C-13 NMR (CMR)
1. NMR experiments may be done with any atom which has a nucleus with a magnetic spin.
2. Examples of various isotopes with nuclear magnetic spin states are listed below.
a) The chart shows the frequency and magnetic field strength for which the nuclei have their
resonance. Note that for a given isotope, resonant frequency is proportional to applied field.
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isotope à
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1H
|
2H
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13C
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19F
|
|||||
|
Ho,
gauss
|
10,000
|
14,100
|
42,276
|
51,480
|
10,000
|
10,000
|
42,276
|
10,000
|
42,276
|
|
Resonant
Frequency MHz |
42.6
|
60.0
|
180
|
220.0
|
6.5
|
10.7
|
45.3
|
40.0
|
169.2
|
3. A significant difference in proton NMR vs CMR is
that 1H is the most abundant isotope of hydrogen.
a) In CMR the lesser abundant isotope is observed. About 1% of all carbon atoms are C-13.
b) The lower abundance means a larger sample must be used in the instrument.
a) In CMR the lesser abundant isotope is observed. About 1% of all carbon atoms are C-13.
b) The lower abundance means a larger sample must be used in the instrument.
Proton NMR requires about .1 mg
of sample. CMR requires about 1-5 mg of sample.
4. The 1H NMR spectrum ranges typically from 0-12 ppm; the CMR spectrum ranges from 0-210 ppm.
a) Even slight differences in a carbon's chemical environment results in a significant chemical
shift so that the number of carbon atoms (of different chemical environment) in the compound
can be determined simply by adding up the number of signals in the spectrum.
5. A great advantage of CMR is that the NMR experiment can be run in two different modes.
a) The "off-resonance decoupling" experiment results in the splitting of each signal for each carbon
atom in the molecule by the hydrogen atoms attached to it.
i) This means that the number of hydrogen atoms attached to the carbon atom can be quickly
deduced simply by counting the peaks of the multiplet and subtracting 1.
4. The 1H NMR spectrum ranges typically from 0-12 ppm; the CMR spectrum ranges from 0-210 ppm.
a) Even slight differences in a carbon's chemical environment results in a significant chemical
shift so that the number of carbon atoms (of different chemical environment) in the compound
can be determined simply by adding up the number of signals in the spectrum.
5. A great advantage of CMR is that the NMR experiment can be run in two different modes.
a) The "off-resonance decoupling" experiment results in the splitting of each signal for each carbon
atom in the molecule by the hydrogen atoms attached to it.
i) This means that the number of hydrogen atoms attached to the carbon atom can be quickly
deduced simply by counting the peaks of the multiplet and subtracting 1.
 |
b) In the proton-decoupled mode each
carbon nucleus appears as a singlet.
i) The protons are kept from coupling with the C-13 nuclei by irradiating the sample with
radio waves which flip the protons: they do not spend enough time in either spin state to
couple with the C-13 nuclei.
i) The protons are kept from coupling with the C-13 nuclei by irradiating the sample with
radio waves which flip the protons: they do not spend enough time in either spin state to
couple with the C-13 nuclei.
 |
ii) Adjacent carbon atoms do
not split the signal in C-13 NMR. Why not?
Answer: Since there are so few C-13 atoms (~1%), the chances of having two C-13 atoms next to each other in a molecule are small.
Answer: Since there are so few C-13 atoms (~1%), the chances of having two C-13 atoms next to each other in a molecule are small.
6. In CMR as in NMR, electronegative atoms attached to
the carbon atom results in a downfield shift.
a) In addition, replacement of a hydrogen atom on a carbon with an alkyl group results in a
downfield shift.
a) In addition, replacement of a hydrogen atom on a carbon with an alkyl group results in a
downfield shift.
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Chemical Shift, ppm
|
10.2 ppm
|
21.8
|
30.9
|
69.0
|
|
Answer
|
C-4
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C-1
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C-3
|
C-2
|
7. In the CMR spectrum of 2-butanol above, assign which
carbons are responsible for each observed
peak.
8. Coupling constants could be used in proton NMR to distinguish cis and trans isomers of alkenes.
a) The chemical shift of the allylic carbons in CMR are lower for the cis isomer than the trans
isomer by about 5 ppm, so this is another way of deciding upon which isomer is being studied.
b) Notice the chemical shifts for the isomers cis- and trans-2-butene.
cis: 11.4 ppm trans: 16.8 ppm
peak.
8. Coupling constants could be used in proton NMR to distinguish cis and trans isomers of alkenes.
a) The chemical shift of the allylic carbons in CMR are lower for the cis isomer than the trans
isomer by about 5 ppm, so this is another way of deciding upon which isomer is being studied.
b) Notice the chemical shifts for the isomers cis- and trans-2-butene.
cis: 11.4 ppm trans: 16.8 ppm
9. Unlike proton NMR, CMR integrals do not yield ratios
of nonequivalent carbons since the peak
intensities in CMR are subject to distortion.
10. Below are shown some CMR chemical shift correlations (ppm).
intensities in CMR are subject to distortion.
10. Below are shown some CMR chemical shift correlations (ppm).
 |
H. Spectral Interpretation
1. It is not necessary to memorize the various chemical shifts for the different types of chemical
environments.
a) Tables can be consulted.
b) It is helpful to remember some generalities such as, thinking of the proton NMR spectrum as a football field:
i) Alkanes resonate around the 10-20 yard line on the right side.
ii) Hydrogens bonded to carbons in turn bonded to highly electronegative groups (OH, Cl, Br) resonate near mid-field.
iii) Inductive shielding effects are cumulative, so the greater the number of Cl's, O's, etc
bonded to a carbon, the more the C-H resonance will move downfield (past mid-field to
the other end zone.)
c) In CMR remember that double-bonded carbons resonate far downfield, C-O and C-X
(X = halogens) moderately far downfield, and C=O very far downfield.
