Upload
others
View
6
Download
0
Embed Size (px)
Citation preview
Lecture 1 on homogeneous hydrogenation and related reactions -emphasis asymmetric hydrogenation
(Rhodium) Asymmetric Hydrogenation was the first successful example of highenantioselectivity using a purely chemical catalyst
Success depends on asymmetry in the product that arises from asymmetry in theligand of cat*
My first lecture traces the evolution of this topic between 1968 and the presentday, and the second provides and understanding of how the reaction happens.
X YX Y X Y
H
H
H
H
Prostereogenicalkene needed
H2, cat* H2, cat*
Product 1 Product 2
Products 1 and 2 are enantiomers
1
The main elements involved as complexes in homogeneoushydrogenation are shown
Wilkinson’s catalyst (1966)- the first practical homogeneoushydrogenation catalyst
RhCl3
Usually as a trihydrate
reflux in C2H5OH
xs. PPh3
ClRh(PPh3)3
NB Change in Rh oxidation state
PPh3
Rh
PPh3
PPh3Cl
idealised square-planarstructure
16e -coordinatively unsaturated
How does Rh(III) become Rh(I)In this procedure; there must be a reducing agent?? How could ethanol participate?
H3CO H
HH
Homogeneous hydrogenation with Wilkinson’s catalyst is selectivefor less substituted double bonds
O O
O O
H2 , cat(PPh3)3RhClO
H2 , cat(PPh3)3RhCl
Fragment of steroid nucleus
H2 , cat(PPh3)3RhCl
carvone dihydrocarvone
What would happen with a heterogeneous catalyst like Pd/C?
The addition of hydrogen to an alkene from a homogeneousrhodium catalyst is stepwise
MLnThe catalyst will be coordinatively unsaturated since it must add both alkene and hydrogen for reaction to occur.
H2C CH3
, H2
MLmH
H
H2C CH3Et
Et
Hydrogen adds cis-; ligands may dissociate to liberate coordination sitesin a multi-step process.
A cis-ligand migration of hydrogen occurs such that a hydridoalkyl complex is formed.The reaction can be reversible leading to isomerization of the alkene.
CH2
CH3
MLmH
Et
H
H H
H
H H
H
MLm
The catalyst is regeneratedin this step
HC CH3Et
H
This step is an elimination of cis-substituents usually termed "reductive elimination".
H2C CH3Et
H HHH
H
H
Dihydride and dihydrogen complexes of known structure - show ushow dihydrogen can add to a transition metal in catalysis
Ph3P Ir PPh3CO
ClPh3P Ir PPh3
H
Cl
H
CO
H2, fast
Vaska's compound 16 e Dihydride 18 e
iPr3P Ir PiPr3
H
Cl
H H2, fastiPr3P Ir PiPr3
H
Cl
H
H
H
16 e Dihydrogen complex 18 e
Metal hydrides may be part bonded to neighbouring carbon
Ru PPri3Pri3P
H
Cl
RhPri3PPri3P
Pri2PCMe2
H
+
X–
Normal hydride Bridged "agostic" hydride
[ Do an “electron count” for both species]!
Cone angles give a qualitative idea of the relative bulk ofphosphines
Bite angles P-M-P provide an important parameter for thecomparison of different chelate ligands
PPh2
PPh2
PPh2
PPh2Fe
PPh2
PPh2
PPh2
PPh2S
O
PPh2 Ph2P
123˚
97˚91˚85˚
112˚
R2P
PR2
M
Bite angle
Different phosphines possess different electronic effects
Increasingdonor ability
The more electron-withdrawing the ligand, the more back-donation from themetal to P. This affects the CO bond:
M C OPR3 M C OPR3
! higher ! lower10
Some critical early developments in asymmetrichydrogenation
NB. Phosphine ligands
Wilkinson's catalyst, ClRh(PPh3)3 demonstrated fast homogeneous hydrogenation of alkenes (1966)
Knowles, Horner showed progress with a simple enantiomericallyenriched phosphine PMe PrPh (15% e.e.) 1968; up to 88% e.e. by 1971
Kagan introduced chelating ligands 1971 DIOP
Knowles claimed high enantiomer excess in enamide reductionCommercial L-DOPA synthesis.(1974).
Trivalent phosphorus is stereochemically far more stablethan trivalent nitrogen
NA B
C N
A
BC
PA B
C P
A
BC
microseconds, 0 ˚C
hours, 100 ˚C
> 109˚
What causes this remarkable difference?
