Mecanismos de Reações Orgânicas - Condensação Aciônica, Substituição Nucleofílica Aromática e Radicais

Suggested solutions for Chapter 37 PROBLEM 1 Give a mechanism for the formation of this silylated enediol and explain why the Me3SiCl is necessary Purpose of the problem Reminder of an important radical reaction Suggested solution This is an acyloin condensation linking radicals derived from esters by electron donation from a dissolving metal here sodium If the esters can form enolates the addition of Me3SiCl protects against that problem by removing the MeO byproduct The first product is a very electrophilic 12dione and it accepts electrons from sodium atoms even more readily than do the original esters The product is an ene diolate that is also silylated under the reaction conditions Details from B M Trost and group J Org Chem 1978 43 4559 428 Solutions Manual to accompany Organic Chemistry PROBLEM 2 Heating the diazonium salt below in the presence of methyl acrylate gives a reasonable yield of a chloroacid Why is this unlikely to be nucleophilic aromatic substitution by the SN1 mechanism p 520 of the textbook Suggest an alternative mechanism that explains the regioselectivity Cl N2 Cl CO2Me heat Cl Cl CO2Me Purpose of the problem Revision of nucleophilic aromatic substitution with diazonium salts and contrasting cations and radicals Suggested solution The cation mechanism is perfectly reasonable as far as the diazonium salt is concerned but it will not do for the alkene Conjugated esters are electrophilic and not nucleophilic alkenes Even if it were to attack the aryl cation we should find the reverse regioselectivity Cl N2 Cl CO2Me Cl CO2Me Cl CO2Me Cl Cl The only way to produce the observed product is to decompose the diazonium salt homolytically To do this we can draw the salt as a covalent compound or transfer one electron from the chloride ion to the diazonium salt The other product would be a chlorine radical Addition to the alkene gives the more stable radical which abstracts chlorine from the diazonium salt and keeps the chain going PROBLEM 3 Suggest a mechanism for this reaction and comment on the ring size formed What is the minor product likely to be Purpose of the problem Activated alkenes are not necessary in radical cyclizations Suggested solution The peroxide is a source of benzoyloxy radicals PhCO2 and these capture hydrogen atoms to give the most stable radical The best one here is stablized by both CN and CO2Et Cyclization onto the alkene gives mainly a secondary radical on a sixmembered ring and this abstracts a hydrogen from starting material to complete the cycle The alternative is to add to the more substituted end of the alkene This gives a less stable primary radical but this 5exo ring closure is often preferred because the orbital alignment is better The minor product has a fivemembered ring Notice that in the last step we have put in only half the mechanismwe shall generally do this from now on as it is clearer There is nothing wrong with putting in another chain of halfheaded arrows going in the other direction PROBLEM 4 Treatment of this aromatic heterocycle with NBS Nbromosuccinimide and AIBN gives mainly one product but this is difficult to purify from minor impurities containing one or three bromine atoms Further treatment with 10 aqueous NaOH gives one easily separable product in modest yield 50 What are the mechanisms for the reactions Purpose of the problem An important radical reaction bromination at benzylic and allylic positions by NBS and an application Suggested solution Two preliminary reactions need to take place NBS is a source of a low concentration of bromine molecules and AIBN initiates the radical chain by forming a nitrilestabilized tertiary radical The new radical abstracts hydrogen atoms from the benzylic positions to make stable delocalized radicals These react with bromine to give the benzylic bromide and release a bromine atom All subsequent hydrogen abstractions are carried out by bromine atoms either of the kind we have just seen or to remove a hydrogen atom from the other methyl group This reaction provides the HBr that generates more bromine from NBS Finally the dibromide reacts with NaOH to give the new heterocycle Both SN2 displacements are very easy at a benzylic centre and the second is intramolecular This product was used to make constrained amino acids by S Kotha and coworkers Tetrahedron Lett 1997 38 9031 Problem 5 Propose a mechanism for this reaction accounting for the selectivity Include a conformational drawing of the product Purpose of the problem Another important radical reaction cyclization of alkyl bromides onto alkenes 432 Solutions Manual to accompany Organic Chemistry Suggested solution This time AIBN abstracts the hydrogen from Bu3SnH and the tin radicals carry the chain along First they remove the bromine atom from the starting material to make a vinyl radical that cyclizes onto the unsaturated ketone to give a radical stabilized by conjugation with the carbonyl group The chain is completed by abstraction of hydrogen from another molecule of Bu3SnH the tin radical formed then allowing the cycle to restart Bu3Sn H CN Bu3Sn CN H Bu3Sn O CO2Me Br O MeO2C O MeO2C H SnBu3 H O MeO2C H Bu3Sn The stereochemistry of the product comes from the requirement of a 13 bridge to be diaxial as this is the only way the bridge can reach across the ring At the moment of cyclization the vinyl radical side chain must be in an axial position O MeO2C MeO2C O H MeO2C O product Problem 6 An ICI process for the manufacture of the diene used to make pyrethroid insecticides involved heating these compounds to 500 C in a flow system Propose a radical chain mechanism for the reaction Purpose of the problem Learning how to avoid a trap in writing radical reactions and to show you that radical reactions can be useful Suggested solution The most likely initiation at 500 C is the homolytic cleavage of the CCl bond to release allyl and chloride radicals The chloride radicals then attack the alkene and abstract a hydrogen atom to give more of the same allylic radical The trap is to form the product by dimerizing the allylic radical Dimerizing radicals does sometimes occur in the acyloin reaction for example but it is a rare process Much more likely is a chain reaction If we add the allylic radical to the alkene part of the allylic chloride we make a stable tertiary radical that can lose chloride radical and propagate the chain The original workers at ICI suggested a different mechanism D Holland and D J Milner Chem and Ind London 1979 707 434 Solutions Manual to accompany Organic Chemistry PROBLEM 7 Heating this compound to 560 C gives two products with the spectroscopic data shown below What are they and how are they formed O Cl 560 C A B A has IR 1640 cm1 mz 138 100 and 140 33 δH ppm 71 4H s 65 1H dd J 17 11 Hz 55 1H dd J 17 2 Hz and 51 1H dd J 11 2 Hz B has IR 1700 cm1 mz 111 45 113 15 139 60 140 100 141 20 and 142 33 δH ppm 99 1H s 775 2H d J 9 Hz and 743 2H d J 9 Hz Purpose of the problem Revision of structure determination and a radical reaction with a difference Suggested solution Compound A contains chlorine mz 138140 31 and that fits C8H7Cl It still has the 14disubstituted benzene ring four aromatic Hs and it is an alkene IR 1640 with three hydrogens on it with characteristic coupling We can write the structure immediately as there is no choice The four aromatic hydrogens evidently have the same chemical shift O Cl 560 C Cl H H H δH 65 δH 51 δH 55 J 11 J 2 J 17 Compound B has mz 140142 31 and a carbonyl group at 1700 cm1 which fits C7H5ClO and looks like an aldehyde δH 99 It still has the disubstituted benzene The structure is even easier this time O Cl 560 C Cl H O H H δH 743 δH 775 δH 99 H H J 9 So how are these products formed At such high temperatures σbonds break and the weakest bonds in the molecule are the CC and CO bonds in Solutions for Chapter 37 Radical reactions 435 the fourmembered ring next to the benzene ring Breaking these bonds releases strain and allows one of the radical products to be secondary and delocalized O Cl b a a Cl O A b Cl O B PROBLEM 8 Treatment of methylcyclopropane with peroxides at very low temperature 150 C gives an unstable species whose ESR spectrum consists of a triplet with coupling of 207 gauss and fine splitting showing dtt coupling of 20 26 and 30 gauss Warming to a mere 90 C gives a new species whose ESR spectrum consists of a triplet of triplets with coupling 222 and 285 gauss and fine splitting showing small ddd coupling of less than 1 gauss Me tBuOOtBu 150 C A 90 C B If methylcyclopropane is treated with tBuOCl various products are obtained but the two major products are C and D At lower temperatures more of C is formed and at higher temperatures more of D Me tBuOCl Cl Cl C D Treatment of the more substituted cyclopropane below with PhSH and AIBN gives a single product in quantitative yield Account for all these reactions identifying A and B and explaining the differences between the various experiments Ph PhSH AIBN PhS Ph Purpose of the problem Working out the consequences of an important substituent effect on radical reactions the cyclopropyl group Suggested solution The peroxide is a source of tBuO radicals and these abstract a hydrogen from the methyl group of the hydrocarbon The first spectrum is that of the cyclopropylmethyl radical The odd electron is in a p orbital represented by a circle and the planar CH2 group is orthogonal to the plane of the ring but the two Ha s are the same because of rapid rotation The odd electron has a large coupling to the two hydrogens Ha on the same carbon a smaller doublet coupling to Hb and small couplings to the two Hc s and two Hd s The coupling to Hb is small because the p orbital containing the odd electron is orthogonal to the CHb bond Warming to 90 C causes decomposition to an openchain radical The odd electron is coupled to the two hydrogens on its own carbon Ha and those on the next carbon Hb each giving a triplet 222 and 285 Coupling to the more remote hydrogens is small Decomposition of the same hydrocarbon with tBuOCl produces the same sequence of radicals but they can now be intercepted by the chlorine atom of the reagent releasing more tBuO radicals and a radical chain is started At lower temperatures the ring opening is slower so more of the cyclopropane is captured The last example also produces a radical next to a cyclopropane ring but this time it can decompose very