Chapter 9
Nuclear Magnetic
Resonance and Mass
Spectrometry
Created by
Professor William Tam & Dr. Phillis Chang
Ch. 9 - 1
About The Authors
These PowerPoint Lecture Slides were created and prepared by Professor
William Tam and his wife Dr. Phillis Chang.
Professor William Tam received his B.Sc. at the University of Hong Kong in
1990 and his Ph.D. at the University of Toronto (Canada) in 1995. He was an
NSERC postdoctoral fellow at the Imperial College (UK) and at Harvard
University (USA). He joined the Department of Chemistry at the University of
Guelph (Ontario, Canada) in 1998 and is currently a Full Professor and
Associate Chair in the department. Professor Tam has received several awards
in research and teaching, and according to Essential Science Indicators, he is
currently ranked as the Top 1% most cited Chemists worldwide. He has
published four books and over 80 scientific papers in top international journals
such as J. Am. Chem. Soc., Angew. Chem., Org. Lett., and J. Org. Chem.
Dr. Phillis Chang received her B.Sc. at New York University (USA) in 1994, her
M.Sc. and Ph.D. in 1997 and 2001 at the University of Guelph (Canada). She
lives in Guelph with her husband, William, and their son, Matthew.
Ch. 9 - 2
1. Introduction

Classic methods for organic structure
determination
●
●
●
●
●
●
●
●
●
Boiling point
Refractive index
Solubility tests
Functional group tests
Derivative preparation
Sodium fusion (to identify N, Cl, Br, I & S)
Mixture melting point
Combustion analysis
Degradation
Ch. 9 - 3

Classic methods for organic structure
determination
● Require large quantities of
sample and are time consuming
Ch. 9 - 4

Spectroscopic methods for organic
structure determination
a) Mass Spectroscopy (MS)
● Molecular Mass & characteristic
fragmentation pattern
b) Infrared Spectroscopy (IR)
● Characteristic functional groups
c) Ultraviolet Spectroscopy (UV)
● Characteristic chromophore
d) Nuclear Magnetic Resonance (NMR)
Ch. 9 - 5

Spectroscopic methods for organic
structure determination
● Combination of these
spectroscopic techniques provides
a rapid, accurate and powerful tool
for Identification and Structure
Elucidation of organic compounds
● Rapid
● Effective in mg and microgram
quantities
Ch. 9 - 6

General steps for structure elucidation
1. Elemental analysis
● Empirical formula
● e.g. C2H4O
2. Mass spectroscopy
● Molecular weight
● Molecular formula
● e.g. C4H8O2, C6H12O3 … etc.
● Characteristic fragmentation
pattern for certain functional
groups
Ch. 9 - 7

General steps for structure elucidation
3. From molecular formula
● Double bond equivalent (DBE)
4. Infrared spectroscopy (IR)
● Identify some specific
functional groups
● e.g. C=O, C–O, O–H, COOH,
NH2 … etc.
Ch. 9 - 8

General steps for structure elucidation
5. UV
● Sometimes useful especially for
conjugated systems
● e.g. dienes, aromatics, enones
6.
1H, 13C
NMR and other advanced
NMR techniques
● Full structure determination
Ch. 9 - 9

Electromagnetic spectrum
cosmic
& -rays

X-rays
0.1nm
ultraviolet
200nm
X-Ray
Crystallography
visible
400nm
infrared
800nm
IR
UV
microwave
radiowave
50m
NMR
1Å = 10-10m
1nm = 10-9m
1m = 10-6m
Ch. 9 - 10
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 - 11
Ch. 9 - 12
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 - 13
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 - 14

Typical 1H NMR spectrum
● Chemical Shift ()
● Integration (areas of peaks  no.
of H)
● Multiplicity (spin-spin splitting) and
coupling constant
Ch. 9 - 15

