Asymmetric synthesis II

reagent controlled allyl/crotylation

reagent controlled reduction

catalyst controlled reduction

catalyst controlled oxidation

O

HTMS

OMOM

B[(-)-ipc]2then K2CO3,

MeOH

74% (2 steps)> 95% de> 90% ee

OHH

H OMOM

MOM = methoxymethyl

(-)-ipc =

(-)-isopinocampheyl

Question 1

Explain both the diastereoselectivity and enantioselectivity of the allylation reaction shown above.

This example is taken from a synthesis of palmerolide A.

O

Angew. Chem. Int Ed. 2007, 46, 5896 & Eur. J. Org. Chem. 2011, 463

OMOM

B

H

R2B

H

≡S

R2B ML

small Smedium M

large L

Answer

When considering the enantioselectivity, the first thing we need to do is determine the conformation of the reagent.

Initially this looks quite challenging (and it is) but we can make some simplifications. Instead of considering the whole pinene unit we condense this moiety to a single stereocentre attached to the boron atom. There are three other substituents based on size; the hydrogen atom is small; the alkyl chain is medium; the branched alkyl chain is large.

X

BX

axax

S LM

R

minimise interactions

≡X

R

X

L

S

M

view

If we assume the reaction will proceed through the standard chair-like transition state then the axial pinene group is the most important as it will suffer the greatest interactions with the substrate and reagents (1,3-diaxial interactions). To minimise 1,3-diaxial interactions the smallest substituent (the hydrogen) will be orientated over the ring of the transition state. The conformation will be staggered so that we minimise torsional strain with the substituents on the boron atom.

Having determined the conformation of the first pinene we can look at the second ...

X

BX

S LM

M L

S

eclipsed

favoured

versus

X

BX

S

L

LM

SM

syn-pentane

disfavoured

The conformation of the second pinene unit should minimise the torsional and non-bonding (steric) strain. No conformation is free of strain so it is a case of minimising the interactions. The left hand structure suffers torsional strain with the aldehyde and allyl unit. The right hand structure has steric strain through something called the syn-pentane interaction.

The latter has been determined (calculations) to be greater.

The equatorial pinene will control the enantioselectivity (the axial pinene controls the orientation of this second pinene).

HS

BL

S LM

M L

S ≡O

BH

R

S

S

LM

LM

H

favoured

BOR

H

H

H

S LM

S

LM

interactiondisfavoured

vs.

Finally, the last variable is which face will the aldehyde approach from, top or bottom in the figure above?

The aldehyde is considered smaller than the allyl unit as there are no substituents on the oxygen compared to the hydrogen atoms on the methylene position of the allyl group. As a result the aldehyde will be on the same face (syn) to the largest substituent on the equatorial pinene.

What does this look like?

O

BH

R

H HH

H

BO

H HH

H

H

R

disfavoured

The transition state on the left is disfavoured as the methyl group (large substituent) and allyl group interact.

OBH

H HH

H

TMS

OMOM

H

OH

H

TMS

OMOM

H

The diastereoselectivity is determined by the geometry of the allyl unit. In the chair-like transition state we can only control which face the aldehyde approaches and its orientation (substituent prefers to be pseudo-equatorial). The position of the allyl substituent is fixed; if we have a Z-alkene then it must be axial (as above) if it is E then it will be equatorial.

The only taxing thing left is unravelling the chair but I hope it is obvious that the two hydrogen atoms are on the same face.

O

C5H11B

(R)-Alpine-borane®

C5H11

H OH

Question 2

This is another example of the use of a chiral reagent in synthesis (massoilactone). Try and determine the rational for the enantioselectivity.

Tetrahedron 1984, 40, 1371

BH

H

HB

Alpine borane® is prepared by the reaction of pinene with 9-BBN (a bulky organoborane). If you remember, hydroboration occurs with a concerted addition to the least sterically demanding face of a double bond (in this case anti to the dimethyl bridge) and with the boron at the least substituted end of the double bond (for both steric and electronic reasons).

If you can’t remember this please revise!

H

HB

R

RB

H

HO

C5H11

minimise 1,3-diaxial

interactions

O

C5H11

HHO

C5H11

The reaction occurs through an elegant mechanism in which the aldehyde coordinates to the boron. Lewis acid activation of the aldehyde increases its electrophilicity. The coordination to the boron also activates the hydride by altering the hybridisation of the boron centre (sp2 to sp3) thus lengthening the bonds and making them weaker. This promotes hydride transfer.

The bicyclic nature of the pinene subunit restricts the conformation of the organoboron species and the reaction must ...

