Transcript Biological Spectroscopy - UCO

Interpretation of Mass Spectra
Part 4
•
Objectives
To describe the main features of EI, CI, ESI spectra
of organic compounds
To indicate how to identify the M+., MH+ or
[M+nH]n+ ion and to suggest a possible molecular
formula
To review the major types of fragmentation
mechanisms and how to recognise them in the mass
spectra of different classes of compounds
To give some practice in the interpretation of mass
spectra of simple unknown organic samples
Relative Molecular Mass
and Isotopes
•
Relative Molecular Mass
Many non-radioactive elements exist in more than one
isotopic form.
The lightest isotopes of the common light elements are
usually the most abundant e.g. 1H, 12C, 14N, 16O, 32S, 35Cl,
79Br
The masses of these isotopes, rounded to the nearest
whole number, are used to calculate the nominal RMM of
a compound. The monoisotopic RMM of a compound is
that calculated using the accurate atomic weights of the
most abundant isotopes of the constituent elements.
Relative Molecular Mass
As the number of H atoms increases, the difference between
the nominal and the mono-isotopic RMM slowly rises and
the nominal RMM is no longer useful, e.g. above about 600
Da.
For example:
Mass:
C20H30
C40H60 C60H90
Nominal
270
540
810
Monoisotopic
270.235
540.470
810.705
Whole Number 270
540
811
Common isotopes of the lighter elements
Isotope
Mass
1H
Relative
Abundanc
e
Isotope
Mass
Relative
Abundanc
e
1.00782 100
2H
2.01410 0.016
12C
12.000
100
13C
13.0033 1.12
14N
14.0031 100
15N
15.0001 0.36
16O
15.9949 100
18O
17.9992 0.2
32S
31.9721 100
33S
32.9715 0.78
34S
33.9679 4.39
35Cl
34.9689 100
37Cl
36.9659 32.7
79Br
78.9183 100
81Br
80.9163 97.5
Relative Molecular Masses
Naturally occurring compounds contain isotopes
in their natural abundances. The inclusion of the
less common isotopes leads to a higher average
RMM.
Hence we can define three different RMMs:
Nominal RMM
Monoisotopic RMM
Average RMM
Which do we measure by mass spectrometry?
Measurement of RMM
For RMMs of < 500, normally the nominal RMM is
obtained at low resolving power
The monoisotopic RMM can be measured if the
resolving power is high enough to resolve isotope
peaks and any other interfering peaks.
If the resolving power is insufficient to resolve
isotope peaks, the average RMM is measured. If
interfering peaks are not resolved, an accurate RMM
cannot be measured.
Isotope Peaks - 1
Ions containing elements that have more than one
isotope exhibit isotope peaks on the high mass side of
the peak due to the lightest ion and may indicate the
elements present in an ion.
The probability that any one carbon atom is a 13C isotope
is 1.1%
If there are n carbon atoms in an ion, the probability that
at least one carbon atom is a 13C isotope is n x 1.1% so
that I[(M+1)+]/I[(M+)] = (n x 1.1)/100
Isotope Peaks - 2
For relatively low RMM samples, this ratio
indicates the number of carbon atoms in the
ion providing that there is no contribution from
MH+ ions, e.g. for alcohols and amines at high
sample pressures.
For example, if M+. is at m/z 112, this could be
due to C8H16+., C7H12O+. or C6H8O2+. for which
the ratio I[(M+1)+]/I[(M+)] would ideally be
0.09, 0.08 and 0.07 respectively. This ratio
exceeds unity once the number of C atoms in
the molecule exceeds 85-90.
Molecular Ion Regions
1121
1217
C80H160
C80H160S3
1402
C100H200
Calculation of Isotope Patterns
Exemplified for Cl and Br atoms in C6H4OClBr
Write out isotopic abundances of Cl and Br isotopes
(including zero abundances) across the top and down the
side of a square.
Construct a matrix by multiplying each member of the
row with each member of the column.
Sum the diagonals (top right to bottom left) to produce
relative abundances.
Isotopic Peaks due to Cl + Br
3
1 |3
0 |0
1 |3
0
0
0
0
1
1
0
1
35Cl, 36Cl, 37Cl
abundances
79Br, 80Br, 81Br
abundances
(vertical)
Relative Abundance.
RMM
Comp.
3 : 0 : 4 : 0 : 1
114
35Cl79Br
116
35Cl81Br
37Cl79Br
118
37Cl81Br
1
1-Bromo-4-Chlorobenzene
Mass Spectrum
192
100
Cl
111
50
75
50
0
12
24
38
Br
44
10
20
30
40
50
(mainlib) Benzene, 1-bromo-4-chloro-
60
96
85
55 61
70
80
90
104
100
128
110
120
130
141
140
155
150
160
170
180
190
The 3:4:1 pattern in M+ region (with 13C peaks between). The
fragment ion at m/z 111, 113 shows the 3:1 pattern indicating
the presence of a Cl atom and that the Br atom has been lost.
