Asymmetric Synthesis Flashcards
How can we tell if a molecule is chiral or not?
A chiral molecule does not have a plane or centre of symmetry when drawn in any conformation.
This definition allows us to deal with molecules containing sulfur, phosphorous and carbon.
How can sulfoxides and phosphines be chiral molecules?
Why are chiral amines rare?
A lone pair on S or P can also be a substitutent and this means that sulfoxides and phosphines can be chiral molecules as well.
For amines, nitrogen inversion occurs too easily so chiral amines are rare.
What are enantiomers?
Enantiomers are mirror images - all the stereogenic centres are inverted.
Two enantiomers have the same properties except in the presence of other chiral molecules, or in their interaction with plane polarised light.
What are diastereoisomers?
It is a stereoisomer that is not an enantiomer. Not all of the stereogenic centres are different.
Diastereoisomers are completely different compounds and have different properties.
What is asymmetric synthesis?
A reaction that selectively creates one configuration of one or more new stereogenic centres by a chiral auxiliary or chiral catalyst on a substrate - the transition states are made diastereoisomeric.
Since disastereoisomers are different, the activtion energy to access each diastereoisomerc TS will be different. This gives the preferential formation.
Why is nucleophilic addition to a C=O group of an aldehyde/ketone a common method in asymmetric synthesis?
Give some examples of typical nucleophiles.
Because it can form a new stereogenic centre at the C of the C=O.
Typical nucleophiles:
- reducing agents = source of H-
- grignard and organolithium reagents = source of R-
- enolates
What is the Burgi-Dunitz angle?
The trajectory of attack of a nucleophile to a C=O is close to the 109o bond angle that results in the product (tetrahedral bond angle).
The angle is approx. 107o and is known as the Burgi-Dunitz angle.
Why is the Burgi-Dunitz angle 107o?
Because this gives optimal overlap of the nucleophile HOMO with the C=O LUMO i.e. Π*C=O is at 107o.
If a chiral ketone has an alpha-stereogenic centre to the carbonyl, how does this change addition of Nu to the C=O?
In principle, addition of a nucleophile to the C=O can ocur from the front and back faces. Either additions gives a different product.
Due to there being a stereogenic centre in the starting material that is unchanged in the product, the two products will be diastereoisomers of each other (one stereogenic centre the same, one different).
Depending on the structure of the group alpha to the C=O, and the nature of Nu, the Nu addition can be highly diastereoselective.
What two things lead to diastereoselectivity in diastereoselective additions to C=O?
- Conformational fix - the starting material has a conformational preference
- Steric hindrance - once the conformation of the ketone is fixed, then steric factors control which face of the C=O is preferentially attacked.
What is a Cram chelation control reaction?
The addition of a grignard reagent to a ketone with an alpha-stereogenic centre.
In a Cram chelation control reaction, how is the conformation of the chiral ketone fixed?
- the ketone fixed - the ketone contains a stereogenic alpha to the C=O group which has a heteroatom attached
- the nucleophilic reagent - the grignard contains an Mg2+ ion
Due to both of the above, a chelation effect can occur to fix the conformation of the starting ketone.
Chelation from Mg2+ occurs to one of the C=O lone pairs and the heteroatom alpha to the C=O.
In a Cram chelation control reaction, how is the direction of attack controlled once the conformation is fixed?
Once conformation is fixed, we can consider the steric hindrance involved on attacking each face of the C=O on the Burgi-Dunitz trajectory (107o).
Attack over the sterically smaller group is favoured and means that the activation energy for accessing this TS is lower than attacking the other face. This will form the major product.
What are the requirements for a Cram chelation control to occur?
- need a heteroatom at the carbon alpha to the C=O group which can donate a lone pair, O and S with alkyl groups are best
- need a metal cation that is a good chelating metal, most commonly Mg2+, Zn2+, sometimes Ti4+
What are the two types of Felkin-Anh model?
- nucleophilic addition of C=O with no heteroatom at the alpha stereogenic centre, with a chelating metal
- nucleophilic addition of C=O with a heteroatom at the alpha stereogenic centre, with no chelating metal
What is the conformational fix for Felkin-Anh (no heteroatom)?
The lowest energy conformations will place the largest of the three substituents at the alpha-stereogenic centre perpendicular to the C=O, as this is the least sterically hindered position - usually Ph.
In a Felkin-Anh model (no heteroatom), what happens once the conformation is fixed?
We consider steric hindrance on attacking the C=O on the Burgi-Dunitz trajectory opposite to the large perpendicular group.
Attacking over the sterically smaller group is favoured and the activation energy for this TS will be lower. This will form the major product.
Describe how to figure out the ‘size’ of groups in the Felkin-Anh model.
The most important aspect is the steric hindrance caused by the first atom in a group and not the overall size of the group.
Aka a more branched carbon will be bigger.
What is the conformational fix for Felkin-Anh (with heteroatom/no chelating metal) and then how does attack of C=O occur?
Conformational fix - place the most electronegative atom at the alpha stereogenic centre perpendicular to the C=O.
Steric hindrance - attack occurs on the Burgi-Dunotz trajectory over the sterically smaller group. This leads the favoured TS with lower activation energy.
How can the TS’s for ketone reduction be made diastereoisomeric?
