Asymmetric synthesis: problems 2

reagent controlled crotylation

substrate control carbonyl addition

substrate controlled aldol reaction

catalyst controlled Diels-Alder reaction

an enantioselective synthesis of kainic acid

1

BO

OiPrO2C

iPrO2C O

OBn 70%88:12 dr

OH

OBn

Question 1

An easy start. Rationalise the observed diastereoselectivity.

For bonus marks is this an example of matched or mismatched substrate/reagent stereoselectivity (sometimes called double asymmetric induction). Or in other words, learn what matched/mismatched means!

2

BO

OiPrO2C

iPrO2C O

OBn 70%88:12 dr

OH

OBn

Answer

This is taken from the synthesis of callipeltoside A and is an example of the Roush asymmetric crotylation. The Roush reagents are more user friendly than the Brown crotylation reagents, being less moisture sensitive and can be stored for longer periods of time. On the downside they often give slightly lower enantioselectivities than the Brown reagents.

Initially it looks like the selectivity may be harder to rationalise than the Brown reagents …

Angew. Chem. Int. Ed. 2012, 51, 9366

O

OH3COH

OOH

O O

NHO

Cl

O

H

H3CO

3

HO

BH

H

O

O

O

OiPr

OOiPr

BO O

O

O

OiPr

OOiPr

HH

OBnH

BnO

disfavoured lone pair repulsion

vs.

anti-Felkin-Anhdisfavoured

OH

HH

H

BnO

≡

BnOH

H OH

H

attractiveinteraction

Felkin-Anh

… with the stereochemical information being further from the reactive centre.

It turns out that the key interaction is an electronic effect and not a steric effect.

The lone pair of electrons on the aldehyde have a disfavoured interaction with the lone pair of electrons on the lower ester group of the tartrate derivative. This means the aldehyde will prefer to be further from this lower ester.

There is also thought to be an attractive interaction … a formyl hydrogen bond as shown in the scheme.

4

OH

OBn

i. OsO4, NMOii. NaIO4

iii. Zn...............

72% (3 steps)85:15 dr

Br OH

OBn

OH

Question 2

This time an example of substrate control (taken from the same synthesis as Q1). Is the reaction under Felkin-Anh control or chelation control? Justify your answer with the appropriate sketch.

5

OH

OBn

i. OsO4, NMOii. NaIO4

iii. Zn...............

72% (3 steps)85:15 dr

Br OH

OBn

OH

Answer

The first step converts the alkene into an aldehyde. The OsO4 causes dihydroxylation of the alkene. The presence of NMO (N-methylmorpholine N-oxide) allows the use of a sub-stoichiometric quantity of osmium (which is expensive and very toxic).

The 1,2-diol is then oxidatively cleaved by treatment with NaIO4. Basically is a practically simpler version of ozonolysis.

6

H

O

O

ZnLn

OH

OBn

O

OLnZnH

H

OLnZnHH

R

H

H

R

H

Hdisfavoured

OLnZn

OO

LnZn O H

R

H

H

ZnLn

H

HR

or

≡

OH

R

OH

H

HOH

OH

HR≡

The second step involves formation of a propargylic zinc reagent. This is the nucleophile. The zinc also coordinates to both the alcohol and the aldehyde. The two possible half-chair conformations are shown on the right. The one with the large substituent in the psuedo-axial position is disfavoured.

Once we have the probably conformation of the chelated substrate the nucleophile can either add from the top or bottom (Si) face. The approach that gives the chair intermediate and not the twist-boat is favoured.

7

BzOO

i. c-hex2BCl, Et3N

ii. EtCHO

94%> 20:1 dr

i. TBSOTfii. xs CH2=CHCH2MgBr

iii. NaIO483%

O OTBS

i. .......

BF3•OEt2ii. HCl

82%10:1 dr

OTMS

OEt

O

OH

O

A C15H20O4

Question 3

Give the structure of intermediate A. Pay special attention to the diastereoselectivity.

Rationalise the observed high selectivity.Lets face it, I give

you the

diastereoselectivity

(as long as you

understand the

second set of

reactions). So this

question is all about

determining why we

observe this

stereochemistry

because this example

does not obey the

normal steric

arguments!

8

BzOO

BzOO

BCl

Cy Cy

HH

:NEt3 BzOO

BCy2

Answer

The first step is formation of the boron enolate. This can give either the E- or Z-boron enolate. It is believed that the bulky chlorodicyclohexyborane exclusively gives the E-enolate as shown.

