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Mass Spectrometry.presentation pharm d programme
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- 4. Only Cations areDetected
- 5. Ionization to RadicalCation Molecular Ion (m+ )
- 6. Glossary • Molecular ion- The ion obtained by the loss of one electron from the molecule (m+ ) • Base peak - The most intense peak in the MS, assigned 100% intensity • Radical cation - positively charged species with an odd number of electrons • Fragment ions - Lighter cations (and radical cations) formed by the decomposition of the molecular ion. These often correspond to stable carbcations. • m/z - mass to charge ratio
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- 8. Octane, m+ = 114 CH3CH2CH2CH2CH2CH2CH2CH3 m+= 114 -15 -29 -43 -57 -71 (base) m-29 m-43 m-57 m-71 Base peak m+
- 9. Isooctane, no molecularion CH3CCH2CHCH3 CH3 CH3 CH3 m+ = 114 loss of (isobutyl) . CH3C CH3 CH3 + m/z = 57
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- 12. Effect of Branchingin Hydrocarbons
- 13. Isotopes • Mass spectrometersare capable of separating and detecting individual ions even those that only differ by a single atomic mass unit. • As a result molecules containing different isotopes can be distinguished. • This is most apparent when atoms such as bromine or chlorine are present (79 Br : 81 Br, intensity 1:1 and 35 Cl : 37 Cl, intensity 3:1) where peaks at "M" and "M+2" are obtained. • The intensity ratios in the isotope patterns are due to the natural abundance of the isotopes. • "M+1" peaks are seen due the the presence of 13 C in the sample.
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- 17. (3-Chloropropyl)benzene CH2 CH2CH2Cl base peakm/e 91 (m-63)
- 18. Chloroacetone CH3CCH2Cl O m+ = 92 m+2= 94 - Cl - CH3 - CH2Cl . . . O CH3CCH2 O CCH2Cl O CH3C + + + m/z = 57 m/z = 77,79 m/z = 43 (base) (3:1 ratio)
- 19. 3-Pentanone C O CH2CH3 CH3CH2 + m/z =86 . CH2CH3 . CH3CH2C=O + loss of m/z = 57 m+ m-29 base
- 20. 2-Pentanone C O CH2CH2CH3 CH3 + m/z =86 . CH2CH2CH3 . CH3C=O + loss of m/z = 43 m-43 base m-15 m+
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- 23. 1-Butanol CH3CH2CH2CH2OH + . m+ = 74 -H2O CH3CH2CH CH2 + . m/z = 56 CH2OH CH2=OH + + m/z = 31 m-18
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- 25. Toluene CH3 + . m/z = 92 lossof H . CH2 m/z = 91 + + tropylium ion m+ m-1
- 26. Propylbenzene CH2CH2CH3 + . loss of CH2CH3 . CH2 + + m/z =91 m/z = 120 m+ m-29
- 27. Isopropylbenzene C H3C CH3 H + . loss of CH3 . CHCH3 +CH3 + m/z = 105 m/z = 120 m+ m-15
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- 31. McLafferty Rearrangement link toSDBS H O H m+ = 86 + . O H H . + O H H . + m/z = 44 + H transfer from carbon results in loss of a neutral alkene
- 32. McLafferty Rearrangements in AlkylBenzenes CH2 CH2 CHCH3 H m+ 134 loss of CH3CH=CH2 . + CH2 H H + . m/e 92 CH2 + m/e 91 - propyl .
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- 34. 2-Octanone O H m+ = 128 O + . + lossof C6H13 . m/z = 43 loss of pentene O H . + m/z = 58 via McLafferty
- 35. 2-Octanone O H m+ = 128 O + . + lossof C6H13 . m/z = 43 loss of pentene O H . + m/z = 58 via McLafferty
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- 41. High Resolution MassSpectrometry Determination of Molecular Formula CO N2 C2H4 CH2N all show m+ at 28 CO 27.9949 N2 28.0062 C2H4 28.0312 CH2N 28.0187 exact mass
- 42. Isotope Ratios CanHelp to Determine Molecular Formula Relative intensities (%) MF MW M M+1 M+2 CO 28.0 100 1.12 0.2 N2 28.0 100 0.76 ---- C2H4 28.0 100 2.23 0.01
- 43. Comparisons of MolecularWeights and Precise Masses MF MW exact mass C3H8O 60.1 60.05754 C2H8N2 60.1 60.06884 C2H4O2 60.1 60.02112 CH4N2O 60.1 60.03242
- 45. Determine the Formula fragmentfinder Molecular mass m+1 m+2 110 111 112 rel. intensity (%) 100 6.96 0.60 exact mass = 110.0376
- 46. Determine the Formula Molecularmass m+1 m+2 118 119 120 rel. intensity (%) 100 7.45 4.55
- 48. Subtract Sulfur’s contribution fragmentfinder Molecular mass m+1 m+2 118 119 120 rel. intensity (%) 100 7.45 4.55 subtract sulfur (32) 86 87 88 100 6.67 0.15
- 49. Determine the Molecular Formula Molecularmass m+1 m+2 154 155 156 rel. intensity (%) 100 15.41 3.77
- 50. Compound gives foursignals in the C-13 NMR spectrum Molecular mass m+1 m+2 m+4 190 191 192 194 rel. intensity (%) 100 6.48 130.77 31.81