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Interpretation of Mass Spectra Part 4

Interpretation of Mass Spectra Part 4. Objectives. To describe the main features of EI, CI, ESI spectra of organic compounds To indicate how to identify the M +. , MH + or [M+nH] n+ ion and to suggest a possible molecular formula

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Interpretation of Mass Spectra Part 4

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  1. Interpretation of Mass Spectra Part 4

  2. Objectives To describe the main features of EI, CI, ESI spectra of organic compounds To indicate how to identify the M+., MH+ or [M+nH]n+ ion and to suggest a possible molecular formula To review the major types of fragmentation mechanisms and how to recognise them in the mass spectra of different classes of compounds To give some practice in the interpretation of mass spectra of simple unknown organic samples

  3. Relative Molecular Mass and Isotopes

  4. Relative Molecular Mass Many non-radioactive elements exist in more than one isotopic form. The lightest isotopes of the common light elements are usually the most abundant e.g. 1H, 12C, 14N, 16O, 32S, 35Cl, 79Br The masses of these isotopes, rounded to the nearest whole number, are used to calculate the nominal RMM of a compound. The monoisotopic RMM of a compound is that calculated using the accurate atomic weights of the most abundant isotopes of the constituent elements.

  5. Relative Molecular Mass As the number of H atoms increases, the difference between the nominal and the mono-isotopic RMM slowly rises and the nominal RMM is no longer useful, e.g. above about 600 Da. For example: Mass: C20H30 C40H60 C60H90 Nominal 270 540 810 Monoisotopic 270.235 540.470 810.705 Whole Number 270 540 811

  6. Common isotopes of the lighter elements

  7. Relative Molecular Masses Naturally occurring compounds contain isotopes in their natural abundances. The inclusion of the less common isotopes leads to a higher average RMM. Hence we can define three different RMMs: Nominal RMM Monoisotopic RMM Average RMM Which do we measure by mass spectrometry?

  8. Measurement of RMM For RMMs of < 500, normally the nominal RMM is obtained at low resolving power The monoisotopic RMM can be measured if the resolving power is high enough to resolve isotope peaks and any other interfering peaks. If the resolving power is insufficient to resolve isotope peaks, the average RMM is measured. If interfering peaks are not resolved, an accurate RMM cannot be measured.

  9. Isotope Peaks - 1 Ions containing elements that have more than one isotope exhibit isotope peaks on the high mass side of the peak due to the lightest ion and may indicate the elements present in an ion. The probability that any one carbon atom is a 13C isotope is 1.1% If there are n carbon atoms in an ion, the probability that at least one carbon atom is a 13C isotope is n x 1.1% so that I[(M+1)+]/I[(M+)] = (n x 1.1)/100

  10. Isotope Peaks - 2 For relatively low RMM samples, this ratio indicates the number of carbon atoms in the ion providing that there is no contribution from MH+ ions, e.g. for alcohols and amines at high sample pressures. For example, if M+. is at m/z 112, this could be due to C8H16+., C7H12O+. or C6H8O2+. for which the ratio I[(M+1)+]/I[(M+)] would ideally be 0.09, 0.08 and 0.07 respectively. This ratio exceeds unity once the number of C atoms in the molecule exceeds 85-90.

  11. Molecular Ion Regions 1402 1121 1217 C80H160 C80H160S3 C100H200

  12. Calculation of Isotope Patterns Exemplified for Cl and Br atoms in C6H4OClBr Write out isotopic abundances of Cl and Br isotopes (including zero abundances) across the top and down the side of a square. Construct a matrix by multiplying each member of the row with each member of the column. Sum the diagonals (top right to bottom left) to produce relative abundances.

  13. Isotopic Peaks due to Cl + Br 3 0 1 35Cl, 36Cl, 37Cl abundances 1 | 3 0 1 0 | 0 0 0 79Br, 80Br, 81Br abundances 1 1 | 3 0 1 (vertical) Relative Abundance. 3 : 0 : 4 : 0 : 1 RMM 114 116 118 Comp. 35Cl79Br 35Cl81Br 37Cl81Br 37Cl79Br

  14. 1-Bromo-4-Chlorobenzene Mass Spectrum The 3:4:1 pattern in M+ region (with 13C peaks between). The fragment ion at m/z 111, 113 shows the 3:1 pattern indicating the presence of a Cl atom and that the Br atom has been lost.

