Exercise plus Solution Quick overview
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1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
1
H NMR spectrum
measured at 250.13 MHz
C
3
H
5
ClO measured in CDCl
3
Deduce the structure!
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
First Considerations
The proton spectrum consists of two signal groups.
There are two chemically distinguishable protons
whose number is not yet known, i.e. n H
a
and m H
b
.
We need the values of n and m.
The molecular formula results in one double bond equivalent.
The structural fragments H-Cl, H-H and H
2
O are not
possible. In each case we would end in a mixture of at
least two components.
n H
a
m H
b
The structural fragment OH is not excluded.
1.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
298.13
305.53
312.89
718.37
725.65
733.05
740.44
Hz
δ =
ν
Sample
ν
Refere𝑛𝑐𝑒
ν
Refere𝑛𝑐𝑒
Chemical shifts
In order to estimate the chemical shifts more accurately than is
possible just by looking at the scale below the spectrum, we
need the peak labels given in Hz together with the well-known
formula
ν
Reference
is 250.13 MHz according to the problem
description, the peak labels do not represent absolute
frequencies, but rather the difference from the
reference signal (ν
Sample
ν
Reference
).