1.MsCl,Et 3 N CH 2 Cl 2,-10 C,97% 2.KOAc,H 2 O acetone, reflux, 82% 3.NaOH(1eq.) MeOH,-20 C,75% H H
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1 Problem Session(4) Yinghua Wang Please provide each reaction mechanism. 1 Ac Ac 1.Ms,Et 3 C 2 2,-10 C,97% 2.KAc, 2 acetone, reflux, 82% 3.a(1eq.),-20 C,75% 4.DMP,C basicAl 2 3,TF 83%(2 steps) 6.SmI 2 (2.1eq.),TF; ,43% 2 1.G-II(5mol%),2-2 C 3,reflux,E/Z=5/1 2.AllylMgBr,TF,0 C 47%(2 steps) 3. acrylic acid, reagent A TIPS Et 3,C 2 2,70% 4.G-II(10mol%) TIPS 2-1 C 2 2,reflux,70% TBS TIPS 5. 2 PhSi,DBU PhCF 3,140 C; 6.TMSC 2,/Et 2 64%(2 steps) 1. TBAF, TF; BS,aAc,aC 3, 2 2.Ac 2,TMP,DMAP,C TMP,C 3 C,155 C 68%(3 steps) TIPS Ac Ac I - reagent A acrylic acid Ac Ac I Ac DMP TMP s u s G-II (oveyda-grubbs2 nd catalyst)
2 Topic: Synthetic Studies of Cyclocitrinol 0. Introduction 0-1. Isolation Isolated as a fungal metabolite from terrestrial P. citrinum Kozlovsky, A. G.; Zhelifonova, V. P.; zerskaya, S. M.; Vinokurova,. G.; Adanin, V. M.; Gräf, U. Pharmazie 2000, 55, Yinghua Wang 0-2. Structual feature riginal proposed structure was assigned as 0-2, which was revised by X-ray crystallographic analysis of 0-3. Amagata, T.; Amagata, A.; Tenney, K.; Valeriote, F. A.; Lobkovski, E.; ardy, J.; Crews, P. rg. Lett. 2003, 5, An unusual C25 steroid: Bicyclo[4.4.1] A/B ring system Bridgehead(anti-Bredt) double bond 8 stereocenters including 2 quarternary carbons C 14D A18 B cyclocitrinol(0-1) 25 proposed structure(0-2) isocyclocitrinol(0-3) 0-3. Biological activity Cyclocitrinol: induce the production of c-amp in GP12-transfected C cells Du,L.;Zhu,T.;Fang,Y.;Gu,Q.;Zhu,W.-J.at,Prod.2008,71, Proposed biosynthesis pathway Marinho,A.M.d..;odrigues-Filho,E.;Ferreira,A.G.;Santos,L.S.J.Braz.Chem.Soc.2005,16, X 19 ergosterol(0-4) [] [] 23 [] [] Problem Session(4)-Answer- -1-
3 0-4. Synthetic study and Total synthesis > Three different approaches to construct the unique bicyclo[4.4.1]undecane A/B ring system are reported Synthetic study by Schmalz's group(problem 1) Formation of cyclopropane Biomimetic reductive fragmentation El Sherikh,S.; ier zu Greffen, A.; Lex, J.; udörfl, J.-M.; Schmalz,.-G. Synlett 2007, 2007, Synthetic study by Leighton's group strain-driven Cope Key: ow to access strained 10-membered ring?( Problem 2) Plummer,C.W.;Wei,C.S.;Yozwiak,C.E.;Soheili,A.;Smithback,S..;Leighton,J.L.J.Am.Chem.Soc.2014, 136, Asymmetric total synthesis by Li's group Type II[5+2] cycloaddition cf) Type I[5+2] cycloaddition 6 Ac 6 Ac steps Br 0-24 TIPS TBS TBS TIPS SnBu Li, C.-C. et. al. J. Am. Chem. Soc. DI: /jacs.8b steps Problem 3 Ac 0-28 TES 9steps Li 0-29 cyclocitrinol(0-1) -2-
4 -Answer- 1 Ac Ac 1.Ms,Et 3 C 2 2,-10 C,97% 2.KAc, 2 acetone, reflux, 82% 3.a(1eq.),-20 C,75% 4.DMP,C basicAl 2 3,TF 83%(2 steps) 6.SmI 2 (2.1eq.),TF; ,43% El Sherikh,S.; ier zu Greffen, A.; Lex, J.; udörfl, J.-M.; Schmalz,.-G. Synlett 2007, 2007, The first construction of the core structure of cyclocitrinols. Keyreaction:SmI 2 -mediatedfragmentationofcyclopropane. Step 1-3. Formation of cyclopropane via homoallylic. S Et3 S Ac Ac ± + Ac Ac Ms S 1 Ms Step1 Ac Ac 1-4a Discussion 1 Ac Ac 1-4b X Ac (X=Ac) 2 from concave face unfavored 2 3 Ac 1 the least crowded 1-5 Step2 a(1eq.) Ac 1-6 Step3-3-
