Orgânica 3

Prova 2 Organica III 1 Dê o nome das reações abaixo Dê os respectivos produtos Demonstre sua formação via mecanismo Determine a configuração de cada produto 2 A reação abaixo ocorre em duas etapas A primeira delas envolve a adição da espécie nucleofílica gerada ao trifenilvinil fosfônio seguido de uma reação de olefinação intramolecular Dê o produto demonstre sua formação via mecanismo 3 Uma forma muito útil de produzir uma amida a partir de uma oxima é através do rearranjo catalisado por ácido dos isômeros de oximas exibidos no esquema abaixo Conforme ilustrado a estrutura do produto amida 2a ou 2b é ditada pela estereoquímica do precursor oxima 1a ou 1b a Escreva um mecanismo detalhado que justifique a transformação de 1a 2a e 1b 2b Ao descrever essa transformação inclua intermediários b Tendo como partida a 2metilciclohexanona preveja a estereoquímica preferencial da oxima gerada E ou Z e a estrutura da amida formada após rearranjo Me Et OH Ph3P Base Me Et OAc SO2Ph NaHg a b OH O H PPh3 I NaH O R2 R1 NH2OH NH2OH R1 R2 N OH R1 R2 N HO H3O H3O H N R2 O R1 H N R1 O R2 1a 2a 1b 2b 4 Considerando a reação a seguir fazendose uso de um Ilídeo de enxofre estabilizado dê o produto esperado Demonstre o mecanismo 5 O dímero pinacol da ciclobutanona sofre rearranjo com expansão de um dos anéis em solução ácida resultando em uma ciclopentanona fundida em espiro ao anel de quatro membros restante Desenhe um mecanismo para essa reação A redução da cetona gera um álcool que também sofre rearranjo para um alceno bicíclico em meio ácido Sugira um mecanismo para essa reação e explique por que esses rearranjos ocorrem 6 Dê o produto de Rearranjo de BayerVilliger para os compostos abaixo demonstre o mecanismo O Me NH2OH H3O I II O Me CO2Et Me2S HO OH O HO redução H H O O O H H a Cl O H b c S S O 3 a C O R1 R2 NHOH CNOH R1 R2 NHOH R1 R2 NHOH R1 R2 1a 2a 4 O Me EtO2C SMe2 5 HO OH HO OH H OH H H H2O H H 6 a O carbono mais substituído migra sem mudar a configuração b c mais substituído c Enantiomericamente puro 3a 1 eq 2 b 2 eq 2 a NH2OH 3eq favorável esternamente I E II 6 a b o carbono mais substituído migra sem mudar a configuração c Enantiomericamente puro Draw the products for the intramolecular Wittig reactions below Refresh page to flip back MOC Quiz ID 0041 CHAPTER 11 NUCLEOPHILIC SUBSTITUTION AT CO WITH LOSS OF CARBONYL OXYGEN carbonyl group by a nucleophile the nucleophile is the carbanion part of the phosphonium ylid This reaction generates a negatively charged oxygen that attacks the positively charged phosphorus and gives a fourmembered ring called an oxaphosphetane formation of the fourmembered ring Now this fourmembered ring like many others is unstable and it can collapse in a way that forms two double bonds Here are the curly arrows the mechanism is cyclic and gives the alkene which is the product of the reaction along with a phosphine oxide decomposition of the fourmembered ring triphenylphosphine oxide The chemistry of some elements is dominated by one particular property and a theme running right through the chemistry of phosphorus is its exceptional affinity for oxygen The PO bond with its bond energy of 575 kJ mol1 is one of the strongest double bonds in chemistry and the Wittig reaction is irreversible and is driven forward by the formation of this PO bond No need here for the careful control of an equilibrium necessary when making acetals or imines THE JULIA OLEFINATION IS REGIOSPECIFIC AND CONNECTIVE 687 The reason for the E selectivity lies in the mechanism of the elimination The details are not fully clear but the first step under the basic conditions of the reduction appears to be the elimination of the acetate or benzoate ester to give a vinyl sulfone The stereochemistry of the vinyl sulfone does not matter because it is immediately reduced by an electron from sodium to give a vinyl radical Much as you saw above in the Birch reduction of alkynes the vinyl radical collects a second electron and becomes a vinyl anion which chooses to adopt the more stable E configuration before being protonated to give the predominantly E alkene We know that there must be an anion intermediate because the elimination is not stereospecificin other words whichever diastereoisomer of the starting material you use all of the examples in this section have been mixtures of diastereoisomers you always get the E alkene product The onestep Julia olefination The Julia reaction is remarkably versatile but it does need three steps to make the alkene addition acylation and reduction A more recent version of the reaction cuts this down to one by using not a phenylsulfone but instead a sulfone carrying an electrondeficient heterocycle for example a tetrazole The anion of the sulfone is made with a strong base here potassium hexamethyldisilazide KHMDSsee p 635 and is added to an aldehyde to give an alkene directly The elimination works because after the addition to the aldehyde the alkoxide that is formed makes itself into a leaving group by grabbing the heterocyclic ring from the sulfur The final elimination is driven by loss of SO2 and typically gives an E alkene although by choosing carefully the base and solvent the selectivity can be tuned to give predominantly Z Tetrazoles look alarmingly unstable but are in fact commonly used in medicinal chemistry You will meet them again shortly in the chapters on heterocyclic chemistry 29 and 30 no text found You can probably apply something of what you know from Chapters 6 and 10 about the reactivity of carbonyl compounds