Elusidasi 4 MS

March 27, 2018 | Author: Deriven Teweng | Category: Mass Spectrometry, Ion, Molecules, Gas Chromatography–Mass Spectrometry, Hydrogen


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Mass SpectrometryThe concept of mass spectrometry The mass spectrometer analyses the masses of cations. A compound is ionised, the ions are separated on the basis of their mass/charge ratio, and the number of ions representing each mass/charge “unit” (m/z) is recorded as a spectrum It is routine to couple a mass spectrometer to some form of chromatographic instrument, such as a gas chromatograph (GC-MS) or a liquid chromatograph (LC-MS). the species that is formed is called a radical cation. • Because M is unstable. • The radical cation M+• is called the molecular ion or parent ion. and symbolized as M+•. it decomposes to form fragments of radicals and cations that have a lower molecular weight than M+•.• When the electron beam ionizes the molecule. 3 . • The mass of M+• represents the molecular weight of M. • For our purposes. . therefore.• Molecular Ion • Molecular ion (M): A radical cation formed by removal of a single electron from a parent molecule in a mass spectrometer = MW. it does not matter which electron is lost. M+• – A dot to show that it has an odd number of electrons. we write the molecular formula of the parent molecule in brackets with: – A plus sign to show that it is a cation. radical cation character is delocalized throughout the molecule. M = 114) 14 (CH2) 29 (CH3CH2) 57 85 71 M+ 114 The parent peak or molecular ion - C8H18 e 70 eV [C8H18] + 2 e + (M ) 5 .Partial MS of octane (C8H18. 40). 57(100) etc. 85(80). 6 . 14 (CH2) 29 (CH3CH2) 57 85 71 M+ 114 114(M+.The largest or most abundant peak is called the base peak and is assigned an intensity of 100%. The relative abundance of all other ions is reported as a % of abundance of the base peak. 71(60). other peaks are then fractions of that. N = 14. C = 12. O = 16. 114) + -CH3CH2CH2 (29+14) Molecular ion (parent peak) [C6H13] Daughter ions (85) [C5H11] (71) 7 . F = 19 etc.Masses are usually rounded off to whole numbers assuming: H = 1. [C8H18] fragmentation -CH CH (29) 3 2 (M . 1 20 30 40 2e- Molecular Ion (Radical Cation) Base Peak 100 90 80 70 60 50 40 30 20 10 0 + M → 50 60 70 m/z 80 90 .(H2O and CH2=CH2) 1-Pentanol . M + Molecule Intensity (% of Base Peak) e- High Energy Electron CH2OH+ + M .H2O Molecular Ion Peak M+ .What’s in a Mass Spectrum? Mass spectrum: A plot of the relative abundance of ions versus their mass-to-charge ratio (m/z).(H2O and CH3) CH3(CH2)3 – CH2OH M .MW 88 M . a stream of solution 6 passes through a strong electric field (10 V/m) .hit the sample with a laser beam • Electrospray Ionization (ESI) .sample M collides with reagent ions present in excess e.+ CH4 + e M + CH5 + CH5+ + CH4 + MH • Fast Atom/Ion Bombardment (FAB) • Laser Desorption & Matrix-Assisted Laser Desorption (MALDI) .g.+ electrons: M+e 2e + M fragments • Chemical ionization (CI) .vapor of sample is bombarded with .WAYS TO PRODUCE IONS • Electron impact (EI) . CH4. atomic mass units 1 amu = 1 dalton (Da) . biomolecules. Chemical Ionization (CI) ionization with very little fragmentation.Ionization Methods 1. Electron Ionization (EI) most common ionization technique. note: amu. Spray ionization (SI) for LC-MS. limited to relatively low MW compounds (<600 amu) 2. still for low MW compounds (<800 amu) 3. etc. Desorption Ionization (DI) for higher MW or very labile compounds 4. Electron Ionization (EI) • vaporized sample is bombarded with high energy electrons (typically 70 eV) • “hard” ionization method leads to significant fragmentation • ionization is efficient but non-selective . versatile and reproducible • fragmentation gives structural information • large databases if EI spectra exist and are searchable Disadvantages • fragmentation at expense of molecular ion • sample must be relatively volatile .Electron Ionization Advantages • inexpensive. ammonia. isobutane. electron capture. – Common CI reagents: methane. adduct formation. methanol • “soft” ionization gives little fragmentation • selective ionization-only exothermic or thermoneutral ion-molecule reactions will occur • choice of reagent allows tuning of ionization . charge exchange. hydrogen.● Chemical Ionization (CI) Vaporized sample reacts with pre-ionized reagent gas via proton transfer. etc. CnHmXxNyOz Index = (n) .The Index of Hydrogen Deficiency (or IHD) is also known as "units of unsaturation".   Hence. acyclic species. The index of hydrogen deficiency is a count of how many molecules of H2 need to be added to a structure in order to obtain the corresponding saturated. the number of double bonds and twice the number of triple bonds.(x/2) + (y/2) +1 . the IHD is the sum of the number of rings.(m/2) . How To Determine Molecular Formulas and Molecular Weights Using Mass Spectrometry Isotopic Peaks & the Molecular Ion 15 . ❖ The presence of oxygen. H. S.+and nitrogen in a compound gives rise to a small M● + 1 peak.❖ The presence of isotopes of carbon. hydrogen. sulfur. N M + 2 Elements: O. or bromine + in a compound gives rise to an M● + 2 peak M + 1 Elements: C. chlorine. Br. Cl 16 . M is even. Cl (33%). thus.+ ● Is M odd or even? According to the nitrogen rule. The compound must have an even number of nitrogen atoms + M ● +2 ❖ The relative abundance of the peak indicates the presence (or absence) of S (4.3%. and Br are absent ❖ The molecular formula can now be established by determining the number of hydrogen atoms and adding the appropriate number of oxygen atoms. Cl.4%). then the compound must contain an even number of nitrogen atoms (zero is an even number) ❖ + ● ● For our unknown. we can assume that S. if necessary 17 . or Br (98%) + ● For our unknown M● +2 = 0. if it is even. Since M. of the unknown.W. number of carbons present is 4 ❖ Using the “nitrogen rule”. this unknown contains zero “N” 18 .W. (12 x 4 = 48).❖ + ● Since M is m/z 72 ⇒ molecular weight = 72 + ● ❖ As determined using the relative abundance of M +1 peak. and there are 4 C present. Thus. = 72. this unknown must have an even number of N. adding 2 “N” will be greater than the M. H and O H = 72 – (4 x 12) – 16 = 8 and thus our unknown has the molecular formula C4H8O 19 .❖ For a molecule composed of C and H only H = 72 – (4 x 12) = 24 but C4H24 is impossible ❖ For a molecule composed of C. BM vs formula 2.Fragment ions and fragment lost C8H8O2 .Index ?? 3.1. . . 23 . .Terima kasih Gambar-gambar yang diambil dari berbagai sumber hanya digunakan untuk mengajar mahasiswa Fak. Farmasi USD.
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