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Mass Spectrometry

Mass Spectrometry. Reading EI Spectra: Recognition of M + Deduction of Formula (3) Fragmentation of M +. Recognition of Molecular Ion. Is h ighest m/z value even? If odd then what? Do highest m/z fragments make sense? M-15 present? M-18? M-31? M-29? M-3  M-14 are impossible

shelby-fry
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Mass Spectrometry

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  1. Mass Spectrometry Reading EI Spectra: Recognition of M+ Deduction of Formula (3) Fragmentation of M+

  2. Recognition of Molecular Ion • Is highest m/z value even? • If odd then what? • Do highest m/z fragments make sense? • M-15 present? M-18? M-31? M-29? • M-3  M-14 are impossible • Even if 1 and 2 are true the highest m/z might not be M+

  3. Nitrogen Rule

  4. Spectrum of an Amine

  5. Spectrum of an Alcohol

  6. M+ present?

  7. Deducing molecular formula • Use %m+1 to get # of Cs • Use Nitrogen rule • M-18 peak indicates oxygen • Doublet M+ peaks indicate Cl (3:1) and Br (1:1) • Peak at 127 = I • Peak at 77 is phenyl group (C6H5) • M+2 peak = 4-5% indicates S • Deduce molecular formula!

  8. Molecular Formula?

  9. Molecular Formula?

  10. Molecular Formula? m/z %abund • 20.2 • 1.2

  11. Index of Hydrogen Deficiency • Units of Unsaturation = r + db • CnHmXxNyOz r + db = n – m/2 – x/2+ y/2 + 1

  12. Fragmentation of M+ • Draw structure of M+ if possible and then draw possible heterolytic cleavages (curved arrows) or homolytic cleavages (fish hooks) • Alternatively, show homolytic or heterolytic cleavage cleavage of the molecule and then remove an electron from one of the fragments. • Fragmentation to give neutral molecule and a radical cation must involve some kind of rearrangement.

  13. General Rules Probablity of bond cleavage depends on stability of fragments produced. (3º > 2º > 1º > CH3) (oxonium > carbocation) (allylic, benzylic > no resonance) • Increase branching  decrease M+ • Add carbons  decrease M+ • Cleave more likely at higher º C • Unsubst cyclic have strong M+ (especially aromatic) • Alkenes cleave to allylic species • R on cycloalkanes α cleave. Cyclohexenes do retro Diels-Alder. • R on aromatic rings β-cleave to benzylic • Bonds β to heteroatoms cleave • Loss of small stable molecules, H2O, NH3, C2H4, HCN, CO, H2S,etc often occur

  14. Alkanes • Straight chain alkanes – homologous series of ions – smooth curve max at 57 or 71. M-15 absent. “picket fence” • Discontinuity = branching. Absence of homolog  no single cleavage possible • Cycloalkanes. Larger M+. More even mass fragments (M-C2H4)

  15. Alkenes • Cleave next to double bond to form allylic fragments. (Double bonds can shift in M+.) • Homologous series is -2 (27, 41, 55, 69..) • Cyclohexenes do retro Diels Alder – give even mass fragment (M-C2H4, M-C3H6 etc.)

  16. Aromatics • No branching = strong M+. Often see M++. • m/z 77, 51 = phenyl (monosubst benzene) • Most prominent peaks are due to beta cleavage of branches to benzylic (really tropylium.)

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