towards nucleophiles to work out what is happening in this reaction between a carbonyl compound and an amine The hydroxylamine first adds to the ketone to form an unstable intermediate similar to a hemiacetal Notice that it is the more nucleophilic nitrogen atom and not the oxygen atom of hydroxylamine that adds to the carbonyl group Like hemiacetals these intermediates are unstable and can decompose by loss of water The product is known as an oxime and it is this compound with its CN double bond that is responsible for the IR absorption at 1400 cm1 We know that the oxime is formed via an intermediate because the 1400 cm1 absorption hardly appears until after the 1710 cm1 absorption has almost completely gone There must really be another curve to show the formation and the decay of the intermediate The only difference is that the intermediate has no double bond to give an IR absorbance in this region of the spectrum The Beckmann fragmentation To introduce the theme of the last section of this chapter a Beckmann rearrangement that is not all that it seems tButyl groups migrate well in the BaeyerVilliger reaction and indeed Beckmann rearrangement of the compound in the margin appears to be quite normal too But when this compound and another compound with a tertiary centre next to the oxime are mixed together and treated with acid it becomes apparent that what is happening is not an intramolecular reaction Each migrating tertiary group must have lost contact with the amide fragment it started out with The molecule must fall apart to give a talkyl cation and a nitrile the Beckmann rearrangement now goes via a fragmentation mechanism Migrating groups have to provide some degree of cation stabilization But if they stabilize a cation too well there is a good chance that fragmentation will occur and the migrating group will be lost as a carbocation Here is a more convincing example of the same fragmentation reaction the conditions but not the results are those of a Beckmann rearrangement In this reaction the ring structure means the cation cannot be trapped by the nitrile and a fragmentation product is isolated The normal mechanism for the Beckmann rearrangement pp 958960 of the textbook involves protonation at OH and migration of the group anti to the NO bond in this case the substituted benzene ring Making nitrile oxides There are two important routes to these compounds both of which feature interesting chemistry Oximes easily made from aldehydes with hydroxylamine NH2OH are rather enollike and can be chlorinated on carbon Treatment of the chlorooxime with base Et3N is strong enough leads directly to the nitrile oxide with the loss of HCl This is an elimination of a curious kind as we cannot draw a connected chain of arrows for it We must use two stepsremoval of the OH proton and then loss of chloride It is a γ elimination rather than the more common β elimination The other method starts from nitroalkanes and is a dehydration Inspect the two molecules and you will see that the nitro compound contains one molecule of H2O more than the nitrile oxide But how to remove the molecule of water The reagent usually chosen is phenyl isocyanate PhNCO which removes the molecule of water atombyatom to give aniline PhNH2 and CO2 This is probably the mechanism although the last step might not be concerted as we have shown The ylid is stabilized by conjugation with the ester groupyou can think of it also as an enolate We can expect reversible addition to the carbonyl group and hence conjugate addition under thermodynamic control The stereochemistry of the ring junction is inevitable only a cis ring can be made a transfused ring would be too strained The interesting centre is that of the ester on the threemembered ring It too is in a more stable configuration on the outside of a folded molecule The intermediate is probably a mixture of diastereoisomers but as the conjugate addition is reversible the cyclopropane may be formed by cyclization of only the diastereoisomer that can give the more stable product Reduction to the alcohol is trivial and then acid treatment allows the loss of water and ring expansion of the remaining fourmembered ring You may well have drawn this as a stepwise process Elimination gives the most substituted alkene Both rearrangements occur very easily because of the relief of strain in going from a four to a fivemembered ring The first reaction is a simple pinacol rearrangement The diol is symmetrical so protonation of either alcohol and migration of either CC bond give the product Migration to oxygen the BaeyerVilliger reaction In 1899 two Germans A Baeyer and V Villiger found that treating a ketone with a peracid RCO3H can produce an ester An oxygen atom is inserted next to the carbonyl group You saw a similar insertion reaction earlier in the chapter and the mechanism here is not dissimilar Both peracids and diazomethane contain a nucleophilic centre that carries a good leaving group and addition of peracid to the carbonyl group gives a structure that should remind you of a semipinacol intermediate with one of the carbon atoms replaced by oxygen Carboxylates are not such good leaving groups as nitrogen but the oxygenoxygen single bond is very weak Once the peracid has added loss of carboxylate is concerted with a rearrangement driven by formation of a carbonyl group BaeyerVilliger reactions are among the most useful of all rearrangement reactions and the most common reagent is mCPBA metachloroperbenzoic acid because it is commercially available