13. Structure Determination:
Nuclear Magnetic Resonance
Spectroscopy
Why This Chapter?
Mass Spec: Molecular size & formula
IR:
Functional Groups
NMR:
Map of Carbons with Hydrogens
 NMR is the most valuable




Used to determine relative location of atoms
Maps carbon-hydrogen framework of molecules
Depends on very strong magnetic fields
More advanced NMR techniques are used in
biological chemistry to study protein structure and
folding
2
13.1 Nuclear Magnetic
Resonance Spectroscopy
Nuclei w/odd # of protons or odd # of
neutrons (~ 1H, 13C, 14N, 2H) spin so act like
tiny magnets randomly oriented.
Parallel: ~ lower E
external magnet
Antiparallel: ~ higher E
Internal magnetic fields align parallel to
or against an aligned external magnet
Parallel orientation is lower in energy making this spin state more populated
3
Nuclear Magnetic Resonance
 Radio energy of exactly correct frequency (resonance) causes
the parallel nuclei to flip to anti-parallel state
Energy needed to flip a spinning parallel nucleus is related to its molecular
environment (proportional to field strength, B)
4
Nuclear Magnetic Resonance
 Radio energy of exactly correct frequency (resonance) causes
the parallel nuclei to flip to anti-parallel state
Energy needed to flip a spinning parallel nucleus is related to its molecular
environment (proportional to field strength, B)
If nucleus is protected (shielded) from the magnet it takes less E to flip.
If nucleus is exposed to the magnet then takes more E to flip.
5
Example:
 I takes 8.0x10-5 kJ/mol to spin-flip a proton at 200 MHz.
Calculate the Energy required to spin-flip a proton (1H) in a
spectrometer operating at 300 MHz.
8.0x10-5 kJ/mol
200 MHz
=
X kJ/mol
300 MHz
X = 1.2x10-4 kJ/mol
6
13.2 The Nature of NMR
•The sample is dissolved in a solvent that does not have a signal itself
Magnet (Applied Field)
Radiofrequency energy is transmitted
and absorption is detected
7
The Nature of NMR Absorptions
NMR of methyl acetate has 2 equivalent kinds of H’s so shows 2 peaks
 Electrons in neighboring bonds shield or expose nuclei from magnetic
field

1H
H’s on C next to electron withdrawing C=O
H’s on C next to electron withdrawing O
 Intensity of 1H NMR peak is proportional to # of equivalent H’s
8
NMR of methyl acetate has 3 kinds of C’s so shows 3 peaks
 Electrons in neighboring bonds shield nuclei from magnetic field

13C
C next door to electron withdrawing C=O
C next to electron withdrawing O
C of C=O
 Intensity of 13C NMR peak is not related to # of equivalent C’s
9
 At room temperature cyclohexane conformations are
interconverting so rapidly that axial and equatorial 1H’s
appear identical.
 When cold cyclohexane conformations intercovert so
slowly that axial and equatorial 1H’s appear different.
10
Example:
 How many signals would you expect each to
have in its 1H and 13C spectra?
1H
CH3
13C
CH3
C C
CH3
CH3
CH3
CH3
O
11
Solution:
 How many signals would you expect each to
have in its 1H and 13C spectra?
