Created by
Professor William Tam & Dr. Phillis Chang
Ch. 9 - 1
Chapter 9
Nuclear Magnetic
Resonance
Ch. 9 - 10
 Electromagnetic spectrum
cosmic
& -rays
X-rays ultraviolet visible infrared micro-
wave
radio-
wave
1Å = 10-10
m
1nm = 10-9
m
1m = 10-6
m
 0.1nm 200nm 400nm 800nm 50m
X-Ray
Crystallography
UV IR NMR
Ch. 9 - 11
2. Nuclear Magnetic Resonance
(NMR) Spectroscopy
 A graph that shows the characteristic
energy absorption frequencies and
intensities for a sample in a magnetic
field is called a nuclear magnetic
resonance (NMR) spectrum
Ch. 9 - 12
Ch. 9 - 13
1. The number of signals in the
spectrum tells us how many different
sets of protons there are in the
molecule
2. The position of the signals in the
spectrum along the x-axis tells us
about the magnetic environment of
each set of protons arising largely
from the electron density in their
environment
Ch. 9 - 14
3. The area under the signal tells us
about how many protons there are in
the set being measured
4. The multiplicity (or splitting pattern)
of each signal tells us about the
number of protons on atoms adjacent
to the one whose signal is being
measured
Ch. 9 - 15
 Typical 1
H NMR spectrum
● Chemical Shift ()
● Integration (areas of peaks  no.
of H)
● Multiplicity (spin-spin splitting) and
coupling constant
Ch. 9 - 16
 Typical 1
H NMR spectrum
Record as: 1
H NMR (300 MHz, CDCl3):
4.35 (2H, t, J = 7.2 Hz, Hc
)
2.05 (2H, sextet, J = 7.2 Hz, Hb
)
1.02 (3H, t, J = 7.2 Hz, Ha
)

chemical
shift ()
in ppm
no. of H
(integration) multiplicity
coupling
constant
in Hz
Ch. 9 - 17
2A. Chemical Shift
 The position of a signal along the x-axis of
an NMR spectrum is called its chemical
shift
 The chemical shift of each signal gives
information about the structural
environment of the nuclei producing that
signal
 Counting the number of signals in a 1
H NMR
spectrum indicates, at a first approximation,
the number of distinct proton environments
in a molecule
Ch. 9 - 18
Ch. 9 - 19
Ch. 9 - 20
 Normal range of 1
H NMR
15 -10
 ppm
"upfield" (more shielded)
"downfield" (deshielded)
(high field
strength)
(low field
strength)
Ch. 9 - 21
 Reference compound
● TMS = tetramethylsilane
as a reference standard (0 ppm)
● Reasons for the choice of TMS as
reference
 Resonance position at higher field
than other organic compounds
 Unreactive and stable, not toxic
 Volatile and easily removed
(B.P. = 28o
C)
Me
Si MeMe
Me
Ch. 9 - 22
 NMR solvent
● Normal NMR solvents should not
contain hydrogen
● Common solvents
 CDCl3
 C6D6
 CD3OD
 CD3COCD3 (d6-acetone)
Ch. 9 - 23
 The 300-MHz 1
H NMR spectrum of
1,4-dimethylbenzene
Ch. 9 - 24
2B. Integration of Signal Areas
Integral Step Heights
 The area under each signal in a 1
H
NMR spectrum is proportional to the
number of hydrogen atoms producing
that signal
 It is signal area (integration), not
signal height, that gives information
about the number of hydrogen atoms
Ch. 9 - 25
H
a
H
b
2 H
a
3 H
b
O
Ha
Ha
Hb
Hb
HbR
Ch. 9 - 26
2C. Coupling (Signal Splitting)
 Coupling is caused by the magnetic
effect of nonequivalent hydrogen
atoms that are within 2 or 3 bonds of
the hydrogens producing the signal
 The n+1 rule
● Rule of Multiplicity:
If a proton (or a set of magnetically
equivalent nuclei) has n neighbors
of magnetically equivalent protons.
It’s multiplicity is n + 1
Ch. 9 - 27
 Examples
Hb C C Cl
Ha
Hb
Hb
Ha
Ha
: multiplicity = 3 + 1 = 4 (a quartet)
Hb
: multiplicity = 2 + 1 = 3 (a triplet)
(1)
Cl C C Cl
Hb
Ha
Cl Hb
Ha
: multiplicity = 2 + 1 = 3 (a triplet)
Hb
: multiplicity = 1 + 1 = 2 (a doublet)
(2)
Ch. 9 - 28
Ch. 9 - 29
 Examples
Note: All Hb
’s are chemically and
magnetically equivalent.