5 Step4-6.SmI 2 -mediatedfragmentationofcyclopropane. Ac Ac I Ac Dess-Martin Periodinane Ac Ac Ac Ac I Ac Ac 1-8 Ac + Ac I Ac Ac I Ac Ac Ac Ac Ac 1-10 Ac I Ac Ac + B Ac I Ac 1-11 Step4 basical 2 3 Ac Sm II I 2 Sm II I 2 I 2 Sm III I 2 Sm III Step5 Sm III I keto-enol tautomerization Discussion Step6 >ProtonationoccursatC3whichisclosertooxygenatom and has higher electron density than C
6 Discussion 1: omoallylic of 19-substituted steroids. KAc 2 S 4 2 /acetone 2 /acetone Ms 19 reflux reflux 19 60% 73% Ms formation of homoallylic cations kinetically favored cations stabilized? 1 cation: disfavored anti-bredt: disfavored a 1-19b(favored) 1-19c When cationic intermediates are stabilized under acidic conditions... thermodynamically favored cations 1 cation: disfavored crowded tetra-substiteted: stable 1-19d 1-19e 1-19f(favored) Details for the, see: Tadanier, J. J. rg. Chem. 1966, 31, Discussion 2: etention of stereocenters during the fragmentiation. I 2 Sm SmI 2 (2eq.) SmI 2 I 2 Sm SmI % 1-19 (>99% ee) 1-21 (>99% ee) > Chirality centers should be disappeared during the fragmentation. > Due to the existence of bridgehead double bonds, 1-20 has inherent non-planar nature and can memorize the absolute stereochemical information. El Sheikh, S.; Kausch,.; Lex, J.; eudörfl, J.-M.; Schmalz,.-G. SYLETT 2006, 10,
7 2 1.G-II(5mol%),2-2 C 3,reflux,E/Z=5/1 2.AllylMgBr,TF,0 C 47%(2 steps) 3. acrylic acid, reagent A TIPS Et 3,C 2 2,70% 4.G-II(10mol%) TIPS 2-1 C 2 2,reflux,70% Ac 5. 2 PhSi,DBU PhCF 3,140 C; 6.TMSC 2,/Et 2 64%(2 steps) TIPS 2-4 Plummer,C.W.;Wei,C.S.;Yozwiak,C.E.;Soheili,A.;Smithback,S..;Leighton,J.L.J.Am.Chem.Soc.2014, 136, Key reactions: Two tandem reactions; cross-metathesis/semipinacol and Ireland claisen/cope. Step 1. Tandem cross-metathesis/semipinacol. u II 2-2 Ac oveyda-grubbs 2 nd catalyst s u s u IV 2-5 Ac u II 2-6 Ac u II ' TIPS 2-1 ' TIPS TS-1-A u II unfavored TIPS ' (Z)-2-7 u IV u II TIPS (Z)-2-8 ' ' u II favored ' u IV ' TIPS TS-1-B TIPS (E)-2-7 u II TIPS (E)-2-8 u II = s s u * Low concentration during cross-metathsis. -6-
8 Step 1. Tandem cross-metathesis/semipinacol.(continued) * After cross-metathesis, u-methylidene exist near epoxide and immediately react with it. s s s s ' u u ' Lewis acid Discussion 1 TIPS TIPS TIPS (E) Step 2-4. Tandem cross-metathesis/semipinacol.(continued) ' Ac TIPS 2-10 Step 1 TIPS ' Ac TIPS 2-10 u u ' or u TIPS u MgBr TIPS MgBr 2-11 BrMg Et 3 I work up I TIPS 2-12 Step 2 +I TIPS 2-13 u II Step3 oveyda-grubbs 2 nd catalyst TIPS u IV TIPS u II TIPS u IV TIPS Step 4-7-
9 Step 5-6. Tandem Ireland aisen/cope. TIPS Ph Si 2-3 Ph Si DBU TIPS Si 2 Ph DBU 2-18 Si 2 Ph TIPS 3 Si Si 3 Ireland aisen TIPS 3 Si Si 3 TS-2- --boat(disfavored) 13,14-epi-2-19 TIPS 3 Si 3 Si Ireland aisen TIPS 3 Si 13 3 Si TS-2- --chair(disfavored) 13-epi-2-19 TIPS 3 Si Si 3 Ireland aisen TIPS 3 Si 14 Si 3 TS-2- --chair(disfavored) 14-epi-2-19 TIPS 3 Si 3 Si Ireland aisen TIPS 3 Si Si TS-2- --boat(favored) 2-19(obtained) oxy-cope TIPS 3 Si Si 2 Ph 3 Si TIPS Si 2 Ph aq. 2 2 PhSi -8-
