Asymmetric synthesis: problems 1

Ph

OH

Br

NaOH

Ph

OH

Br

NaOH

AC12H14O(chiral)

BC12H14O

(achiral; ir = 1715 cm–1)

Question 1

Products A & B are structural isomers. Examine the conformations of the two diastereomeric starting materials and use these to determine the structures of A & B.

Ph

H

O

H

Br

H

H OH

Ph

H

O

H

Br

H

O

Ph

H

Answer

The key to this question is the conformation of the cyclohexane and the necessity for functionality to be trans-diaxial or antiperiplanar for reactions to occur. This structural requirement is the result of orbital overlap and the need for electrons to flow into the C–Br σ* antibonding orbital.

The product is chiral as it has no plane of symmetry (it has no improper rotation axis).

Ph

H

O

H

Br

H

H OH

Ph

H

O

H

Br

H

O

Ph

H

The first reaction involves the formation of an epoxide (oxirane). Deprotonation of the acidic hydroxyl group forms an alkoxide that participates in a substitution reaction with backside attack.

It is fortunate (?) that the phenyl substituent encourages the ring to sit in a conformation that places the hydroxyl group and the bromide axial but it is not essential. Ring-flipping occurs in most systems and as it is the only reactive conformation the energy required to adopt this conformation will comprise some of the activation energy.

Ph

H

H

O

Br

H Ph

H

H

OH

Br

H Ph

H

OH Ph

H

OOH

In the second example it is impossible for the hydroxyl group to be antiperiplanar to the bromide. So epoxide formation is impossible.

In one conformation of the molecule a hydrogen atom is antiperiplanar to the bromide. Thus we can get an E2 elimination (whether elimination occurs with the alcohol or the alkoxide is debatable and, ultimately, unimportant). Elimination gives an enol (or enolate) the tautomeric form of a carbonyl (ir stretch) (or its resonance form).

The molecule is achiral as it has a plane of symmetry running through the middle.

Question 2

With the aid of a Newman projection determine the structure of the product. It is formed as a single diastereoisomer. Note: under the acidic conditions of the reaction the PMB protecting group is lost.

TBDPSO H

O

OPMB

TMSBr TiCl4

85%C24H33BrO3Si

TBDPSO H

O

OPMB

TMSBr TiCl4

85% TBDPSOOH

OHH

Br

Answer

This reaction is taken from a synthesis of amphidinol 3.

It is basically a chance for you to practice manipulating molecules between skeletal and Newman projections.

An extension of this question would have involved the addition of a substituent to the alkene so that you had to deal with the formation of two new stereocentres. How would you approach that problem? (and for once a 6-membered ring would not help you as this reaction proceeds through an open transition state).

TBDPS = is tert-butyldiphenylsiyl (a bulky protecting group) and PMB = para-methoxybenzyl (a substituted benzyl group that is far easier to remove than the standard benzyl group; either acid or an oxidising agent such as DDQ (dichlorodicyanobenzoquinone)).

Org. Biomol. Chem. 2012, 10, 9418

R

HPMBO

RH

O

OPMB

≡O

C

H

PMBO

RH

O

C

H

ClnTi

TMS

Br

TMS

Br

R

H

PMBOC

H

OH

Br

≡

R

HPMBOC

HHO

Br≡ R

H OPMB

H OH

Br

First, convert the skeletal representation into a Newman projection.

This is an example of Cram Chelation control so the two Lewis basic atoms are tethered together and this fixes the conformation of the substrate.

The nucleophile then approaches along the Bürgi-Dunitz angle. One approach is hindered by the R group. The other isn’t.

Finally, convert back to skeletal representation remembering that you haven’t changed one stereocentre.

TBDPSO H

O

OPMB

TMSBr TiCl4

85% TBDPSOOH

OHH

Br

Obviously I’m focusing on the stereochemical outcome of this reaction. I do not have time to discuss the simple chemistry/reactivity as that should be covered in the undergraduate courses.

If you do not know about the chemistry of allylsilanes then I recommend you read up on these valuable nucleophiles. To make life easier I can tell you this is sometimes called the Hosomi-Sakurai reaction.

I have some brief notes on these reagents in “Strategy in Synthesis” lecture 3, a second year paper:

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

OH

C6H13

O

H Ph

O SmI2 (15 mol%)

94%> 99:1 dr

O

C6H13

H OH

O

Ph

Question 3

With the aid of the appropriate drawings, rationalise the stereochemistry of this transformation.

OH

C6H13

OH Ph

OI2Sm

O

C6H13

OPh

HOSmIn

O

C6H13

H OH

O

Ph

In this reaction samarium(II) iodide is behaving as a Lewis acid but it should be remembered that it is a very valuable reagent for single electron transfer (SET).

It is also fun to note that the student on this paper is now a very successful academic in his own right and we might look at some of his work later (and if we don’t you should read it anyways).

Answer

This is an example of what is now known as the Evans-Tishcenko reaction (but the reference below is by Evans and only refers to the reaction as the Tishcenko reaction).

