Chem 117E. Kwan Lecture 3: Coupling Constants
January 31, 2012.
Coupling Constants
Scope of Lecture
Eugene E. Kwan
energy diagrams
for J coupling
the chemical
shift
Helpful References

chemical vs.
magnetic
equivalence
the Fermi
contact
model
first- vs. secondorder
spectra
Hoye's method
what are
typical
couplings?
problem
solving
1. Nuclear Magnetic Resonance Spectroscopy... Lambert, J.B.;
Mazzola, E.P. Prentice-Hall, 2004. (Chapter 3)
2. The ABCs of FT-NMR Roberts, J.D. University Science
Books, 2000. (Chapter 10)
3. Spectrometric Identification of Organic Compounds (7th ed.)
Silverstein, R.M.; Webster, F.X.; Kiemle, D.J. Wiley, 2005.
(useful charts in the appendices of chapters 2-4)
4. Organic Structural Spectroscopy Lambert, J.B.; Shurvell,
H.F.; Lightner, D.A.; Cooks, R.G. Prentice-Hall, 1998.
5. Organic Structure Analysis Crews, P. Rodriguez, J.; Jaspars,
M. Oxford University Press, 1998.
I thank Professor Mazzola (Maryland/FDA) and Professor
Reich (Wisconsin-Madison) for providing useful material.
I thank Professor Reynolds (Toronto) for useful advice.
Key Questions
energy




A - J/2
B - J/2
A + J/2
B + J/2
AB
JJ
spectrum:
(1) What are coupling constants?
(2) How big are they?
H
H
Jortho = +7.5 Hz
H
H
+10 Hz
H F 8.3
(3) How can they be extracted from first-order multiplets?
1 2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
5 Hz
10 Hz
15 Hz
virtual coupling
3.5 3.0 2.5 2.0 1.5 1.0
3.5 3.0 2.5 2.0 1.5 1.0
spectrum A
spectrum B
1.30 1.25 1.20 1.15 1.10 1.05 1.00
Chemical Shift (ppm)
Chem 117E. Kwan Lecture 3: Coupling Constants
Review: What's a Coupling Constant?
Q: Which one of these spectra corresponds to that of ethanol?
Of course, the answer is spectrum B, because spectrum A
doesn't have any proton-proton couplings in it. Now, take a
close look at the bottom of the methyl peak:
Q: What are these tiny peaks off to either side?
These are the carbon-13 satellites. Recall that:
- carbon-13: I=1/2 (natural abundance, 1.11%)
- carbon-12: I=0 (natural abundance, 98.89%)
So what we're seeing is a mixture of two isotopomers (for the
sake of simplicity, let's assume that protium is 100%
abundant, and we can ignore deuterium, tritium, etc.):
~99% 1
H - 12
C and
~1% 1
H - 13
C.
I have bolded the protons, since those are the nuclei whose
magnetic resonance frequency we are observing. Carbon-12
has no magnetic moment, so there is no coupling to it. Carbon-
13 has I=1/2, so there is a doublet which is centered at the
frequency of the 1
H-12
C peak, but is separated by 1
JCH.
(Isotopic substitutions usually only have very small effects on
chemical shifts.)
Q: That's all great, but what exactly do coupling constants
actually correspond to in terms of energy levels and spin
states?
A: It's very complicated, and I defer a more rigorous
discussion till later in the course. For now, take note of this
hand-waving treatment of why line intensities obey Pascal's
triangle for I=1/2 nuclei:

Scenario 1: proton A is adjacent (vicinal) to a proton B.
Only consider the spin states of proton B. It's trivial here:
proton B:
There are only two states and only one way to arrange each
one, so proton A looks like a doublet.
Although this is the simplest case, there's a lot going on here:
(1) Only single-quantum transitions are observable. That
means that the only transitions which flip one of the spins are
allowed. A "double-quantum transition" from  to  does
not result in an observable signal. (This is the weirdness of
quantum-mechanical selection rules.)
(2) Protons A and B are not interacting, so flipping A gives a
transition of frequency A and flipping B gives a transition of
frequency of B. From the diagram, A is smaller than B, so
proton A has a smaller chemical shift than proton B.
(3) Although there are four single-quantum transitions, each is
doubly-degenerate and there will be two singlets of equal
intensity (assuming we don't have to worry about other spins
in the molecule).
Q: What is the diagram if there is a coupling J?
Chem 117E. Kwan Lecture 3: Coupling Constants
Scenario 2: proton A is adjacent (vicinal) to two protons, B
and C, but B and C have the same chemical shift.
Now, we get a triplet:
 


The double intensity for the middle line comes from the fact that
there are two permutations. For a quartet:
 






To me, this is not a very satisfactory description. For one thing,
why don't protons which have the same chemical shift split each
other? Why does the chemical shift difference between proton A
and its neighbors matter at all?
I won't answer these questions now, but here is how to start
thinking about it. Consider the energy diagram for the first
scenario, where we have protons A and B. But let's set the
coupling J to 0, so that A and B don't interact. What's the energy
level diagram for this system?
Remember,  and  are short-hand notations for nuclei which
are in the +1/2 and -1/2 states, respectively. By convention,
the  state is more stable than the  state. If I write , then
it's understood that I mean that proton A is in the  state, and
proton B is in the  state:
energy




AB
spectrum:
A
A
B
B
Here, we must consider the weakly-coupled or first-order case,
where J << |A - B|. Additionally, A and B must be chemically
and magnetically non-equivalent (but I'll tell you more about it
in a moment). The new diagram is (thin solid lines are the new
energy levels, bold lines are the old energy levels):
energy




A - J/2
B - J/2
A + J/2
B + J/2
AB
JJ
spectrum:
every level moves up or down
by J/4 to produce two doublets
Every level is shifted by J/4; every transition is shifted by J/2.
transition
energy
Chem 117E. Kwan Lecture 3: Coupling Constants
Strange stuff happens if we try to construct the same two
diagrams for the case where A = B. You might think the
diagrams should be:
energy


 
 



 
 - J/2
J=0 J>0
 - J/2
 + J/2
However, this would lead to a rather strange spectrum where
the two protons couple to each other, even when they have the
same chemical shift. But why are these diagrams incorrect?
The spins are indistinguishable, but this description allows me to
point to a particular nucleus and tell you what spin it is. Instead:
energy



 
1
2
 
symmetric
 
1
2
 
antisymmetric

If we exchange the nuclei in the symmetric combination, the
wavefunction does not change sign. Conversely, if we exchange
the nuclei in the antisymmetric combination, the wavefunction
gets multiplied by -1. Transitions between symmetric and
antisymmetric states are not allowed, so there are no dashed
lines between them.
What's the diagram look like for J > 0? It turns out that it's:
symmetric
symmetric
energy


