Exercise plus Solution – Quick overview
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2.02.12.22.32.42.52.62.72.82.93.03.13.23.33.43.53.63.7
ppm
1
H
0.68
0.66
1.00
0.67
1.00
Inte-
gral
3.34
3.69
ppm
848.53
854.73
860.94
Hz
596.33
603.79
611.08
Hz
C
6
H
12
O
3
1
H NMR spectrum
recorded at 250.13 MHz
Problem of the Month:
February 2021
Deduce the constitution, assign all nuclei and work out two proton-proton coupling
constants.
After deducing the constitution, turn to the real challenge of this problem (slide 4).
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
1
H/
13
C-HMBC
recorded at 250.13/62.90 MHz
The f
1
projection contains all six carbon signals
of the compound.
There is no separate one dimensional carbon
spectrum given.
~
~
~~
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
1
H
3.4 3.2 3.0 2.6
2.4 2.2
2.0
1
H
ppm
25
30
35
40
50
55
60
65
70
75
13
C
3.8 3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0
1
H
ppm
1
H/
13
C-HSQC
recorded at 250.13/62.90 MHz
1
H/
1
H-DQF-COSY
recorded at 250.13 MHz
And now the challenge
With three chemical shifts and two coupling constants, you can simulate the three multiplets in the proton
spectrum. The simulation for the multiplet at about 3.4 ppm will provide an almost perfect result, but in
the case of the other two multiplet there are small but significant deviations. The multiplet at about 1.9
ppm looks a bit more complex than expected, although there exists a clear base structure. But the
mutiplet at about 2.4 ppm should be a pure triplet.
2.322.342.362.382.402.422.442.462.48
ppm
measured
simulated
Because of the small difference in chemical shift
between the multiplets at 1.9 ppm and 2.4 ppm,
fine splitting is visible in the simulation for which
the experimental resolution is insufficient.
But the real issue are the two "warts" in the
experimental spectrum, for which no evidence is
visible in the simulation.
What might be the reason for those “artifacts”?
Hint:
To answer this extremely difficult question, trawl
through the literature for the proton spectrum of 1-
bromo-2-chloroethane, marvel a little bit about
what you find, and try to understand the theoretical
explanation of this spectrum.
2.02.12.22.32.42.52.62.72.82.93.03.13.23.33.43.53.63.7
ppm
1
H
0.68
0.66
1.00
0.67
1.00
Inte-
gral
Basics
Integration,double bond
equivalents, symmetry
sum of measured integrals: 4.01
number of protons: 12
proportionality coefficient: 2.99
measured integrals * coefficient: 2.99 : 2.00 : 2.99 : 1.98 : 2.03
double bond equivalents: 1
3 22 23
C
6
H
12
O
3
symmetric carbon atoms: no
(all six carbon signals are visible in
the HMBC)
~~
25
30
35
40
50
55
60
65
70
75
13
C
3.8 3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0
1
H
ppm
71.53
58.51
51.49
30.70
24.91
33 2 2 2
2.41 ppm
3.42
ppm
3.34 ppm3.69 ppm
ca. 1.91
ppm
Building blocks
CH
n
-fragments
It is very easy to evaluate a HSQC. The sensitivity, of
course, is less than the sensitivity of a one dimensional
proton spectrum but much higher than a one
dimensional carbon spectrum. Therefore, the
measurement of a HSQC is always recommended, if
possible.
We need some data for the projections, chemical shifts and
integrals from the one dimensional proton spectrum and
the carbon chemical shifts from the one dimensional
carbon spectrum.
The one dimensional carbon spectrum is not explicitely
given here but used as a pseudo projection for the
HMBC. The chemical shifts can be picked there.
~~
25
30
35
40
50
55
60
65
70
75
13
C
3.8 3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0
1
H
ppm
71.53
58.51
51.49
30.70
24.91
33 2 2 2
2.41 ppm
3.42
ppm
3.34 ppm3.69 ppm
ca. 1.91
ppm
Building blocks
CH
n
-fragments
The proton signals at 3.69 ppm and 3.34 ppm could only
belong to methyl groups according to their integral. The
possibility of three symmetric CH-groups could be
immediately excluded by our initial considerations (slide
1).