easily to give a stable secondary benzylic radical This captures a hydrogen atom from PhSH releasing PhS and maintaining an efficient radical chain Ring opening of cyclopropanes is now a standard way of detecting radicals C S Walling and P S Fredericks J Am Chem Soc 1962 91 1877 Solutions for Chapter 37 Radical reactions 437 PhS H CN PhS Ph Ph PhS PhS Ph SPh H PhS Ph PROBLEM 9 The last few stages of Coreys epibatidine synthesis are shown here Give mechanisms for the first two reactions and suggest a reagent for the last step N NHCOCF3 Br Br Cl tBuOK THF 78 C N COCF3 Br N Cl 75 yield N COCF3 N Cl H N N Cl Bu3SnH AIBN benzene reflux 95 yield Purpose of the problem Application of radical reactions in an important sequence plus revision of conformation and stereochemistry Suggested solution The first step involves deprotonation of the rather acidic amide the CF3 group helps and the displacement of the only possible bromidethe one on the opposite face of the sixmembered ring as the SN2 reaction must take place with inversion N N Br Br Cl CF3 O H OtBu N N Br Br Cl CF3 O N COCF3 Br N Cl SN2 with inversion The second step is a standard dehalogenation by Bu3SnH AIBN generates Bu3Sn by hydrogen abstraction from the reagent and this removes the bromine Make sure you complete the chain and do not use H at any point Finally we need to hydrolyse the amide This normally requires strong acid or alkali but the CF3 group makes this amide significantly more electrophilic than most and milder conditions can be used Corey actually used NaOMe in methanol at 13 C for two hours and got a yield of 96 Any reasonable conditions you may have chosen would be fine too Problem 10 How would you make the starting material for this sequence of reactions Give a mechanism for the first reaction that explains its regio and stereoselectivity Your answer should include a conformational drawing of the product What is the mechanism of the last step Attempts to carry out this last step by iodinelithium exchange and reaction with allyl bromide failed Why Why is the alternative shown here successful Purpose of the problem Application of radical reactions when the alternative ionic reactions fail Suggested solution The starting material is an obvious DielsAlder product as it is a cyclohexene with a carbonyl group outside the ring on the opposite side The first step is iodolactonization Iodine attacks the alkene reversibly on both sides but when it attacks opposite the carboxylate anion the lactone ring snaps shut The problem asks for a conformational drawing of the product and indeed that is necessary The 13lactone bridge must be diaxial as that is the only way for the carboxylate to reach across and therefore it must attack from an axial direction too The last step is initiated by AIBN which removes the iodine atom from the compound to make a secondary radical This attacks the allyl stannane and the intermediate loses Bu3Sn and that takes over the job of removing iodine atoms to keep the chain going The radical intermediate has no stereochemistry at the planar radical carbon and attack occurs from the bottom face to avoid the blocking lactone bridge Anionic reactions cannot be used for this allylation If the iodine were metallated the organometallic compound would immediately expel the lactone bridge as carboxylate ion is a good leaving group The radical is stable because the CO bond is strong and not easily cleaved in radical reactions 440 Solutions Manual to accompany Organic Chemistry PROBLEM 11 Suggest a mechanism for this reaction explaining why a mixture of diastereoisomers of the starting material gives a single diastereoisomer of the product Is there any other form of selectivity O OEt Br 1 Bu3SnH AIBN 2 CrVI H2SO4 O O Purpose of the problem A radical ringclosing reaction with a curious stereochemical outcome Suggested solution The abstraction of bromine at first by AIBN and thereafter by Bu3Sn produces a radical that again does not eliminate but adds to an alkene A fivemembered ring is formed this is usually the more favourable closure by attack on the alkene on the opposite side from that occupied by the iPr group The product is a mixture of diastereoisomers as no change occurs at the acetal centre O OEt Br CN SnBu3 initiation thereafter O OEt O OEt Bu3Sn H O O Acidcatalysed oxidation first hydrolyses the acetal and then oxidizes either the hemiacetal or the aldehyde to the lactone Now the molecule is one diastereoisomer as the ambiguous centre is planar The other form of selectivity is the ring size see the textbook p 1000 O OEt H2SO4 O OH H2SO4 O O OH CHO H2SO4 OH CO2H Solutions for Chapter 37 Radical reactions 441 PROBLEM 12 Reaction of this carboxylic acid C5H8O2 with bromine in the presence of dibenzoyl peroxide gives an unstable compound A C5H6Br2O2 that gives a stable compound B C5H5BrO2 on treatment with base Compound B has IR 1735 and 1645 cm1 and NMR δH 618 1H s 500 2H s and 418 2H s What is the structure of the stable product B Deduce the structure of the unstable compound A and mechanisms for the reactions CO2H Br2 PhCO22 A base B Purpose of the problem Revision of structural analysis in combination with an important radical functionalization Suggested solution The starting material is C5H8O2 so the stable compound B has gained a bromine and lost three hydrogens There must be an extra double bond equivalent DBE somewhere in B The IR spectrum shows that the OH has gone and suggests a carbonyl group possibly an ester because of the high frequency and an alkene The NMR shows that both methyl groups have gone and have been replaced by CH2 groups The bromine must be on one of them and the ester oxygen on the other The extra DBE is a ring CO2H O O H Br O O H Br H H H H 1735 cm1 1645 cm1 δH 418 δH 500 δH 618 Since both methyl groups are functionalized unstable A must have one Br on each methyl group The peroxide produces benzoyl radicals that abstract protons from both allylic positions to give stabilized radicals that sttack bromine molecules to give bromide radicals to continue the chain reaction In base the carboxylate cyclizes onto the cis CH2Br group 442 Solutions Manual to accompany Organic Chemistry CO2H H PhCO2 Br Br CO2H CO2H Br initially Br thereafter CO2H Br Br H Br Br CO2H Br CO2H Br Br CO2H Br Br unstable compound A base Br Br O O O O Br stable compound B Suggested solutions for Chapter 38 Problem 1 Suggest mechanisms for these reactions Purpose of the problem Two simple carbene reactions initiated by base Suggested solution Going to the right we must remove the rather acidic proton from CHBr3 to give the carbanion This loses bromide to give dibromocarbene and insertion into cyclohexene gives the product The second reaction is very similar αElimination of HCl gives a carbene that inserts into an alkene These are the simplest reactions of carbenes and are very common PROBLEM 2 Suggest a mechanism for this reaction and explain the stereochemistry Purpose of the problem Another important carbene method used in the synthesis of a natural antibiotic Suggested solution The diazo compound decomposes to gaseous nitrogen and a carbene under catalysis by CuII Insertion into the exposed alkene gives the threemembered ring The stereochemistry partly comes from the tetherthe linkage between the carbene and the rest of the molecule that delivers the carbene to the bottom face of the alkene The rest comes from the inevitable cis fusion between the five and threemembered rings PROBLEM 3 Comment on the selectivity shown in these reactions Purpose of the problem A study in chemoselectivity during carbene insertion into alkenes Suggested solution The first reaction is a variation on SimmonsSmith cyclopropanation Though strictly a carbenoid rather than a carbene it delivers a CH2 group from an organozinc compound bound to an oxygen atom in this case the OMe group Only that alkene reacts The second cyclopropanation occurs at the only remaining alkene with a carbene generated from a diazoester The stereoselectivity comes from attack on the opposite side of the ring from the already established cyclopropane PROBLEM 4 Suggest a mechanism for this ring contraction Purpose of the problem Drawing mechanisms for a rearrangement involving a carbene formed photochemically Suggested solution The carbene formed by loss of nitrogen from the diazoketone rearranges with the migration of either CC bond to give a ketene picked up by methanol PROBLEM 5 Suggest a mechanism for the formation of this cyclopropane Purpose of the problem An unusual type of carbene but it behaves normally Suggested solution There is no doubt that tBuO is a base but which proton does it remove The OH proton perhaps but that doesnt lead to a carbene The proton on the alkyne That molecule has a leaving group but is it too far away Not if you push the electrons through the molecule in a γelimination Normal elimination is βelimination both α and γelimination can produce carbenes The arrows are easy to make sense of if you think of a carbene as a carbon with both a and a charge The carbene is an allenyl carbene with no substituent at the carbene centre It inserts into the alkene in the other molecule Solutions for Chapter 38 Synthesis and reactions of carbenes 447 OH C OH C PROBLEM 6 Decomposition of this diazo compound in methanol gives an alkene A C8H14O whose NMR spectrum contains two signals in the alkene region δH 350 3H s 550 1H dd J 179 79 580 1H ddd J 179 92 and 43 420 1H m and 1327 8H m What is its structure and geometry N2 MeOH A C8H14O When you have done that suggest a mechanism for the reaction using this extra information Compound A is unstable and even at 20 C isomerizes to B If the diazo compound is decomposed in methanol containing a diene compound A is trapped as the adduct shown Account for all these reactions OMe H A OMe B Purpose of the problem Revision of structural analysis alkene geometry and cycloadditions with carbenes as a mechanistic link Suggested solution The starting material is C7H10N2 so it has lost nitrogen and gained CH4O one molecule of methanol We can see the MeO group at δH 350 and the four CH2 groups in the ring are still there 8H m at 1327 All that is left is a multiplet at δH 42 obviously next to OMe and a pair of alkene protons at δH 55 and 58 coupled with J 179obviously a trans alkene That at δH 55 is coupled to one proton and the one at 58 is coupled to two We now have these fragments But these add up to C2H3 too much Clearly the CH attached to OMe and the CH attached