The boron coordinates the carbonyl trans to the bulky stereocentre or cis to the ethyl group to minimise steric interactions. The cyclohexyl substituents then force the methyl group downwards as shown on the right. Deprotonation then gives the E-enolate.

O

R

BCy2Cl

H C

Hvs.

O

R

BCy2Cl

H C

H

9

OO

B

H

H

Et

L

L

HBzO

HOO

H

H

Et

HBzO

≡ BzOO

Et

OH

incorrect

minimisation of A1,3 strain

Directed delivery of

the aldehyde …

Initially, we may think that the stereocentre on the enolate substrate will control the diastereoselectivity as shown above. Here we determine the conformation of the enaolte by minimising A1,3 strain by having the smalles substituent parallel to the enolate double bond.

The aldehyde then approaches the enolate anti to the bulky benzoate ester as shown above. Unfortunately, this gives the wrong diastereoisomer.

Instead it is thought that there is a formyl hydrogen bond between the ester and aldehyde that directs the approach as shown on the right.

C O BL2CH H

O

O PhO

Et H

10

OO

B

H

H

Et

L

L

H O

OPh

OO

B

H

H

Et

L

L

HBzO

OHO

H

H

Et

H OBz

≡BzOO

Et

OH

If we assume hydrogen bonding directs the delivery of the aldehyde then the Zimmerman-Traxler transition state must look like the diagram on the right.

This predicts the correct diastereomer.

11

BzOO OTBS

MgBr

O OTBSO

MgBrPh

O

MgBr

OH OTBSHO

Ph

OHIOO

O O

O OTBSOI

HOHO OO

O OTBS

The next bit is not part of the question but is how the ketone required for the question 4 is formed.

First the Grignard reagent attacks the ketone and removes the ester ‘protecting group’ on the alcohol. This furnishes a diol. The idol can be cleaved with NaIO4 to gibe the necessary ketone.

12

BzOO

i. c-hex2BCl, Et3N

ii. EtCHO

94%> 20:1 dr

i. TBSOTfii. xs CH2=CHCH2MgBr

iii. NaIO483%

O OTBS

i. .......

BF3•OEt2ii. HCl

82%10:1 dr

OTMS

OEt

O

OH

O

A C15H20O4

Question 4

Using the curly arrow notation and appropriate Newman projections, explain how the ketone on the bottom row is transformed into the lactone.

13

O OTBS

OTMS

OEt

BF3•OEt2OTBSO

EtO

OH

R

H

O

≡≡

R

H

O

OTMS

OEt

R

H

HO

OEt

O

Answer

This is simply an example of substrate control so we need to use the Felkin-Anh model to predict/rationalise the diastereochemical outcome.

First draw the Newman projection. Then rotate the C–C bond so that the substituent of the closest stereocentre is perpendicular to the carbonyl group. The silyl ketene acetal then attacks along the Bürgi-Dunitz angle.

14

OTBSO

EtO

OHHCl

OHO

EtO

OHHCl

O

OH

O

Finally, if we treat this molecule with acid we perform two reactions. The first removes the silyl TBS protecting the group. The second is the acid promoted transesterification or lactonisation.

15

PhSOEt

O NCu

N

OO

Ph Ph

22SbF6 cat. 10%,

92%15:1 endo:exo

> 95% ee

SPh

OEtO SPh

O

OEt

endo exo

Question 5

Welcome to the world of asymmetric catalysis. Explain both the enantio- and diastereoselectivity of the Diels-Alder reaction shown above.

16

SPhO

EtOSPh

O

EtO

HOMO

LUMO

EtO

OSPh

HH

SPh

OEtO

Answer

This example comes from the synthesis of norbornenone.

The first task in this question is to explain the diastereoselectivity. In terms of the Diels-Alder reaction this is called endo/exo selectivity. In endo selectivity the dienophile approaches the diene in such a manner that the electron withdrawing group is under the diene. This allows the electron rich diene to interact with the electron poor carbonyl group. This interaction is called secondary orbital overlap. It is shown in three different representations on the right.

The endo product is the kinetically favoured product from the Diels-Alder reaction (but it is sterically disfavoured).

Chem Commun. 1998, 1985

17

PhS

PhS

HHSPh

OOEt

OEt

OO

OEt

The alternative transition state is known as the exo transition state as the dienophile is orientated away from the diene (there is no overlap between the carbonyl group and the diene).

