Exercise plus Solution – Quick overview
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1.01.52.02.53.03.54.0
ppm
1
H
4.104.15
1023.86
1030.99
1038.14
1045.29
Hz
1.251.30
311.31
318.48
325.60
Hz
1
H NMR spectrum
recorded at 250.13 MHz
C
4
H
8
O
2
measured in CDCl
3
Deduce the structure and measure the coupling constant!
For reasons of clarity, all
integrals have been
shifted slightly to the
left.
1.01.52.02.53.03.54.0
ppm
1
H
C
4
H
8
O
2
- one degree of unsaturation
(double bond equivalent)
- three signal groups
- Integration:
low
field multiplet - 5.10 a.u.
singlet
at 2.1 ppm - 7.65 a.u.
high
field multiplet - 7.67 a.u.
all
8 protons - 20.42 a.u.
1
proton
≙
2.55 a.u.
low
field multiplet
≙
2.00 H
singlet
at 2.1 ppm
≙
2.99 H
high
field multiplet
≙
3.01 H
Solution
5.10 a.u. 7.65 a.u. 7.67 a.u.
3H2H 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".
1.01.52.02.53.03.54.0
ppm
1
H
4.104.15
1023.86
1030.99
1038.14
1045.29
Hz
1.251.30
311.31
318.48
325.60
Hz
C
H
H
H
1.27 ppm
CH
H
H
≈2.07 ppm
One of the two methyl signals
appears as a singlet. There are
no neighbouring protons at a
distance of up to 3 bonds.
Two methyl groups may be
immediately identified by
the integral of 3.
3H2H 3H
1.01.52.02.53.03.54.0
ppm
1
H
4.104.15
1023.86
1030.99
1038.14
1045.29
Hz
1.251.30
311.31
318.48
325.60
Hz
C
H
H
H
1.27 ppm
CH
H
H
≈2.07 ppm
3H2H 3H
C
H
H
4.14 ppm
According to the n-1 rule, there are two equivalent
protons in the vicinity (up to three bonds) of this
methyl group.
These protons must be on a methylene group,
because there can only be one neighbouring group.
One of the two methyl signals
appears as a singlet. There are
no neighbouring protons at a
distance of up to 3 bonds.
1.01.52.02.53.03.54.0
ppm
1
H
4.104.15
1023.86
1030.99
1038.14
1045.29
Hz
1.251.30
311.31
318.48
325.60
Hz
C
H
H
H
1.27 ppm
CH
H
H
≈2.07 ppm
3H2H 3H
C
H
H
4.14 ppm
According to the n-1 rule, there are two equivalent
protons in the vicinity (up to three bonds) of this
methyl group.
These protons must be on a methylene group,
because there can only be one neighbouring group.
Of course, due to the three
adjacent methyl protons,
the signal of the
methylene group has to be
a quartet.
1.01.52.02.53.03.54.0
ppm
1
H
4.104.15
1023.86
1030.99
1038.14
1045.29
Hz
1.251.30
311.31
318.48
325.60
Hz
C
H
H
H
1.27 ppm
CH
H
H
≈2.07 ppm
3H2H 3H
C
H
H
4.14 ppm
Of course, due to the three
adjacent methyl protons,
the signal of the
methylene group has to be
a quartet.
We observe a typical
three bond coupling
constant.
7.14 Hz
7.14 Hz
There are still one carbon atom,
two oxygen atoms and a double
bond equivalent to be placed
between the two known
fragments.
There is only one fragment
possible:
But how should the fragment be
placed between the two free
valences?
There are two
possibilities.
C
H
H
H
1.27 ppm
CH
H
H
≈2.07 ppm
C
H
H
4.14 ppm
?
C
O
O