Typical 1H NMR spectrum
Record as:
1
H NMR (300 MHz, CDCl3):
 4.35 (2H, t, J = 7.2 Hz, Hc)
b
2.05 (2H, sextet, J = 7.2 Hz, H )
1.02 (3H, t, J = 7.2 Hz, Ha)
chemical
shift ()
in ppm
coupling
constant
no. of H
(integration) multiplicity in Hz
Ch. 9 - 16
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 1H NMR
spectrum indicates, at a first approximation,
the number of distinct proton environments
in a molecule
Ch. 9 - 17
Ch. 9 - 18
Ch. 9 - 19

Normal range of 1H NMR
"upfield" (more shielded)
"downfield" (deshielded)
15
(low field
strength)
 ppm
-10
(high field
strength)
Ch. 9 - 20

Reference compound
● TMS = tetramethylsilane
Me
Me
Si Me
Me
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. = 28oC)
Ch. 9 - 21

NMR solvent
● Normal NMR solvents should not
contain hydrogen
● Common solvents

CDCl3

C6D6

CD3OD

CD3COCD3 (d6-acetone)
Ch. 9 - 22

The 300-MHz 1H NMR spectrum of
1,4-dimethylbenzene
Ch. 9 - 23
2B. Integration of Signal Areas
Integral Step Heights
The area under each signal in a 1H
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 - 24
Ha
R
Ha
O
Hb
Hb
Hb
a
2H
H
b
b
3H
H
a
Ch. 9 - 25
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 - 26

Examples
(1)
H
Hb
C
b
H
a
C
Ha: multiplicity = 3 + 1 = 4 (a quartet)
Cl
Hb H a
(2) Ha Hb
Cl
C
C
Cl Hb
Hb: multiplicity = 2 + 1 = 3 (a triplet)
Ha: multiplicity = 2 + 1 = 3 (a triplet)
Cl
Hb: multiplicity = 1 + 1 = 2 (a doublet)
Ch. 9 - 27
Ch. 9 - 28

Examples
(3)
H b Ha
Hb
C
H
b
Hb
C
Br
Ha: multiplicity = 6 + 1 = 7 (a septet)
Hb Hb: multiplicity = 1 + 1 = 2 (a doublet)
Hb
Note: All Hb’s are chemically and
magnetically equivalent.
Ch. 9 - 29

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)
doublet (d)
triplet (t)
quartet (q)
quintet
sextet
1
11
121
1331
14641
1 5 10 10 5 1
Ch. 9 - 30

Pascal’s Triangle
H
● For
Br
C
a
H
b
C
Br
Cl Cl
Ha H b
● For
Cl
C
C
Cl Br
Br
Due to
symmetry, Ha
and Hb are
identical
 a singlet
Ha ≠ Hb
 two doublets
Ch. 9 - 31
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 - 32
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 - 33

Example: 1H NMR (300 MHz) of an
unknown compound with molecular
formula C3H7Br
Ch. 9 - 34

Three distinct signals at ~ 3.4, 1.8
and 1.1 ppm
 3.4 ppm: likely to be near an
electronegative group (Br)
Ch. 9 - 35
 (ppm):
3.4
1.8
1.1
Integral:
2
2
3
Ch. 9 - 36
 (ppm):
3.4
Multiplicity: triplet
2 H's on
adjacent C
1.8
1.1
sextet
triplet
5 H's on
adjacent C
2 H's on
adjacent C
Ch. 9 - 37
Complete structure:
most downfield
CH2
signal
Br
• 2 H's from
integration
• triplet
most upfield signal
CH2
CH3
• 2 H's from
integration
• sextet
• 3 H's from
integration
• triplet
Ch. 9 - 38
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 - 39
Ch. 9 - 40
Ch. 9 - 41

1H:
Spin quantum number (I)
I = ½ (two spin states: +½ or -½)
 (similar for 13C, 19F, 31P)
12C, 16O, 32S:
I=0
 These nuclei do not give an NMR
spectrum
Ch. 9 - 42
5. Detecting the Signal: Fourier
Transform NMR Spectrometers
Ch. 9 - 43
Ch. 9 - 44
6. Shielding & Deshielding of Protons