 
1
2
   
1
2
 
3J/4
J/4


(If this seems a bit mysterious, don't panic--I'll derive it later on
in the course. But not today.) In principle, there might be
three possible lines: one at n, one at  - J/2, and one at  + J/2,
but the latter two are not allowed, as they involve transitions
between symmetric and antisymmetric states.
Anyways, there are a few really important messages here:
(1) "Normal behavior" is expected when J << . This means
that a nucleus coupled to n equivalent nuclei will give n+1
lines, with intensities corresponding to Pascal's Triangle.
(2) Equivalent* nuclei don't split each other. Equivalent
means they have to have the same chemical shift (chemical
equivalence) and be completely indistinguishable (magnetic
equivalence). More on this in a moment.
(3) Lines in a spectrum correspond to transitions between
energy levels. The only allowed transitions change the total
spin number by 1. Transitions between symmetric and
antisymmetric states are not allowed.
Chemical vs. Magnetic Equivalence
From this discussion, it's apparent that it's important to know
whether two nuclei are "the same" or not. For example, what
proton spectrum do you expect for 1,1-difluoroethene?
F
F (a) one line (b) two lines
(c) three lines
(d) more than three lines
Chem 117E. Kwan Lecture 3: Coupling Constants
Here is the observed spectrum at 90 MHz in CDCl3 (Lambert
and Mazzola, pg 101):
Uh oh: there are some 10 lines visible! Note that this odd
appearance will not be improved by going to a higher magnetic
field strength. Here's why. This is called an AA'XX' system:
FX'
FXHA
HA'
Jcis
Jtrans
Although symmetry makes the chemical shifts of protons A and
A' the exact same, proton A has different couplings to FX and FX'
than does A'. So A and A' are not the same, and A and A' can
split each other. There are essentially two different protons with
the same chemical shift coupled to two fluorines. Since J is not
less than the chemical shift difference (0), one gets a strange
spectrum. (Later, we'll see where all these extra lines come
from.) This leads to these definitions:
chemically equivalent: same chemical shift; nuclei can be
interchanged by a symmetry operation on the molecule
magnetically equivalent: chemically equivalent and have the
same coupling constant to any other NMR-active nucleus in the
molecule
(This last caveat means that the protons in 1,1-dichloroethene
are chemically and magnetically equivalent. Chlorine-35 and
chlorine-37 are quadrupolar nuclei and the fast relaxation
averages out the various spin states so they are not
considered by this criterion.)
From before, we also have these definitions:
homotopic: two nuclei can be interchanged by rotation;
chemically equivalent in all media
enantiotopic: molecule has no rotational axis of symmetry,
and two nuclei can be interchanged by a plane of reflection;
chemically equivalent in achiral media
diastereotopic: neither homotopic nor enantiotopic; chemically
nonequivalent
If this seems a bit abstract, try these examples. Are the bold
protons:
(a) chemically equivalent / magnetically equivalent?
(b) homotopic / enantiotopic / diastereotopic?
(We care because we want to know if they will split each
other/act as two equivalent nuclei for the n+1 rule.)
H H
H H
These protons are homotopic, chemically
equivalent, and magnetically equivalent.
Chem 117E. Kwan Lecture 3: Coupling Constants
H Me
H H
What about 3-methylcyclopropene?
H Me
H D
H Me
D H
These protons are related by reflection and are enantiotopic.
They are chemically and magnetically equivalent. This is an AX2
system.
How about these acetal protons?
O O
CH3
H3C CH3
HH
These are diastereotopic and chemically and magnetically
non-equivalent, despite the lack of any stereocenters! Suppose
the methyl group is "up" and we label one of the hydrogens:
O O
CH3
H3C CH3
HD Two chiral centers are
simultaneously created!
If the other proton gets labeled, a diasteromer is formed.
To conclude, here are some nomenclature points for describing
spin systems that you will come across:
AB: two protons are not chemically equivalent, but have similar
chemical shifts
AX: two protons have completely different chemical shifts
AA': two protons are chemically equivalent, but magnetically
non-equivalent
Common sense will tell you what happens when you mix and
match these. For example, 1,2-dichlorobenzene is a AA'XX'
system (why is that?).
The Fermi Contact Mechanism
Let's go back to the energy diagram for an AX system with an AX
coupling constant J:
energy




A - J/2
B - J/2
A + J/2
B + J/2
This diagram says that the  state is more stable than the 
state. But what if J < 0? That seems like an odd thing to say if
one interprets J as the spacing between lines, but with the energy
diagram, it can be understood as:
energy




