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## Mass Spectrometry Part 1

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**Mass Spectrometry Part 1**Lecture Supplement: Take one handout from the stage**Determine structure of unknown substance**• Verify purity/identity of known substance Spectroscopy Why bother with spectroscopy?**Mass spectrometry (MS)***molecular formula Infrared spectroscopy (IR) functional groups Nuclear magnetic resonance (NMR) C/H molecular skeleton X-ray crystallography* spatial position of atoms SpectroscopyWhat methods are commonly used? *Not rigorously a type of spectroscopy**Spectroscopy**• Example: unidentified white powder • MS: C10H15N • IR: benzene ring, secondary amine (R2NH) • NMR: has CH2-CH-CH3 • X-ray: not necessary in this case**Fundamental operating principle**Determine mass by manipulating flight path of an ion in a magnetic field Electron gun Magnet Ionization Measure ion sample m/z too small m/z too large mass-to-charge ratio m/z just right introduction (m/z) - + Accelerator plates Detector Detector quiet Detector fires Detector quiet Mass SpectrometryThe Mass Spectrometer Ionization: X + e- X+. + 2 e-**Isotopes**Isotopes: atoms with same number of protons and same number of electrons but different numbers of neutrons • Aston mass spectrum of neon (1919) • Ne empirical atomic weight = 20.2 amu • Ne mass spectrum: predict single peak at m/z = 20.2 Resultsm/zrelative intensity 20.2 no peak 20.0 90% 22.0 10% • Conclusions • Neon is a mixture of isotopes • Weighted average: (90% x 20.0 amu) + (10.0% x 22.0 amu) = 20.2 amu • Nobel Prize in Chemistry 1922 to Aston for discovery of stable element isotopes**m/z = (1 x 12) + (4 x 1) = 16**C H Base peak: most abundant ion The Mass Spectrum Example: methane CH4 + e- CH4+. + 2 e- Relative ion abundance (%) mass-to-charge ratio (m/z)**m/z (amu)Relative abundance (%)**< 0.5 17 1.1 16 100.0 85.0 9.2 3.0 12 1.0 M - 2H M - 3H M - 4H The Mass Spectrum Alternate data presentation... M+2 14C1H4 or 12C3H1H3 or... M+1 13C1H4 or 12C2H1H3 M 12C1H4 M - H Molecular ion (M): intact ion of substance being analyzed Fragment ion: formed by cleavage of one or more bonds on molecular ions**The Mass SpectrumOrigin of Relative Ion Abundances**This table will be provided on an exam. Do not memorize it.**The Mass SpectrumRelative Intensity of Molecular Ion Peaks**Imagine a sample containing 10,000 methane molecules... Molecule# in samplem/zRelative abundance 12C1H4 9889 12 + (4 x 1) = 16 100% 110 13 + (4 x 1) = 17 (110/9889) x 100% = 1.1%* 13C1H4 ~1 14 + (4 x 1) = 18 (1/9889) x 100% = < 0.1%* 14C1H4 *Contributions from ions with 2H are ignored because of its very small natural abundance CH4 mass spectrum m/z = 16 (M; 100%), m/z = 17 (M+1; 1.1%), m/z = 18 (M+2; < 0.1%)**When relative contribution of M = 100% then relative**abundance of M+1/1.1% gives the approximate number of carbon atoms in the molecular formula Formula from Mass SpectrumM+1 Contributors • Comparing many mass spectra reveals M+1 intensity ~1.1% per C in formula • Examples: C2H6 M = 100%; M+1 = ~2.2% • C6H6 M = 100%; M+1 = ~6.6% • Working backwards gives a useful observation... • Other M+1 contributors • 15N (0.37%) and 33S (0.76%) should be considered • 2H (0.015%) and 17O (0.037%) can be ignored**Formula from Mass SpectrumM+2 Contributors**Anything