Content of Lecture 4 - final lecture of set

• Basics of transfer hydrogenation, andapplications in asymmetric synthesis

• Developments of transfer hydrogenationto redox-neutral reactions and C-C bondformation.

1. More detailed analysis of mechanism in(some) asymmetric hydrogenations

2. Scale-up of asymmetric catalysis

Basic principles of a metal-catalysed transferhydrogenation

OH

HMeMe + M

Me Me

O+ MH2

OH

H+ M C

O+ MH2

O O

Transfer hydrogenation from isopropanol or formic acid

reductant oxidant

Consider the formal catalytic cycle using i-PrOH;the reverse reaction will finally lead to racemization

CH3

O

H3C CH3

enantioselectivecatalyst

OH

CH3

OH

H3C CH3

O

H

H

10 M

0.1 M

CH3

OH

H

major minor

Equilibrium is driven by mass action

Noyri and Ikariya found an effective andenantioselective Ru-catalyst family ( also other arenes)

0.5 mol% Ru- catalyst

NH2

RuN

HCO2H, NEt3

Cl

Ph

Ph

99% ; 97% e.e

ArSO2

MeO

CH3

O

MeO

CH3

OH

Ruthenium catalyst derived from: NH2

NH2Ph

Ph RuCl

ClCl

2

and

Previous work had been largely Ir based and less effective

The reaction mechanism could be “classical” ora concerted hydrogen transfer. No evidence for

a Ru-alkoxide has ever been found

NH2

RuN

H

Ph

Ph

ArSO2

H

C O

H

NRuH

NH2

RuN

Cl

Ph

Ph

ArSO2

NH

RuNPh

Ph

ArSO2

H

C O

Ru

OCreduced form

reduction mechanism ?

oxidised form

H2, base

conventional Meerwein-Pondorff-Verleythrough Ru alkoxide

i-PrOH

Me2CO

Compare P2N2Ru catalysts + H2 - any differences?

X√

The range of transfer hydrogenation can bedemonstrated from the selection below

NH2

RuN

Cl

Ph

Ph

RSO2

vary R, arene

Ph CH3

O

Ph CH3

OHcat 1:1000OH

CH3cat 1:200

99%, 96% e.e. 99%, 97% e.e.

Ph

O

CH3

Ph

Ph CH2Cl

O

Ph CH2Cl

OHcat 1:5000OH

CH3cat 1:200

99%, 98% e.e. 99%, 95% e.e.

O

CH3

(epoxides)

N N

Reducing agent may be i-PrOH or NEt3/HCO2H azeotrope

There is a nice example of dynamic kineticresolution with DPEN-Ru and benzoin/benzil

PhPh

O

OPh

PhO

OH

PhPh

O

OH

PhPh

OH

OH

PhPh

OH

OH

PhPh

OH

OH

PhPh

OH

OH

+

+

(R,R) (R,S)

(R,S) (S,S)

starting from Benzil

starting from racemic Benzoin(rapid racemization in base)

k(R)k(S)

55> 95%, 99% e.e.

NH2

RuN

Cl

Ph

Ph

PhSO2Me

Pri

reagent is HCO2H/NEt3 in DMF

How and why do the two enantiomers of benzoin interconvert?

Deracemization. Enzyme as enantiomerically specificacylation catalyst plus Ru racemization catalyst

PhPh

PhPh

O PhPh

PhPh

OH

(CO)2Ru Ru(CO)2HCandida antarctica Lipase BNovozym 435transesterification

Shvo catalyst racemization -base free

Me

OH

Me

O

Me

OH

1 eq "pool"stays constant

Ru Ru

enzyme

Cl

O Me

O

Me

O

O

Me

99% e.e.

+

enzyme

Racemization step

The Ru(DPEN) catalysts can activate C-H in acidicreactants and transfer to electrophilic alkenes

NH

RuNPh

Ph

ArSO2

NH2

RuNPh

Ph

ArSO2

(CH2)CO2Me)2

Me2C=O CH(CO2Me)2

[X-ray structure]oxidised form

Stoichiometric

O(CH2)CO2Me)2

solvent

(S,S)-catalyst

99% e.e.