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
"upfield" (more shielded)
"downfield" (deshielded)
15
(low field
strength)
 ppm
-10
(high field
strength)
Ch. 9 - 45
Ch. 9 - 46

Deshielding by electronegative groups
CH3X
X=
F
Electro4.0
negativity
OH
Cl
Br
I
H
3.5
3.1
2.8
2.5
2.1
 (ppm) 4.26 3.40 3.05 2.68 2.16 0.23
● Greater electronegativity
 Deshielding of the proton
 Larger 
Ch. 9 - 47

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
(lower
2
3
sp < sp < sp
frequency)
frequency)
Ch. 9 - 48
● In fact, protons of terminal alkynes
absorb between  2.0 and  3.0,
and the order is
(higher
(lower
2
3
sp < sp < sp
frequency)
frequency)
Ch. 9 - 49
● 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
H
Shielded
( 2 – 3 ppm)
Ch. 9 - 50
● Aromatic system
Shielded region
Deshielded region
Ch. 9 - 51
● e.g.
Hd
Hc
Hb
H
a
 (ppm)
a
b
c
d
H & H : 7.9 & 7.4 (deshielded)
H & H : 0.91 – 1.2 (shielded)
Ch. 9 - 52
● Alkenes
H
Deshielded
( 4.5 – 7 ppm)
Ch. 9 - 53
● Aldehydes
R
O
H
Electronegativity effect + Anisotropy effect
  = 8.5 – 10 ppm (deshielded)
Ch. 9 - 54
7. The Chemical Shift

Reference compound
● TMS = tetramethylsilane
Me
Me
Si Me
Me
●
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. = 28oC)
Ch. 9 - 55
7A. PPM and the  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 - 56

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 - 57

For example, the chemical shift for benzene
protons is 2181 Hz when the instrument is
operating at 300 MHz. Therefore
=

300 x 106 Hz
= 7.27 ppm
The chemical shift of benzene protons in a
60 MHz instrument is 436 Hz:
=

2181 Hz x 106
436 Hz x 106
60 x 106 Hz
= 7.27 ppm
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)
Ch. 9 - 58
8.
Chemical Shift Equivalent and
Nonequivalent Protons

Two or more protons that are in
identical environments have the same
chemical shift and, therefore, give only
one 1H NMR signal

Chemically equivalent protons are
chemical shift equivalent in 1H NMR
spectra
Ch. 9 - 59
8A. 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 - 60
Br H
same compounds
Br
H
C
C
H
H
H
H
C
C
H
H
H
H
C
C
Br H


H
H
H
H
H
H
C
C
H
H
H
H
Ethane
H
H
Br
C
C
H
H
H
H
C
C
H
H
H
H
C
C
H
Br
H
Br
H
same compounds
H
H
The six hydrogens of ethane are homotopic
and are, therefore, chemical shift equivalent
Ethane, consequently, gives only one
Ch. 9 - 61
signal in its 1H NMR spectrum

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 - 62
same
compounds H
 these 3
H’s of the
CH3 group
Cl
are
homotopic
 the CH3
group gives
only one 1H
NMR signal
H
Cl Br
C
C
H
H
H
Br
C
C
H
H
H
Br
C
C
Cl H
H
H
H
H
H
H
Br
C
C
Cl H
H
Br
C
C
H
H
H
H
These 2 H’s
are also
H
homotopic
to each
other
H
Br
C
C
H
Cl
different compounds
 heterotopic
H
Ch. 9 - 63
H

H
Br
C
C
H
H
H
CH3CH2Br
● two sets of hydrogens that are
heterotopic with respect to each
other
● two 1H NMR signals
Ch. 9 - 64

Other examples
H
(1)
CH3
C
C
H
CH3
H
(2)
 2 1H NMR signals
CH3
H
H
H
 4 1H NMR signals
CH3
Ch. 9 - 65