  15. The Molecular Ion Region For compounds containing only C, H, O and N atoms, 13C isotopes control the isotope pattern. In large molecules, such as proteins, few ions contain less than 2 or 3 heavy isotopes; the peak due to ions containing only light isotopes is of very low intensity. The presence of S, Cl and Br atoms leads to characteristic peaks at M+2, M+4, etc.

  16. Electron Ionization Spectra

  17. Production of EI Spectra 70 eV electrons bombard the sample vapour and deposit typically 0 - 20 eV in a molecule. Typically 8 - 10 eV is needed to produce M+. in its ground state so M+. ions are produced with internal energies of approximately 0 - 10 eV These undergo one or more fragmentations in the ion source and products ejected after about 1 ms are collected to produce the mass spectrum. Discrimination effects differ between instruments so that the spectrum of a given sample varies somewhat from instrument to instrument.

  18. 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. 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).

  19. Hexane [M-C2H5]+ M+.

  20. 2-Hexene [M-C2H4]+. M+. [M-CH3]+

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

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

  23. 1,3-Cyclohexadiene M+.

  24. Benzene M+.

  25. 1,5-hexadien-3-yne M+. [M-C2H2]+.

  26. 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.

  27. 1-Napthalenol M+. [M-CO]+.

  28. Nonanal

  29. C3H7+ [M-C6H13]+ [M-C5H10]+ C2H5+ [M-C2H5]+ C6H13+ M+.

  30. [M-C3H6-C3H6]+. [C4H9CO]+ [M-C3H6]+. M+.

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

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

  33. Identifying the Molecular Ion - 1 The common isotopes of elements C, O, S have even relative atomic masses and even valencies whereas the common isotopes of H, F, Cl, Br, P and Na have odd relative atomic masses and odd valencies. Hence organic compounds that contain only these elements (i.e. no nitrogen atoms) have an even relative molecular mass (RMM) so that M+. occurs at an even value of m/z or the MH+ ion appears at an odd value of m/z .

  34. Identifying the Molecular Ion - 2 The one common element that is an exception to this rule is nitrogen, the most common isotope of which has a relative atomic mass of 14 and a valency of 3. Hence an odd relative molecular mass results when the molecule contains an odd number of N atoms. Thus if M+. has an odd m/z, it suggests a possible amide, amine, nitrile or N-heterocyclic compound.

  35. 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.

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

  37. 81:108:54:12:1 27:27:9:1 9:6:1

  38. 1:4:6:4:1 1:3:3:1

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

  40. Is it Really the Molecular Ion? Check that the supposed M+. loses neutral species that are sensible, e.g. radicals such as alkyl radicals or OH. or molecules such as alkenes, CO, HCl, H2O, etc. If there are losses that cannot be explained, e.g. 3 - 13, 21 - 25 Da, the assignment should be re-examined. If it appears that M+. loses 3 Da, this could arise from losses of CH3 and H2O giving peaks due to the ions [M-15]+ and [M-18]+..

  41. 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. 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.

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

  43. C2H5CO+ C3H7CO+ M+.

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

  45. C7H7+ M+. C5H5+ C3H3+

  46. How to Work Out the Molecular Formula - 1 Start with the RMM and the value of I([M+1])/I([M]) which gives an indication of the number of C atoms present. Suppose the RMM is 136. The maximum possible number of C atoms is found by dividing this by 12. This gives 11 but C11H4 is very unlikely to be correct. Try 10 C atoms, converting the other 12 Da to H atoms, giving C10H16 (e.g. pinene or limonene, etc).

  47. How to Work Out the Molecular Formula - 2 Repeating the process for 9 C atoms gives C9H28 which is unacceptable. Convert 16 H atoms to one O atom giving C9H12O as a possibility, e.g. benzyl ethyl ether. Repeating this for 8 C atoms gives C8H8O2 which could be an aromatic acid or ester. Knowing the number of C atoms present, one can suggest a molecular formula. In this case, suppose C8H8O2 is suggested.

  48. Rings + Double Bonds: 1 For a molecule of molecular formula CxHyNzOa the number of double bond equivalents (DBEs) is given by the expression DBEs = x - (y/2) + (z/2) + 1 A DBE is a unit of unsaturation, e.g. an alkene or carbonyl group is 1 DBE, a saturated ring is 1 DBE, an aromatic ring is 4 DBEs, a triple bond is 2 DBEs. The number of DBEs is independent of the number of O atoms present. Halogen atoms are counted as H atoms, S and P atoms are counted as O and N atoms respectively.

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