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.69 ppm
51.49 ppm
Please continue to the second methyl group …
~~
25
30
35
40
50
55
60
65
70
75
13
C
3.8 3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0
1
H
ppm
71.53
58.51
30.70
24.91
32 2 2
2.41 ppm
3.42
ppm
3.34 ppm
ca. 1.91
ppm
Building blocks
CH
n
-fragments
The proton signals at 3.69 ppm and 3.34 ppm could only
belong to methyl groups according to their integral. The
possibility of three symmetric CH-groups could be
immediately excluded by our initial considerations (slide
1).
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
H
C
H
H
3.34 ppm
58.51 ppm
~~
25
30
35
40
50
55
60
65
70
75
13
C
3.8 3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0
1
H
ppm
71.53
30.70
24.91
2 2 2
2.41 ppm
3.42
ppm
ca. 1.91
ppm
Building blocks
CH
n
-fragments
The remaining cross peaks belong to methylene groups.
Let us extract them step by step.
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
3.42 ppm
71.53
ppm
C
H
H
3.42 ppm
71.53
ppm
~~
25
30
35
40
50
55
60
65
70
75
13
C
3.8 3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0
1
H
ppm
30.70
24.91
2 2
2.41 ppm
ca. 1.91
ppm
Building blocks
CH
n
-fragments
The remaining cross peaks belong to methylene groups.
Let us extract them step by step.
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
3.42 ppm
71.53
ppm
C
H
H
2.41 ppm
30.70
ppm
C
H
H
2.41 ppm
30.70
ppm
~~
25
30
35
40
50
55
60
65
70
75
13
C
3.8 3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0
1
H
ppm
24.91
2
ca. 1.91
ppm
Building blocks
CH
n
-fragments
The remaining cross peaks belong to methylene groups.
Let us extract them step by step.
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
3.42 ppm
71.53
ppm
C
H
H
2.41 ppm
30.70
ppm
H
H
C
ca. 1.91
ppm
24.91
ppm
H
H
C
ca. 1.91
ppm
24.91
ppm
Linking the pieces
part 1 – alkyl chain
First let us reorder the fragments a little bit to make the
next steps easier.
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
H
C
H
H
3.34 ppm
58.51 ppm
H
H
C
ca. 1.91
ppm
24.91
ppm
H
H
C
ca. 1.91
ppm
24.91
ppm
C
H
H
2.41 ppm
30.70
ppm
C
H
H
2.41 ppm
30.70
ppm
C
H
H
3.42 ppm
71.53
ppm
C
H
H
3.42 ppm
71.53
ppm
~
~
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
1
H
3.4 3.2 3.0 2.6
2.4 2.2
2.0
1
H
ppm
Linking the pieces
part 1 – alkyl chain
During our work with the COSY we have no use for the
methyl groups.
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
2.41 ppm
30.70
ppm
C
H
H
3.42 ppm
71.53
ppm
3.42 ppm
2.41 ppm
ca. 1.91
ppm
3.42
ppm
2.41
ppm
ca. 1.91
ppm
H
H
C
ca. 1.91
ppm
24.91
ppm
~
~
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
1
H
3.4 3.2 3.0 2.6
2.4 2.2
2.0
1
H
ppm
Linking the pieces
part 1 – alkyl chain
The first visible proximity in the COSY, is between the
protons with the chemical shifts of ca. 1.91 ppm and
3.42 ppm.
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
2.41 ppm
30.70
ppm
C
H
H
3.42 ppm
71.53
ppm
3.42 ppm
2.41 ppm
ca. 1.91
ppm
3.42
ppm
2.41
ppm
ca. 1.91
ppm
H
H
C
ca. 1.91
ppm
24.91
ppm
~
~
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
1
H
3.4 3.2 3.0 2.6
2.4 2.2
2.0
1
H
ppm
Linking the pieces
part 1 – alkyl chain
The first visible proximity in the COSY, is between the
protons with the chemical shifts of ca. 1.91 ppm and
3.42 ppm.