200
192
100
Cl
111
50
75
50
0
12
38
24
Br
44
55 61
10
20
30
40
50
(mainlib) Benzene, 1-bromo-4-chloro-
60
96
85
70
80
90
104
100
128
110
120
130
155
141
140
150
160
170
180
190
200
192
100
Br
111
Cl
75
50
50
0
25
37
85
55 61
20
30
40
50
60
(mainlib) Benzene, 1-bromo-2-chloro-
70
80
91
90
96
104
100
128
110
120
130
140
140
155
150
160
170
180
190
200
The Molecular Ion Region
For compounds containing only C, H, O and N
atoms, 13C isotopes control the isotope pattern.
In large molecules, such as proteins, few ions
contain less than 2 or 3 heavy isotopes; the
peak due to ions containing only light isotopes
is of very low intensity.
The presence of S, Cl and Br atoms leads to
characteristic peaks at M+2, M+4, etc.
Electron Ionization Spectra
Production of EI Spectra
70 eV electrons bombard the sample vapour and deposit
typically 0 - 20 eV in a molecule.
Typically 8 - 10 eV is needed to produce M+. in its ground
state so M+. ions are produced with internal energies of
approximately 0 - 10 eV
These undergo one or more fragmentations in the ion
source and products ejected after about 1 ms are collected
to produce the mass spectrum. Discrimination effects
differ between instruments so that the spectrum of a given
sample varies somewhat from instrument to instrument.
Initial Inspection of the Spectrum - 1
Look at the overall appearance of the spectrum: try to
identify the molecular ion, M+. and obtain information
from any isotope peaks present.
If the major peaks are at low m/z and M+. is under 20% of
the most intense peaks, the sample is probably aliphatic.
The more intense M+. is, the greater the degree of
unsaturation is present (alkene, carbonyl compound).
Hexane
[M-C2H5]+
57
100
43
41
29
50
M+.
27
42
39
0
2
12
0
10
(mainlib) Hexane
86
28
15
55 58
30
20
30
40
50
71
60
70
84
80
90
100
2-Hexene
[M-C2H4]+.
55
100
[M-CH3]+
42
41
50
27 29
39
26 28 30
38 40
30
40
M+.
84
56
69
43
0
15
10
20
(mainlib) 2-Hexene
44
51
50
53
57
61 63 65 67 70
60
70
74 77 79 81 83 85
80
90
100
Cyclohexane
[M-C2H4]+.
M+.
56
100
[M-CH3]+
41
84
50
55
39 42
27
69
28
0
13
15
10
20
(mainlib) Cyclohexane
26
43
30
30
38
40
44
40
51
50
54 57
61 63 65 67 70
60
70
74 77 79 81
80
83 85
90
100
Cyclohexene
[M-CH3]+
[M-C2H4]+.
67
100
M+.
54
50
39
82
41
27
51 53
14 16
26
29
0
10
15
20
(mainlib) Cyclohexene
25
30
38 40 42
32
35
40
49
45
50
55
55
63
60
65
65
68
74
70
75
77 79 81
80 83
80
85
90
1,3-Cyclohexadiene
79
100
M+.
50
77
39
51
52
27
0
15
10
15
20
25
(mainlib) 1,3-Cyclohexadiene
38 40
28
30
35
40
49
45
50
80
78
63 65
54
55
60
65
74 76
70
75
81
80
85
90
Benzene
M+.
78
100
50
77
51
39
0
26 28
15
10
15
(mainlib) Benzene
20
25
49
40
30
35
40
45
50
61
53
55
60
74 76
63
65
70
75
79
80
85
90
1,5-hexadien-3-yne
M+.
[M-C2H2]+.
100
78
52
50
51
50
77
39
0
12 14
24
10
15
20
25
(mainlib) 1,5-Hexadien-3-yne
27
29 31
30
34
35
74
63
38
40 42 44
40
45
47
49
50
53 55
55
61
60
64
65
67 69
70
73
76
75
79
80
85
90
Initial Inspection of the Spectrum - 2
If peaks due to M+. and other high mass ions dominate the
spectrum, the sample is probably aromatic.
A large number of peaks often indicates a large number of
H atoms are present.
The lack of any dominant peaks suggests the absence of a
hetero-atom.
The simpler the spectrum, the more symmetry is likely to
be present in the sample molecule.
1-Napthalenol
M+.
100
115
144
[M-CO]+.
