Exercise plus Solution – Quick overview
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A
B
C
4
H
8
O
2
in CDCl
3
Problem of the month
December 2022
The structures of the two isomers A and B are easy
to determine with the available spectra and
minimal literature research.
Can you think of an alternative way that would
allow you to make the determination from the
spectra provided without consulting the literature?
13
C{
1
H} NMR spectrum
recorded at 62.9{250.13} MHz
A
B
13
C{
1
H} NMR spectrum
recorded at 62.9{250.13} MHz
(insets with significant expansion)
A
B
1
H NMR spectrum
recorded at 250.13 MHz
Peak-Markierungen
< 10 - [ppm]
> 100 - [Hz]
A B
1
H/
13
C HSQC
recorded at 250.13/62.9 MHz
Problem of the Month
December 2022
Solution
Basics
Integration,double bond
equivalents, symmetry
C
4
H
8
O
2
in CDCl
3
2H Number of protons 3H 3H
Let us take isomer B to get the constitution. The procedure
would be exactly the same for isomer A.
The integration of the proton spectrum of isomer B is easy. We
only have to multiply each integral by a factor of 2.
In both isomers, as well as the same molecular formula, we have:
• the same number of carbon signals and
• the same proton multiplets with the same corresponding
integrals.
The main difference is
• the chemical shift of the quartet.
B
Basics
Integration,double bond
equivalents, symmetry
C
4
H
8
O
2
in CDCl
3
2H Number of protons 3H 3H
We have only two multiplets with a typical vicinal coupling
constant of 7.14 Hz. An ethyl group easily explains a quartet
with two protons and a triplet with 3 protons.
C
H
2
CH
3
4.14
1.27
7.14 Hz
B
B
Basics
Integration,double bond
equivalents, symmetry
C
4
H
8
O
2
in CDCl
3
2H Number of protons 3H 3H
A singlet with three protons has to be a methyl group.
H
3
C
2.06
C
H
2
CH
3
4.14
1.27
7.14 Hz
B
B
Basics
Integration,double bond
equivalents, symmetry
C
4
H
8
O
2
in CDCl
3
2H Number of protons 3H 3H
The sum of both fragments is C
3
H
8
, which means we still have to
assign
• two oxygen atoms,
• one carbon atom and
• one double bond equivalent.
H
3
C
2.06
C
H
2
CH
3
4.14
1.27
7.14 Hz
B
B
Basics
Integration,double bond
equivalents, symmetry
C
4
H
8
O
2
in CDCl
3
2H Number of protons 3H 3H
The sum of both fragments is C
3
H
8
, which means we still have to
assign
• two oxygen atoms,
• one carbon atom and
• one double bond equivalent.
H
3
C
2.06
C
O
O
The result is this fragment with two possibilities to insert it
between our already existing parts of the molecule.
Simply let us try both possibilities.
C
H
2
CH
3
4.14
1.27
7.14 Hz
B
B
Basics
Integration,double bond
equivalents, symmetry
C
4
H
8
O
2
in CDCl
3
What we have to do is to create a copy of all already known pieces of
information with a minor different geometric orientation of the –CO–
O– group.
H
3
C
2.06
C
O
O
O
C
O
C
H
2
CH
3
4.14
1.27
H
3
C
2.06
Now there are two possible final structures.
C
H
2
CH
3
4.14
1.27
7.14 Hz
B
Basics
Integration,double bond
equivalents, symmetry
C
4
H
8
O
2
in CDCl
3
C
O
O
O
C
O
C
H
2
CH
3
4.14
1.27
H
3
C
2.06
H
3
C
2.06
C
H
2
CH
3
4.14
1.27
7.14 Hz
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
C
O
O
C
H
2
CH
3
4.14
1.27
H
3
C
2.06
Now there are two possible final structures.
Now there are two possible final structures.
B
Easy solution
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
C
O
O
C
H
2
CH
3
4.14
1.27
H
3
C
2.06
Now there are two possible final structures.