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
2.41 ppm
30.70
ppm
3.42 ppm
2.41 ppm
ca. 1.91
ppm
3.42
ppm
2.41
ppm
ca. 1.91
ppm
H
H
C
ca. 1.91
ppm
24.91
ppm
C
H
H
3.42 ppm
71.53
ppm
C
H
H
3.42 ppm
71.53
ppm
H
H
C C
H
H
3.42 ppm
ca. 1.91
ppm
24.91
ppm
71.53
ppm
~
~
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
1
H
3.4 3.2 3.0 2.6
2.4 2.2
2.0
1
H
ppm
Linking the pieces
part 1 – alkyl chain
The second proximity, visible in the COSY, is between the
protons with the chemical shifts of ca. 1.91 ppm and 2.41
ppm, which allows us to complete the alkyl fragment.
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
2.41 ppm
30.70
ppm
3.42 ppm
2.41 ppm
ca. 1.91
ppm
3.42
ppm
2.41
ppm
ca. 1.91
ppm
H
H
C
ca. 1.91
ppm
24.91
ppm
H
H
C C
H
H
3.42 ppm
ca. 1.91
ppm
24.91
ppm
71.53
ppm
~
~
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
1
H
3.4 3.2 3.0 2.6
2.4 2.2
2.0
1
H
ppm
Linking the pieces
part 1 – alkyl chain
The second proximity, visible in the COSY, is between the
protons with the chemical shifts of ca. 1.91 ppm and 2.41
ppm, which allows us to complete the alkyl fragment.
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
3.42 ppm
2.41 ppm
ca. 1.91
ppm
3.42
ppm
2.41
ppm
ca. 1.91
ppm
H
H
C C
H
H
3.42 ppm
ca. 1.91
ppm
24.91
ppm
71.53
ppm
C
H
H
2.41 ppm
30.70
ppm
C
H
H
2.41 ppm
30.70
ppm
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
We no longer need the COSY.
To continue let us recall the methyl groups and
rearrange the fragments a little bit for further use.
Something missing?
time for a short inventory
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
Something missing?
time for a short inventory
molecular
formula
C
6
H
12
O
3
known
fragments
C
5
H
12
unassigned
carbon atom
without
attached hydrogen
173.9 ppm
missing
CO
3
one
double bond
equivalent
O
C
173.90
ppm
O
O
As a result let us increase our unordered pile of building
blocks by three hydrogen free fragments.
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
O
C
173.90
ppm
O
O
3.34 ppm
3.42
ppm
Linking the pieces
finalize the puzzle
Two correlation in the HMBC contain very
similar pieces of information.
The first correlation appears between the
signals with chemical shifts of 58.51 ppm
and 3.42 ppm.
H
C
H
H
3.69 ppm
51.49 ppm
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
H
C
H
H
3.34 ppm
58.51 ppm
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
O
C
173.90
ppm
O
O
3.34 ppm
3.42
ppm
Linking the pieces
finalize the puzzle
The second correlation appears between
the signals with chemical shifts of 71.53
ppm and 3.34 ppm.
H
C
H
H
3.69 ppm
51.49 ppm
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
O
C
173.90
ppm
O
O
Linking the pieces
finalize the puzzle
H
C
H
H
3.34 ppm
58.51 ppm
H
C
H
H
3.34 ppm
58.51 ppm
There is one structural arrangement
which explains both peaks.
C
H
H
H
H
C C
H
H
H
CX
H
H
3.42 ppm
2.41 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
58.51 ppm
X is either - O - or - CO -.
H
C
H
H
3.69 ppm
51.49 ppm
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
O
C
173.90
ppm
O
O
Linking the pieces
finalize the puzzle
Would it be possible to replace X by - CO -?
C
H
H
H
H
C C
H
H
H
CX
H
H
3.42 ppm
2.41 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
58.51 ppm
Calculate by yourself the chemical shift of
the methylene protons bonded to the
carbon with the chemical shift of 71.53 ppm
using the Schoolery rules.
H
C
H
H
3.69 ppm
51.49 ppm
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
O
C
173.90
ppm
O
O
Linking the pieces
finalize the puzzle
C
H
H
H
H
C C
H
H
H
CX
H
H
3.42 ppm
2.41 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
58.51 ppm
You get
X = - O - 3.06 ppm
X = - CO - 2.40 ppm
H
C
H
H
3.69 ppm
51.49 ppm
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
O
C
173.90
ppm
O
O
Linking the pieces
finalize the puzzle
C
H
H
H
H
C C
H
H
H
CX
H
H
3.42 ppm
2.41 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
58.51 ppm
There is one more possibility to exclude
X = - CO -.