A - J/2
B - J/2
A + J/2
B + J/2




A - J/2
B - J/2
A + J/2 B + J/2
(1) Positive J values mean that antiparallel states are stabilized
relative to parallel ones. Negative J = opposite.
(2) Spacings in first-order spectra = magnitude of J only.
(3) Second order spectra look different depending on the sign of
J. (Actually, only the relative signs of J matter.)
J > 0 J < 0
Chem 117E. Kwan Lecture 3: Coupling Constants
The Fermi Contact Mechanism
Couplings can be transfered by a number of different
mechanisms, but the most important is the Fermi contact
mechanism. For a more detailed discussion, please see:
"Analyzing and Interpreting NMR Spin-Spin Coupling Constants
Using Molecular Orbital Calculations." Autschbach, J.; Guennic,
B. Le J. Chem. Ed. 2007, 84, 156-170.
Consider a bond A-B. Here, the up and down arrows indicate
the spins of the nuclei and the electrons:
A B
Pauli exclusion
(1) Electron 1 will spend some time in the same space as
nucleus A. Because of the Pauli exclusion principle, the
spins will be opposed. This means that the spins of A and 1
are more likely to be antiparallel than parallel. This is called
"spin polarization."
(2) Similarly, (1 and 2) and (2 and B) should also be antiparallelpolarized.
Overall, this means A and B should be
antiparallel-polarized.
(3) Antiparallel-polarization means a positive J value. Thus,
one-bond coupling constants are almost always positive.
(4) Geminal (two bond) couplings are usually negative, but
numerous exceptions exist.
(5) Vicinal (three bond) couplings are generally positive.
(6) The sign of J can affect 2D NMR spectra.
(7) J coupling is also known as "indirect coupling." If molecular
tumbling is slow, dipolar or "direct" couplings D can also be
observed.
1 2
One Bond Coupling Constants
Because the Fermi contact mechanism depends on overlap
between electrons and the nucleus, the size of one bond
couplings reflects the average degree of s character in a bond's
hybrids.
For nonpolar C-H bonds, the approximate relationship is:
% s = 0.2 x 1
JCH
In general, these couplings are quite large--the closer two
nuclei are, the larger their coupling. There are two places these
are usually observed:
(1) Carbon-13 satellites in proton spectra. However, these can
often get coverd by other signals.
(2) HMBC spectra. This is a common type of 2D NMR
experiment we will look at later. A lot of NMR experiments
seek to sort one-bond from multiple-bond couplings (since
they're much bigger). Errors appear because molecules
have a spread of, rather than particular, 1
JCH values.
Occasionally, an anomalous value of 1
JCH can be a clue that a
bond has an unusual degree of s-character (e.g., cyclopropane
or epoxide). An average value for organic molecules is 140
Hz. Here are some values to calibrate you:
O
Si
CH3 CH3CH3
125 Hz
118 Hz
98 Hz
H3C Li
(For my Chem 106 students, why does MeLi have a very small
one bond C-H coupling constant?)
CH3OH
125 Hz
CH3Cl
142 Hz
150 Hz
O
162 Hz
180 Hz
136 Hz
O
137, 150 Hz
248 Hz
159 Hz
157 Hz
(olefin)
CH2Cl2
177 Hz
Chem 117E. Kwan Lecture 3: Coupling Constants
Geminal (Two Bond) Coupling Constants
These appear as parts of more complex multiplets and depend
a lot on the angle between the three atoms. Here, we will just
consider H-C-H coupling constants 2
JHH. Most values are
negative, although sp2
carbons can have postive values.
Substituents have important effects. In general:
EWG by induction: J is more positive (e.g. fluorine substituent)
EWG by resonance: J is more negative (e.g. carbonyl group)
An explanation involving the ligand group orbitals of the CH2
fragment has been given by Pople and Bothner-By and is
explained by Professor Reich here: http://www.chem.wisc.edu/
areas/reich/nmr/notes-5-hmr-4-gem-coupling.pdf.
Note that coupling in a H-C-D fragment is smaller than the
coupling in an H-C-H fragment (by about 1/6.5) due to the
smaller gyromagnetic ratio of deuterium.
Alkanes: -5 to -20 Hz
Olefins: -3 to +3 Hz
Halogens are both electron-withdrawing by induction and
electron-donating by resonance (J more positive for both):
CH4
-12.4 Hz
CH3Cl
-10.5 Hz
-12.6 Hz
CH2Cl2
-7.5 Hz
These are -conjugation effects:
O
+5.5 Hz-4.4 Hzangle effect
SiMe4
-14.1
O
-5.8 Hz-11.0 Hz
CH4
-12.4 Hz CH3
O
-14.9 Hz
C
CH3
N
-16.9 Hz
Olefin couplings are much smaller (it has been argued that there
are two competing mechanisms which cancel each other out):
H
H
H
H
+2.3 Hz
H
H
H
Br
-1.8 Hz
H
H
H
TMS
+3.8 Hz
H
H
H
Li
+7.4 Hz
H
H
H
Ph
+1.1 Hz
H
H
H
F
-3.2 Hz
A particular anomaly is formaldehyde, which has a 2
JHH of +40
Hz. This is a reinforcing effect from -withdrawl and -donation:
O
H
H
Vicinal (Three Bond) Coupling Constants
(Some material taken here from Professor Reich, lecture
5-HNMR-5; available at http://www.chem.wisc.edu/areas/reich/
chem605/index.htm.)
These are by far the most interesting coupling constants, as
they give stereochemical information. In general, the size of
the coupling constant 3
JHH depends on the H-C-C-H dihedral
angle. When the dihedral angle is 90°, the coupling is
small. When the dihedral angle is 0° or 180°, the coupling is
large. In general 180° gives a larger coupling than 0°. Here
the prototypical cases:
H
H
H
H
H
H
Jax-ax = 8-13 Hz Jax-eq = 1-6 Hz Jeq-eq = 0-5 Hz
(Unfortunately, this means ax/eq and eq/eq are not distinct.)
H
H HH
Jtrans = 12-24 Hz Jax-eq = 3-19 Hz
The trans coupling is
virtually always bigger,
even though the
ranges overlap.
Chem 117E. Kwan Lecture 3: Coupling Constants
Vicinal (Three Bond) Coupling Constants
The Karplus equation relates 3
JHH to the H-C-C-H angle (the
Bothner-By equation is similar):
These curves are qualitatively correct, but are not quantitatively
exact. Electronegative substituents will decrease the coupling
constants. Here are some more numbers:
H
H
Jortho = +7.5 Hz
HH
-8.0 Hz
H H
+11.5 Hz
H
H
+19.0 Hz
H H
+5.6 Hz
note that the cis is larger here:
H H
+9.0 Hz
oxygen decreases J values:
O
H H
+4.4 Hz
O
H H
+3.3 Hz
(The behavior right is because a
dihedral angle of 0° gives a larger
coupling than a dihedral angle of 120°)
Long-Range Couplings
Couplings over four or more bonds are usually small and do not
appear in spectra due to the limitations of linewidth. They can
appear when the coupling is transmitted through -systems:
H
H
allylic coupling
-3 to +1 Hz
(-1 Hz typical)
H H
homoallylic
0 to +3 Hz
(2 Hz typical)
HH
Jmeta = +1.4 Hz
H
Jpara = +0.7 Hz
H
"W couplings" are frequently encountered in polycyclic
frameworks where orbital overlaps are favorable:
OH
H
+1.7 Hz
HH
W coupling
(all bonds anti)
(-1 to 3 Hz)
Ring strain can increase the size of these couplings:
U coupling
(unusual)
(0-1 Hz)
H H
H
H
H
H
H
H
+7 Hz
+18 Hz
+10 Hz
In rare cases, "through space" couplings are possible if one
atom has lone pair electrons and the two nuclei are in van
der Waals contact:
CH3 F <0.5 H F 8.3
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
Chemical Shift (ppm)
First- vs. Second-Order Spectra
Chem 117E. Kwan Lecture 3: Coupling Constants
It's important to be able to determine coupling constants from
multiplet patterns. If there is no problem due to chemical shift
equivalence/magnetic non-equivalence, then a spectrum will be
first-order if the separation between the peaks is much greater
than the coupling:  > 10 J. Here is the transition from firstorder
behavior to second-order behavior in an ABX/AMX system
where JAB = 15, JBX = 7, JAX = 0 (simulated at 500 MHz):
Other multiplet patterns can be much more complicated. Here
is the transition from A3X to A3B:
When A and B get too close, A and X look like they're coupled,
but aren't actually coupled. This is called virtual coupling:
