Exercise plus Solution – Quick PDF overview
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1
H NMR spectrum
recorded at 250.13 MHz
C
3
H
7
NO
2
measured in CDCl
3
Deduce the structure!
Solution
C
3
H
7
NO
2
Basic considerations
Double bond equivalents,
number of signal groups,
chemical shifts
We don’t know about the oxidation number of nitrogen. Therefore, one
cannot determine the number of double bond equivalents beyond doubt.
We observe three signal groups at approx. 4.4 ppm, 2.1 ppm and 1.0
ppm. A more precise indication of the chemical shift is not necessary
here.
C
3
H
7
NO
2
Basic considerations
Integration
measured integrals [a.u.] - 1.00 1.00 1.52
sum of all integrals [a.u.] - 3.52
proportionality coeffcient - 1.99 H/a.u.
a.u. – arbitrary unit
H – number of protons in each signal group
2H 3H2H
Number of protons in each signal group
C
3
H
7
NO
2
Basic considerations
Structural elements
2H 3H2H
Number of protons in each signal group
J = 6.93 Hz
J = 7.09 Hz
J = 7.42 Hz
In each of the three multipletts, the coupling constant -
independent of the multiplicity - is about 7 Hz.
A coupling constant of 7 Hz is observed very often. This is a
coupling via three bonds (technical term: vicinal coupling) along
the chain H - C - C - H under the condition of a free rotation
around the C - C single bond.
C
C
H
H
C
3
H
7
NO
2
2H 3H2H
Number of protons in each signal group
J = 6.93 Hz
J = 7.42 Hz
C
HH
4.4 ppm
H
C
HH
1.0 ppm
H H
C
2.1 ppm
With the assumption we have just given, that all
coupling constants result from the H - C - C - H
pathway, we immediately get three structural
elements.
Basic considerations
Structural elements
C
3
H
7
NO
2
2H 3H2H
Number of protons in each signal group
J = 6.93 Hz
J = 7.42 Hz
C
HH
4.4 ppm
H
C
HH
1.0 ppm
H H
C
2.1 ppm
The signal of the methyl group appears as a triplet.
According to the n+1 rule, this results in 2
neighbouring protons. Neighbouring protons is a
frequently used abbreviation for chemically
equivalent protons bound to the neighbouring
carbon atoms.
There would be two
possibilities.
Basic considerations
Structural elements
C
3
H
7
NO
2
2H 3H2H
Number of protons in each signal group
J = 6.93 Hz
J = 7.42 Hz
C
HH
4.4 ppm
H
C
HH
1.0 ppm
H H
C
2.1 ppm
If the methylene group with the signal at 4.4 ppm
were attached to the methyl group, the multiplet of
this signal group would have to consist of at least
four lines because of the n+1 rule. This methylene
group is ruled out as a neighbour of the methyl
group because we only see 3 lines.
Basic considerations
Structural elements
C
3
H
7
NO
2
2H 3H2H
Number of protons in each signal group
J = 6.93 Hz
J = 7.42 Hz
C
HH
4.4 ppm
H
C
HH
1.0 ppm
H H
C
2.1 ppm
7.42 Hz
H
C
H
H H
H
C
2.1 ppm
1.0 ppm
At 2.1 ppm we observe six lines. In addition to the
methyl group, something else is obviously adjacent.
We have already measured the coupling constant.
Basic considerations
Structural elements
C
3
H
7
NO
2
2H 3H2H
Number of protons in each signal group
J = 6.93 Hz
C
HH
4.4 ppm
We can explain the triplet at 4.4 ppm, if we attach the
methylene group to the open bond of the ethyl group.
The required neighbours for the n+1 rule would be the
two methylene protons at 2.1 ppm.
7.42 Hz
H
C
H
H H
H
C
2.1 ppm
1.0 ppm
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
Basic considerations
Structural elements
C
3
H
7
NO
2
2H 3H2H
Number of protons in each signal group
J = 6.93 Hz
We also already know the coupling constant between
the two methylene groups.