to the alkene are the same atom and the CH2 at the other end of the alkene must be one end of the chain of four CH2s We now have a structure but it doesnt join up This is the test of your belief in spectroscopythe dotted ends must join up to give A Yes this does put an Ealkene in a sevenmembered ring and it is difficult to draw but you were warned that A is unstable The CH2 group next to the CHOMe group is diastereotopic so the coupling constants are different Now that we know the structure of A it is easy enough to find a mechanism Loss of nitrogen produces a carbene that gives an allene in a pericyclic process and this twisted compound the two alkenes are at 90 to each other and protonation gives the trans alkene as a cation that reacts with methanol to give A The twisted alkene is unstable and rotates to the much more stable cis alkene even at 20 C It can rotate because the overlap between the p orbitals is weak as they are not parallel Trapping in a DielsAlder reaction preserves the trans stereochemistry This was the discovery of H Jendralla Angew Chem Int Ed Engl 1980 19 1032 If you were really on the ball youll have noticed that a transcycloheptene is chiral so this compound must be a single diastereoisomer though we dont know which PROBLEM 7 Give a mechanism for the formation of the threemembered ring in the first of these reactions and suggest how the ester might be converted into the amine with retention of configuration Purpose of the problem A routine carbene insertion and a reminder of nitrenes as analogues of carbenes Suggested solution The diazoester gives the carbene under CuI catalysis and insertion into the alkene follows its usual course The only extra is stereoselectivity the insertion happens more easily if the two large groups Ph and CO2Et keep as far apart as possible Conversion of acid derivatives into amines with the loss of the carbonyl group can be done in various ways In chapter 36 we recommended the Curtius and the Hofmann The Hofmann degradation is the easier if we start with an ester converting into the amide with ammonia and then treating with bromine in basic solution The Nbromo amide undergoes αelimination to a nitrene that rearranges to an isocyanate The amine product is an antidepressant discovered by A Burger and W L Yost J Am Chem Soc 1948 70 2198 PROBLEM 8 Explain how this highly strained ketone is formed albeit in very low yield by these reactions How would you attempt to make the starting material Purpose of the problem To show that intramolecular carbene insertion is a powerful way to make cage compounds Suggested solution Oxalyl chloride makes the acid chloride and diazomethane converts this into the diazoketone Now the carbene chemistry Treatment with CuI removes nitrogen and forms the carbene Remarkably this is able to reach across the molecule and insert into the alkene thus forming one three and two new fourmembered rings in one step You will not be surprised at the yield 10 How would you attempt to make the starting material The original workers used another carbene reactionthe CuI catalysed insertion of a diazoester into bistrimethylsilyl acetylene This very strained ketone was used in vain while attempting to make tetrahedrane by G Maier and group Angew Chem Int Ed Engl 1983 22 990 Problem 9 Attempts to prepare compound A by phasetransfer catalysed cyclization required a solvent immiscible with water When chloroform CHCl3 was used compound B was formed instead and it was necessary to use the more toxic CCl4 for success What went wrong Purpose of the problem Carbene chemistry is not always what is wanted how do you avoid it Suggested solution Product B is clearly the adduct of product A and dichlorocarbene which must have come from the chloroform and base The good news is that product A was evidently formed in the basic reaction mixture so if we simply avoid a solvent that is also a carbene source all is well Problem 10 Revision content How would you carry out the first step in this sequence Propose mechanisms for the remaining steps explaining any selectivity Purpose of the problem Revision of specific enol formation rearrangement reactions electrocyclic reactions and conjugate addition plus some carbene chemistry Suggested solution The first step requires a specific enol from an enone Treatment with LDA achieves kinetic enolate formation by removing one of the more acidic hydrogens immediately next to the carbonyl group The lithium enolate is trapped with Me3SiCl to give the silyl enol ether The next step is dichlorocarbene insertion into the more nucleophilic of the two alkenes Dichlorocarbene is an electrophilic carbene so the main interaction is between the HOMO π of the alkene and the empty p orbital of the carbene The carbene is formed by decarboxylation a process that needs no strong base You can draw the ring expansion in a number of ways All start with the removal of the Me3Si group with water You might then simply use a onestep mechanism a but an electrocyclic process via the cyclopropyl cation b might be better This is allowed since the inevitable cis ring junction requires H and OH to rotate outwards Finally a double conjugate addition of MeNH2 to the dienone forms the bicyclic amine Conjugate addition probably occurs first on the more electrophilic chloroenone though it doesnt much matter There is some stereoselectivity in that the remaining chlorine prefers the equatorial position on the new sixmembered ring but this is thermodynamic control as that position is easily enolized Problem 11 How would you attempt to make these alkenes by metathesis 454 Solutions Manual to accompany Organic Chemistry Purpose of the problem Applications of this important and powerful method Suggested solution Metathesis is usually Eselective and these are both Ealkenes so prospects are good We must disconnect each compound at the alkene and add something to the end of each probably just CH2 as the byproduct will then be volatile ethylene OH O HO OH O Each starting material must now be made The stereochemistry of the first tells us that we should add an allyl metal compound to an epoxide The metathesis catalyst will be one of those mentioned in the chapter O Li or allyl Grignard OH PPh3 Ru PPh3 Ph Cl Cl product The second molecule is not symmetrical but this is all right as it will be an intramolecular ringclosing metathesis so we can expect few cross products There are many ways to make the starting material alkylation of a ketone is probably the simplest though conjugate addition would have its advantages The same catalyst can be used and very little would be needed Solutions for Chapter 38 Synthesis and reactions of carbenes 455 PROBLEM 12 Heating this acyl azide in dry toluene under reflux for three hours gives a 90 yield of a heterocycle Suggest a mechanism emphasizing the role of any reactive intermediates N3 O NH2 heat toluene N H H N O Purpose of the problem Demonstrating the practical nature of nitrene chemistry in the context of heterocyclic synthesis Suggested solution Heating an azide liberates nitrogen gas and forms a nitrene In this case rearrangement to an isocyanate is followed by intramolecular nucleophilic attack by the ortho amino group N O NH2 N C NH2 O isocyanate N H N O H H H heat product PROBLEM 13 Give mechanisms for the steps in this conversion of a five into a sixmembered aromatic heterocycle N H Cl3CCO2Na reflux in DME workup in aqueous base N Cl Purpose of the problem It is the turn of carbene chemistry to show its usefulness in that most practical of all subjects heterocyclic synthesis Suggested solution Decomposition of trichloroacetate ion releases the Cl3C carbanion Loss of chloride gives dichlorocarbene and addition to one of the double bonds in the pyrrole gives a bicyclic intermediate Ring expansion can be drawn in various ways There is a direct route from the neutral amine or its anion that doesnt look very convincing or you can ionize one of the chlorides first and open the cyclopropyl cation in an electrocyclic reaction However you explain it this is a good way to make 3substituted pyridines Suggested solutions for Chapter 39 PROBLEM 1 Propose three fundamentally different mechanisms other than variations of the same mechanism with different kinds of catalysis for this reaction How would a D labelling and b 18O labelling help to distinguish the mechanisms What other experiments would you carry out to rule out some of these mechanisms Purpose of the problem Investigating a reaction where there are several reasonable mechanisms Suggested solution The reaction is an ester hydrolysis so the obvious mechanism is to attack the carbonyl group with hydroxide Notice that we draw out each stage of the mechanism and do not use any summary or shorthand But the ester oxygen atom is attached to an aromatic ring with a para nitro group Nucleophilic aromatic substitution would give the same product 458 Solutions Manual to accompany Organic Chemistry Finally the ester can be transformed into an enolate using hydroxide as a base Elimination gives a ketene that can be attacked by hydroxide as a nucleophile to give the product O O O2N H H OH O O O2N O2N C O O OH O2N O CO2H O2N OH CO2 O2N O OH O HO H Mechanism 3 Enolate elimination to give a ketene Mechanism 3 requires the exchange of at least one hydrogen atom with the solvent so if D2O were used as the solvent or better deuterated starting material were used the exchange of one whole deuterium atom would indicate mechanism 3 while no exchange or only minor amounts from the inevitable enolization would show mechanisms 1 or 2 In mechanisms 1 and 3 the added OH group ends up in in CO2H but in mechanism 2 it ends up as the phenol Using H218O as solvent or better labelling the ester oxygen as 18O would separate mechanisms 1 and 3 from 2 O O O2N D D O O O2N O2N OH CO2 D D mech 1 or 2 mech 3 O2N OH CO2 mech 1 or 3 mech 2 O2N OH CO2 D H O2N OH CO2 18O Other experiments we could do might include trying to trap the ketene intermediate in a 2 2 cycloaddition studying the reaction by UV hoping to see the release of pnitrophenolate in mechanism 3 changing the structure of the starting material so that one or other of the mechanisms would be difficult even measuring the effects of the substituent on the benzene ring on the rate or looking for a deuterium isotope effect in the labelled lactone PROBLEM 2 Explain the stereochemistry and labelling pattern in this reaction Purpose of the problem A combination of labelling and