Have a look for this cross peak.
3.69 ppm
H
C
H
H
3.69 ppm
51.49 ppm
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
O
C
173.90
ppm
O
O
Linking the pieces
finalize the puzzle
C
H
H
H
H
C C
H
H
H
CX
H
H
3.42 ppm
2.41 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
58.51 ppm
If you insert - CO - instead of X and keep
everything else in place the length of the
coupling path to the protons with the
chemical shift of 3.69 ppm is at least 6
bonds. Try it!
3.69 ppm
You cannot attach the methyl group with
the chemical shift of 3.69 ppm directly to
the carbon with the chemical shift of 30.70
ppm, because there was no COSY cross peak
between the protons with the chemical
shifts of 2.41 ppm and 3.69 ppm. You would
need an oxygen atom as bridge, resulting in
six bonds between X (-CO-) and H (3.69
ppm).
H
C
H
H
3.69 ppm
51.49 ppm
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
O
C
173.90
ppm
O
Linking the pieces
finalize the puzzle
After changing X to O there are only two
possibilities to finalize the structure.
O
O
C
H
H
H
H
C C
H
H
H
CX
H
H
3.42 ppm
2.41 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
58.51 ppm
C
H
H
H
H
C C
H
H
H
CO
H
H
3.42 ppm
2.41 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
58.51 ppm
H
C
H
H
3.69 ppm
51.49 ppm
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
Linking the pieces
finalize the puzzle
Let us try one of them.
C
H
H
H
H
C C
H
H
H
CO
H
H
3.42 ppm
2.41 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
30.70
ppm
71.53
ppm
58.51 ppm
O
C
173.90
ppm
O
H
C
H
H
3.69 ppm
51.49 ppm
H
C
H
H
3.69 ppm
51.49 ppm
O
O
C
173.90
ppm
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
Linking the pieces
finalize the puzzle
Let us first inspect one HMBC cross peak.
O
H
C O
H
H
C C
H
H
H
H
C C
H
H
H
CO
H
H
3.42 ppm
2.41 ppm
3.69 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
173.90
ppm
30.70
ppm
71.53
ppm
51.49 ppm
58.51 ppm
ca. 1.91
ppm
This is a classical three bond correlation. In
the case of the alternative arrangement of
the -CO-O- group, this would be a much less
likely four-bond correlation.
~
~
40
60
80
100
120
140
160
13
C
3.6 3.4 3.2 3.0 2.6 2.4 2.2 2.0 ppm
1
H
24.91
30.70
51.49
58.51
71.53
173.90
Linking the pieces
finalize the puzzle
O
H
C O
H
H
C C
H
H
H
H
C C
H
H
H
CO
H
H
3.42 ppm
2.41 ppm
3.69 ppm
3.34 ppm
ca. 1.91
ppm
24.91
ppm
173.90
ppm
30.70
ppm
71.53
ppm
51.49 ppm
58.51 ppm
ca. 1.91
ppm
If we estimate the chemical shift of the protons
bound to the carbon with the chemical shift of
30.70 ppm using Shoolery‘s rule we obtain:
2.25 ppm (presented structure)
3.83 ppm (alternative arrangement of –CO-O-)
Good old Shoolery‘s rule is another way
to confirm this structure.
Coupling constants
It looks simple, but it is not
O
H
C O
H
H
C C
H
H
H
H
C C
H
H
H
CO
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
3.69 ppm
3.34 ppm
24.91
ppm
173.90
ppm
30.70
ppm
71.53
ppm
51.49 ppm
58.51 ppm
For the sake of clarity let us remove all
carbon assignments and the proton
assignments of both methyl groups. We
don‘t need them anymore.
O
H
C O
H
H
C C
H
H
H
H
C C
H
H
H
CO
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
Coupling constants
easy start
Both sets of chemically equivalent methylene protons at
2.41 ppm and 3.42 ppm have two chemically equivalent
protons at about 1.91 ppm as the only vicinal coupling
partners. We expect a triplet in both cases.