When the lines get closer together, they get a "roofed" look:
C C
X B
C
A
2.0 1.5 1.0 0.5 0
Chemical Shift (ppm)
7.4
489.0
496.2
504.2
511.0
996.4
1003.7
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
Chemical Shift (ppm)
-7.4
7.4
446.4
453.3
489.4
496.2
504.2
511.4
X A3
B X A
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
Chemical Shift (ppm)
-11.8
0.8
3.2
5.3
20.3
483.9
496.5
499.0
501.1
503.5
516.2
apparent virtual
coupling to A and B
more lines
than expected
X A,B
B A3
Chem 117E. Kwan Lecture 3: Coupling Constants
Hoye's Method
If you have a multiplet, you must first determine if it is first- or
second order:
first-order multiplet: use Hoye's method
second-order multiplet: use spectral simulation
The behavior of second-order multiplets is complex and can be
analyzed with the WINDNMR program (available free of charge
from Professor Reich: http://www.chem.wisc.edu/areas/reich/plt/
windnmr.htm). One can predict the coupling patterns for a
variety of guesses and then "fit" the observed pattern. This is
time-consuming but works very well. (This is different from
using empirical or quantum mechanical methods to predict NMR
spectra; those just predict the values of shifts and couplings,
rather than the appearance of the spectrum directly.)
Fortunately, most of the molecules we work with are relatively
unsymmetrical. In the high-powered magnetic fields of modern
spectrometers, they generally give first-order multiplets. Here is
how to extract the couplings from a first-order multiplet. This
method was developed by Hoye (Minneosta) and co-workers:
"A Practical Guide to First-Order Multiplet Analysis in 1
H NMR
Spectroscopy." Hoye, T.R.; Hanson, P.R.; Vyvyan V. J. Org.
Chem. 1994, 59, 4096-4103.
"A Method for Easily Determining Coupling Constant Values..."
Hoye, T.R.; Zhao, H. J. Org. Chem. 2002, 67, 4014-4016.
(The second paper is the important one.) Before we begin:
(1) First-order multiplets are centrosymmetric. (This is
necessary, but not sufficient.) Without exception, JAB = JBA.
(2) Both of the above multiplets are centrosymmetric, but only
the left one is first-order. Every first-order multiplet has
peak intensities that sum to 2n
. In this fashion, every
multiplet can be considered an n-th order doublet of doublets
of doublets...
For example, the left hand multiplet is a fourth order
multiplet that has 24
=16 "components" (numbered below):
1 2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
line of symmetry
The same cannot be done for the right hand multiplet.
(3) If some of the couplings are the same, a "Pascal's triangle"
pattern becomes embedded in the pattern. For example,
the coupling constants in the multiplet above are 5, 5, 5, and
15 Hz. A three-fold degeneracy corresponds to a quartet by
the n+1 rule. Thus, this is a doublet of quartets (with a
merging of the center lines):
+
1:3:3:1 1:3:3:1 1:3:3:2:3:3:1
overlap in
central line
(4) We will adopt the convention that 1 2 3J J J  
(5) Largely first-order multiplets that display "leaning" can be
analyzed, but the results might not be very accurate.
Simulation is the best strategy there.
Chem 117E. Kwan Lecture 3: Coupling Constants
Hoye's Method
Here is the procedure of Hoye and co-workers:
(0) Verify this is a first-order multiplet.
(1) Determine the order of the multiplet (i.e., determine n in 2n
)
and assign components above.
(2) J1 is the distance from component 1 to component 2; J2 is the
distance from component 1 to component 3.
However, further distances, like 1-4 are not necessarily
couplings.
(3) If there is a component J1 + J2 Hz from component 1, then
remove it from consideration.
(4) The distance from 1 to the next highest remaining component
is J3.
(5) Remove from consideration any components J1+J3, J2+J3,
J1+J2+J3. The distance from 1 to the next highest remaining
component is J4. Etc.
If this seems too abstract, let's try it on the multiplet from the
last page:
1 2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
5 Hz
10 Hz
15 Hz
these are not couplings;
they are merely distances
to orient us
remember to do
all of this in Hz,
not ppm!
Let's go through the checklist:
(0) This is a centrosymmetric multiplet: a dddd, which could
have degenerate couplings.
(1) Components assigned.
(2) J1 = 1-2, which is 5 Hz. J2 = 1-3, which is also 5 Hz. (The
fact that these are the same should not bother you.)
(3) J1+J2 = 10 Hz. This corresponds to components 5, 6, and 7.
However, we only remove one component (5) from
further consideration. If I were doing it on paper, I would
write:
1 2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
circles = a coupling
slash = removed from
consideration
nothing = not yet
considered
(5) J2+J3=J1+J3=10 Hz. Thus, I removed components 6 and 7
from further consideration (once for every duplication):
1 2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
(6) The final coupling is J4 (since
this is a fourth-order doublet.
J4 = 1-8 = 15 Hz. Thus:
J1 = J2 = J3 = 5 Hz
J4 = 15 Hz
This completes the analysis.
(It can be checked by drawing
a coupling tree diagram.)
(4) The next highest component is 4. Thus, J3 is 1-4, which is
5 Hz. (Once again, it doesn't matter this is the same as the
values already extracted.)
Chem 117E. Kwan Lecture 3: Coupling Constants
Think you know what's going on? Here is a question which is
much trickier than it first appears. This question is adapted from
Lambert and Mazzola, page 127.
Please consider cis- and trans-dimethyl 1,2-cyclopropane
dicarboxylate:
Practice Problem: Cyclopropane Stereoisomers
MeO2C
CO2Me
MeO2C CO2Me
The 1D proton NMR spectrum of one of these isomers is shown
below:
Here are expansions of every peak (labels in Hz):
(1) Consider these pairs of protons in each isomer and determine
if they have homo-, enantio-, or diastereotopic relationships:
(2) How many distinct signals should the cis isomer have? What
about the trans isomer?
(3) Determine the coupling constants in the spectra above.
(4) Analyze the couplings to determine if this is the cis or trans
isomer. For reference, these are the couplings in cyclopropane
itself: geminal, -4.3; trans vicinal, 5.6; cis vicinal, 9.
MeO2C CO2Me
H H
MeO2C CO2Me
H H
3.5 3.0 2.5 2.0 1.5 1.0
Chemical Shift (ppm)
1.01.12.25.8
1 2 3 4
2.05 2.04 2.03 2.02 2.01 2.00 1.99
Chemical Shift (ppm)
1002.6
1009.3
1011.1
1017.6
2
1.645 1.640 1.635 1.630 1.625 1.620 1.615 1.610 1.605 1.600 1.595
Chemical Shift (ppm)
800.6
805.8
807.3
812.6
814.1
819.1
3
1.225 1.220 1.215 1.210 1.205 1.200 1.195 1.190 1.185 1.180 1.175
Chemical Shift (ppm)
588.8
594.0
597.3
602.5
605.8
611.0
4
Chem 117E. Kwan Lecture 3: Coupling Constants
(1) Consider these pairs of protons in each isomer and determine
if they have homo-, enantio-, or diastereotopic relationships.
Practice Problem: Cyclopropane Stereoisomers
MeO2C CO2Me
H H
H H
MeO2C CO2Me
D
MeO2C CO2Me
D
These are diastereomers: two different meso compounds. Thus,
these protons area diastereotopic.
MeO2C CO2Me
H H
H H
MeO2C CO2Me MeO2C CO2Me
D D
These compounds are enantiomers so these protons are
enantiotopic.
Cis Isomer
MeO2C H
H CO2Me
H H
The cis compound is meso.
MeO2C
D
Trans Isomer
MeO2C H
H CO2Me
H H
CO2Me
MeO2C
D
CO2Me
MeO2C
CO2Me
Note that this is a C2-symmetric, chiral molecule with two
enantiomers, of which we'll just consider one:
MeO2C
CO2Me
Rotation about the C2-axis simply turns one enantiomer into
itself and does not interconvert them. Now, labelling one or
the other proton in the pair gives these stereoisomers:
The same reasoning applies: these are homotopic and
chemically equivalent, but not magnetically equivalent.
MeO2C
CO2Me