50
OH
0
18
26
10
20
30
(mainlib) 1-Naphthalenol
39 43
40
51
50
58 63
60
72
70
89
80
90
93 97 101
100
125
110
120
130
140
150
160
144
100
115
50
HO
0
15
27
63
39
43
10
20
30
40
(mainlib) 2-Naphthalenol
51 57
50
60
74 79
70
80
89
87 92 97 102
90
100
126
110
120
130
140
150
160
150
160
144
100
115
50
OH
0
18
26
39
43
10
20
30
40
(mainlib) 1-Naphthalenol
50
50
58 63
60
72
70
77
80
89
90
97 102
100
125
110
120
130
140
Nonanal
57
100
41
O
29
50
70
27
98
82
95
0
15 19
10
20
(mainlib) Nonanal
31
30
40
50
60
114
79 85 91
60
70
80
90
100
110
124
120
141
130
140
150
C3H7+
[M-C6H13]+
[M-C5H10]+
C2H5+
[M-C2H5]+
C6H13+
M+.
[M-C3H6-C3H6]+.
[C4H9CO]+
[M-C3H6]+.
M+.
C5H7+
[M-C2H5O]+
M+.
CH2NH2+
Base peak of primary
amines
Found in all amine spectra
and in spectra of amides
M+.
Identifying the Molecular Ion - 1
The common isotopes of elements C, O, S
have even relative atomic masses and even
valencies whereas the common isotopes of H,
F, Cl, Br, P and Na have odd relative atomic
masses and odd valencies.
Hence organic compounds that contain only
these elements (i.e. no nitrogen atoms) have an
even relative molecular mass (RMM) so that
M+. occurs at an even value of m/z or the MH+
ion appears at an odd value of m/z .
Identifying the Molecular Ion - 2
The one common element that is an exception
to this rule is nitrogen, the most common
isotope of which has a relative atomic mass of
14 and a valency of 3.
Hence an odd relative molecular mass results
when the molecule contains an odd number of
N atoms. Thus if M+. has an odd m/z, it
suggests a possible amide, amine, nitrile or Nheterocyclic compound.
Uses of Isotope Peaks
Common elements that give M+2 isotope peaks:
35Cl:37Cl rel. ab. ~ 3 : 1
79Br:81Br rel. ab. ~ 1 : 1
32S:34S
rel. ab. ~ 100 : 4
28Si:30Si rel. ab. ~ 100 : 3.4
Hence peaks at M+2, M+4, etc. indicate the presence of Cl, Br, S, Si;
the absence of these peaks indicates the absence of these elements.
Common elements that give rise to M+1 isotope peaks are
C and
N but only C isotope peaks need be considered:
12C:13C rel. ab. ~ 100 : 1.1
So that I([M+1]+)/I([M+]) = n x 1.1/100 for an ion containing n C
atoms.
Approximate ratio
27:27:9:1
Approximate ratio
9:6:1
Approximate
ratio 3:1
81:108:54:12:1
27:27:9:1
9:6:1
1:4:6:4:1
1:3:3:1
[M-(C6H10)]+
m/z
148 149 150
Intensities 100 7
9
C12H22S2+.
m/z
230 231 232
Intensities 100 13 9
Is it Really the Molecular Ion?
Check that the supposed M+. loses neutral species that are
sensible, e.g. radicals such as alkyl radicals or OH. or
molecules such as alkenes, CO, HCl, H2O, etc.
If there are losses that cannot be explained, e.g. 3 - 13,
21 - 25 Da, the assignment should be re-examined.
If it appears that M+. loses 3 Da, this could arise from
losses of CH3 and H2O giving peaks due to the ions [M15]+ and [M-18]+..
Is it Really the Molecular Ion?
Try to identify the main species lost by M+.. These often
indicate the type of compound to which the sample
belongs.
Rearrangement ions formed by loss of a molecule are
often particularly informative. If no nitrogen is present,
these appear at an even value of m/z.
Identify ions characteristic of a compound type: m/z 105,
77, 51 for benzoyl compounds, m/z 91, 65, 39 for
alkylbenzenes, m/z 30 for amines, etc.
[M-C4H9]+
[M-C3H7]+
[M-CH3]+
[MH2O]+.
M+.
absent
C2H5CO+
C3H7CO+
M+.
[CH3CO]+
[M-C3H6]+.
[M-CH3]+
M+.
C7H7+
M+.
C3H3
+
C5H5+
How to Work Out
the Molecular Formula - 1
Start with the RMM and the value of
I([M+1])/I([M]) which gives an indication of the
number of C atoms present.
Suppose the RMM is 136. The maximum possible
number of C atoms is found by dividing this by
12. This gives 11 but C11H4 is very unlikely to be
correct. Try 10 C atoms, converting the other 12
Da to H atoms, giving C10H16 (e.g. pinene or
limonene, etc).