𝐽
1
=
860.94 Hz − 848.53
2
= 𝟔. 𝟐𝟎 𝐇𝐳
J
2
= 7.37 Hz
J
1
= 6.20 Hz
860.94
854.73
848.53
596.33
603.79
611.08
𝐽
2
=
611.08 Hz − 596.33
2
= 𝟕. 𝟑𝟕 𝐇𝐳
3.42 ppm 2.41 ppm
7.37
Hz
6.20
Hz
O
H
C O
H
H
C C
H
H
H
H
C C
H
H
H
CO
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
Coupling constants
done
For the methylene protons at 1.91 ppm we expect a triplet
of triplets. Because 1.91 ppm and 2.41 ppm are relatively
close in chemical shift, we expect very first signs of higher
order (e.g slight „roofing“) but, in principle the triplet of
triplets looks fine.
7.37
Hz
6.20
Hz
Δδ
𝐽
=
2.41 ppm − 1.91ppm ∗ 250.13 MHz
7.37 Hz
= 𝟏𝟔. 𝟗𝟕
O
H
C O
H
H
C C
H
H
H
H
C C
H
H
H
CO
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
Coupling constants
A last check
The simulation of two multiplets looks fine.
7.37
Hz
6.20
Hz
2.41 ppm
measured
measured
simulated
simulated
1.91 ppm
But … Have a closer look.
Some experimental details are missing in the simulation. Noise?
Coupling constants
Some refinement
8.1Hz
6.70Hz
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
-10.8Hz
8.1Hz
-12.9Hz
-0.2Hz
After changing some coupling constants and adding
geminal coupling constants and a long rang coupling
constant the „warts“ are simulated nearly perfectly.
It is not possible to measure these values directly from
the spectra presented here, but you can repeat the
simulation using the values given and see that they give
the correct result.
The second coupling pathway with a coupling
constant of 6.70 Hz is not shown here for
reason of clarity.
Coupling constants
Some refinement
8.1Hz
6.70Hz
C
H
H
H
H
C C
H
H
3.42 ppm
2.41 ppm
ca. 1.91
ppm
-10.8Hz
8.1Hz
-12.9Hz
-0.2Hz
But …
Why should vicinal coupling constants between chemically
equivalent protons have different values? There is the
possibility of free rotation around the single bond between
the carbon atoms. The vicinal coupling constants, of course,
depend on the dihedral angle following the Karplus equation,
but this effect should be averaged out by the fast rotation
around all possible dihedral angles between 0 and 360
degrees.
Coupling constants
An explanation
C
H
H
H
H
C Y
X
Let us reduce our molecule to a bisubstituted ethane derivative
with two different substituents
X = -CO-O-CH
3
and
Y = -CH2-O-CH
3
.
H1`
H1``
H2`
H2``
C1
C2
X
Y
For the moment don‘t worry about the hydrogen atoms with four
different labels shown here, although you expect that H1` and H1`` or
H2`and H2`` should be equivalent.
H1`
H1``
H2`
H2``
C1
C2
X
Y
Symmetry
Are H2`and H2`` chemically
equivalent?
If you see the static structure of our unsymmetric ethane derivative there
seems to be no question.
There is a symmetry plane inside the molecule, which makes both
H1`/H1`` and H2`/H2`` chemically equivalent.
But there is free rotation around the C1-C2 single bond possible and the
rotamer shown here is not the only one.
Let us introduce some assumptions about the rotation around the C1-C2
bond.
- The first assumption is that even with hydrogens the steric hindrance will
favour the three staggered conformations
- The second assumption is that bond rotation is so much faster than NMR
dwell time and we are seeing the average of the three staggered rotamers.
These assumptions are only necessary to keep the mathematics simple.
H1`
H1``
H2`
H2``
C1
C2
X
Y
C2
120°
C2
120°
C1-C2
180°
change view
to right side
Symmetry
Are H2`and H2`` chemically
equivalent?
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
First let us create the three rotamers (I, II and III)
If we turn C2 in rotamer I clockwise by 120 degree we get rotamer II.
Turning once more by 120 degree results in rotamer III.
For rotamer III it is recommendable to change the viewpoint. Turn the
whole molecule around the C1-C2 bond by 180 degree and have a view
to the molecule from the right side instead from the left side.
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
Symmetry
Are H2`and H2`` chemically
equivalent?
Let us reorder the three rotamers a little bit.
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
As you see there is a mirror plane inside rotamer I
and no symmetry element inside the other two
rotamers.
But on the other hand rotamer II and rotamer III are
mirror images of each other.