MeO2C
CO2Me
D
D
The compound on the right can be rotated about its C2 axis:
MeO2C
CO2Me
D MeO2C
CO2MeD
This is exactly the same as the other compound. Therefore,
these protons are homotopic and chemically equivalent.
However, they are not magnetically equivalent:
MeO2C H
H CO2Me
H
vicinal cis coupling
to circled proton
vicinal trans
coupling to
circled proton
For the other pair:
(2) How many distinct signals should the cis isomer have? What
about the trans isomer?
Cis Isomer: There are four protons, but one pair is enantiotopic,
so there should be three signals overall in an "AMX2" or "ABX2"
system, depending on the chemical shift.
Trans Isomer: There are two sets of homotopic, but magnetically
non-equivalent protons. This means an AA'XX' or AA'BB'
system--two signals, but there will definitely be some secondorder
behavior in the coupling patterns.
(3) Determine the coupling constants in the spectra above.
The spectra are clearly first-order and therefore Hoye's method
is applicable. This also suggests the cis isomer!
2.05 2.04 2.03 2.02 2.01 2.00 1.99
Chemical Shift (ppm)
1002.6
1009.3
1011.1
1017.6
2
1.645 1.640 1.635 1.630 1.625 1.620 1.615 1.610 1.605 1.600 1.595
Chemical Shift (ppm)
800.6
805.8
807.3
812.6
814.1
819.1
3
1.225 1.220 1.215 1.210 1.205 1.200 1.195 1.190 1.185 1.180 1.175
Chemical Shift (ppm)
588.8
594.0
597.3
602.5
605.8
611.0
4
1 2 34 56 7 8
J1 = 1-2 = 6.5
J2 = 1-3 = 8.3
J1 = 1-2 = 5.0
J2 = 1-3 = 6.5
J3 = 1-4 = 6.5
1 2 3 4 5 6 7 8
J1 = 1-2 = 5.2
J2 = 1-3 = 8.5
J3 = 1-4 = 8.5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Cyclopropane Stereoisomers
Applying Hoye's method:
(1) If A is coupled to B with coupling constant J, B must be
coupled to A with the same coupling constant J.
(2) However, spectra are rarely completely first-order or noisefree.
So it's common to see couplings be "out of alignment"
by a little bit.
(3) Here, we have this spin system with these averaged
couplings:
Let's assume this is the cis isomer. 2 integrates to two protons,
and must be next to the ester because those protons are
enantiotopic. This also makes sense based on chemical shift.
Because they are both chemically and magnetically equivalent,
they do not couple to each other. This gives this assignment:
2 3 4
6.5 5.1 5.1
8.4 6.5 8.4
6.5 8.4
MeO2C CO2Me
H H
H H
2 2
34
Why did I assign it this way? The couplings in cyclopropane
are: geminal, -4.3; trans vicinal, 5.6; cis vicinal, 9. Therefore,
the smallest coupling on 3 and 4 is the geminal coupling. That
doesn't tell us anything about stereochemistry, so we can
ignore it:
2 3 4
6.5 5.1 5.1
8.4 6.5 8.4
6.5 8.4
4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2
Chemical Shift (ppm)
1.02.02.11.0
1 2 3 4
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Cyclopropane Stereoisomers
MeO2C CO2Me
H H
H H
2 2
34
Similarly, 4 has two 6.5 Hz couplings to protons 2. This makes
sense for a trans coupling in a cyclopropane. Note that 2 does
not couple to itself, which is why it's only a second-order "dd."
Now, a good spectroscopist always considers the other
possibilities, which, here, is only one possibility: the trans isomer:
2 3 4
6.5 5.1 5.1
8.4 6.5 8.4
6.5 8.4
Now, consider the proton marked tentatively as 3. It has two
protons in a vicinal relationship to it: 2 and the other 2 (which is
the same). Therefore, it has a doubly-degenerate coupling.
The coupling is 6.5 Hz, which makes sense for a trans vicinal
coupling:
Cyclopropane: geminal, -4.3; trans vicinal, 5.6; cis vicinal, 9.
trans: 6.5 Hztrans: 6.5 Hz
MeO2C H
H CO2Me
H H
2
2
34
Now, you can see we have a number of problems. First, 3 and
4 ought to be pseudo-equivalent, but are actually widely
separated. Second, 3 should have a cis coupling to one of the
2's and a trans coupling to the other. But clearly, 3 only has two
different kinds of couplings. And third, why do 3 and 4 couple?
An Unknown Structure
Here's another practice question which doesn't involve a lot of
measuring, but does involve a lot of thinking! (This and the final
problem can be found on page 128 of Lambert and Mazzola).
An unknown compound has the molecular formula C6H12O6
and has the following 1D proton NMR spectrum in D2O (solvent
and exchangeable protons are not shown, 300 MHz):
The carbon-13 spectrum has exactly four resonances, all of
which lie between 71 and 75 ppm. What is its structure?
Hint 1: This compound is commercially available. What type of
compound does the molecular formula suggest to you?
Hint 2: Protons next to two oxygens typically appear around 5
ppm.
Hint 3: Ignore the size of the couplings for now and begin by
considering the connectivity of this unknown.
Couplings (Hz): 1 (2.8x2), 2 (9.6x2), 3 (2.8, 9.6), 4 (9.6x2).
Chem 117E. Kwan Lecture 3: Coupling Constants
An Unknown Structure
Ask any high school student worth his or her salt what molecule
has the formula C6H12O6, and they'll tell you glucose. So let's
start there:
O OH
HO
HO
OH
OH
Now, notice I haven't drawn the stereochemistry. In all of these
problems, I advise you to determine connectivity first and
work out the stereochemistry later. Of course, sometimes
couplings and other stereochemical information can also give
clues to a connectivity pattern, but I find this is a good way to
narrow down a lot of confusing information into what's essential.
Why can't it be glucose? For one thing, it has an anomeric
proton, and our structure has nothing around 5 ppm. Further,
it only has four kinds of carbons, while glucose has relatively low
symmetry.
OK, what else could it be? I know glucose can exist in a bunch
of forms involving aldehyde intermediates:
O OH
HO
HO
OH
OH
OH O
HO
HO
OH
OH
HO
HO
O
HO OH
OH
However, all of these structures have aldehydes (there's no
carbonyl resonance) or anomeric protons. Another crucial clue
comes from the fact that these sugars have seven, not six, nonexchangeable
protons, and we need six.
So, clearly, we need to think a little harder. If you're feeling a bit
lost, let me distract you with an excerpt from one of my favorite
books, Surely You're Joking, Mr. Feynman! Adventures of a
Curious Character:
(Here, Feynman is talking about trying to understand the
complicated theorems mathematicians come up with.)
"I had a scheme, which I still use today when somebody is
explaining something that I'm trying to understand: I keep
making up examples. For instance, the matematicians would
come in with a terrific theorem, and they're all excited. As
they're telling me the conditions of the theorem, I construct
something that fits all the conditions. You know, you have a set
(one ball)--disjoint (two balls). Then the ball turn colors, grow
hairs, or whatever, in my head as they put more conditions on.
Finally they state the theorem, which is a some dumb thing
about the ball which isn't true for my hairy green ball thing, so I
say, 'False!'"
(This is more or less what we're trying to do here.)
"If it's true, they get all excited, and I let them go on for a while.
Then I point out my counterexample.
'Oh. We forgot to tell you it's Class 2 Hausdorff homomorphic.'
'Well, then,' I say, 'It's trivial! It's trivial!" By that time I know
which way it goes, even though I don't know what Hausdorff
homomorphic means."
(Evidently, some degree of fakery is necessary, even for the
best of us.) Now, back to the problem...
If you look at the topology of these structures, you can see the
structure cannot be an ether of any size. The reason is that
if two oxygens get connected, then you lose an exchangeable
proton that has to be made up by a non-exchangeable one, and
then you end up with more than six. (You might be able to get