Other examples
H
(3)
H
H
CH3
H3C
H
H
H
 3 1H NMR signals
Ch. 9 - 66

(1)
Application to
● Examples
H3C
CH3
13C
NMR spectroscopy
1
13C
NMR signal
CH3
(2)
CH3
4
13C
NMR signals
Ch. 9 - 67
(3)
HO
(4)
5
13C
NMR signals
4
13C
NMR signals
OH
HO
OH
Ch. 9 - 68
8B. 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 - 69

Enantiotopic hydrogen atoms have the
same chemical shift and give only one
1H NMR signal:
H
enantiotopic
H
H3C
H3C
G
Br
H
enantiomer
Br
G
H3C
H
Br
Ch. 9 - 70
H
OH
H3C
Hb
Ha
OH
H3C
Hb
G
CH3
CH3
diastereotopic
H
OH
H3C
G
Ha
diastereomers
chirality
centre
H
CH3
Ch. 9 - 71
Br
Ha
Br
Hb
H
Hb
Ha
H
diastereotopic
Br
diastereomers
G
G
H
Ch. 9 - 72
9. Signal Splitting:
Spin–Spin Coupling

Vicinal coupling is coupling between
hydrogen atoms on adjacent carbons
(vicinal hydrogens), where separation
between the hydrogens is by three s
bonds
H
a
Hb
3
J or vicinal coupling
Ch. 9 - 73
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 - 74
9B. Splitting Tree Diagrams and the
Origin of Signal Splitting

Splitting analysis for a doublet
Hb H a
C
C
Ch. 9 - 75

Splitting analysis for a triplet
H b Ha
C
C
Hb
H b Ha H b
C
C
C
Ch. 9 - 76

Splitting analysis for a quartet
b
Hb
a
H
H
C
C
Hb
Ch. 9 - 77

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)
doublet (d)
triplet (t)
quartet (q)
quintet
sextet
1
11
121
1331
14641
1 5 10 10 5 1
Ch. 9 - 78
9C. Coupling Constants – Recognizing
Splitting Patterns
Ha Hb
X
C
C
Hb
Ha Hb
Ch. 9 - 79
9D. The Dependence of Coupling
Constants on Dihedral Angle
 3J
values are related to the dihedral
angle ()
H

H
Ch. 9 - 80

Karplus curve

 ~0o or 180o
 Maximum
3J value

 ~90o
 3J ~0 Hz
Ch. 9 - 81

Karplus curve
● Examples
b
H
b
H
Ha
Ha
(axial, axial)
(equatorial, equatorial)
b
H
a
H
b
a
H
 = 180º
Ja,b = 10-14 Hz
H
 = 60º
Ja,b = 4-5 Hz
Ch. 9 - 82

Karplus curve
● Examples
b
H
Ha
(equatorial, axial)
b
H
Ha
 = 60º
Ja,b = 4-5 Hz
Ch. 9 - 83
9E. Complicating Features

The 60 MHz 1H NMR spectrum of ethyl
chloroacetate
Ch. 9 - 84

The 300 MHz 1H NMR spectrum of
ethyl chloroacetate
Ch. 9 - 85
9F. Analysis of Complex Interactions
Ch. 9 - 86

The 300 MHz 1H NMR spectrum of 1nitropropane
Ch. 9 - 87
10. Proton NMR Spectra and Rate
Processes

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
in high dilution (free OH):
R
O
H
 = ~0.5-1.0 ppm
in conc. solution (H-bonded):
 R
R

H O
O
H
R

H O
proton more deshielded
Ch. 9 - 88

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 - 89
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 - 90

Deuterium Exchange
● To determine which signal in the
NMR spectrum is the OH proton,
shake the NMR sample with a drop
of D2O and whichever peak
disappears that is the OH peak
(note: a new peak of HOD appears)
R
O
H
D2O
R
O
D
+
HOD
Ch. 9 - 91