stereochemistry reveals the details of a surprisingly interesting rearrangement Suggested solution The randomization of the label and the racemization suggest that the carboxylate falls off the allyl cation and then comes back on again at either end While they are detached the distinction between the two ends of both cation and anion disappears as they are delocalized The product is racemic because the two intermediates each have a plane of symmetry and are achiral The retention of relative stereochemistry formation of the trans product from trans starting material could result from stereoselective recombination the two faces of the allyl cation are not the same or from the two ions sticking together as an ion pair so that the acetate slides across one face of the cation An alternative 33 sigmatropic rearrangement would not randomize the labels in the same way PROBLEM 3 The Hammett ρ value for migrating aryl groups in the acidcatalysed Beckmann rearrangement is 20 What does that tell us about the ratedetermining step Purpose of the problem The Hammett relationship gives an intimate picture of the Beckmann rearrangement Suggested solution 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 If this mechanism is correct here we should expect the migration itself to be the slow step The first step is just a proton transfer to oxygen and must be fast The steps after the migration involve attack of water on a carbocation and proton transfers to O and N and these must all be fast The migration breaks a CC bond forms a CN bond and creates an unstable cation But does this agree with the evidence Starting material and product in the migration step are cations so the transition state must be a cation too Any contribution to cation stability made by the migrating group should help and we should therefore expect electrondonating groups to migrate faster This is what we see a ρ value of 20 shows a modest acceleration by electrondonating groups p 1041 ff In the Beckmann rearrangement the anti group migrates but in other rearrangements the migrating group is chosen for a very different reason it is normally the group that is best able to stabilize a positive charge and benzene rings can do this by π participation This would be the participation mechanism The Hammett ρ value of 20 gives very definite evidence that participation does not occur If it did the closure of the unstable threemembered ring would be the slow step and a positive charge would form on the benzene ring itself This would give a much larger ρ value of something like 50 One reason that participation does not occur is that the starting material is planar and the p orbitals in the benzene ring cannot point in the right direction to interact with the σ orbital of the NO bond They are orthogonal to it PROBLEM 4 Between pH 2 and 7 the rate of hydrolysis of this ester is independent of pH At pH 5 the rate is proportional to the concentration of acetate ion AcO in the buffer solution and the reaction goes twice as fast in H₂O as in D₂O Suggest a mechanism for the pHindependent hydrolysis Above pH 7 the rate increases with pH What kind of change is this Purpose of the problem Time for you to try your skill at interpreting pHrate profiles Suggested solution The second part of the question is easily dealt with In alkaline solution the rate of hydrolysis simply increases with pH and we have the normal specific basecatalysed reaction in which hydroxide ion attacks the carbonyl group But this is no ordinary ester The leaving group is a thiol pKa about 8 not the usual alcohol pKa about 16 and so the thiolate anion is a much better leaving group than EtO Also the CF₃ group is very electronwithdrawing so nucleophilic attack on the carbonyl group will be unusually fast This is why there is a region of pHindependent hydrolysis not found with EtOAc You might have suggested that acetate is a nucleophile or a general base catalyst but the solvent deuterium isotope effect suggests that it is a general base The change at pH 7 is a change of mechanism as the faster of two mechanisms appliesa sketch of the pHrate profile will show you the upward curve PROBLEM 5 In acid solution the hydrolysis of this carbodiimide has a Hammett ρ value of 08 What mechanism might account for this Purpose of the problem Interpretation of a small Hammett ρ value Suggested solution The most obvious explanation for a low Hammett ρ value that the aromatic ring is too far away from the reaction will not wash here as the aromatic rings are joined directly to the reacting nitrogen atoms of the carbodiimide The reaction must surely start with the protonation of one of the nitrogens This cannot be the slow step and it would in any case have a large negative ρ value The small ρ value observed suggests that the ratedetermining step must have a large positive ρ value that nearly cancels out the large negative value for the first step Attack by water on the protonated carbodiimide looks about right The expected equilibrium Hammett ρ value for the protonation would be about 25 to 3 so the kinetic Hammett ρ value for the attack of water would have to be about 2 to give a net Hammett ρ value of 08 This looks fine The rest of the mechanism involves proton transfers hydrolysis of an imide and decarboxylation PROBLEM 6 Explain the difference between these Hammett ρ values by mechanisms for the two reactions In both cases the ring marked with the substituent X is varied When R H ρ 03 but when R Ph ρ 51 Purpose of the problem Interpretation of a variation in Hammett ρ value with another structural variation Suggested solution The reaction is obviously nucleophilic substitution at the benzylic centre so we are immediately expecting SN1 or SN2 When R H the reaction occurs at a primary alkyl group and SN2 is expected When R Ph the reaction occurs at a secondary benzylic centre and SN1 is expected Since SN1 produces a cation delocalized round the benzene ring in the slow step a large negative Hammett ρ value is reasonable It is not obvious what sign the Hammett ρ value would have in the SN2 reaction but as there is no buildup of negative charge on the carbon atom in the transition state a small value is reasonable The actual value 03 is very small indeed but if we can read anything into it it suggests a loose SN2 transition state with a small positive charge on carbon PROBLEM 7 Explain how chloride catalyses this reaction Purpose of the problem An extreme example of surprising catalysis Suggested solution At first you might ask how chloride can catalyse anything at all It is a weak base and not a very good nucleophile for the carbonyl group However in polar aprotic solvents like acetonitrile MeCN chloride is not solvated and is both more basic and more nucleophilic In this reaction it cannot be a nucleophilic catalyst as attack on the carbonyl group simply regenerates starting material It cannot be a specific base as it is too weak even in acetonitrile to remove a proton from methanol But it can act as a general base As methanol attacks the carbonyl group its proton becomes more acidic and in the transition state chloride is at last able to act PROBLEM 8 The hydrolysis of this oxaziridine in 01 M sulfuric acid has kH2OkD2O 07 and an entropy of activation of ΔS 76 J mol1 K1 Suggest a mechanism Purpose of the problem Deducing a mechanism from isotope effects and entropy of activation Suggested solution The inverse solvent deuterium isotope effect indicates specific acid catalysis and the modest negative entropy of activation suggests some bimolecular involvement There are various mechanisms you might have proposed and a likely one involves cleavage of the threemembered ring in the protonated amine The second or possibly the third step could be ratedetermining Once the threemembered ring is opened the rest of the mechanism amounts to acidcatalysed hemiacetal hydrolysis The original workers favoured an alternative mechanism that starts with protonation of the oxygen atom and ends up with the hydrolysis of an imine Again the second or third step could be ratedetermining The original work was by J H Fendler and group J Chem Soc Perkin Trans 2 1973 1744 You might also have considered an electrocyclic opening of the threemembered ring PROBLEM 9 Explain how both methyl groups in the product of this reaction come to be labelled If the starting material is reisolated at 50 reaction its methyl group is also labelled Purpose of the problem Exploring a mechanism through labelling Suggested solution The role of silver ion Ag is the removal of the halide to give an acylium ion that reacts not at the carbonyl group but at the methyl group to give CO2 and a methylated benzene ring The simple FriedelCrafts route cannot be the whole story it explains how the added methyl group is labelled but not why it is only partly labelled and how label gets into the other methyl group The only way in which we can explain those extra features is to suggest that methylation initially occurs on the oxygen atom and that a methyl group is transferred from there to the benzene ring We should never have detected this detail without the labelling experiment Alkylation on oxygen provides an alkylating agent that can transfer either CH3 or CD3 and also explains the formation of trideuterotoluene We hope you didnt suggest a methyl cation as an intermediate PROBLEM 10 The pKa values of some protonated pyridines are as follows Can the Hammett correlation be applied to pyridines using the σ values for benzene What equilibrium ρ value does it give and how do you interpret it Why are no 2substituted pyridines included in the list Purpose of the problem Making sure you understand the ideas behind the Hammett relationship Suggested solution The obvious thing to do is to plot the pKa values against the σ values for the substituents using the meta values for the 3substituted and para values for the 4substituted compounds see table on p 1042 of the textbook This gives quite a good straight line and we get a slope Hammett ρ value of 59 The sign is of course positive as the same electronic effects that make benzoic acids more acidic will also make pyridinium ions more acidic The large ρ value may have surprised you but reflect ionization of benzoic acids occurs outside the ring and the charge isnt delocalized round the ring Deprotonation of pyridinium ions occurs on the ring and the charge positive this time is delocalized round the ring This work was done to apply the Hammett relationship to