Alkanes undergo extensive fragmentation
CH3—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH3
43
Relative
intensity
57
100
80
Decane
60
71
40
85
20
0
142
99
20
40
60
80
m/z
100
120
– More stable carbocations will be more abundant.
CH3
Facile
CH3
CH3CH2CH2CHCH3
+
•
CH3CH2CH2 •
+
+ CHCH
3
m/z 43
CH3
Facile
CH3• + CH3CH2CH2CH +
More difficult
m/z 71
m/z 86
CH3
CH3CH2•
+
+ CH
2CHCH3
m/z 57
CH3
|
CH3 - CH2 -- C -- CH3
+
+CH3
m/e = 71
m/e = 15
CH3
|
+C - CH3
|
CH3
m/e = 57
CH3 - CH2 +
m/e = 29
CH3
|
+C - H m/e = 43
|
CH3
– MS of 2,2,4-trimethylpentane
• Branched alkanes have small or absent M+
Propylbenzene fragments mostlyat the benzylic position
Relative
intensity
100
91
80
CH2—CH2CH3
60
40
120
20
0
20
40
60
80
100
m/z
+
91
+
120
 Cycloalkanes
– Loss of side chain
– Loss of ethylene fragments
Ionization followed by fragmentation
Ejection of
electron
Splitting out of ethylene.
fragmentation
Radical/cation
.
– MS of methylcyclopentane:
 Alkene Fragmentation
– Fairly prominent M+
– Fragment ions of CnH2n+ and CnH2n-1+
– Terminal alkenes lose allyl cation if possible to form
resonance-stabilized allylic cations
R
-R
+
•
[CH2=CHCH2 CH2 CH3 ]
CH2 =CHCH2 + +
• CH2 CH3
 Cycloalkenes
– Prominent molecular ion
– Retro Diels-Alder cleavage
+
•
Observed!
+
•
+
+
•
+
Observed!
• Typically you see both.
• More stable cation will predominate
• Also works for hetero-substituted (e.g. make enol)
• Both EI (shown) and in CI. (protonated molecular ion, cleave, then
reprotonation)
– Cyclohexenes give a 1,3-diene and an alkene, a process that
is the reverse of a Diels-Alder reaction.
CH3
+
•
CH3
H3 C
C
H3 C
C
+
•
+
CH2
CH2
Limonene
(m/z 136)
A neutral diene A radical cation
(m/z 68)
(m/z 68)
Mass spectrum of 4-terpineol as a good example
for Retro Diels Alder fragmentation
HO
EI Mass
Spectrum
HO
4-terpineol(MW 154)
+
mz 68
+
+
mz 86
O
Alkyne Fragmentation
– Molecular ion readily visible
– Terminal alkynes readily lose hydrogen atom
– Terminal alkynes lose propargyl cation if possible
(resonance-stabilized propargyl cation or a substituted propargyl cation).
3-Propynyl cation HC C-CH +
2
(Propargyl cation)
+
HC C=CH2
Aromatic Hydrocarbon Fragmentation
– Molecular ion usually strong
– Alkylbenzenes cleave at benzylic carbon tropylium ion formation
– Some molecules undergo very little fragmentation
– Benzene is an example.
The major peak corresponds to the molecular ion.
Relative
intensity
100
m/z = 78
80
60
40
20
0
20
40
60
80
100
120
m/z
Benzene Compounds
CH2OH
OH
+

-CO
-H 
+
m/z 107
m/z 108
H
7 memberd ring
+
H
H
H
-H2
+
H
H
H
Phenyl ion
m/z 77
[C 6 H5 ]+
Benzonium ion
m/z 79
[C6H7]+
CH3
+ 
+ 
H
- CH3 
+
H
CH3
CH3
m/z 91
Tropylium ion
Mass Spectrum of n-Octylbenzene
Tropylium ion
Mass Spectrum of benzyl bromide
Br
Tropylium ion
Bromine pattern
 Alcohol Fragmentation
– One of the most common fragmentation patterns of
alcohols is loss of H2O to give a peak which corresponds to
M-18 and no M peak.
– Molecular ion strength depends on substitution
primary alcohol
weak M+
secondary alcohol VERY weak M+
tertiary alcohol
M+ usually absent
– Another common pattern is loss of an alkyl group from the
carbon bearing the OH (-Cleavage) to give a resonancestabilized oxonium ion and an alkyl radical (largest R group lost
as radical). R'
•
+
R C O H
••
R• +
R"
Molecular ion
(a radical cation)
A radical
+
+
••
R'-C •O
H
R'-C=O-H
•
R"
R"
A resonance-stabilized
oxonium ion
MS of 1-butanol is an example demonstrating both processes.
Elimination of water.
Elimination of propyl radical.
74 – 31 = 43 (C3H7)
74 – 56 = 18 (water).