useful from intensity of M+2? IsotopesNatural abundancesIntensity M : M+2 32S : 34S 95.0 : 4.2 100 : 4.4 35Cl : 37Cl 75.8 : 24.2 100 : 31.9 79Br : 81Br 50.7 : 49.3 100 : 97.2 Conclusion: Mass spectra of molecules with S, Cl, or Br have significant M+2 peaks**C H Cl**M: 36 + 7 + 35 = 78 Relative abundance (%) 78 80 m/z Formula from Mass SpectrumM+2 Contributors C3H7Cl M+2: 36 + 7 + 37 = 80 M:M+2 abundance ~3:1**C H Br**M: 36 + 7 + 79 = 122 122 124 Formula from Mass SpectrumM+2 Contributors C3H7Br Relative abundance (%) M+2: 36 + 7 + 81 = 124 M:M+2 abundance ~1:1 m/z**Identifying the Molecular Ions**• Which peaks are molecular ions? • Highest m/z not always M • M+1 has m/z one more than m/z of M C7H7Br M: m/z = 170**Formula from Mass Spectrum**Summary of Information from Mass Spectrum M: Reveals mass of molecule composed of lowest mass isotopes M+1: Intensity of M+1 / 1.1% = number of carbons M+2: Intensity reveals presence of sulfur, chlorine, and bromine Next lecture: procedure for deriving formula from mass spectrum**Mass Spectrometry Part 2**Lecture Supplement: Take one handout from the stage**Summary of Part 1**• Spectroscopy: Study of the interaction of photons and matter • Useful to determine molecular structure • Types: MS*, IR, NMR, x-ray crystallography* *not really spectroscopy • MS fundamental principle: Manipulate flight path of ion in magnetic field • Charge (z), magnetic field strength are known; ion mass (m) is determined • Isotopes: Natural abundance of isotopes controls relative abundance of ions • Molecular ion (M, M+1, M+2, etc.): Intact ion of substance being analyzed • m/z of M = molecular mass composed of lowest mass isotopes 1H, 12C, 35Cl, etc. • Relative abundance of M+1/1.1% gives approximate number of carbons • M+2 reveals presence of sulfur, chlorine, or bromine • Fragment ion: From decomposition of molecular ion before reaching detector • Analysis of fragmentation patterns not important for Chem 14C**?** C3H7Cl Mass Spectrum Formula Structure How do we derive structure from the mass spectrum? • Not trivial to do this directly • Structure comes from formula; formula comes from mass spectrum**M: m/z = 78**C2H6O3 C3H7Cl C5H4N C6H6 etc. M Mass Spectrum Formula Structure • How do we derive formula from the mass spectrum? • m/z and relative intensities of M, M+1, and M+2 • A few useful rules to narrow the choices**Formula:**NH3 N2H4 C7H5N3O6 C8H10N4O2 m/z (M): 17 32 227 194 } The Nitrogen Rule How Many Nitrogen Atoms? Consider these molecules: NH3 H2NNH2 • Conclusion • When m/z (M) = even, number of N in formula is even • When m/z (M) = odd, number of N in formula is odd**m/z even**odd nitrogen count even nitrogen count even nitrogen count even nitrogen count discarded How Many Nitrogen Atoms?A Nitrogen Rule Example Example: Formula choices from previous mass spectrum M: m/z = 78 C2H6O3 C3H7Cl C5H4N C6H6**C6H12**C6H10 H count = max - 2 H count = max - 4 How Many Hydrogen Atoms? One pi bond Two pi bonds C6H14 max H for 6 C Conclusion: Each pi bond reduces max hydrogen count by two**C6H12**C6H14 max H for 6 C C6H10 H count = max - 2 H count = max - 4 How Many Hydrogen Atoms? One ring Two rings Conclusion: Each ring reduces max hydrogen count by two**C6H15N**C6H14 max H for 6 C C6H16N2 H count = max + 1 H count = max + 2 The Hydrogen Rule How Many Hydrogen