CH 908: Mass SpectrometryLecture 2

Interpreting Electron Impact Mass Spectra

Recommended: Read chapters 3-5 of McLafferty

Prof. Peter B. O’Connor

Objectives for this lecture• The Information contained in isotopic peak distributions• The nitrogen rule• Rings + double bonds• Charge localization, Radical localization• Solving a mass spectrum:

– Where do you start?– Is it the molecular ion?– Small neutral losses– Characteristic peaks– Characteristic series– Loss of the largest alkane radical– Calculating expected peaks

• The odd/even electron rule• What controls peak abundance?

Uses of Isotope Peaks Common elements that give M+2 isotope peaks:

35Cl:37Cl rel. ab. ~ 3 : 1 79Br:81Br rel. ab. ~ 1 : 1 32S:34S rel. ab. ~ 100 : 4 28Si:30Si rel. ab. ~ 100 : 3.4

Hence peaks at M+2, M+4, etc. indicate the presence of Cl, Br, S, Si; the absence of these peaks indicates the absence of these elements.

Common elements that give rise to M+1 isotope peaks are C and N but only C isotope peaks need be considered: 12C:13C rel. ab. ~ 100 : 1.1

So that I([M+1]+)/I([M+]) = n x 1.1/100 for an ion containing n C atoms.

Approximate ratio 27:27:9:1

Approximate ratio9:6:1

Approximate ratio 3:1

81:108:54:12:1

27:27:9:1

9:6:1

1:4:6:4:1

1:3:3:1

C12H22S2+.

m/z 230 231 232 Intensities 100 13 9

[M-(C6H10)]+m/z 148 149 150 Intensities 100 7 9

Initial Inspection of the Spectrum - 1

Look at the overall appearance of the spectrum: try to identify the molecular ion, M+. and obtain information from any isotope peaks present.

Nitrogen Rule

If the major peaks are at low m/z and M+. is under 20% of the most intense peaks, the sample is probably aliphatic.

The more intense M+. is, the greater the degree of unsaturation is present (alkene, carbonyl compound).

If sufficient mass accuracy is available, calculate the possible elemental composition(s).

Calculate the R+DB value(s).

The nitrogen rule

Odd electron ions: a molecule containing the elements C, H, O, N, S or halogen has an odd nominal mass if it contains an odd number of nitrogen atoms.

Even electron ions: a molecule containing the elements C, H, O, N, S or halogen has an odd nominal mass if it contains an Even number of nitrogen atoms.

Caveats: 1. no metals please!2. mass “defects” eventually accumulate to > 1 Da, inverting the rule

Rings plus Double Bonds

What elemental compositions are realistic chemically?

Because of basic valence orbital arrangments, a simple equation can be used to calculate the number of double bonds (or rings) in a molecule.

X - Y/2 + Z/2 + 1 = R+DBX = carbon, siliconY = hydrogen, chlorine, fluorine, etc.Z = nitrogen, phosphorus

Values ending in ½ correspond to even electron ions.Values lower than –½ are not possible chemically.

Is it the Molecular ion?

For Electron Impact:

1.The MI must have the highest mass in the spectrum (except isotopes)

2.The MI must be an odd electron ion, M+●

3.The MI must be able to fragment to generate sensible high mass odd-electron fragments.

Is it the Molecular ion?

Is it the Molecular ion?

C10H15O C10H14OC9H12OC10H13

C8H10O

Example #1:

Is it the Molecular ion?

C10H14 C10H13

C9H11

C8H9

C7H8

C7H7

Example #2:

Is it Really the Molecular Ion?

Try to identify the main species lost by M+.. These often indicate the type of compound to which the sample belongs.

Watch out for common adducts (Na, contaminants, CI reagent gas)

Rearrangement ions formed by loss of a molecule are often particularly informative. If no nitrogen is present, these appear at an even value of m/z.

Identify ions characteristic of a compound type: m/z 105, 77, 51 for benzoyl compounds, m/z 91, 65, 39 for alkylbenzenes, m/z 30 for amines, etc.

M+. absent

[M-CH3]+

[M-H2O]+.

[M-C3H7]+

[M-C4H9]+

M+.

C3H7CO+

C2H5CO+

[M-CH3]+ M+.

[M-C3H6]+.

[CH3CO]+

M+.

C7H7+

C5H5+C3H3+

Fragmentation Patterns (Alkanes)

M+.

Notes:Electron impact, thus…1. Even MI = even number of nitrogens (zero)2. Alkane and Alkene fragment ion series

C3H7C3H5

C2H5C2H3

C4H9

M+.

Notes:Electron impact, thus…1. Even MI = even number of nitrogens (zero)2. Alkene fragment ion series – no alkanes!

Abundance says 6 carbons.

C4H7

C3H5 C3H6

M+.

[M-CH3.]+

[M-C2H4]+.

Notes:Electron impact, thus…1. Even MI = even number of nitrogens (zero)2. Loss of C2H4 implies lack of a saturated terminus.

Abundance says 6 carbons.

C3H5

[M-C3H7.]+

C4H8

C2H3

M+.

[M-CH3]+[M-C2H4]+.

M+.

M+.

M+.

Initial Inspection of the Spectrum - 2

If peaks due to M+. and other high mass ions dominate the spectrum, the sample is probably aromatic.