Mass Spectrum of 3-methyl-1-butanol
 Carbonyl Compounds Fragmentation
Dominant fragmentation pathways:
• a-cleavage
• -cleavage
• McLafferty rearrangement
 Aldehydes and Ketones
Characteristic fragmentation patterns are:
– Cleavage of a bond to the carbonyl group (a-cleavage).
R
C
O
R'
C
O
R'
C
O
+
R
R'
Acylium ion
+
•
O
a-cleavage
m/z 128
O
+
+
•
m/z 43
CH3 •
+
O
+
m/z 113
Note that an a cleavage of an aldehyde could produce a peak at M – 1 by
eliminating H atom.
This is useful in distinguishing between aldehydes and ketones.
– McLafferty rearrangement.
Splitting out an alkene (neutral molecule) and producing a new radical/cation.
H
O
Molecular ion
m/z 114
+
•
McLafferty
rearrangement
H
+
O
m/z 58
+
•
Mass Spec of 2-octanone displays both a cleavage and McLafferty
CO+
CH3
resulting from
a cleavage.
CH3CH2CH2CH2CH2CH2CO+
resulting from a cleavage.
 Carboxylic Acids
Characteristic fragmentation patterns are:
– a-cleavage to give the ion [CO2H]+ of m/z 45.
– McLafferty rearrangement if g H present.
O
a-cleavage
OH
•
O= C-O- H
+
m/z 45
Molecular ion
m/z 88
H
+
+
•
O
McLafferty
rearrangement
OH
Molecular ion
m/z 88
– Loss of water, especially in CI
– Loss of 44 is the loss of CO2
H
+
+
•
O
OH
m/z 60
Fatty Acids
 Esters
a-cleavage and McLafferty rearrangement:
+
•
O
OCH 3
a-cleavage
O
+
m/z 71
Molecular ion
m/z 102
H
OCH3
Molecular ion
m/z 102
McLafferty
rearrangement
H
+
• OCH
O
+
+
+•
O
+
3
OCH3
m/z 59
+
•
O
OCH3
m/z 74
 Ether Fragmentation
– a-cleavage
– C-O cleavage (requires stable cation to lose)
– Ion rearrangement
Ms of 2-Chloroethylphenyl ether
94
100
H
Cl
107
80
94
60
40
156
107
20
20
40
60
80
100
120
140
160
O
 Amines
–The most characteristic fragmentation pattern of 1°, 2°,
and 3° aliphatic amines is -cleavage
CH3
CH3 - CH- CH 2 -CH 2 -N H2
-cleavage
CH3
CH3 - CH- CH 2
•
+
+ CH2 = N H2
m/z 30
 Halide Fragmentation
– Loss of halogen atom
– Elimination of HX
– a-cleavage
 Sulfur Compounds
– Fortunately there is an [M+2]+ of 4% for the natural abundance of
This is diagnostic for S vs 2x16O
34S.
– Aliphatic thiols can split out H2S, [M-34]
– a-cleavage at carbon bearing the sulfur in thiols, thioethers, similar to
ethers, etc.
R
-R•
S
•
+
S
+
 Nitroaromatics
m/z = 93
+
N
Loss
of
•N=O
O
O•
Aromatic!
m/z=65
Good test for aryloxy
(this can form from lots of different origins)
+
O
Loss of
CO
CH+
Clusters of Ions
– Spaced by unit mass
– Each peak is for the same molecular formula
– Different peaks because there are some molecules with 13C, 2H etc.
– Especially significant for Cl, Br
The Nominal mass is m/z
of the lowest member of
the cluster. This is the
isotopomer that has all
the C’s as 12C, all protons
as 1H, all N’s as 14N, etc.
m/z
Analyzing Ion Clusters:
a way to rule candidate structures
– Mass spectrometry “sees” all the isotopomers as
distinct ions
– An ion with all 12C is one mass unit different from
an ion with one 13C and the rest 12C
– Since the isotope distribution in nature is known
for all the elements (13C is 1.1%), the anticipated
range and ratios of ions for a given formula can be
predicted and calculated
Isotopic Clusters
H
H
79
H
H
93.4%
all H are
1H and all
C are 12C
H
H H
H
78
H
H
H
H
79
H H
H
H
6.5%
one C is 13C
H
H
0.1%
one H is 2H
Isotopic Clusters in Chlorobenzene
37Cl
35Cl
Relative
intensity
visible in peaks
for molecular ion
100
112
80
60
40
114
20
0
20
40
60
80
100
120
m/z
Isotopic Clusters in Chlorobenzene
no m/z 77, 79 pair; therefore ion
responsible for m/z 77 peak does
not contain Cl
+
H
H
H
H
H
Relative
intensity
100
80
60
77
40
m/z
20
0
20
40
60
80
100
120
Isotopic Abundance
81Br
 Molecules with Heteroatoms
– Isotopes: present in their usual abundance.
