Mass Spectrometry Part 1. Lecture Supplement: Take one handout from the stage. Determine structure of unknown substance. Verify purity/identity of known substance. Spectroscopy. Why bother with spectroscopy?. Mass spectrometry (MS)* molecular formula. Infrared spectroscopy (IR)
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Determine structure of unknown substance
Why bother with spectroscopy?
molecular formula
Infrared spectroscopy (IR)
functional groups
Nuclear magnetic resonance (NMR)
C/H molecular skeleton
Xray crystallography* spatial position of atoms
SpectroscopyWhat methods are commonly used?*Not rigorously a type of spectroscopy
Fundamental operating principle
Determine mass by manipulating flight path of an ion in a magnetic field
Electron gun
Magnet
Ionization
Measure ion
sample
m/z too small
m/z too large
masstocharge ratio
m/z just right
introduction
(m/z)

+
Accelerator
plates
Detector
Detector quiet
Detector fires
Detector quiet
Mass SpectrometryThe Mass SpectrometerIonization: X + e X+. + 2 e
Isotopes: atoms with same number of protons and same number of electrons but different numbers of neutrons
Resultsm/zrelative intensity
20.2 no peak
20.0 90%
22.0 10%
C
H
Base peak: most abundant ion
The Mass SpectrumExample: methane CH4 + e CH4+. + 2 e
Relative ion abundance (%)
masstocharge ratio (m/z)
m/z (amu)Relative abundance (%)
< 0.5
17 1.1
16 100.0
85.0
9.2
3.0
12 1.0
M  2H
M  3H
M  4H
The Mass SpectrumAlternate data presentation...
M+2
14C1H4 or 12C3H1H3 or...
M+1
13C1H4 or 12C2H1H3
M
12C1H4
M  H
Molecular ion (M): intact ion of substance being analyzed
Fragment ion: formed by cleavage of one or more bonds on molecular ions
This table will be provided on an exam. Do not memorize it.
Imagine a sample containing 10,000 methane molecules...
Molecule# in samplem/zRelative abundance
12C1H4
9889
12 + (4 x 1) = 16
100%
110
13 + (4 x 1) = 17
(110/9889) x 100% = 1.1%*
13C1H4
~1
14 + (4 x 1) = 18
(1/9889) x 100% = < 0.1%*
14C1H4
*Contributions from ions with 2H are ignored because of its very small natural abundance
CH4 mass spectrum
m/z = 16 (M; 100%), m/z = 17 (M+1; 1.1%), m/z = 18 (M+2; < 0.1%)
When relative contribution of M = 100% then relative abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
Formula from Mass SpectrumM+1 ContributorsAnything useful from intensity of M+2?
IsotopesNatural abundancesIntensity M : M+2
32S : 34S
95.0 : 4.2
100 : 4.4
35Cl : 37Cl
75.8 : 24.2
100 : 31.9
79Br : 81Br
50.7 : 49.3
100 : 97.2
Conclusion: Mass spectra of molecules with S, Cl, or Br have significant M+2 peaks
M: 36 + 7 + 35 = 78
Relative abundance (%)
78
80
m/z
Formula from Mass SpectrumM+2 ContributorsC3H7Cl
M+2: 36 + 7 + 37 = 80
M:M+2
abundance
~3:1
M: 36 + 7 + 79 = 122
122
124
Formula from Mass SpectrumM+2 ContributorsC3H7Br
Relative abundance (%)
M+2: 36 + 7 + 81 = 124
M:M+2
abundance
~1:1
m/z
C7H7Br
M: m/z = 170
Summary of Information from Mass Spectrum
M: Reveals mass of molecule composed of lowest mass isotopes
M+1: Intensity of M+1 / 1.1% = number of carbons
M+2: Intensity reveals presence of sulfur, chlorine, and bromine
Next lecture: procedure for deriving formula from mass spectrum
C3H7Cl
Mass Spectrum Formula StructureHow do we derive structure from the mass spectrum?
C2H6O3
C3H7Cl
C5H4N
C6H6
etc.
M
Mass Spectrum Formula StructureNH3
N2H4
C7H5N3O6
C8H10N4O2
m/z (M):
17
32
227
194
}
The Nitrogen Rule
How Many Nitrogen Atoms?Consider these molecules:
NH3
H2NNH2
odd nitrogen count
even nitrogen count
even nitrogen count
even nitrogen count
discarded
How Many Nitrogen Atoms?A Nitrogen Rule ExampleExample: Formula choices from previous mass spectrum
M: m/z = 78
C2H6O3
C3H7Cl
C5H4N
C6H6
C6H10
H count = max  2
H count = max  4
How Many Hydrogen Atoms?One pi bond
Two pi bonds
C6H14
max H for 6 C
Conclusion: Each pi bond reduces max hydrogen count by two
C6H14
max H for 6 C
C6H10
H count = max  2
H count = max  4
How Many Hydrogen Atoms?One ring
Two rings
Conclusion: Each ring reduces max hydrogen count by two
C6H14
max H for 6 C
C6H16N2
H count = max + 1
H count = max + 2
The Hydrogen Rule
How Many Hydrogen Atoms?One nitrogen
Two nitrogens
Conclusion:
m/zMolecular ionRelative abundanceConclusions
102 M 100%
Mass (lowest isotopes) = 102
Even number of nitrogens
103 M+1 6.9%
6.9 / 1.1 = 6.3 Six carbons*
104 M+2 0.38%
< 4% so no S, Cl, or Br
Oxygen?
*Rounding: 6.00 to 6.33 = 6; 6.34 to 6.66 = 6 or 7; 6.67 to 7.00 = 7
Mass (M)  mass (C, S, Cl, Br, F, and I) = mass (N, O, and H)
102  C6
= 102  (6 x 12)
= 30 amu for N, O, and H
OxygensNitrogens30  O  N = HFormulaNotes
0
0
30  0  0 = 30
C6H30
Violates hydrogen rule
1
0
30  16  0 = 14
C6H14O
Reasonable
2
0
30  32  0 = 2
C6H2O2
Not possible
0
30  0  28 = 2
C6H2N2
Reasonable
2*
*Nitrogen rule!
m/zMolecular ionRelative abundanceConclusions
157 M 100%
158 M+1 9.39%
159 M+2 34%
Mass Spectrum FormulaExample #2Mass (lowest isotopes) = 157
Odd number of nitrogens
9.39 / 1.1 = 8.5
Eight or nine carbons
One Cl; no S or Br
*Nitrogen rule!
Mass Spectrum FormulaExample #2Try eight carbons: M  C8  Cl = 157  (8 x 12)  35 = 26 amu for O, N, and H
OxygensNitrogens26  O  N = HFormulaNotes
0
26  0  14 = 12
C8H12ClN
Reasonable
Not enough amu available for one oxygen/one nitrogen or no oxygen/three nitrogens
*Nitrogen rule!
Mass Spectrum FormulaExample #2Try nine carbons: M  C9  Cl = 157  (9 x 12)  35 = 14 amu for O, N, and H
OxygensNitrogens14  O  N = HFormulaNotes
0
14  0  14 = 0
C9ClN
Reasonable
Not enough amu available for any other combination.
nitrogens
H
N
DBE = C 
+
+ 1
2
2
carbons
Example
C8H10ClN
Formula StructureCalculating DBEDBE may be calculated from molecular formula:
DBE = C  (H/2) + (N/2) + 1
= 8  [(10+1)/2] + (1/2) + 1
Four pi bonds and/or ring
Possible benzene ring
= 4
Next lecture: Infrared spectroscopy part 1