Download Reduction reactions of Double bonds

Year 3 CH3E4 notes: Asymmetric Catalysis, Prof Martin Wills
Reorganised to highlight key areas to learn and understand.
You are aware of the importance of chirality. This section 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:
Introductory, no need to revise, but understand concepts.
O
O
H
t-butyl peroxide
(oxygen source)
Ti(OiPr)4
(metal for complex
formation)
OH
Ti
O
O
OH
O
70-90% yield, >90% e.e.
O
HO
CO 2Et
HO
CO 2Et
O
CO 2Et
Ti
The oxygen atom is
directed to the
alkene.
The alkene is
above the peroxide.
Ti
O
CO 2Et
CO 2Et
O O
CO 2Et
Ti
O
O O
O
O
(+)-diethyl
tartrate (source of
chirality)
2) A covalent intermediate may be formed:
Proline catalyses the asymmetric cyclisation of a diketone (known as the Robinson
annelation reaction).
this is not a
chiral centre
O
O
Now this IS a chiral centreS configuration
L-proline
10 mol%:
N
H
CO 2H
O
O
Major product
O
Mechanism is via:
O
M Wills CH3E4 notes
N
O
HO 2C
1
Introductory, no need to revise, but understand concepts.
3) The reaction may take place within an asymmetric environment controlled by an external source:
The chiral counterion controls the asymmetry of the reaction.
E
R
Chiral Acid
H
E
N
H
R
R
R
N
The asymmetric environment
promotes reduction on one face
of cation
H
O O
P
O OH
R
R
N
R
O
P
H
O O
O
N
H
R
The key features of these approaches will be described and examples from the literature will be described.
Some examples of enantiomerically pure drugs:
F3C
NH 2
N
NC
HN
N
Aprepitant
(antiemetic)
Bn
N
O
F
NHtBu
CO 2Et
F
N
O
O
F
HN
H
N
N
N
NH
CF3
O
OH
F
F
H
N
O
CF3
Ly2497282
(Eli Lilly, diabetes)
N
MeN
AZ960
(AZ, anticancer)
NHtBoc
O
Rivastigmine
(Novartis, Alzheimer's)
F
M Wills CH3E4 notes
F
Takada
(Renin inhibitor for
hydpertension)
2
Oxidation reactions of alkenes.
R1
R1
2
2
R1
O
This represents a good way to create chiral centres.
R
OH
R
R3
Dihydroxylation
2
3
R
R3
R1
OH
R
epoxidation
OH
R1
OH
NH2
NH 2
2
R
3
2
R3
R
R
aminohydroxylation
Understand how each enantiomer of ligand gives a different product
enantiomer. No need to memorise which way round it goes.
The Sharpless dihydroxylation reaction employs ligand-acceleration to turn the known dihydroxyation
reaction into an asymmetric version.
This process depends on the use of an amine to accelerate a reaction:
Dihydroxylation
m
R
OsO4
RS
RL
OH
RS
RL
OH
Sharpless et al realised that enantiomericallyenriched amines could change this to an asymmetric
reaction:
OMe
Rm
use of the amine below
speeds the reaction up:
N
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
OMe
N
N
OH
HO
N
AD-mix  contains
DHQ (note both
amine groups are
of the same
OMe
absolute
configuration):
OMe
N
Dihydroquinine (DHQ)
Dihydroquinidine (DHQD)
'psuedo enantiomers'
N
N
N N
N
O
M Wills CH3E4 notes
O
N
3
Understand how each enantiomer of ligand gives a different product
enantiomer. No need to memorise which way round it goes.
RL=large group, RM=medium group, RS=small group.
Rm
OH
DHQD gives:
Rm
OH
RS
RL
DHQ gives:
S
L
R
R
OH
HO
Rm
How to remember::
Ph
OH
HO
Ph
2 x 'OH' added
to lower face.
ADmix-
(DHQ)
Ph
Ph
RL
ADmix-
Ph
OH
(DHQD)
HO
Ph
in this orientation
M Wills CH3E4 notes
2 x 'OH' added
to upper face.
4
Oxidation reactions of alkenes.
Learn the two possible mechanisms. No need to memorise examples.
The mechanism may be one of a number of possibilities:
AD-mix  (DHQ)
A chiral complex
may be formed,
directing the reaction
to one face in [3+2]
cycloaddition:
OMe
N
N
HO
O Os
O
O Ph
amine structure
abbreviated
OMe
N
N
N
N
O
O Os
HO
O
Ph
or it could be a [2+2]
cycloaddition, then
ring-expansion.
O
O
O
O
O
Os
Os
O
Ph
Ph
O
O
O
O
Ph
Ph
Ph
Ph
Ph
HO
OH
Ph
Hydrolysis and
reoxidation.
See if you can work out the mechanism.
Most recent evidence favours the [3+2] addition mechanism:
K. B. Sharpless et al, J. Am. Chem. Soc. 1997, 119, 9907.
M Wills CH3E4 notes
5
Oxidation reactions of alkenes.
No need to memorise the examples, but understand what
the dihydroxylation achieves, and how versatile it can be.
OH
CO 2Et
1 eq. MeSO2NH2
CO 2Et
CO 2Et
CO 2Et
AD-mix-, 0oC
tBuOH/H2O
OH
(DHQD)2PHAL
(AD-mix-)
OH 92% ee
OH 97% ee
OH
OH
1 eq. MeSO2NH2
Cl
AD-mix-, rt
tBuOH/H2O
Cl
OH 98% ee
nC 6H13
(DHQD)2PHAL
Me3Si
OH
AD-mix-, 0oC
tBuOH/H2O
AD-mix-, 0oC
tBuOH/H2O
M Wills CH3E4 notes
OH
AD-mix-, 0oC
tBuOH/H2O
SPh
OH
88% ee
up to 96% ee with
alternative ligand.
OH 98% ee
OH
1 eq. MeSO2NH2
OH
OH 97% ee
Me3Si
1 eq. MeSO2NH2
SPh
nC 6H13
(AD-mix-)
OH
OH 93% ee
1 eq. MeSO2NH2
Ph
AD-mix-, 0oC
tBuOH/H2O
Ph
97% ee
HO
HO
6
Understand the concepts, no need to memorise examples on this slide.