O

CH(CO2Me)2

Also nitroalkenes, other !-dicarbonyl compounds

Two important concepts for catalytic reactionsin synthesis

Atom economy describes the conversionefficiency of a chemical process in terms of allatoms involved. In an ideal chemical process allatoms of the reactants appear in the product.

Redox economy is defined as the minimization ofthe number of non-strategic (those that do notset stereochemistry or are not skeleton-building)or corrective oxidation and reduction steps insynthesis.

The concept of “borrowing hydrogen” uses the samecatalyst for both the oxidation and reduction steps

Knoevenagel reaction

OH O

H

CN

KOH

CNCN

IrCl Cl

dimer; getsreduced to Ir-H

Cl2

LnIrH2LnIr

LnIrH2LnIr

Conditions: 110 ˚C, 2 mol% Ir catalyst, 15 mol% KOH, 15 h.(or 10 minutes µwave, 110 ˚C)

Morken’s reductive coupling, and a possiblemechanism - aldehyde traps the intermediate in

a catalytic hydrosilylation

Ph H

O

CH2

O

OPhH

Ph OPh

OH

Me

O

77% syn, 87% e.e.

+ (R)-BINAP RhClHSiR3

O

RhBINAPOPh

H

H3C

H

OPh

O

RhBINAPOPh

H

H3C

H

OPh

Hydrogen from HSiR3

Reductive Rh-catalysed aldol couplings (Krische) -now trapping hydrogenation intermediates

Can you suggest an outline mechanism?

In the first step, a monohydride is formed

LnRh-X LnRh-XH

H

NR3 LnRh-H + NR3H XH2

coupling catalyst

H

O

CH2

O

PhH Ph

OH

Me

O

Syn:anti = 2:1

+

O2N O2NH2, KOAc

Rh(COD)2 OTfxs. PPh3

OMe

O

Ph

O

H2, K2CO3

Rh(COD)2 OTfxs. PPh3

OPh

O HO Me

with D2,single D here

H

H

H - abstraction

Reductive coupling of a dialkyne (diphosphine =BIPHEP or rac-BINAP)

MeO2C

MeO2C

Ph

Ph H2, 25˚ C

Rh(COD)2 OTfDiphosphine

Ph

Ph

MeO2C

MeO2C

HH

Ph

Ph

MeO2C

MeO2CRhDLn

Ph

Ph

MeO2C

MeO2C RhLn

D

Ph

Ph

MeO2C

MeO2C

DD

RhDLn

D2

Works for cyclisations to 5- or 6-membered rings

Why does simple alkyne hydrogenation not occur?

A transfer hydrogenation catalyst can couplealkynes with primary alcohols

Me

Me

H2COH

(CF3CO2)2Ru(CO)(PPh3)2

thf, 95 ˚CBr

MeMe Br

OHH

H

H2COH

Br

LnRu LnRuH2

CO

Br

H

LnRuH

Me Me

H

Me

Me

Br

O

HH

HRuLn

+

Me Me

MeMe Br

ORuLnHH

H

LnRuH2

Also allenes, dienes, allyl acetates in place of the alkyne

A pincer Ru complex with a labile C-H isactivated to loss of hydrogen - Milstein

Why is the original dehydrogenation product stabilised by H-transfer?

NEt2N PBut

2Ru

CO

H

H

N CH2

Et2N PBut2Ru

CO

heat– H2

NEt2N PBut

2Ru

COH

Complex is stabilised by borrowing hydrogen from the side-chain

18e

16e; isolable

H

The N-Ru bond is also labile leading to easy liganddisplacements; primary alcohol and amine to imine by

a β–elimination route (Step 1)

NEt2N PBut

2Ru

CO

H

H

NEt2N PBut

2Ru

CO

HPhCH2OH N

Et2N

PBut2Ru

CO

H

OPhH2C

Ph O

H PhCH2NH2

Ph N

HCH2Ph

Ph N

HCH2Ph

HOH

hemi-aminal intermediate; not catalysed?

