1. Chemistry 125: Lecture 45 January 29, 2010 Nucleophilic Substitution and Mechanistic Tools: Rate Law, Rate Constant, Structure This For copyright notice see final page of this file

2. Stereochemistry Rate Law Rate Constant Structure X-Ray and Quantum Mechanics Tools for Testing (i.e. Excluding) Mechanisms:

3. Leaving Group Substrate Rate Constant Dependance on Solvent Nu: R-L Nu-R L (+) (-)(-) Product 80 1,000 10,000 16,000 126,000 Nucleophile [1] k rel Br - F-F- H2OH2O HO - Cl - Nu HS - Sec. 7.4d, Table 7.3 I-I- 80,000 -8 -9 7 -10 3.2 15.7 -1.7 pK a (NuH + ) For first-row elements nucleophilicity (attack   C-L ) parallels basicity (attack H + ). Both require high HOMO. But as atoms get bigger, they get better at attacking   C-L (compared to attacking H + )

4. Solvent Leaving Group Substrate Rate Constant Dependance on Nu: R-L Nu-R L (+) (-)(-) Nucleophile 80 1,000 10,000 16,000 126,000 [1] k rel Br - F-F- H2OH2O HO - Cl - Nu HS - Sec. 7.4dg I-I- 80,000 -8 -9 7 -10 3.2 15.7 -1.7 pK a (NuH + ) k rel CH 3 I in H 2 O [1] 14 160 k rel CH 3 Br in Acetone 11 5 [1] harder to break H-bonds to smaller ions Polar solvents accelerate reactions that generate (or concentrate) charge, and vice versa. Sensible Backwards

5. Nucleophile Solvent Rate Constant Dependance on Nu: R-L Nu-R L (+) (-)(-) Sec. 7.4e Substrate goodRSO 2 O - -3 Leaving Group bad good v. bad bad good Br - F-F- H2OH2O HO - Cl - HS - I-I- v. good -8 -9 7 -10 3.2 15.7 -1.7 LpK a (LH + ) Weak bases are good leaving groups (H like R, as expected) R-OH v. bad R-OH 2 + good R-OSO 2 R’ good (Kenyon/Phillips) (acid catalysis)

6. OH Leaving-Group-Trick Lore (sec 7.4f) pK a ~16pK a -1.7 OH 2 + Ether / HBr Good Leaving Group Good Nucleophile

7. OH Leaving-Group-Trick Lore (sec 7.4f) Cf. Kenyon & Phillips (1923) OSO 2 R pK a -3

8. OH Leaving-Group-Trick Lore (sec 7.4f) OSOCl gases

9. OH Leaving-Group-Trick Lore (sec 7.4f) OPX n

10. OH Leaving-Group-Trick Lore (sec 7.4f) OP + 1 2

11. Maximizing Synthetic Speed to support PET scanning (from Loudon, Org. Chem.) e-e-   e+e+ 18 O = + 7 MeV proton - neutron 18 F - t 1/2 110 min http://en.wikipedia.org/wiki/Positron_emission_tomography Need to get 18 F where cancer is or 11 C t 1/2 ~ 20 min 13 N t 1/2 ~ 10 min 15 O t 1/2 ~ 2 min positron Connecting coincident scintillations tells where 18 F’s were. and you have to do it within a couple hours.

12. Yale PET What to synthesize?

13. Problems : K+K+ trifluormethanesulfonate (Triflate) Maybe it would suck up 2-Fluoroglucose as well. Rapid metabolism of tumor sucks up glucose. AcO = CH 3 C-O- (acetate protecting group) O F tied up by H-bonding and by K + cation inversion gives wrong configuration HO a horrid leaving group wrong C-OH could be attacked Glucose start with Mannose 2-Fluoroglucose KF 18 S N 2 ? Cl-SO 2 CF 3 well known for sugars K+K+ F 18- CH 3 CN (aprotic solvent) H 2 O H + Mannose to 2-F 18 Glucose - ASAP F 18

14. Office Hours 3-4:30 this afternoon 181 SCL Loudon, Organic Chemistry

15. Stereochemistry Rate Law Rate Constant Structure X-Ray and Quantum Mechanics Tools for Testing (i.e. Excluding) Mechanisms:

16. So far we’ve just been beating up on the D/A mechanism (trivalent C intermediate) though there are cases (S N 1) where it in fact applies. The tougher problem is to distinguish between concerted and A/D with a very weakly stabilized intermediate. (see supplementary reading on Course website)

17. Might there be Pentavalent A/D Intermediate instead of a Concerted S N 2 Transition State? Pentavalent Intermediate Nu L C L C Transition State

18. Quantum Mechanics says Transition State for OH - attacking less crowded CH 3 OH. 1.88 Å Quantum Mechanics says Transition State for H 2 O attacking protonated t-BuOH. 2.64 Å But neither reaction is practical in the laboratory! What does experiment say?X-ray? Might there be Pentavalent A/D Intermediate instead of a Concerted S N 2 Transition State?