O
C5H11
Diastereoselective reactions:
O
1 eq. MeSO2NH2
O
O
OH
AD-mix-, 0oC
tBuOH/H2O
O
94% ee EtO2C
Asymmetric
dihydroxylation
(CH 2)2N3
OsO4 + oxidant
OH
O
OH
O
Si(tBu)Me2
1 mol% OsO4
Ph 3 eq. K3Fe(CN)6
O
A:B
with no ligand: 2:1
(DHQD)2PHAL: >20:1
(DHQ)2PHAL: 1:10
OH
Ph
(DHQD)2PHAL,
0oC, tBuOH/H2O
A
OH
O
(CH 2)2N3
EtO2C
OH
93% ee
Sharpless aminodihydroxylation is a closely-related process
(CH 2)2N3
EtO2C
B
O
O
HN
CO 2Et
TsN
BnO
OBn
NClNa
4 mol% K2Os2(OH)4
5 mol% (DHQ)2PHAL,
rt, tBuOH/H2O, 45%
CO 2Et
>15:1 regioselectivity,
93% ee
OH
TsN
catalyst 5 mol%
Jacobsen epoxidation of alkenes:
H
H
N
N
Mn
The iodine reagent transfers its oxygen atom
to Mn, then the Mn tranfers in to the alkene
in a second step. The chirality of the catalyst
controls the absolute configuration.
Advantage? You are not limited to allylic
alcohols
But
O
tBu
O
tBu
O
But
I
O
M Wills CH3E4 notes
(hypervalnet iodine
reagent)
Source of oxygen.
7
Reduction reactions of Double bonds (C=C, C=N, C=O).
This is a major area of asymmetric catalysis
- atom efficient, low waste, low energy.
H
R4
H source
R1
R1
R4 H source
O
H
catalyst
catalyst
2
3
2
3
2
R
3
R
R
R
R
R
H source might be H2 gas, hydride, or
an organic molecle (transfer hydrogenation)
OH
H
R2
H source
NR
R3
NHR
H
catalyst
R2
R2
R3
R3
Learn how a chiral environment is created around Rh(I) and how the
enamine substrate co-ordinates.
H
Rh/Diphosphine complex- ligands
create a chiral environment at the metal
PPh2
R
PPh2
PPh2
H
PPh2
P
face
edge
Chiraphos/Rh(I)
R
face
O
Rh
P
P
PPh2
O
R
P
R
P
H
S-BINAP
(often used
with Ru(II)
Chiral environment:
R
R
DuPHOS (R=Me, Et etc)
edge
BPE
R
R
R
P
Rh
R
P
P
PPh2
H
R
R
DIOP /Rh(I)
M Wills CH3E4 notes
8
Reduction reactions of Double bonds (C=C, C=N, C=O).
Learn how a chiral environment is created around Rh(I) and how the
enamine substrate co-ordinates. Understand that there is a difference
in energy between the diastereoisomers which leads to
enantioselectivity.
Examples of generalised applications (specific examples to follow).
MeOCHN
CO 2Me
R3
MeOCHN
H2
catalyst
H
H
H2N
CO 2Me
deprotect
CO 2H
R
2
R
CO 2Me
H
H
CO 2Me
2
R
CO 2H
H
Two diastereoisomers
may be formed, energy
difference influences
stereochemistry
P
Rh
P
H
N
+
CO 2Me
O
H
R
R
H H
N
P
O
Rh
H
R
R3
R
R
This isomer leads
to product, with
hydrogen
transferred to
back face as
drawn.
deprotect H2N
The acyl group co-ordinates to Rh,
reducing flexibility in transition state.
R
R
catalyst
2
R3
amino acid
R3
MeOCHN
H2
MeOCHN
CO 2Me
P
or
MeO2C
P
R3
R
R
R
R3
R
M Wills CH3E4 notes
H H
N
Rh
H
P
The difference in
reactivity may be
due to extra
stability of one
diastereoisomer or
increased activity
of one of them.
O
R
9
Reduction reactions of C=C Double bonds using Rh(I) complexes– representative examples.
No need to memorise examples but understand
that the sense of reduction in each case is relative
to the directing group X.
co-ordinating group
MeOCHN
CO 2Me
H
MeOCHN
CO 2Me
MeOCHN
(NHCOMe, OH, OCOMe etc)
H
X
A
X
A
99.2% ee
H
6.5 atm H2
B
C
MeOCHN
CO 2Me
R3SiO
MeOCHN
H
1 atm H2
1 mol % [Rh(B]+
20oC, ClCH2CH2Cl. R3SiO H
NHCOMe
Fe
MeOCHN
H
6.5 atm H2
H
CO 2Me
Single
diastereoisomer
98.2% ee
CO 2Me
Single
diastereoisomer
95% ee
H NHCOMe
25 at m H2
2 mol % [Rh(C)2]+
20oC, ClCH2CH2Cl.
Ph2P
B
0.2 mol % [Rh(A)]+
25oC, benzene.
97.2% ee
CO 2Me
Fe
CO 2Me
0.2 mol % [Rh(A)]+
25oC, benzene.
MeOCHN
P
A
PPh2
B
H
Addiition of hydrogen is relative
to the co-ordinating group
C
0.2 mol %
[Rh(SS-DuPHOS)]+
(R=Et), 2h, MeOH.
MeOCHN
CO 2Me
6.5 atm H2
P
M Wills CH3E4 notes
O N
P
O
P(C6H11)2
P(C6H11)2
C
D
10
Reduction reactions of C=C Double bonds using Rh(I) complexes– representative examples.
No need to memorise examples but understand
that the sense of reduction in each case is relative
to the directing group X.
O
Directing
O
co-ordinating group
(NHCOMe, OH, OCOMe etc)
H
X
A
X
A
B
CO 2Me
99.8% ee
H
CO 2Me 5 atm H2
EtO2C
C
H
O
0.4 mol %
[Rh(SS-DuPHOS)]+
(R=Et), 12h, DCM.
Directing
B
H
Addiition of hydrogen is relative
to the co-ordinating group
C
O
CO 2Me
4 atm H2
CO 2Me
EtO2C
1.1 mol% [Rh(D)]+
EtOH.
99 % ee
Directing
O
O
F3C
Ph
CF3
P(C6H11)2
P OMe
OMe
4 atm H2
P OMe
OMe
0.8 mol %
Ph
[Rh(SS-DuPHOS)]+
(R=Et), rt MeOH.