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
Symmetry
Are H2`and H2`` chemically
equivalent?
And now let us paint the protons a little bit.
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
Symmetry
Are H2`and H2`` chemically
equivalent?
Different colours mean different chemical shifts, identical
colours represent identical chemical shifts. Alltogether we
have six different chemical shifts for the four protons
inside the three rotamers.
As an example H1` and H1`` in rotamer I are identical due
to the internal mirror plane.
H2`` in rotamer II and H2` in
rotamer III are identical, because
rotamer II and rotamer III are
mirror images.
Symmetry
Are H1`and H1`` chemically
equivalent?
The population of the rotamers is p
I
, p
II
and p
III
with
p
II
= p
III
and
p
I
+ p
II
+ p
III
= 1
To keep the following equations short, we
use single letters for the six different
chemical shifts as follows:
δ
H(red)
= R
δ
H(blue)
= B
δ
H(green)
= G
δ
H(yellow)
= Y
δ
H(purple)
= P
δ
H(white)
= W (you wouldn‘t see a white
letter)
p
I
(rotamer
population)
p
II
p
III
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
Symmetry
Are H2`and H2`` chemically
equivalent?
Now we get for the four protons
δ
H1`
= p
I
* R + p
II
* W + p
III
* Y
δ
H1``
= p
I
* R + p
II
* Y + p
III
* W
δ
H2`
= p
I
* B + p
II
* G + p
III
* P
δ
H2``
= p
I
* B + p
II
* P + p
III
* G
With the boundary condition
p
II
= p
III
we get
δ
H2`
= δ
H2
``
and
δ
H1`
= δ
H1``
p
I
(rotamer
population)
p
II
p
III
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
Symmetry
Are H2`and H2`` chemically
equivalent?
p
I
(rotamer
population)
p
II
p
III
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
We arrive at the result you expected from
the beginning without all those difficult
considerations.
But, just for your curiosity, try to repeat the
calculation after replacing H1`` with a third
substituent Z, different from X and Y. In this
case, there is no symmetry, no mirror plane
inside rotamer I nor a mirror plane
between the rotamers II and III.
But, leaving that aside for the moment, let
us return to the main question:
are H2` and H2``
magnetically equivalent?
Symmetry
Are H2`and H2``
magnetically equivalent?
p
I
(rotamer
population)
p
II
p
III
As we have seen, the protons H2` and H2`` are
chemically equivalent. They are magnetically
equivalent as well, if the condition
3
J
H1`,H2`
=
3
J
H1`,H2``
is fulfilled.
Of course the same has to be valid, if we
replace H1` by H1`` on both sides of the
equation.
Let us see the geometric relations
between H1‘/H2` and H1`/H2`` one
after the other for all three rotamers.
H1`
H1``
H2`
H2``
C1
C2
X
Y
I
H1`
X
Y
C1
C2
H1``
H2`
H2``
II
H1`
X
Y
C1
C2
H1``
H2`
H2``
III
H1`
H2`
C1
C2
I
H1`
C1
C2
H2`
II
H1`
C1
C2
H2`
III
Symmetry
Are H2`and H2``
magnetically equivalent?
p
I
(rotamer
probability)
p
II
p
III
dihedral angle
(H1`-C1-C2-H2`)
60°
dihedral angle
(H1`-C1-C2-H2`)
180°
dihedral angle
(H1`-C1-C2-H2`)
60°
Let us start with the geometry between H1` and
H2`. In all three rotamers H1` is labelled in black
and H2` labelled in red.
We always have to focus on two
planes. The first one is created from
the atoms
H1`, C1 and C2,
the second one from the atoms
H2`, C2 and C1.
The dihedral angles between these planes
are
rotamer I – 60 degree
rotamer II – 180 degree
rotamer III – 60 degree
H1`
H2`
C1
C2
I
H1`
C1
C2
H2`
II
H1`
C1
C2
H2`
III
Symmetry
Are H2`and H2``
magnetically equivalent?
p
I
(rotamer
probability)
p
II
p
III
dihedral angle
(H1`-C1-C2-H2`)
60°
dihedral angle
(H1`-C1-C2-H2`)
180°
dihedral angle
(H1`-C1-C2-H2`)
60°
According to the Karplus equation, the vicinal
coupling constant for a dihedral angle of 180 degree is
significantly larger than the vicinal coupling constant
in the case of a dihedral angle of 60 degrees.