How to Work Out
the Molecular Formula - 2
Repeating the process for 9 C atoms gives
C9H28 which is unacceptable. Convert 16 H
atoms to one O atom giving C9H12O as a
possibility, e.g. benzyl ethyl ether.
Repeating this for 8 C atoms gives C8H8O2
which could be an aromatic acid or ester.
Knowing the number of C atoms present,
one can suggest a molecular formula. In this
case, suppose C8H8O2 is suggested.
Rings + Double Bonds: 1
For a molecule of molecular formula CxHyNzOa
the number of double bond equivalents (DBEs) is
given by the expression
DBEs = x - (y/2) + (z/2) + 1
A DBE is a unit of unsaturation, e.g. an alkene or
carbonyl group is 1 DBE, a saturated ring is 1
DBE, an aromatic ring is 4 DBEs, a triple bond is
2 DBEs. The number of DBEs is independent of
the number of O atoms present. Halogen atoms
are counted as H atoms, S and P atoms are
counted as O and N atoms respectively.
Rings + Double Bonds: 2
If the formula C8H8O2 is suggested, the number of
RDBs is
8 - (8/2) + (0/2) + 1 = 5.
This immediately suggests the presence of an
aromatic ring (4 RDBs) and in view of the
presence of two O atoms, the other is almost
certainly a carbonyl group.
A formula such as C9H12O would have
9 - (12/2) + (0/2) + 1 = 4
This suggests, for example, an aromatic ring (4
RDBs) and a saturated substituent.
Suggesting Possible Structures
C6H5COOCH3 A benzoyl compound so m/z 105, 77
and 51 should be prominent features.
C6H5CH2COOH A benzyl compound so m/z 91, 65
and 39 should be prominent features.
CH3C6H4COOH The o-, m- and p-isomers all give a
prominent peak at m/z 119 (loss of OH), least intense
for the m-isomer. The o-isomer gives an m/z 118 ion
(loss of H2O formed by the OH of the -COOH group
and H from the methyl group. Comparison of the
spectra with those of authentic samples would
confirm the identification.
[M-COOH]+
[M-OH]+
M+.
C7H7+
[M-H2O]+.
M+.
[M-H2O-CO]+.
[M-OH]+
Fragmentation of
+.
M Ions
Unknown mass Spectrum
Odd m/z suggests a N atom present
m/z 105, 77, 51 suggests benzoyl
Since N accounts for 14 of the other 16 Da, 2 H atoms are present
Hence the sample is benzamide, C6H5CONH2.
Common Neutral Losses of Diagnostic Value - 1
15
CH3
Alkyl branching if intense peak, otherwise neglect
16
O
Nitroaromatic, oxime, sulfoxide
16
NH2
RCONH2
18
H2O
Alcohol, (ketone, aldehyde, less common)
20
HF
Alkyl fluoride
26
C2H2
Aromatic hydrocarbon
27
HCN
ArCN, N-heterocylic compounds, ArNH2 rarely
27
C2H3
Ethyl ester (low abundance)
28
CO
Quinones, some phenols
28
C2H4
n-Propyl ketones, ethyl esters, ArOC2H5
29
C2H5
Ethyl ketones, Ar - n-C3H7 compounds
30
CH2O
Aromatic methyl esters
31,32
CH3O,CH3OH
Methyl esters of carboxylic acids
33,34
SH, H2S
RSH
Common Neutral Losses of Diagnostic Value - 2
41
C3H5
Propyl ester
42
C3H6
n-butyl ketone
CH2CO
RCOCH3, ArOCOCH3, ArNHCOCH3
43
C3H7
RCOC3H7, Ar-n-C4H9 compounds
44
CO2
Anhydrides, esters
45
COOH
RCOOH
OC2H5
Ethyl esters of carboxylic acids
46
NO2
Aromatic nitrocompounds
48
SO
Aromatic sulfoxide
55
C4H7
Butyl ester of carboxylic acid
56
C4H8
RCOC5H11, ArOC4H9, Ar-C5H11 (n- or i-)
57
C4H9
RCOC4H9
C2H5CO
RCOC2H5
CH3COOH
Acetate
60
[M-NO2]+
M+.
[M-NO]+
[M-O]+.
[M-C6H12]+.
[CH2C(OH)OC2H5]+.
Characteristic of ethyl ester of
long-chain carboxylic acid
“McLafferty+1” peak
given by ethyl esters
[M-C2H3]+
M+.
Common Characteristic Ions
m/z 105 + 77 + 51
Benzoyl compounds
m/z 91 + 65 + 39
Alkyl benzenes, benzyl compounds
m/z 30
Base peak RNH2 otherwise other amines
m/z 44, 58, 72, . . .
