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

2. m/z (or m/e) proportional to 1/V
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. 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. Ionization Methods Electron impact (EI) Chemical ionization (CI)
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. A MASS SPECTRUM The highest mass peak, corresponds to M+
= Molecular Weight of the compound, BUT

6. In nature there are ISOTOPES!!!
Normal isotopes are: 12C 1H 14N etc.
CH3NH2 MW = = 31
However for any molecule that happens to have
a 13C or a 2H or 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 called ISOTOPE PEAKS

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

8. 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. 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 : 37Cl Br : 81Br S : 34S
3 : : ~4%
this makes chlorides, bromides and sulfides easy to spot

10. RBr exists as equal amounts of
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. Because of the 1:1 ratio of 79Br to 81Br
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
{CH2Br}+
81Br2
79Br2
eg. CH2Br2
M-81Br
M-79Br 3

12. Cl2 M = 35Cl2 = 75% x 75% = 56% M+2 = 35Cl37Cl = 75% x 25% x 2 = 38%
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. In the formula CxHyNzOw
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. F=19 monoisotopic M-19 = fluoride I=127 monoisotopic M-127 = iodide
Other Elements
S: M+2 will be 4% of M+
Sn : M+2 = ~ 4n%
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. 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 OCH3 M-91 C6H5CH2 (Benzyl)
Tropylium cation
(6p aromatic)

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

20. with a HRMS, you essentially analyze which elements are present!
12C= H= N= O=
defined
NO =
CH2O =
N2H =
CH2NH2 =
C2H =
All of these are
mass 30 at low resolution
with a HRMS, you essentially analyze which
elements are present!

21. at least C6 X
–Cl( ) = ( )
p. 222 3

22. p. 222
Proton says we have 7H

23. Proton: 7H, so therefore we have C8H7O2Cl
at least C6 X X
–Cl( ) = ( )
Proton: 7H, so therefore we have
C8H7O2Cl
p. 222

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

25. p. 222
-OCH3

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

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

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

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

32. >CH- -CH2- -CH2- C4H5NOF2 all deshielded -F, -F, -O- J = 2J
p. 226
C4H5NOF2
>CH- -CH2- -CH2-
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