Asymmetric synthesis

Cyclic substrate control Axial attack and chair conformation

Acyclic substrate control Cram chelation control Felkin-Anh control

Acyclic substrate control Allylic A1,3 strain

H

H

OMe2CuLi

96%99:1 dr

H

H

O

Question

Rationalise the stereochemical outcome of the conjugate addition shown above.

HH O

H3C

CH3

HH O

H

H

H

HH OH

H3C

H

H

Answer

The cis junction of a 5,5-fused bicyclic molecule results in a ‘folded’ molecule. There is an inside (endo or concave) face and an outside (exo or convex) face of the molecule. Non-bonding interactions hinder the approach of the nucleophile to the alkene from the inside (Re) face. Additionally, the product of attack from the inside is less stable (but if the reaction is under kinetic control this is unimportant).

You need to start to think about the 3D shape of molecules. It plays an important role in their reactivity.

OMe2CuLi

97%98:2 dr

O

Question 2

Rationalise the stereochemical outcome of the conjugate addition shown above.

O O

CH3CH3

O

CH3

H

≡ HH

CH3O

Answer

The simple rationalisation is that the nucleophile attacks from the opposite face of the enone to the methyl group. This gives the major product. As you can see one substituent is equatorial (good) and one, by necessity, is axial (not favoured).

As we shall see this is not the entire story.

OMe2CuLi

87%92:8 dr

O

Question 3

Rationalise the stereochemistry in this conjugate addition.

O O

CH3

O

CH3CH3

HX

The wrong answer

If we were to apply the same rational as before, the nucleophile attacks anti to the existing substituent we find that we get the wrong product. Why?

OH

O

H

pseudo-axial pseudo-equatorialAnswer First, we need to consider the conformation of the cyclohexenone. It is not a chair, it is closer to the half-chair to the right (but even this is an exaggeration).

As normal, ring-flipping, which has a lower energy barrier in the half-chair compared to a chair, means the substituent can either be pseudo-axial or pseudo-equatorial. The pseudo-equatorial is preferred.

You need to get into the habit of drawing the various conformations of molecules. Practice the chair, the boat, the half-chair and Newman projections.

H

O

HH3C

O

H

H

CH3

H ≡O

H3C

H ≡O

CH3disfavoured

If the nucleophile attacks from the lower (Re) face then a high energy twist-boat conformation is formed. This is disfavoured.

The nucleophile must approach the alkene in an axial manner to maximise overlap with the p orbitals of the alkene. So it approaches either vertically from above or vertically from below. As it attacks the trigonal planar carbon the carbon becomes tetrahedral by moving towards the nucleophile and forming a vertical bond (in the transition state). Approach from below gives the twist-boat.

We initially form an axial bond. This is not good but we have no choice. The nucleophile (HOMO) must overlap with the electrophile (LUMO).

A knowledge of frontier molecular orbitals is very useful.

O

H

H

H3C

H3C

H H

O≡

O

CH3favoured

So what does our new understanding mean to the second question?

Approach from the top face (the same face as the substituent) may initially look disfavoured but if we consider the conformation of the molecule we see that this leads to a chair conformation. This is energetically favourable.

HO

H

O

disfavoured

CH3

HH

H

≡ ≡O

CH3

H3C

H O

H3C

X

Question 2 again We shall assume that the more stable conformation has the methyl substituent in the pseudo-equatorial position. Addition from above (Si) leads to the twist-boat and is disfavoured.

OO

H

H

O

H

H

HH

CH3O≡

O

CH3favouredH3C

In this example the simple ‘intuitive’ explanation is the same as the more complex answer. But that is purely coincidental. Get used to drawing the conformations.

Do you know what Re and Si (as well as re and si) are? If not you have some reading to do ...

Attack from the bottom (Re) face gives a chair and is favoured.

O

NH2

i. TMSOTf, Et3Nii. I2

84%

I

HNO

Question 4

Guess what, I want you to explain the stereochemistry of this reaction.