This transition state is sterically less demanding but is only observed under very forcing conditions which allow the reversibility of the Diels-Alder reaction to be exploited.

Normally Diels-Alder reactions give the previous product, the endo compound.

The Diels-Alder reaction is very powerful and I recommend that you learn more about it. A very brief introduction is given in one of my courses:

http://www.massey.ac.nz/~gjrowlan/stereo2.html

Lecture 10

18

NCu

N

OO

Ph PhO

EtO

SPh

H

H

NCu

N

OO

Ph PhO

EtO

SPh

H

H

vs.

disfavoured favoured

O

EtO

SPh

H

H

SPh

OEtO

≡

Once we have determined that the reaction proceeds through the endo transition state we have to consider the influence of the chiral Lewis acid as this will control the enantioselectivity.

A copper(II) catalyst is most likely to be square planar. As a result of the sulfide substituent the substrate (the dienophile) is bidentate and forms a rigid chelate. The orientation of coordination is unimportant as the box ligand is C2 symmetric.

When the substrate coordinates to the copper ...

19

NCu

N

OO

Ph PhO

EtO

SPh

H

H

NCu

N

OO

Ph PhO

EtO

SPh

H

H

vs.

disfavoured favoured

O

EtO

SPh

H

H

SPh

OEtO

≡

... the phenyl substituent of the sulfide is forced away from the ligand as shown in the diagram. This minimises non-bonding interactions between the phenyl group and the oxazoline ring and/or the phenyl on the ligand.

The diene (in this case cyclopentadiene, a very common test substrate) then approaches anti to the phenyl of the sulfide. This avoids the interaction between the diene and the phenyl. In the drawings on the right this means it approaches from the bottom face and gives the molecule shown.

20

PMBOO

OBnN

RuH2N

PhPh

Ts

Cl HCO2H:Et3N

83%95% ee

PMBO

OBn

OHH

Question 6

Here we are going to go through the synthesis of (–)-α-kainic acid, a neuroexcitatory amio acid used in research into the CNS.

First, investigate the Noyori transfer hydrogenation and see if you can rationalise the stereoselectivity.

NH

CO2H

CO2H

21

N

TsN

PhH

HPhRuH

N

TsN

PhH

HPhRu

OH

H

O

H

N

TsN

PhH

HPhRu

OH

H

O

HN

TsN

PhH

HPhRu

HH

H

Answer

The following questions concern the synthesis of kainic acid.

This first step is an example of a Noyori transfer hydrogenation.

This reaction employs formic acid as a source of hydrogen as this is more practical than the use of hydrogen gas. The mechanism on the right shows how this occurs. Hydrogen bonding between acid and amine positions the formic acid next to the reactive centre. Reductive elimination gives a metal-hydride and carbon dioxide.

J. Org. Chem. 2013, 78, 3355

22

N

TsN

PhH

HPhRu

HH

H

O

R

R

N

TsN

PhH

HPhRu

HH

H

OR

R

HH

OR

R

≡

OHH

OBn

OPMB

vs.

One of the advantages of the Noyori diamine catalysts is that they do not require additional functionality within the substrate to coordinate the catalyst and substrate together. This can be achieved by a simple hydrogen bond between the carbonyl and the ammonium group.

There are two possible orientations for the substrate. Calculations (and the selectivity of the reaction) suggest that substrates with unsaturated functionality (alkynes, alkenes & aromatics) form an attractive interaction ...

23

N

TsN

PhH

HPhRu

HH

H

O

R

R

N

TsN

PhH

HPhRu

HH

H

OR

R

HH

OR

R

≡

OHH

OBn

OPMB

vs.

... between the π-system and a hydrogen atom of the aromatic ring of the catalyst.

This is a simplification of the mechanism, which undoubtedly involves the solvent molecules as well as the substrate and catalyst but it is a useful predictive model.

24

PMBO

OBn

OHHRed-Al®

86% PMBOOHH

OBn

Red-Al® = sodium bis(2-

methoxyethoxy)aluminiu

m hydride

Question 7

This to is a stereoselective reaction (although many people forget that alkenes are stereocentres as soon as they move onto asymmetric synthesis). How does reduction with Red-Al® result in formation of the E (trans) alkene exclusively?

As a clue I will just say that the alcohol is important and that the structure of Red-Al® is given on the right.