Hb C C Br
Ha
Hb
Hb
Ha
: multiplicity = 6 + 1 = 7 (a septet)
Hb
: multiplicity = 1 + 1 = 2 (a doublet)
(3)
Hb
Hb
Hb
Ch. 9 - 30
 Pascal’s Triangle
● Use to predict relative intensity of
various peaks in multiplet
● Given by the coefficient of
binomial expansion (a + b)n
singlet (s) 1
doublet (d) 1 1
triplet (t) 1 2 1
quartet (q) 1 3 3 1
quintet 1 4 6 4 1
sextet 1 5 10 10 5 1
Ch. 9 - 31
 Pascal’s Triangle
● For
● For
Due to
symmetry, Ha
and Hb
are
identical
 a singlet
Ha
≠ Hb
 two doublets
Br C C Br
Hb
Ha
Cl Cl
Cl C C Br
Hb
Ha
Cl Br
Ch. 9 - 32
3. How to Interpret Proton NMR
Spectra
1. Count the number of signals to
determine how many distinct proton
environments are in the molecule
(neglecting, for the time being, the
possibility of overlapping signals)
2. Use chemical shift tables or charts to
correlate chemical shifts with possible
structural environments
Ch. 9 - 33
3. Determine the relative area of each signal,
as compared with the area of other
signals, as an indication of the relative
number of protons producing the signal
4. Interpret the splitting pattern for each
signal to determine how many hydrogen
atoms are present on carbon atoms
adjacent to those producing the signal and
sketch possible molecular fragments
5. Join the fragments to make a molecule in a
fashion that is consistent with the data
Ch. 9 - 34
 Example: 1
H NMR (300 MHz) of an
unknown compound with molecular
formula C3H7Br
Ch. 9 - 35
 Three distinct signals at ~ d3.4, 1.8
and 1.1 ppm
 d3.4 ppm: likely to be near an
electronegative group (Br)
Ch. 9 - 36
d (ppm): 3.4 1.8 1.1
Integral: 2 2 3
Ch. 9 - 37
d (ppm): 3.4 1.8 1.1
Multiplicity: triplet sextet triplet
2 H's on
adjacent C
5 H's on
adjacent C
2 H's on
adjacent C
Ch. 9 - 38
Complete structure:
• 2 H's from
integration
• triplet
• 2 H's from
integration
• sextet
• 3 H's from
integration
• triplet
most upfield signal
most downfield
signal
Br
CH2
CH2
CH3
Ch. 9 - 39
4. Nuclear Spin:
The Origin of the Signal
The magnetic
field associated
with a spinning
proton
The spinning
proton
resembles a tiny
bar magnet
Ch. 9 - 40
Ch. 9 - 41
Ch. 9 - 42
 Spin quantum number (I)
1
H: I = ½ (two spin states: +½ or
-½)
 (similar for 13
C, 19
F, 31
P)
12
C, 16
O, 32
S: I = 0
 These nuclei do not give an
NMR spectrum
Ch. 9 - 43
5. Detecting the Signal: Fourier
Transform NMR Spectrometers
Ch. 9 - 44
Ch. 9 - 45
 All protons do not absorb energy at the
same frequency in a given external
magnetic field
 Lower chemical shift values correspond with
lower frequency
 Higher chemical shift values correspond
with higher frequency
6. Shielding & Deshielding of Protons
15 -10
 ppm
"upfield" (more shielded)
"downfield" (deshielded)
(high field
strength)
(low field
strength)
Ch. 9 - 46
Ch. 9 - 47
 Deshielding by electronegative groups
CH3X
X = F OH Cl Br I H
Electro-
negativity
4.0 3.5 3.1 2.8 2.5 2.1
d (ppm) 4.26 3.40 3.05 2.68 2.16 0.23
● Greater electronegativity
 Deshielding of the proton
 Larger d
Ch. 9 - 48
 Shielding and deshielding by circulation
of p electrons
● If we were to consider only the
relative electronegativities of carbon
in its three hybridization states, we
might expect the following order of
protons attached to each type of
carbon:
(higher
frequency)
sp < sp2
< sp3
(lower
frequency)
Ch. 9 - 49
● In fact, protons of terminal alkynes
absorb between d 2.0 and d 3.0,
and the order is
(higher
frequency)
sp2
< sp < sp3
(lower
frequency)
Ch. 9 - 50
● This upfield shift (lower frequency)
of the absorption of protons of
terminal alkynes is a result of
shielding produced by the
circulating p electrons of the triple
bond
Shielded
(d 2 – 3 ppm)
H
Ch. 9 - 51
● Aromatic system
Shielded region
Deshielded region
Ch. 9 - 52
● e.g.
