1. Mass Spectrometry

3. 3 The GC-MS => A mixture of compounds is separated by gas chromatography, then identified by mass spectrometry.

4. Only Cations are Detected

5. How the mass spectrometer works: 1Sample is vapourised in an evacuated chamber M (l) or M (s) → M (g) 2.The gaseous molecules are then bombarded with a stream of high energy electrons (50-70 eV) to give a radical cation (M + , a molecular ion) in the ionization chamber M (g) + e - → M +  (g) + 2e - Mass Spectroscopy

6. Ionization to Radical Cation Molecular Ion (m + )

7. Definitions Molecular ion(or parent )peak - The ion obtained by the loss of one electron from the molecule (M + ) Base peak - The most intense peak in the MS, assigned 100% intensity Radical cation - positively charged species with an odd number of electrons Fragment ions - Lighter cations (and radical cations) formed by the decomposition of the molecular ion. These often correspond to stable carbocations. Isotopic peaks: due to presence of isotopes of C, H, O, N,... in the sample molecule and appear around the main peaks. m/z - mass to charge ratio

8. The resulting mass spectrum is a graph of the mass of each cation vs. its relative abundance. The peaks are assigned an abundance as a percentage of the base peak. –the most intense peak in the spectrum The base peak is not necessarily the same as the parent ion peak. Background

9. 9 Example =>

10. Isotopes Mass spectrometers are capable of separating and detecting individual ions even those that only differ by a single atomic mass unit. As a result molecules containing different isotopes can be distinguished. This is most apparent when atoms such as bromine or chlorine are present ( 79 Br : 81 Br, intensity 1:1 and 35 Cl : 37 Cl, intensity 3:1) where peaks at "M" and "M+2" are obtained. The intensity ratios in the isotope patterns are due to the natural abundance of the isotopes. "M+1" peaks are seen due the the presence of 13 C in the sample.

11. 11 Molecules with Heteroatoms Isotopes: present in their usual abundance. Hydrocarbons contain 1.1% C-13, so there will be a small M+1 peak. If Br is present, M+2 is equal to M +. If Cl is present, M+2 is one-third of M +. If I is present, peak at 127, large gap. If N is present, M + will be an odd number. If S is present, M+2 will be 4% of M +. =>

12. 12 Isotopic Abundance => 81 Br

14. 14 Mass Spectrum with Chlorine =>

15. 15 Mass Spectrum with Bromine =>

16. Bromomethane

17. 17 Mass Spectrum with Sulfur =>

18. 18 Mass Spectra of Alkanes More stable carbocations will be more abundant. =>

19. Octane, m + = 114 m-29 m-43 m-57 m-71 Base peak m+m+

20. n-Decane

21. Effect of Branching in Hydrocarbons

22. Isooctane, no molecular ion

23. 2-Methylpentane

24. 2-Chloropropane

25. 1-Bromopropane

26. Chapter 1226 Mass Spectra of Alkenes Resonance-stabilized cations favored. =>

27. Mass Spectra of Alkynes

28. –Fragment easily resulting in very small or missing parent ion peak ( M + may not be visible). –May lose hydroxyl radical or water M + - 17 or M + - 18 –Commonly lose an alkyl group attached to the carbinol carbon forming an oxonium ion. 1 o alcohol usually has prominent peak at m/z = 31 corresponding to H 2 C=OH + Mass Spectra of Alcohols

29. MS for 1-propanol M+M+ M + -18 SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/28/09) Mass Spectra of Alcohols

30. 1-Butanol m-18

31. Methanol

32. 3-Methyl-1-penten-3-ol m/z = 71 m + =100

33. 4-Methyl-1-penten-3-ol M + =100 m/z = 57

34. Amines –Odd M + (assuming an odd number of nitrogens are present) –a-cleavage dominates forming an iminium ion Mass Spectra of Amines

36. Ethers –  -cleavage forming oxonium ion –Loss of alkyl group forming oxonium ion –Loss of alkyl group forming a carbocation Mass Spectra of Ethers

