有机波谱分析双语课程 (9).pdf
质谱的应用 有机波谱分析 Organic Spectroscopic Analysis Applications of Mass Spectrometry Introduction to Mass Spectrometry（MS,质谱）Mass spectrometry(MS)can be used to measure the masses of atoms or molecules;identify molecules present in solids,liquids and gases;2.How is Mass Spectrometry used?Introduction to Mass Spectrometry（MS,质谱）Mass spectrometry(MS)can be used to determine the quantity of each type of molecules;determine which atoms comprise a molecule and how they are arranged.e.g.CH3CH2CHO vs CH3C(=O)CH3 2.How is Mass Spectrometry used?Introduction to Mass Spectrometry（MS,质谱）A Propanal（丙醛）B Acetone（丙酮）Detection of melamine in Milk by MS 质谱检测牛奶中的三聚氰胺 三聚氰胺 Applications of MS Mass spectroscopy How to interpret Mass Spectrum 1.Mark the m/z for ion peaks 2.Determine the molecular ion M+3.According to the intensity ratio of(M+1)%vs(M+2)%,determine formula 4.Calculate unsaturation number(UN)5.Interpret the peak relationship 6.Deduce the chemical structure 7.Confirmation Spectrum interpretation(I)Intensity of m/z 96 98 100=67 43 7.0 9 6 1.From the isotopic ratio,we know there are two Cl atoms.96352=26,so except for the two Cl，the molecule may contain 2C&2H，the formula is C2H2Cl2 UN=1,because no ring structure exists,therefore should be C=C.The possible structure is the follows：(1)CHCl=CHCl (2)CH2=CCl2 m/z 26 comes from the loss of Cl2,M70(MCl2).In structure of(2),two Cl are on the same C,which makes the loss of Cl2 impossible.The major fragments come from：CHClC HCl_ Cl.CHClC H_ Cl2CHC Hm/z 96m/z 61m/z 26HCl_ CH CClm/z 36Spectrum interpretation(I)m/z 128 is the molecular ions which losses reasonable fragments with the adjacent peaks:m/z 113(15)、100(28)、m/z 99(29).The molecular mass is even number,thus contain no or even number of N atoms.No N fragmentation pattern m/z 30、44were observed,so there is no N atom.From the series of m/z 43(base peak),57,71,85,99,they are CnH2n+1or CnH2n+1CO.Spectrum interpretation(III)m/z58、86、100(even number)may come from -H rearrangement,indicating unsaturated bond exists.There is no C=C and benzene fragmentation pattern or M1、M45、MOR.Therefore,ketenes,aromatic hydrocarbons,aldehydes,acids or esters are excluded and it may be fatty ketones.From M28(m/z100),we can determine there is CH3CH2CH2CO-;From M42(m/z86)，we can also determine there may have the following structure：CH3CH2CH2CH2CO or (CH3)2CHCH2CO CH3CH2CH2CCH2CH2OCH2CH3CH3CH2CH2CCH2OCHCH3CH3(1)(2)Spectrum interpretation(III)From the loss of CH3,m/z 113(M15)，structure(2)is more reasonable.The major fragments come from：29 43 71 8599 85 57 43CH3CH2CH2CCH2CHOCH3CH3Spectrum interpretation(III)C4H9CCH2OCH2CH2H_ CH2=CH2C CCH2OHCCC3H7CCH2OCH2CHCH3H_ CH2=CHCH3C3H7CCH2OHH2CCOCH3H2CH2CH_ CH2=CH2H2CCCH3OHm/z 100m/z 86m/z 58M+128M+128CSpectrum interpretation(III)Thanks for your attention!The ENDThe END