1H
CH3
CH3
1
CH3
C C
CH3
CH3
2 CH3
C C
CH3
CH3
CH3
O
13C
3
C C
CH3
CH3
H H
H
H
CH3
CH3
O
H
H
H H
CH3
CH3
CH3
5
C
O
C
C
C
CH3
C CH3
C
12
13.3 Chemical Shifts
Shift = relative energy of resonance
Downfield = deshielded
(more exposed to
magnet)
Upfield = shielded
(more protected
from magnet)
tetramethylsilane [TMS]
Reference point
CH3
H3C Si
CH3
CH3
13
13.3 Chemical Shifts
Other signals measured in ppm relative to TMS
tetramethylsilane [TMS]
Reference point
CH3
H3C Si
CH3
CH3
14
Measuring Chemical Shift
 Numeric value of chemical shift: difference between
strength of magnetic field at which the observed
nucleus resonates and field strength for resonance of
a reference
 Difference is very small but can be accurately
measured
 Taken as a ratio to the total field and multiplied by
106 so the shift is in parts per million (ppm)
 Absorptions normally occur downfield of TMS, to the
left on the chart
 Calibrated on relative scale in delta () scale
 Independent of instrument’s field strength
15
13.4 Signal Averaging & FT-NMR
Carbon-13: (only carbon
isotope with a nuclear spin)
Natural abundance =1.1%
of C’s so sample is very
dilute in this isotope




Sample measured using
repeated accumulation of data
and averaging of signals,
incorporating pulse and the
operation of Fourier transform
(FT-NMR)
All signals are obtained
simultaneously using a broad
pulse of energy and
resonance recorded
Frequent repeated pulses give
many sets of data that are
averaged to eliminate noise
Fourier-transform of averaged
pulsed data gives spectrum
Single run
Average of 200 runs
16
13.5 13C NMR Spectroscopy
 Each signal shows different types of environments of carbon
 13C resonances are 0 to 220 ppm downfield from TMS
 C’s shift downfield (deshield) if next to electron-withdrawing
 Like O, N, X (halogens)
sp2 C ~
sp3 C signal ~
 110 to 220
 0 to 90
C(=O) at low field,
 160 to 220
17
13C
NMR Example: 2-butanone
18
13C
NMR Example: p-bromoacetopheone
19
Learning Check:
Assign resonances in the given 13C spectrum of methyl propanoate
1
CH3
O
2
3
4
O C CH2 CH3
20
Solution:
Assign resonances in the given 13C spectrum of methyl propanoate
1
CH3
O
2
3
4
O C CH2 CH3
21
13.6 DEPT 13C NMR
 DEPT (distortionless enhancement by polarization transfer)
 Normal spectrum shows all C’s then:



Obtain spectrum of all C’s except quaternary
(broad band decoupled)
Change pulses to obtain separate information for
CH2, CH
Subtraction reveals each type
22
8
DEPT
13C
NMR
7
CH3
Normal spectrum shows all C’s:
(Difficult to Assign so many C’s)
5
CH3
C
5
6
C
H
HH
C
4
H
3
OH
C
C
H H
2
CH3
1
2
6
DEPT-90: shows only CH’s
Quaternary C’s don’t show
(Can now narrow our assignments)
23
8
DEPT
13C
NMR
7
CH3
Normal spectrum shows all C’s:
(Difficult to Assign so many C’s)
5
CH3
C
5
6
C
H
HH
C
4
H
3
C
C
H H
2
OH
2
CH3
1
7,8
1
6
DEPT-135:
Positive =shows CH’s and CH3’s
Negative =shows CH2’s
(Can narrow assignments even further)
24
8
DEPT
13C
NMR
7
CH3
Normal spectrum shows all C’s:
(Difficult to Assign so many C’s)
5
CH3
C
5
6
C
H
HH
C
4
H
3
C
C
H H
2
OH
2
CH3
1
4
7,8
2
1
6
DEPT-135:
Positive =shows CH’s and CH3’s
Negative =shows CH2’s
(Can narrow assignments even further)
25
13.7 Uses of13C NMR: Example
 Evidence for product of elimination of 1-chloro-methyl cyclohexane
Cl
CH3
CH2
CH3
KOH
ethanol
or
Expect 5 different C’s;
Expect 7 different C’s;
3 sp3 resonances  20-50
2 sp2 resonances  100-150
5 sp3 resonances  20-50
2 sp2 resonances  100-150
CH3
26
13.8 1H NMR & Proton Equivalence
 Proton NMR is much more sensitive than
13C
and the
active nucleus (1H) is nearly 100 % of the natural
abundance
 Shows how many kinds of nonequivalent hydrogens
are in a compound
 Theoretical equivalence can be predicted by seeing if
replacing each H with “X” gives the same or different
outcome
 Equivalent H’s have the same signal while
nonequivalent are different

There are degrees of nonequivalence
27
Nonequivalent H’s
 If replacement of each H with “X” gives a different constitutional
isomer then the H’s are in constitutionally heterotopic
environments and will have different chemical shifts
 – they are nonequivalent under all circumstances
28
Equivalent H’s
 Two H’s that are in identical environments (homotopic) have the
same NMR signal
