1
Chem 3500-NMR Spectroscopy-Lab 2 solutions
Problem 1
What is the nuclear spin number of the following nuclei? Explain.
29
14
Si
7
3
Li
19
9
16
8
F
O
6
3
Li
An isotope is defined by it mass number (A) and its atomic number (Z):
A
Z
X
A= number of protons + number of neutrons
Z= number of protons
Each unpaired particle possesses a spin quantum number 1/2. The nuclear spin number
I is the sum of the unpaired proton and neutron spin quantum number.
Nucleus
Nb protons
Nb of neutrons
I
29Si
14
15
1/2
7Li
3
4
1/2
19F
9
10
½
16O
8
8
0
6Li
3
3
1
2
Chem 3500-NMR Spectroscopy-Lab 2 solutions
Problem 2
In a magnetic field B0, the Larmor frequency of a 1H nucleus,
0(
1
H), is 600 MHz.
a) Calculate B0.
2
B0
( X)
(X)
0
600.10 6
2.675.10 8
14. 09 T
2
B0
B0
b) What is the Larmor frequency of a 13C nucleus,
13
C
13
C
13
C
0
0
0
0(
13
C), in the same magnet?
(13 C) B 0
2
6.7283 10 7 14.09
2
150.91MHz
c) What will be those frequencies in an 11.74T magnet?
13
C
13
C
0
0
1
0
H
150.91 11.74
14.09
125.73MHz
600 11.74
14.09
499.92MHz
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Chem 3500-NMR Spectroscopy-Lab 2 solutions
Problem 3
In a 1H spectrum recorded in an 11.74T magnet, two peaks are separated by 5 ppm.
a) What would be the chemical shift difference between the two peaks if the spectrum
was acquired at 18.79T?
The chemical shift in ppm is independent of the magnetic field. The difference between the
2 peaks will be the same: 5 ppm.
b) What is the frequency difference (in Hz) between the two peaks at 11.74T?
The chemical shift is given by the following equation:
ref
10 6
0
So the difference of chemical shift between 2 peaks a and b is:
a
b
10 6
0
So:
a
b
0
10
6
We can calculate the Larmor frequency of the 1H at 11.74T,
frequency difference:
0
B0
2
a
b
a
b
499.98MHz
0
2499 Hz
10
6
5 499.98 10 6 10
6
0
and then calculate the
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Chem 3500-NMR Spectroscopy-Lab 2 solutions
c) What magnetic field strength (in Tesla) would produce a frequency difference of 2000Hz
between the two peaks?
At 11.74T, the frequency difference is 2499Hz. To get a 2000Hz difference, the magnetic
field should be:
B0
2000
11.74
2499
9.4T
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Chem 3500-NMR Spectroscopy-Lab 2 solutions
Problem 4
1
HC
2
HC
HC
3
6
CH
5
C
OH 7
CH
4
a. Match the chemical shifts with the correct proton atoms
H7=3.7
H6=H4=6.75
H1=H3=7.10
H2=6.83
b. Justify your answers to ‘a’. Draw resonance structures, state which nuclei are shielded
or deshielded relative to one another and to benzene and why.
In the 1H spectrum, the signals around 7 ppm are due to the aromatic
protons, as in benzene.
The hydroxyl signal is at 3.7 ppm, and is deshielded relative to aliphatic
protons due to the attached (electronegative) oxygen atom.
Resonance structures can be drawn (below) where the lone pair of
electrons on the oxygen is donated to the C-O bond, thus localizing a
negative charge at either the ortho (4 or 6) or para (2) positions. Because
the charge density, and thus the shielding, at the ortho and para positions
will be greater than at the meta position, the 1H at the meta positions will
be expected to be larger: (meta) > (para/ortho). Thus, (H1/3)=7.10.
Finally, one would expect that the charge density at the ortho position
would be larger than at the para position because of its proximity to the
oxygen and the C=O double bond, and thus would be more shielded. So,
the 1H at the ortho position ( (H4/6)=6.75, is smaller than at the para
position ( (H2)=6.83.
