CH 908: Mass Spectrometry Lecture 4 Interpreting Electron Impact Mass Spectra – Continued… Recommended: Read chapters 8-9 of McLafferty Prof. Peter B. O’Connor

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

Cyclic species require ≥2 bond cleavages to generate fragments.

Retro Diels-Alder

Example: Predict the fragmentations for this molecule: p-dioxane

p-dioxane

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

M+. [M-C2H2]+.

M+. [M-HCN]+.

M+. [M-HCO]+

M+. [M-C2H2]+. [M-C3H3]+ [M-CHS]+

H – rearrangements H migration

Types of Hydrogen Rearrangements

Double H migration “Mclafferty +1” rearrangement

Long-range rearrangements

Displacements

Eliminations

Even-Electron Ions, CI,

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.

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.

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. 

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

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([M+1]+)/I([M]+.) to estimate number of C atoms present  Postulate a molecular formula and estimate the double bond equivalents

 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. General Hints for Solving Spectra - 2

 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. General Hints for Solving Spectra - 3

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

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. Practical problems with “real world” spectra - 1

Example 1: Assign each peak

Example 2: Assign each peak

Example 3: Assign each peak

Self Assessment Explain how C 60 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 H 2 common? What is a ‘distonic ion’? In MS of oligosaccharides, it’s common to lose several residues due to an internal rearrangement. Why is this a problem in interpreting the spectra?

Fini.

[-C2H5]+ M+. -C2H6 Does this assignment make sense?