Step 2. Another oxidative step converts the hemiaminalinto an amide; cycle is completed by H2 loss

NEt2N PBut

2Ru

CO

H

H

NEt2N PBut

2Ru

CO

H NEt2N

PBut2Ru

CO

H

O

Ph N

HCH2Ph

HOH

Ph N

OCH2Ph

H

Ph NH

CH2Ph

H

Amide - formed catalytically from alcohol and amine

How many different moleculesof catalyst are involved in theoverall cycle?

Overall catalytic reaction and conditions

NEt2N PBut2Ru

CO

H

catalyst 0.01 mMol

RNH2

10 mMol

+ R"CH2OH

10 mMol

catalysttoluene, 100 ˚C R

HN

O

R"

Argon flow

PhCH2 96%

PhCH2 MeOCH2 99%

n-C5H11

MeOCH2 99%

Et(Me)CHPhCH2 70%

New topic. Experimental work on the mechanism ofasymmetric hydrogenation of dehydro AA’s(kinetics,

spectroscopy) does not explain the origin of asymmetry

RhP P

Phospolane quadrant

bulky

bulky

open

open

Me

Me

Me

Me

(S)-product

Several empirical models for correlation of catalyst and product configuration exist. The most useful (for C2 symmetric ligands) is the quadrant rule.

Look into the P-Rh-P plane.

(S,S)-phospholane

There is general agreement between different groupson the broad features of the mechanism

1. In the ground state, one face of the alkene binds more strongly than the other. The major diastereomer is not the one that leads to the main reaction pathway.

2. The first step is the rapid reversible formation of a molecular hydrogen (!2-H2) complex

3. The turnover limiting step is either formation of the dihydride or the insertion of Rh-H into the coordinated alkene; the enantiomer-determining step is insertion.

4. The pathway from the minor diastereomer is the preferential one

5. The best DFT calculations of transition-state energies give a good correlation with experimental enantiomer excesses

Understanding of reaction pathways for asymmetriccatalysis requires GS and TS information!

I3

GroundState

I2I1

Transition state

Product

E

Pre-equilibrium

Post rate-determining

Only the energies of the Ground State and the Transition State for I1 to I2 need to be considered

Let’s assume that the pathway to the enantiomer has thesame profile and the ground-state is in rapid equilibrium

TS(S)

I2

I3I1

Reactant pool; the grey surface now represents the ground-state energy

I3

I2I1

TS(R)

(R)- Product

(S)-Product

The enantiomer excess arises directly from the(small) difference in transition-state energies

At 300K:

!!G – RT ln (kfast/kslow)

e.e. !!G

80

90

95

99

5.57.3

9.1

13.1

So 95 - 99 e.e is a bigger energy jump than 80 - 95 e.e. What about 99 - 99.9 e.e.?

The steps in Rh asymmetric hydrogenation according toseveral DFT calculations; all intermediates are cationic

H2

Product

H2 addition Migratory insertion

PRh

P

S

S

PRh

P

ONH

RO2C

R" PRh

P

ONH

RO2C

R"H H

P

RhP

ONH

RO2C

R"H

H

P

RhP

O

NH

CO2R

R"H H

Si bound

Computation indicates that the high-energy steps are H2 addition andmigratory insertion; one of these is turnover-limiting.

Completion of the analysis requires consideration ofboth reaction pathways (to (R) or (S) products)

H2

Product

H2 addition Migratory insertion

PRh

P

S

S

PRh

P

ONH

RO2C

R" PRh

P

ONH

RO2C

R"H H

P

RhP

ONH

RO2C

R"H

H

P

RhP

O

NH

CO2R

R"H H

all species cationic Rh+

PRh

P

OHN

CO2R

R"

Si bound

Re bound

PRh

P

OHN

CO2R

R"

PRh

P

OHN

CO2R

R"

P

RhP

OHN

CO2R

R"H

HH

HH H

H2

Preferred binding

Preferred reaction

ent-Product

Enantioselectivity does not always arise in turnoverlimiting step - example from conjugate addition!

