1 Year 3 CH3E4 notes: Asymmetric Catalysis, Prof Martin Wills 2012-2013 You are aware of the...
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Transcript of 1 Year 3 CH3E4 notes: Asymmetric Catalysis, Prof Martin Wills 2012-2013 You are aware of the...
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Year 3 CH3E4 notes: Asymmetric Catalysis, Prof Martin Wills 2012-2013
You are aware of the importance of chirality. This course will focus on asymmetric catalysis, i.e. the use of a catalyst to create new enantiomerically pure molecules. This can be achieved in several ways:
M Wills CH3E4 notes
OOH
OO
H
OH
CO2EtHO
HO CO2Et
t-butyl peroxide(oxygen source)
Ti(OiPr)4 (metal for complex formation)
(+)-diethyl tartrate (source of chirality)
70-90% yield, >90% e.e.
Introductory, no need to revise, but understand concepts.
O
CO2EtO
O CO2Et
Ti
Ti
O O
O
O
CO2EtO
O CO2Et
Ti
Ti
OO
O
The oxygen atom isdirected to the alkene.The alkene is above the peroxide.
2) A covalent intermediate may be formed – a catalytic unit binds in a temporary process to the substrate:
O O
O
NH
CO2H O
O
O
NO
HO2C
10 mol%:
Proline catalyses the asymmetric cyclisation of a diketone (known as the Robinson annelation reaction).
this is not a chiral centre L-proline
Now this IS a chiral centre-S configuration
Major product
Mechanism is via:
1) A metal atom may ‘template’ the reaction in some way e.g. Sharpless epoxidation of alkenes:
2M Wills CH3E4 notes
N
R
The chiral counterion controls the asymmetry of the reaction.
O
OP
O
O
R
RO
OP
O
OH
R
R
Chiral Acid
N
R
H
NR
HH
E
E
N
R
H
H
The asymmetric environmentpromotes reduction on one face of cation
3) The reaction may take place within an asymmetric environment controlled by an external source:
The key features of these approaches will be described and examples from the literature will be described.
N
O O
HN
HN N
O F
F3C
CF3
Aprepitant(antiemetic)
F
HN
N
HN
NH
N
F
NC
AZ960(AZ, anti-cancer)
MeNO N
O
Rivastigmine(Novartis, Alzheimer's)
NF
F
CF3
O NHtBu
OH
NH2
Ly2497282(Eli Lilly, diabetes)
F
F
NBn
CO2Et
NHtBoc
Takada(Renin inhibitor for hydpertension)
Some examples of enantiomerically pure drugs:
understand concepts.
3M Wills CH3E4 notes
No 1 Lipitor (Atorvastatin)PfizerCholesterol regulator
NPh
O
NHPh
F
HO2C
HOHO
N
OMe
S
O
HN
N
OMe
No. 2 Nexium(Esomeprazole)AstraZenecaAntiulcerant.
No. 3 Plavix(Clopidogrel)Bristol-Myers SquibbPlatemet aggregationinhibitor.
N
S
ClCO2Me
N
OH
S CO2H
Cl
No. 4 Singulair(Montelukast)Merck Anti-Asthmatic
O
NMe2
NC
F
No 5 Lexapro(Escitalapram)Forest LaboratoriesAntidepressant.
No 6 Crestor RosuvastatinAstrazenecaCholesterol regulator
HO2C
HOHO
N
N
F
N
SO2Me
O
CO2H
NH2
I
I
I
I
HO
No. 7 Synthroid (Levothryoxine)Abbott. Thyroid preparation.
HO
HO
OH
NHtBu
No 8 ProAir HFA (TEVA), Ventolin HFA(GlaxoSmithKline)SalbutamolB2 stimulant, antiasthmatic*sold in racemic form.
No. 9 Advair Diskus(Fluticasone and Salmeterol)GlaxoSmithKlineCorticoids.
No. 10 Cymbalta(Duloxetine)LillyAntidepressantO
S
NHMe
HO
HO
OH
NH(CH2)6(CH2)4Ph
O
F
F H
O
HO
H
O
S
F
O
+
9 out of the top ten US prescribed drugs in 2010 are in single enantiomer form http://cbc.arizona.edu/njardarson/group/sites/default/files/Top 200 Brand-name Drugs by Total US Prescriptions in 2010sm_0.pdf
For information only. No need to memorise.
4M Wills CH3E4 notes
Oxidation reactions of alkenes.
R2R3
This represents a good way to create chiral centres.
R1
R2R3
R1
R2R3
R1
R2R3
R1
O
OH
OH
OH
NH2
R2R3
R1 OH
NH2
epoxidation Dihydroxylation aminohydroxylation
The Sharpless dihydroxylation reaction employs ligand-acceleration to turn the known dihydroxyation reaction into an asymmetric version.
RSRL
This process depends on the use of an amine to accelerate a reaction:
Rm
RSRL
Rm OH
OH
Dihydroxylation
OsO4
N
use of the amine belowspeeds the reaction up:
N
N
OMe
OH
N
N
OMe
HO
Dihydroquinine (DHQ) Dihydroquinidine (DHQD)
Sharpless et al realised that enantiomerically-enriched amines could change this to an asymmetricreaction:
'psuedo enantiomers'
ADmix- contains a dimer of quinine '(DHQ)2PHAL'ADmix- contains a dimer of quinidine '(DHQD)2PHAL)'(also a small amount of osmium salt + stoichiometric K3Fe(CN)6
N
N
OMe
O
AD-mix contains DHQ (note both
amine groups are of the same
absolute configuration):
N
N
OMe
ONN
Understand how each enantiomer of ligand gives a different product enantiomer.
