CH3E9 Asymmetric Catalysis Practice Problems 2023-2024 PDF

CH3E9 Course. M Wills component on asymmetric catalysis practice problems. And answers… 1) Given the following reaction: Ph OH H Ph H ADmix- contains a dimer of quinine '(DHQ)2PHAL' (also a small amount of osmium salt + stoichiometric K3Fe(CN)6 ADmix- OH (DHQ) OMe N OH N Dihydroquinine (DHQ) Propo...

CH3E9 Course. M Wills component on asymmetric catalysis practice problems. And answers… 1) Given the following reaction: Ph OH H Ph H ADmix- contains a dimer of quinine '(DHQ)2PHAL' (also a small amount of osmium salt + stoichiometric K3Fe(CN)6 ADmix- OH (DHQ) OMe N OH N Dihydroquinine (DHQ) Propose a synthesis of: MeO MeO O from H H , Br , Br O OH and other common reagents Answer: H H MeO Mg Br MeO H O H H H H H MeO H MeO O MgBr MeO OH2 OH H dihydroxylation Br MeO MeO MeO OH H NaH O H O H OH OH OH Assumption here is that secondary (least hindered) alcohol reacts with benzyl bromide. 2) Predict the product of hydrogenation of of A using i) catalyst B and ii) catalyst C. OH A Ph2 P Ru P Ph2 P(Cy)3 N Ir C B Answer: Reduces only the bond adjacent to the OH, as a directing effect is needed. Ph2 P OH Ru P Ph2 B Reduces all the double bonds. OH P(Cy)3 N Ir C 3) Explain, with the aid of an appropriate diagram, how the racemic substrate A is converted into an enantiomerically pure product B in the reaction shown below. O Asymmetric hydrogenation catalyst Solvent, H2 O OH OEt Cl racemic Answer: O OEt Cl single enantiomer single diastereoisomer O OH O OH OEt Cl Cl Cl Asymmetric hydrogenation catalyst Solvent, H2 O OEt OEt reduced qucikly O O reduced slowly O OEt Cl single enantiomer single diastereoisomer The answer is that the substrate is rapidly isomerisation through the enol form. The catalyst then reacts with a single enantiomer of substrate to give a single enantiomer of product in diastereoisomerically-pure form – which means that the whole molecule is enantiomerically pure. Eventually all of the substrate ends up as the product because it is drawn over to the product side as the correct enantiomer forms. 4) Given the reaction above (in q 3), propose a synthesis of 1 from 2 (max 3 steps). Answer: 5) Ligand 3 forms a complex with Rh(I) which acts as a catalyst for asymmetric hydrogenation of alkenes containing a nearby co-ordinating group. i) Draw the complex which will form between 1 and Rh(I). What is the charge on it? See below. H P P P Rh P Rh H ii) Draw the hydride complex formed when the complex from part 1 reacts with hydrogen gas. See above ii) Draw the two diastereoisomeric complexes which will be formed when the acylamino acrylate substrate binds to the Rh(I).1 complex. Hint – the NCO is the co-ordinating group. H N H Ph P Rh O H R Me H N H Me P P Rh H R Ph P O iii) Which of the complexes leads to the product? Hint – think about which face the H is on. It is the left hand complex, because that has the Rh-H on the back face with the substrate in the correct orientation to give the observed product. The Hs are transferred to the back face. iv) Given the result shown above, predict the major product which will form upon reduction of the following substrates using a Rh(I).1 as the catalyst (co-ordinating group indicated by *): H N Ph CO2Me H N Me O * Me Ph Ph H H N CO2Me O * Me H Ph O Ph Me Ph Ph H H N Ph * O * CO2Me Me Ph Ph H Ph O * Ph O * OH * O CO2Me Me O Ph Ph H OH * 6) Draw a mechanism for the reaction shown below and explain how the chiral phosphoric acid is able to direct the reduction (don’t try to work out why it directs to the face shown, just broadly how it exerts an effect). Why is the ketone not reduced in the same reaction? 10 mol% Catalyst 2 O O P OH O Ar OH N Ar EtO2C N Me This forms: H Me H P O O Ar H CO2Et N Et MeO OMe The homochiral counterion shields one face and hydride adds to other side. OH O O N H Et 1.4 eq. Hantzsch ester Ar OH Et NH CO2Et Et MeO EtO2C O NH enol -> ketone H Non-aromatic product therefore ketone form is favoured. O NH H OMe 7) Draw the intermediate which forms in this reaction, and illustrate the addition step to diethylazodicarboxylate (DEAD). Nb sorry I forgot to put in the sodium borohydride in the question. OMe 8) Given the following transformation: 10 mol% O O + Ph H N H Ph O H OH Propose a synthesis of: PhS O O O SPh O H from , O HS O and other common reagents (you can use any reagent more than once) (although not asked for specifically above, make sure you know the mechanisms) Answer: , 9) Given the following transformation: 10 mol% O Ph2P AcO N H NH Pd H2N Design a synthesis of: NH AcO from OMe Answer: MeO , OMe NH2 , Grubbs metathesis catalyst, and other common reagents NH2 P N Pd MeO OMe NH NH2 AcO MeO OMe catalyst above MeO Be aware of the mechanisms of the reactions including the Grubbs metathesis catalyst and the Pd catalysed substitution. LnRu MeO OMe NH NH LnRu LnRu OMe OMe MeO product OMe NH OMe MeO NH LnRu NH MeO OMe Grubbs metathesis catalyst