CH 908: Mass Spectrometry Lecture 4 Interpreting Electron Impact Mass Spectra

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CH 908 Mass SpectrometryLecture 4
Interpreting Electron Impact Mass Spectra
ContinuedRecommended Read chapters 8-9 of
McLafferty

• Prof. Peter B. OConnor

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outline

• Ring fragmentations
• Radical site migration
• H-rearrangement saturated
• H-rearrangement unsaturated
• 2H rearrangement
• Displacement (similar to a long-range alpha
  cleavage)
• Elimination
• PE diagrams

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Cyclic species require 2 bond cleavages to
generate fragments.
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Retro Diels-Alder
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Retro Diels-Alder
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Example Predict the fragmentations for this
molecule p-dioxane
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p-dioxane
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88
58
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Fragmentation of Aromatic Molecular Ions

• Many simple aromatic molecular ions fragment by
  elimination of a small, unsaturated molecule by
  breaking the aromatic ring but giving a further,
  stable cyclic ion as a product.
• Examples of small molecules lost include -
• Benzene, C2H2, Pyridine, HCN, Thiophene, HCS,
  Furan, HCO, Phenols, CO, Anilines, HCN

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M.
M-C2H2.
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M-HCN.
M.
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M-HCO
M.
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M.
M-C2H2.
M-C3H3
M-CHS
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H rearrangementsH migration
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Types of Hydrogen Rearrangements
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Double H migrationMclafferty 1 rearrangement
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Long-range rearrangements
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Displacements
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Eliminations
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Even-Electron Ions, CI,

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Even-Electron Ions - 1

• Under EI conditions, M. ions are formed and a
  major fragmentation process is the loss of a
  radical, R., producing an even-electron ion.
• Once a radical has been lost, all subsequent
  fragmentations involve the loss of a molecule to
  form further even-electron ions.
• Under CI conditions, an even-electron ion, such
  as MH, is formed subsequent fragmentations
  involve the loss of a molecule to form further
  even-electron ions.

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Sites of Protonation

• In order to rationalise the fragmentation of MH
  ions, one must consider at which sites in the
  sample molecule the proton is attached. The
  spectrum may then be rationalised in terms of the
  fragmentation of the different types of MH ions.
• In general, protonation occurs on heteroatoms
  having lone pairs of electrons, such as O, N and
  Cl. This frequently followed by charge-induced
  elimination of a molecule containing the
  hetero-atom. Other possible protonation sites
  are aromatic rings and regions of unsaturation.

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Even Electron Ions

• Ephedrine ionised by methane CI may protonate for
  example on the O atom of the OH group
• Protonation on the N atom leads to the loss of
  CH3NH2 by a similar mechanism, yielding an ion of
  m/z 135. Both m/z 148 and 135 are observed in
  the CI spectrum, indicating the presence of OH
  and HNCH3 groups in the molecule.

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Ephedrine EI and CI Spectra

• Ephedrine, RMM 165, gives an EI spectrum
  dominated by the m/z 58 fragment ion and no M.
  ion giving an RMM. Methane CI gives an MH ion
  at m/z 166 and fragments at m/z 148, 135 and 58
  due to protonation on the OH and NHCH3 groups or
  on the aromatic ring respectively

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General Hints for Solving Spectra - 1

• Aromatic or aliphatic? Provisionally identify
  M.
• Check that proposed neutral losses are sensible
• Is N present? Is assignment of M. incorrect?
• Check for isotope peaks for Cl, Br, S, heavy
  metals
• Use I(M1)/I(M.) to estimate number of C
  atoms present
• Postulate a molecular formula and estimate the
  double bond equivalents

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General Hints for Solving Spectra - 2

• Inspect higher mass ions, possibly formed from
  M. in one step, e.g. even mass fragments formed
  in a rearrangment process
• Look for characteristic neutral losses such as 16
  Da, O from ArNO2 or NH2 from an amide, 30 Da,
  CH2O from ArOCH3 and characteristic ions, m/z 30,
  amines, m/z 74 methyl esters, 105/77/51, 91/65/39
  for benzoyl and alkyl benzene compounds
• Do not assume adjacent peaks are due to
  sequential losses of neutrals two or more charge
  sites lead to competing fragmentation routes.

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General Hints for Solving Spectra - 3

• Do not try (initially) to interpret every small
  peak, especially those at low m/z which result
  from sequential fragmentation
• Never postulate the loss of a radical from an
  even electron ion without very good reason
• Use negative evidence as well as positive
  evidence e.g. if there is no peak at m/z 91,
  the sample is unlikely to be an alkyl benzene.

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Practical problems with real world spectra - 1

• Beware of spurious peaks such as the following
• Background peaks from previous samples, pump oil
  or from an air leak, e.g. m/z 40, 32, 28, 18 etc.
• Peaks arising from incomplete removal of common
  solvents such as m/z 83, 85, 87 from CHCl3, m/z
  58, 43 from acetone
• Peaks present due to incomplete reaction leaving
  traces of starting materials in the sample
• Peaks due to homologues, e.g. at 14 m/z units
  above or below the true molecular ion peak

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Practical problems with real world spectra - 1

• Compare your spectrum with that of an authentic
  sample obtained by use of the same ionisation
  technique (probably from a database) but remember
  that an exact match of relative intensities is
  unlikely to be found because of varying mass
  discrimination effects.

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Example 1 Assign each peak
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Example 2 Assign each peak
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Example 3 Assign each peak
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Self Assessment

• Explain how C60 can lose 24 Da. How can benzene
  lose 26 Da?
• Hydrogen atom rearrangments are usually promoted
  by
• Can multiple hydrogen atoms rearrange to generate
  the observed fragments? How?
• Can you generate neutral radical losses from an
  even electron ion? Why?
• Is loss of H2 common?
• What is a distonic ion?
• In MS of oligosaccharides, its common to lose
  several residues due to an internal
  rearrangement. Why is this a problem in
  interpreting the spectra?

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Fini.
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-C2H5
Does this assignment make sense?
-C2H6
M.