Here is the basic mechanism (badly drawn). The samarium acts as a Lewis acid and activates the aldehyde allowing the alcohol to add. The samarium then activates the ketone (not shown) and mediates the internal hydride transfer.

J. Am. Chem. Soc. 1990, 112, 6447

OHHOO SmII

C6H13 OHHOOH

C6H13PhPh ≡

C6H13

HHO H O

O

Ph

To understand the diastereoselectivity of the reaction we need to look at the probable conformation. The drawing above shows the samarium tethering the ketone to the hemi-acetal-like oxygen. As you can see the hydride is 5-atoms away from the carbonyl (like a 1,5-hydride shift) so we can use a 6-membered transition state and so we can model this with chair conformation. The two oxygen atoms adopt the axial position to allow the substituents to be in the pseudo-equatorial position.

OHHOO SmII

C6H13 OHHOOH

C6H13PhPh ≡

C6H13

HHO H O

O

Ph

The facial selectivity (Si in this case) of the hydride approach is controlled by the existing stereocentre. Approach from the top (Re) face is only possible if the isopropyl group adopts the pseudo-axial position. This is, of course, disfavoured. So approach is from the bottom face (away from us or Si) as this places the isopropyl group pseudo-equatorial.

The tetrahedral intermediate collapses to reform the carbonyl group and reduce the ketone in an analogous fashion to the Cannizzaro reaction or the Meerwein-Ponndorf-Verley oxidation.

O

i. LDAii. TBSCliii. heat

49% > 90% ee

OH

OO

O

i. LDAii. TBSCliii. heat OH

OO

➎➊ ➋ ➌ ➍

➎

➊ ➋ ➌ ➍

Question 4

With the aid of the appropriate drawings explain why the two diastereomers of the starting material both undergo rearrangement to give the same enantiomer of product.

O

i. LDAii. TBSCliii. heat

49% > 90% ee

OH

OO

O

i. LDAii. TBSCliii. heat OH

OO

➎➊ ➋ ➌ ➍

➎

➊ ➋ ➌ ➍

Answer

Hopefully these reactions are not too hard. They are an example of the Ireland-Claisen rearrangement. With a bit of luck my numbering might have given you a clue as to the nature of the reaction.

In steps i & ii a silyl ketene acetal is formed (silyl enol ether). Warming the reaction then promotes the rearrangement (often warming means returning the reaction to rt).

J. Org. Chem. 1993, 58, 4589

O

OTBS

O

HHOTBS

O

HHOTBS

OH

O

The Ireland-Claisen rearrangement is an example of pericyclic (sigmatropic) reaction. It involves 6 atoms and the movement of 6 electrons (3 curly arrows). As such you should instantly be thinking about using a chair transition state to model the reaction.

The key control element is the methyl substituent. This can either be pseudo-equatorial or pseudo-axial with each different conformation resulting in a different enantiomer.If you ever spot two double bonds whose ends are 6 atoms apart think about a rearrangement.

O

OTBS

O

HHOTBS

O

HHOTBS

OH

O

Once we have the chair conformation we draw the reaction, rearranging the two double bonds.

This gives us the product drawn in a 3D manner. Yet again, the easiest place to make an error is taking this representation and converting it to the normal skeletal depiction.As before, if you are ever in doubt as to whether you have drawn the right stereochemistry or have inverted the stereocentre simply assign the stereochemical descriptor to both drawings (in this case the product is S).

O

OTBS O

HOTBS

H

O

HOTBS

HOH

O

If we apply the same principles to the second diastereomer we find we form the same enantiomer.

Again we place the methyl substituent of the the stereocentre in the pseudo-equatorial position. The methyl substituent of the alkene must be axial - we have no choice as the alkene cannot be rotated.

The rearrangement generates the acid and we have to unwrap the drawing.

O

N O

O

Ph

F

i. LiHMDSii. BrCH2CN

80%O

N O

O

Ph

F

NC

Question 5

Rationalise the diastereoselectivity of the alkylation reaction shown above.

O

N O

O

Ph

F

H NLiO

N

O

Ph O

Si

Si

HF

vs.

H NLiO

N

O

Ph O

Si

SiH

FO

N O

O

Ph

F Li

H

H H

Answer

This is your introduction to the use of chiral auxiliaries (unless you have read my notes in which case it is revision) and this example is taken from a synthesis of PNP405 a drug used to prevent transplant rejection.

To rationalise the diastereochemical outcome of this reaction we first need to determine the geometry of the enolate formed (E or Z).

The Ireland model suggest that this occurs through a 6-membered ring. The two competing factors are repulsion between the aryl substituent and the oxazolidinone auxiliary or 1,3-diaxial repulsion between the base and the aryl substituent. With imides such as this the auxiliary is considered larger than the trimethylsilyl group and so the top conformation is disfavoured and we get the Z-enolate.

J. Org. Chem. 2002, 67, 6612

O

N O

O

Ph

F Li

NC

Br

H

O

N O

O

Ph

F

HNC

Once we know the geometry of the enolate the rest is easy.

The lithium cation coordinates to the oxygen of the enolate and the carbonyl of the oxazolidinone auxiliary. The chelate prevents rotation of the C–N bond and so fixes the conformation of the auxiliary.