O

NH2

OSiMe3S

F3CO

O

O

N

SiMe3

H

HNEt3

O

NH

SiMe3

NH

O

II

ISiMe3

II

HNO

Answer

This is from the synthesis of Tamiflu®. Above is the mechanism of the reaction. The TMSOTf activates the amide, making it a better nucleophile during the cyclisation reaction. It is then treated with iodine. The iodine can add to either face of the molecule to reversibly form an iodonium ion. The amide then attacks in an SN2-like reaction (backside attack). But what about the stereochemistry?

J. Am. Chem. Soc. 2006, 128, 6310

HOTMS

NH

H

OHNSiMe3

HOTMS

NHI

HOTMS

NH

I

no reaction

no reaction

H

OHNSiMe3

I

no reaction

H

OHNSiMe3

I

NH

HI

O≡

I

HNO

The starting material can exist as two different conformations. As normal the one with the equatorial substituent will be more stable. But this is not the one that reacts. The amide cannot approach the C–I σ* anti-bonding orbital (180° to the bond) so there is no reaction.

The iodine can add reversibly to either face of the alkene. Only the iodonium species that is anti to the amide will react to give the product.

HOTMS

NH

H

OHNSiMe3

HOTMS

NHI

HOTMS

NH

I

no reaction

no reaction

H

OHNSiMe3

I

no reaction

H

OHNSiMe3

I

NH

HI

O≡

I

HNO

Iodonium ring opening (or any small ring such as an epoxide) will always occur to give the axial product initially. This is called trans-diaxial ring opening.

The reason is yet again to do with orbital overlap. The nucleophile cannot approach the σ* antibonding orbital if it is equatorial. The nucleophile would have to be in the ring to be 180° behind the leaving group.

Normally, the ring flips to give the more stable equatorial conformation. In this case it cannot because of the bridge.

TBDPSO SOO

tolDIBAL-H-78°C

99%TBDPSO S

OOH

tol

Question 5

Explain the stereoselectivity exhibited by this reduction.

Of course, you remember the difference between stereoselective and stereospecific? No? Then look it up!

AlH

SO

tolO

R

HSO

tolOH

R

disfavoured

≡ TBDPSO SOOH

tol

Note: I often draw the products in different orientations to the products. Practice manipulating them so that you are confident they are the same each time.

See, drawing the chair conformation of cyclohexane is important ...

Wrong answer

This is taken from a synthesis of amphidinol 3.

The reaction shown on this page forms the wrong diasteroisomer. It shows that we have internal delivery of the hydride as the DIBAL coordinates with the sulfoxide. This both activates and positions the reductant.

This conformation is disfavoured as it places the sulfoxide substituent in the pseudo-axial position and this is unstable due to 1,3-diaxial interactions.

Org. Biomol. Chem. 2012, 10, 9418

H Al OS tolR

O

H OS tolR ≡ TBDPSO S

OOH

tol

OH

Correct answer

The favoured conformation of the substrate has the tolyl group pseudo-equatorial. The hydride is then delivered intramolecularly. Note the hydride is more or less attacking the carbonyl along the Bürgi-Dunitz angle as we would expect.

There is a second conformation the substrate could adopt that has an equatorial tolyl group shown on the right. This is disfavoured as it has the long R group axial and hence 1,3-diaxial interactions with DIBAL.

H Al OS tolO

R

disfavoured

tolSO OH OO

i. CSA, acetoneii. Et2BOMe, NaBH4

80% tolSO OH OH

Question 6

Guess what? Explain why we get the syn diol.

tolSO OH OO

step i.

tolSO OH O

O O

Answer

This is taken from later in the same synthesis of ampidinol 3.

First some simple chemistry to unmask the ketone. I had to leave this step in as the authors did not report the yield for each separate step. Hopefully you could all draw a curly arrow mechanism for this reaction, it is simply acetal hydrolysis or more accurately, transacetalisation (but that doesn’t look like a real word).

