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Mass Spectrometry Part 1

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Mass Spectrometry Part 1 Lecture Supplement: Take one handout from the stage Spectroscopy Spectroscopy What methods are commonly used? Spectroscopy Mass Spectrometry ... – PowerPoint PPT presentation

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Title: Mass Spectrometry Part 1


1
Mass Spectrometry Part 1
  • Lecture Supplement
  • Take one handout from the stage

2
Spectroscopy
Why bother with spectroscopy?
3
SpectroscopyWhat methods are commonly used?
Not rigorously a type of spectroscopy
4
Spectroscopy
  • Example unidentified white powder
  • MS C10H15N
  • IR benzene ring, secondary amine (R2NH)
  • NMR has CH2-CH-CH3
  • X-ray not necessary in this case

5
Mass SpectrometryThe Mass Spectrometer
Ionization X e- ? X. 2 e-
6
Isotopes
Isotopes atoms with same number of protons and
same number of electrons but different numbers of
neutrons
  • Aston mass spectrum of neon (1919)
  • Ne empirical atomic weight 20.2 amu
  • Ne mass spectrum predict single peak at m/z
    20.2

Results m/z relative intensity
20.2 no peak
20.0 90 22.0
10
  • Conclusions
  • Neon is a mixture of isotopes
  • Weighted average (90 x 20.0 amu) (10.0 x
    22.0 amu) 20.2 amu
  • Nobel Prize in Chemistry 1922 to Aston for
    discovery of stable element isotopes

7
The Mass Spectrum
Example methane CH4 e- ? CH4. 2 e-
Relative ion abundance ()
mass-to-charge ratio (m/z)
8
The Mass Spectrum
Alternate data presentation...
M2
14C1H4 or 12C3H1H3 or...
M1
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
9
The Mass SpectrumOrigin of Relative Ion
Abundances
M contributors M contributors M1 contributors M1 contributors M2 contributors M2 contributors
Isotope Natural Abundance Isotope Natural Abundance Isotope Natural Abundance
1H 99.9855 2H 0.015 3H ppm
12C 98.893 13C 1.107 14C ppm
14N 99.634 15N 0.366
16O 99.759 17O 0.037 18O 0.204
19F 100.0
32S 95.0 33S 0.76 34S 4.22
35Cl 75.77 37Cl 24.23
79Br 50.69 81Br 49.31
127I 100.0
This table will be provided on an exam. Do not
memorize it.
10
The Mass SpectrumRelative Intensity of Molecular
Ion Peaks
Imagine a sample containing 10,000 methane
molecules...
Molecule in sample m/z
Relative 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 lt 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
(M1 1.1), m/z 18 (M2 lt 0.1)
11
Formula from Mass SpectrumM1 Contributors
  • Comparing many mass spectra reveals M1 intensity
    ? 1.1 per C in formula
  • Examples C2H6 M 100 M1 2.2
  • C6H6 M 100 M1 6.6
  • Working backwards gives a useful observation...
  • Other M1 contributors
  • 15N (0.37) and 33S (0.76) should be considered
  • 2H (0.015) and 17O (0.037) can be ignored

12
Formula from Mass SpectrumM2 Contributors
Anything useful from intensity of M2?
Isotopes Natural abundances
Intensity M M2
32S 34S
100 4.4
95.0 4.2
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 M2 peaks
13
Formula from Mass SpectrumM2 Contributors
C3H7Cl
M2 36 7 37 80
MM2 abundance 31
14
Formula from Mass SpectrumM2 Contributors
C3H7Br
Relative abundance ()
M2 36 7 81 124
MM2 abundance 11
m/z
15
Identifying the Molecular Ions
  • Which peaks are molecular ions?
  • Highest m/z not always M
  • M1 has m/z one more than m/z of M

C7H7Br M m/z 170
16
Formula from Mass Spectrum
Summary of Information from Mass Spectrum
M Reveals mass of molecule composed of
lowest mass isotopes M1 Intensity of M1 / 1.1
number of carbons M2 Intensity reveals
presence of sulfur, chlorine, and bromine
Next lecture procedure for deriving formula from
mass spectrum
17
Mass Spectrometry Part 2
  • Lecture Supplement
  • Take one handout from the stage

18
Summary of Part 1
  • Spectroscopy Study of the interaction of photons
    and matter
  • Useful to determine molecular structure
  • Types MS, IR, NMR, x-ray crystallography
    not really spectroscopy
  • MS fundamental principle Manipulate flight path
    of ion in magnetic field
  • Charge (z), magnetic field strength are known
    ion mass (m) is determined
  • Isotopes Natural abundance of isotopes controls
    relative abundance of ions
  • Molecular ion (M, M1, M2, etc.) Intact ion of
    substance being analyzed
  • m/z of M molecular mass composed of lowest mass
    isotopes 1H, 12C, 35Cl, etc.
  • Relative abundance of M1/1.1 gives approximate
    number of carbons
  • M2 reveals presence of sulfur, chlorine, or
    bromine
  • Fragment ion From decomposition of molecular ion
    before reaching detector
  • Analysis of fragmentation patterns not important
    for Chem 14C

