Lecture5: 123.312


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Finishing oxidation by looking at the Baeyer-Villiger reaction and then turning our attention to reduction. Once again we will see the usual suspects with a who is who of hydride sources.

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Lecture5: 123.312

  1. 1. FUNCTIONALGROUP 123.312 INTERCONVERSIONS functional group CHAPTER7 interconversions H H O H H Cr O H H R O S OH CHAPTER seven R O reduction previously we had looked at oxidations now we turn our attention to E the opposite, reduction 1 2 3 need to be able to control OH oxidation now adding hydrogen or Text H chemoselectivity N removing oxygen (or other HO R H Clheteroatom) R R N O C O HO salmefamol OMe O O O R OH R R1/H R OH H2N NH2 O O O Cl Cl O Cl Cl vs vs R NH2 R2 R R R OR1 Cl Cl R1 H R1 R2 R1 O R1O OR1 O R Cl R R R NH2 lets approach this by example & look at the synthesis of this Reduction anti-asthma drug 4 5 6 starting material contains many hydrogenolysis & reductive hydrogenolysis & reductive reducible groups amination amination OMe OMe O OH O OH O O O OH H2, Pd / C, O OH H2, Pd / C, O OH MeO H+, ketone MeO H+, ketone MeO NaBH4 MeO N MeO N MeO HO HO HN HN N N Ph Ph HO Ph Ph HO HO HO Ph Ph Ph Ph O O hydrogenation reductively OMe OMe it also reduces imine formed sodium borohydride only by condensation of amine & reacts with the ketone & cleaves benzyl groups (as seen earlier) but... ketone in a process called not the ester reductive amination 7 8 9
  2. 2. ester reduction look at various reagents many different forms of reduction in organic synthesis OMe OMe O OH OH OH LiAlH4 MeO HN HN HO HO finally lithium aluminium hydride reduces the ester ©status frustration@flickr ©golbog@flickr 10 11 12 Reduction of Aldehydes/ketones mechanism of borohydride mechanism of borohydride reduction reduction H Na H Et H Na H Et Et H O O Et H O O NaBH4 O H B H O H B H O or LiAlH4 H OH H R1 R2 H R1 R2 solvent, reductant & cation I doubt mechanism is are all important concerted but all these R1 R2 R1 R2 steps must occur H H Et H OH Et H OH O B H O B H relatively easy reduction H R1 R2 H R1 R2 13 14 15 by-product is also a (less) powerful reductant sodium borohydride was developed during the war but H not reported until 1953 H Et Et B O B H Et H H Et is not a reducing agent, add electron donating it is a base (eg sodium groups & get more powerful hydride) reductant (superhydride LiBH4) ©alifaan@flickr 16 17 18
  3. 3. Reduction of esters NaBH4 R1 O O R2 LiAlH4 R1 H H OH H O R2 LiAlH4 O O O O O O O O R1 H R1 R2 R1 Cl R1 OR2 R1 NR2 R1 H R1 R2 R1 Cl lithium aluminium hydride reduces esters all the way only reduces reactive c=o bonds down to alcohols reduces nearly all carbonyls 19 20 21 mechanism of lithium mechanism of lithium by-product is also a (less) aluminium hydride reduction aluminium hydride reduction powerful reductant... H Li Li H Li Li O O O O H Al H R2 R1 R2 H Al H R2 R1 R2 H R1 O O H R1 O O H H H H H O R2 H H H O R2 AlH3 Li Al H Li Li Al H Li O O H O O H O R1 H Al H R1 H Al H H H H R1 H H H H R1 H R1 H H AlH3 H AlH3 H O H O O H O cation important; again, note H– is remove it & R1 not the reductant R1 ...could be termed a R1 H R1 H reaction stops H H H H H H more selective reagent! 22 23 24 Reduction of amides mechanism of lithium aluminium amine normally a poor leaving group hydride reduction of amides Li H O Li O H Al H R2 R1 N NR22 H R1 H O H H R2 Li AlH3 O O R1 N R2 LiAlH4 R1 N R2 H Al H H NR22 O AlH3 R1 NR22 X R1 H R 2 N R2 R1 NR22 H R2 R2 H H R1 H second reduction (of the lithium aluminium H H iminium cation) does not Amine anions often used as bases hydride can perform R2 require metal cation as it (think LDA) they have high pka so R1 N are poor leaving groups this reduction is already charged R2 25 26 27
