6.5 1.56.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0
1
H
ppm
1044.07
1036.91
1029.78
1022.66
Hz
4.2 4.1
311.47
304.33
297.19
Hz
1.2
step
step
by
Exercise plus Solution Quick overview
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6.5 1.56.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0
1
H
ppm
1044.07
1036.91
1029.78
1022.66
Hz
4.2 4.1
1.00
0.97
1.03
2.02
3.09
Inte
-
gral
311.47
304.33
297.19
Hz
1.2
1588.91
1587.25
1571.60
1569.94
1522.65
1512.30
1505.35
1495.00
1437.73
1436.07
1427.38
1425.72
Hz
6.3 5.76.2 6.1 6.0 5.9 5.8
Deduce the structure. Analyze the multiplets between 5.5 and 6.5
ppm and extract all coupling constants.
Make a stereochemically correct assignments of the protons
belonging to these signals.
C
5
H
8
O
2
measured in CDCl
3
1
H NMR spectrum
recorded at 250.13 MHz
6.5 1.56.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0
1
H
ppm
1.00
0.97
1.03
2.02
3.09
Inte
-
gral
The integration is simple. It was arbitrarily defined that the multiplet
at approx. 6.3 ppm is caused by one proton. The assumption was
correct, as can be easily determined by simply rounding the
remaining four integrals.
1044.07
1036.91
1029.78
1022.66
Hz
4.2 4.1
1588.91
1587.25
1571.60
1569.94
1522.65
1512.30
1505.35
1495.00
1437.73
1436.07
1427.38
1425.72
Hz
6.3 5.76.2 6.1 6.0 5.9 5.8
311.47
304.33
297.19
Hz
1.2
C
5
H
8
O
2
measured in CDCl
3
Step-by-step-solution
6.5 1.56.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0
1
H
ppm
1.00
0.97
1.03
2.02
3.09
Inte
-
gral
The integration is simple. It was arbitrarily defined that the multiplet
at approx. 6.3 ppm is caused by one proton.The assumption was
correct, as can be easily determined by simply rounding the
remaining four integrals.
1H1H 1H 2H 3H
1044.07
1036.91
1029.78
1022.66
Hz
4.2 4.1
1588.91
1587.25
1571.60
1569.94
1522.65
1512.30
1505.35
1495.00
1437.73
1436.07
1427.38
1425.72
Hz
6.3 5.76.2 6.1 6.0 5.9 5.8
311.47
304.33
297.19
Hz
1.2
C
5
H
8
O
2
measured in CDCl
3
6.5 1.56.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0
1
H
ppm
The quartet and the triplet are due to an ethyl group. The step-by-
step derivation will not be discussed. The focus here is on the signals
at about 6 ppm.
1H1H 1H 2H 3H
1044.07
1036.91
1029.78
1022.66
Hz
4.2 4.1
1588.91
1587.25
1571.60
1569.94
1522.65
1512.30
1505.35
1495.00
1437.73
1436.07
1427.38
1425.72
Hz
6.3 5.76.2 6.1 6.0 5.9 5.8
311.47
304.33
297.19
Hz
1.2
H
2
C CH
3
7.14 Hz
4.13
ppm
1.22
ppm
C
5
H
8
O
2
measured in CDCl
3
How to understand the multipletts at about 6 ppm - especially the
middle one?
1588.91
1587.25
1571.60
1569.94
1522.65
1512.30
1505.35
1495.00
1437.73
1436.07
1427.38
1425.72
Hz
H
2
C CH
3
7.14 Hz
4.13
ppm
1.22
ppm
6.3 5.76.2 6.1 6.0 5.9 5.8
6.3 5.76.2 6.1 6.0 5.9 5.8
1588.91
1587.25
1571.60
1569.94
1522.65
1512.30
1505.35
1495.00
1437.73
1436.07
1427.38
1425.72
Hz
C
5
H
8
O
2
measured in CDCl
3
What is still missing?
H
2
C CH
3
7.14 Hz
4.13
ppm
1.22
ppm
6.3 5.76.2 6.1 6.0 5.9 5.8
1588.91
1587.25
1571.60
1569.94
1522.65
1512.30
1505.35
1495.00
1437.73
1436.07
1427.38
1425.72
Hz
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
On the one hand, we got 2 double bond equivalents (DBE) from the
molecular formula, none of which has been assigned so far, and on
the other hand, a residue C
3
H
3
O
2
remains after the ethyl group has
been subtracted.
missing: 2DBE + C
3
H
3
O
2
C
5
H
8
O
2
measured in CDCl
3