O
O
O H
O 96% ee
PPh2
P
P(C6H11)2
D
CF3
Fe
E
CF3
Directing
ButO2C
O
B
O
ButO2C
35 at m H2
5 mol% [Rh(E)]+
-5oC, toluene.
M Wills CH3E4 notes
H
O
B
O
94 % ee
11
Reduction reactions of Double bonds using catalysts derived from Ru(II) (C=C).
Learn that Ru(II) complexes of diphosphine ligands can also direct
hydrogenations. No need to memorise examples.
1 atm H2
H3C
CO 2H 0.5 mol%
[(R-BINAP)Ru(OAc)2]
MeOH/DCM (5:1)
MeO
directing group
MeO
Directing groups on the substrate help to
improve rates and enantioselectivity:
(BINAP or similar biaryl ligands are generally
favoured)
H
CO 2H
>97% ee
H3C
H
100 a tm H2
N
O
MeO
135 a tm H2
N
H
MeO
Ar
O
0.5 mol%
[(S-BINAP)Ru(OAc)2]
MeOH
directing group
O
1 mol%
[(R-BINAP)Ru complex]
MeOH, 50oC.
O
N
O
O
MeO
directing group
N
H
MeO
O
H
Ar
>99.5% ee
100 a tm H2
H
O
O
C2H5
F
C3H7
F
5 atm H2
CO 2H
directing group
1 mol%
[(R-BINAP)Ru complex]
MeOH, 50oC.
O
H
C3H7
O
0.2 mol%
[(R-BINAP)Ru complex]
DCM, 50oC.
directing group
C2H5
O
95% ee
CO 2H
90% ee
M Wills CH3E4 notes
12
Reduction reactions of Double bonds using catalysts derived from Ru(II) (C=C).
Learn that Ru(II) complexes of diphosphine ligands can also direct
hydrogenations. No need to memorise examples.
Allyliic alcohols provide a good example of
how the directing group works.
alcohol is directing group
OH
OH
OH
0.2 mol%
Ru(S-BINAP)
0.2 mol%
Ru(S-BINAP)
H2
Hydrogen on front face
relative to OH
H2
H
H3C
OH
H3C H
OH
OH
H CH 3
Hydrogen on front face
relative to OH
M Wills CH3E4 notes
13
Reduction reactions of isolated C=C double bonds can be achieved with variants of Crabtree’s catalyst.
Crabtree's catalyst works well on isolated (i.e. no nearby co-ordinating group) C=C, bonds:
+
N
+
PCy3
PF6-
4
R
R
PCy3
N
Ir
1
Ir
H2
catalyst
2
R3
R
P(oTo l)2
Ir(COD)
O
CH 3
97% ee
H CH 3
CH 3
N
1 mol% catalyst C
CH 3
MeO
89% ee
50 at m H2
rt, CH2Cl2
MeO
CH 3
S
N
B((3,5-C6H3(CF3)2)4(BARF-)
H CH 3
CH 3
B((3,5-C6H3(CF3)2)4(BARF-)
+
C
CH 3
rt, CH2Cl2
A
PPh2
Ir(COD)
92% ee
50 at m H2
CH 3
Ir(COD)
tBu
+
iPr
H
0.1 mol% catalyst A
PPh2
B
N
R3
R
1 mol% catalyst B
B((3,5-C6H3(CF3)2)4-
O
H
2
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.
50 at m H2
rt, CH2Cl2
+
O
O
R4
CH 3
+
Ph
H
No need to memorise examples.No directing group required
Asymmetric versions of the Crabtree catalyst
(prepared as COD complexes, but with the
COD left off for clarity):
N
R1
Ph
PPh2
Ir(COD)
D
B((3,5-C6H3(CF3)2)4-
OH
CH 3
M Wills CH3E4 notes
0.5 mol% catalyst D
50 at m H2
rt, CH2Cl2
OH
H CH 3 99% ee
Steric control - not OH group.
14
Reduction reactions of isolated C=C double bonds can be achieved with variants of Crabtree’s catalyst.
Understand that Ir(I) complexes with P and N donors can reduce
double bonds without a directing group in substrate. No need to
memorise examples.
+
O
N
Ph
Particularly challenging application:
P(oTo l)2
Ir(COD)
B
B((3,5-C6H3(CF3)2)4Vitamin E precursor
1 mol% catalyst B
AcO
R
O
AcO
50 at m H2
rt, CH2Cl2
R
R
R
O
>98% RRR enantiomer. Each reduction is controlled by the
catalyst i.e. it is not diastereocontrol.
M Wills CH3E4 notes
15
Reduction reactions of C=O Double bonds using organometallic complexes.
Understand that a C=O group can be reduced by a Ru or
Rh complex as well. No need to memorise examples.
The same principle regarding directing groups also applies to C=O reduction, Ru and Rh are most commonly used:
O
O
H3C
H OH
0.1 mol% [(R-BINA P)Ru(OAc)2]
OMe
86 atm H2
51h, 20oC, EtOH,100%
O
H3C
OMe
99% ee
directing group
directing group
O
H3C
O
P
4 atm H2
OMe
72h, 25oC, MeOH,99%
OMe
H OH
H3C
O
P
OMe
OMe
>95% ee
bromine is directing group
O
H3C
Br
0.1 mol% [(R-BINA P)Ru(OAc)2]
86 atm H2
H OH Br
H3C
o
62h, 20 C, EtOH,97%
>92% ee
M Wills CH3E4 notes
16
Reduction reactions of C=O Double bonds using organometallic complexes.
Understand that a C=O group can be reduced by a Ru or
Rh complex as well. No need to memorise examples.
The same principle regarding directing groups also applies to C=O reduction, Ru and Rh are most commonly used:
PPh2
RuBr2
O
PPh2
1-2 m ol%
H3C
SPh
H OH
30 atm H2
30h, rt, 100%
H3C
SPh
94% ee
directing group
0.25m ol%
N
(C5H9)2P
Rh(OCOCF3)
O
H3C
OP(C5H9)2
NMe 2.HCl
50 atm H2
18h, 20oC, PhMe, 100%
2
H OH
H3C
NMe 2.HCl
99% ee
0.5mol%
directing group
H
O
NHMe.HCl
H
P Rh(I). P
tBu But
10 atm H2
H OH
complex
18h, 50oC, MeOH, 92%
NHMe.HCl
99% ee
directing group
M Wills CH3E4 notes
17
Reduction reactions of C=O Double bonds using organometallic complexes.
