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Mass Spectrometry Part 1 PowerPoint PPT Presentation


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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Mass Spectrometry Part 1

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Mass spectrometry part 1 l.jpg

Mass Spectrometry Part 1

Lecture Supplement:

Take one handout from the stage


Spectroscopy l.jpg

  • Determine structure of unknown substance

  • Verify purity/identity of known substance

Spectroscopy

Why bother with spectroscopy?


Spectroscopy what methods are commonly used l.jpg

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

SpectroscopyWhat methods are commonly used?

*Not rigorously a type of spectroscopy


Spectroscopy4 l.jpg

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


Mass spectrometry the mass spectrometer l.jpg

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

mass-to-charge ratio

m/z just right

introduction

(m/z)

-

+

Accelerator

plates

Detector

Detector quiet

Detector fires

Detector quiet

Mass SpectrometryThe Mass Spectrometer

Ionization: X + e- X+. + 2 e-


Isotopes l.jpg

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

Resultsm/zrelative 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


The mass spectrum l.jpg

m/z = (1 x 12) + (4 x 1) = 16

C

H

Base peak: most abundant ion

The Mass Spectrum

Example: methane CH4 + e- CH4+. + 2 e-

Relative ion abundance (%)

mass-to-charge ratio (m/z)


The mass spectrum8 l.jpg

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 Spectrum

Alternate 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


The mass spectrum origin of relative ion abundances l.jpg

The Mass SpectrumOrigin of Relative Ion Abundances

This table will be provided on an exam. Do not memorize it.


The mass spectrum relative intensity of molecular ion peaks l.jpg

The Mass SpectrumRelative Intensity of Molecular Ion Peaks

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%)


Formula from mass spectrum m 1 contributors l.jpg

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 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...

  • Other M+1 contributors

  • 15N (0.37%) and 33S (0.76%) should be considered

  • 2H (0.015%) and 17O (0.037%) can be ignored


Formula from mass spectrum m 2 contributors l.jpg

Formula from Mass SpectrumM+2 Contributors

Anything 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


Formula from mass spectrum m 2 contributors13 l.jpg

C H Cl

M: 36 + 7 + 35 = 78

Relative abundance (%)

78

80

m/z

Formula from Mass SpectrumM+2 Contributors

C3H7Cl

M+2: 36 + 7 + 37 = 80

M:M+2

abundance

~3:1


Formula from mass spectrum m 2 contributors14 l.jpg

C H Br

M: 36 + 7 + 79 = 122

122

124

Formula from Mass SpectrumM+2 Contributors

C3H7Br

Relative abundance (%)

M+2: 36 + 7 + 81 = 124

M:M+2

abundance

~1:1

m/z


Identifying the molecular ions l.jpg

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

C7H7Br

M: m/z = 170


Formula from mass spectrum l.jpg

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


Mass spectrometry part 2 l.jpg

Mass Spectrometry Part 2

Lecture Supplement:

Take one handout from the stage


Summary of part 1 l.jpg

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


Mass spectrum formula structure l.jpg

?

C3H7Cl

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


Mass spectrum formula structure20 l.jpg

M: m/z = 78

C2H6O3

C3H7Cl

C5H4N

C6H6

etc.

M

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

  • A few useful rules to narrow the choices


How many nitrogen atoms l.jpg

Formula:

NH3

N2H4

C7H5N3O6

C8H10N4O2

m/z (M):

17

32

227

194

}

The Nitrogen Rule

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


How many nitrogen atoms a nitrogen rule example l.jpg

m/z even

odd nitrogen count

even nitrogen count

even nitrogen count

even nitrogen count

discarded

How Many Nitrogen Atoms?A Nitrogen Rule Example

Example: Formula choices from previous mass spectrum

M: m/z = 78

C2H6O3

C3H7Cl

C5H4N

C6H6


How many hydrogen atoms l.jpg

C6H12

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


How many hydrogen atoms24 l.jpg

C6H12

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


How many hydrogen atoms25 l.jpg

C6H15N

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:

  • Each nitrogen increases max H count by one

  • For C carbons and N nitrogens, max number of H = 2C + N + 2


Mass spectrum formula l.jpg

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


Mass spectrum formula example 1 l.jpg

Given information

Mass Spectrum FormulaExample #1

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 spectrum formula example 128 l.jpg

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

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

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


Mass spectrum formula example 2 l.jpg

m/zMolecular ionRelative abundanceConclusions

157 M 100%

158 M+1 9.39%

159 M+2 34%

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


Mass spectrum formula example 230 l.jpg

1*

*Nitrogen rule!

Mass Spectrum FormulaExample #2

Try 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


Mass spectrum formula example 231 l.jpg

1*

*Nitrogen rule!

Mass Spectrum FormulaExample #2

Try 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.


Formula structure l.jpg

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


Formula structure calculating dbe l.jpg

hydrogens and halogens

nitrogens

H

N

DBE = C -

+

+ 1

2

2

carbons

Example

C8H10ClN

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 - [(10+1)/2] + (1/2) + 1

Four pi bonds and/or ring

Possible benzene ring

= 4


Formula structure common math errors l.jpg

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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