Amines, amides
m/z 31
Primary alcohol; low intensity, other alcohols, ethers
m/z 31, 45, 59, . . .
Ethers
m/z 74
Methyl esters of carboxylic acids
m/z 60
Straight chain carboxylic acids
m/z 77 or 76
Mono- or di-substituted benzene (low intensity)
[CH2NH2]+
Suggests amine or amide
[CH3CO]+
[C2H4NH2]+ [M-C3H6]+.
[M-CH3]+
M+.
[M-C8H16]+. [CH2C(OH)OCH3]+.
Characteristic of methyl ester of a
long-chain carboxylic acid
CH3(CH2)8CO.OCH3
Characteristic H-rearrangement
ions of straight chain methyl esters
143
[M-OCH3]+
M+.
The Odd-Even Electron Rule - 1
For molecules that do not contain an odd
number of N atoms, M+. is an even mass,
odd electron ion.
If it loses a molecule in a rearrangment
process, the resulting fragment ion is again
an even mass, odd electron ion.
If it loses a radical, which is an odd mass,
odd electron species, this produces an odd
mass, even electron fragment ion.
The Odd-Even Electron Rule - 2
Once a radical has been lost to produce an even
electron, closed shell ion, further fragmentations can
occur only by the loss of molecules to produce further
odd mass, even electron ions.
Successive loss of two radicals NEVER occurs.
Do not assume that an ion is always formed from the
next highest mass fragment ion. Ions may fragment
by several routes so that adjacent peaks may not
belong to ions of the same fragmentation sequence.
(CH3)2CH-C6H4COOH
All fragment ions are odd
mass, even electron ions
M+.
Charge Localisation - 1
Although the charge on a molecular ion may
be delocalised, it is useful to consider it
formally as localised.
Where on the molecular ion is the charge
located?
Which is the easiest (lowest energy) electron
to remove?
These are usually
(a) lone pair electrons on heteroatoms
(b) p-electrons in unsaturated systems
Charge Localisation - 2
If there is a choice of electrons that could be
removed, the formal charge may be placed on
one of several atoms.
Hence, formally, one can think of M+. ions as
consisting of a mixture of ions with the formal
charge being on one of several possible sites.
Each type of molecular ion can give rise to a
different type of fragmentation and the spectrum
observed will be the weighted sum of the
products of these.
Examples of Charge Localisation
Carbonyl compounds are assumed to lose a
lone pair electron from the carbonyl oxygen
Ionized toluene is assumed to have lost a ring
p-electron
Charge Localisation at Several Sites
Here there are three possible sites for charge localisation.
Fragmentation may be rationalised in terms of the
decomposition of three different types of molecular ion
NH2+.
NH2
NH2
+.
O
R
O+.
O
R
R
[M-CH3]+
a-cleavage
[M-CH3CO]+
inductive cleavage
M+.
Factors Influencing Ion Abundance - 1
Eint required for decomposition: in general, low
energy processes will predominate but different
ionization methods yield different internal
energy distributions and hence different mass
spectra from a particular sample.
Stability of the product ion
Factors Influencing Ion Abundance - 2
Stability of the neutral product
Delocalization of electron e.g. in allyl radical
Placing of electron on electronegative atom e.g. .OH
Loss of small stable molecule containing multiple
bonds, e.g. CO, C2H2, HCN
Stevenson’s Rule
AB+.  A+ + B. or A. + B+
Preference for formation of ion from fragment
having lower IE (except largest R. is lost
preferentially)
Radical Site Initiation (a-Cleavage) - 1
This is particularly important for ions that
contain N or O atoms.
Electron pairing occurs by the transfer of an
odd electron from the bond alpha to the atom
carrying the charge and the transfer of the odd
lone pair electron to form a new bond.
The remaining electron from the alpha bond is
lost on the radical that is eliminated as a result
of the bond breaking.
Radical Site Initiation (a-Cleavage) - 2
In this type of fragmentation, the charge site does not
move but the radical site moves as a result of the alpha
bond breaking. In the example below, C - X is the alpha
bond broken.
An example is the fragmentation of carbonyl compounds:
R - C - X RCO+ + X.
||
O+.
(X = H, R, OH, OR, Cl, NH2) aldehydes, ketones, acids,
esters, acid halides, amides
Mechanisms of Alpha Cleavage
X = H, R, OH, OR, Cl, NH2
for aldehydes, ketones, acids,
esters, acid halides and amides.
m/z 30 is the base peak in
primary amine spectra
Primary alcohols give the m/z
31, ethers often give m/z 45,
59 . . by this reaction
 The ease of loss of alkyl
groups is R1>R2>R3 where
this is also the order of
decreasing size.
