1 The Chlorine Rule: An Analysis of Isotope Patterns of Compounds Containing Multiple Bromine and Chlorine Atoms Ray A. Gross, Jr. With an Introduction to the Isotope-Pattern Analyzer

2 My Reasons for this Presentation Present results obtained at PGCC Show that content found in textbooks can be improved Motivate students

3 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) :1m Cl 35 (75.78) 37 (24.22) :1n

4 Mass Spectrometer

5 Schematic diagram of a mass spectrometer

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8 Why Br and Cl? Molecular-ion peaks of C 10 H 20 Br 1 Cl 1, C 10 H 19 Br 2 Cl 1 and C 10 H 18 Br 3 Cl 1.

9 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).

10 Herbert C. Brown Nobel Laureate Hydroboration-oxidation with BH 3 (CHM 201) Reduction with NaBH 4 (CHM 202/204)

11 Lillian Berg NVCC-Annandale

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20 Chlorine Constant

21 Bromine Constant I M = 3 n

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24 Theoretical Considerations Ideal Compounds Br (a:b) = 1:1 Cl (a:b) = 3:1 13 C and 2 H negligible

25 Bromine Binomial Ratio (a:b) = 1:1 (1a + 1b) m for Br m (1a + 1b) 1 = 1a + 1b = 1:1 (1a + 1b) 2 = 1a 2 + 2ab + 1b 2 = 1:2:1

26 Chlorine Binomial Ratio (a:b) = 3:1 (3a + 1b) n for Cl n (3a + 1b) 1 = 3a + 1b = 3:1 (3a + b) 2 = 9a 2 + 6ab + 1b 2 = 9:6:1

27 Ideal Model = Binomial Pair (1a + 1b) m (3a + 1b) n Br 1 Cl 1 3a 2 + 4ab + 1b 2 = 3:4:1

28 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, (article available at front desk)

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

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

31 Structure Begets Properties Let’s examine structures. Assume 3:1 and 1:1 isotopic abundances of chlorine and bromine. Consider Br m, Cl n and Br m Cl n compounds.

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35 Results N = 2 m 4 n N = 2 m 2 n 2 n N = 2 A 2 n Chem. Educ. 2003, 8,

36 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

37 Follow-on to the Chlorine Rule An automated A + 2 isotope-pattern analyzer (IPA) IPA is on my website J. Chem. Educ., in press

38 Example of a Print Out of a Mass Spectrum in the Molecular-Ion Region MassPercent

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

40 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.

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

42 Ende

43 Lecture attended by hordes of students eager to learn.

44 Gross and Friends