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Chem 3500-NMR Spectroscopy-Lab 2 solutions
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Chem 3500-NMR Spectroscopy-Lab 2 solutions
Problem 5
Write the structures corresponding to the following spectra:
C2H4Cl:
C3H5O2Cl:
H
O
H
Cl
H3C
Cl
H
C
O
H
C
H2
C9H12:
C10H14:
CH3
HC
CH
C
H
H3C
H3C
CH3
C
H
CH3
CH3
H3C
C4H10O2:
H
H3C
H
C
H
O
O
H
H
C6H8O:
CH3
H3C
HC
CH
O
CH3
Cl
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Chem 3500-NMR Spectroscopy-Lab 2 solutions
Problem 6
Knowing that X and Y have a spin=0, draw the 1H spectrum of the following compounds:
a) C2H2XY: the 2 protons are not magnetically equivalent. They each see X and Y differently. X
and Y are not NMR active, therefore they cannot be coupled with anything. The protons are
in a AX system. This means that they will have different chemical shifts and they will also be
coupled by 3JHH. Each of their NMR signal will be a doublet.
b,c,d) C2H2XX: in each case, the protons are magnetically equivalent, since they interact with the
X nuclei the same way. Therefore they are not coupled. The spectra will display a single peak.
The only difference between b, c and d will be the chemical shift.
e)
13
CCH2X2: Because of the presence of one
13
13
C, the protons are not equivalent anymore and
they are coupled with each other and with C. The coupling constants are 3JHH, 1JCH and 2JCH. We
are in a AXM system.
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Chem 3500-NMR Spectroscopy-Lab 2 solutions
Problem 7
a)
Jac
Jab
Ha is coupled with 1 Hb and 3 equivalent Hc. This is a AX3M system. The coupling between Ha
and Hb will result in a doublet (1:1) where each of the multiplet peaks are separated by J ab. The
coupling between Ha and Hc will result in a quartet (1:3:3:1) with multiplet peaks separated by J ac.
The signal of Ha is a quartet of doublet. Jab is the distance between the doublet peaks and J ac the
distance between the quartet peaks. In that case Jab < Jac.
If Jab> Jac, we will see a doublet of quartet.
If Jab=Jac, then it will be a AX4 system and we will see a quintuplet.
b) Ha is coupled with spin-1/2 nuclei. The Ha peak is a triplet of doublet. There is therefore 2
coupling constants and 2 type of nuclei involved: 2X with JAX (triplet) and 1M with J AM
(doublet). JAX>JAM
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Chem 3500-NMR Spectroscopy-Lab 2 solutions
Problem 8
Draw the spectrum of a nuclei Ha coupled with 3 other spin-1/2 nuclei such that:
-Jab=10Hz
-Jac=8Hz
-Jad=7Hz
10Hz
8Hz
7Hz
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Chem 3500-NMR Spectroscopy-Lab 2 solutions
Problem 9
This is the 1H spectrum of C3H7NO2. Write the structure. Explain the multiplets and measure the
J couplings.
Index of hydrogen deficiency for C3H7NO2 is (3*2+2-7-1)/2=0. The carbons are all saturated.
There are 3 groups of protons on the spectrum A, B, C:
A. The peak at 1.1 ppm is a triplet of intensity 3. This indicates a CH3 seeing a CH2. The J coupling is
526.76-520.17 = 520.17-513.61 = 6.59 Hz
B. The peak at 4.4 ppm is a triplet of intensity 2. This indicates a CH2 seeing another CH2. The J
coupling is 2203.25-2195.46 = 2195.46-2187.68 = 7.78Hz. B is also the most deshielded of the
peaks. It may be close from NO2 group.
C. The peak at 2 ppm is a multiplet with 8 peaks of intensity 2. This indicates a CH2 coupled with
other 1Hs, but with overlapping couplings. The J couplings are the following:
1024.31-1017.75 = 6.56 Hz
1017.75-1016.68 = 1.07 Hz
1016.68-1009.97 = 6.71 Hz
1009-97-1003.25=6.72Hz
1003.25-996.69=6.56Hz
996.69-995.62=1.07Hz
995.62-988.90=6.72Hz.
We can then deduce that A is seeing C which is seeing B.
The structure will then be: CH3(A)-CH2(C)-CH2(B)-NO2