reactions of pyridines with acid chlorides R B Moody and group J Chem Soc Perkin Trans 2 1976 68 There are no 2substituted pyridines on the list since like orthosubstituted benzenes they cannot be expected to give a good correlation because of steric effects PROBLEM 11 These two reactions of diazo compounds with carboxylic acids give gaseous nitrogen and esters as products In both cases the rate of reaction is proportional to diazo compoundRCO2H Use the data for each reaction to suggest mechanisms and comment on the difference between them Purpose of the problem Application of contrasting isotope effects to detailed mechanistic analysis Suggested solution The first reaction has a normal kinetic isotope effect RCO2H reacts faster than RCO2D while the second has an inverse deuterium isotope effect RCO2H reacts slower than RCO2D This suggests that there is a ratedetermining proton transfer in the first reaction but specific acid catalysis in the second ie fast equilibrium proton transfer followed by slow reaction of the protonated species Protonation occurs at carbon in both reactions and this can be a slow step Solutions for Chapter 39 Determining reaction mechanisms 469 Ar Ar N N H O O R rate determining step Ar Ar N N H N2 Ar Ar H O R O Ar Ar O R O The second reaction follows much the same pathway except that loss of nitrogen is now difficult because the cation would be very unstable primary and next to a CO2Et group so the second step is SN2 and rate determining EtO2C N N H fast EtO2C N N O R O rate determining step EtO2C O R O N2 PROBLEM 12 Suggest mechanisms for these reactions and comment on their relevance to the Favorskii family of mechanisms O 1 Br2 2 EtO EtOH CO2Et Ph Me Ph Br O MeO MeOH MeO MeOH Ph O Ph CO2Me Ph bromoketone added to base base added to bromoketone Purpose of the problem Extension of a section of the chapter pp 10613 of the textbook into new reactions with internal trapping of intermediates Suggested solution In the first reaction the bromination must occur on the alkene to give a dibromide We cannot suggest stereochemistry at this stage and it is better to continue with the standard Favorskii mechanism and see what happens Everything follows until the very last step when the opening of the cyclopropane provides electrons in just the right place to eliminate the second bromide and put the alkene back where it was This alternative 470 Solutions Manual to accompany Organic Chemistry behaviour of a proposed intermediate gives us confidence that the intermediate really is involved O Br2 O Br Br EtO EtOH O Br Br O Br twoelectron disrotatory electrocyclic O Br EtO Br O OEt gives product The stereochemistry of the initial bromination turns out to be irrelevant as it disappears when the oxyallyl cation is formed We know the stereochemistry of the final product so we know the stereochemistry of the cyclopropanone it must be on the opposite face of the fivemembered ring to the methyl group The disrotatory closure of the oxyallyl cation evidently goes preferentially one way with the H and the CMe2Br substituents going upwards and the carbonyl group going down O Br O Br H EtO Br O OEt H CO2Et The second reaction to the right is a normal Favorskii The only point of interest is the way the threemembered ring breaks up The more stable carbanion is the doubly benzylic one so that leaves Ph Me Ph Br O MeO MeOH Ph Ph base added to bromoketone Ph Ph Br O Ph Ph O twoelectron disrotatory electrocyclic O Ph Ph MeO Ph Ph MeO O MeO H CO2Me The reaction with excess bromoketone starts the same way but the oxyallyl cation is intercepted by one of the benzene rings in a fourelectron conrotatory electrocyclic reaction like the Nazarov reaction p 927 of the textbook You may wonder how excess MeO stops this from happening It doesnt The oxyallyl cation and the cyclopropanone are in equilibrium and excess MeO captures the cyclopropanone and drives the normal Favorskii onwards If there is no excess MeO the oxyallyl cation lasts long enough for the fivemembered ring to be the main product This work was part of a thorough investigation into the mechanism of the Favorskii rearrangement by F G Bordwell and group J Am Chem Soc 1970 92 2172 PROBLEM 13 A typical Darzens reaction involves the basecatalysed formation of an epoxide from an αhaloketone and an aldehyde Suggest a mechanism consistent with the data below a The rate expression is rate k3PhCOCH2ClArCHOEtO b When Ar is varied the Hammett ρ value is 25 c The following attempted Darzens reactions produced unexpected results Purpose of the problem Trying to get a complete picture of a reaction using physical data and structural variation Suggested solution The ethoxide is not incorporated into the product but appears in the rate expression Its role must be as a base and there is only one set of enolizable protons We start by making the enolate of the chloroketone This cannot be the slow step as the aldehyde appears in the rate expression Then we can attack the aldehyde with the enolate and finally close the epoxide ring by nucleophilic displacement of chloride ion If this mechanism is right the kinetic data show that the second step is ratedetermining a reasonable deduction as it is a bimolecular step and that the first step is a preequilibrium We can write rate k2enolateArCHO And we know from the preequilibrium that K1 enolate PhCOCH2ClEtO So the rate expression becomes when we substitute for enolate rate K1k2PhCOCH2ClEtOArCHO and this matches the observed rate expression though the apparently third order rate constant is revealed as the product of an equilibrium constant and a second order rate constant The Hammett ρ value shows a modest gain of electrons near the Ar group in the ratedetermining step We must not take the preequilibrium into account as ArCHO is not involved in this step In fact a Hammett ρ value of 25 is typical of nucleophilic attack on a carbonyl group conjugated to the benzene ring The unexpected products come from variations in this mechanism paraMethoxybenzaldehyde is conjugated and unreactive so the enolate ignores it and reacts with the unenolized version of itself With salicylaldehyde the second example the phenolic OH group will exist as an anion under the reaction conditions Alkylation by the Solutions for Chapter 39 Determining reaction mechanisms 473 chloroketone allows enolate formation leading to an intramolecular aldol reaction CHO OH EtO CHO O Ph O Cl CHO O O Ph EtO O O Ph H O O Ph O O Ph O OH PROBLEM 14 If you believed that this reaction went by elimination followed by conjugate addition what experiments would you carry out to try and prove that the enone is an intermediate Ph Cl O NaCN H2O EtOH Ph CN O Purpose of the problem Turning the usual question backwards what evidence do you want rather than how to interpret what you are given Suggested solution The suggested mechanism of elimination followed by conjugate addition might be contrasted with direct SN2 to see what evidence is needed mechanism 1 simple SN2 displacement mechanism 2 eliminationaddition a elimination b addition There are many types of evidence you might suggest here are some of them Exchange of protons in D2OEtOD would suggest eliminationaddition Kinetic evidence tricky as you cannot be sure which is the slow step A Hammett plot with substituted benzene rings The SN2 mechanism would have a small ρ as the benzene ring is a long way from the action Base catalysis mechanism 2 is base catalysed mechanism 1 isnt Kinetic isotope effect might be found in mechanism 2 Stereochemistry If a substituent were added to make the terminal carbon chiral inversion would be expected for mechanism 1 and racemization for mechanism 2 But choose a small substituent otherwise it would be a very different compound Suggested solutions for Chapter 40 PROBLEM 1 Suggest mechanisms for these reactions explaining the role of palladium in the first step Purpose of the problem Revision of enol ethers and bromination the Wittig reaction and of course first steps in palladium chemistry Suggested solution The first step is a reaction of an enol with an allylic acetate catalysed by palladium0 via an η3 allyl cation There is no regiochemistry to worry about as the diketone and allylic acetate are both symmetrical NBS in aqueous solution is a polar brominating agent ideal for reaction with an enol ether The intermediate is hydrolysed to the ketone by the usual acetal style mechanism You might have drawn the η³ allyl cation complex in various satisfactory wayssome are mentioned on p 1089 of the textbook Finally an intramolecular Wittig reaction This is a slightly unusual way to do what amounts to an aldol reaction but the 55 fused enone system is strained and the Wittig went under very mild conditions K2CO3 in aqueous solution The stereochemistry of the new double bond is the only one possible and Wittig reactions with stabilized ylids generally give the most stable of the possible alkene This process is a general way to make 55 fused systems devised by B M Trost and D P Curran J Am Chem Soc 1980 102 5699 PROBLEM 2 This Heckstyle reaction does not lead to regeneration of the alkene Why not What is the purpose of the formic acid HCO2H in the reaction mixture Purpose of the problem Making sure you understand the steps in the mechanism of the Heck reaction Suggested solution The reaction must start with the oxidative addition of Pd0 into the PhI bond The reagent added is PdII so one of the reduction methods on page 1081 of the textbook must provide enough Pd0 to start the reaction going The oxidative addition gives PhPdI and this does the Heck reaction on the alkene Addition occurs on the less hindered top exo face and the phenyl group is transferred to the same face Normally now the alkyl palladiumII species would lose palladium by βelimination This is impossible in this example as there is no hydrogen atom syn to the PdI group Instead an external reducing agent is needed and that is the role of the formate anion it provides a hydride equivalent by transfer hydrogenation when it loses CO2 A heterocyclic version of this reaction was part of a synthesis of the natural analgesic epibatidine by S C Clayton and A C Regan Tetrahedron Lett 1993 34 7493 PROBLEM 3 Cyclization of this unsaturated amine with catalytic PdII under an atmosphere of oxygen gives a cyclic unsaturated amine in 95 yield How does the reaction work Why is the atmosphere of oxygen necessary Explain the stereochemistry and regiochemistry of the reaction How would you