19. Problem: Neither Transition State nor Intermediate would hold together long enough to study. Pentavalent Intermediate Nu L C L C Transition State

20. HOYWAT JACS 4354-4371 (2005) Nu LC Held in place by molecular framework +

21. HOYWAT + JACS 4354-4371 (2005) CH 3 O O C + CH 3 -O(CH 3 ) 2 BF 4 - : : ARE THERE BONDS HERE? OCH 3 + NOT elongated to reflect superposed average of two “bell-clapper” structures. F 3 BF - BF 3 2.64 Å Powerful alkylating agent like “Meerwein’s. Reagent” Et 3 O + BF 4 -

22. 5.02 Å 4.86 Å shortened by 0.16 Å BUT without central C + etc. shortened by 0.21 Å! 4.96 Å 4.75 Å 125° 114° 125° 113° Pentavalent C attraction? Eclipsed repulsion bent in etc. Pentavalence seemed to be a safe inference

23. 5.02 Å 4.86 Å 5.08 Å 5.00 Å O SiO 2 CF 3 124° 116° 125° 113° Central O only slightly repulsive compared to C +. Eclipsed repulsion

24. 126° 112° 4.92 Å 4.56 Å BF 2 BF 2 does seem to suck in CH 3 O groups. Eclipsed repulsion

25. 125° 113° 5.02 Å 4.86 Å Double minimum with stronger nucleophile O - (higher HOMO & lower LUMO) 127° 109° 4.84 Å 1.47 Å 2.99 Å CF 3 - O K+K+ and. nearby 1.88 Å

26. Range Bonded O (or S) seems to “use up” the vacant AO. C CH 3 O OCH 3 + C CH 3 O OCH 3 + Higher neighbor HOMOs favor tetravalence. For F  withdrawal dominates  donation. Compared to what?

27. AXAX A-X distances (Å) Compound Short & Long X A X Distances H nonbond reference B “tight” Bonded? B tetracoordinate C+C+ C+C+ ~ equal symmetrical very different unsymmetrical B “loose” like H No sign of stability for pentavalent S N 2 “intermediate ”  transistion state (as calculated by q. mech.) Pressed in by H  CH 3 repulsion

28. F CH 2 CH 2 H:OH "E2 Elimination" ABN AON F H OH CH 2 E2  -Elimination Text sec. 7.9 Rate influenced by [base], nature of leaving group, H isotope (kinetic isotope effect) C H H O C H O D D D k H > k D but only if bond is weakened in rate-determining transition state Heavier atom, lower ZPE see Lecture 8: frames 21-22 ZPE (kinetic)

29. E2  -Elimination Text sec. 9.5 Rate influenced by [base], nature of leaving group, H isotope (kinetic isotope effect) Stereochemistry (Anti) sec 9.5G Regiochemistry (Zaitsev/Hammond) sec 9.5F E2 vs. S N 2 (Sterics & Base Strength) sec 9.5G

30. N C and RC C as Nucleophiles Cyanide Table 9.6 p. 396 Acetylide Sec. 14.7B pp. 665-666 Ethylene Oxide as C 2 Source Synthesis Games Study Problem 14.2 p. 666-667 Problems 14.18-14.23 p. 666-668

31. 0 1 -2 -3 -4 log (fraction of R-Br converted to HOR/min) (CH 3 ) 3 C (CH 3 ) 2 CH SN1SN1 Hughes Ingold (1933-1940) CH 3 CH 3 CH 2 EtOH/H 2 O (4:1) 55°C NaOH + R-BrHO-R + NaBr k 2 (M -1 min -1 ) concerted displacement slowed by crowding k 1 (min -1 ) D/A accelerated by crowding, (CH 3 ) 3 C + cation stabilization, polar solvent (0.01 M) plus ~19% E2 Rate extrapolated from lower temperature. Depends on [OH - ]

32. S N 1 and E1 sec. 9.6 Product Determined After Rate 9.6B By Competition for Short-Lived Cation Rearrangement of Short-Lived Cation 9.6C Net Inversion from Short-Lived Ion Pair 9.6D

33. EtOH/H 2 O (4:1) 55°C NaOH + tBu-BrHO-t-Bu + NaBr (0.01 M) + CH 2 =C(CH 3 ) 2 E2 or E1? How do you tell? Overall rate (and % alkene) depends on [OH - ] Kinetic Isotope Effect shows whether H is being transferred in rate-determining step.

34. 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

35. 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

36. A lesson from E2 Elimination

37. 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

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

39. Smith-Lakritz

40. 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

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

42. 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:

43. 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

44. 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.

45. End of Lecture 45 Jan. 29, 2010 Copyright © J. M. McBride 2010. Some rights reserved. Except for cited third-party materials, and those used by visiting speakers, all content is licensed under a Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0).Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0) Use of this content constitutes your acceptance of the noted license and the terms and conditions of use. Materials from Wikimedia Commons are denoted by the symbol. Third party materials may be subject to additional intellectual property notices, information, or restrictions. The following attribution may be used when reusing material that is not identified as third-party content: J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0