 Test by replacing each with X if they give the identical result, they
are equivalent (homotopic)
29
Enantiotopic Distinctions
 If H’s are in environments that are mirror images of each other, they
are enantiotopic
 Replacement of each H with X produces a set of enantiomers
 The H’s have the same NMR signal (in the absence of chiral
materials)
30
Diastereotopic Distinctions
 In a chiral molecule, paired hydrogens can have different
environments and different shifts
 Replacement of a pro-R hydrogen with X gives a different
diastereomer than replacement of the pro-S hydrogen
 Diastereotopic hydrogens are distinct chemically and
spectrocopically
*
31
Learning Check:
 Identify sets of H’s as
Unrelated (U), homotopic (H), enantiotopic (E), or diasterotopic (D)
32
Solution:
 Identify sets of H’s as
Unrelated (U), homotopic (H), enantiotopic (E), or diasterotopic (D)
E
D
D
D
D
H
33
13.9 Chemical Shifts in 1H NMR
 Proton signals range from  0 to  10
 Electronegative atoms cause downfield shift
H’s on sp3 C
H’s on sp2 C
Lower field
Higher field
34
Shifts in 1H NMR
35
13.10 Integration of 1H NMR
Absorptions: Proton Counting
 The relative intensity of a signal (integrated area) is proportional
to the number of protons causing the signal
 For example in ethanol (CH3CH2OH), the signals have the
integrated ratio 3:2:1
 For narrow peaks, the heights are the same as the areas and
can be measured with a ruler
3
1
36
13.11 Spin-Spin Splitting in 1H NMR
 Peaks are often split into multiple peaks due to
interactions between nonequivalent protons on
adjacent carbons, called spin-spin splitting
 The splitting is into one more peak than the number
of H’s on the adjacent carbon (“n+1 rule”)
 The relative intensities are in proportion of a binomial
distribution and are due to interactions between
nuclear spins that can have two possible alignments
with respect to the magnetic field
 The set of peaks is a multiplet
(2 = doublet, 3 = triplet, 4 = quartet)
37
Simple Spin-Spin Splitting
 In bromoethane see 2 kinds of H’s
 One at 3.42 and one at 1.68
 Each signal split by neighbors
H H
H C C Br
H H
38
Simple Spin-Spin Splitting
An adjacent
CH2 gives a
ratio of 1:2:1
H H
An adjacent
CH3 group can
have four
different spin
alignments as
1:3:3:1
H C C Br
H H
J (coupling constant) = The separation of
peaks in a multiplet is a constant, in Hz
39
Rules for Spin-Spin Splitting
Equivalent
protons do not
split each other
The signal of a
proton with n
equivalent
neighboring H’s
is split into n + 1
peaks
Protons farther than 2 C’s
apart do not split each other
40
n+1
41
Spin-Spin Splitting Example:
Integration shows ~6:1 ratio
 Shift ~1.7 (shielded)
 6 H’s see 1 neighbor
(1+1=2 doublet)
 Shift ~4.3 (deshielded)
 1 H see’s 6 neighbors
(6+1=7 septuplet)
6
1
42
Spin-Spin Splitting Example:
Integration shows ratio
Singlet at 3.8
(deshielded)
3 H’s see 0
neighbors
Shift ~7.8 & 6.8
(deshielded)
2 H’s see 1 neighbor
2x (1+1=2 doublet)
•
a
2
b
b
b
Triplet ~1.2
(shielded)
3 H’s see 2
neighbors
(typical
CH3-CH2-c=o)
(typical CH3-CH2)
3
(typical CH3-O)
(typical para pattern)
a
Quartet ~2.8
(~deshielded)
2 H’s see 3
neighbors
2
3
2
a
43
Learning Check:
From the 1H NMR of C4H10O propose a structure.
44
Solution:
From the 1H NMR of C4H10O propose a structure.
CH3
CH2
O CH2
CH3
45
13.12 More Complex Spin-Spin
Splitting Patterns: trans-cinnamaldehyde
 Spectra more complex if overlapping signals, multiple
nonequivalence
Doublet of
Doublets (dd)
~6.7
(~deshielded)
1 H see’s 2
different
neighbors
Doublet ~9.8
(deshielded)
1 H see 1
neighbor
•(typical aldehyde)
1
3
3
b
(typical
CH-CH-CH)
a c
1
a
1
2
1
b
2
c
a
b
46
47
13.12 More Complex Spin-Spin
Splitting Patterns: trans-cinnamaldehyde
 Spectra more complex if overlapping signals, multiple
nonequivalence
Doublet ~9.8
(deshielded)
1 H see 1
neighbor
Doublet at 7.5
(deshielded)
1 H see’s 1
neighbor
a=d
2 H’s see 1
neighbor
(typical CH-CH)
(large J value
typical trans alkene H)
•(typical aldehyde)
1
3
3
a c
b
b = dd
2 H’s see 2
different
neighbors
(typical CH-CH-CH)
1
1
(typical
CH-CH-CH)
a
1
2
Doublet of
Doublets (dd)
~6.7
(~deshielded)
1 H see’s 2
different
neighbors
b
2
c
a
b
48
13.13 Uses of 1H NMR Spectroscopy
 Determine the regiochemistry of hydroboration/oxidation of
methylenecyclohexane. Which structure gives this 1H NMR?
