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Mass spectrometry (MS) 2
Dr. Cristina Minguillón
Mass Spectrometer
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Ion sources
Ionization
method
Typical
Analytes
Sample
Introduction
Mass
Range
Method
Highlights
Electron
El
t
I
Impact
t
(EI)
Relativelyy
small
volatile
GC or
liquid/solid
probe
to
1,000
Daltons
Hard method
til
versatile
provides
structure info
Chemical Ionization
(CI)
Relatively
small
volatile
GC or
liquid/solid
probe
to
1,000
Daltons
Soft method
molecular ion
peak [M+H]+
Electrospray (ESI)
Peptides
Proteins
nonvolatile
Liquid
Chromatography
or syringe
or syringe
to
200,000
Daltons
Soft method
ions often
multiply
charged
Fast Atom
Bombardment
(FAB)
Carbohydrates
Organometallics
Peptides
nonvolatile
Sample mixed
in viscous
matrix
to
6,000
Daltons
Soft method
but harder
than ESI or
MALDI
Matrix Assisted
Laser Desorption
(MALDI)
Peptides
Proteins
Nucleotides
Sample mixed
in solid
matrix
to
500,000
Daltons
Soft method
very high
mass
Mass analyzers
Analyzer
System Highlights
Quadrupole
Unit mass resolution, fast scan,
low cost
Sector (Magnetic and/or
Electrostatic)
High resolution, exact mass
Time-of-Flight (TOF)
Theoretically, no limitation for m/z
maximum, high throughput
Ion Cyclotron Resonance (ICR)
Very high resolution, exact mass,
perform ion chemistry
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Examples
Mass Spectrometry
¾Molecular ion (M .+): If the molecular ion appears, it will be the highest mass in an
EI spectrum (except for isotope peaks). This peak will represent the molecular
weight of the compound.
Its appearance depends on the stability of the
compound. Double bonds, cyclic structures and aromatic rings stabilize the
molecular ion and increase the probability of its appearance.
¾Reference Spectra: Mass spectral patterns are reproducible. The mass spectra of
many compounds have been published and may be used to identify unknowns.
Instrument computers generally contain spectral libraries which can be searched for
matches.
¾Fragmentation: General rules of fragmentation exist and are helpful to predict or
interpret the fragmentation pattern produced by a compound.
compound Functional groups and
overall structure determine how some portions of molecules will resist fragmenting,
while other portions will fragment easily. Further information may be found in mass
spectrometry reference books.
¾Isotopes: Isotopes occur in compounds analyzed by mass spectrometry in the
same abundances that they occur in nature.
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Examples
Relative isotopic abundance of elements
Element Isotope
Carbon
Hydrogen
Nitrogen
Oxygen
Sulfur
Chl i
Chlorine
Bromine
12C
1H
14N
16O
32S
35Cl
79Br
Relative
Relative
Relative
Isotope
Isotope
Abundance
Abundance
Abundance
13C
100
1.11
2
100
H
0.016
15N
100
0.38
17
18O
100
O
0.04
0.20
33
34
100
S
0.78
S
4.40
37Cl
100
32 5
32.5
81
100
Br
98.0
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Examples: alcohols
3-Pentanol
C5H12O
MW = 88.15
An alcohol's molecular ion is small or nonexistent. Cleavage of the C-C bond next to
the oxygen usually occurs. A loss of H2O
may occur
Examples: aldehydes
3-Phenyl-2-propenal
C9H8O
MW = 132.16
Cleavage of bonds next to the carboxyl
group results in the loss of hydrogen
(molecular ion less 1) or the loss of CHO
(molecular ion less 29).
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Examples: alkanes
Hexane
C6H14
MW = 86.18
Molecular ion peaks are present, possibly
with low intensity. The fragmentation
pattern contains clusters of peaks 14 mass
units apart (which represent loss of
(CH2)nCH3).
Examples: amides
3-Methylbutyramide
C5H11NO
MW = 101.15
Primary amides show a base peak due to
the McLafferty rearrangement
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Examples: amines
n-Butylamine
C4H11N
MW = 73.13
Molecular ion peak is an odd
number. Alpha-cleavage dominates
aliphatic amines
Examples: amines
n-Methylbenzylamine
C8H11N
MW = 121.18
Another example is a secondary amine
shown below. Again, the molecular ion peak
is an odd number. The base peak is from
the C-C cleavage adjacent to the C-N bond.
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Examples: aromatic
Naphthalene
C10H8
MW = 128.17
Molecular ion peaks are strong due to the
stable structure.
Examples: carboxilic acids
2-Butenoic acid
C4H6O2
MW = 86.09
In short chain acids, peaks due to the loss
of OH (molecular ion less 17) and COOH
(molecular ion less 45) are prominent due to
cleavage of bonds next to C=O.
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Examples: esters
Ethyl acetate
C4H8O2
MW = 88.11
Fragments appear due to bond cleavage
next to C=O (alkoxy group loss, -OR) and
hydrogen rearrangements.
Examples: ethers
Ethyl methyl ether
C3H8O
MW = 60.10
Fragmentation tends to occur alpha to the
oxygen atom (C-C bond next to the oxygen).
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Examples: halides
1-Bromopropane
C3H7Br
MW = 123.00
The presence of chlorine or bromine atoms
is usually recognizable from isotopic peaks.
Examples: ketones
4-Heptanone
C7H14O
MW = 114.19
Major fragmentation peaks result from
cleavage of the C-C bonds adjacent to the
carbonyl.
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Steps to interpret a MS
1. Look for the molecular ion peak.
This peak (if it appears) will be the highest mass peak in the spectrum, except for
isotope peaks.
Nominal MW (meaning
(meaning=rounded
rounded off) will be an even number for compounds
containing only C, H, O, S, Si.
Nominal MW will be an odd number if the compound also contains an odd number
of N (1,3,...).
2. Try to calculate the molecular formula:
The isotope peaks can be very useful. Carbon 12 has an isotope, carbon 13. Their
abundances are 12C=100%, 13C=1.1%. This means that for every 100 (12)C atoms
there are 1.1
1 1 ((13))C atoms.
atoms If a compound contains 6 carbons,
carbons then each atom
has a 1.1% abundance of (13)C. Therefore, if the molecular ion peak is 100%,
then the isotope peak (1 mass unit higher) would be 6x1.1%=6.6%.
If the molecular ion peak is not 100% then you can calculate the relative
abundance of the isotope peak to the ion peak. For example, if the molecular
ion peak were 34% and the isotope peak 2.3%: (2.3/34)x100 = 6.8%. 6.8% is
the relative abundance of the isotope peak to the ion peak. Next, divide the
relative abundance by the isotope abundance: 6.8/1.1=6 carbons.
Steps to interpret a MS
Follow this order when looking for information provided by isotopes:
Look for A+2 elements: O, Si, S, Cl, Br Look for A+1 elements: C, N "A" elements: H, F, P, I 3. Calculate the total number of rings plus double bonds:
For the molecular formula: CxHyNzOn
rings + double bonds = x - (1/2)y + (1/2)z + 1
4. Postulate the molecular structure consistent with abundance and m/z of
fragments.
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