remove the CO2Bn group from the product Purpose of the problem Introducing you to aminopalladation like oxypalladation nucleophilic attack on a palladium πcomplex Suggested solution The πcomplex between the alkene and PdII permits nucleophilic attack by the amide on its nearer end and in a cis fashion because the nucleophile is tethered by a short chain of only two carbon atoms Nucleophilic attack and elimination of Pd0 occur in the usual way The removal of the CO2Bn group would normally be done by hydrogenolysis but in this case ester hydrolysis by say HBr would be preferred to avoid reduction of the alkene The free acid decarboxylates spontaneously PROBLEM 5 Explain why enantiomerically pure lactone gives syn but racemic product in this palladiumcatalysed reaction Purpose of the problem Helping you to understand the details of palladiumcatalysed allylation Suggested solution Following the usual mechanism the palladium complexes to the face of the alkene opposite the bridge The ester leaves to give an allyl cation complex This is attacked by the malonate anion from the opposite face to the palladium So the overall result is retention of configuration the syn starting material giving the syn product The racemization comes from the structure of the allyl cation complex It is symmetrical with a plane of symmetry running vertically through the complex as drawn Attack by the malonate anion occurs equally at either side of the plane giving the two enantiomers of the syn diastereoreisomer in equal amounts PROBLEM 4 Suggest a mechanism for this lactone synthesis Purpose of the problem Introducing you to carbonyl insertion into a palladium II σcomplex Suggested solution Oxidative insertion into the aryl bromide carbonylation and nucleophilic attack on the carbonyl group with elimination of Pd0 form the catalytic cycle No doubt the palladium has a number 1 or 2 of phosphine ligands complexed to it during the reaction and these keep the Pd0 in solution between cycles PROBLEM 6 Explain the reactions in this sequence commenting on the regioselectivity of the organometallic steps Purpose of the problem Revision of allylic Grignard reagents the synthesis of pyridines and the mechanism of the Wacker oxidation Suggested solution The allylic Grignard reagent does direct addition from the end remote to the magnesium atom as often happens Hydrolysis of the silyl enol ether reveals an aldehyde Now the Wacker oxidation by whatever detailed mechanism you prefer must involve the addition of water to a PdII πcomplex of the alkene and βelimination of palladium to give Pd0 which is recycled by oxidation with oxygen mediated by copper Finally the pyridine synthesis is simply a double enamineimine formation between ammonia and the two carbonyl groups Probably the aldehyde reacts first M A Tius Tetrahedron Lett 1982 23 2819 PROBLEM 7 Give a mechanism for this carbonylation reaction Comment on the stereochemistry and explain why the yield is higher if the reaction is carried out under a carbon monoxide atmosphere Hence explain this synthesis of part of the antifungal compound pyrenophorin Purpose of the problem More carbonylation with a Stille coupling Suggested solution The tinpalladium exchange transmetallation occurs with retention of configuration at the alkene The exchange of the benzyl group for the benzoyl group is necessary to get the reaction started Now the coupling can take place on the palladium atom producing the product and Pd0 which can insert oxidatively into the CCl bond Transmetallation sets up a sustainable cycle of reactions It is better to have an atmosphere of carbon monoxide because the acyl palladium complex can give off CO and leave a PdPh σcomplex The atmosphere of CO reverses this reaction The second sequence starts with a radical hydrostannylation chapter 37 giving the Evinyl stannane preferentially if a slight excess of Bu3SnH is used Now the coupling with the acid chloride takes place as before though this time we have an aliphatic carbonyl complex There is no problem with βelimination as that would give a ketene Again the stereochemistry of the vinyl stannane is retained in the product Solutions for Chapter 40 Organometallic chemistry 483 Bu3Sn CO2Bn COCl OSiPh2tBu Pd Ph3P Ph3P OSiPh2tBu O CO2Bn OSiPh2tBu O O OBn Ph32PdClBn PROBLEM 8 A synthesis of an antifungal drug made use of this palladiumcatalysed reaction Give a mechanism explaining the regio and stereochemistry NHMe OAc Ph3P4Pd N Me tBu Purpose of the problem A simple example of amine synthesis using palladium Suggested solution The palladium forms the usual allyl cation complex and the nitrogen nucleophile attacks the less hindered end thus also retaining the conjugation Attack at the triple bond would give an allene The E stereochemistry of the palladium complex is retained in the product OAc Ph3P4Pd PdL2 RNHMe product 484 Solutions Manual to accompany Organic Chemistry PROBLEM 9 Work out the structures of the compounds in this sequence and suggest mechanisms for the reactions explaining any selectivity CHO OH heat with catalytic acid CHO A PdII CuCl O2 B KOH H2O THF C B has IR 1730 1710 cm1 δH 94 1H s 26 2H s 20 3H s and 10 6H s C has IR 1710 cm1 δH 73 1H d J 55 Hz 68 1H d J 55 Hz 21 2H s and 115 6H s Purpose of the problem An intramolecular aldol reaction p 636 of the textbook and a Wacker oxidation p 1096 of the textbook Suggested solution B clearly has aldehyde and ketone functional groups with nothing but singlets in the NMR On the other hand C has a cis disubstituted alkene with a small and therefore cis J value and is a cyclopentenone CHO A PdII CuCl O2 CHO O B KOH H2O THF C O Solutions for Chapter 40 Organometallic chemistry 485 PROBLEM 10 A synthesis of the BristolMyers Squibb antimigraine drug Avitriptan a 5HT receptor antagonist involves this palladiumcatalysed indole synthesis Suggest a mechanism and comment on the regioselectivity of the alkyne attachment S NHMe O O I NH2 SiEt3 N N OMe S NHMe O O N H SiEt3 N N MeO cat PdOAc2 Ph3P LiCl Na2CO3 H2O MeCN Purpose of the problem A new reaction for you to trya palladiumcatalysed indole synthesis Suggested solution Although palladiumII is added to the solution the aryl iodide tells you that this is an oxidative insertion of Pd0 produced by one of the methods described on p 1081 of the textbook The resulting PdII species complexes to the alkyne and the amine can now attack the triple bond This gives a heterocycle with the PdII in the ring Coupling of the two organic fragments extrudes Pd0 to start a new cycle The nitrogen attacks the more hindered end of the alkyne so that the palladium can occupy the less hindered end This is the Larock indole synthesis R C Larock and E K Yum J Am Chem Soc 1991 113 6689 and its use in the synthesis of Avitriptan is described in P D Brodfuehrer et al J Org Chem 1997 62 9192 Suggested solutions for Chapter 41 Problem 1 Explain how this synthesis of amino acids starting with natural proline works Explain the stereoselectivity of each step after the first Purpose of the problem A simple exercise in the creation of a new stereogenic centre via a cyclic intermediate Suggested solution Nothing exciting happens until the hydrogenation step The stereoselectivity of the reaction with ammonia is interesting but not of any consequence as that stereochemistry disappears in the elimination This gives the Eenone as expected since the alkene and the carbonyl group are in the same plane This method was invented by B W Bycroft and G R Lee J Chem Soc Chem Commun 1975 988 488 Solutions Manual to accompany Organic Chemistry N CO2Me R O O NH3 N R O O NH2 O N NH O O H HO R CF3CO2H N NH O O H H2O R N NH O O H R H N NH O O H R The new stereogenic centre is created in the hydrogenation step The molecule is slightly folded and the catalyst interacts best with the outside convex face so that it adds hydrogen from the same face as the ring junction hydrogen All that remains is to hydrolyse the product without racemization Did you notice that the configuration of the new amino acid S is the same as that of the natural amino acids PROBLEM 2 This is a synthesis of the racemic drug tazodolene If the enantiomers of the drug are to be evaluated for biological activity they must be separated At which stage would you recommend separating the enantiomers and how would you do it O Ph O Ph O O BF3 O Ph O 1 HO NH2 2 H2 catalyst H N OH Ph OH H HCl H N OH Ph Ph3PBr2 N Ph Purpose of the problem First steps in planning an asymmetric synthesis by resolution Suggested solution You need to ask which is the first chiral intermediate Can it be conveniently resolved Will the chirality survive subsequent steps The first intermediate is chiral but it enolizes very readily and the enol is achiral so thats no good The second intermediate is chiral but it has three chiral centres and these are evidently not controlled We would have to separate the diastereoisomers before resolution and that would be a waste of time and material since all of them give the next intermediate anyway The next intermediate the amino alcohol is ideal it has only one chiral centre and that is not affected by the last reaction It has two handles for resolutionthe amine and the alcohol We might make a salt with tartaric acid or an ester of the alcohol with some chiral acid Alternatively we could resolve tazadolene itself it still has an amino group and we could form a salt with a suitable acid Problem 3 How would you make enantiomerically enriched samples of these compounds either enantiomer Purpose of the problem First steps in planning an asymmetric synthesis Suggested solution There are many possible answers here What we had in mind was some sort of asymmetric DielsAlder reaction for the first an asymmetric aldol for the second or else opening an epoxide made by Sharpless epoxidation asymmetric dihydroxylation for the third and perhaps asymmetric dihydroxylation of a Zalkene for the fourth Of course you might have used resolution or asymmetric hydrogenation or the chiral pool or any other strategy from chapter 41 This synthesis is from the Upjohn company and is in only the patent literature Chem Abstr 1984 100 6311 490 Solutions Manual to accompany Organic Chemistry O N O O O N O O LiBH4 OH H tBuOOH iPrO4Ti DDET O Me3Al OH OH OH OH OH OH OsO4 K3FeCN6 K2CO3 DHQD ligand OH OH OsO4 K3FeCN6 K2CO3 DHQD ligand Lewis acid PROBLEM 4 In the following reaction sequence the stereochemistry of mandelic acid is transmitted to a new hydroxyacid by stereochemically