1 Chapter 9 Nuclear Magnetic Resonance and Mass Spectrometry About The Authors These PowerPoint Lecture Slides were created and prepared by Professor William Tam and his wife Dr Phillis Chang ID: 273602
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Slide1
Ch. 9 - 1
Chapter 9
Nuclear Magnetic
Resonance and Mass
SpectrometrySlide2
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 -
2Slide3
Ch. 9 - 3
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
DegradationSlide4
Ch. 9 - 4
Classic methods for organic structure determination
Require large quantities of sample and are time consumingSlide5
Ch. 9 - 5
Spectroscopic methods for organic structure determination
Mass Spectroscopy (MS)
Molecular Mass & characteristic fragmentation
patternInfrared Spectroscopy (IR)Characteristic functional groups
Ultraviolet Spectroscopy (UV)
Characteristic chromophore
Nuclear Magnetic Resonance (NMR)Slide6
Ch. 9 - 6
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 quantitiesSlide7
Ch. 9 - 7
General steps for structure elucidation
Elemental analysis
Empirical formula
e.g. C2H4OMass spectroscopy
Molecular weight
Molecular formula
e.g. C
4
H
8
O
2
, C
6
H
12
O
3
… etc.
Characteristic fragmentation pattern for certain functional groupsSlide8
Ch. 9 - 8
General steps for structure elucidation
From molecular formula
Double bond equivalent (DBE)
Infrared spectroscopy (IR)Identify some specific functional groups
e.g. C=O, C–O, O–H, COOH, NH
2
… etc.Slide9
Ch. 9 - 9
General steps for structure elucidation
UV
Sometimes useful especially for conjugated systems
e.g. dienes, aromatics, enones1H, 13
C NMR and other advanced NMR techniques
Full structure determination Slide10
Ch. 9 - 10
Electromagnetic spectrumSlide11
Ch. 9 - 11
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) spectrumSlide12
Ch. 9 - 12Slide13
Ch. 9 - 13
The number of signals in the spectrum tells us how many different sets of protons there are in the molecule
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 environmentSlide14
Ch. 9 - 14
The area under the signal tells us about how many protons there are in the set being measured
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 measuredSlide15
Ch. 9 - 15
Typical
1
H NMR spectrum
Chemical Shift ()Integration (areas of peaks
no. of H)
Multiplicity (spin-spin splitting) and coupling constantSlide16
Ch. 9 - 16
Typical
1
H NMR spectrumSlide17
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 moleculeSlide18
Ch. 9 - 18Slide19
Ch. 9 - 19Slide20
Ch. 9 - 20
Normal range of
1
H NMRSlide21
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. = 28
o
C)Slide22
Ch. 9 - 22
NMR solvent
Normal NMR solvents should not contain hydrogen
Common solvents
CDCl3
C
6
D
6
CD
3
OD
CD
3
COCD
3
(d
6
-acetone)Slide23
Ch. 9 - 23
The 300-MHz
1
H NMR spectrum of 1,4-dimethylbenzeneSlide24
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 atomsSlide25
Ch. 9 - 25
H
a
H
b
2 H
a
3 H
bSlide26
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
+ 1Slide27
Ch. 9 - 27
ExamplesSlide28
Ch. 9 - 28Slide29
Ch. 9 - 29
Examples
Note: All
H
b’s are chemically and magnetically equivalent.Slide30
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)
nsinglet (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 1Slide31
Ch. 9 - 31
Pascal’s Triangle
For
For
Due to symmetry,
H
a
and
H
b
are identical
a singlet
H
a
≠
H
b
two doubletsSlide32
Ch. 9 - 32
How to Interpret Proton NMR
Spectra
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)
Use chemical shift tables or charts to correlate chemical shifts with possible structural environmentsSlide33
Ch. 9 - 33
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
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
Join the fragments to make a molecule in a fashion that is consistent with the dataSlide34
Ch. 9 - 34
Example:
1
H NMR (300 MHz) of an unknown compound with molecular formula C
3H7BrSlide35
Ch. 9 - 35
Three distinct signals at
~
d
3.4, 1.8 and 1.1
ppm
d
3.4
ppm: likely to be near an
electronegative
group (Br)Slide36
Ch. 9 - 36
d
(ppm):
3.4
1.8 1.1
Integral:
2
2
3Slide37
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 CSlide38
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
signalSlide39
Ch. 9 - 39
Nuclear Spin:
The Origin of the Signal
The magnetic field associated with a spinning proton
The spinning proton resembles a tiny bar magnetSlide40
Ch. 9 - 40Slide41
Ch. 9 - 41Slide42
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 spectrumSlide43
Ch. 9 - 43
Detecting the Signal: Fourier
Transform NMR SpectrometersSlide44
Ch. 9 - 44Slide45
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
Shielding & Deshielding of ProtonsSlide46
Ch. 9 - 46Slide47
Ch. 9 - 47
Deshielding by electronegative groups
CH
3
X
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
dSlide48
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 < sp
2
< sp
3
(lower frequency)Slide49
Ch. 9 - 49
In fact, protons of terminal alkynes absorb between
d
2.0 and
d 3.0, and the order is(higher frequency)
sp
2
< sp < sp
3
(lower frequency)Slide50
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)Slide51
Ch. 9 - 51
Aromatic system
Shielded region
Deshielded regionSlide52
Ch. 9 - 52
e.g.