5M Wills CH3E4 notes
DHQD gives: DHQ gives:
RSRL
Rm OH
HORS
RL
Rm OH
OH
How to remember::
Ph
Ph
Ph
Ph OH
HO
ADmix- ADmix-
(DHQD)(DHQ)Ph
Ph OH
HO
RL=large group, RM=medium group, RS=small group.
2 x 'OH' addedto lower face.
2 x 'OH' addedto upper face.in this orientation
Rm
RL
Understand how each enantiomer of ligand gives a different product enantiomer. Be aware and learn which enantiomer is formed relative to the substituents using each form of ‘ADmix’.
6M Wills CH3E4 notes
Oxidation reactions of alkenes.
The mechanism may be one of a number of possibilities:
NN
N
OMe
HO
A chiral complexmay be formed,directing the reactionto one face in [3+2] cycloaddition:
OsPh
Ph
OO
OO
AD-mix (DHQ)
Os Ph
Ph
O
O
O
O
amine structure abbreviated
Os
Ph
OO
O
O
N
N
OMe
HO
Hydrolysis andreoxidation.See if you can work out the mechanism.
Ph
Ph
HO
OH
N
Os Ph
Ph
O
O
O
O
Ph
or it could be a [2+2] cycloaddition, then ring-expansion.
Evidence favours the [3+2] addition mechanism: K. B. Sharpless et al, J. Am. Chem. Soc. 1997, 119, 9907.
Learn the two possible mechanisms for the oxidation,The means by which chirality transfer is achieved is not fully understood.
7M Wills CH3E4 notes
Oxidation reactions of alkenes.
CO2Et 1 eq. MeSO2NH2
AD-mix-, 0oCtBuOH/H2O
CO2Et
OH
OH 97% ee
1 eq. MeSO2NH2
AD-mix-, rttBuOH/H2O
OH
OH 98% ee
Cl Cl
1 eq. MeSO2NH2
AD-mix-, 0oCtBuOH/H2O
OH
OH 98% ee
SPh SPh
1 eq. MeSO2NH2
AD-mix-, 0oCtBuOH/H2O
OH
OH 93% ee
OH OH
CO2Et CO2Et
OH
OH 92% ee
nC6H13(DHQD)2PHAL
(AD-mix-)nC6H13
OH
OH 97% eeMe3Si
Me3Si
(DHQD)2PHAL
(AD-mix-)
AD-mix-, 0oCtBuOH/H2O
OH
OH
88% eeup to 96% ee withalternative ligand.
Ph1 eq. MeSO2NH2
AD-mix-, 0oCtBuOH/H2O
Ph
HO
HO
97% ee
No need to memorise the examples, but understand what the dihydroxylation achieves, and how versatile it can be.
8M Wills CH3E4 notes
1 eq. MeSO2NH2
AD-mix-, 0oCtBuOH/H2O
O
OC5H11
OH
O
O
94% ee
1 mol% OsO43 eq. K3Fe(CN)6
(DHQD)2PHAL, 0oC, tBuOH/H2O
OSi(tBu)Me2
Ph
O
93% ee
PhOH
Diastereoselective reactions:
EtO2C(CH2)2N3
OEtO2C
(CH2)2N3O
OH
OH
EtO2C(CH2)2N3
OOH
OH
A
B
with no ligand:(DHQD)2PHAL:(DHQ)2PHAL:
Asymmetricdihydroxylation
OsO4 + oxidant
A:B2:1
>20:11:10
Understand the concepts, no need to memorise examples.
9
Zaragozic acid synthesis – key asymmetric dihydroxylations.
OTMS
O
O
O
O
BnO2C
O
CO2BnOH
O
O
O
O
OSEM
OHHO
oxidationsesterifications
OH
O
O
O
O
OSEM
OH
O
O
O
O
OSEM
OH
HO
PMBO
O
O
O
OSEM
PMBO
MeO
PMBO
O
OSEM
PMBO
MeOAD-mix
(performs anasymmetricdihydroxylation
Then use 2-methoxypropene and acid toform acetal.
DDQ, H2O
OsO4, NMO
NMO=
N
O
O
Zaragozic acid A/Squalestatin S1Chlesterol-lowering.
OO
O
HO2CHO2C
CO2HOH
Ph
O OH
AcO
K. C. Nicolaou. E. W. Yue, Y. Naniwa, F. DeRiccardis, A. Nadin, J. E. Leresche. S. LaGreca. Z. Yang, Angew. Chem. Int. Ed. 1994, 33, 2184
Understand the concepts, no need to memorise examples on this slide.
10M Wills CH3E4 notes
Reduction reactions of Double bonds (C=C, C=N, C=O).
R2R3
This is a major area of asymmetric catalysis- atom efficient, low waste, low energy.
R1
R2R3
R1
H
R4 R4H
H source
catalyst
R2R3
O
R2R3
OHH
H source
catalyst
R2R3
NR
R2R3
NHRH
H source
catalyst
H source might be H2 gas, hydride, oran organic molecle (transfer hydrogenation)
11M Wills CH3E4 notes
Reduction reactions of Double bonds (C=C, C=N, C=O).
Ph
HO2C NH
O
H2
Ph
HO2C NH
O
H
S
N-acylated amine acid.
Rh. catalyst
-acylamino acrylate
<1 mol%
P P P Rh P
S S.. ..
OMe
MeOOMe
MeO
RR-DiPAMP = a homochiral ligand DiPAMP coordinated to Rh(I)
Addition of hydrogen to an acylamino acrylate results in formation of an amino acid precursor.