Org. Biomol. Chem. 2012, 10, 9418

OOB

HS(O)tol

R

Et

Et

OOB

Et

EtH

S(O)tol

R

H

H

BOO

HH

R

Et

Et S(O)tol

BO

O R

H

H

Et

Et

S(O)tol

tolSOOHHHO

tolSOOHHHO

This is an example of external delivery of the reductant. The boron Lewis acid ties the ketone and alcohol together. This provides a 6-membered ring with a double bond present so we can model the conformation as a half-chair. There are two conformations of the half-chair, one has a pseudo-axial sulfoxide and the other a pseudo-equatorial group. The latter is, of course, preferred. Addition then occurs from the top (Re) face or the bottom (Si) face. The latter results in the disfavoured twist-boat conformation while the former furnishes a chair and the major product.

H

O

OBn

SnCl4

89%> 30:1 dr OBn

HO H

Question 7

Explain the stereoselectivity in this example of a carbonyl-ene reaction.

H

O

OBn

H3C

HBnO

O

C

H

≡

CH3H

BnOO

C

H

ClnSn

H

CH3OOH

CH

Bn

H

CH3

O

HOC

HBn ≡

HO H

OBnH

A

B

Answer

This is taken from a synthesis of (+)-peloruside A.

In the carbonyl-ene reaction the alkene acts as the nucleophile (it is a pericyclic reaction).

On a simplistic level we can justify the stereochemical outcome through a simple Newman projection. The first step is draw the Newman projection without altering the stereocentre. I have achieved this by keeping the methyl group of the stereocentre eclipsing the aldehyde (A).

This is not the reactive conformation of the molecule. In this example the tin Lewis acid coordinates the carbonyl and ether thus locking the conformation of the molecule as shown in (B).

The nucleophile will attack the carbonyl along the Bürgi-Dunitz angle. There are two approaches ...

Org. Lett. 2012, 14, 178

H

O

OBn

H3C

HBnO

O

C

H

≡

CH3H

BnOO

C

H

ClnSn

H

CH3OOH

CH

Bn

H

CH3

O

HOC

HBn ≡

HO H

OBnH

A

B

... the red approach is disfavoured due to non-bonding interactions with the methyl group. The green approach (Si face of aldehyde) is favoured as it just passes a small hydrogen atom.

The next problem is once we have decided which face is attacked. How do we get back to a classic skeletal representation? My advice is to rotate the Newman projection until the carbon chain is in one plane (luckily in this example the nucleophile is in the same plane as the methyl group). This representation shows us the relationship of the new stereocentre to the old (unchanged) stereocentre. Here we see the alcohol and ether are on the same face. As we haven’t changed the initial stereocentre we know that the alcohol must be upwards.

This is an example of Cram chelation control.

O H

H

O

H

SnCl Cl

Bn

O HH

O

H

SnCl Cl

Bnvs. O H

H

O

H

Bn

An alternative representation is ...

This shows the pericyclic nature of the addition (and the 1,5-hydrogen shift). The tin Lewis acid coordinates the two oxygen atoms of the substrate. The reaction can be depicted as proceeding through a chair-like transition state with the alkene either approaching from the same face as the methyl (disfavoured) or anti to the methyl substituent (favoured).

This is still Cram Chelation control just depicted in a different fashion.

If you are ever worried that you have inverted a stereocentre when manipulating your drawings remember that you can check by simply assigning the correct stereochemical descriptor (R or S) and ensuring that it is constant in all your drawings.

TBSO O

HSnBu3

BF3•OEt2

81%> 95:5 dr TBSO H OH

Question 8

Rationalise the stereochemical outcome of another nucleophilic addition to a chiral acyclic aldehyde.

H

O

C

H

≡TBSO O

HTBSO BF3

H

TBSO O

C

H

BF3

TBSO H OH

Bu3Sn SnBu3

H

TBSO

OH

CH

HH

OTBS

HOC

H

≡

Answer

This is from one of many syntheses of the epothilones.