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 - 92

Phenols
● Intramolecular H-bonding causes
downfield shift
O
12.1 ppm
H
O
Ch. 9 - 93
11. Carbon-13 NMR Spectroscopy
11A. Interpretation of 13C NMR
Spectra

Unlike 1H with natural abundance
~99.98%, only 1.1% of carbon,
namely 13C, is NMR active
Ch. 9 - 94
11B. One Peak for Each Magnetically
Distinct Carbon Atom
 13C 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 13C
spectra are 6000 times weaker than 1H
spectra, thus require a lot more scans
for a good spectrum)
Ch. 9 - 95

Note for a 200 MHz NMR (field strength
4.70 Tesla)
● 1H NMR  Frequency = 200 MHz
●
13C
NMR  Frequency = 50 MHz
Ch. 9 - 96

Example:
● 2-Butanol
H
CH3
C
CH2
CH3
OH
Proton-coupled
13C NMR spectrum
Ch. 9 - 97

Example:
● 2-Butanol
H
CH3
C
CH2
CH3
OH
Proton-decoupled
13C NMR spectrum
Ch. 9 - 98
11C. 13C 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 - 99

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 p
electrons)
In 1H NMR, (i) and (iii) most significant;
in 13C NMR, (ii) most significant (since
chemical shift range >> 1H NMR)
Ch. 9 - 100

Electronegative substituents cause
downfield shift

Increase in relative atomic mass of
substituent causes upfield shift
X
Electronegativity Atomic Mass
13
C NMR: CH3X
Cl
2.8
35.5
23.9 ppm
Br
2.7
79.9
9.0 ppm
I
2.2
126.9
-21.7 ppm
Ch. 9 - 101

Hybridization of carbon
● sp2 > sp > sp3
e.g.
H2C
CH2
123.3 ppm
HC
CH
71.9 ppm
H3C
CH3
5.7 ppm
Ch. 9 - 102

Anisotropy effect
● Shows shifts similar to the effect
in 1H NMR
e.g.
C
C
C
shows large
upfield shift
Ch. 9 - 103
Ch. 9 - 104
Ch. 9 - 105
Cl
(a)
CH2
(b)
CH
(c)
CH3
OH
1-Chloro-2-propanol
(c)
(b) (a)
Ch. 9 - 106
11D. Off-Resonance Decoupled Spectra
NMR spectrometers can differentiate among carbon
atoms on the basis of the number of hydrogen
atoms that are attached to each carbon
 In an off-resonance decoupled 13C NMR spectrum,
each carbon signal is split into a multiplet of peaks,
depending on how many hydrogens are attached to
that carbon. An n + 1 rule applies, where n is the
number of hydrogens on the carbon in question.
Thus, a carbon with no hydrogens produces a
singlet (n = 0), a carbon with one hydrogen
produces a doublet (two peaks), a carbon with two
hydrogens produces a triplet (three peaks), and a
methyl group carbon produces a quartet (four
peaks)
Ch. 9 - 107

9
Off-resonance
decoupled 13C NMR
2
N
1
8
3
N
7
6
4
5
O
Broadband proton-decoupled
13C
NMR
Ch. 9 - 108
11E. DEPT 13C Spectra

DEPT 13C NMR spectra indicate how
many hydrogen atoms are bonded to
each carbon, while also providing the
chemical shift information contained in
a broadband proton-decoupled 13C
NMR spectrum. The carbon signals in a
DEPT spectrum are classified as CH3,
CH2, CH, or C accordingly
Ch. 9 - 109
Cl
(a)
CH2
(b)
CH
(c)
CH3
(b)
(a)
(c)
OH
1-Chloro-2-propanol
Ch. 9 - 110

The broadband proton-decoupled 13C
NMR spectrum of methyl methacrylate
Ch. 9 - 111
12. Two-Dimensional (2D) NMR
Techniques