OAl

H

O HO

ONa

25

PMBO

R2

OHH

OAl

H

O HO

ONa

PMBO

R2

OH Al(OR)2

OAl

H

O HO

O

PMBOAlO

R2

OROR

H

PMBOOHH

OBn

H

H

H

The Red-Al serves two functions in this reaction; it activates the alkyne and acts as the hydride source. It is thought that the alcohol reacts with Red-Al first, the aluminium Lewis acid is then tethered to the substrate and this is essential for stereoselectivity. The electron deficient aluminium activates the electron rich alkyne. A second equivalent of Red-Al attacks the activated alkyne, delivering a hydride and forcing the alkyne to attack the aluminium to give a five-membered ring. The formation of the ring forces the hydride and the aluminium onto opposite faces of the alkene. The C–Al bond can be protonated (or halogenated) with retention of stereochemistry.

26

PMBOOHH

OBn

H

H

HO

O

DCC, DMAP

80% PMBOOBn

O

O

i. LiHMDS, TMSCl, –78°C to rtii. CH2N2

79%

O

O

OBnPMBO

DCC =

dicyclohexylcarbodiimide

DMAP =

dimethylaminopyridine

LiHMDS = lithium

hexamethyldisilazide

[lithium

bis(trimethylsilyl)amide]

Question 8

If you completed last weeks problems then this sequence should offer no challenge (if you haven’t looked at last weeks questions then you can either cheat and look at the answers (thus spoiling the enjoyment) or have a go now).

Reagents are given on the right.

NC

N

N

N

27

PMBOOBn

O

O

PMBOOBn

O OTMS

Answer

The first step is esterification and I will not go through the mechanism of this reaction (if you don’t know it or can’t work it out go through your undergraduate notes, it will in be in there somewhere).

The second step involves stereoselective enolate/silyl ketene acetal formation. It is essential that we control the geometry of the ‘enolate’ as this effects the relative stereochemistry of the two new stereocentres formed in the rearrangement.

Under standard conditions such as this esters invariably give the (OSi)-E-enolate. One simplistic interpretation of the Ireland model says this is due to the fact the ester substituent can rotate out of the way so is smaller than the substituents of the base and hence the enolate substituent will be better accommodated on the same face as the alkoxy substituent.

28

O

HHPMBOOTMS

HOBn

O

HHPMBOOTMS

HOBn

≡

H H

O

O

OBnPMBO

H

H

Once the silyl ketene acetal has formed the molecule is set-up for the Ireland-Claisen rearrangement. This proceeds through the standard Zimmerman-Traxler-like 6-membered transition state. The existing stereocentre controls the conformation of the transition state with the larger substituent adopting the pseudo-equatorial position. All other substituents positions are fixed as they are on alkenes.

The representations on the right show the reaction and the stereochemical outcome.

29

O

O

OBnPMBO

i. DIBALii. Ac2O,

DMAP, Et3Niii. DDQ

73% (3 steps)O

OBnHO

O

DDQ = 2,3-dichloro-5,6-

dicyano-1,4-

benzoquinone

DDQ is a mild oxidant

Simple functional group transformations. I have only included it so that you know how we get to each of the key intermediates (and so that you can practise your synthesis since that is the point of the course!).

O

ONC

NC Cl

Cl

30

O

O

OBnPMBO

i. DIBALii. Ac2O,

DMAP, Et3Niii. DDQ

73% (3 steps)O

OBnHO

O

DDQ = 2,3-dichloro-5,6-

dicyano-1,4-

benzoquinone

DDQ is a mild oxidant

DIBAL reduces the ester to an alcohol.

Acetic anhydride with dimethylaminopyridine as a catalyst and triethylamine as a base forms a new ester (an acetate).

The DDQ deprotects the PMB ether. It acts as an oxidant, accepting a hydride from the PMB group and effectively, oxidising the PMB ether to an aldehyde and releasing the free alcohol. This only works for the electron rich PMB group and not normal benzyl ethers.

O

ONC

NC Cl

Cl

31

AcO

OBnHO

Ti(iOPr)4, (–)-DET, t-BuOOH

82%100% de

AcO

OBnHO

O

Question 9

This is another example of the Sharpless Asymmetric Epoxidation. See if you can determine the product, obviously paying special attention to the chemoselectivity (or is it regioselectivity?) (and not looking at the next slide).