The combination of an enantiomerically-pure (homochiral) ligand with rhodium(I) results in formation of a catalyst for asymmetric reactions.
Understand how a chiral environment is created around Rh(I) and how the enamine substrate co-ordinates.
M Wills CH3E4 notes 12
Rh-diphosphine complexes control asymmetric induction by controlling the face of the alkene which attaches to the Rh. Hydrogen is transferred, in a stepwise manner, from the metal to the alkene. The intermediate complexes are diastereoisomers of different energy.
Rh/DiPAMP
P Rh P
OMe
OMe
Ph
HO2C NH
O P Rh P
OMe
OMe
Ph
CO2HNH
O
More stable,but less reactivecomplex
Less stable, but more reactive - leads to product
Ph
CO2HNH
O
H2
HH
H
S
Ph
HO2C NH
O
Ph
CO2HNH
O
Using Rh(DIPAMP) complexes, asymmetric reductions may be achieved in very high enantioselectivity.
Understand how a chiral environment is created around Rh(I) and how the enamine substrate co-ordinates.
M Wills CH3E4 notes 13
Other chiral diphosphines are not chiral at P, but contain a chiral backbone which ‘relays’ chirality to conformation of the arene rings.
P Rh P
PPh2
PPh2
O
O
PPh2
PPh2H
H
S-BINAP (often used with Ru(II)
PPh2
PPh2H
H
DIOP/Rh(I)
face
face
edge
edge
Chiraphos/Rh(I)
Rh/Diphosphine complex- ligandscreate a chiral environment at the metal
PR
R
PR
R
DuPHOS (R=Me, Et etc)
P
R
R
P
R
R
Rh
Chiral environment:
P
R
R
P
R
R
BPE
Understand how a chiral environment is created around Rh(I).
14M Wills CH3E4 notes
Reduction reactions of C=C Double bonds using Rh(I) complexes– representative examples.
MeOCHN MeOCHNCO2Me CO2MeH
6.5 atm H2
0.2 mol % [Rh(SS-DuPHOS)]+(R=Et), 2h, MeOH.
99.2% ee
C
X A
C
X A
B B
H
HAddiition of hydrogen is relative to the co-ordinating group
MeOCHN MeOCHNCO2Me CO2MeH
6.5 atm H2
0.2 mol % [Rh(A)]+25oC, benzene.
97.2% ee
P PA
MeOCHN MeOCHNCO2Me CO2MeH
6.5 atm H2
0.2 mol % [Rh(A)]+25oC, benzene. 98.2% eeH
co-ordinating group (NHCOMe, OH, OCOMe etc)
Singlediastereoisomer
MeOCHN MeOCHNCO2Me CO2MeH
1 atm H2
1 mol % [Rh(B]+20oC, ClCH2CH2Cl. R3SiOR3SiO 95% eeH
Singlediastereoisomer
Fe Fe
PPh2
Ph2P
B
OP
O
N
C
NHCOMe
25 atm H2
2 mol % [Rh(C)2]+20oC, ClCH2CH2Cl.
NHCOMeH
P(C6H11)2
P(C6H11)2
D
No need to memorise examples but understand that the sense of reduction in each case is relative to the directing group X.
15M Wills CH3E4 notes
Reduction reactions of C=C Double bonds using Rh(I) complexes– representative examples.
C
X A
C
X A
B B
H
HAddiition of hydrogen is relative to the co-ordinating group
co-ordinating group (NHCOMe, OH, OCOMe etc)
OO
CO2Me CO2MeH
4 atm H2
0.4 mol % [Rh(SS-DuPHOS)]+(R=Et), 12h, DCM.
99.8% ee
O O
P(C6H11)2
P(C6H11)2
CO2Me CO2MeH
5 atm H2
1.1 mol% [Rh(D)]+ EtOH.
99 % ee
EtO2C EtO2C
P4 atm H2
0.8 mol % [Rh(SS-DuPHOS)]+(R=Et), rt MeOH.
96% ee
OPh
O
O
OMeOMe
P
OPh
O
O
OMeOMe
H
D
B BH
35 atm H2
5 mol% [Rh(E)]+ -5oC, toluene.
94 % ee
ButO2CButO2C
O
O
O
O
Fe
E
PPPh2
F3C
CF3
CF3
CF3
Directing
Directing
Directing
Directing
No need to memorise examples - understand that the sense of reduction in each case is relative to the directing group X – different ligands give different product enantiomers.
16M Wills CH3E4 notes
Reduction reactions of Double bonds using catalysts derived from Ru(II) (C=C).
Directing groups on the substrate help to improve rates and enantioselectivity:(BINAP or similar biaryl ligands are generally favoured)CO2H
1 atm H2
0.5 mol%[(R-BINAP)Ru(OAc)2]MeOH/DCM (5:1)
CO2H
HH3C
directing groupMeO MeO
>97% ee
135 a tm H2
0.5 mol%[(S-BINAP)Ru(OAc)2]MeOH
NMeO
ArO
H NMeO
ArO
H
H
MeO MeO
>99.5% ee
CO2HC3H7
F 5 atm H2
1 mol%[(R-BINAP)Ru complex]MeOH, 50oC.
CO2HC3H7
F H
90% ee
N O
OO
N O
H3C
OO
H
1 mol%[(R-BINAP)Ru complex]MeOH, 50oC.
100 a tm H2
O
OC2H5
O
OC2H5
H100 a tm H2
0.2 mol%[(R-BINAP)Ru complex]DCM, 50oC.
95% ee
directing group
directing group
directing group
directing group
Learn that Ru(II) complexes of diphosphine ligands can also direct hydrogenations. No need to memorise examples.