HCOSY
● 1H–1H correlation spectroscopy

HETCOR
● Heteronuclear correlation
spectroscopy
Ch. 9 - 112

HCOSY of 2-chloro-butane
H2
H3
H1
H4
H4
H1
H3
H2
Ch. 9 - 113

HETCOR of 2-chloro-butane
C3
C1
C4
C2
H4
H1
H3
H2
Ch. 9 - 114
13. An Introduction to Mass
Spectrometry

Partial MS of octane (C8H18, M = 114)
14 (CH2)
29 (CH3CH2)
57
85
71
M+
114
Ch. 9 - 115

The M+ peak at 114 is referred to as
the parent peak or molecular ion
-
C8H18

e
70 eV
[C8H18] + 2 e(M+)
The largest or most abundant peak is
called the base peak and is assigned
an intensity of 100%, other peaks are
then fractions of that e.g. 114(M+,40),
85(80), 71(60), 57(100) etc.
Ch. 9 - 116

Masses are usually rounded off to
whole numbers assuming:
H = 1, C = 12, N = 14, O = 16, F = 19 etc.
[C8H18]
fragmentation
(M , 114) -CH3CH2 (29)
+
[C6H13]
Daughter
ions
(85)
-CH3CH2CH2 (29+14)
[C5H11]
Molecular ion (parent peak)
(71)
Ch. 9 - 117
14. Formation of Ions: Electron
Impact Ionization


In the mass spectrometer, a molecule in the
gaseous phase under low pressure is
bombarded with a beam of high-energy
electrons (70 eV or ~ 1600 kcal/mol)
This beam can dislodge an electron from a
molecule to give a radical cation which is
called the molecular ion, M+ or more
accurately
M
70 eV e-
M
Ch. 9 - 118

M
This molecular ion has considerable
surplus energy so it can fly apart or
fragment to give specific ions which
may be diagnostic for a particular
compound
- m1º
A
- m2º
B
- m3º
C
etc.
mº = neutral fragment radical
Ch. 9 - 119
15. Depicting the Molecular Ion
CH3CH2 CH3
Radical cations from ionization
of nonbonding on p electron
H3C
OH
H3C
N
CH3
H2C
CHCH 2CH3
CH3
Methanol
Trimethylamine
1-Butene
Ch. 9 - 120

Ionization potentials of selected
molecules
Compound
Ionization
Potential (eV)
CH3(CH2)3NH2
8.7
C6H6 (benzene)
9.2
C2H4
10.5
CH3OH
10.8
C2H6
11.5
CH4
12.7
Ch. 9 - 121
16. Fragmentation
1.
2.
3.
The reactions that take place in a mass
spectrometer are unimolecular, that is, they
do not involve collisions between molecules
or ions. This is true because the pressure is
kept so low (10-6 torr) that reactions
involving bimolecular collisions do not occur
We use single-barbed arrows to depict
mechanisms involving single electron
movements
The relative ion abundances, as indicated by
peak intensities, are very important
Ch. 9 - 122
16A. Fragmentation by Cleavage at a
Single Bond
 When a molecular ion fragments, it will
yield a neutral radical (not detected) and
a carbocation (detected) with an even
number of electrons
 The fragmentation will be dictated to
some extent by the fragmention of the
more stable carbocation:
+
ArCH2
>
+
CH2=CHCH2
>
3o
>
2o
>
1o
>
+
CH3
Ch. 9 - 123

e.g.
R+
R
CH
X
● Site of ionization:
R
+
+
CH3
+
CH3
n>p>s
non-bonding
Ch. 9 - 124
As the carbon skeleton becomes more
highly branched, the intensity of the
molecular ion peak decreases
 Butane vs. isobutane

70eV
e-
a
b+
M (58)
70eV
e
M+(58)
a
(43)
+ CH3
+ CH2CH3
b
(29)
(43)
+ CH3
Ch. 9 - 125
16B. Fragmentation of Longer Chain
and Branched Alkanes
 Octane vs. isooctane
(85) +
(71) +
(57) +
+
M (114)
(43) +
+
M+(114)
(57)
Ch. 9 - 126
16C. Fragmentation to Form
Resonance-Stabilized Cations
 Alkenes
● Important fragmentation of terminal
alkenes
 Allyl carbocation (m/e = 41)
R
R
+
(41)
Ch. 9 - 127
Carbon–carbon bonds next to an atom
with an unshared electron pair usually
break readily because the resulting
carbocation is resonance stabilized
 Ethers