The easiest way to select the correct one is the estimation of the
chemical shifts of the methylene protons. That‘s possible using the
good old Schoolery rules.
(CH
2
) = (0.23 + 0.47 + 3.13) ppm = 3.83 ppm
(CH
2
) = (0.23 + 0.47 + 1.55) ppm = 2.25 ppm
Apparently this estimation best fits to the upper structure.
The lower structure appently belongs to isomer A, but of
course all chemical shifts derived for isomer B are wrong.
Let‘s remove all of them.
B
A
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
C
O
O
C
H
2
CH
3
H
3
C
We didn‘t assign the carbon chemical shifts so far. For
this task, of course, our spectrum of choice is the HSQC.
Full assignment
Carbon signals
B
A
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
C
O
O
C
H
2
CH
3
H
3
C
Full assignment
Carbon signals
We didn‘t assign the carbon chemical shifts so far. For
this task, of course, our spectrum of choice is the HSQC.
The chemical shifts for the pseudo projections come
from the one dimensional proton and carbon spectra.
4.14 1.272.06
20.05
59.57
13.53
B
A
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
C
O
O
C
H
2
CH
3
H
3
C
Full assignment
Carbon signals
Let‘s select one cross peak to demonstrate the carbon
signal assignment in detail.
The assignment procedure for both methyl groups is the
same.
4.14 1.272.06
20.05
59.57
13.53
59.57
B
A
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
C
O
O
C
H
2
CH
3
H
3
C
Full assignment
Carbon signals
The assignment procedure for both methyl groups is the
same.
4.14 1.272.06
20.05
13.53
59.57
20.05
13.53
B
A
C
O
O
C
H
2
CH
3
H
3
C
Full assignment
Carbon signals
The chemical shift for the carbonyl group carbon atom
comes from the one dimensional carbon spectrum.
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
We do not repeat the identical assignment procedure for
isomer A here. Let us take the final result only.
B
A
The alternative way
Don‘t use a textbook
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
But how can we distinguish between isomer A and isomer B
without using Schoolery‘s rule?
There has to be a way to differentiate between the two
isomers using the carbon-carbon coupling pattern. The
signal-to-noise ratio of the carbon spectrum is good enough
to show these couplings.
But how to use these couplings without refering to a
textbook about the typical size of such coupling constants?
B
A
C
O
C
H
2
CH
3
O
H
3
C
3.69
2.35
1.16
50.57
26.61
8.35
173.86
7.59 Hz
The alternative way
Don‘t use a textbook
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
Do you see any difference between the isomers, which
might be helpful in using the carbon-carbon satellites
without any prior knowledge?
It is a question of pattern recognition. But, what on earth
might be the pattern?
B
A
C
O
C
H
2
CH
3
O
H
3
C
3.69
2.35
1.16
50.57
26.61
8.35
173.86
7.59 Hz
C
O
C
H
2
CH
3
O
H
3
C
3.69
2.35
1.16
50.57
26.61
8.35
173.86
7.59 Hz
The alternative way
Don‘t use a textbook
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
Do you see any difference between the isomers, which
might be helpful in using the carbon-carbon satellites
without any prior knowledge?
It is a question of pattern recognition. But, what on earth
might be the pattern?
B
A
Let us label the eight carbon atoms using three different
colours. What could distinguish carbon atoms that have
been marked with the same colour?
And now?
C
O
C
H
2
CH
3
O
H
3
C
3.69
2.35
1.16
50.57
26.61
8.35
173.86
7.59 Hz
The alternative way
Don‘t use a textbook
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
B
A
no neighbouring carbon atom with a distance of one bond
one neighbouring carbon atom with a distance of one bond
two neighbouring carbon atom with a distance of one bond
It is a reasonable assumption, that all one bond carbon-
carbon coupling constants are of comparable size.
Each of the three cases results in a different pattern for the
satellite signals due to carbon-carbon coupling.
Let us assume a model compund and think about what the
pattern might look like.