32

AcO

OBnHO

Ti(iOPr)4, (–)-DET, t-BuOOH

82%100% de

AcO

OBnHO

O

Answer

The chemoselectivity is easy to understand. For the Sharpless Asymmetric Epoxidation to proceed it must coordinate to the substrate. It does this through an alcohol group and as a result the SAE invariably occurs at allylic alcohols (although I realise there is an example in my notes that is not allylic). So the allylic alcohol reacts considerably faster than the isolated alkene.

33

OH

"O" D-(–)-DET unnatural isomer

R

You are not expected to draw the postulated transition state. All you need to say is that the Sharpless Asymmetric Epoxidation is predictable and that the (–)-enantiomer of the tartrate ligand would be expected to cause epoxidation from the top face.

The mnemonic that shows the selectivity of each ligand is given on the right. Remember, the alcohol goes in the bottom right corner and the alkene goes backwards into the plane of the page.

34

AcO

OBnHO

O

TBSCl, imidazole

94%AcO

OBnTBSO

O

i. K2CO3, MeOHii. DIAD, Ph3P, (PhO)2P(O)N3

45% (2 steps)N3

OBnTBSO

O

DIAD is the ispropyl

version of DEAD. Due

to its increased

steric bulk it is less

prone to side

reactions that result

in the formation of

hydrazides (attack at

the carbonyl group).

More functional group interconversions. See if you can work out a mechanism for step ii.

As a clue this is an example of the Mitsunobu reaction that was covered in the third year.

35

O N

ON O

OPh3P

OiPr N

ON OiPr

OPh3P

PO

N3PhOPhO

OiPr N

ON OiPr

OPh3P

P(O)(OPh)2

R OH

R O

N3

N3PPh3

R N3

Ph3P=O

The order of events is protection of the primary alcohol with the silyl group.

The acetate is then removed by base-mediated hydrolysis.

The primary alcohol is substituted with an azide possibly by the mechanism given on this page.

First the triphenylphosphine is activated by reaction with DIAD. This creates an anion that can displace the azide anion form DPPA. The alcohol is then activated by reaction with the cationic phosphine. Finally, SN2 displacement gives the product.

36

N3

OBnTBSO

O

i. TBAFii. PPh3, H2O

iii. Boc2O, Et3N

61% (3 steps)N H

Ot-BuO

OH

OBn

OH

Question 10

This is a straight forward, SN2-like, substitution reaction. Suggest a reason why the reaction gives the pyrrolidine (5-membered) ring preferentially over formation of the piperidine ring (6-membered).

Make sure you are happy that this is the only diastereomeric product.

37

R NN

N:PPh3

R NNN

Ph3PR N

N NPPh3

R NPPh3

HO

H

N2

R NH

PPh3

OH

R NH

PPh3

OH

R NH2

P3P=O

Answers

The use of triphenylphosphine and water to reduce an azide to an amine is a classic reaction. For those of you that are interested the mechanism is given above. As is frequently the case the driving force for this reaction is the formation of triphenylphosphine oxide and, of course, nitrogen.

38

HNHBnO

H HOH

H

O

HNHBnO

H HOH

H

HO

The transition state of the cyclisation can be approximated as the diagram above. This is based on a chair-like transition state with two of the three substituents in the pseudo-equatorial position. It is impossible to get all of them equatorial; this is one of the reasons the synthesis of kainic acid is so challenging.

In the cyclisation the amine has to approach the C–O σ* antibonding orbital of the epoxide. This is 180° to the C–O bond. Hopefully the drawing above shows that this should be possible for the formation of the pyrrolidine (5-membered ring) but would be more challenging for the 6-ring.

This transition state also explains the diastereoselectivity.

39

N HOt-BuO

OH

OBn

OH

i. NaIO4ii. NaBH4

iii. Li•naphthalene

71%N H

Ot-BuO

OH

OH

i. Jones reagentii. TFA

62%NH

CO2HH

CO2H

No more stereochemistry to worry about. These are just the ‘end-game’ transformations to get to the final product.

The diol is cleaved with the periodate and the resulting aldehyde is reduced to an alcohol with sodium borohydride.

The lithium naphthalenide is a reductant (source of electrons) and is used to cleave the benzyl ether.

The two primary alcohols are then oxidised to carboxylic acids with Jones reagent. Surprisingly, this does not cleave the Boc group so deprotection is achieved with TFA (trifluoroacetic acid).

40