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
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
=
1588.91 Hz + 1569.94 Hz
2 ∗ 250.13 MHz
= 6.314 ppm
𝐽 =
(1437.73 Hz + 1436.07 Hz)
2
−
1427.38 Hz + 1425.72 Hz
2
𝐽 = 10.35 Hz
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
Even if the intensity distribution of the four lines does not perfectly
correspond to the ratio 1 : 1 : 1 : 1, the multipletts at approx. 6.3 ppm and
approx. 5.7 ppm should each be doublets of doublets (the two calculations are
examples for extracting the values).
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
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
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
If we measure the distance between the first two and the last two lines of the
middle multiplet, we find there the 10.35 Hz from the doublet of doublets at
5.724 ppm.
10.35 Hz
10.35 Hz
Are that two doublets despite the somewhat unexpected intensity
distribution? Let's try it out (the two new frequency data are mean values).
1517.48
1500.17
10.35 Hz
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
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
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
Now we have two lines exactly 17.30 Hz apart. Despite the unusual intensity
distribution, there is also a doublet of doublets at about 6 ppm.
1517.48
1500.18
6.033 ppm
10.35 Hz
17.30 Hz
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
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
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
And the strange intensity distribution?
6.033 ppm
10.35 Hz
17.30 Hz
The roof effect, because due to the small chemical difference compared to the
coupling constants, a consideration according to the 1st order rules is no
longer valid without restrictions.
1 72 3 4
5
6 98
10
1211 13 14
150
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
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
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
And the strange intensity distribution?
6.033 ppm
10.35 Hz
17.30 Hz
The roof effect, because due to the small chemical difference compared to the
coupling constants, a consideration according to the 1st order rules is no
longer valid without restrictions.
1 72 3 4
5
6 98
10
1211 13 14
150
1 72 3 4
5
6 98
10
1211 13 14
150
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
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
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
And the strange intensity distribution?
6.033 ppm
10.35 Hz
17.30 Hz
The roof effect, because due to the small chemical difference compared to the
coupling constants, a consideration according to the 1st order rules is no
longer valid without restrictions.
1 72 3 4
5
6 98
10
1211 13 14
150
1 72 3 4
5
6 98
10
1211 13 14
150
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
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
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
To demonstrate the roof effect between the signals at 6.314 ppm and 6.033
ppm, we had used the lines with a distance of 17.31 Hz each.
6.033 ppm
10.35 Hz
17.30 Hz
For the second roof effect, the lines with a distance of 10.35 Hz each are now
of interest.
1 72 3 4
5
6 98
10
1211 13 14
150
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
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
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
To demonstrate the roof effect between the signals at 6.314 ppm and 6.033
ppm, we had used the lines with a distance of 17.31 Hz each.
6.033 ppm
10.35 Hz
17.30 Hz
For the second roof effect, the lines with a distance of 10.35 Hz each are now
of interest.
1 72 3 4
5
6 98
10
1211 13 14
150
1 72 3 4
5
6 98
10
1211 13 14
150
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
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
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
To demonstrate the roof effect between the signals at 6.314 ppm and 6.033
ppm, we had used the lines with a distance of 17.31 Hz each.
6.033 ppm
10.35 Hz
17.30 Hz
For the second roof effect, the lines with a distance of 10.35 Hz each are now
of interest.
1 72 3 4
5
6 98
10
1211 13 14
150
1 72 3 4
5
6 98
10
1211 13 14
150
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
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
From the chemical shift of 6.033 ppm and the two coupling constants of
10.35 Hz and 17.30 Hz, a structural fragment can be derived
immediately from the components still unassigned.
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
6.033 ppm
10.35 Hz
17.30 Hz
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
6.033 ppm
10.35 Hz
17.30 Hz
C C
H
H H
C
5
H
8
O
2
molecular formula: found so far: - C
2
H
5
missing: 2DBE + C
3
H
3
O
2
From the chemical shift of 6.033 ppm and the two coupling constants of
10.35 Hz and 17.30 Hz, a structural fragment can be derived
immediately from the components still unassigned.
17.31 Hz 10.35 Hz
5.724 ppm6.314 ppm
1.66 Hz 1.66 Hz
C C
H
H H
10.35 Hz
17.30 Hz
6.033 ppm
The two coupling constants are perfect textbook values for protons in E-
(17.30 Hz) and Z-position (10.35 Hz) to each other, bound to sp
2
-
hybridised carbon atoms.
10.35 Hz
- C
2
H
5
missing: 2DBE + C
3
H
3
O
2
found so far:C
5
H
8
O
2
molecular formula: missing: 1DBÄ + CO
2
- C
2
H
5
+ CH
2
=CH-
The chemical shifts of the protons in geminal position to each other can
now be easily assigned via the coupling constants.
17.31 Hz 10.35 Hz
1.66 Hz 1.66 Hz
C C
H
H H
10.35 Hz
17.30 Hz
6.033 ppm
6.314 ppm
5.724 ppm
1.66 Hz
H
H
The value of 1.66 Hz is the geminal coupling constant between those
protons.
missing: 1DBÄ + CO
2
- C
2
H
5
+ CH
2
=CH-found so far:C
5
H
8
O
2
molecular formula:
C C
H
H H
10.35 Hz
17.30 Hz
6.033 ppm
6.314 ppm
5.724 ppm
1.66 Hz
H
H
The new structural fragment contains one double bond equivalent, two
carbon atoms and 3 protons.
For the complete structure, a partial structure with the molecular
formular CO
2
including two open bonds and one double bond equivalent
is then missing.
missing: 1DBÄ + CO
2
- C
2
H
5
+ CH
2
=CH-found so far:C
5
H
8
O
2
molecular formula:
C C
H
H H
10.35 Hz
17.30 Hz
6.033 ppm
6.314 ppm
5.724 ppm
1.66 Hz
H
H
H
2
C CH
3
7.14 Hz
4.13
ppm
1.22
ppm
C
O
O
O
O
C
There are only two structures which fulfil all
three conditions at the same time.
Let's just try one of them and estimate the
chemical shift of the methylene protons using
the Schoolery rules.
O
O
C
δ = 0.47 ppm + 1.55 ppm
= 2.12 ppm
The estimated value differs
significantly from the
measured value 4.13 ppm.
missing: 1DBÄ + CO
2
- C
2
H
5
+ CH
2
=CH-found so far:C
5
H
8
O
2
molecular formula: - C
2
H
5
C C
H
H H
10.35 Hz
17.30 Hz
6.033 ppm
6.314 ppm
5.724 ppm
1.66 Hz
H
H
H
2
C CH
3
7.14 Hz
4.13
ppm
1.22
ppm
C
O
O
We still have one shot left.
C
O
O
δ = 0.47 ppm + 3.13 ppm
= 3.60 ppm
Now the estimated value
differs significantly less from
the measured value 4.13 ppm.
✓
C C
H
H H
10.35 Hz
17.30 Hz
6.033 ppm
6.314 ppm
5.724 ppm
1.66 Hz
H
H
H
2
C CH
3
7.14 Hz
4.13
ppm
1.22
ppm
C
O
O
It's not too hard now to get to the final solution …
