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

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

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 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 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 e - m/z too large Detector quiet

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 % % 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 The Mass Spectrum Example: methane CH 4 + e -  CH 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 The Mass Spectrum Alternate data presentation... m/z (amu) Relative abundance (%) 18 < 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 The Mass Spectrum Origin of Relative Ion Abundances M contributorsM+1 contributorsM+2 contributors Isotope Natural Abundance Isotope Natural Abundance Isotope Natural Abundance 1H1H % 2H2H0.015% 3H3Hppm 12 C C Cppm 14 N N O O O F S S S Cl Cl Br Br I100.0 This table will be provided on an exam. Do not memorize it.

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 x 1) = 16100% 13 C 1 H (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 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 Formula from Mass Spectrum M+2 Contributors Anything useful from intensity of M+2? 32 S : 34 S Isotopes Natural abundances Intensity M : M : : Cl : 37 Cl75.8 : : Br : 81 Br50.7 : : 97.2 Conclusion: Mass spectra of molecules with S, Cl, or Br have significant M+2 peaks

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

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

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

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 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 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 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 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 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

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 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 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 Mass Spectrum  Formula Example #1 m/z Molecular ion Relative abundance Conclusions 102 M 100% 103 M+1 6.9% 104 M % 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 Mass Spectrum  Formula Example #1 Mass (M) - mass (C, S, Cl, Br, F, and I) = mass (N, O, and H) C 6 = (6 x 12) = 30 amu for N, O, and H Oxygens Nitrogens 30 - O - N = H Formula Notes = 30C 6 H 30 Violates hydrogen rule = 14C 6 H 14 OReasonable = -2C 6 H -2 O 2 Not possible 0 2* *Nitrogen rule! = 2C6H2N2C6H2N2 Reasonable Other data (functional groups from IR, NMR integration, etc.) further trims the list

29 Mass Spectrum  Formula Example #2 m/z Molecular ion Relative abundance Conclusions 157 M 100% 158 M % 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 = 12C 8 H 12 ClNReasonable1* *Nitrogen rule! Try eight carbons: M - C 8 - Cl = (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 Mass Spectrum  Formula Example #2 Oxygens Nitrogens 14 - O - N = H Formula Notes = 0C 9 ClNReasonable1* *Nitrogen rule! Try nine carbons: M - C 9 - Cl = (9 x 12) - 35 = 14 amu for O, N, and H Not enough amu available for any other combination.

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 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 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 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