Change the reducing agent - need a chiral, enantiomerically pure verson of NaBH4 - CBS catalyst.
Describe the mechanism of asymmetric ketone reduction using the CBS catalyst.
The reaction requires 1.0 molar equivalent of the reducing agent, BH3, as each BH3 transfers just one hydride to each ketone molecule - BH3 is the stoichiometric reducing agent.
Typically 10 mol% of the chiral catalyst is used.
What is the difference between NaBH4 (BH4-) and BH3 in asymmetric ketone reduction?
- BH4- is negatively charged and can deliver a hydride to C=O quickly
- BH3 is neutral (Lewis acid) and needs to complex to a lone pair on a heteroatom (Lewis acid-Lewis base complex) before it will deliver a hydride to C=O
- BH3 will only reduce the ketone very slowly, but the amine of CBS donates its lone pair to BH3 to give a very reactive borohydride-like species that carries out the reduction
Draw the mechanism for asymmetric reduction of a ketone with CBS.
Draw the transition state model for the (S)-CBS transition state shown below, and therefore for (R)-CBS.
What is the CBS mnemonic?
What can the alcohol product from asymmetric ketone reduction be used for?
- first activated as a leaving group by conversion into e.g. tosylate group
- SN2 nucleophilic substitution reactions can then be used to introduce new groups at the stereogenic centre
In a racemic alpha-alkylation of enolates reaction, why is a 50:50 mixture of enantiomers formed?
Draw a transition state to help explain the preferred formation of the major diastereoisomer of the molecule below.
Chelation of Li to enolate and oxazolidonone provies the conformational fix. The iPr group blocks the bottom face = steric factors, causing attack of the top face to be favoured to give the favoured product.
How is the chiral oxazolidonone auxiliary synthesised?
What are the different ways of removing the chiral auxiliary in enolate alpha-alkylation?
What is the predictive mnemonic for asymmetric enolate alpha-alkylation?
Describe a racemic aldol reaction using LDA.
Alpha-deprotonation of an amide using LDA followed by reaction with an aldehyde as the electrophile is an aldol reaction. It forms a bond between the C-2 and C-3 carbons in the product, which is a 1,3-hydroxy amide.
Aldol reactions with lithium enolates give bad diastereoselectivity - actually gives all four diastereoisomers (2 of both syn/anti).
Syn:anti ratio not necessarily 50:50 as the transition states have different energies.
Describe a racemic aldol reaction using boron enolate.
The boron enolate version is much better as it only forms the syn aldol products. The starting material, aldehyde and product is the ‘same’ but the reagents are different.
In a racemic boron enolate aldol reaction, why is a 50:50 racemic mixture of enantiomers formed?
How can the chiral auxiliary be removed in a syn-diol asymmetric reaction?
Reaction conditions for producing the products below are similar to alpha-alkylation. The hydroxyl group in the syn-aldol product can cause a problem for deprotection reactions as the OH group can be deprotonated - sometimes better to protect it as a TBDMS either before removing the chiral auxiliary.
Describe the boron enolate geometry in an asymmetric syn-aldol reaction and how it controls diastereoselectivity.
Key thing is that the R group alpha to the C=O will be cis to the OB(Bu)2 group in the boron enolate, to avoid steric hindrance.
In an asymmetric syn-aldol reaction, does chelation cause the conformational fix?
The oxygen lone pair on the C=O of the oxazolidonone can chelate tothe boron which has an empty p-orbital. This gives an unreactive boron enolate. This means that, here, chelation does not provide the conformational fix.
To react with the aldehyde, the oxygen lone pair of the aldehyde must coordinate to the boron.
In an asymmetric syn-aldol reaction, if chelation doesn’t cause the conformational fix then what does?
Dipoles.
The lowest energy conformation of the non-chelated boron enolate has to C-O bond of the enolate and the C=O bond of the oxazolidonone opposite to each other. This minimises the dipole repulsion. It is achieved by bond rotation.
Therefore, the conformational fix comes from the dipoles pointing in opposite directions.
Draw the transition state for the molecule shown below containing (S)-oxazolidonone, and therefore draw the transition state for (R)-oxazolidonone.
What is the predictive mnemonic for asymmetric syn-aldol reactions?
Draw the mechanism for asymmetric anti-aldol reactions.
The reactions proceeds via an iminium ion and an enamine.
Draw the TS for the molecule below containing (S)-proline, and therefore for (R)-proline.
What is the predictive mnemonic for aysymmetric anti-aldol reactions?
What transformations can the anti-aldol product undergo?
The aldehyde functionality can undergo a range of further reactions. Often the free hydroxyl group will be protected first.
- reduction to primary alcohol via NaBH4
- nucleophilic addition to aldehyde via Grignard reagent
Describe a racemic dihydroxylation of an alkene reaction.
Why is it racemic?
It is a stereospecific reaction where the two OH groups are delivered to the same face of the alkene - syn delivery.
Describe the catalytic asymmetric dihydroxylation reaction (AD reaction).
Describe the mechanism of asymmetric dihydroxylation reactions.
What is the predictive mnemonic for asymmetric dihydroxylation of alkenes?
What reactions can asymmetric dihydroxylation products undergo?
Terminal diols can be converted into terminal epoxides then into amino alcohols.
