Exercise plus Solution – Quick PDF overview
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3.50 1.001.502.002.503.00
1
H
1.19
1.81
Inte-
gral
1716.28
1709.19
1702.20
1695.19
577.27
570.29
563.21
Hz
C
4
H
10
O measured in CDCl
3
Deduce the structure.
Where could the small signal at about
1.6 ppm come from?
1
H NMR spectrum
recorded at 500.13 MHz
Solution
Let us start with the easiest part, the signal
at about 1.6 ppm.
C
4
H
10
O measured in CDCl
3
The sample is measured in CDCl
3
. Typically, two
proton signals are found in CDCl
3
, although this is
not at all to be expected from the molecular
formula CDCl
3
.
First, there is a signal from CHCl
3
at about 7.25 ppm. This is due
to the non-100% deuteration and is therefore often referred to
as the "residual proton signal". In this example, the signal is
outside the spectral range shown.
On the other hand, deuterated solvents nearly always contain a small
amount of water as an impurity. In CDCl
3
, its signal appears at about 1.6
ppm.
With high probability, HOD is present.
HOD
Solution
=
577.27 Hz + 563.21 Hz
2 ∗ 500.13 MHz
= 𝟏. 𝟏𝟒 𝐩𝐩𝐦
There are no double bond equivalents.
Although simple integration is usually a good
starting point for spectral evaluation, we
ignore it here for now.
Let's start with the multiplet at about 1.1 ppm.
The exact chemical shift is 1.14 ppm.
As a reminder, ppm is not a unit of
measurement, but simply a ratio of frequencies
x 10
-6
, but is used like a unit of measurement.
HOD
1.14 ppm
C
4
H
10
O measured in CDCl
3
Solution
The signal group is a triplet.
Number of lines - 1 = 2.
Which means, there are two equivalent nuclei with I = ½ in
neighbourhood.
The only possible neighbours are the protons with the signal at
approx. 3.4 ppm. Of course, one could also think of nuclei such as
19
F
or
31
P, but they do not appear in the molecular formula.
HOD
1.14 ppm
2 Neigh-
bors
Solution
𝐽 =
577.27 Hz − 563.21 Hz
2
= 𝟕. 𝟎𝟑 𝐇𝐳
The value of the coupling constant is 7.03 Hz.
A coupling constant of about 7 Hz is observed
very often and is typical for coupling via three
bonds (technical term: vicinal coupling) along
the chain H - C - C - H under the condition of a
free rotation around the C - C single bond.
HOD
1.14 ppm
J = 7.03 Hz
H X
C
C
YH
2 Neigh-
bors
HOD
1.14 ppm
J = 7.03 Hz
H X
C
C
YH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
We have just determined the assignment of one of the two protons of this
fragment and measured the vicinal coupling constant between the two
protons.
The chemical shift of the second proton is still missing. The coupling
constant must be identical, a quick check ((1716.28 Hz - 1695.19 Hz) / 3)
returns an identical value.
2 Neigh-
bors
HOD
1.14 ppm
J = 7.03 Hz
H X
C
C
YH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
Because of the triplet structure of the proton signal at 1.14 ppm, the signal
at 3.41 ppm must come from 2 equivalent protons.
3.41 ppm
2 Neigh-
bors
This condition is fulfilled if we replace X with H.
HOD
1.14 ppm
J = 7.03 Hz
H H
C
C
YH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
1
What does the proton spectrum of this hypothetical molecule look like
now?
The triplet is due to the two equivalent protons bound to the
neighbouring carbon atom.
3.41 ppm
At 1.14 ppm we observe a triplet with the integral of 1.
HOD
1.14 ppm
J = 7.03 Hz
H H
C
C
YH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
2
1
What does the proton spectrum of this hypothetical molecule look like
now?
The doublet structure is due to a proton bound to the first carbon atom.
3.41 ppm
At 3.41 ppm we observe a doublet with the integral of 2.
HOD
1.14 ppm
J = 7.03 Hz
H H
C
C
YH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
2
1
The triplet is fine. But for the multiplet at 3.41 ppm we should have a
quartet instead of a doublet.
The number of neighbouring nuclei (in this case protons) is always
responsible for the multiplet structure.
3.41 ppm
What do we need to change?
To change the multiplet structure at 3.41 ppm, we need to change the
number of protons bound to the carbon on the right side.
HOD
1.14 ppm
J = 7.03 Hz
H H
C
C
YH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
2
1
How much protons do we need to get a quartet at 3.41 ppm?
3.41 ppm
The general rule for nuclei with I = ½ (e.g. protons) would be
multiplicity = number of neighbours + 1.
We would need three equivalent protons on the right C atom, in other
words, there would have to be a methyl group.
HOD
1.14 ppm
J = 7.03 Hz
H H
C
C
HH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
3.41 ppm
2
3
What does our multiplet at 1.14 ppm look like now?
H
The number of neighbours remains unchanged at 2, i.e. we continue to see a
triplet. The integral is now 3, of course.
1
HOD
1.14 ppm
J = 7.03 Hz
H H
C
C
HH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
3.41 ppm
2
3
And what about 3.41 ppm now?
H
The number of neighbours has increased from 1 to 3, accordingly we now see a
quartet instead of a doublet. The integral remains unchanged.
HOD
1.14 ppm
J = 7.03 Hz
H H
C
C
HH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
Finally we have an ethyl group.
H
What about the integrals?
We expect a ratio of 2 : 3 between the number of methylene protons at
3.41 ppm and the number of methyl protons at 1.14 ppm.
The measured integral ratio 1.19 : 1.81 corresponds almost perfectly to the
expected value.
3.41 ppm
HOD
1.14 ppm
J = 7.03 Hz
H H
C
C
HH
1.14 ppm
7.03 Hz
3.41 ppm
J = 7.03 Hz
Great!
But we found a fragment of C
2
H
5
and the molecular formula is C
4
H
10
O?
There are no other signals in the proton spectrum.
H
Let's do a little math.
molecular formula - C
4
H
10
O
known fragment - C
2
H
5
still to be assigned - C
2
H
5
O
3.41 ppm
HOD
1.14 ppm
J = 7.03 Hz
3.41 ppm
J = 7.03 Hz
Let's do a little math.
molecular formula - C
4
H
10
O
known fragment - C
2
H
5
still to be assigned - C
2
H
5
O
Apart from the oxygen, we see the already known ethyl fragment a second time.
The second ethyl group must of course be chemically equivalent to the ethyl
group already found.
The oxygen is then the bridge between the two ethyl
groups and finally we get diethyl ether..
H H
C
C
HH
H
H H
C
C
HH
H
HH
C
C
H H
H
1.14 ppm
7.03 Hz
3.41 ppm
O
Let's do a little math.
molecular formula - C
4
H
10
O
known fragment - C
2
H
5
still to be assigned - C
2
H
5
O
