MASS SPECTRA measure a compound’s Mol. Wt.

1. p. 213 MASS SPECTRA measure a compound’s Mol. Wt. M+ e- + Molecule Molecule + + cule Mole + + 2 e- Mol ecule + + Mo lecule This ionization type is called: electron impact MS

2. p. 213 where m = ion mass n = ion velocity e (or z) = ion charge V = potential diff. of accelerating plates r = radius of curvature of the path B = magnetic field strength when B is constant and r is constant m/z (or m/e) proportional to 1/V z is normally +1, but sometimes eject 2 electrons, so get M2+, so get peak at M/2 By changing V slowly, can collect fragments in increasing order of mass

3. p. 214 A MASS ‘SPECTRUM’ Amount of charge (= # ions) received by collector measured for each mass, and plotted as graph of m/e (strictly m/z) vs relative abundance

4. p. 221 Ionization Methods Electron impact (EI) M+ from electron beam impact – lots of fragmentation Chemical ionization (CI) M+ and M+H+ from CH4 collisions – relatively little fragmentation Fast atom bombardment (FAB) M+ and M+H+ from Argon beam collisions – low fragmentation Matrix-assisted laser desorption (MALDI) M+H+ from matrix protonation Electrospray ionization (ESI) M+ or M- but works best if molecule already charged – little fragmentation

5. p. 214 A MASS SPECTRUM The highest mass peak, corresponds to M+ = Molecular Weight of the compound, BUT

6. p. 214 In nature there areISOTOPES!!! Normal isotopes are: 12C 1H 14N etc. CH3NH2 MW = 12 +3 +14 +2 = 31 However for any molecule that happens to have a 13Cor a 2Hor a 15N, its MW = 32 and if molecule happens to have a 13C and a 2H then its MW = 33 these of course will be seen (BUT ARE SMALL) and are calledISOTOPE PEAKS

7. p. 214 and 218 Nitrobenzene = C6H5NO2 = 123 BASE PEAK Molecular Ion, M+ ISOTOPE PEAK How large are the isotope peaks?

8. p. 214 How large are isotope peaks? For most elements we use, not very! 13C = 1.1%; 2H = 0.02%; 15N = 0.4%; 18O = 0.2%

9. p. 214 How large are isotope peaks? For most elements we use, not very! 13C = 1.1%; 2H = 0.02%; 15N = 0.4%; 18O = 0.2% However for some, that is not true: 35Cl : 37Cl79Br : 81Br 32S : 34S 3 : 11 : 1~4% this makes chlorides, bromides and sulfides easy to spot

10. p. 215 Bromides: RBr exists as equal amounts of R79Br and R81Br so MW = [R+79] AND [R+81] we call these M+ and M+2 R=CH3CH2 = 29 108 110 fragment 3

11. p. 215 Brn Because of the 1:1 ratio of 79Br to 81Br higher bromides show ‘NMR’-like patterns i.e. Br2 = 1:2:1 (M) : (M+2) : (M+4) Br3 = 1:3:3:1 M, M+2, M+4, M+6 Br4 = 1:4:6:4:1 etc 79Br81Br 81Br79Br 81Br2 {CH2Br}+ 79Br2 eg. CH2Br2 M-81Br M-79Br 3

12. p. 216 Cl2 M = 35Cl2 = 75% x 75% = 56% M+2 = 35Cl37Cl = 75% x 25% x 2 = 38% M+4 = 37Cl2 = 25% x 25% = 6% = 100 : 67 : 11

13. p. 216 {CH2Cl}+ CH2Cl2 = 100 : 67 : 11 M-35Cl 1

14. p. 217 OTHER ELEMENTS {C,H, N, O} In the formula CxHyNzOw 17O Chance of seeing the M+1 and M+2 peaks? 18O Essentially C isotopes are the only ones to worry about here M+1 = (# Carbons) x 1.1%

15. p. 218 Nitrobenzene = C6H5NO2 = 123 So, the relative size of M+1 for an organic can be used to establish the # of C M+1 = 6 x 1.1% = 7%

16. p. 217 Other Elements Sn : M+2 = ~ 4n% S: M+2 will be 4% of M+ F=19 monoisotopic M-19 = fluoride I=127 monoisotopic M-127 = iodide Some elements have lots of isotopes eg. Sn (TIN) has 10! more complex but computable patterns

18. p. 219 Fragmentation patterns (we aren’t focusing on this much but it is useful to know some common fragments) M-1 H M-43 C3H7 (i-Pr) M-15 CH3 (Me) M-44 CO2 M-17 OH M-45 CO2H M-28 CO M-57 C4H9 (t-Bu) M-29 C2H5 (Et)M-77 C6H5 (Ph) M-31 OCH3M-91 C6H5CH2 (Benzyl) Tropyliumcation (6p aromatic)

19. p. 221 HIGH RESOLUTION MASS SPECTRA HRMS instruments can measure accurately to 4 decimal places of mass! so need to use these values, not integers!

20. p. 221 12C=12.00001H=1.00783 14N=14.0031 16O=15.9949 defined NO = 29.99800 CH2O = 30.01056 N2H2 = 30.02186 CH2NH2 = 30.03442 C2H6 = 30.04698 All of these are mass 30 at low resolution with a HRMS, you essentially analyze which elements are present!

21. at least C6 X 170.014 –Cl(34.9689) = 135.045(+- 0.004) p. 222 3

22. p. 222 Proton says we have 7H

23. at least C6 X X 170.014 –Cl(34.9689) = 135.045(+- 0.004) Proton: 7H, so therefore we have C8H7O2Cl p. 222

24. p. 222 C8H7ClO 2 DBE = {(2x8 + 2)-(7+1)}/2 = 5 = aromatic + 1 170s 155s 135s 130d 110d 56q -OCH3 }C8 benzene = C6 -CO X acid chloride , possibly C=C-O–C=O IR = 1770 840 para-benzene 5

25. p. 222 -OCH3

26. p. 222 DBE = {(2x8 + 2)-(7+1)}/2=5 = aromatic + 1 C8H7ClO 2 170s155s 135s 130d 110d 56q -OCH3 }C8 benzene = C6 -CO X acid chloride , (possibly C=C—O–C=O ) IR = 1770 840 para-benzene 110 155

27. p. 223 E = F X X 154.042-18.998 = 135.044 (+-0.002) But from 13C: >4C, so X and from 1H have 7H, so X C8H7O2F {(2x8 +2)-(7+1)}/2 = 5 (aro + 1) 7

28. p. 224 C8H7FO2 174s 133s 130d 129d 126d 89d F-decoupled -COO benzene = C6 dd 1J=186Hz } C8 >CHF IR -COOH not conj mono-benzene 10

29. p. 224 3

30. p. 225 = F2 121- 83 = 38 121.034 – 2(18.998) = 83.038 (4) but at least 4C so C4H5NOF2 DBE = {(2x4 +2 +1) – (5 + 2)}/2 =2 p. 225 5

31. p. 225 C4H5NOF2 } C4H5N >CH- -XCH2- Y-CH2- 3

32. p. 226 >CH- -CH2- -CH2- C4H5NOF2 all deshielded -F, -F, -O- J = 2J triplet (of t) -CHF2 (of t) next to –CH2- J -CH2-CHF2 td 6

33. p. 226 -CH2-CHF2 X --O-- --CH2-- -OCH2- singlet F2CH—CH2—O—CH2—CN 2

34. ASSIGNMENT 9