The essential elements of the synthesis of anenantiomerically pure phosphine
MeP(O)OH
Ph
Me
OPri
SOCl2, then l-menthol P
O
PhMe
Me
OPri
PO
MePh
+
RP SP
separate by recrystallisation
Me
OPri
PO
PhMe
RP
2-MeOC6H4MgBr PO
PhMe
Inversion
OMe H2PO
MeOMe
(i) Basethen Cu(II) (ii) HSiCl3
Bu3N
PPhOMe
PPh OMe
(i) Retention(ii) Inversion
(R,R)-DIPAMP
The initial observation (above) and the first successful development(below) in asymmetric hydrogenation
CH2
CO2H
CH3CO2HH
"15% optical purity"
H2, pressure,60 ˚C, MeOH
Cl3RhP3, 0.15%
PMe
Me Me
"69% optical purity"
CO2H
H
MeOCHN
MeO
AcOH2, 1 atm20 ˚C, MeOHalkene2RhP2Cl
CO2H
H
MeOCHN
MeO
AcOH
H
88% (S)-enantiomer
PMe
OMe
≥ 95% optical purity
Reactions Ligands
Synthesis of dehydroamino acids from aldehydes. The productsare prostereogenic (prochiral)
O
H
HO2C
H2C
NHCOMe
AcOH, Ac2O90˚ O
N
O
MeO
HNMe
CO2H
NaOH
O
N
O
Me
H
H
H
CO2H
HNMe
HOHBr, AcOH
HBr,AcOH
(Z)-isomer
(E)-isomer
Can you suggest a mechanism for these steps?
Substrate
Kagan’s catalyst for the synthesis of N-acetylphenylalanine - firstuse of chelating diphosphine
NHCOMeHO2C NHCOMeHO2C
O
O
CH2PPh2
CH2PPh2
Me
Me
0.2 mol%[ClRh(C8H14)2]2
H2, EtOH, C6H6
1 bar, 20˚C
72% e.e. (R)
R,R-DIOP
H
HC2 axis
HO
HO
CO2H
CO2H
H
HTartaric acidDIOP
The Monsanto process for the synthesis of dehydroaminoacids (WS Knowles) post 1975
MeO
MeO
Ph
P
Ph
P
Rh+
OAc
NHCOMe
OMe
HO2C
OAc
NHCOMe
OMe
HO2C
H2, MeOH
3 bar, 50˚C
1/ >10000 catalyst/ substrate
96% e.e; 100% e.e. after recrystallisation / MeOHBF4
-
OH
NH2
OH
HO2CL-DOPA
Chelating ligandMedical use of L-DOPA?
How to make cationic rhodium catalysts for asymmetrichydrogenation - use in MeOH
Rh X–
Rh X–
+
+
RhCl3
refluxin i-PrOH
refluxin i-PrOH
X = Clthenion-exchange with e.g.BF4
–, PF6–, CF3SO3
–
RhP
P
+
– 1 alkene X–
RhP
P
+
X–
– 1 alkene
How does the oxidation state of Rh change from (III) to (I) in the first step?
chelatediphosphine
chelatediphosphine
What happens to the dialkene during catalytic hydrogenation?
The product configuration is fixed by cis-ligand transfer ofhydrogen from rhodium to the alkene
H H
N CO2H
O
Me
H
H H
N CO2H
O
Me
H
H H
N CO2H
O
Me
H
H H
N CO2H
O
Me
H
H H
N CO2H
O
Me
H
RhH2
RhH2
H
H
H
H
1 2
3
Si-face aboveRe-face below
Re-coordination
Si-coordination
(S)-product
(R)-product
H2 transfer
H2 transfer
Successful asymmetric hydrogenation with rhodium complexes= polar functional group that can also bind to the metal to
form a chelate
CH2
Ph EtCH3
Ph EtHlow e.e.
CH2
Ph CO2HCH3
Ph CO2HH fair e.e.
CH2
Ph NHCOMe
CH3
Ph NHCOMeHhigh e.e.
H2
H2H2
CH2
Ph NH
O
Me
H2Rh+
Strong binding
CH2
Ph
H2Rh+
Less strong binding
HO
O
20
Ligand development 1 - BINAP
PPh2PPh2
C2 symmetry, only racemises > 150˚ C
synthesisedfrom:
OHOH
BINOL availableas either enantiomer
Why is it possible for BINAP, BINOL to exist as stable enantiomers?What about the corresponding biphenyls?
Ligand development 2 - DUPHOS and BPE. Now the chiral element is inthe substituents around P, not the chelate.
P
Me
Me
P
Me
Me
P
Me
Me
P
Me
Me
(S,S)-DUPHOS (S,S)-BPEmay vary alkyl groups
Me
Me
HO
HOsource of 5-ring chirality
Ligand development 3 - BisPP*; even simpler NB. The products are isolated as borane complexes
PCH3CH3
nasty, reactive
P PBut
Me
MeBut
BH3
reagents
PCH3CH3
BH3
nice, stable
P PBut
But
MeBut
reagents
DescribeP-B bonding?
What about the P-CH2-P analogue - any good?
Ligand development 4 - Phanephos. This provides anexample of planar chirality
PPh2
Ph2P
Br
Br
achiral chiral(one enantiomer shown)
X
X X
X
Enantiomers
Any other simple examples of planar chirality?
