Mass Spectrometry (MS)

Transcription

1 Mass Spectrometry (MS) MW Molecular formula Structural information GC-MS LC-MS

2 To Do s Read Chapter 7, and complete the endof-chapter problem 7-4. Answer Keys are available in CHB204H

3 MS Principles Molecule Ions Ion separation Ion collector Molecular ion + fragments Based on m/z values (1) Hard ionization (2) Soft ionization EI (electron impact) CI(chemical) DI(desorption FAB (fast-atom-bombardment) MALDI (matrix-assisted laser desorption) ESI(electrospray) Magnet Magnet + electric field (double focusing) Quadruple mass filter Ion trap Time-of-flight Ion flux m/z

4 Simplest MS EI-MS Use high energy electron beam for ionization The energy of electron beam is high enough to kick out a bonding electron, kcal/mol (or 8-13 ev) Molecular ion Or parent ion

5 EI-MS Instrument -1 E = 1/2 mυ 2 = zv m = ion mass z = ion charge υ = velocity of ion V = accelerating field potential υ 2 = 2zV/m

6 EI-MS Instrument -2 In magnetic fields, ions travel in a circular path of different radius(r), depending on their mass (m) and charges(z). r = mυ zh υ 2 = H2 z 2 r 2 m 2 H : magnetic field strength

7 EI-MS Instrument - 3 υ 2 = 2zV/m υ 2 = H2 z 2 r 2 m 2 H 2 r 2 m/z = 2V r (instrument) is fixed, and usually, V is kept constant. H corresponds to a collection of ions with the same m/z values.

8 MS of Methane MW M molecular ion Ion current fragments or daughter ions isotope peaks 13 CH 4, CH 3 D, etc m/z Structural info Molecular formula

9 Molecular Ion Molecular ion is the ion of the highest mass in the spectrum (except for isotope peaks).

10 EI-MS of Anisole M Isotope peaks Note: significant fragmentation

11 EI-MS of Methyl Hexyl Ether M+ EI is called a hard ionization, and significant fragmentation is generally observed. EI-MS is commonly used for small molecules with MW < 600.

12 To Observe Molecular Ion or ESI

13 Soft Ionizations (CI) CI : chemical ionization Electron beam M + excess CH 4 (carrier)

14 EI- and CI-MS of THF EI CI

15 CI-MS of Pentylbenzene Carrier gas = isobutane (C 4 H 10 ) [M + H] + C 11 H 16 = 148 [M + C 3 H 7 ] + M+

16 Soft Ionizations (FAB and MALDI) FAB : Fast-Atom-Bombardment MALDI : Matrix-assisted Laser Desorption MALDI Preferred MALDI matrix or Xenon beam (FAB) Sample and matrix must form good crystals - intimate mixing may be required.

17 MALDI-MS of Cytochrome-c

18 MALDI-MS of Transferrin

19 Soft Ionizations (ESI) ESI : Electrospray Ionization Very mild, no high energy collision is involved Commonly used for large molecules Different ionization mechanism from others Multiply charges species are commonly observed

20 ESI-MS of Cytochrome-c Isotope peaks For [M + nh] n+, the spacing is 1/n

21 ESI-MS of Transferrin

22 ESI-MS of Transferrin (deconvoluted)

23 Determining MW by MS [M+H] + = H 3 C H 2 N O S O N H N N CH 3 = C 12 H 14 N 4 O 2 S MW = [M+H] = Close, but not the same.

24 Averaged MW and Monoisotopic MW To determine MW, find the molecular ion and use m/z value. If z = 1, m/z = MW. MW for methane (CH 4 ) x x 4 = This is called averaged MW, and NOT what MS will give you. CH 4 in nature may contain isotopes of carbon and hydrogen according to their natural abundances. Since MS sorts out ions one by one, [ 12 CH 4 ] + and [ 13 CH 4 ] + would show up at different m/z values. 12 CH 4 = 16, 13 CH 4 = 17, 12 CH 3 D = 17 etc D = 2 H

25 Monoisotopic MW of Methane mass(amu) abundance 12 C % 13 C % mass(amu) abundance 1 H % 2 H(D) % CH 4 = x x 4 = amu (exact mass of CH 4 ) It rounds to 16 (nominal mass). cf. averaged MW = Most MS instruments can determine only nominal mass, except for high-resolution MS (expensive).

26 Molecular Ion and Isotope Peaks of Methane m/z

27 Calculating the Exact Mass Find the most abundant isotopes, and use their exact masses for calculation. C 6 H 6 O = x x x 1 = C 3 H 7 NO 2 S = x x x x x 1 = * ChemDraw can calculate exact mass.

