### An Analysis of Mass Spectral Isotope Patterns of Compounds

```The Chlorine Rule: An Analysis of
Isotope Patterns of Compounds
Containing Multiple Bromine and
Chlorine Atoms
With an Introduction to the Isotope-Pattern Analyzer
Ray A. Gross, Jr.
1
My Reasons for this Presentation
• Present results obtained at
PGCC
• Show that content found in
textbooks can be improved
• Motivate students
2
Isotopes of Br and Cl
a
Low
mass
b
High
mass
Ratio
(a/b)
Rounde Variable
d
# atoms
ratio
Br
79
(50.69)
81 (49.31)
1.028
1:1
m
Cl
35 (75.78)
37
(24.22)
3.129
3:1
n
3
Mass Spectrometer
4
100
77
Br
79
77
61 59
Bromobenzene
Br
81
M = 156 158 = M + 2
Schematic diagram of a mass spectrometer
5
100
77
Br
79
77
61 59
Bromobenzene
Br
81
M = 156 158 = M + 2
6
52
77
Cl
100 32
Chlorobenzene
35
77
Cl
35
3:1
112 114
7
Why Br and Cl?
Br1Cl1
M +2 +4
Br2Cl1
M +2 +4 +6
Br3Cl1
M +2 +4 +6 +8
Molecular-ion peaks of C10H20Br1Cl1, C10H19Br2Cl1 and
C10H18Br3Cl1.
8
Premise
In lieu of pattern matching, it should be
possible to determine the number of Br
and Cl atoms in a molecular formula of a
compound by analyzing the molecularion cluster (i.e., by cluster analysis).
9
Herbert C. Brown
Nobel Laureate
Hydroboration-oxidation
with BH3 (CHM 201)
Reduction with NaBH4
(CHM 202/204)
10
Lillian Berg
NVCC-Annandale
11
100
77
Br
79 or 81
ArBrmCln
61 59
100 98
m = 1, n = 0
A=m+n=1
T=A+1=2
1:1
M = 156 158 = M + 2
12
51:100:49
Br
Br
ArBrmCln
m=2
n=0
A=2
T=3
1:2:1
234 238
13
34:100:97:32
Br
Br
Br
ArBrmCln
m=3
n=0
A=3
T=4
1:3:3:1
312
318
14
18:68:100:66:16
Br
Br
Br
Br
ArBrmCln
m=4
n=0
A=4
T=5
1:4:6:4:1
390 398
15
52
Cl
ArBrmCln
m=0
n=1
A=1
T=2
100 32
3:1
112 114
16
100:66:11
Cl
Cl
m=0
n=2
A=2
T=3
9:6:1
ArBrmCln
146 150
17
27:27:9:1
Cl
Cl
Cl
ArBrmCln
m=0
n=3
A=3
T=4
180
186
18
81:108:54:12:1
Cl
Cl
Cl
Cl
ArBrmCln
m=0
n=4
A=4
T=5
214
222
19
Chlorine Constant
3 1
3 1
Br0Cl1
M +2
3 4 1
3 4 1
Br1Cl1
M +2 +4
3 7 5 1
3 7 5 1
Br2Cl1
M +2 +4 +6
3 10 12 6 1
3 11 13 6 1
Br3Cl1
M +2 +4 +6 +8
20
Bromine Constant
1 1
1 1
3 4 1
3 4 1
Br1Cl0
M +2
Br1Cl1
M +2 +4
9 15 7 1
10 16 7 1
Br1Cl2
M +2 +4 +6
27 54 36 10 1
30 59 38 10 1
Br1Cl3
M +2 +4 +6 +8
IM = 3n
21
190
114 = Br + Cl
76 = benzene ring
a disubstituted benzene
78 100 24
T=3
A=2
*L /R = 78/24 = 3
n=1
m=1
ArBr1Cl1
190
194
190 194
22
T=4
A=3
*L /R = 63/7 = 9
n=2
L
m=1
ArBr1Cl2
224
Br = 79
2Cl = 70
149
Ar = 75
224
L
230
a bromodichlorobenzene
R
224
R
230
23
Theoretical Considerations
Ideal Compounds
Br (a:b) = 1:1
Cl (a:b) = 3:1
13C and 2H negligible
24
Bromine Binomial
• Ratio (a:b) = 1:1
• (1a + 1b)m for Brm
• (1a + 1b)1 = 1a + 1b = 1:1
• (1a + 1b)2 = 1a2 + 2ab + 1b2 = 1:2:1
25
Chlorine Binomial
• Ratio (a:b) = 3:1
• (3a + 1b)n for Cln
• (3a + 1b)1 = 3a + 1b = 3:1
• (3a + b)2 = 9a2 + 6ab + 1b2 = 9:6:1
26
Ideal Model = Binomial Pair
(1a +
m
1b) (3a
+
n
1b)
Br1Cl1
2
3a
+ 4ab +
2
1b
= 3:4:1
27
Results
(1a + 1b)m(3a + 1b)n = 1m3na(m + n) + …. +
1m1nb(m + n)
I(L/R) = 1m3n/1m1n
IM = 3n
Chlorine Rule: When I equals 1, 3, 9,
27 or 81; n is 0, 1, 2, 3, or 4,
respectively, where n = number of
chlorine atoms.