– Hydrocarbons contain 1.1% C-13, so there will
be a small M+1 peak.
– If Br is present, M+2 is equal to M+.
– If Cl is present, M+2 is one-third of M+.
– If iodine is present, peak at 127, large gap.
– If N is present, M+ will be an odd number.
– If S is present, M+2 will be 4% of M+.
Molecular Formula
as a
Clue to Structure
Molecular Weights
– One of the first pieces of information
we try to obtain when determining a
molecular structure is the molecular
formula.
– We can gain some information about
molecular formula from the molecular
weight.
– Mass spectrometry makes it relatively
easy to determine molecular weights.
Exact Molecular Weights
O
CH3(CH2)5CH3
Heptane
CH3CO
Cyclopropyl acetate
Molecular formula
C7H16
C5H8O2
Molecular weight
100
100
Exact mass
100.1253
100.0524
– Mass spectrometry can measure exact masses.
– Therefore, mass spectrometry can be used to
distinguish between molecular formulas.
The “Nitrogen Rule”
– Molecules containing atoms limited to C,H,O,N,S,X,P of
even-numbered molecular weight contain either NO
nitrogen or an even number of N.
– This is true as well for radicals as well.
– Not true for pre-charged, e.g. quats, (rule inverts) or
radical cations.
– In the case of Chemical Ionization, where [M+H]+ is
observed, need to subtract 1, then apply nitrogen rule.
– Example, if we know a compound is free of nitrogen and
gives an ion at m/z=201, then that peak cannot be the
molecular ion.
The Nitrogen Rule
– A molecule with an odd
number of nitrogens has
an odd molecular weight.
– A molecule that contains O N
2
only C, H, and O or
which has an even number
of nitrogens has an even
molecular weight.
O2N
NH2
93
NH2
138
NO2
NH2
183
Index of Hydrogen Deficiency
Degree of Unsaturation
– Relates molecular formulas to multiple bonds and
rings
– For a molecular formula, CcHhNnOoXx, the
degree of unsaturation can be calculated by:
index of hydrogen deficiency = ½ (2c + 2 - h - x + n)
 Molecular Formulas
– Knowing that the molecular formula of a substance is C7H16
tells us immediately that is an alkane because it
corresponds to CnH2n+2
– C7H14 lacks two hydrogens of an alkane, therefore contains
either a ring or a double bond
 Index of Hydrogen Deficiency
index of hydrogen deficiency =
1
2
(molecular formula of alkane – molecular formula of compound)
Example 1
C7H14
Index of hydrogen deficiency
= 1 (molecular formula of alkane –
2
molecular formula of compound)
= 1 (C7H16 – C7H14)
2
= 1 (2) = 1
2
Therefore, one ring or one double bond.
Example 2
C7H12
= 1 (C7H16 – C7H12)
2
= 1 (4) = 2
2
Therefore, two rings, one triple bond,
two double bonds, or one double bond + one ring.
Oxygen has no effect
CH3(CH2)5CH2OH (1-heptanol, C7H16O) has same
number of H atoms as heptane
Index of hydrogen deficiency =
1
2
(C7H16 – C7H16O) = 0
No rings or double bonds
O
CH3CO
Cyclopropyl acetate
Index of hydrogen deficiency =
1 (C H – C H O )
5 12
5 8 2
2
=2
One ring plus one double bond
If halogen is present
Treat a halogen as if it were hydrogen.
H
Cl
C
H
C
CH3
C3H5Cl
same index of hydrogen
deficiency as for C3H6
Rings versus Multiple Bonds
– Index of hydrogen deficiency tells us the sum of
rings plus multiple bonds.
– Using catalytic hydrogenation, the number of
multiple bonds can be determined.
Interpretation of Mass Spectra
 Select a candidate peak for the molecular ion (M+)
 Examine
spectrum
for
peak
characteristic isotopic patterns
clusters
of
 Test (M+) peak candidate by searching for other
peaks correspond to reasonable losses
 Look for characteristic low-mass fragment ions
 Compare spectrum to reference spectra
Isotopic
Abundances
Formula Matching Basics
• Atomic weights are not integers (except
–
–
14N
12C)
= 14.0031 amu; 11B = 11.0093 amu; 1H = 1.0078 amu
16O = 15.9949 amu; 19F = 18.9984 amu; 56Fe = 55.9349 amu
• Sum of the mass defects depends on composition
– Hydrogen increases mass defect, oxygen decreases it
• Accurate mass measurements narrow down the possible
formulae for a particular molecular weight
– 301 entries (150 formulae) in NIST’02 with nominal MW = 321
– 4 compounds within 0.0016 Da (5 ppm) of 321.1000.
• Mass spectrum and user info complete the picture
– Isotope distributions indicate/eliminate elements (e.g. Cl, Br, Cu)
– User-supplied info eliminates others (e.g. no F, Co, I in reaction)
• Isomers are not distinguished in this analysis