17M Wills CH3E4 notes
Reduction reactions of Double bonds using catalysts derived from Ru(II) (C=C).
alcohol is directing group
OH
H20.2 mol%Ru(S-BINAP)
OH
HH3C
OH
H20.2 mol%Ru(S-BINAP)
OH
CH3H
Allyliic alcohols provide a good example of how the directing group works.
OH
Hydrogen on front facerelative to OH
OHH3CH
Hydrogen on front facerelative to OH
Learn that Ru(II) complexes of diphosphine ligands can also direct hydrogenations of allylic alcohols. No need to memorise examples.
18M Wills CH3E4 notes
Reduction reactions of isolated C=C double bonds can be achieved with variants of Crabtree’s catalyst.
R2R3
Crabtree's catalyst works well on isolated (i.e. no nearby co-ordinating group) C=C, bonds:
R1
R2R3
R1
H
R4 R4H
H2
catalystNIr
PCy3
+
N
Ir(COD)P(oTol)2
Asymmetric versions of the Crabtree catalyst(prepared as COD complexes, but with the COD left off for clarity):
O
PhO N
PPh2
tBu
Ir(COD)
+ +
No directing group required
NIr
PCy3
+The catalyst is prepared with a cycloactadiene (COD) ligand but this is hydrogenated at the start of the catalytic cycle. The 'parent' Crabtree catalyst is, of course, non-chiral.
0.1 mol% catalyst ACH3CH3H
50 atm H2
rt, CH2Cl2 97% ee
B((3,5-C6H3(CF3)2)4-
PF6-
B((3,5-C6H3(CF3)2)4-
(BARF-)
CH3
CH3MeO
CH3
CH3MeO
AB
O N
OPPh2
iPr
Ir(COD)
+
B((3,5-C6H3(CF3)2)4-
(BARF-)
C
1 mol% catalyst C
50 atm H2
rt, CH2Cl2
89% ee
H
0.5 mol% catalyst D
50 atm H2
OH
rt, CH2Cl2
99% ee
S
NIr(COD)
PPh2
Ph
+
B((3,5-C6H3(CF3)2)4-
D
CH3
OH
CH3H
CH3
CH3
CH3
CH3
1 mol% catalyst B
50 atm H2
rt, CH2Cl2
92% ee
H
Steric control - not OH group.
No need to memorise examples.
19M Wills CH3E4 notes
Reduction reactions of isolated C=C double bonds can be achieved with variants of Crabtree’s catalyst.
N
Ir(COD)P(oTol)2
O
Ph
+
B((3,5-C6H3(CF3)2)4-
B
O
AcO
O
AcO
RR R R
1 mol% catalyst B
50 atm H2
rt, CH2Cl2>98% RRR enantiomer. Each reduction is controlled by the catalyst i.e. it is not diastereocontrol.
Vitamin E precursorParticularly challenging application:
Understand that Ir(I) complexes with P and N donors can reduce double bonds without a directing group in the substrate, i.e. sterically-driven. No need to memorise examples.
20M Wills CH3E4 notes
Reduction reactions of C=O Double bonds using organometallic complexes.
H3C
O 0.1 mol% [(R-BINAP)Ru(OAc)2]
86 atm H2
O
OMeH3C
OH O
OMe
H
directing group
H3C
O
H3C
OHH
bromine is directing group
Br Br
The same principle regarding directing groups also applies to C=O reduction, Ru and Rh are most commonly used:
51h, 20oC, EtOH,100% 99% ee
H3C
O
4 atm H2P
O
OMeH3C
OH
P
O
OMe
H
72h, 25oC, MeOH,99% >95% eeOMe
0.1 mol% [(R-BINAP)Ru(OAc)2]
86 atm H2
62h, 20oC, EtOH,97%>92% ee
OMe
directing group
Understand that a C=O group can be reduced by a chiral Ru or Rh complex as well. No need to memorise examples.
21M Wills CH3E4 notes
Reduction reactions of C=O Double bonds using organometallic complexes.
The same principle regarding directing groups also applies to C=O reduction, Ru and Rh are most commonly used:
H3C
O 1-2 mol%
30 atm H2SPhH3C
OH
SPh
H
30h, rt, 100% 94% ee
PPh2
RuBr2
PPh2
H3C
O
0.25mol%
50 atm H2NMe2.HCl H3C
OH
NMe2.HCl
H
18h, 20oC, PhMe, 100% 99% ee
N OP(C5H9)2(C5H9)2P
Rh(OCOCF3)2
O
10 atm H2NHMe.HCl
OH
NHMe.HCl
H
18h, 50oC, MeOH, 92%99% ee
P
tBu
P
But
H H0.5mol%
Rh(I).complex
directing group
directing group
directing group
Understand that a C=O group can be reduced by a Ru or Rh complex as well. No need to memorise examples.
22M Wills CH3E4 notes
Reduction reactions of C=O Double bonds using organometallic complexes.
Principle: The substrate is rapidly racemising and one enantiomer is selectively reduced:
R2
O
catalyst
O
OMe
R2
OH O
OMe
H
R1
R1H
R2
O O
OMe
R1
R2
O O
OMe
R1
R2
O O
OMe
R1
H
enol
fast
H2reduced very slowly
Racemic!
Enantiomerically Pure
Dynamic kinetic resolution can result in formation of two chiral centres:
Learn that a beta-keto ester can epimerise rapidly and that one enantiomer is more quickly reduced. Be able to draw the mechanism of this. No need to memorise examples.
23M Wills CH3E4 notes
Reduction reactions of C=O Double bonds using organometallic complexes.