Dynamic kinetic resolution can result in formation of two chiral centres:
Understand 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.
O
Racemic!
O
R2
OMe
1
R
reduced very slowly
H2 catalyst
O
O
O
R2
OMe
R1
R2
H
O
O
OMe
R1 enol
O
R2
OMe
1
R
fast
H OH
O
2
R
OMe
Principle: The substrate is rapidly racemising and one
enantiomer is selectively reduced:
1
H R
Enantiomerically Pure
M Wills CH3E4 notes
18
Reduction reactions of C=O Double bonds using organometallic complexes.
Dynamic kinetic resolution can result in formation of two chiral centres:
No need to memorise examples.
O
H OH O
4 atm H2
P OMe
P OMe
H3C
0.17 mol%
OMe
Ru/R-BINAP
OMe
H NHAc
65h, 25oC, MeOH
94% de
>98% ee
O
H3C
AcHN
88% de, 98% ee
e.g.
O
O
H3C
Cl
90 at m H2
H OH
0.5 mol%
H C
OMe Ru/R-BINAP 3
5h, 80oC, DCM
O
OMe
Cl H
98% de
99% ee
O
100 a tm H2
O
H3C
OMe
NHCO 2Ph
1 mol%
Ru/R-BINAP
20h, 50oC, DCM
H OH
H3C
O
OMe
H
NHCO 2Ph
88% de
98% ee
M Wills CH3E4 notes
19
Ketone reduction by pressure hydrogenation (i.e. hydrogen gas) can be achieved using a
modified catalyst containing a diamine, which changes the mechanism.
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.
Ph2
P
H
H2
N
Ph
N
H2
Ph
Ru
P
Ph2
O
H
HO H
Very high e.e.
from very low
catalyst loadings
H2 , solvent
Mechanism
Ph
O
Me
Ph2
P
P
Ph2
H
Ru
H
H
H
N
N
H2
Ph
OH
Me
H
Ph
Ph2
P
P
Ph2
Ph
Ru
H
N
N
H2
H
Ph
Ph
H2
M Wills CH3E4 notes
20
Ketone reduction by pressure hydrogenation (i.e. hydrogen gas) can be achieved using a
modified catalyst containing a diamine, which changes the mechanism.
No need to memorise the examples.
OH
O
O
0.2 mol% catalyst
2.5 mol% KOH
5 atm H2, EtOH, 5h, 100%
H OH
97% e.e.
M Wills CH3E4 notes
0.2 mol% catalyst
0.24 mol% KOH
5 atm H2, iPrOH,
3.5h, 100%
cis:trans 100:1
21
The use of hydride type reagents.
H source
O
R2
catalyst
R3
R
R2
H source
NR
2
OH
H
catalyst
3
R
R3
Oxazaborolidines require a relatively high catalyst loading of 10%,
But are effective in several applications.
NHR
H
R2
H
R3
Ph
Ph
Ph
O
N
O
B
i) Borane (BH3), Me
oxazaborolidine catalyst
ii) hydrolysis (work up)
How it works:
HO H
O
H Ph
N
B
Me
O
B
H
Concave molecule H
H
hydride directed to one face.
Understand that hydride reagents can also be used in reductions.
Understand how the mechanism of hydride transfer relates to the
previous slide. Be able to draw the mechanism of the hydride
transfer step.
M Wills CH3E4 notes
22
The use of hydride type reagents.
Understand that hydride reagents can also be used in reductions.
Understand how the mechanism of hydride transfer relates to the
previous slide. Be able to draw the mechanism of the hydride
transfer step.
More contemporary focus is on asymmetric transfer hydrogenation and on
organocatalysis.
Transfer hydrogenation – Ru catalysts.
O
edge/face interaction
R
activation
Ru
TsN
Cl
substrate
HCO 2H, Et3N
N
Ph
H
Ru
TsN
H
H
N
Ph
H
Ph
Ru
TsN
N
Ph
H
H
H
O
H
N
Ph
Ph
Ph
Ru
TsN
H
O
H
H
R
Ph
HCO 2H, (-CO 2)
Product
R
ee >96%
full conversion
Catalyst prepared by combining:
Cl
Ph
NHTs
Ru
and
Ph
NH 2
Cl
Cl
Ru
Cl
in iPrOH/KOH
or
HCO2H/Et3N
Rh
TsN
H
N
Ph
H
Ph
M Wills CH3E4 notes
H
Ir
TsN
H
N
Ph
H
Rhodium and
iridium complexes
are isoeletronic with
Cp' on metal in place
of arene.
H
Ph
23
Examples of reductions using transfer hydrogenation with metal complexes: add C=O and C=N reductions.
These are examples to provide an appreciation of the scope, No
need to memorise examples.
O
X
H OH
0.5 mol% SS-Ru catalyst
X=H, 98% ee
X=Cl, 95% ee
X=OMe 96% ee
HCO2H/TEA
28oC,
TsN
X
A wide range of substituents can be tolerated, except for orthogroups, which result in reduced selectivities.
O
Cl
99% ee
0 or 1
0 or 1
H
Ph
O
HCO2H/TEA
28oC,
Cl
N
Ph
H OH
0.5 mol% SS-Ru catalyst
Improved catalyst with a
link between amine and
arene ring.
Following reactions are
with this catalyst.
Ru
0.1 mol% RR-Ru catalyst
above
X
X
X=H, Cl, OMe
typically 96% ee
0.1 mol% RR-Ru catalyst
above
O
Cl
H OH
0.5 mol% SS-Ru catalyst
N
S
O2
HCO2H/TEA
28oC,
HCO2H/TEA
28oC,
S
O2
Precursor to jknown drug
C4H9
0.5 mol% SS-Ru catalyst
0.6 mol% KOH, iPrOH
28oC,
C4H9
N
S
H
OH
>99% ee
up to 97% ee
Other reduction products:
H
O
H OH
Cl
98% ee
S
Cl
HCO2H/TEA
28oC,
(i.e. fused five or six-membered ring)
O
H OH
OH
X
98% ee X=O
97% ee X=S
M Wills CH3E4 notes
H OH
N
H OH
OPh
97% ee
N
97% ee
24
These are examples to provide an appreciation of the scope, No
need to memorise examples.