28 The Nitrogen Rule Note: use nominal mass (1) A molecule with even # of nitrogen, including zero, has even molecular mass. (2) A molecule with odd # of nitrogen has odd molecular mass. The nitrogen rule works because nitrogen has an even mass (14) and has and odd-numbered valency (3). CH 4 = 16, CH 3 OH = 32, C 6 H 5 OH = 94 C 5 H 6 N 2 (4-aminopyridine) = 94 C 2 H 7 N (ethylamine) = 45, C 9 H 7 N (quinoline) = 129

29 Benzamide

30 How about this compound?

31 Even # of Nitrogens (M+H) Exact Mass 425 (M+H) Nominal Mass 424 (M) Even # of nitrogen!

32 MF from Exact Mass Exact mass is so accurate that, for small molecules, only a particular combination of elements gives the observed mass. Exact mass Molecular Formula CO N C 2 H 4

33 For Larger Molecules, Not So Straightforward Exact mass Molecular Formula ± C 3 H 7 NO 2 S ± C 3 H 7 NO 2 S H 3 N 5 O 3 CH 5 N 4 OS C 4 HN 4 O C 6 H 3 NO ± possible molecular formula

34 Using Isotope Peaks M + m/z = 128, 100% MW = 128 C 6 H 5 OCl m/z = 130 (M+2), 32.2% m/z = 131 (M+3), 2.1% M + = 128 (100%) m/z = 129 (M+1), 6.9% M+1 = 6.9% M+2 = 32.2% M+3 = 2.1 % Relative intensities of isotope peaks Elemental composition of molecule (molecular formula)

35 Isotope Peaks for Different MF

36 The Table 1 1 H = = (deviation from nominal mass) 2 H(D) = (M+1 isotope of 1 H) Nominal mass Natural abundances relative to M ( )

37 Calculating Isotope Peaks of Methane M + M+1 (%) M+2 (%) 13 CH 4 CH 3 D = ( x 4) x 100 = 1.17% = { x x 4 4! x x } x 100 2!x2! = 0.07% 13 CH 3 D CH 2 D 2 Balls (m) Boxes (n) # of combinations = n! m!(n-m)!

38 Isotope Peaks of Propanol (C 3 H 8 O) (1) M+1 (%) 13 C x 3 x 100 = 3.33 D x 8 x 100 = O x 1 x 100 = % (2) M+2 (%) 13 C 2 {(0.0111) 2 x 3!/2!1!} x 100 = O x 1 x 100 = CD x 3 x x 8 x 100 = D 2 {( ) 2 x 8!/2!6!} X 100 = C 17 O x 3 x x 1 x 100 = D 17 O ~ 0 Major contributor 0.24% Major contributors

39 C 6 H 5 ClO M+1 13 C, D, 17 O M+2 13 C 2, 18 O, 37 Cl, 13 CD, D 2, 13 C 17 O, D 17 O M+3 13 C 3, 13 C 2 D, 13 C 17 2 O, 13 CD 2 etc M+4 many Major contributors

40 Calcd vs Obsd C 10 H 10 O 2 S M + = 194 (100%) M % calcd 12.12% M+2 5.8% calcd 5.35% M+3 0.7% calcd 0.53%

41

42 A Source of Error Multiple molecular formula are typically involved in an isotope peak - A challenge in MS simulation. - Higher resolution instruments can resolve these peaks.

43 # of Carbons from M+1 Peak The major contributor for M+1 peak is the species that contain one 13 C in the molecule. Each carbon in the molecular formula adds 1.11% to M+1 peak. # of carbons ~ M+1(%) 1.11

44 Naphthalene & 1,3- dimethylnaphthalene Naphthalene C 10 H 8 1,3-dimethylnaphthalene C 12 H (M) 100% (M+1) 11% 11/1.11 = (M) 100% (M+1) 13% 13/1.11 = 11.7

45 Thioanisole & Styrylthioacetic acid C 7 H 8 S C 10 H 10 O 2 S 124 (M) 100% 125 (M+1) 8.9% 126 (M+2) 4.4% 8.9 / 1.11 = (M) 100% 125 (M+1) 8.9% 126 (M+2) 4.4% 12.9 / 1.11 = 11.6

46 Unknown Example #1 Unknown hydrocarbon M + = 84 (100%) M+1 = 6.8% M+2 = 0.2%

47 Hand Calculation of MF (Example 1) Unknown hydrocarbon - only C & H M + = 84 (100%) M+1 = 6.8% M+2 = 0.2% 6.8 / 1.11 = 6.13 C 5, C 6 or C 7? Calculated M+2. Major contributor is 13 C 2. If C 5, M+2 = {(0.0111) 2 x 5!/2!3!} x 100 = 0.12 If C 6, M+2 = {(0.0111) 2 x 6!/2!4!} x 100 = 0.18 A fit! If C 7, M+2 = {(0.0111) 2 x 7!/2!5!} x 100 = 0.26 Thus, C 6 H x. MW = 84. X = 12 C 6 H 12 M+1 (calcd) = 6.85% M+2 (calcd) = 0.19%

