lecture 1 on homogeneous hydrogenation and related...
TRANSCRIPT
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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
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The main elements involved as complexes in homogeneoushydrogenation are shown
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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
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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?
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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
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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
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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]!
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Cone angles give a qualitative idea of the relative bulk ofphosphines
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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
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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
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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).
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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?
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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
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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
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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
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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
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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?
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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?
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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
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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
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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?
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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
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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?
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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?
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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
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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
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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
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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
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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
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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
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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.
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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!!
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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
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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
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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
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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