C
O
C
H
2
CH
3
O
H
3
C
3.69
2.35
1.16
50.57
26.61
8.35
173.86
7.59 Hz
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
B
A
Carbon-carbon
couplings
Probability and spectra
~
––
13
C ––
12
C ––
12
C ––≈ 1%
~
––
12
C ––
12
C ––
13
C ––≈ 1%
~
––
12
C ––
13
C ––
12
C ––≈ 1%
~
~ ~
Sum of all components
––
13
C ––
13
C ––
12
C ––≈ 0.01%
1
J
GR
1
J
GR
1
J
GR
––
12
C ––
13
C ––
13
C ––≈ 0.01%
1
J
RB
1
J
RB
1
J
RB
–– C –– C –– C ––
We should remember, that every NMR spectrum is not the result of
one molecule, but of a huge ensemble of molecules.
Let‘s simulate the carbon spectrum of the
single (asymmetric) model compound
(equivalent to the right part of isomer A)
––
12
C ––
12
C ––
12
C ––≈ 97%
(no signal at all)
C
O
C
H
2
CH
3
O
H
3
C
3.69
2.35
1.16
50.57
26.61
8.35
173.86
7.59 Hz
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
B
A
Carbon-carbon
couplings
Probability and spectra
~
~ ~
–– C –– C –– C ––
Please note, that the carbon-carbon multiplet pattern for
the red carbon atom is neither a triplet nor a doublet of
doublets. Rather, it is two nested independent doublets.
For a triplet or a doublet of doublets we would need
three
13
C atoms within the same molecule. The
probability is very, very low: about 0,0001%
–– C –– C –– C ––
C
O
C
H
2
CH
3
O
H
3
C
3.69
2.35
1.16
50.57
26.61
8.35
173.86
7.59 Hz
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
B
A
Carbon-carbon
couplings
Practical use
~
~ ~
Which satellite pattern due to carbon-carbon coupling
do we expect for the three types of carbon atoms shown
before?
Apparently the coupling pattern for the carbon atom
marked in mauve is unique and we have to look wether
this pattern is visible in the carbon spectra of isomer A
or isomer B.
–– C –– C –– C ––
no satellites due to carbon-carbon coupling
one doublet due to carbon-carbon coupling
two nested doublets due to carbon-carbon coupling
Due to the use of Schoolery‘s rule we already know the
solution. Let us forget this solution to demonstrate the
use of the carbon-carbon coupling pattern.
A
~
~ ~
C
O
C
H
2
CH
3
O
H
3
C
3.69
2.35
1.16
50.57
26.61
8.35
173.86
7.59 Hz
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
Carbon-carbon
couplings
Practical use
–– C –– C –– C ––
two nested doublets due to carbon-carbon coupling
The pattern we are looking for is part of the carbon
spectrum for isomer A and not visible anywhere else.
A
A
C
O
C
H
2
CH
3
O
H
3
C
3.69
2.35
1.16
50.57
26.61
8.35
173.86
7.59 Hz
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
Carbon-carbon
couplings
Practical use
The pattern we are looking for is part of the carbon
spectrum for isomer A and not visible anywhere else.
no satellites due to carbon-carbon coupling
And there is no carbon-carbon splitting here.
A
In the case of isomer B we should see only doublets as
satellites due to carbon-carbon coupling.
Let‘s check.
B
one doublet due to carbon-carbon coupling
Carbon-carbon
couplings
Practical use
one doublet due to carbon-carbon coupling
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
7.14 Hz
59.57
20.05
13.53
O
C
H
2
CH
3
C
H
3
C
O
4.14
1.27
2.06
20.05
59.57
13.53
169.97
7.14 Hz
B
B
Apart from three spikes of unknown source, we see
exactly the four expected doublets.
Of course, in practice, one would not solve this simple
question with this enormous measurement effort. The
objective here was to motivate the search for solutions
off the beaten track.
Contributions
Measurements
Rainer Haeßner
Spectrometer time
TU Munich
Discussions and
native English
language support
Alan M. Kenwright
Compilation
Rainer Haeßner
