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Chiral Auxiliaries

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Total cards: 24•Content version: v4•Public deck•Updated: 8/1/2026

Card preview (24)

#1
Front (Question)

What is a chiral auxiliary

Back (Answer)

• A prochiral substrate attached to a chiral non-racemic group • Chiral auxiliary is the chiral non-racemic group

#2
Front (Question)

How are chiral auxiliaries used

Back (Answer)

• A reaction is conducted which results in diastereomeric products which may be readily separated • Cleavage of the auxiliary from the purified reaction mixture yields the chiral, non racemic products

#3
Front (Question)

What are the requirements of a good chiral auxiliary

Back (Answer)

• Enantiomerically pure and available in both enantiomers • Cheap and available in quantity • Easy to introduce into the substrate • Gives high and predictable diastereocontrol • Easy to purify the major diastereomer • Easy to remove the chiral auxiliary from product.

#4
Front (Question)

What are used in diastereoselective enolate alkylations

Back (Answer)

• Generally ise amides as substrates- they give complete control over enolate geometry • Oxazolidinone auxiliaries of Evans are widely used for asymmetric alkylation

#5
Front (Question)

Describe how diastereoselective enolate alkylations work -Evans

Back (Answer)

• First chiral auxiliary is attached to ketone- forms imide • LDA (or NaHDMS) , THF, -78 degrees forms z-enolate • Z-enolate formed due to avoidance of steric clashing and chelation with Li (Na) to ketones of oxazolidinone and ketone • Then electrophile added and one face is blocked by R-group of oxazolidinone

#6
Front (Question)

What are the three oxazolidinone

Back (Answer)

• Valine derived • Ephedrine-derived • Phenylalanine-derived

#7
Front (Question)

What do oxazolidinone enolates react readily with

Back (Answer)

• Variety of reactive electrophiles • MeI • BnBr • AllylBr • NBS • trysl azide • oxaziridines • Azodicarboxylates

#8
Front (Question)

What are the different auxiliary cleavage methods

Back (Answer)

• LiBH4- produces alcohol form of substrate separated from chiral auxiliary • LiOOH, H2O produces carboxylic acid form of substrate • MeONHMe.HCl, Me3Al- forms Weinreb amide- some of chiral auxiliary still attached

#9
Front (Question)

What is Weinreb amide useful for

Back (Answer)

• Useful as treatment with bulky reducing agents • Produce aldehyde

#10
Front (Question)

How is an oxazolidinone produced

Back (Answer)

• From amino acid • Add NaBH4/H2SO4 (used to make BH3 in reaction) and reduces acid to amino-alcohol • Add C1 carbonyl equivalent- ethylchloroformate + K2CO3 • Amino group reacts with carbonyl displacing the chloride • under action of base hydroxyl group undergoes second attack to displace the ethoxy group- forms ring system

#11
Front (Question)

How can you get an aldehyde from a Weinreb amide

Back (Answer)

• Add DIBAL-H (bulky-reducing agent) • Al Coordinates with both O- stabilises transition state - 5 Membered • H is added to C=O • Reaction is worked up with H+/H2O • Excess DIBAL-H is destroyed to form aluminium hydroxide species- fast reaction • Aluminium in 5-membered transition state is replaced by H giving hemiaminal species- slow • Hemiaminal species is unstable so generates aldehyde and parent amine • Aldehyde is never exposed to reducing agent as is destroyed quick

#12
Front (Question)

Why is it hard to reach aldehyde OS from ester

Back (Answer)

• If add reducing agent to ester can’t stop at aldehyde • As aldehyde is more reactive than the ester- as soon as formed is reduced further to alcohol

#13
Front (Question)

What is the aldol reaction important for

Back (Answer)

• Exceptionally important in organic synthesis • Particularly for synthesis of biologically active polypropionate natural product

#14
Front (Question)

Describe Evans aldol reaction

Back (Answer)

• Very effective stereocontrol • Forms z-enolate exclusively due to Boron coordination of both C=O • Very high syn selectivity with addition of OH to R • 6-membered Zimmerman-Traxler TS • Aldehyde R group goes equatorial • Aldehyde approaches away from large alkyl group on auxiliary • The dipoles of the carbonyl of the auxiliary and aldehyde are opposed

#15
Front (Question)

Describe the diastereoselective enolate alkylation Myers method

Back (Answer)

• Based upon pseudoephedrine amides - Cheap and available in both enantiomeric forms • Amide enolates much more reactive than imide enolates (have a higher pKa) and hence unactivated alkyl halides may be used

#16
Front (Question)

Describe process of Diastereoselecive enolate alkylation Myers method

Back (Answer)

• Add acid chloride to pseudoephedrine forms pseudoephedrine amide • Add LDA, LiCL, R’X • Solvent forms OLi with Alcohol and C=O which blocks one face of the enolate • Less hindered approach used - highly diastereoselective

#17
Front (Question)

What can be produced from the product of Myers enolate alkylation

Back (Answer)

• Carboxylic acid • Ketone • Aldehyde • Alcohol

#18
Front (Question)

How is a carboxylic acid produced from the product of Myers enolate alkylation

Back (Answer)

• H2SO4, dixane reflux • Or Bu4NOH, tBuOH, water reflux

#19
Front (Question)

How is an aldehyde produced from the product of Myers enolate alkylation

Back (Answer)

• LiAlH(OEt)3, 0 degrees

#20
Front (Question)

How is an alcohol produced from the product of Myers enolate alkylation

Back (Answer)

• LiH2N

#21
Front (Question)

How is a ketone produced from the product of Myers enolate alkylation

Back (Answer)

• MeLi, 0 Degrees

#22
Front (Question)

What is used to carry out the asymmetric synthesis of alpha-amino acids

Back (Answer)

• Schollkopf’s auxiliary • Diastereo enolate alkylation reaction

#23
Front (Question)

Describe how Schollkopf’s auxiliary is produced

Back (Answer)

• Take amino acid valine - has sterically demanding isopropyl unit • Protect NH2 using Boc2O, NaHCO3 and Base THF, H2O • Acid group of valine couples with amino group of Glycine methyl ester- • Then heat to 180 degrees which removes boc groups freeing NH2 group allowing cyclisation • Add Me3O+ BF4- - electrophilic alkylating agent - both C=O change to OMe

#24
Front (Question)

Describe Schollkopf’s auxilliary reaction to produce an alpha amino acid

Back (Answer)

• Add strong base e.g. BuLi which deprotonates Schollkopf’s auxiliary to produce anion • Add BnBr which approaches from the less hindered face- opposite to large iPr group • Add H3O+ (mild acid) to hydrolyse the two N=C bonds which results in cleavage of heterocycle leaving 2 amino acid units- one is valine- recycle other is new one with Bn side chain