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.

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My Reasons for this

Presentation

•

Present results obtained at PGCC

•

Show that content found in textbooks can be improved

•

Motivate students

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Isotopes of Br and Cl

a Low mass b High mass

Ratio (

a

/

b

) Rounde d ratio Variable # atoms Br 79 (50.69) 81 (49.31) 1.028

Cl 35 (75.78) 37 (24.22) 3.129

1:1 3:1

m n

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Mass Spectrometer

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77 Br 79 77 Br 81 100 61 59 Bromobenzene M = 156 158 = M + 2 Schematic diagram of a mass spectrometer

5

77 Br 79 77 Br 81 100 61 59 Bromobenzene M = 156 158 = M + 2

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77 Cl 35 77 Cl 35 52 Chlorobenzene 100 32 3:1 112 114

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Why Br and Cl?

Br 1 Cl 1 Br 2 Cl 1 Br 3 Cl 1 M +2 +4 M +2 +4 +6 M +2 +4 +6 +8 Molecular-ion peaks of C 10 H 20 Br 1 Cl 1 , C 10 H 19 Br 2 Cl 1 C 10 H 18 Br 3 Cl 1 .

and

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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 molecular ion cluster (i.e., by cluster analysis).

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Herbert C. Brown Nobel Laureate

Hydroboration-oxidation with BH 3 (CHM 201) Reduction with NaBH 4 (CHM 202/204)

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Lillian Berg NVCC-Annandale

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77 Br 79 or 81 ArBr

m

Cl

n

100 61 59 m = 1, n = 0 A = m + n = 1 T = A + 1 = 2 100 98 1:1 M = 156 158 = M + 2

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Br Br ArBr

m

Cl

n

51:100:49 m = 2 n = 0 A = 2 T = 3 1:2:1 234 238

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Br Br Br ArBr

m

Cl

n

34:100:97:32 m = 3 n = 0 A = 3 T = 4 1:3:3:1 312 318

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Br Br Br Br ArBr

m

Cl

n

m = 4 n = 0 A = 4 T = 5 18:68:100:66:16 1:4:6:4:1 390 398

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Cl ArBr

m

Cl

n

52 m = 0 n = 1 A = 1 T = 2 100 32 3:1 112 114

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Cl Cl ArBr

m

Cl

n

m = 0 n = 2 A = 2 T = 3 100:66:11 9:6:1 146 150

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Cl Cl Cl ArBr

m

Cl

n

m = 0 n = 3 A = 3 T = 4 27:27:9:1 180 186

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Cl Cl Cl Cl ArBr

m

Cl

n

81:108:54:12:1 m = 0 n = 4 A = 4 T = 5 214 222

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Chlorine Constant

3 1 3 1 Br 0 Cl 1 3 4 1 3 4 1 Br 1 Cl 1 3 7 5 1 3 7 5 1 Br 2 Cl 1 3 10 12 6 1 3 11 13 6 1 Br 3 Cl 1 M +2 M +2 +4 M +2 +4 +6 M +2 +4 +6 +8

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1 1 1 1

Bromine Constant

3 4 1 3 4 1 Br 1 Cl 0 Br 1 Cl 1 9 15 7 1 10 16 7 1 Br 1 Cl 2 27 54 36 10 1 30 59 38 10 1 Br 1 Cl 3 M +2 M +2 +4 M +2 +4 +6 M +2 +4 +6 +8

I

M = 3

n

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190 114 = Br + Cl 76 = benzene ring a disubstituted benzene T = 3 78 100 24 A = 2 *L 190 /R 194 = 78/24 = 3 n = 1 m = 1 ArBr 1 Cl 1 190 194

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T = 4 A = 3 *L 224 /R 230 = 63/7 = 9 L Br = 79 2Cl = 70 149 Ar = 75 224 n = 2 m = 1 ArBr 1 Cl 2 a bromodichlorobenzene L R 224 R 230

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Theoretical Considerations

Ideal Compounds Br (

a

:

b

) = 1:1 Cl (

a

:

b

) = 3:1 13 C and 2 H negligible

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Bromine Binomial

•

Ratio (

a

:

b

) = 1:1

•

(1

a

+ 1

b

)

m

for Br

m

•

(1

a

+ 1

b

) 1 = 1

a

+ 1

b

= 1:1

•

(1

a

+ 1

b

) 2 = 1

a

2 + 2

a b

+ 1

b

2 = 1:2:1

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Chlorine Binomial

•

Ratio (

a

:

b

) = 3:1

•

(3

a

+ 1

b

)

n

for Cl

n

•

(3

a

+ 1

b

) 1 = 3

a

+ 1

b

= 3:1

•

(3

a

+

b

) 2 = 9

a

2 + 6

a b

+ 1

b

2 = 9:6:1

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Ideal Model = Binomial Pair

(

1a

+

1b

)

m

(

3a

+

1b

)

n

Br 1 Cl 1

3a

2 + 4

a b

+

1b

2 =

3

:4:

1

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Results ( 1a + 1b )

m

( 3a + 1b )

n

= 1

m

3

n a

(m + n) 1

m

1

n b

(m + n) + …. +

I

( L / R ) = 1

m

3

n

/ 1

m

1

n I

M = 3

n

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)

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Roald Hoffmann-Nobel Laureate Conservation of orbital symmetry “Oxygen” Priestley vs Sheele Hoffmann Djerassi Woodward

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Gross giving lecture with Hoffmann, Djerassi and Woodward looking on.

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Structure Begets Properties

• •

Let’s examine structures.

Assume 3:1 and 1:1 isotopic abundances of chlorine and bromine.

•

Consider Br

m

, Cl

n

compounds.

and Br

m

Cl

n

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Br

=

m = 1, N = 2 Br Br

=

m = 2, N = 4

*

Br

*

Br 1 (156) Br

*

Br Br 1 (234) 2 (236)

*

Br 1 (158)

*

Br Br

*

Br Br 1 (238) Br Br Br m = 3, N = 8

= *

Br Br Br

*

Br Br Br

*

Br Br Br

*

Br Br Br

*

Br Br

*

Br Br

*

Br Br Br 1 (312) Br 3 (314) 3 (316) 1 (318) Br

*

Br Br Br

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Cl

*

Cl

*

Cl

*

Cl

*

Cl

=

n = 1, N = 4 Cl Cl

=

n = 2, N = 16

*

Cl Cl 3 (112) 1 (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 9 (146) 6 (148) 1 (150)

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m = 1, n = 0 N = 2 m = 1, n = 1 N = 8 m = 2, n = 1 N = 16 m = 2, n = 2 N = 64 Br Br 1:1 BrCl BrCl BrCl Br Cl Br Cl Br Cl Br Cl BrCl 3:4:1 BrClBr BrClBr BrClBr Br Cl Br Br ClBr Br ClBr Br ClBr BrCl Br BrCl Br BrCl Br BrCl Br Br ClBr Br Cl Br Br Cl Br Br Cl Br BrClBr 3:7:5:1 BrClBrCl BrClBrCl BrClBrCl BrClBrCl BrClBrCl BrClBrCl BrClBrCl BrClBr BrClBrCl BrClBr Cl Cl BrClBrCl BrClBr Cl Br Cl BrCl Br Cl BrCl Br Cl BrCl Br Cl Br C l Br ClBrCl Br ClBrCl Br ClBrCl Br ClBr Cl Br ClBrCl Br ClBrCl Br ClBrCl Br ClBr Cl Br ClBrCl Br ClBrCl Br ClBrCl Br ClBr Cl BrCl BrCl BrCl BrCl BrCl BrCl BrCl Br C l BrCl Br Cl BrCl Br Cl BrCl Br Cl BrCl BrCl BrCl Br Cl BrCl Br Cl BrCl Br Cl BrCl BrCl BrCl Br Cl BrCl Br Cl BrCl Br Cl BrCl BrCl Br ClBr Cl Br ClBr Cl Br ClBr Cl Br ClBrCl Br Cl Br Cl Br Cl Br Cl Br Cl Br Cl Br Cl BrCl Br Cl Br Cl Br Cl Br Cl Br Cl Br Cl Br Cl BrCl Br Cl Br Cl Br Cl Br Cl Br Cl Br Cl Br Cl BrCl BrClBr Cl BrClBr Cl BrClBr Cl BrClBrCl 9:24:22:8:1

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Results

N = 2

m

4

n

N = 2

m

2

n

2

n

N = 2

A

2

n

Chem. Educ. 2003, 8, 182-186

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Summary Part I for Br

m

Cl

n

Compounds

• • •

Derived a chlorine-rule equation, I M = 3

n

Applied it to find gross structures of unknowns Derived a unit-sample equation, N = 2

A

2

n

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Follow-on to the Chlorine Rule

•

An automated A + 2 isotope-pattern analyzer (IPA)

•

IPA is on my website J. Chem. Educ., in press

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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

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Molecular-Ion Data is Entered into the IPA The Excel program returns the A + 2 (Cl, Br, S) composition of the molecular formula

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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., Br 1 Cl 2 ) and record it on your slip • Write your name on the slip and turn it in next Tuesday.

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Acknowledgement: Mass Spectra from the Spectral Data Base System (SDBS)

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Ende 42

Lecture attended by hordes of students eager to learn.

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Gross and Friends

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