19
Mass Spectrum ? Formula ? Structure
How do we derive structure from the mass spectrum?
  • Not trivial to do this directly
  • Structure comes from formula formula comes from
    mass spectrum

20
Mass Spectrum ? Formula ? Structure
  • How do we derive formula from the mass spectrum?
  • m/z and relative intensities of M, M1, and M2
  • A few useful rules to narrow the choices

21
How Many Nitrogen Atoms?
Consider these molecules
NH3
H2NNH2
  • Conclusion
  • When m/z (M) even, number of N in formula is
    even
  • When m/z (M) odd, number of N in formula is odd

22
How Many Nitrogen Atoms?A Nitrogen Rule Example
Example Formula choices from previous mass
spectrum
M m/z 78 C2H6O3 C3H7Cl C5H4N C6H6
23
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
24
How Many Hydrogen Atoms?
One ring
Two rings
Conclusion Each ring reduces max hydrogen count
by two
25
How Many Hydrogen Atoms?
One nitrogen
Two nitrogens
Conclusion
  • Each nitrogen increases max H count by one
  • For C carbons and N nitrogens, max number of H
    2C N 2

26
Mass Spectrum ?? Formula
  • Procedure
  • Chem 14C atoms H C N O F S Cl Br I
  • M molecular weight (lowest mass isotopes)
  • M1 gives carbon count
  • M2 presence of S, Cl, or Br
  • No mass spec indicator for F, I Assume absent
    unless otherwise specified
  • Accounts for all atoms except O, N, and H
  • MW - mass due to C, S, Cl, Br, F, and I mass
    due to O, N, and H
  • Systematically vary O and N to get formula
    candidates
  • Trim candidate list with nitrogen rule and
    hydrogen rule

27
Mass Spectrum?? FormulaExample 1
m/z Molecular ion Relative abundance
Conclusions
102 M 100
Mass (lowest isotopes) 102 Even number of
nitrogens
103 M1 6.9
6.9 / 1.1 6.3 Six carbons
104 M2 0.38
lt 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
28
Mass Spectrum?? FormulaExample 1
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
Oxygens Nitrogens 30 - O - N H
Formula Notes
0
0
30 - 0 - 0 30
C6H30
Violates hydrogen rule
1
0
30 - 16 - 0 14
C6H14O
Reasonable
2
0
30 - 32 - 0 -2
C6H-2O2
Not possible
0
30 - 0 - 28 2
C6H2N2
Reasonable
2
Nitrogen rule!
  • Other data (functional groups from IR, NMR
    integration, etc.) further trims the list

29
Mass Spectrum?? FormulaExample 2
Mass (lowest isotopes) 157 Odd number of
nitrogens
9.39 / 1.1 8.5 Eight or nine carbons
One Cl no S or Br
30
Mass Spectrum?? FormulaExample 2
Try eight carbons M - C8 - Cl 157 - (8 x 12) -
35 26 amu for O, N, and H
Oxygens Nitrogens 26 - O - N H
Formula Notes
0
26 - 0 - 14 12
C8H12ClN
Reasonable
Not enough amu available for one oxygen/one
nitrogen or no oxygen/three nitrogens
31
Mass Spectrum?? FormulaExample 2
Try nine carbons M - C9 - Cl 157 - (9 x 12) -
35 14 amu for O, N, and H
Oxygens Nitrogens 14 - O - N H
Formula Notes
0
14 - 0 - 14 0
C9ClN
Reasonable
Not enough amu available for any other
combination.
32
Formula?? Structure
  • What does the formula reveal about molecular
    structure?
  • Functional groups
  • Absent atoms may eliminate some functional groups
  • Example C7H9N has no oxygen-containing
    functional groups
  • Pi bonds and rings
  • Recall from previous one pi bond or one ring
    reduces max H count by two
  • Each two H less than max H count double bond
    equivalent (DBE)
  • If formula has less than full H count, molecule
    must contain one pi bond or ring

33
Formula?? StructureCalculating DBE
DBE may be calculated from molecular formula
  • One DBE one ring or one pi bond
  • Two DBE two pi bonds, two rings, or one of each
  • Four DBE possible benzene ring

DBE C - (H/2) (N/2) 1 8 - (101)/2
(1/2) 1
Four pi bonds and/or ring Possible benzene ring
4
34
Formula?? StructureCommon Math Errors
  • Small math errors can have devastating effects!
  • No calculators on exams
  • Avoid these common spectroscopy problem math
    errors
  • Divide by 1.1 ? divide by 1.0
  • DBE cannot be a fraction
  • DBE cannot be negative

Next lecture Infrared spectroscopy part 1
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