H

HB

R

RB

H

HO

C5H11

minimise 1,3-diaxial

interactions

O

C5H11

HHO

C5H11

... occur through a boat-like transition state. The facial selectivity is determined by the orientation of the aldehyde and the minimisation of 1,3-diaxial interactions. The alkyne is smaller than an alkyl group is as it restricted to being a linear cylinder of carbon instead of a rotating chain of atoms (and it has no hydrogen atoms).

The transition state can be depicted in an alternative manner ...

HOBC5H11

RR

HOHC5H11

≡

C5H11

H OH

... this shows the same salient features with the methyl substituent of the reagent controlling the approach of the aldehyde.

Personally, I have find this drawing easier.

O

OCO2Et

N BO

HPh Ph

0.25eq

BH3•THF 0.65eq

94%93% ee

OCO2Et

HHO

Question 3

This is our first example of asymmetric catalysis. This is the use of the CBS reductant in a synthesis of forskolin. With the basic principles we have already covered you should be able to devise a reasonable rational for the enantioselectivity.

Tetrahedron Lett. 1989, 30, 7297 & Tetrahedron Lett. 1988, 29, 6409

RL RS

H OH

NB O

HPhPh

H3B

RL RS

O

NB O

HPhPh

BH3•THF

Ph

Ph

OB NB

HOH

H

RL

RSPh

Ph

OB NB

HOH

H

RL

RS

The mechanism for the reaction is quite elegant. The catalyst is derived from proline (so easily accessed). The coordination of the stoichiometric reductant, BH3, to the nitrogen lone pair has to effects; firstly, it activates the reductant converting borane into a borohydride anion. Secondly, the positive charge on the nitrogen atom increases the polarity of the N–B bond and thus makes the ring boron (endocyclic) more Lewis acidic. This allows better activation of the ketone.

RL RS

H OH

NB O

HPhPh

H3B

RL RS

O

NB O

HPhPh

BH3•THF

Ph

Ph

OB NB

HOH

H

RL

RSPh

Ph

OB NB

HOH

H

RL

RS

Coordination of the ketone increases the electrophilicity of the carbonyl group and tethers the ketone to the reductant. As is so often the case we now have an arrangement of 6 atoms and so the favoured transition state is thought to resemble a chair-like conformation. The enantioselectivity arises from the minimisation of 1,3-diaxial interactions with the small substituent of the ketone being pseudo-equatorial. For the real system this looks like ...

Ph

Ph

OB NB

HOH

H

EtO2CO

HHPh

Ph

OB NB

HOH

H

HH

EtO2CO

disfavoured

... here are the two chair-like transition states for the CBS reduction of the cyclic ketone. The one on the right is disfavoured due to 1,3-diaxial interactions between the two methyl substituents.

TBSO OTBS

OH

(–)-DIPT, Ti(OiPr)4, t-BuO2H

88%13:1 dr

TBSO OTBS

OHO

Question 4

This is an example of the archetypal asymmetric catalytic reaction, the Sharpless Asymmetric Epoxidation (SAE). It is taken from a synthesis of psymberin. Can we rationalise the diastereoselectivity?

Org. Lett. 2008, 10, 5625

Sometimes we have to accept that it is very hard to predict a reactions outcome by simple conformational analysis (or even in-depth computational calculation). Often the intermediates/species involved in a reaction simply are not known and without such basic information speculation seems pointless.

When this occurs with a useful transformation, such as the SAE, then empirical rules are often drawn up so that we can still use the reaction. With the SAE there is a useful mnemonic that frequently predicts the stereochemical outcome. It is acceptable to just quote this ...

OH

"O" D-(–)-DET unnatural isomer

"O" L-(+)-DET natural isomer

OTBSH

... with this mnemonic the alcohol is placed in the lower righthand corner and the allylic alkene is placed back into the plane of the paper. The (–)-enantiomer of the tartrate reagent delivers oxygen to the top face (as required in this example) while the (+)-enantiomer delivers the oxygen form the bottom face.

There is believed to be an experimental basis for this mnemonic. Studies on the active species have determined that there are 8 possible titanium species present with ...

E

O O

O

TiO

OO

O

O

TiO

O

CO2iPr

iPrO2C

iPr

iPr

OiPrt-Bu

E

O O

O

TiO

OO

O

O

TiO

O

CO2iPr

iPrO2C

iPr

iPr

OiPrt-Bu

R

OTBSH

H

H

R

OTBSH

... the dimer above the most likely candidate to be responsible for enantio- (or diastereo-) selectivity.