Some imines can also be reduced by asymmetric transfer hydrogenation:
Ru
TsN
Ph
MeO
MeO
N
Cl
N
H
Ph
0.5 mol%
RR-Ru catalyst
HCO2H/TEA
28oC,
Typically >96% ee
MeO
N
NH
MeO
H
Ar
N
H
H
Ph
Typically >98% ee
O
Other ligands can be used with ruthenium(II)
in asymmetric catalysis (and also with Rh and Ir), e.g.
NH
Ph
N
H
Ar
as above
BocHN
OH
NH 2
NH
H2N
Ph
N
H
OH
NH HN
PPh2
Ph2P
NHTs
OH
Challenging substrates:
Excellent ligand for
transfer hydrogenation.
O
H OH
0.5 mol% [RuCl2(arene)]2
1.25 mol% ligand
5 mol% NaOH, iPROH
28oC, 16h, 96%
O
O
P
O
P
O
O
O
0.5 mol% [RuCl2(arene)]2
90% ee
H OH
1.25 mol% ligand
5 mol% NaOH, iPROH
28oC, 22h, 99%
M Wills CH3E4 notes
99% ee
25
O
Asymmetric transfer hydrogenation by organocatalysis.
H H
Inspired by Nature's NADH;
a coenzyme which transfers hydride
H
H H
CONH 2
CONR 2
CONR 2
R2NOC
OH
H
N
R
N
N
R
Me
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.
Use combination of a chiral acid with a hydride source:
R
H H
O
Homochiral acid (directs reaction)
R=aryl ring, trialkylsilyl etc., usually a
bulky group. Catalytic amount needed.
O
O
O
O
P
*OP
OH
+
OH
H H
CONR 2
R2NOC
Mechanism:
O
*OP
H
R1
O
R2
+
N
H
R3
O
+
OH
H
1
condensation
R
N
Source of hydride - stoichiometric
amount needed. Similar to NADH used in
biological transformations. Known as
'Hantzsch ester'.
H
CONR 2
R2NOC
N
H
R
(Either use a preformed imine or via reductive amination)
CONR 2
R2NOC
O
*O P
R3
O
O
H
1
R
N
H
H H
R2NOC
R2
close ion pair formed
M Wills CH3E4 notes
N
R2
CONR 2
O
O
*O P
O
R3
Proton
can be
reused
N
H
H
N
R1
H
N
H
26
R2
R3
Asymmetric transfer hydrogenation by organocatalysis.
No need to memorise examples, but
understand the concept.
fully heteroaromatic rings can be reduced:
Some examples of reductions:
OMe
OMe
N
1 mol% cat
where R=H
HN
1.4 eq. Hantzsch ester
Toluene, 35oC, 71h, 91%
N
H
93% ee
OMe
N
1 mol% cat
where R=H
1.4 eq. Hantzsch ester
Toluene, 35oC, 60h, 80%
HN
MeO
90% ee
90% ee
MeO
Ts
N
Asymmetric reductive amination:
Ts
N
O
H
N
H
2.4eq. Hantzsch ester
toluene, 60oC, 12h
MeO
95%
MeO
OMe
H
1 mol% cat where
R=bulky aryl
+
H2N
M Wills CH3E4 notes
10 mol% cat where
R= SiR3
HN
H
1.2 eq. Hantzsch ester
Toluene, 40oC, 48h, 90%
+ molecular seives
27
93% ee
Formation of chiral centres by nucleophilic additions to unsaturated bonds.
This slide is for information only and does not need to be memorised
EtO C
for the exam. EtO C
Nu
NHR
Nu
Nu
OH
NR
O
2
2
R2
R2
R3
Nu
R3
Results:
Diethylzinc additions
NMe 2
O
HO
Et
OH
Et2Zn, toluene (solvent)
H
R2
R2
R3
H
(-)-DAIB (see below)
R2
R3
Nu
mol% DAIB used
(relative to Et2Zn
Yield
E.e.
0 (i.e. none)
0%
-
2 (0.02 eq.)
97%
98
100 (1.0 eq.)
0%
-
Et2Zn
NMe 2
-EtH
OH
O
O
With excess Et2Zn, a dimer is formed. the
dimer splits and enters the catalytic cycle:
Me2
N
Et2Zn
Me2
N
O
PhCHO
Zn
O
O
Zn
N
acid workup
H
Zn
O
Me2
N
Zn
PhCHO
H
Ph
Zn
O
O
Zn
Ph
H
Me2
N
O
Zn
O
M Wills CH3E4
notes
+
Zn
reduction
product!
H
Et2Zn
2 DAIB of 15% ee will give a
Another interesting Me
fact:
product of 95% ee! This is because the dimer made
from one of each enantiomer is more stable, and does
not split up to enter the catalytic cycle.
H
Ph
H
Zn
O
H
O
O
Me2
N
H
Ph
Zn
H
H
- CH2=CH2
R3
How come a little bit
of amino alcohol
catalyses the
reaction, but a lot of
it doesn't?
Answers - the stoichiometric (100% ligand) reaction forms the wrong reactive species:
Me2
N
R2
Nu
R3
Ph
Ph
Ph
H
Ph
H acid workup
H
HO
28
Ph
H
Ph
More applications of organocatalysis.
Examples of common
organocatalysts:
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.
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
O
O
10 mol%:
N
H
or pyrrolidines:
Ph
N
H
Ph
Ph
N
H
or other N-heterocycles:
O
NMe
O
Ph
Major product
CO 2H
N
H
Now this IS a chiral centreS configuration
L-proline
CO 2H
L-proline
N
H
CO 2H
O
The enantiomeric
compound is:
O
O
O
Mechanism is via:
N
O
O
HO 2C
M Wills CH3E4 notes
29
More applications of organocatalysis.
No need to memorise examples.
This is now the basis for many other reactions e.g.:
O
L-proline
Aldols:
O
O
10 mol%: N
H
H
H
Me
Me
O
CO 2H
90% yield
OH
4:1 anti:syn
H
Me
DMF
O
N
Me
anti product e.e.: 99%
O
H
H
Me
Me
OtBu
and even more complex ones:
O
O
H2N
H
TBSO
OTB S
20 mol%
O
CO 2H
O
O
OH
3 mol% water, rt 2 days.
68%, major product: D-fructose
precursor
O
TBSO
OTB S O
These reactions take place via formation of an enamine which then reacts with the other reagent e.g.
M Wills CH3E4 notes
30
More applications of organocatalysis which proceed via formation of an enamine
– bonds to C atoms.