10 TIPS 1) tautomelization TMSC 2 TIPS TIPS Step TMS TMS 1) TMS C 2 2 TIPS TIPS TIPS 2-22 C C Kuhnel,E.;Laffen,D.D.P.;Lloyd-Jones,G.C.;Campo,T.M.;Shepperson,I..;Slaughter,J.L. Angew. Chem. Int. Ed. 2007, 46, Discussion 1. The order of cross-metathesis and semipinacol. The model experiment of a tandem cross-metathesis(cm)/semipinacol rerrangement reaction Step6 TIPS G-II(5 mol%) C 3,reflux + Ac 75% (E/Z=5/1) TIPS 2-25 Control experiments for the tandem reaction. Ac TIPS G-II(5 mol%) Ac C 3,reflux + Ac no rection TIPS 2-25 > CM didn't proceed between product 2-26 and 2-24, which means CM occured first. Ac C 3,reflux no rection Ac TIPS 2-27 G-II(5 mol%) C 3,reflux no rection TIPS 2-25 >eating2-27inrefluxc 3 resultedinnoreaction,whichmeanstraceinc 3 wasn'tthecatalyst. > Treatment of isolated 2-27 with G-II resulted in no reaction, which means G-II itself wasn't the catalyst. Plummer,C.W.;Soheili,A.;Leighton,J.L.rg.Lett.2012,14,
11 Discussion: Driving force of Cope. The ground state calculation of Cope substrates Another possible strain-relieving pathway of 10-membered ring (MacroModel using the PLS 2005 force field in vacuum) >Calculationresultof2-28indicatetheteriminiofthealkenesaresofar(~4.5Å)thataCope of 2-28 may be disfavored. >Tetheringthetwovinylgroupsintoa10-memberedringsuchas2-29,whichintroducesastraininthesubstrate anddecreasesthedistanceoftheterminiofthealkenes(~3.4å). In order to release the strain of 10-membered ring, a Cope might be accelerated. TES, DBU C,125 C 2 s 2 a 14 [3,3] Cope C 2 TES TES TES TES [ 2 a + 2 a ]:forbidden [1,3] sigmatropic [ 2 rerrangement a + 2 s ]:allowed inversion of C14 occurs 1M 2 a TES TES =TES: 2-36 =: % DBU = 2 =TES: =, 2 =: % 2-33(desired) 33% C 2 About[1,3] sigmatropic, see: Berson, J. A. Acc. Chem. es. 1968, 1, 152. TIPS 2-38(=) 2-3(=) 2 PhSi DBU,PhCF C [3,3] TIPS aq.; TMSC 2 [3,3] 3 Si 3 Si TIPS 2-39 Thebulkygroupmight depress[1,3]- due to its steric hinderence 12-membered byproducts C (=): 38% 2-3(=): 64% -10-
12 3 TBS TIPS 1. TBAF, TF; BS,aAc,aC 3, 2 2.Ac 2,TMP,DMAP,C TMP,C 3 C,155 C 68%(3 steps) Ac Liu,J.;Wu,J.;Fan,J.-.;Yan,X.;i,G.;Li,C.-C.J.Am.Chem.Soc.DI: /jacs.8b02629 The first total synthesis of cyclocitrinol Key reaction: intramolecullar[5+2] cycloaddition. Step 1-2. Achmatowicz reaction. F Si Si F 3-1 TBAF Br Br Br Br Achmatowicz reaction Br Br Br Ac ± + Ac 3-9 Step1 2DMAP 2Ac Ac 3-10 Ac Step2 Ac Ac About Achmatowicz reaction, see: _PS_Kengo_MASUDA -11-
13 Ac Ac Ac -12- Step 3. intramolecullar[5+2] cycloaddition. Ac 3-10 Ac Ac Ac B Ac formation of oxidopyrylium zwitter ion Ac Ac Ac Ac Step far 3-TS1 3-TS3 3-TS2 3-TS4 Ac Ac Ac Ac Ac strained 3-2(obtained) 3-13c(not obtained) 3-13b(not obtained) 3-13a(not obtained) strained
14 Problem 2. Step 5. aisen from(e)-enolate 3 Si DBU Si 3 TIPS TIPS ' TIPS 3 Si Si 3 Ireland aisen TIPS 3 Si Si 3 TS-2'- --chair(disfavored) 13,14-epi-2-19 TIPS Ireland aisen TIPS 14 3 Si Si 3 3 Si Si 3 TS-2'- --boat(disfavored) 14-epi-2-19 TIPS 3 Si Si 3 TS-2- --chair(favored) seems to be relatively favored Ireland aisen TIPS 3 Si 2-19(obtained) Si 3 TIPS 3 Si Si 3 TS-2'- --boat(disfavored) Ireland aisen TIPS 13 Si 3 3 Si 13-epi
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