6.93 Hz
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
Basic considerations
Structural elements
C
3
H
7
NO
2
Final structure
Two choices
6.93 Hz
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
499.47
506.71
513.80
521.13
528.19
535.60
Hz
If we compare the fragment found with the
molecular formula, NO
2
remains. This could be
a nitro group or a nitrite. Possibly, using an
increment scheme for the methylene group at
4.4 ppm, a decision might be possible, but it
will not be made here. Nitropropane is much
more likely.
NO
2
Multiplet structure
A pseudo sextet
6.93 Hz
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
499.47
506.71
513.80
521.13
528.19
535.60
Hz
What is the correct explanation of the signal at 2.1
ppm?
NO
2
2.1 ppm
In a first step the neighbouring 3 equivalent
protons of the methyl group cause a quartet
splitting.
499.47
506.71
513.80
521.13
528.19
535.60
Hz
The integral ratio of the four quartet lines is
1 : 3 : 3 : 1.
6.93 Hz
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
NO
2
13 31
2.1 ppm
A triplet is formed from each of the four lines
of the quartet by the two equivalent protons at
4.4 ppm. Let's start with the left line of the
quartet.
Multiplet structure
A pseudo sextet
499.47
506.71
513.80
521.13
528.19
535.60
Hz
The integral ratio of the three triplet lines is
1 : 2 : 1.
6.93 Hz
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
NO
2
13 31
2.1 ppm
1 2 1
The second line of the quartet also forms a triplet in
the integral ratio 1 : 2 : 1.
Multiplet structure
A pseudo sextet
499.47
506.71
513.80
521.13
528.19
535.60
Hz
But since we startet from a triple-intensity line of
the quartet we also have to triple the ratio from
1 : 2 : 1 to 3 : 6 : 3.
6.93 Hz
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
NO
2
13 31
2.1 ppm
1 2 1
3 6 3
We can proceed analogously with the two other
lines of the quartet.
Multiplet structure
A pseudo sextet
499.47
506.71
513.80
521.13
528.19
535.60
Hz
6.93 Hz
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
NO
2
13 31
2.1 ppm
1 2 1
3 6 3
3 6 3
1 2 1
Because of the very similar coupling constants,
some of the 12 lines almost coincide. Because of
the natural linewidth of the NMR signals, these
superimposed lines cannot be observed separately.
But we can sum the integrals of the overlapping
lines and get an integral ratio of
1 : 5 : 10 : 10 : 5 : 1.
1 510 1510
Multiplet structure
A pseudo sextet
499.47
506.71
513.80
521.13
528.19
535.60
Hz
6.93 Hz
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
NO
2
2.1 ppm
Only the two outermost lines of this pseudo sextet
are not superimposed. The frequency difference of
these two lines should be 3 * 7.42 + 2 * 6.93 Hz =
36.12 Hz.
The measured frequency difference is 36.13 Hz
(535.60 Hz - 499.47 Hz). The distances between
the inner lines are linear combinations of 6.93 Hz
and 7.42 Hz because of the superposition.
1 510 1510
6.93 6.937.42 7.42 7.42 Hz
Multiplet structure
A pseudo sextet
Final remark
The symmetry is tricky
6.93 Hz
H
C
H
H H
H
C
C
HH
7.42 Hz
4.4 ppm
2.1 ppm
1.0 ppm
NO
2
1088.10
1095.03
1101.98
Hz
One addition remark for – really very advanced – readers:
The multiplet at 4.4 ppm looks like a perfect triplet.
In fact, the symmetry of the molecule is more subtle than is
apparent at first (and even the second, third, …) sight. The
explanation, which is by no means simple, is not to be given here.
An example of how much the multiplet in unsymmeric achiral
compounds with -CH
2
-CH
2
- substructure can deviate from the
expected triplet structure is 1-Br-2-Cl-ethane (next page).
Cl
H H
C
C
HH
Br
300.08 MHz
Source: Maik Icker, University Leipzig
Final remark
The symmetry is tricky