C4H3+
C6H5+
[M-H]+ C6H5CO+
M+.
[M-R.]+ a-cleavage,
57+, 71+
[R]+ inductive cleavage
29+, 43+
M+.
M+.
[M-16]+.
characteristic of amides
[CH2OH]+
[M-H2O-C3H6]+.
[M-H2O]+.
M+. absent
[CH2OH]+
[C2H5OCH2]+
3 examples of [CH2OR]+ ions
at m/z 31, 59 and 87
[C4H9OCH2]+
M+.
[CH2NH2]+ base peak of
primary amines
M+.
Loss of the
Largest Alkyl Radical -1
 Where there is a choice of alkyl radicals
that can be lost in an alpha-cleavage, the
largest radical is lost preferentially followed
by the next largest, etc.
Loss of the
Largest Alkyl Radical - 2
The loss of H or an alkyl radical gives rise to
the following series of ions for different
classes of compounds
Aldehydes and ketones m/z 29, 43, 57, 71 . . .
Aldehydes usually give a fairly prominent m/z
29 peak (CHO+) and a weaker [M - 1]+ peak.
Methyl ketones often give m/z 43 (CH3CO+)
as the base peak.
Loss of the
Largest Alkyl Radical - 3
Alcohols and ethers m/z 31, 45, 59, 73 . . .
Primary alcohols typically give a m/z 31 peak
(CH2OH+), often of fairly low intensity.
Methyl ethers give m/z 45 peaks, again often of
low intensity (charge induced fragmentation
usually predominates).
Amines m/z 30, 44, 58, 72 . . . n-alkyl amines
often give m/z 30 as the base peak whereas
secondary and tertiary amines often give m/z 44
and 58 as base peaks together with m/z 30
Loss of the
Largest Alkyl Radical – 4 (amines)
CH3
C3H7
H
C
H
A
NH2
H3C
H
C
H
N
H
H
C
H
B
CH3
H3C
C
NH2
CH3
C
[A] C3H7 loss predominates to give an intense m/z 30 peak;
little H loss evident, m/z 72 negligible; M+. m/z 73 (10%)
[B] CH3 loss is preferred to H loss giving m/z 58 as the base
peak; some H loss giving m/z 72 (20%); M+. m/z 73 (30%)
[C] CH3 loss is only possible alpha-cleavage so that m/z 58 is
again the base peak but there is no peak at either m/z 72 (H
loss) or 73, M+.
[CH2NHC2H5]+
-C2H4
M+.
[M-C5H11]+
[M-C3H7]+
[M-CH3]+
M+.
C3H7 loss
C2H5 loss
CH3 loss
M+. absent
Allylic Cleavage
This is a major type of fragmentation for alkenes leading
to alkyl radical loss. Unfortunately, migration of the
double bond occurs before fragmentation so that the
observation of ions of this type is of little structural value.
The mass spectra of many alkenes, especially polyenes,
tend to be independent of the position of the double bond
so that isomers cannot be distinguished.
The m/z 41 ion is the most common ion observed in the
mass spectra of aliphatic compounds, together with
homologues of m/z 55, 69, 83 . . .
Allylic ions at m/z 41, 55, 69, 83
M+.
Charge Site Initiation
(Inductive Cleavage)
[HOCH2]+
[C2H5OCH2]+
C3H7+
C2H5+
M+.
[CH2SC2H5]+
[C3H7]+
[C2H3
M+.
]+
[CH2SC3H7]+
Rearrangement Reactions
McLafferty Rearrangement - 1
McLafferty Rearrangement - 2
Deuterium labelling studies show that a g-H atom
is transferred quite specifically through a sixmembered transition state. If there are no g-H
atoms, the rearrangement does not occur.
 Note that substituents on the a-carbon atom are
retained in the ion but substituents on the b- and
g-carbon atoms are lost as part of the alkene useful in locating site of branching in an alkyl
chain.
[C2H5CO]+
No alkyl chain 3 or more C atoms
long so no McLafferty
Rearrangement leading to alkene loss
[C2H5]+
M+.
[C3H7]+,
[CH3CO]+
C2H4 loss from C3H7 alkyl chain
CH3 loss
M+.
Loss of C4H8 from C5H11
alkyl chain (McLafferty
Rearrangement)
[M-C5H11]+
[M-OCH3]+
M+.
Loss of C4H8 from
C5H11 alkyl chain
[M-NH2]+
M+.
McLafferty Rearrangement - 3
If an even mass fragment ion is found which
could be formed from the molecular ion by the
loss of 28, 42, 56, 70, . . . Da., always suspect that
it results from a McLafferty rearrangement or of a
related process.