37. MS of diethylether (CH 3 CH 2 OCH 2 CH 3 ) Mass Spectra of Ethers

38. Aldehydes (RCHO) –Fragmentation may form acylium ion –Common fragments: M + - 1 for M + - 29 for Mass Spectra of Aldehydes

39. MS for hydrocinnamaldehyde M + = 134 105 91 SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/28/09) Mass Spectra of Aldehydes

40. Ketones –Fragmentation leads to formation of acylium ion: Loss of R forming Loss of R’ forming Mass Spectra of Ketones

41. MS for 2-pentanone M+M+ SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/28/09) Mass Spectra of Ketones

42. Esters (RCO 2 R’) –Common fragmentation patterns include: Loss of OR’ –peak at M + - OR’ Loss of R’ –peak at M + - R ’ Mass Spectra of Esters

43. M + = 136 105 77 SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/28/09) Mass Spectra of Esters

44. Examples of Mass Spectra of Carbonyl compounds

45. 3-Pentanone m+m+ m-29 base

46. Cyclohexanone

47. Chloroacetone

48. McLafferty Rearrangement for carbonyl compounds

49. 2-Octanone

50. Decanoic Acid

51. Methyl Octanoate

52. Mass Spectra of Aromatics

53. –Fragment at the benzylic carbon, forming a resonance stabilized benzylic carbocation (which rearranges to the tropylium ion) M+M+ Mass Spectra of benzene

54. Aromatics may also have a peak at m/z = 77 for the benzene ring. M + = 123 77

55. Toluene m+m+ m-1

56. (3-Chloropropyl)benzene

57. Propylbenzene M+M+ M-29

58. Isopropylbenzene M+M+ M-15

59. n-Butylbenzene

60. McLafferty Rearrangements in Alkyl Benzenes

61. Benzamide

62. p-Chloroacetophenone

63. 2,4-Dimethoxyacetophenone

64. High Resolution Mass Spectrometry Determination of Molecular Formula

65. Isotope Ratios Can Help to Determine Molecular Formula Relative intensities (%) MFMW M M+1M+2 CO28.0100 1.120.2 N 2 28.0100 0.76---- C 2 H 4 28.0100 2.230.01

66. The Nitrogen Rule The nitrogen rule states that an odd number of nitrogen atoms will form a molecular ion with an odd mass number. An even number of nitrogen atoms (or none at all) will produce a molecular ion with an even mass number. This occurs because nitrogen has an odd-numbered valence. Examples: C 6 H 5 CH 2 NH 2 MW = 107 H 2 NCH 2 CH 2 NH 2 MW = 60

67. The “Rule of Thirteen” can be used to identify possible molecular formulas for an unknown hydrocarbon, C n H m. –Step 1: n = M + /13 (integer only, use remainder in step 2) –Step 2: m = n + remainder from step 1 Rule of Thirteen

68. Example: The formula for a hydrocarbon with M + =106 can be found: –Step 1: n = 106/13 = 8 (R = 2) –Step 2: m = 8 + 2 = 10 –Formula: C 8 H 10 Rule of Thirteen

69. Rule of 13 Heteroatom Equivalents ElementCH Equivalent ElementCH Equivalent 1 H 12 C 31 PC2H7C2H7 16 OCH 4 32 SC2H8C2H8 14 NCH 2 16 O 32 SC4C4 16 O 14 NC2H6C2H6 35 ClC 2 H 11 19 FCH 7 79 BrC6H7C6H7 28 SiC2H4C2H4 127 IC 10 H 7

70. Candidate molecular formulas for M +. = 108 108/13 = 8, R = 4  C 8 H 12 O = CH 4 ; therefore, C 8 H 12 – CH 4 + O = C 7 H 8 O or C 8 H 12 – 2(CH 4 ) + 2O = C 6 H 4 O 2 or C 8 H 12 – C 6 H 7 + Br = C 2 H 5 Br Calculate three candidate molecular formulas for C 10 H 18.