1. Mass Spectrometry Part 1 Lecture Supplement page 94 Sample introduction Measure ion mass-to-charge ratio (m/z) Detector Ionization + - Accelerator plates Magnet m/z just right

2. Spectroscopy and Spectrometry Why bother with spectroscopy? Determine structure of unknown substanceVerify purity/identity of known substance Definitions Spectroscopy : Study of the interaction of electromagnetic radiation with matter Spectrometry : Use of spectroscopy as a quantitative tool

3. Spectroscopy and Spectrometry 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. Spectroscopy and Spectrometry MS: C 10 H 15 N IR: Benzene ring, secondary amine (R 2 NH) NMR: Has CH 2 -CH-CH 3, Ph, and CH 3 Example: Unidentified white powder

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 Ionization: X + e -  X +  + 2 e - m/z too large Detector quiet m/z just right Detector fires

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 90.0% 22.0 10.0% Conclusions Neon is a mixture of isotopes Weighted average: (90.0% 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. m/z = (1 x 12) + (4 x 1) = 16 C H The Mass Spectrum Example: Methane CH 4 + e -  CH 4 +  + 2 e - Mass-to-charge ratio (m/z) Relative ion abundance

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 of 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 Also M+1, M+2, etc.

9. Identifying the Molecular Ions and Base Peak Highest m/z not always M M: m/z = 170 Which peaks are molecular ions?C 7 H 7 Br Most abundant ion in spectrum Not always a molecular ion Not always 100% relative abundance Base peak: m/z Relative ion abundance

10. The Mass Spectrum Origin of Relative Ion Abundances This table will be provided on an exam. Do not memorize it. M contributorsM+1 contributorsM+2 contributors Isotope Natural abundance Isotope Natural abundance Isotope Natural abundance 1H1H99.985 % 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 %

11. The Mass Spectrum Relative Intensity of Molecular Ion Peaks Imagine a sample containing 10,000 methane molecules... *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%) Molecule # in samplem/zRelative 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%*

12. Formula from Mass Spectrum M+1 Contributors Examples:C 2 H 6 M = 100%; M+1 = 2.26% C 6 H 6 M = 100%; M+1 = 6.66% Working backwards gives a useful observation... When relative abundance 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%) are often ignored Comparing many mass spectra reveals M+1 intensity  ~1.1% per C in formula

13. Formula from Mass Spectrum M+2 Contributors Anything useful from intensity of M+2? Conclusion: Mass spectra of molecules with S, Cl, or Br have significant M+2 peaks IsotopesNatural abundances (%)Intensity ratio M/(M+2) 32 S : 34 S95.0 : 4.2100 : 4.4 35 Cl : 37 Cl75.8 : 24.2100 : 31.9 79 Br : 81 Br50.7 : 49.3100 : 97.2

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

15. 122 Relative ion 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

16. Formula from Mass Spectrum M: Reveals mass of molecule composed of lowest mass isotopes M+1: Intensity of M+1/1.1% = approximate number of carbons M+2: ~4% intensity reveals presence of sulfur ~33% intensity reveals presence of chlorine ~100% intensity reveals presence of bromine Next lecture : Procedure for deriving formula from mass spectrum Summary of Information from Mass Spectrum

17. Mass Spectrometry Part 2 Lecture Supplement page 95 122 Relative ion abundance m/z 124

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 (~4%), chlorine (~33%), or bromine (~100%) Fragment ion: From decomposition of molecular ion before reaching detector Analysis of fragmentation patterns not important for Chemistry 14C

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

20. Mass Spectrum  Formula  Structure How do we derive formula from the mass spectrum? 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 m/z and relative intensities of M, M+1, and M+2

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. 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, so # N is even odd nitrogen count discarded even nitrogen count

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 Pi bonds

24. How Many Hydrogen Atoms? Conclusion: Each ring reduces max hydrogen count by two One ring Two 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 Rings

25. How Many Hydrogen Atoms? One nitrogen Two 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 Same conclusions apply for F, Cl, Br, and I instead of H For C carbons and N nitrogens: Max number of hydrogens + halogens = 2C + N + 2 Hydrogen/Halogen Rule (H Rule) Nitrogen atoms

26. Mass Spectrum  Formula 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 reveals 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/halogen rule Procedure

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 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 Better: Try both six carbons and seven carbons < 4% so no S, Cl, or Br Even number of nitrogens Oxygen? Given information

28. Mass Spectrum  Formula Example #1 Mass (M) - mass (C, S, Cl, Br, F, and I) = mass (N, O, and H) 102 - 102 (C 6 ) = 30 amu for N, O, and H *Nitrogen rule! Other data (functional groups from IR, NMR integration, etc.) further trims the list OxygensNitrogens30 - O - N = HFormulaNotes 0030 - 0 - 0 = 30C 6 H 30 Violates H-rule 1030 - 16 - 0 = 14C 6 H 14 OReasonable 2030 - 32 - 0 = -2C 6 H -2 O 2 Not possible 02*30 - 0 - 28 = 2C6H2N2C6H2N2 Reasonable

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. 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 Oxygens Nitrogens 14 - O - N = H Formula Notes 014 - 0 - 14 = 0C 9 ClNReasonable 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. 1* *Nitrogen rule!

31. Formula  Structure What does the formula reveal about molecular structure? 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 pair of H less than max H count = double bond equivalent (DBE) If formula has less than full H count, molecule must contain at least one pi bond or ring Functional groups

32. 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 rings Possible benzene ring Example C 8 H 10 ClN = 4 DBE

33. Formula  Structure Common Math Errors Divide by 1.1  divide by 1.0 DBE cannot be a fraction DBE cannot be negative Small math errors can have devastating effects! No calculators on exams Avoid these common spectroscopy problem math errors: