Exercise plus Solution – Quick PDF overview
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3.5 3.0 2.5 2.0 1.5 1.0
ppm1
H
3.65 3.60 3.55
922.1
916.0
885.4
891.5
897.6
903.7
909.0
Hz
1.101.121.14 1.08 1.06
278.3
272.1
Hz
1
H NMR spectrum
recorded at 250.13 MHz
C
6
H
14
O measured in CDCl
3
Deduce the structure!
3.5 3.0 2.5 2.0 1.5 1.0
ppm1
H
C
6
H
14
O
- no degree of unsaturation
(double bond equivalents)
- fourteen protons and two signal
groups only – there has to be
some kind of symmetry
- Integration:
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".
Solution
low
field multiplet - 1.7 a.u.
high
field
multiplet
- 9.61 a.u.
all 14
protons - 11.31 a.u.
1
proton
≙
0.81 a.u.
low
field multiplet
≙
2.1 H
high
field
multiplet
≙
11.9 H
1.7 a.u.
9.61 a.u.
2H
12H
3.5 3.0 2.5 2.0 1.5 1.0
ppm1
H
C
6
H
14
O
Solution
2H
12H
3.65 3.60 3.55
922
.
1
916
.
0
885
.
4
891
.
5
897
.
6
903
.
7
909
.
0
H
z
Let us start with the septet.
- no degree of unsaturation
(double bond equivalents)
- fourteen protons and two signal
groups only – there has to be
some kind of symmetry
- Integration:
low
field multiplet - 1.7 a.u.
high
field
multiplet
- 9.61 a.u.
all 14
protons - 11.31 a.u.
1
proton
≙
0.81 a.u.
low
field multiplet
≙
2.1 H
high
field
multiplet
≙
11.9 H
3.65 3.60 3.55
922
.
1
916
.
0
885
.
4
891
.
5
897
.
6
903
.
7
909
.
0
H
z
2H
According to the n-1 rule, we need 6 equivalent
neighbour protons to observe a septet.
H
C
H
2
C
H
2
C
H
2
C
C
H
2
CH
3
H
3
C
H
C
CH
3
H
3
C
third possibility
first possibility
Because of the integral of
2 the fragment is needed
twice, resulting in 8 C-
atoms.
second possibility
Where should the
remaining C
3
H
6
O be
bound?
C
6
H
14
O
3.65 3.60 3.55
922
.
1
916
.
0
885
.
4
891
.
5
897
.
6
903
.
7
909
.
0
H
z
2H
H
C
H
2
C
H
2
C
H
2
C
C
H
2
CH
3
H
3
C
C
6
H
14
O
And why do we see an integral of 2?
H
C
CH
3
H
3
C
H
C
CH
3
H
3
C
According to the n-1 rule, we need 6 equivalent
neighbour protons to observe a septet.
Because of the integral of
2 the fragment is needed
twice, resulting in 8 C-
atoms.
Where should the
remaining C
3
H
6
O be
bound?
third possibility
3.65 3.60 3.55
922
.
1
916
.
0
885
.
4
891
.
5
897
.
6
903
.
7
909
.
0
H
z
2H
C
6
H
14
O
And why do we see an integral of 2?
H
C
CH
3
H
3
C
The integral of 2 is
obtained via an identical
second isopropyl fragment.
C
CH
3
H
3
C
H
H
C
CH
3
H
3
C
And now only one
oxygen atom is missing
from the molecular
formula. The two
isopropyl fragments can
be connected using the
oxygen atom.
O
3.65 3.60 3.55
922
.
1
916
.
0
885
.
4
891
.
5
897
.
6
903
.
7
909
.
0
H
z
2H
C
6
H
14
O
What about the methyl groups?
H
C
CH
3
H
3
C
C
CH
3
H
3
C
H
O
1.101.12 1.08
278
.
3
272
.
1
H
z
12H
All six methyl protons within the isopropyl group are chemically
equivalent and are three bonds separated from the same
neighbour proton.
We expect a doublet.
The integral of 12
results from two
identical isopropyl
groups.
3.65 3.60 3.55
922
.
1
916
.
0
885
.
4
891
.
5
897
.
6
903
.
7
909
.
0
H
z
2H
C
6
H
14
O
One final remark.
H
C
CH
3
H
3
C
C
CH
3
H
3
C
H
O
1.101.12 1.08
278
.
3
272
.
1
H
z
12H
In principle there is a second
way for the scalar coupling
across 5 bonds. Normally,
only couplings over up to
three bonds may be seen.
Here we are not able to see
this coupling.