It is though that the tartrate ligands create a pocket for TBHP activation and that the allylic alcohol must coordinate to the same titanium centre (and this is why the SAE only works reliably for the epoxidation of allylic alcohols). Internal delivery of the oxygen then leads to the observed product.

Cl

OSnBu3

Cl

BF3•OEt2

> 20:1 drCl

OH

Cl

Cl

O

Cl

OH

ClCl

B

Question 5

Can you rationalise the change in diastereoselectivity between these two allylation reactions with the same chiral substrate? Both examples are taken from studies on a class of molecule known as the chlorosulfolipids.

J. Org. Chem. 2014, 79, 2226

HCl

O

H H

ClO

H SnBu3

Cl

H

ClHO

H

Cl

≡

Cl

OH

Cl

F3B F3B

Answer

Basically this is just another way of asking you to justify the stereochemical outcome of each reaction and see why they are different.

The first task is to determine which face of the aldehyde the reagents would prefer to react with. If we start with the allyl stannane (top reaction). This is simply an example of Felkin-Anh addition as can be demonstrated with the Newman projection.

If we draw the conformation with the large isopropyl group perpendicular to the carbonyl we get ...

HCl

O

H H

ClO

H SnBu3

Cl

H

ClHO

H

Cl

≡

Cl

OH

Cl

F3B F3B

... the correct stereochemistry for the alcohol.

Note: I have not drawn the stereochemistry for the second chloride as we need to determine that in a separate step.Warning

Arguably, this reaction

could have proceeded

through a different

transition state with

the chloride

perpendicular to the

carbonyl. This would

agree with the Polar

Felkin-Anh model which

states that

electronegative atoms

tend to be

perpendicular to the pi

system to maximise

orbital overlap.

O

RH

HCl

SnBu3

OH

RH

HClOH

RHHCl

≡ ≡

OH

RH

H

Cl

≡

R

OH

Cl

F3B

Now we have to determine the diastereoselectivity of the second chloride substituent. The reaction of allyl stannanes proceed through what is know as an open transition state (non-cyclic, no chair). Initially, it was proposed that an anti transition state (as shown above with the allyl group anti to the carbonyl) was responsible for the selectivity. The chloride substituent was orientated away from the aldehyde. This is quite attractive for a number of reasons (not least it is easy to draw). But, this being chemistry, it is probably not always (ever?) the case. Calculations have shown that the allyl group is normally gauche to the carbonyl (this is favoured by orbital overlap apparently) as shown on the next slide ...

O

RHH

Cl

Bu3SnOH

RHH

Cl ≡

OH

RH

H

Cl

≡

R

OH

Cl

F3B

... in this transition state it is argued that the allyl group is anti to the aldehyde substituent (minimising interactions) with the chloride bisecting the O–C–R area. This minimises interactions between the chloride and the R and H groups of the aldehyde. It should be remembered that nucleophiles approach along the Bürgi-Dunitz angle and so we want the smallest group orientated between R & H (see the structure on the right).

Felkin-Anh and this open transition state explain the observed product stereochemistry.

OR

H BF3HCl

SnBu3

minimise interactions

BO

HH

ClCl

H

Felkin-Anh (Re)disfavoured

syn-pentanedipole-dipole

vs. OB

Cl

HH

Cl

Hfavoured

OHCl

HH

Cl

H

≡OHCl

HH

Cl

H

≡

Cl

OH

Cl

The allyl borane reagent reacts through a closed (cyclic) transition state and so we need to draw a chair-like arrangement of atoms. The first representation shows the reaction proceeding through attack predicted by the Felkin-Anh model. This does not occur due to the syn-pentane interaction (sterics) or the destabilising alignment of the C–Cl dipoles (electronics). Instead the reaction must proceed through anti-Felkin-Anh attack on the aldehyde ...

BO

HH

ClCl

H

Felkin-Anh (Re)disfavoured

syn-pentanedipole-dipole

vs. OB

Cl

HH

Cl

Hfavoured

OHCl

HH

Cl

H

≡OHCl

HH

Cl

H

≡

Cl

OH

Cl

... in this arrangement we minimise the syn-pentane interaction (C–Cl eclipses C–H). Once again the hardest part is unravelling the 3D representation into a standard skeletal representation. I have tried to make it easier by highlighting in bold all the atoms in the plane of the paper. Thus you should see that two hydrogens and a chloride are on the top face.

Note: the reaction could proceed through the polar-Felkin-Anh transition state with the chloride perpendicular to the carbonyl but this is unlikely as the isopropyl group would eclipse the other chloride.