d
(ppm)
Ha
& H
b
: 7.9 & 7.4 (deshielded)
H
c
& H
d
: 0.91 – 1.2 (shielded)Slide53
Ch. 9 - 53
Alkenes
Deshielded
(
d
4.5 – 7 ppm)Slide54
Ch. 9 - 54
Aldehydes
Electronegativity effect + Anisotropy effect
d
= 8.5 – 10 ppm (deshielded)Slide55
Ch. 9 - 55
Reference compound
TMS = tetramethylsilane
as a reference standard (0 ppm)
Reasons for the choice of TMS as referenceResonance position at higher field than other organic compounds
Unreactive and stable, not toxic
Volatile and easily removed
(B.P. = 28
o
C)
The Chemical ShiftSlide56
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 fieldSlide57
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 TMSSlide58
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)Slide59
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 1H NMR spectra
Chemical Shift Equivalent and
Nonequivalent ProtonsSlide60
Ch. 9 - 60
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 equivalentSlide61
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
same compounds
same compoundsSlide62
Ch. 9 - 62
If replacing hydrogens by a different atom gives
different compounds
, the hydrogens are said to be
heterotopicHeterotopic atoms have different chemical shifts and are not chemical shift equivalentSlide63
Ch. 9 - 63
These 2 H’s are also
homotopic
to each other
different compounds
heterotopic
same compounds
these 3 H’s of the CH
3
group are
homotopic
the CH
3
group gives only one 1H NMR signalSlide64
Ch. 9 - 64
CH
3
CH
2
Br
two sets of hydrogens that are heterotopic with respect to each other
two
1
H NMR signalsSlide65
Ch. 9 - 65
Other examples
2
1
H NMR signals
4
1
H NMR signalsSlide66
Ch. 9 - 66
Other examples
3
1
H NMR signalsSlide67
Ch. 9 - 67
Application to
13
C NMR spectroscopy
Examples
1
13
C NMR signal
4
13
C NMR signalsSlide68
Ch. 9 - 68
5
13
C NMR signals
4
13
C NMR signalsSlide69
Ch. 9 - 69
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
enantiotopicSlide70
Ch. 9 - 70
Enantiotopic hydrogen atoms have the same chemical shift and give only one
1
H NMR signal:
enantiomer
enantiotopicSlide71
Ch. 9 - 71
diastereomers
diastereotopic
chirality
centreSlide72
Ch. 9 - 72
diastereomers
diastereotopicSlide73
Ch. 9 - 73
Vicinal coupling
is coupling between hydrogen atoms on adjacent carbons (vicinal hydrogens), where separation between the hydrogens is by three
s
bondsSignal Splitting:Spin–Spin CouplingSlide74
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 enantiotopicSlide75
Ch. 9 - 75
9B.
Splitting Tree Diagrams and the
Origin of Signal Splitting
Splitting analysis for a doubletSlide76
Ch. 9 - 76
Splitting analysis for a tripletSlide77
Ch. 9 - 77
Splitting analysis for a quartetSlide78
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)
nsinglet (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 1Slide79
Ch. 9 - 79
9C.
Coupling Constants
–
Recognizing
Splitting PatternsSlide80
Ch. 9 - 80
9D.
The Dependence of Coupling
Constants on Dihedral Angle
3
J values are related to the dihedral angle (
)Slide81
Ch. 9 - 81
Karplus curve
f
~0
o or 180o
Maximum
3
J value
f
~90
o
3
J ~0 HzSlide82
Ch. 9 - 82
Karplus curve
ExamplesSlide83
Ch. 9 - 83
Karplus curve
ExamplesSlide84
Ch. 9 - 84
9E.