H3C
O
0.5 mol%Ru/R-BINAP
O
OMeH3C
OH O
OMe
H
Cl Cl H
e.g.
H3C
O 100 atm H2
1 mol% Ru/R-BINAP
O
OMe H3C
OH O
OMe
H
HNHCO2Ph NHCO2Ph
90 atm H2
88% de, 98% ee
H3C
O
0.17 mol% Ru/R-BINAP
P
O
OMe H3C
OH
P
O
OMe
H
AcHN NHAcH
4 atm H2
98% de99% ee
OMe
OMe
20h, 50oC, DCM
5h, 80oC, DCM
65h, 25oC, MeOH
94% de>98% ee
88% de98% ee
Dynamic kinetic resolution can result in formation of two chiral centres:
No need to memorise examples – these illustrate the diversity of the process.
24
Ketone reduction by pressure hydrogenation (i.e. hydrogen gas) can be achieved using a modified catalyst containing a diamine, which changes the mechanism.
Ph2P
PPh2
Ru N
NH2
Ph
Ph
H
H
Mechanism
H
H
OMe
Ph
Ph2P
PPh2
Ru N
NH2
Ph
PhH
H
H
OHMe
Ph
H2
OHO H
H2 , solvent
Ph2P
PPh2
Ru
H2N
NH2
Ph
Ph
H
H
Very high e.e.from very lowcatalyst loadings
M Wills CH3E4 notes
Understand that the mechanism changes when a diamine is added to a Ru(II)/diphosphine complex, and this allows C=O bonds to be reduced without a nearby directing group present. Be able to draw the mechanism of this.
25
Ketone reduction by pressure hydrogenation (i.e. hydrogen gas) can be achieved using a modified catalyst containing a diamine, which changes the mechanism.
M Wills CH3E4 notes
O0.2 mol% catalyst2.5 mol% KOH
5 atm H2, EtOH, 5h, 100%
OHH
97% e.e.
O OH
cis:trans 100:1
0.2 mol% catalyst0.24 mol% KOH
5 atm H2, iPrOH, 3.5h, 100%
No need to memorise the examples.
26M Wills CH3E4 notes
The use of hydride type reagents.
Transfer hydrogenation – Ru catalysts.
Ru
N
TsN
Ph
Ph
HH
Cl
activation
HCO2H, Et3N Ru
N
TsN
Ph
Ph
HH
H
OR
edge/face interaction
Ru
N
TsN
Ph
PhH
H
H
O
R
HCO2H, (-CO2)
Product
R
O
Ru
N
TsN
Ph
Ph
HH
H
substrate
Catalyst prepared by combining:
NH2
NHTsPh
Ph
andRuCl
Cl
Cl
Ru
Cl
in iPrOH/KOHorHCO2H/Et3N
ee >96%full conversion
Rh
N
TsN
Ph
Ph
HH
HIr
N
TsN
Ph
Ph
HH
H
Rhodium andiridium complexesare isoeletronic withCp' on metal in placeof arene.
Understand that hydride reagents can also be used in reductions. Be able to draw the mechanism of the hydride transfer step.
27M Wills CH3E4 notes
Examples of reductions using transfer hydrogenation with metal complexes: add C=O and C=N reductions.
O OHH
X
0.5 mol% SS-Ru catalyst
HCO2H/TEA28oC,
X
X=H, 98% eeX=Cl, 95% eeX=OMe 96% ee
A wide range of substituents can be tolerated, except for ortho-groups, which result in reduced selectivities.
0.5 mol% SS-Ru catalyst
HCO2H/TEA28oC,
99% ee
O OHH
0 or 10 or 1
(i.e. fused five or six-membered ring)
0.5 mol% SS-Ru catalyst
HCO2H/TEA28oC,
98% ee
SO2
SO2
O OHH
S S
Precursor to jknown drug
0.5 mol% SS-Ru catalyst
0.6 mol% KOH, iPrOH28oC, >99% ee
O H
C4H9
OH
O OHH0.1 mol% RR-Ru catalystabove
HCO2H/TEA28oC,
XX=H, Cl, OMetypically 96% ee
Cl Cl
Ru
NTsN
Ph
Ph
H
Cl
Improved catalyst with alink between amine and arene ring.Following reactions arewith this catalyst.
NHCO2H/TEA28oC,
up to 97% ee
XOHH
98% ee X=O97% ee X=S
O
X
Cl NClH OH
OHH
OPh
OHH
N
N
97% ee 97% ee
Other reduction products:
C4H9
0.1 mol% RR-Ru catalystabove
These are examples to provide an appreciation of the scope, No need to memorise examples.
28M Wills CH3E4 notes
0.5 mol% RR-Ru catalyst
HCO2H/TEA28oC,
Ru
NTsN
Ph
PhH
Cl
Some imines can also be reduced by asymmetric transfer hydrogenation:
N
Ar
NH
Ar
Typically >96% ee
H
MeO
MeO
MeO
MeONH
N
Ph NH
NH
PhH
Typically >98% ee
as above
5 mol% NaOH, iPROH28oC, 16h, 96%
Other ligands can be used with ruthenium(II) in asymmetric catalysis (and also with Rh and Ir), e.g.
90% ee
OPO
O
OP
O
Excellent ligand fortransfer hydrogenation.
O 0.5 mol% [RuCl2(arene)]2
1.25 mol% ligand
OHH
5 mol% NaOH, iPROH28oC, 22h, 99%
O 0.5 mol% [RuCl2(arene)]2
1.25 mol% ligand
OHH
99% ee
Challenging substrates:
NH HN
Ph2PPPh2H2N NHTs
OH
NH2
NH
OH
BocHNNH
OH
PhO
These are examples to provide an appreciation of the scope, No need to memorise examples.