  4. 4. the reduction of Reduction of acids Reduction of acids Text amides really isn’t this simple to be honest! X O LiAlH4 H H O LiAlH4 O H H H R1 O R1 O R1 O R1 O Li just get salt formation along lithium aluminium hydride with the evolution of hydrogen does not reduce acids under gas (h2) standard conditions 28 29 30 Reduction of acids Reduction of acids big Text difference between lithium aluminium hydride & borane O BH3 H H O BH3 O BH3 H H H B H H H R1 O R1 O 3 R1 O R1 O R1 O reactive the reduction can be achieve with a different first step is to ‘protect’ the acid the second reagent, borane (or equivalent of borane diborane) does the reduction ©jnthnhys@flickr 31 32 33 importance: liAlH4 is nucleophilic importance: BH3 is electrophilic Text H O H O B R2 H Al H R2 H H R1 N R1 !+ O H R2 more !+ve the borane needs carbonyl the faster it activation reacts with weaker should react more ‘electron rich’ borane reverses carbonyls normal chemoselectivity attacks electron poor carbonyls attacks electron rich carbonyls ©j heffner@flickr 34 35 36
  5. 5. borane reduction of amides chemoselectivity (enantiospecific) chemoselectivity (enantiospecific) H H O B H BH2 H H H H H H H H B O LiBH4 BH3 LiBH4 BH3 R2 O H R1 N R1 R2 R2 N CO2H EtO2C CO2H EtO2C CO2H EtO2C CO2H EtO2C R2 R1 N H HO OH HO OH R2 H+ H+ H+ H+ R2 H H H H H H H H H BH2 H H B R2 2 H O O start with a single O O O O select the correct O O R1 N R1 N R R2 enantiomer of starting reagent & we can form R1 N material either enantiomer of R2 R2 R2 lactone 37 38 39 partial reduction very hard theoretically it is diisobutylaluminium hydride (dibal) Text possible...practice can be more problematic O O O R1 OR2 R1 H R1 NR22 Al difficult to prevent full H reduction to either alcohol or amine dibal’s reactivity is more similar to bh3 than liaiH4 ©0olong@flickr 40 41 42 must be cold mechanism summary starting iBu iBu material Al iBu iBu NR2 O O O O O O Al H R1 R1 R1 R2 R1 OR2 R1 NR2 R1 O H H H OH R1 OR2 NaCNBH3 reduced slow no!reaction DIBAL R1 OR2 O –78°C O NaBH4 LiBH4 LiAlH4 R1 OR2 R1 H O H3O O AliBu2 BH3 NHR2 OH OH OH NR2 OH react ester at low R1 H R1 OR2 temperature & you can the crucial H R1 R1 R1 R2 R1 R1 R1 isolate the aldehyde bit is the stability of the stable at tetrahedral intermediate low temp product ©Rachel's flickrs@flickr 43 44 45
  6. 6. want to achieve the following problem: no reagent suitable transformation... nabh4 only reduces ketone chemoselectivity OH O O O O O O reduction reduction OEt OEt OH OH OEt which OH functional LiAlh4 reduces group will everything react? ©amortize@flickr 46 47 48 solution: Acetal protecting group solution: Acetal protecting group O O reduction O O O reduction O diol reacts OEt OH OEt OH with more reactive ketone to give acetal then we can do the HO HO reduction OH H2O OH H2O H H H H O LiAlH4 O LiAlH4 O O O O O O O O OEt OH OEt OH use a protecting group 49 50 51 hydrogenation is the catalytic hydrogenation can be addition of hydrogen chemoselective... O H2, O H H H H Pd / C hydrogenation H H C X C X hydrogen can be added across hydrogenation normally most double or triple bonds & requires a catalyst can cleave some single bonds this permits exquisite chemoselectivity 52 53 54