48 Unknown Example #2 M + = 108 M+1 = 7.95% M+2 = 0.45%

49 Hand Calculation of MF (Example 2) M + = 108 M+1 = 7.95% M+2 = 0.45% 7.95 / 1.11 = 7.2 Use the nitrogen rule - even #, including zero, of nitrogen C 7? M+2 (calcd, only 13 C 2 ) = {(0.0111)2 x 7!/2!5!} x 100 = 0.26 < 0.45 (obsd) Perhaps, contains one oxygen ( 18 O = 0.2%) (M+1) corr = ( 17 O) = 7.91, 7.91 / 1.11 = 7.1 Still C 7 M+2 (calcd, 13 C 2 and 18 O) = {(0.0111) 2 x 7!/2!5!} x = 0.46% So, perhaps, C 7 H x O, and MW = 108. x = 8 C 7 H 8 O M+1(calcd) = 7.94% M+2(calcd) = 0.46%

50 # of Hydrogens from Exact Mass The fractional part of exact mass comes from H and other non-carbon elements. Except for carbon, the most abundant element in organic molecule is usually hydrogen. Thus, the fractional part of exact mass could be used to determine the # of hydrogen.

51 # of Hydrogens from Exact Mass - 2 C 60 = (no hydrogen!) CH 4 = C 6 H 6 = C 3 H 8 O = / = 4 (four hydrogens) / = 6 (six hydrogens) / = 7.3 ~ 8H If you know that there is one oxygen prior to the calculation, ( ) = 8 (eight hydrogens) Use additional information to increase the accuracy of calculation

52 Unknown Example #3 M + = ± M+1 = 20.6 ± 0.2% M+2 = 8.8 ± 0.2%

53 Are there any heteroatoms? M + = ± M+1 = 20.6 ± 0.2% M+2 = 8.8 ± 0.2% 20.6/1.11 = C 19? 19 x 12 = 228 > 218 Cannot be a hydrocarbon!

54 Hand Calculation of MF (Example 3) M + = ± M+1 = 20.6 ± 0.2% M+2 = 8.8 ± 0.2% nominal mass = 218 Even #, including zero, of nitrogens From M+2, S 2, SiS or Si 2 (a) If S 2, (M+1)corr = ( 33 S) x 2 = / 1.11 = C 17? C 17 S 2 = 268 > 218, impossible! (b) If SiS, (M+1)corr = ( 29 Si) = / 1.11 = 13.2 C 13? M+2(calcd, 13C2, 34S and 30Si) = {(0.0111)2 x 13!/2!11!} x = 8.71 But, C 13 SSi = 216. Only 2 hydrogens?? A fit! C 13 H 2 SSi = (does not match )

55 Hand Calculation of MF (Example 3), cont. (c) So, it must be Si 2 (M+1) corr = x 2 = / 1.11 = 9.37 C 9? Now, we use exact mass to estimate the # of hydrogens x # of Hs = = 20.7 H ? ΔMW = x 28-9 x 12-1 x 21 = 33 Try different combinations of elements; 2O + 1H = 33 C 9 H 22 O 2 Si 2 = N + 5H = 33 C 9 H 26 N 2 Si 2 = O + F - 2H = 33 C 9 H 20 OFSi 2 = A fit! (M+1) calcd = 20.6% (M+2) calcd = 8.92%

56 MF Calculation - Summary (1) M+ nitrogen rule M+2 hetero atom? (2) M+1 # of carbons (3) Calculate expected M+2 Include 13 C 2 and (M+2) heteroatoms (4) Exact mass # of hydrogens (5) Guess other elements (6) Verify MF by calculating exact mass, M+1, M+2 etc

57 An Alternative Strategy - The Rule of 13 - Start with C n H n, or (CH) n for molecular ion, CH = 13 amu Add hydrogens to obtain the initial MF. m/z = 92, 92/13 = 7 and R = 1 (CH) 7 + 1H = C 7 H 8 Add heteroatoms and adjust the number of carbon and hydrogen. N = 14 is equivalent of CH 2 O = 16 is equivalent of CH 4

58 Unknown#4 86/13 = (CH) 6 + 8H = C 6 H 14 M+ = 86

59 Unknown #4, IR OH C 6 H 14 C 5 H 10 O (add O and remove CH 4 )

60 Unknown #5 M + = 171 M+2

61 Unknown #5 Significant M+2 peak suggests a Cl. Odd MW suggests an odd number of nitrogen = = (CH) 10 +6H = C 10 H 16 Cl Assume one nitrogen, C 9 H 14 NCl (add N and remove CH 2 )

62 Unknown#5, IR NO2 C 9 H 14 NCl C 7 H 6 NO 2 Cl (add 2 x O and remove 2 x CH 4 )

63 Lithium and Boron Lithium 6 Li 7.4% 7 Li 92.6% Boron 10 B 19.7% 11 B 80.3%

64 The Table 1 - B and Li

65 Boron-containing Molecules C 12 H 11 BO 2 = 198 C 22 H 28 B 2 O 4 = % (M-1) 198 (M + ) 100% % % % (M-2) % (M-1) 378 (M + ) 100% % %

66 Lithium-containg Molecules C 5 H 7 LiO 2 = % (M-1) 106 (M + ) 100% %