around this if you relax some of the other conditions, like not
having an oxygen on every carbon, but you can't do it with the
given functional group pattern.)
But there's clearly one degree of unsaturation!
Chem 117E. Kwan Lecture 3: Coupling Constants
An Unknown Structure
The open chain that we have to add a degree of unsaturation to
is this:
HO
OH
OH
OH
OH
OH
If connecting oxygens to form an ether are no good, then the
logical alternative is to connect carbons. (You might also think
about making peroxide-type structures. However, this will not
solve the problem of having only six non-exchangeable protons.
Also, remember what I said about horses vs. zebras!)
Here are two possibilities (there are more). We are connecting
the carbons marked by asterisks:
HO
HO
OH
OH
OH
OH
OH
HO
HO OH
OH
OH
HO
OH
OH
OH
OH
OH
*
*
*
*
HO
OH
OH
OH
OH
OH
*
*
*
*
However, the carbon-13 says that we don't have any
quaternary centers with oxygens on them. (My Ph.D. advisor,
Professor Evans, says one cannot refer to tertiary carbinols as
"quaternary." True enough, but in NMR, using the term
quaternary to designate any non-protonated carbon is very
useful.) Thus, the six-membered ring is the only possible ring
size.
We have arrived at inositol, which looks sort of like a sugar,
but isn't really very sweet. It's thought to be an important
messaging compound in biology. It has 9 stereoisomers. Now,
let's look at the couplings and overall symmetry to determine
which one it is. Recall there are four kinds of carbons.
This implies some sort of symmetry. The simplest kind of
symmetry is a plane of symmetry. Such a plane can either pass
through two vertices or between two vertices:
HO
HO
OH
OH
OH
OH
through two vertices: between two vertices:
HO
HO
OH
OH
OH
OH
The latter is bad news, because that would mean only three
kinds of carbons. Here are all the possibilities. Which of these
is viable?
OH
OH
OHHO
HO
OH
OH
OH
OHHO
HO
OH
OH
OH
OHHO
HO
OH
OH
OH
OHHO
HO
OH
OH
OH
OHHO
HO
OH
OH
OH
OHHO
HO
OH
OH
OH
OHHO
HO
OH
OH
OH
OHHO
HO
OH
OH
OH
OHHO
HO
OH
myo scyllo muco
neo allo epi
D-chira L-chira cis
Chem 117E. Kwan Lecture 3: Coupling Constants
An Unknown Structure
We could sort through the stereoisomers on symmetry grounds,
but it's easier to just guess a structure that has a plane of
symmetry that passes through two vertices. I'll need to
consider the couplings, too:
Couplings (Hz): 1 (2.8x2), 2 (9.6x2), 3 (2.8, 9.6), 4 (9.6x2).
As a reminder, 1 and 4 are worth 1H, while 2 and 3 are worth
2H. This means 1 and 4 lie along the symmetry plane:
HO
OH
I've just arbitrarily drawn them in the equatorial positions, which
is always a good place to start. However, in this case, it's
wrong, because the coupling suggest that 1 is equatorial,
while 4 is axial (if they were both equatorial, then neither of
them would have trans-diaxial couplings of 10 Hz):
HO
OH
H
H
H
H
HO
HO
Because 4 has to have two diaxial couplings, it's adjacent
hydroxyls must be equatorial. But what about the hydroxyls
adjacent to 1? Here are the two possibilities, given that
axial-equatorial and equatorial-equatorial couplings are
relatively similar in size:
trans couplings
are large: about
10 Hz
4
1
HO
OH
H
H
H
H
HO
HO
OH
OH
H
H
HO
OH
H
H
H
H
HO
HO
H
H
OH
OH
(The two other ax/eq possibilities don't obey the symmetry
requirements.)
The key clue here is that 2 and 3 both have large couplings,
meaning that the hydrogens must be axial:
HO
OH
H
H
H
H
HO
HO
OH
OH
H
H4
1
3
2
How did I make these assignments, despite the very similar
chemical shifts? 2 has two axial couplings, while 3 has only
one. This completes the problem: this is myo-inositol.
Note that in more complicated systems involving complex
multiplet patterns in stereochemically rich and conformationally
rigid environments (e.g., steroids), a wide range of short- and
long-range couplings can be seen. The key there is to focus on
the large vicinal ones. We'll see more of this later.
Practice Problem: Carbon-Fluorine Couplings
Fluorine-19 is 100% abundant, has a spin of 1/2, and has a
gyromagnetic ratio which is almost as high as proton's (94%),
so it's almost as sensitive as proton (about 83%). Because
carbon-13 spectra are not fluorine-19 decoupled, fluorine
couplings can and do appear in 1D carbon-13 spectra. The
couplings range in size. Here it is for fluorobenzene:
F
1
JCF = 245 Hz
2
JCF = 21
3
JCF = 8
4
JCF = 3
The size of a heteronuclear coupling is proportional to the
product of their gyromagnetic ratios, so fluorine behaves very
similarly to proton (e.g., carbon-fluorine and carbon-proton
couplings are similar in size).
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
In the next two problems, we'll look at the carbon-13 spectra of
some biaryls (courtesy Dr. Jaclyn Henderson). Please assign
the spectrum of this compound. There is no reason to "fly blind"
here, so the predicted spectra (ChemDraw 9) are provided right.
F
1
JCF = 245 Hz
2
JCF = 21
3
JCF = 8
4
JCF = 3 111.6
130.6
111.6
160.0
113.7
160.0 154.6
149.2
120.8
137.2
124.2F
F
N
F
F
N
100 MHz, CDCl3
predicted carbon-13 shifts
160 155 150 145 140 135 130 125 120 115
Chemical Shift (ppm)
80.0 77.5 75.0
Chemical Shift (ppm)
160 158 156 154 152 150
Chemical Shift (ppm)
16010.3
16017.6
16260.6
16267.9
136 135 134 133 132 131 130 129 128 127 126 125 124 123
Chemical Shift (ppm)
13082.2
13092.4
13102.7
119.0 118.5 118.0 117.5 117.0
Chemical Shift (ppm)
11863.7
11882.0
11899.5 112.0 111.5
Chemical Shift (ppm)
11232.8
11239.4
11251.9
11258.5
160 155 150 145 140 135 130 125 120 115
Chemical Shift (ppm)
80.0 77.5 75.0
Chemical Shift (ppm)
160 158 156 154 152 150
Chemical Shift (ppm)
16010.3
16017.6
16260.6
16267.9
136 135 134 133 132 131 130 129 128 127 126 125 124 123
Chemical Shift (ppm)
13082.2
13092.4
13102.7
119.0 118.5 118.0 117.5 117.0
Chemical Shift (ppm)
11863.7
11882.0
11899.5 112.0 111.5
Chemical Shift (ppm)
11232.8
11239.4
11251.9
11258.5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
F
F
N
100 MHz, CDCl3
(1) How many distinct carbon signals are there? There are 9, due to the symmetry plane in the
difluoroaryl ring.
(2) Does the spectrum have 9 signals in it? Clearly, there are more than 9 signals here due to
carbon-fluorine couplings. First, recognize that the peak near 77 ppm is deuterochloroform,
which is showing a 1:1:1 triplet due to coupling to an I=1 nucleus, deuterium. Next, label the
carbons (blue numbers at the bottom).
(3) Why are there ten signals, not nine?
1 2 3 4 5 6 7 8 9 10 CDCl3
160 155 150 145 140 135 130 125 120 115
Chemical Shift (ppm)
80.0 77.5 75.0
Chemical Shift (ppm)
160 158 156 154 152 150
Chemical Shift (ppm)
16010.3
16017.6
16260.6
16267.9
136 135 134 133 132 131 130 129 128 127 126 125 124 123
Chemical Shift (ppm)
13082.2
13092.4
13102.7
119.0 118.5 118.0 117.5 117.0
Chemical Shift (ppm)
11863.7
11882.0
11899.5 112.0 111.5
Chemical Shift (ppm)
11232.8
11239.4
11251.9
11258.5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
F
F
N
100 MHz, CDCl3
(3) Why are there ten signals, not nine? The numbering isn’t right. Recall the couplings in
fluorobenzene:
At 100 MHz, a 245 Hz coupling is 2.45 ppm. There are four or five notches between the peaks,
which is about right (0.5 ppm/notch). More accurately, this is a doublet of doublets with a
coupling of 16267.9-16017.6 = 250.3 Hz. Thus, this is a one-bond
coupling. The smaller coupling is 7.3 Hz, which is a three-bond/vicinal
coupling:
1 1 2 3 4 5 6 7 8 9 CDCl3
F
1
JCF = 245 Hz
2
JCF = 21
3
JCF = 8
4
JCF = 3
F
F
N
F
F
N
1 1
1 1
one-bond coupling vicinal coupling
160 155 150 145 140 135 130 125 120 115
Chemical Shift (ppm)
80.0 77.5 75.0
Chemical Shift (ppm)