Complicating Features
The 60 MHz
1
H NMR spectrum of ethyl chloroacetateSlide85
Ch. 9 - 85
The 300 MHz
1
H NMR spectrum of ethyl chloroacetateSlide86
Ch. 9 - 86
9F.
Analysis of Complex InteractionsSlide87
Ch. 9 - 87
The 300 MHz
1
H NMR spectrum of 1-nitropropaneSlide88
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
Proton NMR Spectra and Rate
ProcessesSlide89
Ch. 9 - 89
Why don’t we see coupling with the O–H proton, e.g. –CH
2
–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) Slide90
Ch. 9 - 90
Trick:
Run NMR in d
6
-DMSO where H-bonding with DMSO’s oxygen prevents H’s from exchanging and we may be able to see the coupling Slide91
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 D
2
O and whichever peak disappears that is the OH peak (note: a new peak of HOD appears)Slide92
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 Slide93
Ch. 9 - 93
Phenols
Intramolecular H-bonding causes downfield shift
12.1 ppmSlide94
Ch. 9 - 94
Unlike
1
H with natural abundance ~99.98%, only 1.1% of carbon, namely
13C, is NMR activeCarbon-13 NMR Spectroscopy
11A.
Interpretation of
13
C NMR
SpectraSlide95
Ch. 9 - 95
11B.
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)Slide96
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 MHzSlide97
Ch. 9 - 97
Example:
2-Butanol
Proton-coupled
13
C NMR spectrumSlide98
Ch. 9 - 98
Example:
2-Butanol
Proton-decoupled
13
C NMR spectrumSlide99
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 spectrumSlide100
Ch. 9 - 100
Factors affecting chemical shift
Diamagnetic shielding due to bonding electrons
Paramagnetic shielding due to low-lying electronic excited state 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) Slide101
Ch. 9 - 101
Electronegative substituents cause downfield shift
Increase in relative atomic mass of substituent
causes upfield shiftSlide102
Ch. 9 - 102
Hybridization of carbon
sp
2
> sp > sp
3
123.3 ppm
71.9 ppm
5.7 ppmSlide103
Ch. 9 - 103
Anisotropy effect
Shows shifts similar to the effect in
1
H NMR
shows large
upfield shiftSlide104
Ch. 9 - 104Slide105
Ch. 9 - 105Slide106
Ch. 9 - 106
(a)
(b)
(c)Slide107
Ch. 9 - 107
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
13
C 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)Slide108
Ch. 9 - 108
Off-resonance decoupled
13
C NMR
Broadband proton-decoupled
13
C NMRSlide109
Ch. 9 - 109
11E.
DEPT
13
C Spectra
DEPT
13
C 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
13
C NMR spectrum. The carbon signals in a DEPT spectrum are classified as CH
3
, CH
2
, CH, or C accordinglySlide110
Ch. 9 - 110
(a)
(b)
(c)Slide111
Ch. 9 - 111
The broadband proton-decoupled
13
C NMR spectrum of methyl methacrylateSlide112
Ch. 9 - 112
HCOSY
1
H–
1H correlation spectroscopyHETCOR
Heteronuclear correlation spectroscopy
Two-Dimensional (2D) NMR
TechniquesSlide113
Ch. 9 - 113
HCOSY of 2-chloro-butane
H
2
H
1
H
1
H
3
H
3
H
4
H
4
H
2Slide114
Ch. 9 - 114
HETCOR
of 2-chloro-butane
H
1
H
2
H
3
H
4
C
1
C
2
C
3
C
4Slide115
Ch. 9 - 115
Partial MS of octane (C
8
H
18, M = 114)An Introduction to Mass
SpectrometrySlide116
Ch. 9 - 116
The M
+
peak at 114 is referred to as the
parent peak or molecular ion
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. Slide117
Ch. 9 - 117
Masses are usually rounded off to whole numbers assuming:
H = 1, C = 12, N = 14, O = 16, F = 19 etc.
Molecular ion (parent peak)
Daughter
ionsSlide118
Ch. 9 - 118
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
Formation of Ions: Electron
Impact IonizationSlide119
Ch. 9 - 119
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 compoundSlide120
Ch. 9 - 120
Depicting the Molecular Ion
Radical cations from ionization
of nonbonding on
p
electronSlide121
Ch. 9 - 121
Compound
Ionization
Potential (eV)
CH
3
(CH
2
)
3
NH
2
8.7
C
6
H
6
(benzene)
9.2
C
2
H
4
10.5
CH
3
OH
10.8
C
2
H
6
11.5
CH
4
12.7
Ionization potentials of selected moleculesSlide122
Ch. 9 - 122
Fragmentation
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 importantSlide123
Ch. 9 - 123
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:
ArCH
2
+
> CH
2
=CHCH
2
+
> 3
o
> 2
o
> 1
o
> CH
3
+
Slide124
Ch. 9 - 124
e.g.