The driving force for the rearrangement is the
formation of a strong bond between the H atom
and the unsaturated heteroatom carrying the
charge. Similar reactions occur with H-transfer to
other heteroatoms.
McLafferty Rearrangement - 4
Peaks due to alkene losses at
m/z 112, 98, 84, 70, 56, 42
M+.
[M-C3H6]+.
M+.
[M-C2H4]+.
M+.
H-Atom Rearrangement to a
Saturated Heteroatom - 1
[M-HCl]+
M+.
H-Atom Rearrangement to a
Saturated Heteroatom - 2
In these reactions, an unpaired electron on the
heteroatom is donated to form a new bond with
a H atom with cleavage of the original bond to
that H atom.
A second radical site reaction leading to H2O+.
formation is not favoured since this is an
energetic process that does not lead to
particularly stable products.
H-Atom Rearrangement to a
Saturated Heteroatom - 3
Instead, a charge site reaction occurs leading to
the loss of H2O (very common for primary
alcohols, much less so for secondary and tertiary
alcohols because of α-cleavage competition)
followed by a further charge site reaction in
which C2H4 is lost.
In general, the loss of a small neutral species is
more energetically favorable than formation of a
small ionic species.
Fragmentation of Aromatic
Molecular Ions
Many simple aromatic molecular ions
fragment by elimination of a small,
unsaturated molecule by breaking the aromatic
ring but giving a further, stable cyclic ion as a
product.
Examples of small molecules lost include: Benzene, C2H2, Pyridine, HCN, Thiophene, HCS,
Furan, HCO, Phenols, CO, Anilines, HCN
M+.
[M-C2H2]+.
[M-HCN]+.
M+.
[M-HCO]+
M+.
[M-C2H2]+.
[M-C3H3]+
[M-CHS]+
M+.
H-Atom Rearrangement in Phenols
and Anilines
•
 Phenols expel CO to give an [M-28]+. peak often accompanied by an
[M-29]+ peak due to the loss of a H atom.
Replacing the O atom by NH to give aniline, the loss of HCN to give an
[M-27]+. ion can be similarly rationalized.
Deuterium labeling shows considerable H/D scrambling indicates that
the mechanisms are more complicated than indicated above.
[M-CO]+.
M+.
M+.
[M-HCN]+.
Phenol
Aniline
H-Atom Rearrangement to a
Saturated Heteroatom - 4
THE ORTHO EFFECT
requires a labile H atom (OH,
NH2, CHO) and formation of a
stable ionic product by charge
migration, eliminating a small
saturated molecule e.g. H2O,
CH3OH, HCl. These products
are more energetically
favourable than those given by
charge retention reactions.
2-Hydroxybenzaldehyde
M+.
[M-H]+
122
100
[M-CO-H2O]+.
O
50
[M-CO]+.
[M-H2O]+.
65
39
50
29
0
13 17
93
76
OH
42
10
20
30
40
(mainlib) Benzaldehyde, 2-hydroxy-
47
104
63
53
71
50
60
70
74
87
80
90
100
110
120
130
2-Hydroxybenzoic acid ethyl ester
Loss of
C2H5OH
Loss of CO from
m/z 120
120
100
M+.
O
50
92
O
39
65
27
0
43
53
20
30
40
50
60
70
(mainlib) Benzoic acid, 2-hydroxy-, ethyl ester
166
OH
76 81
80
138
104 109
90
100
110
120
130
140
150
160
170
180
Even-Electron Ions, CI,
Even-Electron Ions - 1
Under EI conditions, M+. ions are formed and a major
fragmentation process is the loss of a radical, R.,
producing an even-electron ion.
Once a radical has been lost, all subsequent
fragmentations involve the loss of a molecule to form
further even-electron ions.
Under CI conditions, an even-electron ion, such as MH+,
is formed; subsequent fragmentations involve the loss of
a molecule to form further even-electron ions.
Sites of Protonation
In order to rationalize the fragmentation of MH+ ions, one
must consider at which sites in the sample molecule the
proton is attached. The spectrum may then be
rationalized in terms of the fragmentation of the different
types of MH+ ions.
In general, protonation occurs on heteroatoms having lone
pairs of electrons, such as O, N and Cl. This frequently
followed by charge-induced elimination of a molecule
containing the hetero-atom. Other possible protonation
sites are aromatic rings and regions of unsaturation.
Even Electron Ions
 Ephedrine ionized by methane CI may protonate on the O
atom of the OH group:
 Protonation on the N atom leads to the loss of CH3NH2 by
a similar mechanism, yielding an ion of m/z 135. Both
m/z 148 and 135 are observed in the CI spectrum,
indicating the presence of OH and HNCH3 groups in the
molecule.
