Transcript Mass Spectrometry - Science @ Stanislaus

Mass Spectrometry Part 1
Lecture Supplement:
Take one handout from the stage
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Spectroscopy
Why bother with spectroscopy?
•Determine structure of unknown substance
•Verify purity/identity of known substance
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Spectroscopy
What methods are commonly used?
Mass spectrometry (MS)*
molecular formula
Infrared spectroscopy (IR)
functional groups
Nuclear magnetic resonance (NMR)
C/H molecular skeleton
X-ray crystallography*
spatial position of atoms
*Not rigorously a type of spectroscopy
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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
H
N
CH3
CH3
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Mass Spectrometry
The Mass Spectrometer
Fundamental operating principle
Determine mass by manipulating flight path of an ion in a magnetic field
Electron gun
Ionization: X + e-  X+. + 2 eMagnet
Ionization
m/z
m/z too
too
justlarge
small
right
sample
introduction
Measure ion
mass-to-charge ratio
(m/z)
+ -
Accelerator
plates
Detector
fires
Detector
Detector
quiet
Detector
quiet
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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
20.2
20.0
22.0
relative intensity
no peak
90%
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
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The Mass Spectrum
Example: methane CH4 + e-  CH4+. + 2 e-
Relative ion abundance (%)
Base peak: most abundant ion
m/z = (1 x 12) + (4 x 1) = 16
C
H
mass-to-charge ratio (m/z)
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The Mass Spectrum
Alternate data presentation...
m/z (amu) Relative abundance (%)
18
17
16
15
14
13
12
< 0.5
1.1
100.0
85.0
9.2
3.0
1.0
M+2
M+1
M
M-H
M - 2H
M - 3H
M - 4H
14C1H
or 12C3H1H3 or...
13C1H or 12C2H1H
4
3
12C1H
4
4
Molecular ion (M): intact ion of substance being analyzed
Fragment ion: formed by cleavage of one or more bonds on molecular ions
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The Mass Spectrum
Origin of Relative Ion Abundances
M contributors
M+1 contributors
M+2 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.
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The Mass Spectrum
Relative Intensity of Molecular Ion Peaks
Imagine a sample containing 10,000 methane molecules...
Molecule
12C1H
4
13C1H
4
14C1H
4
# in sample
m/z
Relative abundance
9889
12 + (4 x 1) = 16
100%
110
13 + (4 x 1) = 17
(110/9889) x 100% = 1.1%*
~1
14 + (4 x 1) = 18
(1/9889) x 100% = < 0.1%*
*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%)
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Formula from Mass Spectrum
M+1 Contributors
Comparing many mass spectra reveals M+1 intensity  ~1.1% per C in formula
•Examples: C2H6 M = 100%; M+1 = ~2.2%
C6H6 M = 100%; M+1 = ~6.6%
Working backwards gives a useful observation...
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
Other M+1 contributors
•15N (0.37%) and 33S (0.76%) should be considered
•2H (0.015%) and 17O (0.037%) can be ignored
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Formula from Mass Spectrum
M+2 Contributors
Anything useful from intensity of M+2?
Isotopes
Natural abundances
Intensity M : M+2
: 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
32S
Conclusion: Mass spectra of molecules with S, Cl, or Br have significant M+2 peaks
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Formula from Mass Spectrum
M+2 Contributors
CH3
H
C
Cl
C3H7Cl
CH3
C H Cl
M: 36 + 7 + 35 = 78
M+2: 36 + 7 + 37 = 80
78
80
M:M+2
abundance
~3:1
m/z
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Formula from Mass Spectrum
Relative abundance (%)
M+2 Contributors
Br
C3H7Br
C H Br
M: 36 + 7 + 79 = 122
M+2: 36 + 7 + 81 = 124
122
124
M:M+2
abundance
~1:1
m/z
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Identifying the Molecular Ions
Which peaks are molecular ions?
•Highest m/z not always M
•M+1 has m/z one more than m/z of M
Br
C7H7Br
M: m/z = 170
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Formula from Mass Spectrum
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
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Mass Spectrometry Part 2
Lecture Supplement:
Take one handout from the stage
17
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, M+1, M+2, 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 M+1/1.1% gives approximate number of carbons
•M+2 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
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Mass Spectrum  Formula  Structure
How do we derive structure from the mass spectrum?
?
CH3

