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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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
C H Cl abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
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
C H Br abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
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
Lecture Supplement:
Take one handout from the stage
? abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
C3H7Cl
Mass Spectrum Formula StructureHow do we derive structure from the mass spectrum?
M: m/z = 78 abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
C2H6O3
C3H7Cl
C5H4N
C6H6
etc.
M
Mass Spectrum Formula StructureFormula abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula:
NH3
N2H4
C7H5N3O6
C8H10N4O2
m/z (M):
17
32
227
194
}
The Nitrogen Rule
How Many Nitrogen Atoms?Consider these molecules:
NH3
H2NNH2
m/z even abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
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
C abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula6H12
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
C abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula6H12
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
C abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula6H15N
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:
Given information abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
Mass Spectrum FormulaExample #1m/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/z abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formulaMolecular 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
1* abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
*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
1* abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
*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.
hydrogens and halogens abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula
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