Parallel Kinetic Resolution. Group Meeting Timothy Chang
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1 Parallel Kinetic Resolution (PKR) Group Meeting Timothy Chang
2 Vedejs, E.; Chen, X. J. Am. Chem. Soc. 1997, 119, Tanaka, K.; Fu, G. C. J. Am. Chem. Soc. 2003, 125, KR versus PKR
3 The Basic of KR (1) S R S S For k R > k S + Reagent + Reagent Cat (R) k R Cat (R) k S P R P S Efficiency i (relative rate or selectivity it factor) s=k rel =k fast /k slow =k R /k S =exp( G /RT) Why perform KR: 1. Racemate is cheap 2. No reasonable enantioselective approach 3. Classical resolution (stoichiometric) does not provide high ee P R G R G = G S G R S R G S "selectivity-determining diastereomeric transition states" S S P S KR Consideration: 1. S and P are easily separated 2. High yield (~50%), high ee 3. Short reaction time 4. Scalibility 5. Low cat. loading 6. Inexpensive cat. 7. Minimal waste 8. Reproducibility 9. Broad scope 10. Functional group compatibility Walsh P. J.; Kozlowski, M. C. Fundamentals of Asymmetric Catalysis Chapter 7
4 The Basic of KR (2) Cat (R) S R + Reagent k R P R Cat (R) S S + Reagent P S k S s=k rel = ln[(1 - C)(1- ee) / ln[(1 - C)(1 + ee)] (1) C = conversion ee = ee of S s can be measured experimentally by knowing C and ee s = k rel = ln[1 - C(1 + ee')] / ln[1 - C(1 - ee')] (2) ee' = ee of P Combine (1) and (2): ee / ee' = C / (1-C) Realistically, s factor is often moderate. It is impossible to obtain both good yield and high ee at the same time with moderate s. KR is useful if s is at least 10. However, the yield of S is sacrified. Kagan, H. B.; Fiaud, J. C. Topics in Stereochemistry Vol.18, Kinetic Resolution
5 A Problem of Relative Rate Kagan, H. B.; Fiaud, J. C. Topics in Stereochemistry Vol.18, Kinetic Resolution.
6 The Basic of PKR Solution: Minimize built up of the less reactive substrate (S S ) by a simultaneous transformation of S S. P 2R S R P 1R Cat (R) Cat (R) k 2R k 1R P 2S Cat (R) k 2S S S Cat (R) k 1S P 1S Ideal situation: k 1R = k 2S >> k 1S = k 2R P 1R / P 2S is constant during the course of resolution [S R ]/[S S ] = 1 ee = 0 S R / S S is constant during the course of resolution Example: If s 1 =s 2 =49 P 1R :P 1S =49:1 P 2S :P 2R =49:1 ee(p 1R )=96% ee(p 2S ) = 96% 2S 2R ( 2S) ee = (49-1) / (49 + 1) To achieve the same result in KR, s needs to be 200: s = ln[1 - C(1 + ee')] / ln[1 - C(1 + ee')] C = 0.5, ee = 96%, s =200 s(ork rel ) can be lower in PKR than in KR to achieve high ee Vedejs, E.; Chen, X. J. Am. Chem. Soc. 1997, 119, 2584.
7 PKR Mathematical Treatment (1) Kagan, H. B. Tetrahedron 2001, 57, Kagan, H. B.; Fiaud, J. C. Topics in Stereochemistry Vol.18, Kinetic Resolution.
8 PKR Mathematical Treatment (2) conversion C (R)-1 diastereomers (R)-1 (R,R')-2 (R,S') equiv. chiral catalyst t ee = 0 or reagent + + (S)-1 (S)-1 (S,S')-2 (S,R')-3 enantiomers 0.5 equiv. C=x 2 +x 3 0 C 1 x 1 +x 2 +x 3 =1 fractional amounts: ee 1 ee 2 ee 3 x 1 x 2 x 3 Kagan, H. B. Tetrahedron 2001, 57, Kagan, H. B.; Fiaud, J. C. Topics in Stereochemistry Vol.18, Kinetic Resolution.
9 PKR Mathematical Treatment (3) ee of SM versus product fraction KR ee PKR 0 0,1 0,2 0,3 0,4 0,5 0,6 X2<X3 X2>X X 2 KR Kagan, H. B. Tetrahedron 2001, 57, Kagan, H. B.; Fiaud, J. C. Topics in Stereochemistry Vol.18, Kinetic Resolution.
10 Relative Energy Considerations
11 Proof of Principle using Quasienantiomeric Electrophiles Z* = stoichiometric chiral reagents (quasienantiomers) P 2R Z* 2 k 2R S R P 1R k 1R Z* 1 P 2S Z* 2 S S k 2S Z* 1 k 1S P 1S P 1R and P 2S are quasienantiomers s 2 =42 s 1 =41 Basic criteria for a successful PKR: 1. Minimal mutual interference wrt catalyst or reagents 2. Have similar rates 3. Have opposite enantiocontrol wrt S R and S S 4. P 1R and P 2S are easily separated "leakage" Vedejs, E.; Chen, X. J. Am. Chem. Soc. 1997, 119, 2584.