160 158 156 154 152 150
Chemical Shift (ppm)
16010.3
16017.6
16260.6
16267.9
136 135 134 133 132 131 130 129 128 127 126 125 124 123
Chemical Shift (ppm)
13082.2
13092.4
13102.7
119.0 118.5 118.0 117.5 117.0
Chemical Shift (ppm)
11863.7
11882.0
11899.5 112.0 111.5
Chemical Shift (ppm)
11232.8
11239.4
11251.9
11258.5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
F
F
N
100 MHz, CDCl3
(4) Height of Peaks. Recall that quaternary carbons are often shorter than protonated ones (this is
due to differential relaxation and a lack of nOe enhancment; to be discussed in detail later).
According to the ChemDraw predictions, the only carbons which should be really downfield are
the ones next to nitrogen. One of them is quaternary, and one of them is protonated, which
makes sense if this assignment is made:
1 1 2 3 4 5 6 7 8 9 CDCl3
2 3
111.6
130.6
111.6
160.0
113.7
160.0 154.6
149.2
120.8
137.2
124.2F
F
N
predicted carbon-13 shifts
F
F
N
1
1
3
2
(4) Fluorine couplings.
150.1 150.0 149.9 149.8 149.7 149.6 149.5 149.4 149.3 149.2 149.1
Chemical Shift (ppm)
2 3
3 has small vicinal couplings to two
fluorines, while 2 doesn’t, which is
why it looks fatter (look at the peak
height at half maximum).
160 155 150 145 140 135 130 125 120 115
Chemical Shift (ppm)
80.0 77.5 75.0
Chemical Shift (ppm)
160 158 156 154 152 150
Chemical Shift (ppm)
16010.3
16017.6
16260.6
16267.9
136 135 134 133 132 131 130 129 128 127 126 125 124 123
Chemical Shift (ppm)
13082.2
13092.4
13102.7
119.0 118.5 118.0 117.5 117.0
Chemical Shift (ppm)
11863.7
11882.0
11899.5 112.0 111.5
Chemical Shift (ppm)
11232.8
11239.4
11251.9
11258.5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
F
F
N
100 MHz, CDCl3
(5) Other Fluorine-Coupled Peaks. What are the other peaks? Carbon 8 is a true 1:2:1 triplet with
J = 17.5x2 Hz, which can only happen if it has two geminal (two-bond) couplings to two equal
fluorines. Therefore, it must be the ipso carbon on the fluorinated ring. It also has a very small
height, which makes sense for a quaternary carbon.
1 1 2 3 4 5 6 7 8 9 CDCl3
4 5 6 7
8 9
F
F
N
1
1
3
28
160 155 150 145 140 135 130 125 120 115
Chemical Shift (ppm)
80.0 77.5 75.0
Chemical Shift (ppm)
160 158 156 154 152 150
Chemical Shift (ppm)
16010.3
16017.6
16260.6
16267.9
136 135 134 133 132 131 130 129 128 127 126 125 124 123
Chemical Shift (ppm)
13082.2
13092.4
13102.7
119.0 118.5 118.0 117.5 117.0
Chemical Shift (ppm)
11863.7
11882.0
11899.5 112.0 111.5
Chemical Shift (ppm)
11232.8
11239.4
11251.9
11258.5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
F
F
N
100 MHz, CDCl3
(6) Other Fluorine-Coupled Peaks. Carbon 9 is quite intense, especially for a peak that has a
splitting in it. It must be ortho to the fluorines. This is an AA’XX’ system, where A and A’ are
these ortho carbons and X and X’ are the fluorines—the carbons and fluorines are chemically
equivalent but magnetically inequivalent and therefore this is a second-order multiplet pattern. It
is reminiscent of an “AB quartet.” The coupling is approximately 6 Hz, but note that the line
spacing and the actual coupling no longer coincide directly for a second-order
spectrum.
1 1 2 3 4 5 6 7 8 9 CDCl3
4 5 6 7
8 9
F
F
N
1
1
3
28
9
9
160 155 150 145 140 135 130 125 120 115
Chemical Shift (ppm)
80.0 77.5 75.0
Chemical Shift (ppm)
160 158 156 154 152 150
Chemical Shift (ppm)
16010.3
16017.6
16260.6
16267.9
136 135 134 133 132 131 130 129 128 127 126 125 124 123
Chemical Shift (ppm)
13082.2
13092.4
13102.7
119.0 118.5 118.0 117.5 117.0
Chemical Shift (ppm)
11863.7
11882.0
11899.5 112.0 111.5
Chemical Shift (ppm)
11232.8
11239.4
11251.9
11258.5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
F
F
N
100 MHz, CDCl3
(7) Other Fluorine-Coupled Peaks. The only other peak with coupling is 5, which must be the
carbon meta to the fluorines. The couplings are 2x10.3 Hz, which are right for an anti vicinal
coupling.
1 1 2 3 4 5 6 7 8 9 CDCl3
4 5 6 7
8 9
F
F
N
1
1
3
28
9
9
5
160 155 150 145 140 135 130 125 120 115
Chemical Shift (ppm)
80.0 77.5 75.0
Chemical Shift (ppm)
160 158 156 154 152 150
Chemical Shift (ppm)
16010.3
16017.6
16260.6
16267.9
136 135 134 133 132 131 130 129 128 127 126 125 124 123
Chemical Shift (ppm)
13082.2
13092.4
13102.7
119.0 118.5 118.0 117.5 117.0
Chemical Shift (ppm)
11863.7
11882.0
11899.5 112.0 111.5
Chemical Shift (ppm)
11232.8
11239.4
11251.9
11258.5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
F
F
N
100 MHz, CDCl3
(8) Pyridine Ring. What remains is to assign carbons 4, 6, and 7. Here is where chemical shift
predictions come in handy again: they show that the carbon para to the pyridine nitrogen should
be the most downfield. But how can 6 and 7 be distinguished?
1 1 2 3 4 5 6 7 8 9 CDCl3
4 5 6 7
8 9
111.6
130.6
111.6
160.0
113.7
160.0 154.6
149.2
120.8
137.2
124.2F
F
N
predicted carbon-13 shifts
F
F
N
1
1
3
28
9
9
5
4
160 155 150 145 140 135 130 125 120 115
Chemical Shift (ppm)
80.0 77.5 75.0
Chemical Shift (ppm)
160 158 156 154 152 150
Chemical Shift (ppm)
16010.3
16017.6
16260.6
16267.9
136 135 134 133 132 131 130 129 128 127 126 125 124 123
Chemical Shift (ppm)
13082.2
13092.4
13102.7
119.0 118.5 118.0 117.5 117.0
Chemical Shift (ppm)
11863.7
11882.0
11899.5 112.0 111.5
Chemical Shift (ppm)
11232.8
11239.4
11251.9
11258.5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
F
F
N
100 MHz, CDCl3
(9) Pyridine Ring. Chemical shifts predict that 6 should be on the left and 7 should be on the right.
Examining an expansion reveals that 6 is much wider than 7, consistent with two unresolved,
small four-bond couplings to fluorine, confirming the predictions. This completes the assignment.
1 1 2 3 4 5 6 7 8 9 CDCl3
4 5 6 7
111.6
130.6
111.6
160.0
113.7
160.0 154.6
149.2
120.8
137.2
124.2F
F
N
predicted carbon-13 shifts
126.0 125.5 125.0 124.5 124.0 123.5 123.0
Chemical Shift (ppm)
6 7F
F
N
1
1
3
28
9
9
5
4
76
final answer
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
(10) Proton Spectrum. We won’t bother interpreting this, but I just wanted to point out that although
the proton spectrum is very well resolved, it is a very complex series of partially overlapping
second-order multiplets due to the symmetry, and therefore magnetic inequivalence, of the
fluorines. (This compound came from a Pd-catalyzed cross-coupling of the difluorarylboronic acid
and the 2-chloropyridine, so the sample is more or less regioisomerically pure.
But a mixture of isomeric pyridines, for example, might look like total gibberish!)
F
F
N
400 MHz, CDCl3
8.7 8.6 8.5 8.4 8.3 8.2 8.1 8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7
Chemical Shift (ppm)
2.02.01.01.01.0
155.0 154.5 154.0 153.5 153.0 152.5 152.0 151.5 151.0 150.5 150.0 149.5 149.0 148.5 148.0 147.5 147.0
Chemical Shift (ppm)
14974.0
14980.6
15221.3
15228.7
15361.9
15367.0
15601.9
15607.8
152 144 136 128 120 112 104 96 88 80 72 64 56 48 40 32 24 16
Chemical Shift (ppm)
145 144 143
Chemical Shift (ppm)
14483.7
14486.6
14495.4
14499.0
121 120 119 118 117 116 115 114 113 112 111 110
Chemical Shift (ppm)
11060.1
11074.8
11088.7
11403.4
11413.6
12003.5
12020.3
12024.0
12040.8
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
112.2
116.5
142.4
148.3
118.8
151.9 129.5
115.6
136.8
146.3
115.6
118.9
68.5
31.8 19.0
14.1
55.8
NH2
O
F
F
O
Here is a harder example (also courtesy Dr. Jaclyn Henderson).
Please assign this carbon-13 NMR spectrum (100 MHz, CDCl3).
predicted shifts
all peaks
right of arrow
are singlets
155.0 154.5 154.0 153.5 153.0 152.5 152.0 151.5 151.0 150.5 150.0 149.5 149.0 148.5 148.0 147.5 147.0
Chemical Shift (ppm)
14974.0
14980.6
15221.3
15228.7
15361.9
15367.0
15601.9