Atoms? One nitrogen Two nitrogens Conclusion: • Each nitrogen increases max H count by one • For C carbons and N nitrogens, max number of H = 2C + N + 2**Mass Spectrum Formula**• Procedure • Chem 14C atoms: H C N O F S Cl Br I • M = molecular weight (lowest mass isotopes) • M+1: gives carbon count • M+2: presence of S, Cl, or Br • No mass spec indicator for F, I Assume absent unless otherwise specified • Accounts for all atoms except O, N, and H • MW - mass due to C, S, Cl, Br, F, and I = mass due to O, N, and H • Systematically vary O and N to get formula candidates • Trim candidate list with nitrogen rule and hydrogen rule**Given information**Mass Spectrum FormulaExample #1 m/zMolecular ionRelative abundanceConclusions 102 M 100% Mass (lowest isotopes) = 102 Even number of nitrogens 103 M+1 6.9% 6.9 / 1.1 = 6.3 Six carbons* 104 M+2 0.38% < 4% so no S, Cl, or Br Oxygen? *Rounding: 6.00 to 6.33 = 6; 6.34 to 6.66 = 6 or 7; 6.67 to 7.00 = 7**Mass Spectrum FormulaExample #1**Mass (M) - mass (C, S, Cl, Br, F, and I) = mass (N, O, and H) 102 - C6 = 102 - (6 x 12) = 30 amu for N, O, and H OxygensNitrogens30 - O - N = HFormulaNotes 0 0 30 - 0 - 0 = 30 C6H30 Violates hydrogen rule 1 0 30 - 16 - 0 = 14 C6H14O Reasonable 2 0 30 - 32 - 0 = -2 C6H-2O2 Not possible 0 30 - 0 - 28 = 2 C6H2N2 Reasonable 2* *Nitrogen rule! • Other data (functional groups from IR, NMR integration, etc.) further trims the list**m/zMolecular ionRelative abundanceConclusions**157 M 100% 158 M+1 9.39% 159 M+2 34% Mass Spectrum FormulaExample #2 Mass (lowest isotopes) = 157 Odd number of nitrogens 9.39 / 1.1 = 8.5 Eight or nine carbons One Cl; no S or Br**1****Nitrogen rule! Mass Spectrum FormulaExample #2 Try eight carbons: M - C8 - Cl = 157 - (8 x 12) - 35 = 26 amu for O, N, and H OxygensNitrogens26 - O - N = HFormulaNotes 0 26 - 0 - 14 = 12 C8H12ClN Reasonable Not enough amu available for one oxygen/one nitrogen or no oxygen/three nitrogens**1****Nitrogen rule! Mass Spectrum FormulaExample #2 Try nine carbons: M - C9 - Cl = 157 - (9 x 12) - 35 = 14 amu for O, N, and H OxygensNitrogens14 - O - N = HFormulaNotes 0 14 - 0 - 14 = 0 C9ClN Reasonable Not enough amu available for any other combination.**Formula Structure**• What does the formula reveal about molecular structure? • Functional groups • Absent atoms may eliminate some functional groups • Example: C7H9N has no oxygen-containing functional groups • Pi bonds and rings • Recall from previous: one pi bond or one ring reduces max H count by two • Each two H less than max H count = double bond equivalent (DBE) • If formula has less than full H count, molecule must contain one pi bond or ring**hydrogens and halogens**nitrogens H N DBE = C - + + 1 2 2 carbons Example C8H10ClN Formula StructureCalculating DBE DBE may be calculated from molecular formula: • One DBE = one ring or one pi bond • Two DBE = two pi bonds, two rings, or one of each • Four DBE = possible benzene ring DBE = C - (H/2) + (N/2) + 1 = 8 - [(10+1)/2] + (1/2) + 1 Four pi bonds and/or ring Possible benzene ring = 4**Formula StructureCommon Math Errors**• Small math errors can have devastating effects! • No calculators on exams • Avoid these common spectroscopy problem math errors: • Divide by 1.1 divide by 1.0 • DBE cannot be a fraction • DBE cannot be negative Next lecture: Infrared spectroscopy part 1