These are examples to provide an appreciation of the scope, No need
to memorise examples.
C-N bond formation:
L-proline
CO 2Bn
O
CO 2H
i) 10 mol%: N
o
H MeCN, 0 C
N
+
H
via:
N
CO 2Bn
nBu
O
CO 2Bn
OH
CO 2Bn
N
H
nBu
97% ee
N
ii) NaBH4, EtOH, 94%
HO
CO 2Et i) 10 mol%:
N
DCM, rt.
O
+
H
N
CO 2Et
N
H
Ar
Ar
NH 2
N
N
nBu CO 2Bn
H
nBu
amino acids.
Ar= 3,5-(CF3)2C6H3
CO 2H CO 2Bn
N
O
97% ee
O
NH
OTM S
ii) NaBH4, EtOH, 83%
O
via:
MeN
O
C-Halide bond formation:
O
Cl
Cl
+
H
C5H11
1.2 eq.
O
5 mol%
MeN
Cl
Cl
Cl
N
92% ee
O
DCM, -24oC, 71%
N
H
Ph H
ii) 20 mol%:
tBu
+
H
tBu
Br
Br
Cl
N
DCM, -24oCH
Ar
Ar
OTM S
ii) NaBH4, MeOH
OH
95% ee
C5H11
O
O
etc
Cl
Cl
1.2 eq.
O
But
Cl
Cl
Cl
O
H2O
C5H11
Cl
Z-enamine,
O
orientated away
MeN
from dimethyls
phenyl ring
N
blocks lower face
Cl
Cl
C5H11
Br
tBu
M Wills CH3E4 notes
31
Ph
Cl
C5H11
C=C reduction by organocatalysis.
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.
Asymmetric catalysis of C=C bonds can be catalysed by organocatalysts, if they are conjugated to a C=O:
H
General mechanism:
R1
H
R1
O
2
R
H
N
+ H
N
R
H H
2
H
R
CONR 2
R2NOC
+
H
R
CONR 2
R2NOC
NR 2
R2
N
N
H
R1
H
H
H2O
R
N
H
R1
O
H
R2
H
H
H
H H
CONR 2
R2NOC
Source of hydride.
N
H
20
C=C reduction by organocatalysis.
No need to memorise examples.
Examples:
O
H
Ph
O
O
H
10 mol%
Bn
Me
N
H
Ph
CO 2Et
1.02 eq.
H
tBu
H H
EtO2C
O
O
NMe
H
Me
90% ee
H
t-Bu
O
NMe
5 mol%
H
Me
H
tBu
N
H
t-Bu
H H
CO 2Et
EtO2C
1.2 eq.
Me
H
90% ee
N
H
N
H
O
NMe
20 mol%
O
Bn
N
H
O
O
H H
tBu
CO 2tBu
ButO2C
1.1 eq.
tBu
N
H
20
Additions to C=O – aldol reactions are a very important class of synthetic reaction.
Catalytic asymmetric aldol reactions can be directed by a chiral Lewis acid (ML*n), which initially binds to the electrophilic component
the silyl enol ether is required as nucleophile (known as the 'Mukaiyama aldol' reaction:
Me3
SIMe 3
Si
SiMe
3
O
OH
SiMe3
ML*n
ML*n
O
O
O
O
hydrolysis
O
O
O
R2
R3
3
R2
R
H
2
R2
3
R2
R
H
H
R
R3
R1
1
H
1
R
1
1
R
R
ML*n
R
Silyl enol ether
These slides are for information only and do not need to be
memorised for the exam. However you should understand that
addition of a silyl enol ether to an aldehyde can by catalysed by a
Lewis acid.
22 mol%
H
O
BnO
SiMe3
OBn +
N
NMe H
O
H
SiMe3
20 mol% Sn(OTf)2
20 mol% SnO
MeCN, -78oC
Open transitions states operate in this case, rather
than chair-like cyclic ones:
LnM
Favoured approach
leading to major
BnO O
product.
H
O HO
Me3Si
O
OBn SiMe3
MLn
MeO
BnO O
SiMe3
BnO
Disfavoured by
H
steric clash.
97:3 syn:anti, 91% ee.
Me3Si
OBn
O
SiMe3
M Wills CH3E4 notes
34
Additions to C=O – aldol reactions are a very important class of synthetic reaction.
This slide is for information only and do not need to be
memorised for the exam. However you should understand that
addition of a silyl enol ether to an aldehyde can by catalysed by a
Lewis acid.
22 mol%
O
SiMe3
+
EtS
Me
N
Me
O
H
R
N
H
20 mol% Sn(OTf)2
MeCN, -78oC
O
OSiMe3
EtS
R
Me
Open transitions states operates
again in this case and ligand determines overall
face selectivity:
LnM
R=Ph, 93:7 syn:anti,
O
Favoured approach
BnO
90% ee (syn)
leading to major
H
R
product.
R=nC7H15, 100:0 syn:anti
Me3Si
O
OBn
>98% ee
M Wills CH3E4 notes
35
Other examples of metal/ligand-catalysed asymmetric aldol reactions.
This slide is for information only and do not need to be memorised
for the exam. However you should understand that addition of a silyl
enol ether to an aldehyde can by catalysed by a Lewis acid.
O
O
N BBu
Ts
20 mol%
O
SiMe3
+
Ph
N
H
O
H
R
O
MeCN, 14h, -78oC
OSiMe3
Ph
R
R=Ph, 89% ee
R=nC6H13, 93% ee
No relative stereocontrol
in this case, just absoluteby the catalyst.
O
20 mol%
OiPr O HO 2C
O
O
t-Bu
SiMe3
+
OiPr
O
H
R
EtO
O
O
O B
3,5-(CF 3)2C6H3
MeCN, -78oC
O
t-Bu
SiMe3
+
20 mol%
O
N BBu
Ts
O
H
O
R
MeCN, 1h, -78oC
O
OSiMe3
EtO
R
91-99% ee
OSiMe3
R
93:7 syn:anti, 94% ee
Although in some cases, the relative stereochemistry can be changed by a different combination of reagents:
H
OMe H Bn
O
N
Si(tBu)Me2
MLn
OMe
OMe
O
O HO
NMe
O
MeO
O
+
OMe
H
OMe
MeO
OMe
MeO
O
[(nBu)2Sn(OAc)2]
BnO
Favoured approach
OBn
Me3Si
Sn(OTf)2
leading to major
major product, 87% ee.
