Stereoselective Synthesis PDF

lOMoARcPSD|6077384 Stereoselective synthesis Advanced Organic Chemistry and Laboratory (The University of Warwick) Studocu is not sponsored or endorsed by any college or university Downloaded by Anushaya Jeyabalan ([email protected]) lOMoARcPSD|6077384 David Fox Lecture 1  3       Nucleoph...

lOMoARcPSD|6077384 Stereoselective synthesis Advanced Organic Chemistry and Laboratory (The University of Warwick) Studocu is not sponsored or endorsed by any college or university Downloaded by Anushaya Jeyabalan ([email protected]) lOMoARcPSD|6077384 David Fox Lecture 1  3       Nucleophile attacks carbonyl carbon at an angle greater than 90 degrees, Burgi Dunitz Angle Carbonyl and alkene MO diagrams Electrophile attacks an alkene carbon at an angle of 90 degrees 1-2 additions  felkin control/ sterics o Large substituent at 90 degrees to the carbonyl o Nucleophile will attack at an angle slightly greater than 90 degrees to the carbonyl, so that there will be some interaction with the group 120 degrees from the carbonyl. 1-2 additions  polar model  reagent: Me-Li/Nu, Me-MgCl o Enhanced stereoselectivity based on the polarity/repulsion of Cl and Nuo Once drawn out the representations of the molecules, rotate them so that the APP positions are shown on the diagrams and you can check the alignment Chelation control  reagent: BuMgCl o BnO  lone pairs on oxygen can coordinate to the magnesium and the carbonyl oxygen o Difference between the polar model/90 degree angle is that this is the chelation model o Rules for chelation, that govern the nature of the chelating group and the nucleophile/metal  Polar groups must have lone pairs to chelate, but not halogen (p. electronegative and wont donate LP)  Metal needs to be M2+/M3+  bi/tri/tetra  to be good Lewis acid. Can be Li if reaction is a) not aldol  Li needs to coordinate to incoming aldehyde in that case b) polar group is O, N (hard) but not S (not good ligand for lithium) Lecture 4 and 5   1,3-directed additions o Aldol reaction + BF3 coordinates to aldehyde + silyl ether attacks carbonyl carbon o Move the R group to the back o BF3 increases the dipole of the carbonyl + dipole of the C-O bond too o Dipoles oppose is favourable without a nonpolar solvent (electronic) o Dipoles reinforcing each other is less stable o Nucleophile attacks from bottom (sterics at the top) o Ratio is dependent on the R group o When there is an additional stereocentre, this poses a new problem 1-3-directed reductions  reagants: 1) Et2B-OMe, NaBH4 or 2) NaBH(OAc)3 with AcOH o Can either treat with the above reagents to give either 1,3 syn or 1,3 anti-product respectively o Why does BH4 attack from the top in the 1st reaction with EtB-OMe? o Produces a chair like structure from the top Lecture 6-9   C=C hydroboration of chiral alkenes o Boron ends up on less hindered end, kind of also the less likely to hold a positive charge o Oxidation occurs with retention of configuration o BH3 gives 80:20 over the products and 93:7 diastereoselecitivity o BBN is more hindered, more regioselectivity  95:5 overall products and 99:1 diastereoselectivity. Add EtOH/H2O2 o Bh3 reacts with hydroxy groups o Retention maintained with BH3, NaOH/H2O2 C=C cycloproponation o CH2I2 and AlMe3  backside attack, formation of an epoxide? Like a -O- ring o Mcpba  backside attack  same product as above Downloaded by Anushaya Jeyabalan ([email protected]) lOMoARcPSD|6077384   o OsO4  formation of a diol C=C SE2’ o Tertiary haloalkane + TiCl4 C=C chelation control o Mcpba o Backside attack o For cyclohexene, a half chair forms Lecture 10-14        Chiral template o Molecule temporarily attached to a pro-chiral centre and controls stereochemistry of the reaction  then removed o High ee because selectivity in alkylation was greater than 95:5, so final product will also be high o Removal of a cbz protecting group from an amine o Problems with synthesis: bond cleavage (in orange) replaced with H leads tothe byproduct losing its initial stereochemistry o Would be better if the byproduct could maintain the stereochemistry so it could be recycled - chiral auxiliary Evans auxiliary o Made from alpha amino acid, single enantiomers from nature o Following alkylation with an evans auxiliary, we see that the enolate is z geometry, the Na coordinates to the oxygens to stabilise and keeps it flat+rigid and stereodirecting group is benzyl o Electrophile comes from the top face o Only the z geometry?  