This not an example of Cram Chelation control but the Felkin-Anh model (there are many different models for the addition of nucleophiles to carbonyls. This is one of the most general as long as you remember to consider both steric and electronic factors).

J. Org. Chem. 2008, 73, 9675

H

O

C

H

≡TBSO O

HTBSO BF3

H

TBSO O

C

H

BF3

TBSO H OH

Bu3Sn SnBu3

H

TBSO

OH

CH

HH

OTBS

HOC

H

≡

First draw a Newman projection of the substrate. This is shown in the top line of the diagram above. Next rotate the stereocentre until the largest substituent is perpendicular to the carbonyl group. There are two such conformations; the one above and the one to the right.

H OTBSO

C

H

BF3

H

O

C

H

≡TBSO O

HTBSO BF3

H

TBSO O

C

H

BF3

TBSO H OH

Bu3Sn SnBu3

H

TBSO

OH

CH

HH

OTBS

HOC

H

≡

Remember that the nucleophile approaches along the Bürgi-Dunitz angle. In the conformation to the right this is disfavoured by non-bonding interactions with either the largest substituent or the methyl group. In the conformation above the nucleophile can approach close to the hydrogen substituent. This is favoured.

H OTBSO

C

H

BF3

H

O

C

H

≡TBSO O

HTBSO BF3

H

TBSO O

C

H

BF3

TBSO H OH

Bu3Sn SnBu3

H

TBSO

OH

CH

HH

OTBS

HOC

H

≡

Once we have determined which face of the aldehyde is attacked all that remains is to convert the Newman projection back to a skeletal representation. H OTBSO

C

H

BF3

SiMe2Ph

C6H13

OTBS

H23C11 i. 9-BBNii. H2O2, NaOH

68%> 95:5 dr

SiMe2Ph

C6H13

OTBSC11H23

OH

Question 9

Explain the diastereoselectivity of this hydroboration/oxidation sequence.

BH

9-BBN = R2BH =

C11H23

H HR2

Si

H

Ph

≡ C11H23

H HH

R2

SiMePhC11H23

H HH

R2

SiMePh

HBRR

R2B H

HC11H23

H HSiMePh

R2R2

SiMe2Ph

C11H23

R2BH2O2, NaOH

R2

SiMe2Ph

C11H23

OH

A

Answer

An old example taken from a synthesis of tetrahydrolipstatin.

The selectivity can be explained by inspecting the various conformations of the substrate. The ground state conformation probably resembles A. This minimises allylic or A1,3 strain; the strain between C11H23 and the allylic stereocentre.

Tetrahedron Lett. 1990, 31, 3645

C11H23

H HR2

Si

H

Ph

≡ C11H23

H HH

R2

SiMePhC11H23

H HH

R2

SiMePh

HBRR

R2B H

HC11H23

H HSiMePh

R2R2

SiMe2Ph

C11H23

R2BH2O2, NaOH

R2

SiMe2Ph

C11H23

OH

A

The organoborane approaches the alkene anti to the bulky silyl group. Arguably this would be at 90° to the alkene to maximise orbital overlap but calculations by Houk have shown that the transition state is close to the staggered product so the approach is directly opposite to the C–Si bond. The regiochemistry of the hydroboration can be explained by considering the electronic influence of the silyl group. These can stabilise a β cation so the γ carbon is more nucleophilic (and this aligns with the electrophilic boron.

C11H23

H HR2

Si

H

Ph

≡ C11H23

H HH

R2

SiMePhC11H23

H HH

R2

SiMePh

HBRR

R2B H

HC11H23

H HSiMePh

R2R2

SiMe2Ph

C11H23

R2BH2O2, NaOH

R2

SiMe2Ph

C11H23

OH

A

Remember your undergraduate chemistry? Oxidation of the organoborane occurs with retention of stereochemistry.

The concept of A1,3 strain is very important. Read about it.