d (ppm)
H
a
& H
b
: 7.9 & 7.4 (deshielded)
H
c
& H
d
: 0.91 – 1.2 (shielded)
Hd Hc
Hb
Ha
Ch. 9 - 53
● Alkenes
Deshielded
(d 4.5 – 7 ppm)
H
Ch. 9 - 54
● Aldehydes
Electronegativity effect + Anisotropy effect
 d = 8.5 – 10 ppm (deshielded)
O
R
H
Ch. 9 - 55
 Reference compound
● TMS = tetramethylsilane
as a reference standard (0 ppm)
● Reasons for the choice of TMS as
reference
 Resonance position at higher field
than other organic compounds
 Unreactive and stable, not toxic
 Volatile and easily removed
(B.P. = 28o
C)
7. The Chemical Shift
Me
Si MeMe
Me
Ch. 9 - 56
7A. PPM and the d Scale
 The chemical shift of a proton, when
expressed in hertz (Hz), is
proportional to the strength of the
external magnetic field
 Since spectrometers with different
magnetic field strengths are commonly
used, it is desirable to express chemical
shifts in a form that is independent of
the strength of the external field
Ch. 9 - 57
 Since chemical shifts are always very small
(typically 5000 Hz) compared with the total
field strength (commonly the equivalent of
60, 300, or 600 million hertz), it is
convenient to express these fractions in
units of parts per million (ppm)
 This is the origin of the delta scale for the
expression of chemical shifts relative to TMS
 =
(observed shift from TMS in hertz) x 106
(operating frequency of the instrument in hertz)
Ch. 9 - 58
 For example, the chemical shift for benzene
protons is 2181 Hz when the instrument is
operating at 300 MHz. Therefore
 The chemical shift of benzene protons in a
60 MHz instrument is 436 Hz:
 Thus, the chemical shift expressed in ppm is
the same whether measured with an
instrument operating at 300 or 60 MHz (or
any other field strength)
 =
2181 Hz x 106
300 x 106
Hz
= 7.27 ppm
 =
436 Hz x 106
60 x 106
Hz
= 7.27 ppm
Ch. 9 - 59
 Two or more protons that are in
identical environments have the same
chemical shift and, therefore, give only
one 1
H NMR signal
 Chemically equivalent protons are
chemical shift equivalent in 1
H NMR
spectra
8. Chemical Shift Equivalent and
Nonequivalent Protons
Ch. 9 - 60
Homotopic and Heterotopic Atoms
 If replacing the hydrogens by a
different atom gives the same
compound, the hydrogens are said to
be homotopic
 Homotopic hydrogens have identical
environments and will have the same
chemical shift. They are said to be
chemical shift equivalent
Ch. 9 - 61
 The six hydrogens of ethane are homotopic
and are, therefore, chemical shift equivalent
 Ethane, consequently, gives only one
signal in its 1
H NMR spectrum
samecompounds
same
compounds
H
C CH
H
H
H
H
Ethane
H
C CH
H
H
H
Br
H
C CH
H
Br
H
H
H
C CH
H
H
Br
HH
CC H
H
H
Br
H
H
CC H
H
Br
H
H
H
CC H
H
H
H
Br
Ch. 9 - 62
 If replacing hydrogens by a different
atom gives different compounds,
the hydrogens are said to be
heterotopic
 Heterotopic atoms have different
chemical shifts and are not
chemical shift equivalent
Ch. 9 - 63
These 2 H’s
are also
homotopic
to each
other
different compounds
 heterotopic
same
compounds
 these 3
H’s of the
CH3 group
are
homotopic
 the CH3
group gives
only one 1H
NMR signal
H
C CH
H
H
H
Br
H
C CH
H
H
Cl
BrBr
CC H
H
H
Cl
H
Br
CC H
H
H
Cl
H
Br
CC H
H
Cl
H
H
Br
CC H
H
H
H
Cl
Ch. 9 - 64
 CH3CH2Br
● two sets of hydrogens that are
heterotopic with respect to each
