1. 1 Mass Spectrometry Part 1 Lecture Supplement: Take one handout from the stage

2. 2 Spectroscopy Why bother with spectroscopy? Determine structure of unknown substanceVerify purity/identity of known substance

3. 3 Spectroscopy What methods are commonly used? *Not rigorously a type of 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

4. 4 Spectroscopy Example: unidentified white powder MS: C 10 H 15 N IR: benzene ring, secondary amine (R 2 NH) NMR: has CH 2 -CH-CH 3 X-ray: not necessary in this case

5. 5 Detector quiet m/z too small Mass Spectrometry The Mass Spectrometer Fundamental operating principle Determine mass by manipulating flight path of an ion in a magnetic field sample introduction Measure ion mass-to-charge ratio (m/z) Detector Ionization Electron gun + - Accelerator plates Magnet m/z just right Detector fires Ionization: X + e -  X +. + 2 e - m/z too large Detector quiet

6. 6 Isotopes Aston mass spectrum of neon (1919) Ne empirical atomic weight = 20.2 amu Ne mass spectrum: predict single peak at m/z = 20.2 Results m/z relative 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 Isotopes: atoms with same number of protons and same number of electrons but different numbers of neutrons Nobel Prize in Chemistry 1922 to Aston for discovery of stable element isotopes

7. 7 The Mass Spectrum Example: methane CH 4 + e -  CH 4 +. + 2 e - mass-to-charge ratio (m/z) Relative ion abundance (%) Base peak: most abundant ion m/z = (1 x 12) + (4 x 1) = 16 C H

8. 8 The Mass Spectrum Alternate data presentation... m/z (amu) Relative abundance (%) 18 < 0.5 17 1.1 16 100.0 15 85.0 14 9.2 13 3.0 12 1.0 M+2 M+1 M Molecular ion (M): intact ion of substance being analyzed Fragment ion: formed by cleavage of one or more bonds on molecular ions 12 C 1 H 4 13 C 1 H 4 or 12 C 2 H 1 H 3 14 C 1 H 4 or 12 C 3 H 1 H 3 or... M - H M - 2H M - 3H M - 4H

9. 9 The Mass Spectrum Origin of Relative Ion Abundances M contributorsM+1 contributorsM+2 contributors Isotope Natural Abundance Isotope Natural Abundance Isotope Natural Abundance 1H1H99.9855% 2H2H0.015% 3H3Hppm 12 C98.893 13 C1.107 14 Cppm 14 N99.634 15 N0.366 16 O99.759 17 O0.037 18 O0.204 19 F100.0 32 S95.0 33 S0.76 34 S4.22 35 Cl75.77 37 Cl24.23 79 Br50.69 81 Br49.31 127 I100.0 This table will be provided on an exam. Do not memorize it.

10. 10 The Mass Spectrum Relative Intensity of Molecular Ion Peaks Imagine a sample containing 10,000 methane molecules... Molecule # in sample m/z Relative abundance 12 C 1 H 4 988912 + (4 x 1) = 16100% 13 C 1 H 4 11013 + (4 x 1) = 17(110/9889) x 100% = 1.1%* 14 C 1 H 4 ~114 + (4 x 1) = 18(1/9889) x 100% = < 0.1%* *Contributions from ions with 2 H are ignored because of its very small natural abundance CH 4 mass spectrum m/z = 16 (M; 100%), m/z = 17 (M+1; 1.1%), m/z = 18 (M+2; < 0.1%)

11. 11 Formula from Mass Spectrum M+1 Contributors Comparing many mass spectra reveals M+1 intensity  ~1.1% per C in formula Examples: C 2 H 6 M = 100%; M+1 = ~2.2% C 6 H 6 M = 100%; M+1 = ~6.6% Working backwards gives a useful observation... 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 Other M+1 contributors 15 N (0.37%) and 33 S (0.76%) should be considered 2 H (0.015%) and 17 O (0.037%) can be ignored

12. 12 Formula from Mass Spectrum M+2 Contributors Anything useful from intensity of M+2? 32 S : 34 S Isotopes Natural abundances Intensity M : M+2 100 : 4.495.0 : 4.2 35 Cl : 37 Cl75.8 : 24.2 100 : 31.9 79 Br : 81 Br50.7 : 49.3100 : 97.2 Conclusion: Mass spectra of molecules with S, Cl, or Br have significant M+2 peaks

13. 13 m/z Relative abundance (%) C 3 H 7 Cl 80 Formula from Mass Spectrum M+2 Contributors M+2: 36 + 7 + 37 = 80 C H Cl M: 36 + 7 + 35 = 78 M:M+2 abundance ~3:1 78

14. 14 122 Relative abundance (%) m/z C 3 H 7 Br Formula from Mass Spectrum M+2 Contributors M+2: 36 + 7 + 81 = 124 C H Br M: 36 + 7 + 79 = 122 M:M+2 abundance ~1:1 124

15. 15 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 C 7 H 7 Br M: m/z = 170

