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.
0
1
2
3
0.61.01.41.82.22.63.03.4 ppm
1
H
20
25
30
35
40
13
C
Solution
Part 2 – Building blocks
18.22
30.60
31.38
39.42
3.3 … 3.4 1.80 1.63 0.85
34 2
18.22
0.85
CH
3
18.22
0.85
CH
3
Undoubtedly, the first fragment to be taken from
the HSQC is a methyl group.
2
3
0.61.01.41.82.22.63.03.4 ppm
1
H
20
25
30
35
40
13
C
Solution
Part 2 – Building blocks
18.22
30.60
31.38
39.42
3.3 … 3.4 1.80 1.63 0.85
04 2
18.22
0.85
CH
3
According to the sign of the cross peak, the next
fragment could be both a methyl or a methine group.
In the HSQC there is another cross peak with the same
proton chemical shift visible. This means that less than 3
protons (which is the integral of the proton spectrum) are
available for this cross peak. A methyl group is therefore
excluded.
30.60
1.63
C
H
30.60
1.63
C
H
0
1
2
3
4
0.61.01.41.82.22.63.03.4 ppm
1
H
20
25
30
35
40
13
C
Solution
Part 2 – Building blocks
18.22
30.60
31.38
39.42
3.3 … 3.4 1.80 1.63 0.85
2 02
18.22
0.85
CH
3
30.60
1.63
C
H
According to the sign of the cross peak in the edited HSQC,
the complex proton signal at about 3.4 ppm can only
belong to a CH
2
group.
31.38
3.3 .. 3.4
3.3 .. 3.4
C
H
H
31.38
3.3 .. 3.4
3.3 .. 3.4
C
H
H
Four protons, but only one carbon signal?
There are two equivalent methylene groups in the molecule!
2 *
0.61.01.41.82.22.63.03.4 ppm
1
H
20
25
30
35
40
13
C
Solution
Part 2 – Building blocks
18.22
30.60
31.38
39.42
3.3 … 3.4 1.80 1.63 0.85
2 00 2
18.22
0.85
CH
3
30.60
1.63
C
H
31.38
3.3 .. 3.4
3.3 .. 3.4
C
H
H
2 *
39.42
1.80
1.63
C
H
H
39.42
1.80
1.63
C
H
H
There are two further equivalent
methylene groups.
This time the proton signals are
clearly separated.
2 *
1.0
1.5
2.0
2.5
3.0
1
H
0.61.01.41.82.22.63.03.4
ppm
1
H
3 34 2
Solution
Part 3 – Connect the
building blocks
18.22
0.85
CH
3
30.60
1.63
C
H
31.38
3.3 .. 3.4
3.3 .. 3.4
C
H
H
2 *
39.42
1.80
1.63
C
H
H
Two COSY cross peaks show the
same connectivity.
2 *
1
2
2
3
2
3
4
1.0
1.5
2.0
2.5
3.0
1
H
0.61.01.41.82.22.63.03.4
ppm
1
H
3
Solution
Part 3 – Connect the
building blocks
18.22
0.85
CH
3
30.60
1.63
C
H
2 *
2 *
31.38
3.3 .. 3.4
3.3 .. 3.4
C
H
H
39.42
1.80
1.63
C
H
H
31.38
3.3 .. 3.4
3.3 .. 3.4
C
H
H
39.42
1.80
1.63
C
H
H
The resulting fragment exists twice
(by virtue of symmetry).
0
1
e
x
1
2
0
1
2
1.0
1.5
2.0
2.5
3.0
1
H
0.61.01.41.82.22.63.03.4
ppm
1
H
3
Solution
Part 3 – Connect the
building blocks
18.22
0.85
CH
3
30.60
1.63
C
H
31.38
39.42
3.3 .. 3.4
1.80
1.63
3.3 .. 3.4
C
H
H
C
H
H
C
H
H
C
H
H
According to the number of
remaining protons 1 + 3
protons have to be
connected. In principle you
don‘t need the COSY cross
peak for this task, but of
course the cross peak exists.
1.0
1.5
2.0
2.5
3.0
1
H
0.61.01.41.82.22.63.03.4
ppm
1
H
Solution
Part 3 – Connect the
building blocks
31.38
39.42
3.3 .. 3.4
1.80
1.63
3.3 .. 3.4
C
H
H
C
H
H
C
H
H
C
H
H
18.22
0.85
CH
3
30.60
1.63
C
H
30.60
1.63
C
H
18.22
0.85
CH
3
30.630.831.031.231.4
31.390
31.378
30.600
Solution
Part 4 – What is the halogene
31.38
39.42
3.3 .. 3.4
1.80
1.63
3.3 .. 3.4
C
H
H
C
H
H
C
H
H
C
H
H
18.22
30.60
0.85
1.63
CH
3
C
H
Either a really carefully done measurement or sometimes an
appropriate postprocessing shows an isotope pattern (1:1) at
31.38 ppm.
The natural isotopes of which halogene show the isotope
ratio of about 1:1?
19
F - 100 %
35
Cl /
37
Cl - 76/24 %
79
Br /
81
Br - 51/49%
127
I - 100%
Solution
Part 5 – Nearly done
18.22
30.60
0.85
1.63
CH
3
C
H
C
H
H
C
H
H
Br
31.38
39.42
3.3 .. 3.4
1.80
1.63
Br C
H
H
C
H
H
Even if it is impossible to evaluate the crowded
region between 1.63 and 1.80 ppm in the COSY, there
is only one final combination of all components
possible.
0.84
ppm
3.3 … 3.4
ppm
1.63
ppm
1.80
ppm
30.60
ppm
31.38
ppm
18.22
ppm
39.42
ppm
H
H
H
3
C
H
C
H
H
Br
C
C
H
H
C
C
Br
H
H
≈1.78
ppm
Solution
Part 6 – Stereochemistry
But why are the
methylene protons not
equivalent?
There is no center of
chirality anywhere.
H
H
H
3
C
H
C
H
H
Br
C
C
H
H
C
C
Br
H
H
H
H
H
3
C
H
C
H
H
Br
C
C
H
H
C
C
Br
X
H
H
H
H
3
C
H
C
H
H
Br
C
C
H
H
C
C
Br
H
X
H
H
H
3
C
H
C
H
H
Br
C
C
H
H
C
C
Br
H
H
Solution
Part 6 – Stereochemistry
Let us tentatively exchange, one after the other, both protons of any
methylene group using a fictive nucleus X.
H
H
H
3
C
H
C
H
H
Br
C
C
H
H
C
C
Br
X
H
H
H
H
3
C
H
C
H
H
Br
C
C
H
H
C
C
Br
H
X
Solution
Part 6 – Stereochemistry
We create two centers of chirality at
once. It is impossible to make both
molecules coincide by any symmetry
operation (including mirroring). The
molecule on the left and the molecule
on the right side are chemically
different, which finally means, all four
methylene group protons are
diasterotopic.
*
*
*
*
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
0.84
ppm
3.32
ppm
1.63
ppm
1.80
ppm
30.60
ppm
31.38
ppm
18.22
ppm
39.42
ppm
H
H
H
3
C
H
C
H
H
Br
C
C
H
H
C
C
Br
H
H
≈1.78
ppm
3.38
ppm
Measuring the proton NMR spectrum at 900
MHz definitely shows two signals for the
protons of the diastereotopic methylene
groups in positions 1 and 5.
The quality of this spectrum is a little bit
low, because the measurement was done
during helium refill to avoid blocking the
spectrometer for more challenging samples.