•
Ephedrine EI and CI Spectra
Ephedrine, RMM 165,
gives an EI spectrum
dominated by the m/z 58
fragment ion and no M+.
ion. Methane CI gives an
MH+ ion at m/z 166 and
fragments at m/z 148,
135 and 58 due to
protonation on the OH
and NHCH3 groups or on
the aromatic ring
respectively
Field’s Rule - 1
In the decomposition of even electron ions , if more than
one neutral can be eliminated, the lower the proton
affinity of the neutral, the greater the tendency to leave.
Hence under CI conditions, a protonated chloroalkane has
a high tendency to lose HCl (PA 5.8 eV), a protonated
alcohol has a moderate tendency to lose H2O (PA 7.2 eV)
and a protonated amine has a low tendency to lose NH3
(PA 9.0 eV), consistent with the [M-18]+ and [M-31]+
relative abundances in the CI mass spectrum of
ephedrine.
Field’s Rule - 2
In the case of unsaturated ions which could lose
unsaturated neutrals, the higher proton affinities of
these species often result in other more complex Hrearrangement modes of fragmentation being
more favourable. For example,
C2H5O+=CH2  CH2O + C2H5+ (9% CID) (inductive cleavage)
C2H5O+=CH2  C2H4 + HO+=CH2 (58% CID)
(H rearrangement) predominates since the PAs of
C2H4 and CH2O are respectively 7.0 and 7.4 eV.
Charge Site Rearrangements - 1
Even-electron ions undergo H-transfer to the
charge site; all electrons remain paired (low
energy process), the charge site does not move
and an unsaturated or cyclic neutral species is
lost
Charge Site Rearrangements - 2
Very important in the fragmentation of amines.
After an initial α-cleavage, this fragmentation
requires the presence of at least one ethyl group
or larger so that an alkene may be lost.
[CH2NH2]+
[M-CH3]+ a-cleavage
-C2H4
M+.
Violations of Even Electron Rule - 1
Usually R - Y+ undergoes heterolytic cleavage
to give R+ + Y, all species being even-electron
species.
This is a much lower energy process than
homolytic cleavage with charge retention
giving R. + Y+. so producing two oddelectron species.
Violations of Even Electron Rule - 2
 The most common exceptions are in the spectra
of aromatic molecules containing electronegative
substituents, Cl, CN, NO2, etc.
 In the spectra of 1,3- and 1,4-dinitrobenzenes, the
ions C6H4+. and C6H3+ are of almost equal
abundance (losses of 2 NO2 and NO2 + HNO2
respectively) not only because IE(benzyne) <
IE(NO2) but also because other reactions giving
even-electron products are not energetically more
competitive. Other examples of this behaviour
are common.
M+.
NO+
-C2H2
[M-NO2]+
-O
-NO
Practical Problems - 1
Beware of spurious peaks such as the following:
Background peaks from previous samples, pump oil or
from an air leak, e.g. m/z 40, 32, 28, 18 etc.
Peaks arising from incomplete removal of common
solvents such as m/z 83, 85, 87 from CHCl3, m/z 58, 43
from acetone
Peaks present due to incomplete reaction leaving traces of
starting materials in the sample
Peaks due to homologues, e.g. at 14 m/z units above or
below the true molecular ion peak
Practical Problems - 2
• Compare your spectrum with that of an
authentic sample obtained by use of the same
ionisation technique but remember that an
exact match of relative intensities is unlikely to
be found because of varying mass
discrimination effects.
General Hints - 1
Aromatic or aliphatic? Provisionally identify M+.
Check that proposed neutral losses are sensible
Is N present? Is assignment of M+. incorrect?
Check for isotope peaks for Cl, Br, S, heavy metals
Use I([M+1]+)/I([M]+.) to estimate number of C atoms
present
Postulate a molecular formula and estimate the double bond
equivalents
General Hints - 2
Inspect higher mass ions, possibly formed from M+. in one
step, e.g. even mass fragments formed in a rearrangment
process
Look for characteristic neutral losses such as 16 Da, O
from ArNO2 or NH2 from an amide, 30 Da, CH2O from
ArOCH3 and characteristic ions, m/z 30, amines, m/z 74
methyl esters, 105/77/51, 91/65/39 for benzoyl and alkyl
benzene compounds
Do not assume adjacent peaks are due to sequential losses
of neutrals; two or more charge sites lead to competing
fragmentation routes.
General Hints - 3
Do not try to interpret every small peak,
especially those at low m/z which result from
sequential fragmentation
Never postulate the loss of a radical from an
even electron ion without very good reason
Use negative evidence as well as positive
evidence: e.g. if there is no peak at m/z 91, the
sample is unlikely to be an alkyl benzene.