controlled reactions Give mechanisms for each reaction and state whether it is stereospecific or stereoselective Offer some rationalization for the creation of new stereogenic centres in the first and last reactions HO2C Ph OH Smandelic acid tBuCHO O O O Ph tBu 1 LDA 2 PrBr O O O Ph tBu H H2O HO2C Ph OH Purpose of the problem Your chance to examine an ingenious method of asymmetric induction Suggested solution The first reaction amounts to cyclic acetal formation except that one of the alcohols is a carboxylic acid The reaction is stereospecific no change at the original chiral centre and stereoselective at the new one The second reaction creates a lithium enolate and alkylates it It is again stereospecific at the unchanged chiral centre and stereoselective at the new one Finally acetal hydrolysis preserves the new quaternary centre unchanged stereospecific by a mechanism that is the reverse of the first step Now as far as the rationalization is concerned the first step takes place through a sequence of reversible reactions and therefore under thermodynamic control so the most stable product will be formed It may seem surprising that this should be the cis compound but the conformation of this chairlike fivemembered ring prefers to have the two substituents pseudoequatorial The alkylation is under kinetic control and as a lithium enolate has more or less a flat ring the alkyl halide approaches the opposite face to the tBu group It has to approach orthogonally to the ring as it must overlap with the p orbital of the enolate This is Seebachs clever method of preserving the knowledge of a chiral centre while it is destroyed in a reaction First a temporary centre at the tbutyl group is created in a stereoselective reaction the original centre is destroyed by enolization but the temporary centre can be used to recreate it D Seebach et al J Am Chem Soc 1983 105 5390 PROBLEM 5 This reaction sequence can be used to make enantiomerically enriched amino acids Which compound is the origin of the chirality and how is it made Suggest why this particular enantiomer of the product amino acid might be formed Suggest reagents for the last stages of the process Would the enantiomerically enriched starting material be recovered Purpose of the problem Stepbystep discusssion of a simple but useful sequence Suggested solution The amine phenylethylamine is the origin of the chirality It is easily made by resolution for example by crystallizing the salt of the racemic amine with tartaric acid This means that both enantiomers are readily available This particular enantiomer of the amino acid product belongs to the natural S series The unnatural R enantiomer would also be valuable and can be made from the other enantiomer of the starting material The last stages of the process require cleavage of one CN bond and hydrolysis of the nitrile It will be important to do this without racemizing the newly created centre The CN bond can be cleaved reductively by hydrogenation as it is an Nbenzyl bond This would also hydrogenate the nitrile so that must first be Solutions for Chapter 41 Asymmetric synthesis 493 hydrolysed using acid or base as weak as possible The starting material is not recovered and the chirality is lost as the byproduct is just ethyl benzene the nitrogen atom being transferred to the product N H Ph H Me R CN acid or base N H Ph H Me R HO2C H2PdC NH R CO2H Me Ph PROBLEM 6 Explain the stereochemistry and mechanism in the synthesis of the chiral auxiliary 8phenylmenthol from pulegone After the reaction with Na in iPrOH what is the minor 13 component of the mixture O Rpulegone PhMgBr CuBr O Ph KOH EtOH reflux O Ph 5545 mixture 8713 mixture with other diastereoisomer Na iPrOH toluene OH Ph 8713 mixture with another diastereoisomer Cl Cl O 1 2 crystallize 3 KOH H2O EtOH OH Ph one diastereoisomer one enantiomer Purpose of the problem A combination of conformational analysis stereoselective reactions and resolution to get a single enantiomer Suggested solution The first reaction is a conjugate addition that evidently goes without any worthwhile stereoselectivity The stereochemistry is not fixed in the addition but in the protonation of the enolate in the workup Equilibration of the mixture by reversible enolate formation with KOH in ethanol gives mostly the allequatorial compound Reduction by that smallest of reagents an electron gives the allequatorial product Since the stereochemical ratio in the product is the same as in the starting materials 8713 the reduction must be totally stereoselective The allequatorial ketone gives 100 allequatorial alcohol and the minor isomer must give one other diastereoisomer we cannot say which The mixture still has to be separated and as it is a mixture of diastereoisomers it can be separated by physical means The chloroacetate is just a convenient crystalline derivative PROBLEM 7 Suggest syntheses for single enantiomers of these compounds Purpose of the problem Devising your own asymmetric syntheses Suggested solution The first compound is an ester derived from a cyclic secondary alcohol that could be made from the corresponding enone by asymmetric reduction Reduction with Coreys CBS reducing agent gave the alcohol in 93 ee The second compound could be made by a Wittig reaction with a stabilized ylid and the required diol aldehyde derived from an epoxyalcohol and hence an allylic alcohol by Sharpless epoxidation The first part of the synthesis gives an intermediate that had been used in the synthesis of the antibiotic methymycin In practice the Wittig was carried out on the epoxyaldehyde and treatment of the last intermediate with aqueous acid gave the target molecule 496 Solutions Manual to accompany Organic Chemistry PROBLEM 8 This compound is a precursor to a Novartis drug used for the control of inflammation How might it be made from a chiral pool starting material CO2H OH Purpose of the problem Spotting in a target structural features of available chiral pool compounds Suggested solution The hydrocarbon skeleton of the target is that of the amino acid phenylalanine The configuration is S the same as the natural amino acid so we can use the standard amino acid to hydroxy acid conversion via diazotization described on p 1105 of the textbook which goes with retention of configuration The aromatic ring needs hydrogenating too Ph CO2H NH2 CO2H NH2 H2 Ni cat CO2H OH NaNO2 HCl H2O Lphenylalanine PROBLEM 9 Propose catalytic methods for the asymmetric synthesis of these four precursors to drug molecules Purpose of the problem Identifying reliable catalytic reactions that give desired structural features Suggested solution The sertraline precursor is a chiral alcohol with the stereogenic centre adjacent to an aromatic ring An obvious approach is to make the hydroxyl group by asymmetric reduction of the corresponding ketone CBS reduction is a possibility as is a rutheniumcatalysed hydrogenation using the ligand TsDPEN p1115 of the textbook The second compound is a chiral sulfide Although there are direct asymmetric ways of making chiral sulfur compounds a reliable approach to sulfides is to use SN2 substitution of a more readily made chiral precursor because a thiolate is usually a good nucleophile The SN2 reaction goes with inversion so we need the chiral alcohol shown below converted to a derivative such as a tosylate capable of undergoing substitution Care will be needed to avoid elimination but thiolates are excellent nucleophiles and not too basic so you would expect a successful outcome 498 Solutions Manual to accompany Organic Chemistry CO2Me O CO2Me OH H2 Ru cat chiral ligand CO2Me OTs TsCl py CO2Me OTs ArSH base The third compound contains a 123trifunctionalized arrangement that should prompt you to think of asymmetric epoxidation Azide is a good nucleophile for opening epoxides so we can start with the allylic alcohol shown here carry out an asymmetric epoxidation and convert to the target with inversion of configuration HO OMe OMe HO O OMe OMe HO OH N3 OMe OMe NaN3 tBuOOH TiOiPr4 DDET The final compound is a diol so asymmetric dihydroxylation is a possible approach The precursor is a rather unreactive alkene but asymmetric dihydroxylation is a versatile reaction which can still perform well on challenging substrates O MeO K3FeCN6 DHQ2PHAL K2OsO2OH4 K2CO2 MeSO2NH2 MeO OH OH O PROBLEM 10 The triatomine bug which causes Chagas disease can be trapped by using synthetic samples of its communication pheromone which consists of a 41 mixture of the enantiomers of this heterocycle How would you synthesize the required mixture of enantiomers Why would the other diastereoisomer of this compound be more of a challenge to make O O Purpose of the problem Identifying structural features that can be made by asymmetric synthesis Suggested solution To make a 41 mixture of enantiomers you need either to mix them in the right proportions or to mix equal amounts of racemic mixture and a single enantiomer In either case you need an asymmetric synthesis The target compound is an acetal that can be made from a chiral diol so you should immediately consider asymmetric dihydroxylation The advantage of Sharpless asymmetric dihydroxylation is that it can very easily give either enantiomer in fact it is one reaction where the enantioselective version is better than the racemic one so you would be advised to make the two enantiomers using the two alternative chiral ligands mix them in the correct proportions then form the acetal Note that the starting alkene is trans Making the other diastereoisomer would require the cis alkene This is not a problem in itself but more of a challenge for the catalyst because now it has to distinguish between two similar groups Et and Me in order to oxidize one face of the alkene enantioselectively for the trans alkene the selection is between either Et and H or Me and H switching Et for Me makes no difference to the outcome 500 Solutions Manual to accompany Organic Chemistry PROBLEM 11 This compound was developed by the Nutrasweet company as an artificial sweetener Propose a strategy for its synthesis Would your proposed approach still be suitable if the compound had turned out to be a successful product required in multitonne quantities O N H O H N N CN CO2H NC00637 Purpose of the problem Proposing an efficient synthetic route to a chiral target molecule a common challenge in the pharmaceutical and related industries Suggested solution The target can be best disconnected into three fragments at the amide