The number of bromine atoms m
equals A – n.
J.Chem.Educ. 2004, 81, 1161-1168 (article available at front desk)
28
Roald Hoffmann-Nobel Laureate
Conservation of orbital symmetry
“Oxygen” Priestley vs Sheele
Hoffmann
Djerassi
Woodward
29
Gross giving lecture with Hoffmann, Djerassi and
Woodward looking on.
30
Structure Begets Properties
• Let’s examine structures.
• Assume 3:1 and 1:1 isotopic abundances
of chlorine and bromine.
• Consider Brm, Cln and BrmCln
compounds.
31
Br
*
Br
*
1
(156)
1
(158)
Br
=
m = 1, N = 2
Br
Br
=
m = 2, N = 4
Br
*
Br
Br
*
Br
1
(234)
Br
*
Br
Br
Br
Br
*
Br
*
Br
Br
*
Br
1
(238)
2
(236)
Br
Br
Br
*
Br
Br
*
Br
Br
Br
Br
=
Br
m = 3, N = 8
Br
*
Br
Br
Br
*
Br
*
Br
Br
1
(312)
3
(314)
Br
Br
3
(316)
1
(318)
Br
*
Br
Br
32
Cl
Cl
*
Cl
*
Cl
*
Cl
*
=
n = 1, N = 4
3
1
(112) (114)
Cl
*
Cl
*
Cl
Cl
Cl
*
Cl
Cl
*
Cl
Cl
*
Cl
Cl
*
Cl
Cl
*
Cl
Cl
*
Cl
Cl
*
Cl
Cl
*
Cl
Cl
Cl
*
Cl
Cl
Cl
*
Cl
Cl
Cl
*
Cl
Cl
=
n = 2, N = 16
Cl
*
Cl
*
Cl
Cl
*
9
6
1
(146) (148) (150)
33
m = 1, n = 0
N=2
Br
Br
1:1
m = 1, n = 1
N=8
BrCl BrCl BrCl BrCl
BrCl BrCl BrCl BrCl
3:4:1
m = 2, n = 1
N = 16
BrClBr
BrClBr
BrClBr
BrClBr
m = 2, n = 2
N = 64
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBr
BrClBr
BrClBr
BrClBr
3:7:5:1
BrClBr
BrClBr
BrClBr
BrClBr
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBr
BrClBr
BrClBr
BrClBr
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrC l
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrC l
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
BrClBrCl
34
9:24:22:8:1
Results
N = 2m4n
N = 2m2n2n
N = 2A2n
Chem. Educ. 2003, 8, 182-186
35
Summary Part I
for BrmCln Compounds
• Derived a chlorine-rule equation, IM = 3n
• Applied it to find gross structures of
unknowns
• Derived a unit-sample equation, N = 2A2n
36
Follow-on to the Chlorine Rule
• An automated A + 2 isotope-pattern
analyzer (IPA)
• IPA is on my website
J. Chem. Educ., in press
37
Example of a Print Out of a Mass
Spectrum in the Molecular-Ion Region
Mass
224
225
226
227
228
229
230
Percent
64.4
4.3
100.0
6.9
45.6
3.2
6.4
38
Molecular-Ion Data is Entered
into the IPA
The Excel program returns the A
+ 2 (Cl, Br, S) composition of the
molecular formula
39
Homework Assignment
for
Selected Students
• Pick up slip from front desk
• Enter data from your slip into IPA
• Obtain the Cl, Br, S composition (e.g.,
Br1Cl2) and record it on your slip
• Write your name on the slip and turn it in
next Tuesday.
40
Acknowledgement: Mass
Spectra from the Spectral
Data Base System (SDBS)
41
Ende
42
Lecture attended by hordes of
students eager to learn.
43
Gross and Friends
44
```