CH2Cl2, -23oC
product.
Mukaiyama, Chem. Eur. J. 1999, 5, 121-161
Cycloaddition reactions can be catalysed by Lewis acid/chiral ligands.
The ligand and metal choice can have a dramatic effect:
Understand how a copper complex of the bis(oxazolidine) ligand
can control the Diels-Alder reaction. Be able to draw the complex of
Cu and Mg and illustrate which face the cyclic diene adds from. Be
able to draw the product, which is of endo stereochemistry. Do not
memorise examples.
Cu complex is square planar
tBu blocks lower face, Cp adds from other side
10 mol%
O
O
X
+
X=CH2 98:2 endo:exo
>98% ee
Cu
O
O
X
N
N
TfO
OTf
O
O
Cu
O
N
O
o
18h, -78 C, CH2Cl2
O
N
X=O, 80:20 endo:exo
97% ee (endo)
M Wills CH3E4 notes
N
N
O
O
O
O
N
X
37
Cycloaddition reactions can be catalysed by Lewis acid/chiral ligands.
The ligand and metal choice can have a dramatic effect:
No need to memorise examples, but understand how the
selectivity is controlled.
M Wills CH3E4 notes
38
There are many other similar catalysts for Lewis-acid catalysed Diels-Alder reactions.
Be able to draw the Cu complex and how it controls the reaction.
No need to memorise examples.
Organocatalysts can be applied to Diels-Alder reactions, by forming a cationic intermediate:
Intramolecular versions also work:
O
5 mol%
O
N
O
O
O
Cu
O
N
TB DMSO
N
N
O
TB DMSO
O
O
H
SbF6
Me
H
N
4
4
24h, rt, CH2Cl2 81%
>99:1 endo:exo
96% ee
H
(-)-Isopulo'upone
(natural product)
H
24
There are many other similar catalysts for Lewis-acid catalysed Diels-Alder reactions.
For the organocatalysis part you should be able to draw a
mechanism for imine formation, for the cycloaddition
(understanding that it is endo and with addition from the
unhindered face) and the product, as well as the hydrolysis step.
No need to memorise examples.
O
O
O
O
NMe
H
+
O
R
Ph
N
H
Ph
Ph
Ph
H
NMe
20 mol%
N
H
Ph
2
R
O
H
Product =
O
O
H
+
R1
R1
R1
O
NMe
H
H
R1
R2 O
+
2N
R
+
R2 N
R2
+
N
NMe
1
NMe
NMe
R1
H
O
N
H
Ph
tBu
O
+
Et
Ph
10 mol%
NMe
20 mol%
NHCbz O
N
H
20:1, 90% ee
CbzHN
N
H
Ar
O
Et
O
>100:1, 98% ee
+
O
O
NH
O
O
O
Bn
NH 2
CHO
H
92% ee
24
Allylic substitution reactions are powerful methods for forming C-C bonds.
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.
PdLn
Ph
Nucleophile adds
trans to PdLn group
Ph
Nu
AcO
Attack at the other
end of allylic system
gives alternative
enantiomer:
Nu
Pd(0), Nu
Ph
Ph
Ph
Ph
Nu
AcO
Ph
PdLn
Ph
Ph
Ph
PdLn
Nu
Nu
PdLn
Ph
Ph
The Pd is behind the
allylic group.
PdLn
Chiral ligand
Nu
Ph
PdLn
LnPd
LnPd
Ph
Ph
Ph
Ph
Ph
41
M Wills CH3E4 notes
Nu
Allylic substitution reactions are powerful methods for forming C-C bonds.
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.
PdLn
Example ligand:
0
Ph
Nucleophile adds
trans to PdLn group
O
Ph
Ph2P
N
Pd
Nu
R
Favoured conformation: the allyl group is in
front of the Pd complex.
H
O
Ph2P
N
Pd
Ph
R
Ph
or
OAc
(racemic)
Ph
Ph
0
O
Ph2P
N
Pd
Ph
-AcO
O
Ph2P
Ph
N
Pd
OAc R
Ph
H
O Ph
Ph
Ph2P
R
Ph
Pd
Nu
Ph2P
Ph
Ph
Trans effect favours
addition to end opposite
the P atom.
H
N
disfavoured by
steric clash with equatorial H
or slower to react.
N
Pd
Nu
R
Ph
Ph
(enantiomerically enriched)
and catalyst is released to re-enter cycle.
42
M Wills CH3E4 notes
Allylic substitution reactions – examples of ligands and reactions.
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.
5 mol%
Ph2P
O
Other ligands commonly used:
AcO
Ph
2.5 mol% [Pd(allyl)Cl]2
Ph
Ph
NaCH( CO 2Et)2
EtO2C
O
O
Ph
Ph
CO 2Et
O
N
97% ee
Ph
O
N
NH
PPh2
M Wills CH3E4 notes
Ph
N
PCy2
PPh2
Ph2P
and many more...
PPh2
Ph
HN
Trost Ligand
N
Ph2PO
Me
(t-Bu)S
iPr
Fe
43
Allylic substitution reactions – examples of ligands and reactions.
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.
Trost ligand creates a chiral environment
through the phenyl rings on the phosphines.
CO 2Et
OAc
O
O
NH
HN
PPh2
Trost
ligand (7.5 mol%)
CO 2Et
NaCH( CO 2Et)2
AcO
>98% ee
AcO
2.5 mol% [Pd(allyl)Cl]2
Ph2P
O
In this example (below) the catalyst displaces one OAc selectively,
and also controls the regio and stereochemistry of the reaction. O
Pd
AcO
Trost ligand
and palladium
OAc
Ph
Trost
ligand (7.5 mol%)
O
2.5 mol% [Pd(allyl)Cl]2
N
iPr
AcO
O
O
P
Pd
Ph
O
98% ee
O
P
N
O
AcO
AcO
Ph
Ph
Trost
ligand (7.5 mol%)
2.5 mol% [Pd(allyl)Cl]2
NaCMe (CO 2Et)2
Pri
>95% ee
AcO
EtO2C
M Wills CH3E4 notes
Ph
AcO
Ph
90% ee
CO 2Et
44
Allylic substitution reactions – examples of ligands and reactions.
These are examples to provide an appreciation of the scope,
No need to memorise examples.