16. 16 Formula 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 Summary of Information from Mass Spectrum Next lecture: procedure for deriving formula from mass spectrum

17. 17 Mass Spectrometry Part 2 Lecture Supplement: Take one handout from the stage

18. 18 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 1 H, 12 C, 35 Cl, 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

19. 19 Mass Spectrum  Formula  Structure Not trivial to do this directly Structure comes from formula; formula comes from mass spectrum How do we derive structure from the mass spectrum?  ?   C 3 H 7 Cl

20. 20 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 M: m/z = 78 C 2 H 6 O 3 C 3 H 7 Cl C 5 H 4 N C 6 H 6 etc. M

21. 21 How Many Nitrogen Atoms? Consider these molecules: NH 3 H 2 NNH 2 Formula:NH 3 N2H4N2H4 32 C7H5N3O6C7H5N3O6 227 C 8 H 10 N 4 O 2 194 Conclusion When m/z (M) = even, number of N in formula is even 17m/z (M): The Nitrogen Rule } When m/z (M) = odd, number of N in formula is odd

22. 22 How Many Nitrogen Atoms? A Nitrogen Rule Example M: m/z = 78 C 2 H 6 O 3 C 3 H 7 Cl C 5 H 4 N C 6 H 6 Example: Formula choices from previous mass spectrum m/z even odd nitrogen count discarded even nitrogen count

23. 23 How Many Hydrogen Atoms? C 6 H 14 max H for 6 C One pi bondTwo pi bonds Conclusion: Each pi bond reduces max hydrogen count by two C 6 H 12 H count = max - 2 C 6 H 10 H count = max - 4

24. 24 How Many Hydrogen Atoms? Conclusion: Each ring reduces max hydrogen count by two One ringTwo rings C 6 H 14 max H for 6 C C 6 H 12 H count = max - 2 C 6 H 10 H count = max - 4

25. 25 How Many Hydrogen Atoms? One nitrogenTwo nitrogens C 6 H 15 N H count = max + 1 C 6 H 16 N 2 H count = max + 2 Conclusion: C 6 H 14 max H for 6 C Each nitrogen increases max H count by one For C carbons and N nitrogens, max number of H = 2C + N + 2 The Hydrogen Rule

26. 26 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

27. 27 Mass Spectrum  Formula Example #1 m/z Molecular ion Relative abundance Conclusions 102 M 100% 103 M+1 6.9% 104 M+2 0.38% Mass (lowest isotopes) = 102 Even number of nitrogens 6.9 / 1.1 = 6.3 Six carbons* *Rounding: 6.00 to 6.33 = 6; 6.34 to 6.66 = 6 or 7; 6.67 to 7.00 = 7 < 4% so no S, Cl, or Br Oxygen? Given information

28. 28 Mass Spectrum  Formula Example #1 Mass (M) - mass (C, S, Cl, Br, F, and I) = mass (N, O, and H) 102 - C 6 = 102 - (6 x 12) = 30 amu for N, O, and H Oxygens Nitrogens 30 - O - N = H Formula Notes 0030 - 0 - 0 = 30C 6 H 30 Violates hydrogen rule 1030 - 16 - 0 = 14C 6 H 14 OReasonable 2030 - 32 - 0 = -2C 6 H -2 O 2 Not possible 0 2* *Nitrogen rule! 30 - 0 - 28 = 2C6H2N2C6H2N2 Reasonable Other data (functional groups from IR, NMR integration, etc.) further trims the list

29. 29 Mass Spectrum  Formula Example #2 m/z Molecular ion Relative abundance Conclusions 157 M 100% 158 M+1 9.39% 159 M+2 34% Mass (lowest isotopes) = 157 Odd number of nitrogens 9.39 / 1.1 = 8.5 Eight or nine carbons One Cl; no S or Br

30. 30 Mass Spectrum  Formula Example #2 Oxygens Nitrogens 26 - O - N = H Formula Notes 026 - 0 - 14 = 12C 8 H 12 ClNReasonable1* *Nitrogen rule! Try eight carbons: M - C 8 - Cl = 157 - (8 x 12) - 35 = 26 amu for O, N, and H Not enough amu available for one oxygen/one nitrogen or no oxygen/three nitrogens

31. 31 Mass Spectrum  Formula Example #2 Oxygens Nitrogens 14 - O - N = H Formula Notes 014 - 0 - 14 = 0C 9 ClNReasonable1* *Nitrogen rule! Try nine carbons: M - C 9 - Cl = 157 - (9 x 12) - 35 = 14 amu for O, N, and H Not enough amu available for any other combination.

32. 32 Formula  Structure What does the formula reveal about molecular structure? Functional groups Absent atoms may eliminate some functional groups Example: C 7 H 9 N 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

33. 33 Formula  Structure Calculating 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 + + 1 N 2 nitrogens carbons hydrogens and halogens DBE = C - (H/2) + (N/2) + 1 = 8 - [(10+1)/2] + (1/2) + 1 Four pi bonds and/or ring Possible benzene ring Example C 8 H 10 ClN = 4

34. 34 Formula  Structure Common 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