29M Wills CH3E4 notes
Asymmetric transfer hydrogenation by organocatalysis.
R1
R2
NR3
H
Use combination of a chiral acid with a hydride source:
R1
R2
O NR3
H
H+
O
OP
O
OH
R
R
O
OP
O
OH
O
OP
O
O
*
*
Homochiral acid (directs reaction)R=aryl ring, trialkylsilyl etc., usually abulky group. Catalytic amount needed.
+
N
CONR2
H
R2NOCH H
Source of hydride - stoichiometric amount needed. Similar to NADH used inbiological transformations. Known as'Hantzsch ester'.
O
OP
O
OH*+
condensation
N
CONR2
H
R2NOCH H
R1
R2
NR3
H N
CONR2
H
R2NOCH
H
R1
R2
HN
R3
O
OP
O
O*
close ion pair formed
N
CONR2
H
R2NOCH H
Mechanism:
(Either use a preformed imine or via reductive amination)Protoncan be reused
N
CONR2
R
H H
O
N
CONH2
Inspired by Nature's NADH;a coenzyme which transfers hydride
R
H H
N
CONR2
Me
R2NOC
H
HOH
Understand that Hantzsch esters are used as reagents for reduction of C=N bond in organocatalysis reactions. Be able to draw the mechanism of the hydride transfer step and the imine formation. No need to memorise examples.
30M Wills CH3E4 notes
Asymmetric transfer hydrogenation by organocatalysis.
Some examples of reductions:
N
OMe
1 mol% cat where R=H
1.4 eq. Hantzsch esterToluene, 35oC, 71h, 91%
HN
OMe
H
93% ee
N
OMe
1 mol% cat where R=H
1.4 eq. Hantzsch esterToluene, 35oC, 60h, 80%
HN
OMe
H
90% ee
N
MeO
MeO
NH
MeO
MeO
H1 mol% cat where R=bulky aryl
2.4eq. Hantzsch estertoluene, 60oC, 12h95%
90% ee
fully heteroaromatic rings can be reduced:
Asymmetric reductive amination:
O
+
TsN
10 mol% cat where R= SiR3
1.2 eq. Hantzsch esterToluene, 40oC, 48h, 90%+ molecular seives
HN
TsN
H
H2N 93% ee
No need to memorise examples, but understand the concepts.
31
More applications of organocatalysis.
O O
O
NH
CO2H O
O
O
O
O
NO
HO2C
10 mol%:
Some time ago, it was found that proline catalyses the asymmetric cyclisation of a diketone (known as the Robinson annelation reaction).
this is not a chiral centre L-proline
Now this IS a chiral centre-S configuration
The enantiomericcompound is:
Major product
Mechanism is via:
M Wills CH3E4 notes
NH
CO2H
NH
Ph
NH
PhPh
NH
NMe
CO2H
O
Ph
Examples of commonorganocatalysts:
L-proline
or pyrrolidines:
or other N-heterocycles:
Understand that the combination of a chiral amine and a ketone or aldehyde forms an enamine which directs a subsequent aldol reaction. Be able to draw the mechanism of the enamine formation, the reaction with a ketone or aldehyde and the subsequent hydrolysis step. No need to memorise examples.
32
More applications of organocatalysis.
M Wills CH3E4 notes
H
O ONH
CO2H
Me
H
Me DMF
H
O OH
Me Me
O O
OTBS
H
O 3 mol% water, rt 2 days.TBSO
O
OtBu
CO2HH2N
O OH
OTBS OTBSOO
10 mol%:
This is now the basis for many other reactions e.g.:
Aldols: L-proline90% yield
4:1 anti:syn
anti product e.e.: 99%
and even more complex ones: 20 mol% 68%, major product: D-fructose precursor
These reactions take place via formation of an enamine which then reacts with the other reagent e.g.
H
Me
NO
Me
O
OH
No need to memorise examples – these illustrate the diversity of the process.
33M Wills CH3E4 notes
More applications of organocatalysis which proceed via formation of an enamine – bonds to C atoms.
H
O
NH
CO2H
nBu
MeCN, 0oC
OH
nBu
i) 10 mol%:
C-N bond formation:
L-proline
97% ee
NN
CO2Bn
CO2Bn
+
ii) NaBH4, EtOH, 94%
NNH
CO2Bn
CO2Bn
HO
O
nBu
NH2
amino acids.
H
nBu
NCO2H
NN
CO2Bn
CO2Bn
via:
H
O i) 10 mol%: 97% ee
NN
CO2Et
CO2Et
+
ii) NaBH4, EtOH, 83%
NHDCM, rt. OTMS
ArAr
Ar= 3,5-(CF3)2C6H3
NHO
O
H
O
C5H11
C-Halide bond formation:
+
OCl
ClCl
Cl
ClCl
NH
MeN
Ph
O
5 mol%1.2 eq.
DCM, -24oC, 71%H
O
C5H11
Cl
N
MeN
O
92% ee
C5H11
via:
O
ClCl
ClCl
Cl
Cl
Z-enamine,orientated away from dimethylsphenyl ringblocks lower face N
MeN
Ph
O
C5H11
ClH2O
C5H11
Cl
O
H
O
tBuii) NaBH4, MeOH
NHDCM, -24oC OTMS
ArAr
ii) 20 mol%:
+
OBut tBu
Br
1.2 eq.
Br
OH
tBu
Br
95% ee
etc
These are examples to provide an appreciation of the scope, No need to memorise examples.
