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CHM 151 (PS 9 F07)

CHM 151 (PS 9 F07) - Br NaOEt OEt 55 45 Br NaOEt OEt 82 18...

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CHEMISTRY 151L.001 Fall 2007 Problem Set 7 1. Considering the following table and what you know about the unimolecular and bimolecular substitution and elimination mechanisms predict the products of the following reactions and pay attention to stereochemistry: STRONG BASES WEAK BASES GOOD NUCELOPHILES HO , RO RC C I - , Br - , RS - , CN - , N 3 - POOR NUCLEOPHILES (CH 3 ) 3 CO ( t -BuOK) , other bulky alkoxides Cl - , F - , RCO 2 , H 2 O, ROH, PR 3 , OTs NaOH OTs NaO t- Bu OTs H 2 O HOH HO Bu Ph Br NaCN HOCH 3 Cl NaI H 3 C C O CH 3 Br CH 3 OH
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CHM 151L.001; Problem Set 7 2 Br NaN 3 DMSO Cl NaOH HOH, heat OMs O O DMSO 2. 2-Bromopropane reacts with sodium ethoxide to give a 55/45 ratio of elimination to substitution products while with 2-bromobutane the corresponding ratio is 82/18. Explain.
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Unformatted text preview: Br NaOEt OEt + 55% 45% Br NaOEt OEt + + 82% 18% CHM 151L.001; Problem Set 7 3 3. Primary alkyl tosylate 1 reacts in a bimolecular reaction with sodium methoxide to give the expected products 2 and 3 in the expected ratio ( 2 / 3 = 70/30). Similar reaction between tosylate 4 and sodium methoxide affords the different expected products 5 and 6 in the new ratio of 5 / 6 = 85/15. What are the structures of 2 , 3 , 5 , and 6 ? (Place the structures in the correct boxes.) OTs OTs CH 3 + NaOCH 3 + NaOCH 3 1 2 + 3 5 + 6 (70%) (30%) (85%) (15%) 4 2 3 5 6...
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CHM 151 (PS 9 F07) - Br NaOEt OEt 55 45 Br NaOEt OEt 82 18...

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