Ligand summary. Types of chelate ligand, all with twofoldsymmetry axes (C2)
P
Me
Me
P
Me
Me
(S,S)-DUPHOS
O
O
CH2PPh2
CH2PPh2
Me
Me
H
H
PPhOMe
PPh OMe
(R,R)-DIPAMP
PPh2PPh2
P PBut Me
MeBut
(S)-BINAP
PPh2
Ph2P(R,R)-DIOP
(S,S)-PHANEPHOS
(R,R)-BisPP*
First chelating chiral biphosphine;backbone chirality
Monsanto ligand for L-DOPAsynthesis; P-chirality
Best known of all chiral ligands; axial chirality
Very effective use of alpha-phospholanesubstituents
Example of planar chirality
Simple concept; high enantioselectivity
Phosphoramidites in Asymmetric Hydrogenation.This challenges the 25 yearassumption that chelates are best for asymmetric Rh hydrogenation
AcHN CO2Me
HPh CH2Ph
AcHN CO2Me
H2, L2Rh+
P
P
Me
Me
Me
Me
L2 =O
OP NH
PhMe
L2 = 2 x
89% e.e.94% e.e.
MeOH
Faster
Phosphoramidites in Asymmetric Hydrogenation again
P
P
Me
Me
Me
Me
L2 =O
OP NH
PhMe
L2 = 2 x
H CO2Me
NHAcMe
CO2Me
H2, L2Rh+ NHAcMe
MeOH
94% e.e.67% e.e.
Slower
Hydrogenation of simple enamides with a rhodium catalystand DUPHOS-type ligand
NHAc NHAc
R
H2, 4 atm
iPrOH, -10˚C
95- 97% e.e
R
E and Z mixture
P
P
Me
MeMe
Me
as Rh ligand
NHAc
RN
NMgBr
R
Ac2ORCH2MgBr
Synthesis
Hydrogenation
H
catalystH
enamide
Enol acetates are like enamides and form chelates to the metal -example shows alkene relative reactivities
O
CH2CH3
C5H11 O
C5H11
O
CH2 CH3C5H11
O
2 atm H2, L2Rh+
thf or MeOH
98.5% e.e.
94% e.e.
2 atm H2, L2Rh+
thf or MeOH
P
P
Me
MeMe
Me
L2 =
CH3
O
CH3
O
O
CH3 O
CH3
Asymmetric hydrogenations of tetrasubstituted alkenes;stereospecific at both reacting centres
CO2Me
NHAc
Me
CO2Me
NHAcMe
MeOH, RTH2, 4 atm
MeOH, RTH2, 4 atm
CO2Me
CO2Me
NHAc
Me
NHAc
MeP
P
Me
MeMe
Meas Rh ligand
Rh cat
Rh cat
Single diastereomer of product in each case- what does thistell us about the H2 addition process?
30
Examples of drug precursors synthesised by asymmetrichydrogenation. Need to vary ligand to optimise results
PPh2
Ph2PN
NBoc
AcCO2Me N
NBoc
AcCO2Me
H2, MeOH
86% e.e.
Rh catalyst
Phanephos
CO2HCN
CO2HCN
HH
Rh catalyst
H2, MeOHP P
Me
Trichickenfootphos98% e.e.
The Quadrant Rule for prediction of the hydrogenation of dehydroamino-acids shown with the Monsanto ligand DIPAMP
Ph
RhP Po-An
Ph
o-An
Modified arylphosphine quadrant
bulky
bulky
open
open
(S)-product
Hard to see why Ph is more bulkythan MeO-Ph!!
The quadrant rule for prediction of the stereochemical course ofhydrogenation of dehydroamino-acids phospholanes
RhP P
Phospolane quadrant
bulky
bulky
open
open
R
R
R
R
(s)
It’s an empirical rule - just helps topredict which enantiomer ofdehydroamino acid is formed but offersno understanding of why!
Look into the P-RhP plane from theopposite side to the ligand and splitthe space occupied into quadrants
The Quadrant Rule for prediction of the hydrogenation ofdehydroamino-acids bis-PP*
RhP P
(General) alkylphosphine quadrant L = large; S = small.
bulky
bulky
open
open
S
LS(s)
LSee previous slide
One stereogenic centre is enough! Probably the best ligand for Rhasymmetric hydrogenation
RhP P
L = large; S = small.
bulky
bulky
open
bulky
L
LS(s)
L
P PMe
Me = S
Tert-butyl = L
Selections from IgNobel Prize awards
IgNobel PEACE Prize : Claire Rind and Peter Simmons ofNewcastle University, in the U.K., for electrically monitoring theactivity of a brain cell in a locust while that locust was watchingselected highlights from the movie "Star Wars.”
IGNobel prizes: Peace: Howard Stapleton of Merthyr Tydfil,Wales, for inventing an electromechanical teenager repellant -- adevice that makes annoying high-pitched noise designed to beaudible to teenagers but not to adults; and for later using thatsame technology to make telephone ringtones that are audible toteenagers but probably not to their teachers