Asymmetric catalysis of C=C bonds can be catalysed by organocatalysts, if they are conjugated to a C=O:
N
CONR2
H
R2NOC
H H
R1
R2
H
N
H
N
CONR2
H
R2NOC
H
R1
R2
H
NR2
HH
R1
R2
H
O
H
+NH
RR
H2OR1
R2
O
HH
+NH
R
N
CONR2
H
R2NOCH H
Source of hydride.
General mechanism:
C=C reduction by organocatalysis.
34
Understand that a chiral amine can direct a conjugate reduction reaction. Be able to draw the mechanism of the hydride transfer step and the imine formation and hydrolysis. No need to memorise examples.
Ph Me
H
O
H 10 mol%NH
NMe
tBu
O
Bn
Ph Me
O
H
N
CO2Et
H
EtO2CH H
Examples:
H
90% ee
t-Bu Me
H
O
H5 mol%
NH
NMe
tBu
O
t-Bu H
O
H
N
CO2Et
H
EtO2CH H
Me
90% ee1.02 eq. 1.2 eq.
O
tBu
20 mol%
NH
NMeO
N
CO2tBuButO2CH H
1.1 eq.
BnO
O
tBu
H
C=C reduction by organocatalysis.
35
No need to memorise examples.
M Wills CH3E4 notes
36
M Wills CH3E4 notes
Allylic substitution reactions are powerful methods for forming C-C bonds.
Ph Ph
AcO
Pd(0), Nu
Chiral ligandPh Ph
Nu
Ph Ph
AcO PdLn
Ph Ph
PdLn
Nu
Ph Ph
Nu PdLn
PdLn
PhPh
Nu
Ph Ph
LnPdNu
PhPh
NuLnPd
PdLn
Attack at the other end of allylic systemgives alternative enantiomer:
The Pd is behind theallylic group.
Understand that a flat allyl complex is formed and that the ligand directs a nucleophile to one end by a combination of steric and electronic factors. No need to memorise examples.
Ph
Ph
PdLn
Nucleophile addstrans to PdLn group
Nu
37
M Wills CH3E4 notes
Allylic substitution reactions are powerful methods for forming C-C bonds.
Ph
Ph
PdLn
Nucleophile addstrans to PdLn group
Example ligand:
N
O
Ph2P
RPd
Ph2PO
H
R
Ph2P Pd NO
H
R
PhPh
Favoured conformation: the allyl group is in front of the Pd complex.
PhPh
disfavoured bysteric clash with equatorial Hor slower to react.
Ph2P Pd N
PhPh
Nu
Trans effect favours addition to end oppositethe P atom.
PhPh
Nu
Pd N
N
O
Ph2P
RPd
Ph Ph
OAc
Ph Ph
OAc
0
N
O
Ph2P
RPd
Ph Ph
0
-AcO
or
Nu
H
(racemic)
(enantiomerically enriched)and catalyst is released to re-enter cycle.
Understand that a flat allyl complex is formed and that the ligand directs a nucleophile to one end by a combination of steric and electronic factors. No need to memorise examples.
38M Wills CH3E4 notes
Allylic substitution reactions – examples of ligands and reactions.
N
O
Ph2P
PhPh
PhAcO
PhPh
PhCO2EtEtO2C
5 mol%
2.5 mol% [Pd(allyl)Cl]2
NaCH(CO2Et)2
97% ee
Other ligands commonly used:
NH
PPh2
HN
Ph2P
OON
O
Ph
N
O
Ph
Trost Ligand
N
PPh2
PPh2
PCy2
Fe
(t-Bu)S
Ph2PO
iPr
Me
and many more...
These are examples to provide an appreciation of the scope, No need to memorise examples. Just understand that a Pd/chiral ligand combination is required.
39M Wills CH3E4 notes
Allylic substitution reactions – examples of ligands and reactions.
Trost ligand creates a chiral environmentthrough the phenyl rings on the phosphines.
NH
PPh2
HN
Ph2P
OO
Pd
P Pd P
OAcTrostligand (7.5 mol%)
NaCH(CO2Et)2
2.5 mol% [Pd(allyl)Cl]2
CO2Et
CO2Et
>98% ee
AcO OAc
Trost ligand and palladium
OO
AcO
O
O
98% ee
In this example (below) the catalyst displaces one OAc selectively,and also controls the regio and stereochemistry of the reaction.
PhAcO
AcO
Trostligand (7.5 mol%)
2.5 mol% [Pd(allyl)Cl]2N
O
O
iPr
Ph
Ph
AcO
NO
O
Ph
Pri
90% ee
These are examples to provide an appreciation of the scope, No need to memorise examples. Understand that a Pd/chiral ligand combination is required.
PhPh
AcO
PhPh
N
Other transformations which can be achieved by allylic substitution - soft nucleophiles generally favoured, otherwise the only limits are yourown imagination...
Pd(0) + chiral ligand
OO
N
O
O
K
PhPh
AcO
PhPh
HN
Pd(0) + chiral ligand
H2N
OAcPd(0) + chiral ligand
N
O
O
K
N
O
O
Ph OAc
Ph
N
HN
Pd(0) + chiral ligand
O O
TsHNNHTs
O OO
NHTs
O
Pd(0) + chiral ligand
MeO2CO OCO2Me
Pd(0) + chiral ligand
H2N Ph N
Ph
Allylic substitution reactions – examples of ligands and reactions.
These are examples to provide an appreciation of the scope, No need to memorise examples.
M Wills CH3E4 notes 41
Asymmetric catalysis – Isomerisation
Ph2P
PPh2
[Rh/S-BINAP]
Rh
NMe2 NMe2
Isomerisation (not a reduction!)