OH

OHRh Rh 2

H2O

HORh

H2O

PhRh

PhB(OH)2slow

fast

O

PhRh

O

fast

H2O, fast

O

Ph

Rate ! [Rh]0.5[PhB(OH)2]1[ketone]0

H2O/dioxan

Model reaction for asymmetric catalysis with chiral dienes as ligands, e.g.

Ph

PhWhich is the turnover-limiting andwhich the enantiomer-determining step?

New topic. Relative ease of application of differentmethodologies for large-scale enantiomer production

In large-scale asymmetric synthesis the price of the(expensive) metal may not be the main factor

Turnover efficiency/ recovery often more important than cost of ligand or metal

High e.e. / good separation from the catalyst means avoidance of a purification step

"Classical" resolution of an intermediate often competes with asymmetric catalysis- especially if the "wrong" hand can be recovered and easily racemised.

Sometimes the chiral pool can solve the problems of asymmetric synthesis - availability of natural source, ease of synthetic transformations....

Rounded spot prices of precious metals in in € / gram, Nov 2009

Ru 2 Rh 50 Pd 7 Ir 10 Pt 30

The ligand may be more expensive than the metal (companies don't pay Aldrich prices for bulk)!

General considerations:

Hydrogenation is dominant among industrialprocesses involving asymmetric catalysis

The original asymmetric catalytic process -Monsanto asymmetric hydrogenation

Ruthenium has provided asymmetric catalystsfor both ketone and alkene reduction

Iridium catalysis has provided the most substantial scale-upin asymmetric hydrogenation Metolachlor; herbicide

The asymmetric synthesis of menthol by alkeneisomerization with a rhodium BINAP catalyst

Case History 1 - the large scale synthesis ofCandoxatril by asymmetric hydrogenation. A

RuCl2n

MgCl

Me

Ru(Meallyl)2

catalyst precursor

CO2Na

OMeO

ButO2C

1. 2.

Candoxatril precursor

3.CO2Na

OMeO

ButO2C

CO2Na

OMeO

ButO2C

CO2Na

OMeO

ButO2C

H

> 90% e.e.

isomerised alkene 22%Early hydrogenations MeO

MeO

PPh2PPh2

H2, Ru complex

Preliminaries

Discovery chemists prepare a compound with cardiovascular activity - eight steps, 4% overall yield

Challenge for process chemistry - make 2 tons for Phase 3 testing as the (R)-enantiomer

Case history 1 - the large scale synthesis ofCandoxatril by asymmetric hydrogenation B

4. Pressures higher than 4 atm. decrease the e.e.

5.CO2Na

OMeO

ButO2CCO2Na

OMeO

ButO2C H

MeOMeO

PPh2PPh2

H2, Ru complex

Optimisation

Key discovery mixed water / thf as solvent, 50 ˚C gives much reduced - isomerisation and higher e.e.

99.4% e.e., 96.5% pure

Make cyclohexylammonium salt,recrystallise

CO2–

OMeO

ButO2C H

99.7% e.e., 99.4% pure

NH3+O

MeO

ButO2C H O

HN

CO2H

Candoxatril

Case history 2 - the multi kiloton synthesis of (R)-metolachlor by asymmetric imine hydrogenation

1.

Preliminaries

Racemic compound established as an effective herbicide "Dual"; decision made to synthesis theenantiopure 1-(S) isomers. Atropisomerism about the C-N bond is not important for bioactivity

NMe

O

CH2Cl

MeO

NH2C

O

CH2Cl

MeO

Rh asymmetrichydrogenationunsuccessful

NH3C

MeO

Ir asymmetric hydrogenation successful!

Best previous e.ein imine hydrogenation 22%

2.NHH3C

MeO

NH3C

MeO

H2PPh2

Fe

Me

P

Me

Me2

Optimum "Josiphos" ligand

80 atm, 50 ˚C

needs 30% AcOH, also Bu4N I80% (S)