Other transformations which can be achieved by allylic substitution - soft nucleophiles generally favoured, otherwise the only limits are your
own imagination...
Ph
Ph
Ph
Ph
O
Pd(0) + chiral ligand
AcO
O
N
O
Pd(0) + chiral ligand
O
OAc
K
N
O
N
K
O
N
O
O
Ph
Ph
Ph
Ph
Ph
Pd(0) + chiral ligand
AcO
HN
H2N
Pd(0) + chiral ligand
Ph
HN
Pd(0) + chiral ligand
Pd(0) + chiral ligand
O
O
O
O
O
NHTs
TsHN
N
OAc
O
N
HTs
H2N
MeO2CO
OCO 2Me
Ph
N
Ph
Uses of enzymes in asymmetric synthesis.
this can
Invert an alcohol overall.
Understand that asymmetric reactions can be done by an enzyme.
By racemising the substrate, the reaction can give 100% of a chiral
product (see analogy with slide 11).
O
OH
R2
R1
OH
Enzyme
Acylating agent
e.g. AcOCH=CH2
O
+
R1
R2
Dynamic kinetic resolution can produce 100% yield.
O
OH
R1
2
R
R1
one enantiomer formed
selectively 50% max yield.
OH
R2
1
R
Enzyme
2
R
Acylating agent
e.g. AcOCH=CH2
O
R1
R2
one enantiomer formed selectively 100% max yield.
M Wills CH3E4 notes
46
Uses of enzymes in asymmetric synthesis.
Understand that asymmetric reactions
can be done by an enzyme. By
racemising the substrate, the reaction
can give 100% of a chiral product (see
analogy with slide 11). No need to
memorise mechanism of racemisation.
R1
O
2
R
R1
R2
Ph
Ph
Ph
Ph
OC
OC
Ru
Ph
Ph
Ru
OC
OC
H
Cl
N
O
H
N
R
ROH
OR
O
O
rotate
O
O
H
R2
Ph
Ph
Ph
Ru
OC
O
OC
H
R1
Ph
Ph
Ph
R2
R1
Ph
Ph
Ph
Ph
Ph
Ph
Ru
OC
O
OC
H
R2
M Wills CH3E4 notes
Ph
Ph
Ph
Enzyme
Ru
OC
OC
Ph
Ph
R
O
Ph
Ph
R
Ru
R
OH Ph
R2
R1
Cl
2
Selective ring opening of a heterocycle:
R
R1
O
O
N
N
OH
O
H
Ph
1
Ph
OC
OC
Ph
Ph
Ph
R2
R1
Ph
Ph
Ph
OH
Mecahnism of inversion by oganometallic complex:
Ph
Racemisation may be achieved via oxidation/reduction:
OH
this can
Invert an alcohol overall.
OC Ru
OC
H OH
R1
+
R2
R1
enantiomer
47
Uses of dehydrogenase enzymes in synthesis.
These are examples to provide an appreciation of the scope, No need
to memorise examples.
Amine 'deracemisation' using an enzyme.
O
OH
Enzyme
Dehydrogenase
1
2
R
R
+
2
1
R
R
NH
Enzymes can be 'evolved towards
particular substrates - Reetz etc.
N
R
NH
Over several cycles,
all in situ, almost complete
conversion to product is
achieved.
R H
R
NaBH4
Enzyme catalysis: amine oxidation.
Chem. Commun. 2010, 7918-7920.
H
H
Step 2: multicomponent coupling.
N
H
step 1:
monoamine
oxidase M
(enzyme)
37oC,
OH
O
N
N
H
O
N
H
94% ee
H
N
O
O
O
O
AcO
C
N
H
N
N
H
H
AcO
H
N
N
O
O
N
H
H
H
N
H
N
N
N
Step 3: remove OAc to give OH,
then oxidise to -keto amide)
O
Telepravir (Hepatitis CNS3 protease inhibitor)
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.
M Wills CH3E4 notes
48
These are examples to provide an appreciation of the scope, No
need to memorise examples.
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
Optimised mutant
enzyme
O
Optimised mutant
enzyme
O
O
O
>99% ee
O
H
H
94% ee
O
O
H
H
Optimised mutant
enzyme
Cl
O
Optimised mutant
enzyme
O
Cl
O
>99% ee
O
H
H
99% ee
O
O
H
H
Optimised mutant
enzyme
Optimised mutant
enzyme
O
O
O
>99% ee
O
H
H
HO
HO
91% ee
Optimised mutant
R
enzyme
R
O
O
Optimised mutant
enzyme
O
O
O
O
R= nBu 97% ee
R=CH2Ph 78% ee
R=Ph 96% ee
99% ee
Me OH
M Wills CH3E4 notes
Me
OH
49
Other asymmetric reactions – for interest.
Concluding material, non examinable.
Asymmetric hydroboration:
L=
L=
O
H OH
N
BH
Ph2P
O
Asymmetric hydroformylation:
Ph2P
Ph2P
1 mol%[Rh(COD)L]BF4
MeO
20oC, THF, then
H2O2, NaOH
MeO
Product formed in 88% ee.
Ligand =
O
H
H
H2, CO, 0.1-0.2 mol% ligand
0.05 mol% Rh(acac)(CO)2,
PPh2
O
88:12
O
P
H
+
60oC, >99% conversion
Product formed in 96% ee.
O
94% ee
Asymmetric catalysis – Isomerisation.
Ph2
P
Rh
PPh2
H
NMe 2
H
NMe 2
[Rh/S-BINAP]
Isomerisation (not a reduction!)
R-citronellal, 96-99% e.e.
M Wills CH3E4 notes
H
H
O
ZnBr2
then H2,
Ni cat (to
reduce alkene)
OH
(-)-menthol
50
There are many other reactions which have been converted into asymmetric processes.
Catalytic Strecker synthesis:
R
NH2
+
O
R2
Concluding material, non
examinable.
Nu
NR
R3
R2
3
R
NHR
R2
NC
H+/H2O
R3
NHR
R2
HO2C
R3
N
Other reactions:
Hetero Diels-Alders
Hydrosilylation
1,3-dipolar cycloadditions.
[2+2] cycloadditions
Hydroacylation
Cyclopropanation
Hydrocyanation
Cross coupling reactions
Epoxidation using iminium salts
Conjugate addition reactions
Asymmetric allylation
Etc. etc.
M Wills CH3E4 notes
51