other
● two 1
H NMR signals
H
C CH
H
H
H
Br
Ch. 9 - 65
 Other examples
 2 1
H NMR signals
 4 1
H NMR signals
(1) C C
H
H
CH3
CH3
(2) H
CH3H
H
H CH3
Ch. 9 - 66
 Other examples
 3 1
H NMR signals
(3)
H3C
CH3
H H
H
H
H H
Ch. 9 - 67
 Application to 13
C NMR spectroscopy
● Examples
 1 13
C NMR signal
 4 13
C NMR signals
(1) H3C CH3
(2)
CH3
CH3
Ch. 9 - 68
 5 13
C NMR signals
 4 13
C NMR signals
(3)
OHHO
(4)
OH
HO
Ch. 9 - 69
B. Enantiotopic and Diastereotopic
Hydrogen Atoms
 If replacement of each of two
hydrogen atoms by the same group
yields compounds that are
enantiomers, the two hydrogen atoms
are said to be enantiotopic
Ch. 9 - 70
 Enantiotopic hydrogen atoms have the
same chemical shift and give only one
1
H NMR signal:
enantiomer
enantiotopic
H3C Br
H H
H3C Br
H G
H3C Br
G H
Ch. 9 - 71
diastereomers
diastereotopic
chirality
centre
CH3
H OH
H3C
Ha
Hb
CH3
H OH
H3C
GHb
CH3
H OH
H3C
Ha
G
Ch. 9 - 72
diastereomers
diastereotopic
Hb
Br
Ha
H
Hb
Br
G
H
G
Br
Ha
H
Ch. 9 - 73
 Vicinal coupling is coupling between
hydrogen atoms on adjacent carbons
(vicinal hydrogens), where separation
between the hydrogens is by three s
bonds
9. Signal Splitting:
Spin–Spin Coupling
Ha
Hb
3
J or vicinal coupling
Ch. 9 - 74
9A. Vicinal Coupling
 Vicinal coupling between heterotopic
protons generally follows the n + 1
rule. Exceptions to the n + 1 rule can
occur when diastereotopic hydrogens
or conformationally restricted systems
are involved
 Signal splitting is not observed for
protons that are homotopic
(chemical shift equivalent) or
enantiotopic
Ch. 9 - 75
B. Splitting Tree Diagrams and the
Origin of Signal Splitting
 Splitting analysis for a doublet
C C
Ha
Hb
Ch. 9 - 76
 Splitting analysis for a triplet
C
Hb
C
Ha
Hb
C C C
Ha
Hb
Hb
Ch. 9 - 77
 Splitting analysis for a quartet
Hb C
Hb
C
Ha
Hb
Ch. 9 - 78
 Pascal’s Triangle
● Use to predict relative intensity of
various peaks in multiplet
● Given by the coefficient of
binomial expansion (a + b)n
singlet (s) 1
doublet (d) 1 1
triplet (t) 1 2 1
quartet (q) 1 3 3 1
quintet 1 4 6 4 1
sextet 1 5 10 10 5 1
Ch. 9 - 79
Coupling Constants – Recognizing
Splitting Patterns
X C
Ha
C
Hb
Hb
Hb
Ha
Ch. 9 - 80
9D. The Dependence of Coupling
Constants on Dihedral Angle
 3
J values are related to the dihedral
angle ()
H
H

Ch. 9 - 81
 Karplus curve
 f ~0o
or 180o
 Maximum
3
J value
 f ~90o
 3
J ~0 Hz
Ch. 9 - 82
 Karplus curve
● Examples
Hb
Ha
Hb
Ha
 = 180º
Ja,b = 10-14 Hz
(axial, axial)
Hb
Ha
Hb
Ha
 = 60º
Ja,b = 4-5 Hz
(equatorial, equatorial)
Ch. 9 - 83
 Karplus curve
● Examples
Hb
Ha
Hb
Ha
 = 60º
Ja,b = 4-5 Hz
(equatorial, axial)
Ch. 9 - 84
9E. Complicating Features
 The 60 MHz 1
H NMR spectrum of ethyl
chloroacetate
Ch. 9 - 85
 The 300 MHz 1
H NMR spectrum of
ethyl chloroacetate
Ch. 9 - 86
. Analysis of Complex Interactions
Ch. 9 - 87
 The 300 MHz 1
H NMR spectrum of 1-
nitropropane
Ch. 9 - 88
 Protons of alcohols (ROH) and amines may
appear over a wide range from 0.5 – 5.0 ppm
● Hydrogen-bonding is the reason for this
range
10. Proton NMR Spectra and Rate
Processes
in high dilution (free OH):
 = ~0.5-1.0 ppm
in conc. solution (H-bonded):
H O
R
H
O
R
H O
R


proton more deshielded
R O H
Ch. 9 - 89
 Why don’t we see coupling with the
O–H proton, e.g. –CH2–OH (triplet?)
