Mass Spectrometry Reading EI Spectra: Recognition of M+

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Presentation transcript:

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

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+

Nitrogen Rule

Spectrum of an Amine t-BuNH2

Spectrum of an Alcohol 2,5-dimethyl-2-hexanol

M+ present? 2,2-dimethylhexane

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!

Molecular Formula? C8H18

Molecular Formula? C7H15Br

Molecular Formula? m/z %abund 20.2 1.2 5.7% m+1

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

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.

General Rules Probablity of bond cleavage depends on stability of fragments produced. (3º > 2º > 1º > CH3) (oxonium > carbocation) (allylic, benzylic > no resonance) 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 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

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)

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.)

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.)