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