2. From CMR one can find the number of nonequivalent carbons and in the off-resonance mode the
number of hydrogens on each carbon.
3. From proton NMR probably the most useful aspect is the spin-spin splitting patterns.
4. The integrations in proton NMR yield proton ratios in each chemical environment and should be
evaluated.
5. The chemical shift of ~7 is usually indicative of the presence of the benzene ring.
a) A signal between 9.5 and 10 usually indicates the presence of an aldehyde
b) A signal between 10 and 12 usually indicates the presence of a carboxylic acid.
I. Sample Problems
1. 3-hexanol is treated with boiling sulfuric acid forming a mixture of isomeric alkenes:
1. It is not necessary to memorize the various chemical shifts for the different types of chemical
environments.
a) Tables can be consulted.
b) It is helpful to remember some generalities such as, thinking of the proton NMR spectrum as a football field:
i) Alkanes resonate around the 10-20 yard line on the right side.
ii) Hydrogens bonded to carbons in turn bonded to highly electronegative groups (OH, Cl, Br) resonate near mid-field.
iii) Inductive shielding effects are cumulative, so the greater the number of Cl's, O's, etc
bonded to a carbon, the more the C-H resonance will move downfield (past mid-field to
the other end zone.)
c) In CMR remember that double-bonded carbons resonate far downfield, C-O and C-X
(X = halogens) moderately far downfield, and C=O very far downfield.
2. From CMR one can find the number of nonequivalent carbons and in the off-resonance mode the
number of hydrogens on each carbon.
3. From proton NMR probably the most useful aspect is the spin-spin splitting patterns.
4. The integrations in proton NMR yield proton ratios in each chemical environment and should be
evaluated.
5. The chemical shift of ~7 is usually indicative of the presence of the benzene ring.
a) A signal between 9.5 and 10 usually indicates the presence of an aldehyde
b) A signal between 10 and 12 usually indicates the presence of a carboxylic acid.
I. Sample Problems
1. 3-hexanol is treated with boiling sulfuric acid forming a mixture of isomeric alkenes:
 |
|
|
The
mixture was then separated by gas chromatography and the CMR spectra taken
of each fraction.
The following CMR information resulted (in ppm): |
|
|
Isomer
|
d,
ppm
|
|
A
|
12.3,
13.5, 23.0, 29.3, 123.7, 130.6
|
|
B
|
13.4,
17.5, 23.1, 35.1, 124.7, 131.5
|
|
C
|
14.3,
20.6, 131.0
|
|
D
|
13.9,
25.8, 131.2
|
Identify each isomer.
Answer. Isomers
A and B have six signals and therefore must have six nonequivalent carbons (III
and IV). Isomer B has two chemical shifts about 5 ppm greater than Isomer A
(17.5 vs 13.5, 35.1 vs 29.3), which indicates isomer B must be compound III (the
trans isomer) and A must be compound IV.
By
similar analysis, Isomer C must be compound I (the cis isomer) and Isomer D must
be compound II.
2. A 1H NMR experiment is done on a
compound of formula C4H7Cl3. Use the following
proton NMR data to determine the structure of the compound. (s = singlet, d =
doublet; t = triplet; m = multiplet).
d
(ppm): 0.9 (t); 1.7 (m); 4.3 (m);
5.8 (d)
Integration ratio: 3 : 2 : 1 : 1
Integration ratio: 3 : 2 : 1 : 1
Answer.
The triplet at .9 (upfield) with an integration of 3 (3
H's) is a methyl group next to a methylene group. CH3CH2-
The doublet at 5.8 is so downfield that it is most likely -CHCl2. Being a doublet there must be only one hydrogen on the adjacent carbon.
The multiplet at 4.3 with an integration of 1 could account for this adjacent carbon with 1 hydrogen atom; the fact that it is downfield can be explained by the presence of a Cl atom and being next to a carbon with two Cl atoms.
The doublet at 5.8 is so downfield that it is most likely -CHCl2. Being a doublet there must be only one hydrogen on the adjacent carbon.
The multiplet at 4.3 with an integration of 1 could account for this adjacent carbon with 1 hydrogen atom; the fact that it is downfield can be explained by the presence of a Cl atom and being next to a carbon with two Cl atoms.
-CHCl-
The structure of the compound is likely to be
1,1,3-trichlorobutane
3. A compound of formula C3H7NO2
(an -NO2 group is present – thank IR for that) has the
following proton NMR spectrum. Determine the structure of this compound.
following proton NMR spectrum. Determine the structure of this compound.
 |
Answer: CH3CH2CH2NO2
Ha: 1.0 ppm (t); Hb: 2.0 ppm (m); Hc: 4.4 ppm (t)
4. A compound has the formula C4H7O2Br.
The signal at 10.97 ppm was moved onto the chart since the
chart paper only runs from 0-10 ppm. Determine the structure based on the proton NMR spectrum.
chart paper only runs from 0-10 ppm. Determine the structure based on the proton NMR spectrum.
 |
Answer. CH3CH2CHBrCO2H
Ha: 1.1 ppm
(t); Hb: 2.0 ppm (m); Hc: 4.2 ppm (t); Hd: 10.97 ppm (s)
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