Exercise plus Solution – Quick overview
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0.81.01.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
0.76
0.50
0.75
1.00
Inte-
gral
C
6
H
12
X
2
measured in CDCl
3
Deduce the structure without taking care about the
weird coupling pattern.
Replace „X“ by the correct
halogen.
Count the number of chemically different
protons.
1
H NMR spectrum
measured at 500.13 MHz
171819202122232425262728303132333435363738394041424344 ppm
13
C
18.22
30.60
31.38
39.42
ppm
30.630.831.031.231.4
31.390
31.378
30.600
To process the area with two signals around 31 ppm
some kind of „magic“ (the correct name is reference
deconvolution) was used.
What might be the reason for the splitting visible at
31.8 ppm?
13
C{
1
H} NMR spectrum
measured at 125.83{500.13} MHz
1.0
1.5
2.0
2.5
3.0
1
H
0.61.01.41.82.22.63.03.4 ppm
1
H
0.61.01.41.82.22.63.03.4 ppm
1
H
20
25
30
35
40
13
C
1
H/
1
H COSY
measured at 500.13 MHz
multiplicity edited
1
H/
13
C HSQC
measured at 500.13/125.83 MHz
CH oder CH
3
CH
2
auxiliary linies
0.81.01.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
0.75
0.50
0.75
1.00
Inte-
gral
1
H NMR spectrum
measured at 900.14 MHz
Finally, compare your solution with the
1
H NMR
spectrum measured using a higher frequency.
This spectrum is not necessary to solve the problem.
0.81.01.21.41.61.82.02.22.42.62.83.03.23.4 ppm
1
H
0.76
0.50
0.75
1.00
Inte-
gral
Solution
Part 1 - Integration
In this example the integration procedure is not described in detail.
There are four signal groups and the integral ratio is
4 : 3 : 2 : 3
4 3 2 3
Solution
Part 2 – Building blocks
If available, the HSQC/HMQC is nearly
always the best starting point to collect all or
at least a large number of partial structures
as an unordered pile of building blocks.
The integrals from the proton spectrum have
just been determined, the chemical shifts of
the carbon signals may be taken from the one-
dimensional carbon spectrum.
For the chemical shifts of the proton multiplets, only
estimated values are possible due to the absence of
peak labels.