● Cleavage a (to ether oxygen) C–C bonds
O
CH3
+
O
O
(m/e = 59)
Ch. 9 - 128

Alcohols
● Most common fragmentation: - loss
of alkyl groups
a
CH3 +
OH
OH
(m/e = 59)
a OH b
M+(74)
b
CH3CH2 +
OH
OH
(m/e = 45)
Ch. 9 - 129

Carbon–carbon bonds next to the
carbonyl group of an aldehyde or
ketone break readily because
resonance-stabilized ions called
acylium ions are produced
Ch. 9 - 130

Aldehydes
● M+ peak usually observed but may
be fairly weak
● Common fragmentation pattern
 a-cleavage
H
+ R C O
acylium ion
R
+ H C O
(m/e = 29)
O
R
H
Ch. 9 - 131

Ketones
● a-cleavage
O
+
a
b
O
(m/e = 71)
a
b
O
+
(m/e = 99)
Ch. 9 - 132

Alkyl-substituted benzenes ionize by
loss of a π electron and undergo loss
of a hydrogen atom or methyl group
to yield the relatively stable tropylium
ion (see Section 14.7C). This
fragmentation gives a prominent peak
(sometimes the base peak) at m/z 91
Ch. 9 - 133

Aromatic hydrocarbons
● very intense M+ peaks
● characteristic fragmentation pattern
(when an alkyl group attached to
the benzene ring): - tropylium
cation
CH3
CH3 +
CH2 rearrangement
benzyl cation
tropylium cation
(m/e = 91)
Ch. 9 - 134
16D. Fragmentation by Cleavage of
Two Bonds

Alcohols frequently
show
a
prominent
+
●
peak at M - 18. This corresponds to
the loss of a molecule of water
● May lose H2O by 1,2- or 1,4elimination
Ch. 9 - 135
1,2-elimination:
OH
+
(M - 18)
M
1,4-elimination:
M
H2O
OH
H
(M - 18)
OH + CH3CH2
+
H2O
Ch. 9 - 136

Cycloalkenes show a characteristic
fragmentation pattern which
corresponds to a reverse Diels-Alder
reaction
retro Diels-Alder

+
e.g.
+
Ch. 9 - 137

Aromatic hydrocarbons
● e.g.
H
McLafferty Rearrangement
CH2
H
H
(m/e = 92)
+
Ch. 9 - 138

Ketones
● McLafferty rearrangement
H
OH
O
OH
+
1st McL. Rearr.
H
O
OH
(m/e = 86)
2
nd
McL. Rearr.
OH
+
(m/e = 58)
Ch. 9 - 139
OH
i
H
ii
O
2º radical
H
OH
(m/e = 86)
observed
i
ii
OH
1º radical
OH
(m/e = 114)
NOT observed
Ch. 9 - 140

Characteristic of McLafferty
rearrangement
1. No alkyl migrations to C=O, only H
migrates
H
O
H
O
R
R
X
R
O
H
Ch. 9 - 141

Characteristic of McLafferty
rearrangement
2. 2o is preferred over 1o
H
H
O
ii
i
OH
2º radical
OH
not
1º radical
Ch. 9 - 142
17. How To Determine Molecular
Formulas and Molecular Weights
Using Mass Spectrometry
17A. Isotopic Peaks & the Molecular Ion
Ch. 9 - 143
The presence of isotopes of carbon,
hydrogen, and nitrogen+ in a compound
●
gives rise to a small M + 1 peak
 The presence of oxygen, sulfur,
chlorine, or bromine
in a compound
+
●
gives rise to an M + 2 peak