A large number of peaks often indicates a large number of H atoms are present.

The lack of any dominant peaks suggests the absence of a hetero-atom.

The simpler the spectrum, the more symmetry is likely to be present in the sample molecule.

M+.

[M-CO]+.

M+.[M-H]+

M+.

[M-C2H5]+C6H13+

[M-C5H10]+

[M-C6H13]+

C3H7+

C2H5+

andC4H9+

M+.[M-C3H6]+.

[C5H9O]+

[M-C6H12]+.

C4H9+

M+.

[M-C2H5O]+

C5H7+

CH2NH2+ Base peak of primary aminesFound in all amine spectra and in spectra of amides

M+.

Common Neutral Losses - 115 CH3 Alkyl branching16 O, NH2 Nitroaromatic, oxime, sulfoxide or amines/amides17 NH2 RCONH218 H2O Alcohol, (ketone, aldehyde, less common)20 HF Alkyl fluoride26 C2H2 Aromatic hydrocarbon27 HCN ArCN, N-heterocylic compounds, ArNH2 rarely

C2H3 Ethyl ester (low abundance)

28 CO Quinones, some phenols

C2H4 n-Propyl ketones, ethyl esters, ArOC2H5

29 C2H5 Ethyl ketones, Ar - n-C3H7 compounds30 CH2O Aromatic methyl esters

31,32 CH3O,CH3OH Methyl esters of carboxylic acids

33,34 SH, H2S RSH

For a more extensive list, see McLafferty, Table A.5, pp. 348-350.

Common Neutral Losses - 241 C3H5 Propyl ester42 C3H6 n-butyl ketone

CH2CO RCOCH3, ArOCOCH3, ArNHCOCH3

43 C3H7 RCOC3H7, Ar-n-C4H9 compounds44 CO2 Anhydrides, esters45 COOH RCOOH

OC2H5 Ethyl esters of carboxylic acids

46 NO2 Aromatic nitrocompounds48 SO Aromatic sulfoxide55 C4H7 Butyl ester of carboxylic acid56 C4H8 RCOC5H11, ArOC4H9, Ar-C5H11 (n- or i-)57 C4H9 RCOC4H9

C2H5CO RCOC2H5

60 CH3COOH Acetate

For a more extensive list, see McLafferty, Table A.5, pp. 348-350.

Common Characteristic Ions

m/z 105 + 77 + 51 Benzoyl compounds

m/z 91 + 65 + 39 Alkyl benzenes, benzyl compounds

m/z 30 Base peak RNH2 otherwise other amines

m/z 44, 58, 72, . . . Amines, amides

m/z 31Primary alcohol; low intensity, other alcohols, ethers

m/z 31, 45, 59, . . . Ethers

m/z 74 Methyl esters of carboxylic acids

m/z 60 Straight chain carboxylic acids

m/z 77 or 76 Mono- or di-substituted benzene (low intensity)

The Odd-Even Electron Rule

Once a radical has been lost to produce an even electron, closed shell ion, further fragmentations can occur only by the loss of molecules to produce further odd mass, even electron ions.

Successive loss of two radicals NEVER occurs.

Do not assume that an ion is always formed from the next highest mass fragment ion. Ions may fragment by several routes so that adjacent peaks may not belong to ions of the same fragmentation sequence.

Charge Localisation - 1 Although the charge on a molecular

ion may be delocalised, it is useful to consider it formally as localised.

Where on the molecular ion is the charge located?

Which is the easiest (lowest energy) electron to remove?

These are usually (a) lone pair electrons on heteroatoms (b) -electrons in unsaturated systems

Charge Localisation - 2 If there is a choice of electrons that could be removed,

the formal charge may be placed on one of several atoms.

Hence, formally, one can think of M+. ions as consisting of a mixture of ions with the formal charge being on one of several possible sites.

Each type of molecular ion can give rise to a different type of fragmentation and the spectrum observed will be the weighted sum of the products of these.

Examples of Charge Localisation

Carbonyl compounds are assumed to lose a lone pair electron from the carbonyl oxygen

Ionised toluene is assumed to have lost a ring p-electron

[M-CH3]+ -cleavage

[M-CH3CO]+ inductive cleavage M+.

Factors Influencing Ion Abundance - 1

Eint required for decomposition: in general, low energy processes will predominate but different ionisation methods yield different internal energy distributions and hence different mass spectra from a particular sample.

Stability of the product ion

Factors Influencing Ion Abundance - 2

Stability of the neutral product Delocalisation of electron e.g. in allyl radical Placing of electron on electronegative atom e.g. .OH Loss of small stable molecule containing multiple

bonds, e.g. CO, C2H2, HCN Stevenson’s Rule

AB+. A+ + B. or A. + B+ Preference for formation of ion from fragment

having lower IE (except largest R. is lost preferentially)

Self assessment• Electron impact generates many fragments, why?• How can I calculate the number of sulfur atoms in my molecule from

the isotope distribution alone (assuming high enough resolution)?• A peak at m/z 30 indicates what moiety? 91? 43? 15? 24?• A peak at 58 immediately suggests what?• Calculate R+DB for C6H6, C60, C16H34O2.• State the nitrogen rule.

Fini…

CH908: Mass spectrometryLecture 2

(CH3)2CH-C6H4COOH

All fragment ions are odd mass, even electron ions M+.