12 Proof of Principle Vedejs, E.; Chen, X. J. Am. Chem. Soc. 1997, 119, 2584.
13 PKR using Quasienantiomeric Nucelophiles Fox et. al. J. Am. Chem. Soc. 2004, 126, 4490.
14 How to Spot a Potential PKR Ligand survey <Expt 1> Using a racemic ligand: O O n-bu [Rh(cod) 2]BF 4 n-bu H N rac-binap N + C + Me Me O O n-bu racemic n-bu 58% yield Result: poor yield Action: bad reaction, bad ligand, discard this reaction --> no publication :-( A better action: identify the side product Me O n-bub <Expt 2> Subject the enantiopure ligand to the reaction Me O H n-bu racemic + N C O n-bu [Rh(cod) 2 ]BF 4 (5mol%) (R)-BINAP (5 mol%) CH 2 Cl 2,rt 20 h Me n-bu O N 58% yield 49% ee n-bu O + Me O 24% yield 81% ee "Serendipity" discovery of PKR <Expt 3> Confirm PKR by using enantioenriched SM Expect the formation of one product (out of the two possibilities) <Expt 4...> Screen more enantiopure ligands, reaction optimization... Publication in JACS or ACIEE :-) (happy graduate student) n-bu
15 Chemodivergent PKR (1) Tanaka et. al. ACIEE 2006, 45, 2734.
16 Chemodivergent PKR (2) Dolye and Martin et. al. J. Am. Chem. Soc. 1995, 117,
17 Application of Chemodivergent PKR in Synthesis C 12 H 25 S N O 2 O Rh O Rh 4 Rh 2 (R-DOSP) 4 Davies et. al. J. Am. Chem. Soc. 2006, 128, 2485.
18 Model/Rational Behind the Divergent Reactivities Davies et. al. J. Am. Chem. Soc. 2006, 128, 2485.
19 Regiodivergent PKR (1) Group Question Coming Up O Me (R) (S) racemic [Rh(CO) 2 Cl 2 ] 2 (2.5 mol%) MeOH : TFE (1:1) 60 MeO Me (S) (S) OH 66% yield racemic HO Me MeO (R) (R) not observed Webster, R.; Böing, C.; Lautens, M. J. Am. Chem. Soc. 2009, 131, 444.
20 Regiodivergent PKR (1), Group Question O R [Rh(cod) 2 OTf] (5 mol%) R (R,S) or(s,r)-ppf-pt-bu 2 (6 mol%) Nucleophile X THF, 60 Nu racemic X OH X + X Nu HO R X X R OH side product, 1c 1. Find a general trend in yield and ee betweeneen the two products in the table. 2. Provide a rational for this trend. 3. Propose a energy diagram to correlate with your hypothesis in 2. Webster, R.; Böing, C.; Lautens, M. J. Am. Chem. Soc. 2009, 131, 444.
21 Group Question Answers Webster, R.; Böing, C.; Lautens, M. J. Am. Chem. Soc. 2009, 131, 444.
22 Group Question Answers, Cont. Webster, R.; Böing, C.; Lautens, M. J. Am. Chem. Soc. 2009, 131, 444.
23 Group Question Answers, Cont. k 1(R,R) >k 1(R,S) >k 2(S,R) >k 2(S,R) E 1 E2 E 2 B' S' S B A' E 1 >E 2 A Webster, R.; Böing, C.; Lautens, M. J. Am. Chem. Soc. 2009, 131, 444.
24 Regiodivergent PKR (2) Tanaka, K.; Fu, G. C. J. Am. Chem. Soc. 2009, 131, 444.
25 Regiodivergent PKR (3) Jana, C. K.; Studer, A. ACIEE 2007, 46, 6542.
26 Stereodivergent PKR in Total Synthesis Sarpong et. al. ACIEE 2009, 48, 2398.
27 Stereodivergent PKR in Total Synthesis Sarpong et. al. ACIEE 2009, 48, 2398.
28 Apparent PKR 2 3a 4a Using rac-1 gave S N 2':S N 2=98:2 Feringa et. al. ACIEE 2001, 40, 930.
29 A Hypothesis for Apparent PKR k B' >k B k 1 >> k 4 k 3 >> k 2 k B' k B Feringa et. al. ACIEE 2001, 40, 930.
30 Using Two Catalysts in a Three Phase System Vedejs, E.; Rozners, E. J. Am. Chem. Soc. 2001, 123, 2428.
31 Using Two Catalysts in a Three Phase System Calculation based on s = 23 Conversion (%) SM ee (%) Max. yield (%) Product ee (%) Max. yield (%) Vedejs, E.; Rozners, E. J. Am. Chem. Soc. 2001, 123, 2428.
32 Summary Pseudo Stereodivergent Stoichiometric Resolving Agent Quasienantiomeric Nucleophiles (Fox) Quasienantiomeric Electrophiles (Vedejs) PKR Biological or Enzymatic Approach Chemodivergent (Tanaka) (Doyle) (Davies) One of the two products can be achiral Chiral Catalyst + (Reagent) Regiodivergent (Lautens) (Fu) (Studer) Stereodivergent (Sarpong)
33 Summary PKR minimizes the built up of the slower reacting enantiomer. The s factor can be significantly less to achieve comparable results to KR. If s factor is greater than 125, it is not worthwhile to perform a PKR. Rti Rational ldesign of a PKR is very challenging. hll Discovery of a PKR depends on careful analysis of the products. For an ideal PKR, the chiral catalyst should have complete control for the Regio or Stereoselelctivity over substrate control.
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