step
step
by
4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1139.03
1133.69
1128.36
1123.03
Hz
4.55 4.50
311.25
305.91
Hz
1.29 1.25 1.21
Exercise plus Solution – Quick overview
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4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1139.03
1133.69
1128.36
1123.03
Hz
4.55 4.50
311.25
305.91
Hz
1.29 1.25 1.21
1.00
6.16
3.04
Inte
-
gral
1
H NMR spectrum
recorded at 250.13 MHz
C
4
H
10
O
2
measured in CDCl
3
Deduce the structure!
4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1139.03
1133.69
1128.36
1123.03
Hz
4.55 4.50
311.25
305.91
Hz
1.29 1.25 1.21
1.00
6.16
3.04
Inte
-
gral
C
4
H
10
O
2
recorded in CDCl
3
Step-by-step-solution
𝒏
𝐃𝐁𝐄
=
πŸπ’
𝐂
βˆ’ 𝒏
𝐇
+ 𝟐
𝟐
𝐽 =
(1139.03Hz + 1123.03Hz)
3
𝐽 = 5.33Hz
There is no double bond equivalent.
The chemical shifts are not particularly important in this example.
An estimation is sufficient.
There are only two multipletts, i.e. they can only be "pure"
multiplets – in thuis case a doublet and a quartet.
The coupling constant is 5.33 Hz.
4.5 ppm 3.3 ppm 1.2 ppm
5.33 Hz
(d)
5.33 Hz
(q)
4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
1.00
6.16
3.04
Inte
-
gral
C
4
H
10
O
2
recorded in CDCl
3
There is no double bond equivalent.
4.5 ppm 3.3 ppm 1.2 ppm
5.33 Hz
(d)
5.33 Hz
(q)
1H 6H 3H
Integration is simple, the proportionality factor between
the measured integral (in arbitrary units) and the proton
number is just 1.
4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
C
4
H
10
O
2
recorded in CDCl
3
There is no double bond equivalent.
4.5 ppm 3.3 ppm 1.2 ppm
5.33 Hz
(d)
5.33 Hz
(q)
1H 6H 3H
C
4
H
10
O
2
C
4
H
10
O
2
H
3
C
CH
Let us first assume that all three signal groups originate
from CH
n
fragments. The reason why OH is out of question
follows in a few steps.
Then we would have a CH group at 4.5 ppm and a CH
3
group at 1.2 ppm.
4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
4.5 ppm 3.3 ppm 1.2 ppm
5.33 Hz
(d)
5.33 Hz
(q)
1H 6H 3H
C
4
H
10
O
2
H
3
C CH
H
3
C
CH
If the carbon atoms of the two fragments are joined
together, the coupling patterns of both multiplets can be
well explained.
In the case of CH, of course, two free valences are missing.
5.33 Hz
A coupling constant of 5.33 Hz is somewhat small for
vicinal coupling. 7 Hz would be ideal. On the other hand, a
wide-range coupling across four single bonds would be
much smaller, about 0.5 Hz.
Let us now replace CH
with OH on a trial basis.
4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
4.5 ppm 3.3 ppm 1.2 ppm
5.33 Hz
(d)
5.33 Hz
(q)
1H 6H 3H
C
4
H
10
O
2
H
3
C OH
This would give us methanol. Further connections are not
possible, but the part of the molecule responsible for the
singlet at approx. 3.3 ppm still has to be connected
somehow.
Lets return to CH
3
–CH<.
4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
4.5 ppm 3.3 ppm 1.2 ppm
1H 6H 3H
C
4
H
10
O
2
A singlet with 6 protons might be due to two equivalent
CH
3
or three equivalent CH
2
groups.
H
3
C CH
5.33 Hz
4.5 ppm
1.2 ppm
Two out of a total of four C atoms are already assigned.
Three more carbon atoms are impossible. The signal at 3.3
ppm comes from two equivalent methyl groups.
H
3
C
H
3
C
The methyl groups cannot be directly
attached to the CH group. If they were
directly attached, we would see a doublet at
3.8 ppm.
X
X
4.5 4.0 3.5 3.0 2.5 2.0 1.5 ppm
1
H
3.3 ppm
1H 6H 3H
C
4
H
10
O
2
A short inventory is sufficient for a complete structural
elucidation.
H
3
C CH
5.33 Hz
4.5 ppm
1.2 ppm
H
3
C
H
3
C
molecular formula : C
4
H
10
O
2
known fragments : C
4
H
10
missing : O
2
This results in a single option for the final
structure.
O
O
H
3
C
H
3
C
3.3 ppm
Contributions
Measurements
Rainer Haeßner
Spectrometer time
TU Munich
Discussions and
native English
language support
Alan Kenwright
Compilation
Rainer Haeßner
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