H
O
H H
R-citro-nellal, 96-99% e.e.
ZnBr2
then H2, Ni cat (to reduce alkene)
H
OH
(-)-menthol
Understand that this is an isomerisation.
42M Wills CH3E4 notes
Uses of enzymes in asymmetric synthesis.
R1 R2
OH Enzyme
Acylating agente.g. AcOCH=CH2 R1 R2
OH
R1 R2
O
O
+
one enantiomer formed selectively 50% max yield.
Dynamic kinetic resolution can produce 100% yield.
R1 R2
OH Enzyme
Acylating agente.g. AcOCH=CH2
R1 R2
O
O
one enantiomer formed selectively 100% max yield.
R1 R2
OH
this canInvert an alcohol overall.
Understand that asymmetric reactions can be achieved using an enzyme. By racemising the substrate, the reaction can give 100% of a chiral product.
43M Wills CH3E4 notes
Uses of enzymes in asymmetric synthesis.
Racemisation may be achieved via oxidation/reduction:
R1 R2
O
R1 R2
OH
R1 R2
OH
RuOCOC
Cl
Ph
Ph
Ph
PhPh
R1 R2
OH
RuOCOC
Cl
Ph
Ph
Ph
PhPh
R1R2
ORu
OCOC
Ph
Ph
Ph
PhPh
H
R1R2
ORu
OCOC
Ph
Ph
Ph
PhPh
H
R2R1
ORu
OCOC
Ph
Ph
Ph
PhPh
H
R2 R1
ORu
OCOC
Ph
Ph
Ph
PhPh
H
RuOCOC
Ph
Ph
Ph
PhPh
+
rotate
R2 R1
OHH
H
enantiomer
Mecahnism of inversion by oganometallic complex:
this canInvert an alcohol overall.
N
O
O
R1
N
O
O
R
N
O
OH
REnzyme
ROH
HN
OR
OR
O
Selective ring opening of a heterocycle:
Understand that asymmetric reactions can be done by an enzyme. By racemising the substrate, the reaction can give 100% of a chiral product. No need to memorise mechanism of racemisation.
44M Wills CH3E4 notes
Enzyme catalysis: amine oxidation.Chem. Commun. 2010, 7918-7920.
step 1:monoamine oxidase M(enzyme)37oC,
NH
H H
N
H H
94% ee
N
H H
O
HN
HN
AcO
O
O
HN
NH
N
O
O
N
O
HN
NH
N
O
O
N
OH
CN
HN
AcO
OTelepravir (Hepatitis CNS3 protease inhibitor)
Step 3: remove OAc to give OH,then oxidise to -keto amide)
Step 2: multicomponent coupling.
R1 R2
O
R1 R2
OHDehydrogenase
Enzymes can be 'evolved towards particular substrates - Reetz etc.
Amine 'deracemisation' using an enzyme.
NH
R
N
RNaBH4
+NH
R H
Over several cycles,all in situ, almost completeconversion to product isachieved.
Enzyme
Uses of dehydrogenase enzymes in synthesis.
For a nice example of use of an enzyme in dynamic kinetic resolution to make side chain of taxol see: D. B. Berkowitz et al. Chem. Commun. 2011, 2420-2422.
These are examples to provide an appreciation of the scope, No need to memorise examples.
45M Wills CH3E4 notes
Review on directed evolution by Reetz: M. T. Reetz, Angew. Chem. Int. Ed. 2011, 50, 138-174.By undertaking cycles of directed evolution, highly selective enzymes can be prepared, as shown by the example of desymmetrisation (Baeyer-Villiger reaction) shown below:
O
H
H
H
H
Optimised mutant enzyme
O
O
O
H
H
H
H
Optimised mutant enzyme
O
O
ClCl
O
H
H
H
H
Optimised mutant enzyme
O
O
OR
Optimised mutant enzyme
O
OR
O
Optimised mutant enzyme
O
O
O
Optimised mutant enzyme
O
O
O
Optimised mutant enzyme
O
O
HO HO
O
Optimised mutant enzyme
O
O
OHMe Me OH
94% ee
99% ee
91% ee
R= nBu 97% eeR=CH2Ph 78% eeR=Ph 96% ee
>99% ee
>99% ee
>99% ee
99% ee
These are examples to provide an appreciation of the scope, No need to memorise examples.
46M Wills CH3E4 notes
Other asymmetric reactions – for interest.
Asymmetric hydroboration:
MeO
OBH
O
1 mol%[Rh(COD)L]BF420oC, THF, thenH2O2, NaOH
MeO
OHHN
Ph2P
L=
Product formed in 88% ee.
Ph2P
L=
Product formed in 96% ee.
Ph2P
Asymmetric hydroformylation:
H2, CO, 0.1-0.2 mol% ligand0.05 mol% Rh(acac)(CO)2,
60oC, >99% conversion
H PPh2
P
O
HO
H+94% ee
88:12
O
O
Ligand =
Concluding material, non examinable.
47M Wills CH3E4 notes
R2R3
NR
R2R3
NHR
NC
Nu
R2R3
OR NH2 +
N
H+/H2O
R2R3
NHR
HO2C
Catalytic Strecker synthesis:
There are many other reactions which have been converted into asymmetric processes.
Other reactions:
Hydrosilylation
Hydroacylation
Hydrocyanation
Epoxidation using iminium salts
Asymmetric allylation
Hetero Diels-Alders
1,3-dipolar cycloadditions.
[2+2] cycloadditions
Cyclopropanation
Cross coupling reactions
Conjugate addition reactions
Etc. etc.
Concluding material, non examinable.