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Chem 125 Lecture 57 3/4/09 This material is for the exclusive use of Chem 125 students at Yale and may not be copied or distributed further. It is not.

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Presentation on theme: "Chem 125 Lecture 57 3/4/09 This material is for the exclusive use of Chem 125 students at Yale and may not be copied or distributed further. It is not."— Presentation transcript:

1 Chem 125 Lecture 57 3/4/09 This material is for the exclusive use of Chem 125 students at Yale and may not be copied or distributed further. It is not readily understood without reference to notes or the wiki from the lecture.

2 CH 3 -Br + OH - 5. (5 min) Give a real example of the influence of a change of reactant structure on the ratio of S N 2 to E2 products. Be as specific and quantitative as you can. (You will need to show the ratios for two different reactants.) (CH 3 ) 3 C-Br + OH - Perspectives on Drastic Product Ratios Synthetic Organic Chemist : Reliable High-Yield Tool Physical-Organic Chemist : Definitive E a Difference Unambiguous interpretation of cause e.g. syn- vs. anti-hydrogenation of acetylene e.g. Steric retardation of S N 2 / 10 5 acceleration for t-Butyl via S N 1

3 Perspectives on 50:50 Product Ratios Physical-Organic Chemist : Valuable “Borderline” Reference Synthetic Organic Chemist : Deadly Influence on 12-Step Synthesis (1/2) 12 = 0.02% Yield (Might provide optimizable lead) Allows Sensitive Tests of Subtle Influences. e.g. isotope effect by competition

4 A lesson from E2 Elimination

5 If Step 1 (motion) is rate-limiting, H- and D-transfer products should form in equal amounts. (because their motions should be equally fast) If Step 2 (atom shift) is rate-limiting, more H-transfer product should form. k H /k D > 1 (kinetic “isotope effect”) In a Very Viscous Solvent Can Short-Range Motion Constitute a Rate- (and Product-) Determining Step? Generates steric hindrance & requires moving radicals past N 2 N N CH 3 CH 3 H 3 C CD 3 CD 3 CD 3 UV Light CH 3 CH 3 H 3 C CD 3 CD 3 CD 3 Radical-Pair Combination CH 3 CH 3 H 3 C CD 2 CD 3 CD 3 D D (1) Rotate N 2 + C 4 D 9 (2) Shift D atom exothermic/easy/fast N N Radical-Pair “Disproportionation” (1) Rotate N 2 + C 4 H 9 (2) Shift H atom exothermic/easy/faster CD 3 CD 3 CD 3 CH 3 H 3 C CH 3 H CH 2 Jo David’s Question: N N N N

6 t-Butylhydrazine (prepare from) ? To do his project, Jo David needed to prepare this compound. E2 >> S N 2 CD 3

7 Smith-Lakritz

8 It is very common to change a C=X double bond into C=O and H 2 X (we ’ ll discuss this soon) C=N-R  C=O + H 2 N-R

9 - + t-Butylhydrazine ??? Jo David Fine April-October 1971 O CD 3

10 Jo David Fine Jo David Fine Notebook p. 91 (October 1971) Jo David is now a respected professor of dermatology at Vanderbilt University, whose son has graduated from Yale. Happy Ending:

11 Crucial Lesson (from S. Nelsen, U. Wisc.) 95% 5% S N 1 When you need a compound, % yield isn’t everything! HCl salt easily purified by crystallization E1 / E2 Major product a gas, just “goes away” CD 3

12 Happy Ending: Jo David Fine’s successor found that in fluid solvents, there was more H- than D-transfer (atom transfer is rate- limiting), but that in very viscous solvents at low temperature this “kinetic isotope effect” disappeared (there were equal amounts of H- and D-transfer), because motion had indeed become rate-limiting.

13 Chapter 10 Alcohols and Thiols Sections 10.1-10.3C

14 6th Exam Mean 76.2 1/3 > 83 2/3 > 74 Sum of Two Exams Mean 133 1/3 > 144 2/3 > 126

15 End of Lecture 57 March 4, 2009


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