The phenyl group blocks the bottom face and so the electrophile must approach from the top face giving us the product shown.

ON

OO i. Bu2BOTf, EtNiPr2

ii. aldehydeO

i. KOHii. CH2N2

iii. (EtCO)2CO, Et3N

OCH3

OO

OLDA,

TMSCl

OCH3

OO

OTMSi. heatii. HCl

CC13H21NO4

DC11H18O4

Question 6

Using the appropriate drawings determine the structure of C & D. It goes without saying that you should pay close attention to the stereochemistry.

ON

OO

BBu

OTf

BuON

OO

H

BBu Bu

iPr2EtN ON

OOB

Bu Bu

Answer

This was taken from a 3rd year exam paper I set when I was teaching in the UK. I can’t find the reference to the original paper (naughty me).

The first step is formation of the boron enolate. Remember that a tertiary amine is not a strong enough base to directly deprotonate an imide; coordination of boron and carbonyl activates the α-position. Once again the Z-enolate is favoured due to the bulk of the auxiliary/amide.

ON

OB

O

H Bu Bu

vs. ON

OOB

O

H Bu Bu

O

The reaction follows a different pathway to the alkylation in the previous question. The aldehyde must coordinate with boron. Without this Lewis acid activation the boron enolate is insufficiently nucleophilic to attack the weakly electrophilic aldehyde.

When the aldehyde coordinates to the boron it prevents the boron from interacting with the auxiliary (boron is already coordinatively saturated) yet we still get a favoured conformation for the auxiliary (as demonstrated by high diastereoselectivities).

ON

OB

O

H Bu Bu

vs. ON

OOB

O

H Bu Bu

Ofavoured disfavoured

The conformation on the right is disfavoured due to dipole-dipole interactions. The alignment of dipoles is unstable as the build-up of negative charges close to one-another repel (like aligning to magnets in the same direction). Alternatively you could argue that the lone pairs of electrons on each oxygen atom are in close proximity, which is disfavoured. The conformation on the left is favoured as the dipole-dipole interaction is minimised (as close to cancelling out as possible) or the lone pairs have maximum separation.

BBu

N

O BO

O

H

Bu

Bu

OH

O O

HNH

Bu

O

Ovs.

OH O

HNH

O

O≡N

OOH

OO

Once again we have 6 atoms reacting with the movement of 6 electrons. The stereochemistry can be rationalised with a chair-like transition state. This is known as the Zimmerman-Traxler transition state.

We have two choices; the orientation of the aldehyde and the facial selectivity on the enolate.

The aldehyde will adopt the position that places its substituent in the pseudo-equatorial position.

The facial selectivity is the result of the minimisation of 1,3-diaxial-like interactions.

BBu

N

O BO

O

H

Bu

Bu

OH

O O

HNH

Bu

O

Ovs.

OH O

HNH

O

O≡N

OOH

OO

If the aldehyde approaches from the top face of the enolate (Si face) (top left picture) then the isopropyl substituent of the auxiliary will clash with the boron substituents.

Attack from the lower (Re) face of the enolate avoids such destabilising interactions.

Once again, the hardest part is probably unravelling the 3D representation and drawing our skeletal representation.

My only tip is to say it should be obvious that the two hydrogen atoms are on the same face (down).

ON

OO i. Bu2BOTf, EtNiPr2

ii. aldehydeO

i. KOHii. CH2N2

iii. (EtCO)2CO, Et3N

OCH3

OO

OLDA,

TMSCl

OCH3

OO

OTMSi. heatii. HCl

CC13H21NO4

DC11H18O4

The product of the aldol reaction is C.

The next three steps are simple functional group transformations. Potassium hydroxide hydrolyses the imide to give a carboxylic acid. Mild esterification with diazomethane (CH2N2) is followed by a second esterification, this time of the secondary alcohol.

Formation of the silyl ketene acetal proceeds to give the E-enolate.Please note: the nomenclature for the enolates of esters is annoying. Simply swapping the silyl group for a lithium changes the name to Z-enolate without changing the shape.

NH

OLiH

OR

vs.N

HOLi

OR

H

disfavoured

Why is the E-enolate favoured (the enolate with the methyl cis to the alkoxy substituent)?

Again we can use the Ireland model of deprotonation to determine this. In this example the repulsion between base and methyl group is key. The ester substituent can rotate out of the way (unlike in the amide/imide example earlier). Of course, this is a simplification, lithium compounds tend to form aggregates which complicate analysis.More information can be found at an old version of my notes (lecture 4):

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

OCH3

OO

OTMS

OH

TMSO H O

OCH3

OH

TMSO H O

OCH3≡HO

O

OCH3

OH

The final step is another example of the Ireland-Claisen rearrangement. This is a useful reaction as it permits the readily formed C–O bond (esterification) to be converted to a C–C bond with communication of stereochemical information.

The reaction proceeds through a chair-like transition state with the largest substituent at the stereocentre (not the alkene) adopting the pseudo-equatorial position and controlling the diastereoselectivity.

The hardest task is just drawing this without inverting any stereocentres.