling on pyridines is going to come up won the homework
and already I think we see that very soon.
Now, all of this is given in more detail in Prech [phonetic].
So Prech has wonderful examples for pyridines, for thiophenes,
for all these types of systems off of real compounds and off
of typical examples, but here distilled
into really 1 little appendix and really 1 page
of the appendix is so much different good stuff that's
going to help you out with some of the problems
that you're working on.
So, okay, I think this is where I'd like to wrap it up today.
We're going to talk more about first order splitting next time
and we're going to walk through some examples of doublets
of doublets and triplets of doublets and doublets
of triplets and doublets of doublets of doublets
and that'll prepare you
for a workbook assignment that comes later on. ------------------------------bb79b27d2c62--