H
C
Cl
CH3
•Not trivial to do this directly
•Structure comes from formula; formula comes from mass spectrum
CH3

C3H7Cl

H
C
Cl
CH3
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Mass Spectrum  Formula  Structure
How do we derive formula from the mass spectrum?
•m/z and relative intensities of M, M+1, and M+2
M
M: m/z = 78
C2H6O3
C3H7Cl
C5H4N
C6H6
etc.
•A few useful rules to narrow the choices
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How Many Nitrogen Atoms?
Consider these molecules:
O
CH3
O2N
NH3
NO2
H3 C
N
N
H2NNH2
O
NO2
CH3
N
N
CH3
Formula:
NH3
N2H4
C7H5N3O6
C8H10N4O2
m/z (M):
17
32
227
194
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
}The Nitrogen Rule
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How Many Nitrogen Atoms?
A Nitrogen Rule Example
Example: Formula choices from previous mass spectrum
discarded
M: m/z = 78
C2H6O3
m/z even
even nitrogen count
C3H7Cl
even nitrogen count
C5H4N
odd nitrogen count
C6H6
even nitrogen count
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How Many Hydrogen Atoms?
One pi bond
Two pi bonds
HC
C6H14
max H for 6 C
C6H12
H count = max - 2
C
C6H10
H count = max - 4
Conclusion: Each pi bond reduces max hydrogen count by two
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How Many Hydrogen Atoms?
C6H14
max H for 6 C
One ring
Two rings
C6H12
H count = max - 2
C6H10
H count = max - 4
Conclusion: Each ring reduces max hydrogen count by two
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How Many Hydrogen Atoms?
One nitrogen
Two nitrogens
NH2
NH2
NH2
CH3
N
H3C
N
N
CH3
CH3
C6H14
max H for 6 C
C6H15N
H count = max + 1
C6H16N2
H count = max + 2
Conclusion:
•Each nitrogen increases max H count by one
•For C carbons and N nitrogens, max number of H = 2C + N + 2
The Hydrogen Rule
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Mass Spectrum  Formula
Procedure
•Chem 14C atoms: H C N O F S Cl Br I
•M = molecular weight (lowest mass isotopes)
•M+1: gives carbon count
•M+2: 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
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Mass Spectrum Formula
Given information
Example #1
m/z Molecular ion Relative abundance
Conclusions
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
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Mass Spectrum Formula
Example #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
1
0
30 - 16 - 0 = 14
C6H14O
Reasonable
2
0
30 - 32 - 0 = -2
C6H-2O2
Not possible
0
2*
30 - 0 - 28 = 2
C6H2N2
Reasonable
Violates hydrogen rule
*Nitrogen rule!
•Other data (functional groups from IR, NMR integration, etc.) further trims the list
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Mass Spectrum Formula
Example #2
m/z Molecular ion Relative abundance
Conclusions
157
M
100%
Mass (lowest isotopes) = 157
Odd number of nitrogens
158
M+1
9.39%
9.39 / 1.1 = 8.5
Eight or nine carbons
159
M+2
34%
One Cl; no S or Br
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Mass Spectrum Formula
Example #2
Try eight carbons: M - C8 - Cl = 157 - (8 x 12) - 35 = 26 amu for O, N, and H
Oxygens
0
Nitrogens
1*
26 - O - N = H
Formula
Notes
26 - 0 - 14 = 12
C8H12ClN Reasonable
*Nitrogen rule!
Not enough amu available for one oxygen/one nitrogen or no oxygen/three nitrogens
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Mass Spectrum Formula
Example #2
Try nine carbons: M - C9 - Cl = 157 - (9 x 12) - 35 = 14 amu for O, N, and H
Oxygens
0
Nitrogens
1*
14 - O - N = H
Formula
Notes
14 - 0 - 14 = 0
C9ClN
Reasonable
*Nitrogen rule!
Not enough amu available for any other combination.
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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
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Formula Structure
Calculating DBE
DBE may be calculated from molecular formula:
hydrogens and halogens
nitrogens
DBE = C - H + N + 1
2
2
carbons
•One DBE = one ring or one pi bond
•Two DBE = two pi bonds, two rings, or one of each
•Four DBE = possible benzene ring
Example
C8H10ClN
DBE = C - (H/2) + (N/2) + 1
= 8 - [(10+1)/2] + (1/2) + 1
=4
Four pi bonds and/or ring
Possible benzene ring
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Formula Structure
Common 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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