15607.8
152 144 136 128 120 112 104 96 88 80 72 64 56 48 40 32 24 16
Chemical Shift (ppm)
145 144 143
Chemical Shift (ppm)
14483.7
14486.6
14495.4
14499.0
121 120 119 118 117 116 115 114 113 112 111 110
Chemical Shift (ppm)
11060.1
11074.8
11088.7
11403.4
11413.6
12003.5
12020.3
12024.0
12040.8
all peaks
right of arrow
are singlets
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
68.5
31.8 19.0
14.1
55.8
NH2
O
F
F
O
predicted shifts
(1) Number of Peaks. There is no symmetry in this molecule, so there should be 17
signals total. There are a lot of peaks in this molecule, so I’ll label the carbons
below the chloroform peaks A-E. These are trivial to assign, particularly if you have
ChemDraw handy:
A B C D E
NH2
O
F
F
O
A
B
C
D
E
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
112.2
116.5
142.4
148.3
118.8
151.9 129.5
115.6
136.8
146.3
115.6
118.9
NH2
O
F
F
O
predicted shifts
155.0 154.5 154.0 153.5 153.0 152.5 152.0 151.5 151.0 150.5 150.0 149.5 149.0 148.5 148.0 147.5 147.0
Chemical Shift (ppm)
14974.0
14980.6
15221.3
15228.7
15361.9
15367.0
15601.9
15607.8
152 144 136 128 120 112 104 96 88 80 72 64 56 48 40 32 24 16
Chemical Shift (ppm)
145 144 143
Chemical Shift (ppm)
14483.7
14486.6
14495.4
14499.0
121 120 119 118 117 116 115 114 113 112 111 110
Chemical Shift (ppm)
11060.1
11074.8
11088.7
11403.4
11413.6
12003.5
12020.3
12024.0
12040.8
(2) Number of Peaks. There are two aryl rings, so there should be 12 aryl signals.
However, after accounting for the one-bond coupling in the ipso-fluorocarbons,
there only seem to be 11 carbons! (Don’t forget to count 5, which is not part of
any expansion.) Can you resolve this inconsistency?
A B C D E
1 1 2 2 3 4
6 7 8 9 10 11
5
146 144 142 140 138 136 134 132 130 128 126 124 122
Chemical Shift (ppm)
14499.0
14495.4
14486.6
14483.7
110.5 110.0 109.5
Chemical Shift (ppm)
11088.7
11084.3
11074.8
11064.5
11060.1
119.5 119.0 118.5 118.0 117.5 117.0 116.5 116.0 115.5 115.0 114.5 114.0 113.5 113.0
Chemical Shift (ppm)
12040.8
12024.0
12020.3
12003.5
11413.6
11403.4
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
112.2
116.5
142.4
148.3
118.8
151.9 129.5
115.6
136.8
146.3
115.6
118.9
NH2
O
F
F
O
predicted shifts
3 4 5 6 7
(2) Ipso-Fluorocarbons. Obviously, these are 1 and 2. We can guess that 1 is on top.
(3) Fluorine-Carbon Couplings. 4 and 8 are easy: 4 (3.7, 12.4 Hz); 8 (16.8, 20.5). 11
seems to have one 11088.7-11084.3 = 4.4 Hz and two 11088.7-11074.8 = 13.9 Hz
couplings. However, you will see this doesn’t really fit the multiplet very well (left). A
better fit (for a first-order multiplet) is 4.4, 12.1x2 Hz, but in both cases, there should
be two central peaks, not one, even accounting for line width. What’s wrong?
8 9 10 11
146 144 142 140 138 136 134 132 130 128 126 124 122
Chemical Shift (ppm)
14499.0
14495.4
14486.6
14483.7
110.5 110.0 109.5
Chemical Shift (ppm)
11088.7
11084.3
11074.8
11064.5
11060.1
119.5 119.0 118.5 118.0 117.5 117.0 116.5 116.0 115.5 115.0 114.5 114.0 113.5 113.0
Chemical Shift (ppm)
12040.8
12024.0
12020.3
12003.5
11413.6
11403.4
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
112.2
116.5
142.4
148.3
118.8
151.9 129.5
115.6
136.8
146.3
115.6
118.9
NH2
O
F
F
O
predicted shifts
3 4 5 6 7
(4) Accidental Overlap. In this molecule, there are only two fluorines, so no carbon
can have more than two fluorine-carbon couplings! The only explanation is that the
relatively intense central peak is really a protonated carbon, and there’s a doublet of
doublets (J = 4.4, 24.1 Hz) sandwiching it. This is where the missing twelfth carbon
is. (I usually get worried when I can’t find the right number of carbons. You should
be too—it’s one of the first things to check. One exception is carbamate carbons,
which are usually part of N-acyl protecting groups like BOC. Chemical exchange
and quadrupolar broadening can make those resonances hard to detect.)
(5) Difluoraryl Ring. The full list of couplings is now: 4 (3.7, 12.4 Hz); 8 (16.8, 20.5); 10
(10.2); 11 (4.4, 24.1). 8 clearly has two geminal couplings and has the right
chemical shift to be the carbon between the fluorines. 11 has one geminal coupling,
which means it has to be ortho to a fluorine:
8 9 10
11
12
11
NH2
O
F
F
OR
1
2
11?
11?
8
146 144 142 140 138 136 134 132 130 128 126 124 122
Chemical Shift (ppm)
14499.0
14495.4
14486.6
14483.7
110.5 110.0 109.5
Chemical Shift (ppm)
11088.7
11084.3
11074.8
11064.5
11060.1
119.5 119.0 118.5 118.0 117.5 117.0 116.5 116.0 115.5 115.0 114.5 114.0 113.5 113.0
Chemical Shift (ppm)
12040.8
12024.0
12020.3
12003.5
11413.6
11403.4
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
112.2
116.5
142.4
148.3
118.8
151.9 129.5
115.6
136.8
146.3
115.6
118.9
NH2
O
F
F
O
predicted shifts
3 4 5 6 7
(6) Difluoraryl Ring. Couplings: 4 (3.7, 12.4 Hz); 8 (16.8, 20.5); 10 (10.2); 11 (4.4,
24.1). By chemical shift, 11 must be the protonated carbon. 4 seems to have a
smaller geminal coupling, but must be the carbon next to oxygen:
8 9 10
11
12
11
NH2
O
F
F
OR
1
2
4
11
8
146 144 142 140 138 136 134 132 130 128 126 124 122
Chemical Shift (ppm)
14499.0
14495.4
14486.6
14483.7
110.5 110.0 109.5
Chemical Shift (ppm)
11088.7
11084.3
11074.8
11064.5
11060.1
119.5 119.0 118.5 118.0 117.5 117.0 116.5 116.0 115.5 115.0 114.5 114.0 113.5 113.0
Chemical Shift (ppm)
12040.8
12024.0
12020.3
12003.5
11413.6
11403.4
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
112.2
116.5
142.4
148.3
118.8
151.9 129.5
115.6
136.8
146.3
115.6
118.9
NH2
O
F
F
O
predicted shifts
3 4 5 6 7
(7) Ipso Carbons. 3 and 5 clearly do not have any couplings and by chemical shift,
must be ipso to the methyl ether and the arylamine:
8 9 10
11
12
11
NH2
O
F
F
OR
1
2
4
11
8
3
5
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
NH2
O
F
F
OnBu
112.2
116.5
142.4
148.3
118.8
151.9 129.5
115.6
136.8
146.3
115.6
118.9
(7) What’s Left? The carbons left are 6, 7, 9, 10, and 12. This is tricky, but we can use
the line widths and heights to help us make an educated guess. 12, as a sharp
singlet, must be ortho to the methyl ether. 10 has a lot of unresolved couplings and
must be either meta to the fluorines or at the ipso ring junction. Based on chemical
shift, it is likely on the difluoroaryl ring. 7 and 9 have some unresolved long-range
couplings, and based on chemical shift 7 is para to the amine. This puts 6, as the
shortest peak, at the ipso ring junction. The only trouble is that 6 should have some
vicinal couplings. (Professor Reynolds suggests this could be a cancellation of
through-space and through-bond couplings.) Thus, further experiments would be
needed to confirm this set of assignments (any inconsistency is always a concern).
6 7 8 9 10 11
12
121.5 121.0 120.5 120.0 119.5 119.0 118.5 118.0 117.5 117.0 116.5 116.0 115.5 115.0 114.5 114.0 113.5 113.0 112.5 112.0 111.5 111.0 110.5 110.0
Chemical Shift (ppm)
120.75 120.50 120.25
Chemical Shift (ppm)
116.9 116.8 116.7116.6
Chemical Shift (ppm)
6 Hz wide
20 Hz wide
(8) Spectrum Processing. The previous spectra were
processed with standard line broadening apodization
(LB=1) but the spectrum below has not been
apodized (LB=0) and shows a corresponding
enhancement in resolution. I’ll discuss how this works
in a few lectures.
NH2
O
F
F
OR
1
2
4
11
8
3
5
10
6
12
7
9
final answer
Chem 117E. Kwan Lecture 3: Coupling Constants
Practice Problem: Carbon-Fluorine Couplings
NH2
O
F
F
O
(9) Proton Spectrum. With the removal of the symmetry in the difluoroaryl group, the magnetic
inequivalence problem disappears. Can you identify all the peaks?
7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0
Chemical Shift (ppm)
3.02.02.05.02.04.9