  7. 7. catalytic hydrogenatation reduces poison the catalyst... partial hydrogenation... alkynes H2, H2, Pd / H H CaSO4, Pb H H Pd / C H H R1 R2 R1 R2 R1 R2 Lindlar R1 R2 catalyst can we stop this the reaction is stereoselective giving this blocks some of the the cis product (we’ll find out how to reaction halfway & form reactive sites on the an alkene? make the trans isomer later) catalyst ©furryscaly@flickr 55 56 57 FUNCTIONALGROUP 123.312 hydrogenation can cleave single hydrogenolysis of benzyl ethers bonds INTERCONVERSIONS H H C X H H C X R1 O H2, Pd / C H R1 O H CHAPTER8 seen this before during protection/deprotection section hydrogenolysis can also 58 cleave alkyl halides 59 E 60 looked at substrate (R-LG)... C–C bond formation is foundation of organic synthesis functional group interconversions Nuc R LG R Nuc CHAPTER eight previously we looked at c–c bond formation the substrate & which leaving groups were good in substitution reactions now look at c-based nucleophiles ©Ricketts Fish@flickr 61 62 63
  8. 8. Need... Need... ...a carbanion Cnuc Celec virtually all C-C bond forming reactions come down to this...(not necessarily charged species but certainly polarised C C 64 Cnuc Celec C C we are now going to concentrate on carbanions C– 65 C do not really exist (just ask an inorganic chemist) just behave like this 66 ...a carbanion 1. organometallics C metal 1. organometallics M R MX R X or R M insert a metal into a carbon halide bond two ways to make carbanions 67 68 69 behave as... grignard reagents made from organolithium reagents made halides from chlorides (normally) R Mg 2 x Li R Li R Br R MgBr R Cl LiCl but more much more reactive complex insertion occurs via single electron than grignard reagents R X Sol transfer (SET) process and the final structure in solution depends Mg Mg structure is solvent dependent on r. normally oligomeric (dimer, Sol X R trimer etc) ©jurvetson@flickr 70 71 72
  9. 9. methyl lithium is a tetramer in the many other solid state (& solution) organometallic reagents... Cl R Zn R Zr 2. deprotonation R Cu Li R R all with their own Copyright: Ben Mills (2007) pros & cons 73 74 75 2. deprotonation base R H R H base this can be considered but, how easy is it is removal of a proton to give a carbanion but reality more complex & we to selectively are invariably forming an organometallic reagent 76 remove a proton ©smeerch@flickr ? 77 ©dhammza@flickr revision 78 we can use pKa to assess ease of we can use pKa to assess ease of deprotonation... deprotonation... low pKa H O H X H H X H O H R H conjugate base R H base more acidic proton base conjugate acid acid base easier to form remember, pKa indicates where the equilibrium lies or how easy it is to with deprotonations to form carbanions we are effectively looking at (carb)anion remove the proton from the acid removing the proton from the conjugate acid 79 80 81
  10. 10. conjugate conjugate http://www2.lsdiv.harvard.edu/labs/evans/ acid pKa value acid pKa value base base CH4 CH3 49 OH O 17 O O O O NH3 NH2 36 11 OEt H2C CH2 H2C CH 36 OEt O O H H H 26 O O O O 9 23 EtO CH3 EtO CH2 O O O O 4.8 19 OH O best pKa H3C CH3 H3C CH2 O O table I S OH S O -0.6 know 19 O O OH O HCl Cl -7 82 83 84 You do not need to learn these values But, you do need to understand electron withdrawing groups what factors effect them... stabilise anion & make H+ more acidic H3C O F3C O H H pKa = 15.9 pKa = 12.4 F3C O F3C O H H CF3 F3C CF3 pKa = 9.3 pKa = 5.4 this is the inductive effect, electrons ©Graham Johnson, Graham Johnson Medical Media, Boulder, Colorado being pulled through ! bonds 85 86 87 delocalisation stabilises anion remember, many groups involved in resonance... you all love the other effect that stabilises a H N carbanion... H H R C N R C pKa = 51 O resonance O O O R N O O R N H H H H O pKa = 28.3 O O O O the ability to spread the charge over more atoms causes a large degree of anion R S R S stabilisation & hence the big differences observed Ph Ph ©stefan linecker@Flickr in pKa 88 89 90
  11. 11. we’ll concentrate on... enolates O ©alfred sim@flickr 91 92