49
50
Which of the following nuclei does not show
magnetic behavior?
20%
1.
1H
2.
2H
3.
12C
4.
13C
5.
17O
1
20%
2
20%
20%
3
4
20%
5
Which of the following is true of 13C NMR
spectra?
1.
2.
3.
4.
5.
The number of carbon atoms
in a molecule can be
ascertained.
The number of hydrogen
atoms in a molecule can be
ascertained.
Certain functional groups can
be deduced from the
locations of the peaks.
Both the number of carbon
atoms and the number of
hydrogen atoms in a
molecule can be ascertained.
All of these.
20%
1
20%
2
20%
20%
3
4
20%
5
Chemically equivalent nuclei always show a
single absorption.
50%
50%
True
2. False
1.
1
2
How many signals will appear in the 13C NMR
spectrum of the following molecule?
20%
1.
2.
3.
4.
5.
20%
20%
20%
3
4
20%
1
2
3
4
5
1
2
5
How many signals will appear in the 13C NMR
spectrum of the following molecule?
OCH3
1.
2.
3.
4.
5.
20%
20%
20%
20%
3
4
20%
3
4
5
6
7
1
2
5
Which of the following molecules best fits the
following 13C NMR data?
13C
NMR data: 20, 22, 32, 44, 67 ppm
25%
1.
CH3
HO
25%
25%
2.
HO
CH3
H3C
H3C
3.
4.
HO
HO
H3C
25%
CH3
H3C
CH3
1
2
3
4
What is the relationship between Ha and Hb in
the following compound?
Cl
20%
20%
20%
20%
3
4
20%
Ha
Hb
1.
2.
3.
4.
5.
chemically unrelated
homotopic
enantiotopic
diastereotopic
none of these
1
2
5
What is the relationship between Ha and Hb in
the following compound?
20%
OH
20%
20%
20%
3
4
20%
Hc
Ha
1.
2.
3.
4.
5.
Hb
chemically unrelated
homotopic
enantiotopic
diastereotopic
none of these
1
2
5
Order the following protons from lowest to
highest chemical shift value.
20%
O
Hb
Hc
20%
20%
20%
3
4
20%
Hd
Ha
1.
2.
3.
4.
5.
Ha < Hc < Hb < Hd
Ha < Hc < Hd < Hb
Hc < Ha < Hd < Hb
Hc < Ha < Hb < Hd
Hc < Hd < Ha < Hb
1
2
5
1H
NMR will allow one to distinguish between
the following two molecules:
H Br
Br
H
50%
50%
True
2. False
1.
1
2
The painkiller Demerol has the structure shown below. How
many peaks would you expect to see in the 13C NMR spectrum of
this substance?
20%
CH3
N
20%
20%
20%
3
4
20%
O
O
1.
2.
3.
4.
5.
10
11
12
14
15
1
2
5
The spectrum shown could represent the
molecule in the illustration.
50%
True
2. False
50%
1.
1
2
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
3.
20%
20%
20%
3
4
20%
2.
4.
5.
1
2
5
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
20%
20%
20%
3
4
20%
2.
OCH3
OCH3
3.
O
4.
O
5.
O
1
2
5
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
O
2.
OH
3.
OH
4.
5.
HO
HO
HO
HO
20%
20%
20%
3
4
20%
O
OH
OH
1
2
5
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
20%
20%
3
4
20%
2.
OH
Cl
3.
20%
5.
4.
NH2
NH2
1
2
5
Which of the following molecules best fits the
following NMR spectrum?
20%
1.
20%
20%
20%
3
4
20%
2.
O
O
OCH3
3.
5.
4.
O
O
O
O
OH
OH
1
2
5