X
Site of ionization:
n >
p
>
s
non-bondingSlide125
Ch. 9 - 125
As the carbon skeleton becomes more highly branched, the intensity of the molecular ion peak decreases
Butane vs. isobutane
a
bSlide126
Ch. 9 - 126
16B.
Fragmentation of Longer Chain
and Branched Alkanes
Octane vs. isooctaneSlide127
Ch. 9 - 127
16C.
Fragmentation to Form
Resonance-Stabilized Cations
Alkenes
Important fragmentation of terminal alkenes
Allyl carbocation (m/e = 41) Slide128
Ch. 9 - 128
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 bondsSlide129
Ch. 9 - 129
Alcohols
Most common fragmentation: - loss of alkyl groups
a
bSlide130
Ch. 9 - 130
Carbon–carbon bonds next to the carbonyl group of an aldehyde or ketone break readily because resonance-stabilized ions called
acylium ions
are producedSlide131
Ch. 9 - 131
Aldehydes
M
+
peak usually observed but may be fairly weakCommon fragmentation patterna
-cleavageSlide132
Ch. 9 - 132
Ketones
a
-cleavage
a
a
b
bSlide133
Ch. 9 - 133
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 91Slide134
Ch. 9 - 134
Aromatic hydrocarbons
very intense M
+
peaks characteristic fragmentation pattern (when an alkyl group attached to the benzene ring): - tropylium cation Slide135
Ch. 9 - 135
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 H
2
O by 1,2- or 1,4-eliminationSlide136
Ch. 9 - 136Slide137
Ch. 9 - 137
Cycloalkenes show a characteristic fragmentation pattern which corresponds to a reverse Diels-Alder reaction
e.g.Slide138
Ch. 9 - 138
Aromatic hydrocarbons
e.g.Slide139
Ch. 9 - 139
Ketones
McLafferty rearrangement Slide140
Ch. 9 - 140Slide141
Ch. 9 - 141
Characteristic of McLafferty rearrangement
No alkyl migrations to C=O, only H migrates
XSlide142
Ch. 9 - 142
Characteristic of McLafferty rearrangement
2
o
is preferred over 1oSlide143
Ch. 9 - 143
How To Determine Molecular
Formulas and Molecular Weights
Using Mass Spectrometry
17A.
Isotopic Peaks & the Molecular IonSlide144
Ch. 9 - 144
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 peakSlide145
Ch. 9 - 145
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 presentThe isotopic peaks, in general, give us one method for determining molecular formulasSlide146
Ch. 9 - 146
Example
Consider 100 molecules of CH
4
C12: 100 C
13
: 1.11
H
1
: 100 H
2
: 0.016Slide147
Ch. 9 - 147
1.11 molecules contain a
13
C atom
4x0.016 = 0.064 molecules contain a
2
H atom
Intensity of
M
+ 1 peak:
1.11+0.064=1.174% of the
M
peakSlide148
Ch. 9 - 148
≈
100
1.17
m
/z
relative ion abundance
M
M +1Slide149
Ch. 9 - 149
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 = 0.3Slide150
Ch. 9 - 150
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 atomsSlide151
Ch. 9 - 151
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 unknownSlide152
Ch. 9 - 152
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 necessarySlide153
Ch. 9 - 153
Since M is m
/z 72
molecular weight = 72
As determined using the relative abundance of M +1 peak, number of carbons present is 4Using 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 contains zero “N”Slide154
Ch. 9 - 154
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 C
4
H
8
OSlide155
Ch. 9 - 155
17C.
High-Resolution Mass SpectrometrySlide156
Ch. 9 - 156
Example 1
O
2
, N2H4 and CH3OH all have M.W. of 32 (by MS), but accurate masses are differentO2 = 2(15.9949) = 31.9898
N
2
H
4
= 2(14.0031) + 4(1.00783) = 32.0375
CH
4
O = 12.00000 + 4(1.00783) + 15.9949 = 32.0262Slide157
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Example 2
Both C
3
H8O and C2H4O2 have M.W. of 60 (by MS), but accurate masses are different
C
3
H
8
O = 60.05754
C
2
H
4
O
2
= 60.02112Slide158
Ch. 9 - 158
Mass Spectrometer Instrument
DesignsSlide159
Ch. 9 - 159
19. GC/MS AnalysisSlide160
Ch. 9 - 160
END OF CHAPTER 9