Let us write for the coupling constants
between H1` and H2`
J
L
(arge)
if the dihedral angle is 180 degree and
J
S
(mall)
in the case of 60 degree.
J
L
J
S
J
S
J
H1`,H2`
=
p
I
* J
S
+ p
II
* J
L
+ p
III
* J
S
H1`
H2``
C1
C2
I
H1`
C1
C2
H2``
III
H1`
C1
C2
H2``
II
Symmetry
Are H2`and H2``
magnetically equivalent?
p
I
(rotamer
probability)
p
II
p
III
dihedral angle
(H1`-C1-C2-H2``)
180°
dihedral angle
(H1`-C1-C2-H2``)
60°
dihedral angle
(H1`-C1-C2-H2``)
60°
J
S
J
S
J
L
J
H1`,H2``
=
p
I
* J
L
+ p
II
* J
S
+ p
III
* J
S
Let us repeat the same considerations for
H2``, labelled in green.
H1`
C1
C2
H2`
II
Symmetry
Are H2`and H2``
magnetically equivalent?
dihedral angle
(H1`-C1-C2-H2`)
180°
dihedral angle
(H1`-C1-C2-H2`)
180°
J
L
J
L
H1`
H2``
C1
C2
I
J
H1`,H2`
= p
I
* J
S
+ p
II
* J
L
+ p
III
* J
S
J
H1`,H2``
= p
I
* J
L
+ p
II
* J
S
+ p
III
* J
S
Finally we have
Now, please keep in mind the already known relations (p
II
= p
III
and p
I
+ p
II
+ p
III
= 1) and play around a little bit with
the population of rotamer I. Start with p
I
= 0.333.
You will see how the two coupling constants vary in opposite directions with the population p
I
.
The two coupling constants would only be identical in
the case of p
I
= p
II
= p
III
, and with the two couplings we
previously labelled as J
s
being identical. But......
We made some simplifications. In principle no
pair of the three rotamers result in identical
coupling constants for either J
L
or J
S
. As an
example see the environment for the two
rotamers with dihedral angles of 180 degree
between the coupling protons (J
L
). In spite of
an idential dihedral angle the coupling
pathway is clearly different.
So if we did happen to find identical coupling
constants in such a system it would be purely
by luck!
Symmetry
Conclusion
C
H
H
H
H
C Y
X
As soon as an asymmetrically substituted ethane is
recognized as a structural fragment within an achiral
compound, the methylene protons of this ethane fragment
are always chemically equivalent and always magnetically
non-equivalent.
Why achiral?
That‘s very simple. Within chiral compounds the methylene protons are
chemically non-equivalent, which means, the question of magnetic
equivalence doesn‘t appear.
A
A`
X
X`
0.70.80.91.01.11.21.31.41.51.61.81.92.02.12.22.32.42.52.62.72.82.9
ppm
1
H
Symmetry
Keep your eyes open
Magnetic non-equivalence in alkyl chains is often
not visible at a first glance. But with open eyes, you
can see the effect almost everywhere, such as for
example in the methylene group of propylamine.
As you see the intensity ratio of the “triplet” lines
at 2.6 ppm deviates from 1:2:1 and the center line
shows deviations from the ideal line shape,
especially in the lower area. On the other hand, the
triplet of the methyl group is almost perfect, except
for the roofing effect.
Contributions
Measurements
Rainer Haeßner
Spectrometer time
TU Munich
Discussions and
native English
language support
Alan M. Kenwright
Nils Schlörer
Compilation
Rainer Haeßner
Contributions
Some special thanks.
This problem of the month is the result of an exciting discussion within the AMMRL mailing list.
It is not possible to mention all of the valuable feedback here - sorry - but some special contributions
should be mentioned, I believe.
Svetlana Simowa provided an easy to understand explanation.
Novruz Akhmedov extracted the coupling constants used in the simulations from the raw data.
Hsin Wang contributed some text building blocks for the explanation using only a few words to focus on
the essential details.
Karel Klika pointed out that, in principle, there is no perfect average even in the case of idential
populations of all three rotamers.
Lukas Hintermann always provided immediate response to stereochemical questions of all kind.