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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
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2 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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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) 1.0281:1m Cl 35 (75.78) 37 (24.22) 3.1293:1n
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4 Mass Spectrometer
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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.
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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).
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10 Herbert C. Brown Nobel Laureate Hydroboration-oxidation with BH 3 (CHM 201) Reduction with NaBH 4 (CHM 202/204)
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11 Lillian Berg NVCC-Annandale
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20 Chlorine Constant
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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
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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
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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
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27 Ideal Model = Binomial Pair (1a + 1b) m (3a + 1b) n Br 1 Cl 1 3a 2 + 4ab + 1b 2 = 3:4:1
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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, 1161-1168 (article available at front desk)
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29 Roald Hoffmann-Nobel Laureate Conservation of orbital symmetry “Oxygen” Priestley vs Sheele Hoffmann Djerassi Woodward
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30 Gross giving lecture with Hoffmann, Djerassi and Woodward looking on.
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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, 182-186
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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
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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
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38 Example of a Print Out of a Mass Spectrum in the Molecular-Ion Region MassPercent 224 64.4 225 4.3 226100.0 227 6.9 228 45.6 229 3.2 230 6.4
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39 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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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.
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41 Acknowledgement: Mass Spectra from the Spectral Data Base System (SDBS)
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42 Ende
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43 Lecture attended by hordes of students eager to learn.
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44 Gross and Friends
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