M
+ 1 Elements:
C, H, N
M
+ 2 Elements:
O, S, Br, Cl
Ch. 9 - 144

+
●
The M + 1 peak can be used to
determine the number of carbons in a
molecule
+
●

The M + 2 peak can indicate whether
bromine or chlorine is present

The isotopic peaks, in general, give us
one method for determining molecular
formulas
Ch. 9 - 145

Example
● Consider 100 molecules of CH4
H
H
1
1
12
C
H
H
1
H
1
H1
C
H1
1
13
H
1
H
1
C12
H2
H1
H1
M + 1 = 17
M : 16
C12: 100
C13: 1.11
H1: 100
H2: 0.016
Ch. 9 - 146
H1
H
1
C12
H1
H1
H1
H
1
H1
C13
H1
H1
C12
H2
H1
H1
M + 1 = 17
M : 16
1.11 molecules
contain a 13C atom
4x0.016 = 0.064 molecules
contain a 2H atom
+
●
Intensity of M + 1 peak:+
1.11+0.064=1.174% of the M● peak
Ch. 9 - 147
relative ion abundance
M
+
●
100
+
●
≈
m/z
M +1
1.17
Ch. 9 - 148
17B. How To Determine the Molecular
Formula
m/z
Intensity
+
(% of M● )
72
73.0/73 x 100 = 100
73
3.3/73 x 100 = 4.5
74
0.2/73 x 100 = 0.3
Ch. 9 - 149

+
●
Is M odd or even? According to
the nitrogen rule, if it is even, then
the compound must contain an
even number of nitrogen atoms
(zero is an even number)
+
●
● For our unknown, M is even. The
compound must have an even
number of nitrogen atoms
Ch. 9 - 150

The
relative
abundance
of
the
+
●
M +1 peak indicates the
number of carbon atoms.
Number of C atoms
=
relative
+
●
abundance of (M +1)/1.1
● For our unknown
Number of C atoms =
4.5
1.1
~4
Ch. 9 - 151
+
●

The relative abundance of the M +2
peak indicates the presence (or
absence) of S (4.4%), Cl (33%), or Br
(98%)
+
●
● For our unknown M +2 = 0.3%; thus,
we can assume that S, Cl, and Br are
absent

The molecular formula can now be
established by determining the
number of hydrogen atoms and adding
the appropriate number of oxygen
atoms, if necessary
Ch. 9 - 152
+
●
Since M is m/z 72
 molecular weight = 72
 As determined using the relative
+
●
abundance of M +1 peak, number of
carbons present is 4
 Using the “nitrogen rule”, this unknown
must have an even number of N.
Since M.W. = 72, and there are 4 C
present, (12 x 4 = 48), adding 2 “N”
will be greater than the M.W. of the
unknown. Thus, this unknown
Ch. 9 - 153
contains zero “N”


For a molecule composed of C and H
only
H = 72 – (4 x 12) = 24
but C4H24 is impossible

For a molecule composed of C, H and O
H = 72 – (4 x 12) – 16 = 8
and thus our unknown has the
molecular formula C4H8O
Ch. 9 - 154
17C. High-Resolution Mass Spectrometry
Ch. 9 - 155

Example 1
● O2, N2H4 and CH3OH all have M.W. of 32
(by MS), but accurate masses are
different
 O2 = 2(15.9949) = 31.9898

N2H4 = 2(14.0031) + 4(1.00783) =
32.0375

CH4O = 12.00000 + 4(1.00783) +
15.9949 = 32.0262
Ch. 9 - 156

Example 2
● Both C3H8O and C2H4O2 have M.W. of 60
(by MS), but accurate masses are
different

C3H8O = 60.05754

C2H4O2 = 60.02112
Ch. 9 - 157
18. Mass Spectrometer Instrument
Designs
Ch. 9 - 158
19. GC/MS Analysis
Ch. 9 - 159
 END OF CHAPTER 9 
Ch. 9 - 160