Me favoured in axial position (unusual, but there is steric clash with the auxiliary when in eqt pos, which is greater than the 1,3 diaxial steric clash as the ring is slightly flatter/bond lengths to sodium are long and so there is less clash) Aldol auxiliary o SM will arrange itself so that the carbonyls are opposing dipoles o Base removes hydrogen  weaker base o OTf = OSO2CF3  like a tosylate but better LG  this is removed o Forms a z enolate Me wants to avoid auxiliary group so sits away o Stereoselectivity very good o Boron chelates to incoming aldehyde through a 6mem transition state, not the auxiliary o Enolate C-O and auxiliary C=O have opposing dipoles o Aldehyde attacks from least hindered side Evans conjugated addition and DA reactions o Can control conjugate addition stereochemistry Asymmetric diels-alder o Et2AlCl  powerful lewis acid + cyclopentene o 4 new stereocentres Evans auxiliary bond cleavage o Use LiOH, H2O2 o BuLi, PhCH2OH o EtSH, BuLi o Me3Al + MeNHOME Hydrolysis of a cyclic acetal: HCl, H2O Lecture 15-18  achiral alkenes and carbonyls with chiral boron reagents (derived from alpha pinene) Lecture 15 Downloaded by Anushaya Jeyabalan ([email protected]) lOMoARcPSD|6077384      Tri-subsititued alkene w/ chiral hydroborating agent and oxidateive workup (NaOh, H2O2) to give chiral secondary alcohol in high ee Initial hydroboration  boron on less hindered  B and H same face as double bond In alpha pinene only different due to the 2 methyl groups  boron will avoid this So, after drawing the newman projection (redrawn for clarity), we can see that there would be steric clash if we attack from aabove due to the large group, and steric clash from the bottom right-hand side due to the medium group. Therefore, favoured position is bottom left-hand attack Now we must consider the position of the groups on the alkene  tri-subsititued, only 1 H and this goes in the position with potential clash with medium group Lecture 16        0.5 eq of BH3  More hindered product whereby we have (BH) 2  secondary More suited for asymmetric hydroboration for less substituted alkenes  not tri After NaOh, H2O2 we get a secondary alcohol, with 87% ee Doesn’t matter which end boron ends up (symmetric alkene) but does matter the face of attack View A: Position same again as last time, however the alkene is different. Due to the large R* group, it blocks the entirety of the front face  hence H goes at the front now and CH3 to the back View B: H at back, CH3 at the front 3 different pathways after the formation of the boron: o H2O2, OH-  standard oxidative workup to give the alcohol o H2N – OSO3H  Electrophilic nitrogen source o NaOME, I2  SE2, Inversion  expect middle of C-B to be where the electron density is , but some of the bonding orbital sticks out at the back and can overlap with the iodine Lecture 17   Reagent with 2x pinene groups  react with methanol + Hydrogen  substitute methoxy group on the boron with an allyl Grignard reagent  add a simple aldehyde  98% e.e. chiral secondary alcohol Consider that the boron will react in a 6-mem transition state: o Front face methyl group on boron R groups o Hydrogen over the ring o to minimise the interactions with the 6 mem ring we have the R1 hydrogen facing the ring, and R2 pushes away from the axial pinene group/R1 o CH2 is at the back CH2>O  can see clearly in the second, disfavoured approach that the CH2 clashes with the R2 group o Large group at front and medium group at back Lecture 18      React alkene with 9-BBN  bicyclic product/tertiary borane The reaction scheme shown has just flipped the relative bonds  Reduce the carbonyl, break CH bond, pick up boron, reform double bond  reform SM and then we can oxidise to give us secondary alcohol Key feature: tertiary CH bond< primary CH bond  better reducing agent than the secondary CH ones  hence most reactive and attacks carbonyl OH + CH3SOC2CL, base  OSO2CH3 Downloaded by Anushaya Jeyabalan ([email protected])