1. CH 908: Mass Spectrometry Lecture 2 Interpreting Electron Impact Mass Spectra Recommended: Read chapters 3-5 of McLafferty Prof. Peter B. O’Connor

2. 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?

3. 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.

4. Approximate ratio 27:27:9:1 Approximate ratio 9:6:1 Approximate ratio 3:1

5. 81:108:54:12:1 27:27:9:1 9:6:1

6. 1:4:6:4:1 1:3:3:1

7. C12H22S2+. m/z 230 231 232 Intensities 100 13 9 [M-(C6H10)]+ m/z 148 149 150 Intensities 100 7 9

8. 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).

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

10. 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+DB X = carbon, silicon Y = hydrogen, chlorine, fluorine, etc. Z = nitrogen, phosphorus Values ending in ½ correspond to even electron ions. Values lower than –½ are not possible chemically.

11. 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.

12. Is it the Molecular ion?

13. C 10 H 15 O C 10 H 14 O C 9 H 12 O C 10 H 13 C 8 H 10 O Example #1:

14. Is it the Molecular ion? C 10 H 14 C 10 H 13 C 9 H 11 C 8 H 9 C 7 H 8 C 7 H 7 Example #2:

15. 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.

16. M+. absent [M-CH3]+ [M-H2O]+. [M-C3H7]+ [M-C4H9]+

17. M+. C3H7CO+ C2H5CO+

18. [M-CH3]+M+. [M-C3H6]+. [CH3CO]+

19. M+. C7H7+ C5H5+ C3H3+

20. Fragmentation Patterns (Alkanes)

21. M+. Notes: Electron impact, thus… 1. Even MI = even number of nitrogens (zero) 2. Alkane and Alkene fragment ion series C3H7C3H7 C3H5C3H5 C2H5C2H5 C2H3C2H3 C4H9C4H9

22. M+. Notes: Electron impact, thus… 1. Even MI = even number of nitrogens (zero) 2. Alkene fragment ion series – no alkanes! Abundance says 6 carbons. C4H7C4H7 C3H5C3H5 C3H6C3H6

23. 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. C3H5C3H5 [M-C3H7. ]+ C4H8C4H8 C2H3C2H3

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

25. M+.

28. 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.

29. M+. [M-CO]+.

30. M+. [M-H]+

31. M+. [M-C2H5]+ C6H13+ [M-C5H10]+ [M-C6H13]+ C3H7+ C2H5+ and C4H9+

32. M+. [M-C3H6]+. [C5H9O]+ [M-C6H12]+. C4H9+

33. M+. [M-C2H5O]+ C5H7+

34. CH2NH2+ Base peak of primary amines Found in all amine spectra and in spectra of amides M+.

35. Common Neutral Losses - 1 15CH3Alkyl branching 16O, NH2Nitroaromatic, oxime, sulfoxide or amines/amides 17NH2RCONH2 18H2OAlcohol, (ketone, aldehyde, less common) 20HFAlkyl fluoride 26C2H2Aromatic hydrocarbon 27HCNArCN, N-heterocylic compounds, ArNH2 rarely C2H3Ethyl ester (low abundance) 28COQuinones, some phenols C2H4n-Propyl ketones, ethyl esters, ArOC2H5 29C2H5Ethyl ketones, Ar - n-C3H7 compounds 30CH2OAromatic methyl esters 31,32 CH3O,CH3O H Methyl esters of carboxylic acids 33,34SH, H2SRSH For a more extensive list, see McLafferty, Table A.5, pp. 348-350.

36. Common Neutral Losses - 2 41C3H5Propyl ester 42C3H6n-butyl ketone CH2CORCOCH3, ArOCOCH3, ArNHCOCH3 43C3H7RCOC3H7, Ar-n-C4H9 compounds 44CO2Anhydrides, esters 45COOHRCOOH OC2H5Ethyl esters of carboxylic acids 46NO2Aromatic nitrocompounds 48SOAromatic sulfoxide 55C4H7Butyl ester of carboxylic acid 56C4H8RCOC5H11, ArOC4H9, Ar-C5H11 (n- or i-) 57C4H9RCOC4H9 C2H5CORCOC2H5 60CH3COOH Acetate For a more extensive list, see McLafferty, Table A.5, pp. 348-350.

37. Common Characteristic Ions m/z 105 + 77 + 51Benzoyl 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 31 Primary 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)

38. 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.

39. 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

40. 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.

41. 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

42. [M-CH3]+  -cleavage [M-CH3CO]+ inductive cleavage M+.

43. Factors Influencing Ion Abundance - 1  E int 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

44. 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)

45. 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.

46. Fini… CH908: Mass spectrometry Lecture 2

47. (CH3)2CH-C6H4COOH All fragment ions are odd mass, even electron ions M+.