Exercise plus Solution Quick overview
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1.01.52.02.53.03.54.04.55.05.56.06.57.0
ppm
1
H
Which alkylbromide was used in the etherisation of
Phenol?
Assign, with the help of the DEPT subspectra, as many of
the carbon peaks as possible.
How is the multiplet displayed in blue made up?
For the multiplet shown in red und the one at highest
field, shown in green, work out the theoretically expected
number of lines.
1
H NMR spectrum
measured at 300.13 MHz
1252.05
1258.11
1264.15
1270.20
1276.24
1282.28
4.25 4.20
27 4.43
28 1.90
28 9.34
36 7.78
37 3.87
1.2 1.1
1.0 0.9
1.7 1.6 1.5
1
H NMR spectrum
measured at 300.13 MHz
Enlarged sections of the overview spectrum
on the previous page.
Scale division: [ppm]
Peak label: [Hz]
160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm
13
C
9.63
19.10
29.11
74.63
115.77
120.27
129.29
158.17
ppm
13
C{
1
H} NMR spectrum
measured at 75.47{300.13} MHz
DEPT CH only
DEPT CH
2
only
DEPT CH
3
only
C C
C
CC
C O
C
n
H
2n+1
HH
H
H H
Br C
n
H
2n+1
C C
C
CC
C O
H
HH
H
H H
Step-by-step-solution
+
2n+1 H
δ ≈ 1 … 4 ppm
5 H
δ ≈ 6 … 8 ppm
You might express one part of the task directly into
chemical reaction equation.
We expect signals in two
well-separated spectral
regions.
1.01.52.02.53.03.54.04.55.05.56.06.57.0
ppm
1
H
C C
C
CC
C O
C
n
H
2n+1
HH
H
H H
5 H 2n+1 H
Distribution of the integrals
1.01.52.02.53.03.54.04.55.05.56.06.57.0
ppm
1
H
5 H 2n+1 H
7.3
a.u.
11.0
a.u.
5 H 11.0 a.u. + 7.3 a.u.
1 H 3.7 a.u.
3.7
a.u.
1H
7.5
a.u.
2H
11.1
a.u.
3H
11.3
a.u.
3H
a.u. ???
arbitrary units
This depends on the output
device and your home country.
For example, using a tablet in
Europe, you could think about
"centimeters".
2n+1 = 9
n = 4
Butyl bromide was used
for etherification.
Determination of n
160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10
ppm
13
C
DEPT CH only
DEPT CH
2
only
DEPT CH
3
only
Four different isomers of the butyl moiety are possible.
The easiest way to determine which of these is present is the use of DEPT subspectra.
160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10
ppm
13
C
DEPT CH only
DEPT CH
2
only
DEPT CH
3
only
In the individual isomers we expect the following CH
n
groups.
0 * CH
3 * CH
2
1 * CH
3
1 * CH
1 * CH
2
1 * CH
3
(symmetry!)
1 * CH
1 * CH
2
2 * CH
3
0 * CH
0 * CH
2
1 * CH
3
(sym-
metry)
160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10
ppm
13
C
DEPT CH only
DEPT CH
2
only
DEPT CH
3
only
1 * CH
1 * CH
2
2 * CH
3
Only the 2-butyl residue is compatible with the
DEPT subspectra.
2 * CH
3
1 * CH
2
1 * CH
(The =CH- signals between 110 and 130
ppm belong to the phenyl group.)
1.7
1.6 1.5
4.25 4.20
1252.05
1258.11
1264.15
1270.20
1276.24
1282.28
Hz
1.2 1.1
1.0 0.9
274.43
281.90
289.34
367.78
373.87
Hz
C
C
C
C
C
C
H
C
H
3
C C
CH
3
HH
H
H
H
H
H
Analysis of the proton multiplets
methyl protons attached to carbon 1
Only the proton in the 2-position is adjacent.
We expect a doublet.
6.09 Hz
The coupling constant is 6.09 Hz.
1.7
1.6 1.5
4.25 4.20
1252.05
1258.11
1264.15
1270.20
1276.24
1282.28
Hz
1.2 1.1
1.0 0.9
274.43
281.90
289.34
367.78
373.87
Hz
The five protons separated via three bonds are not
chemically equivalent, but the size of the coupling
constants should be comparable. The result is a
pseudo sextet.
The average coupling constant is 6.05 Hz.
C
C
C
C
C
C
H
C
H
3
C C
CH
3
HH
H
H
H
H
H
6.09 Hz
Analysis of the proton multiplets
methin proton attached to carbon 2
1.7
1.6 1.5
4.25 4.20
1252.05
1258.11
1264.15
1270.20
1276.24
1282.28
Hz
1.2 1.1
1.0 0.9
274.43
281.90
289.34
367.78
373.87
Hz
Adjacent are the two protons of the methylene
group. We expect a triplet.
The coupling constant is 7.46 Hz.
C
C
C
C
C
C
H
C
H
3
C C
C
HH
H
H
H
H
H
6.09 Hz
Note: Both the coupling constant and the triplet
structure will need to be reviewed a little later.
H
3
7.46 Hz
Analysis of the proton multiplets
methyl protons attached to carbon 4
1.7
1.6 1.5
4.25 4.20
1252.05
1258.11
1264.15
1270.20
1276.24
1282.28
Hz
1.2 1.1
1.0 0.9
274.43
281.90
289.34
367.78
373.87
Hz
C
C
C
C
C
C
H
C
H
3
C C
CH
3
HH
H
H
H
H
H
*
6.09 Hz
7.46 Hz
Why is the signal of the two remaining
methylene protons that crowded?
The compound has a center of chirality at C-2,
so we have not one but two signals from two
diastereotopic protons.
H-3`
H-3``
Analysis of the proton multiplets
methylene protons attached to carbon 3
1.7
1.6 1.5
4.25 4.20
1252.05
1258.11
1264.15
1270.20
1276.24
1282.28
Hz
1.2 1.1
1.0 0.9
274.43
281.90
289.34
367.78
373.87
Hz
C
C
C
C
C
C
H
C
H
3
C C
CH
3
HH
H
H
H
H
H
*
6.09 Hz
Because of the similarity of the chemical shifts
of H-3' and H-3'', the whole spectrum cannot be
properly evaluated according to 1st order rules.
The signal of the methyl group in the 4-position
is in any case not a triplet, but more likely a
doublet of doublets.
H-3`
H-3``
Analysis of the proton multiplets
methyl protons attached to carbon 4
1.01.52.02.53.03.54.04.55.05.56.06.57.0
ppm
1
H
C
C
C
C
C
C
H
C
H
3
C C
CH
3
HH
H
H
H
H
H
d
ddq
ddq
ddq
dd
*
Summary
160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10
ppm
13
C
C
C
C
C
C
C
H
C
H
3
C C
CH
3
HH
H
H
H
H
H
Summary
Contributions
Measurements
Gottfried Zimmermann
Spectrometer time
Goethe University Frankfurt
Discussions and
native English
language support
Alan Kenwright
Compilation
Rainer Haeßner
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