● Because the acidic protons are
exchangeable about 105
protons
per second (residence time 10-5
sec), but the NMR experiment
requires a time of 10-2
– 10-3
sec.
to “take” a spectrum, usually we
just see an average (thus, OH
protons are usually a broad
singlet)
Ch. 9 - 90
Trick:
● Run NMR in d6
-DMSO where Hbonding
with DMSO’s oxygen
prevents H’s from exchanging and
we may be able to see the coupling
Ch. 9 - 91
 Deuterium Exchange
● To determine which signal in the
NMR spectrum is the OH proton,
shake the NMR sample with a drop
of D2
O and whichever peak
disappears that is the OH peak
(note: a new peak of HOD appears)
D2O
+ HODR O H R O D
Ch. 9 - 92
 Phenols
● Phenol protons appear downfield at
4-7 ppm
● They are more “acidic” - more H+
character
● More dilute solutions - peak
appears upfield: towards 4 ppm
OH O H
Ch. 9 - 93
 Phenols
● Intramolecular H-bonding causes
downfield shift
12.1 ppm
O
H
O
Ch. 9 - 94
 Unlike 1
H with natural abundance
~99.98%, only 1.1% of carbon,
namely 13
C, is NMR active
11. Carbon-13 NMR Spectroscopy
11A. Interpretation of 13
C NMR
Spectra
Ch. 9 - 95
B. One Peak for Each Magnetically
Distinct Carbon Atom
 13
C NMR spectra have only become
commonplace more recently with the
introduction of the Fourier Transform
(FT) technique, where averaging of
many scans is possible (note 13
C
spectra are 6000 times weaker than 1
H
spectra, thus require a lot more scans
for a good spectrum)
Ch. 9 - 96
 Note for a 200 MHz NMR (field strength
4.70 Tesla)
● 1
H NMR  Frequency = 200 MHz
● 13
C NMR  Frequency = 50 MHz
Ch. 9 - 97
 Example:
● 2-Butanol
Proton-coupled
13
C NMR spectrum
CH3 C CH2 CH3
H
OH
Ch. 9 - 98
 Example:
● 2-Butanol
Proton-decoupled
13
C NMR spectrum
CH3 C CH2 CH3
H
OH
Ch. 9 - 99
11C. 13
C Chemical Shifts
 Decreased electron density around an
atom deshields the atom from the
magnetic field and causes its signal to
occur further downfield (higher ppm,
to the left) in the NMR spectrum
 Relatively higher electron density
around an atom shields the atom from
the magnetic field and causes the
signal to occur upfield (lower ppm, to
the right) in the NMR spectrum
Ch. 9 
Factors affecting chemical shift
i. Diamagnetic shielding due to bonding
electrons
ii. Paramagnetic shielding due to low-lying
electronic excited state
iii. Magnetic Anisotropy – through space
due to the near-by group (especially 
electrons)
In 1
H NMR, (i) and (iii) most significant;
in 13
C NMR, (ii) most significant (since
chemical shift range >> 1
H NMR)
Ch. 9 
Electronegative substituents cause
downfield shift
 Increase in relative atomic mass of
substituent causes upfield shift
X
Cl
Br
I
Electronegativity
2.8
2.7
2.2
Atomic Mass
35.5
79.9
126.9
13
C NMR: CH3X
23.9 ppm
9.0 ppm
-21.7 ppm
Ch. 9 
Hybridization of carbon
● sp2
> sp > sp3
123.3 ppm 71.9 ppm 5.7 ppm
H2C CH2 HC CH H3C CH3
e.g.
Ch. 9 
Anisotropy effect
● Shows shifts similar to the effect
in 1
H NMR
shows large
upfield shift
C C
e.g.
C
Ch. 9 -
Ch. 9 -
Ch. 9 -
(a)(b)
(c)
Cl CH2 CH CH3
OH
(a) (b) (c)
1-Chloro-2-propanol