bonds The aminopyridine can be made by the standard methods of heterocycle synthesis chapter 30 so we are more interested in the other two chiral fragments The middle one is an amino acid and you should recognize it as a member of the chiral pool Sglutamic acid so this poses no problem of synthesis Though it will need to be appropriately protected to form the correct amide O N H O H N N CN CO2H NC00637 O OH disconnect amides HN N CN H2N O CO2H OH Sglutamic acid The final fragment is a simple chiral carboxylic acid so we need a method for its asymmetric synthesis The most obvious choice is probably an asymmetric alkylation using Evans oxazolidinone auxiliary formation of the appropriate derivative of hexanoic acid is simple and the enolate will be alkylated diastereoselectively by methyl iodide You would probably take this approach if you need to make a few grams for initial studies Solutions for Chapter 41 Asymmetric synthesis 501 N O O O 1 LDA 2 MeI N O O O LiOOH O OH recyclable auxiliary HN O O NaH RCOCl If this compound were needed on the tonne scale then auxiliary chemistry is no good however efficient recycling may be A good alternative for the synthesis of compounds with unfunctionalized chiral centres adjacent to carboxylic acids or alcohols is the use of rutheniumcatalysed hydrogenation O OH O OH H2 cat RuClSBINAP PROBLEM 12 The two aldehydes below are valuable products in the perfumery industry Tropional is a component of Issey Miyakes LEau dIssey and Florhydral is a component of Allure by Chanel How would you make them as single enantiomers CHO O O CHO Tropional Florhydral Purpose of the problem Designing a synthesis where absolute stereochemistry must be controlled Suggested solution Both targets have a single simple chiral centre carrying a methyl group so we need to devise a synthesis passing through an achiral precursor For Tropional you might imagine alkylating a derivative of Evans auxiliary followed by reduction to the aldehyde but a more economical approach would be to use asymmetric reduction of an unsaturated carboxylic acid since the compound required is readily made using an aldoltype condensation of the available aldehyde piperonal Florhydral has the methyl group β to the aldehyde One possible approach is an asymmetric conjugate addition but again asymmetric reduction of the acid or allylic alcohol is preferable since the required alkene is easy to make by aldol chemistry Here we show one example with the acid and one with the alcohol but either are possibilities in both cases Suggested solutions for Chapter 42 PROBLEM 1 Do you consider that thymine and caffeine are aromatic compounds Explain Purpose of the problem Revision of aromaticity and exploration of the structures of nucleic acid bases Suggested solution Thymine a pyrimidine has an alkene and lone pair electrons on two nitrogens making six in all for an aromatic structure You may have shown this by drawing delocalized structures Caffeine a purine is slightly more complicated as it has two rings You might have said that each ring is aromatic if you counted all the lone pairs on nitrogen except those on the pyridinelike nitrogen see p 741 of the textbook for what we mean here in the fivemembered ring Or you might have drawn a delocalized structure with ten electrons around its periphery six electrons in sixmembered ring six electrons in fivemembered ring ten electrons in two rings together PROBLEM 2 Human hair is a good source of cystine the disulfide dimer of cysteine Hair is boiled with aqueous HCl and HCO2H for a day the solution concentrated and a large amount of sodium acetate added About 5 of the hair by weight crystallizes out as pure cystine αD 216 How does the process work Why is such a high proportion of hair cystine Why is no cysteine isolated by this process Make a drawing of cystine to show why it is chiral How would you convert the cystine to cysteine Purpose of the problem Some slightly more complicated amino acid chemistry including stereochemistry and the SH group Suggested solution Prolonged boiling with HCl hydrolyses the peptide linkages shown as thick bonds below in a generalized structure and breaks the hair down into its constituent amino acids The cystine crystallizes at neutral pHs and the mixture of HCl and NaOAc provides a buffer Hair is much crosslinked by disulfide bridges and these are not broken down by hydrolysis No cysteine is isolated because i most of it is present as cystine in hair and ii any cysteine released in the hydrolysis will be oxidized in the air to cystine The stereochemistry of cysteine is preserved in cystine which has C2 symmetry and no plane or centre of symmetry so either of the diagrams below will suit It is not important whether you draw the zwitterion or the uncharged structure Reduction of the SS bond by NaBH4 converts cystine to cysteine PROBLEM 3 The amide of alanine can be resolved by pig kidney acylase Which enantiomer of alanine is acylated faster with acetic anhydride In the enzymecatalysed hydrolysis which enantiomer hydrolyses faster In the separation why is the mixture heated in acid solution and what is filtered off How does the separation of the free alanine by dissolution in ethanol work If the acylation is carried out carelessly particularly if the heating is too long or too strong a byproduct is formed that is not hydrolysed by the enzyme How does this happen Purpose of the problem Rehearsal of some basic amino acid and enzyme chemistry plus revision of stereochemistry and asymmetric synthesis Suggested solution The acylation takes place by the normal mechanism for the formation of amides from anhydrides that is by nucleophilic attack on the carbonyl group and loss of the most stable anion acetate from the tetrahedral intermediate The two isomers of alanine are enantiomers and enantiomers must react at the same rate with achiral reagents In the enzymecatalysed reaction the acylase hydrolyses the amide of one enantiomer but not the other This time the two enantiomers do not react at the same rate as the reagent or catalyst if you prefer is the single enantiomer of a large peptide Not surprisingly the enzyme cleaves the amide of natural alanine and leaves the other alone The purification and separation first requires removal of the enzyme This is soluble in pH 8 buffer but acidification and heating denature the enzyme this is rather like what happens to egg white on heating and destroy its structure The solid material filtered off is this denatured enzyme The separation in ethanol works because the very polar amino acid is soluble only in water but the amide is soluble in ethanol Overheating the acid solution causes cyclization of the amide oxygen atom onto the carboxylic acid This reaction happens only because the formation of a fivemembered ring an azlactone These compounds are dreaded by chemists making peptides because they racemize easily by enolization the enol is achiral Solutions for Chapter 42 Organic chemistry of life 507 HN O OH OH H N O H HO OH H2O N O O H N O OH aromatic achiral enol PROBLEM 4 A patent discloses this method of making the antiAIDS drug d4T The first few stages involve differentiating the three hydroxyl groups of 5methyluridine as we show below Explain the reactions especially the stereochemistry at the position of the bromine atom O N HN OH HO HO O O 1 MsCl pyridine 2 NaOH 3 PhCO2Na 4 HBr O N HN Br MsO PhOCO O O Suggest how the synthesis might be completed O N HN Br MsO PhOCO O O O N HN HO O O Purpose of the problem A chance for you to explore nucleoside chemistry particularly the remarkable control the heterocyclic base can exert over the stereochemistry of the sugar Suggested solution There is a remarkable regio and stereochemical control in this sequence How are three OH groups converted into three different functional groups with retention of configuration The first step must be the formation of the trimesylate Then treatment with base brings the pyrimidine into play and allows replacement of one mesylate by participation through a fivemembered ring Now the weakly nucleophilic benzoate can replace the only primary mesylate and the participation process is brought to completion with HBr Opening the ring gives a bromide with double inversionthat is retention To complete the synthesis of the drug some sort of elimination is needed removing both Br and Ms in a syn fashion You might do this in a number of ways probably by metallation of the bromide and loss of mesylate It turns out that the twoelectron donor zinc does this job well Finally the benzoate protecting group must be removed There are many ways to do this but butylamine was found to work well PROBLEM 5 How are phenyl glycosides formed from phenols in nature or in the laboratory Why is the configuration of the glycoside not related to that of the original sugar Purpose of the problem Revision of the mechanism of acetal formation and the anomeric effect Suggested solution The hemiacetal gives a locally planar oxonium ion that can add the phenol from the top or bottom face The bottom face is preferred because of the anomeric effect and acetal formation is under thermodynamic control PROBLEM 6 Caustic soda NaOH was used to clean ovens and blocked drains Many commercial products for these jobs still contain NaOH Even concentrated sodium carbonate Na2CO3 does quite a good job How do these cleaners work Why is NaOH so dangerous to humans especially if it gets into the eye Purpose of the problem Relating the structure of fats to everyday things as well as to everyday chemical reactions Suggested solution The grease in ovens and blockages in drains are generally caused by hard fats that solidify there Fats are triesters of glycerol p 1148 of the textbook and are hydrolysed by strong base giving liquid glycerol and the watersoluble sodium salts of the acids 510 Solutions Manual to accompany Organic Chemistry R O O R O O R O O NaOH HO OH OH O R O liquid glycerol water soluble sodium salt water soluble Sodium hydroxide is dangerous to humans because it not only hydrolyses esters but attacks proteins It damages the skin and is particularly dangerous in the eyes as it quickly destroys the tissues there Strong bases are more dangerous to us than are strong acids though they are bad enough The sodium salts from fats as well as glycerol are used in soaps PROBLEM 7 Draw all the keto and enol forms of ascorbic acid the reduced form of vitamin C Why is the one shown here the most stable O HO OH H OH HO O ascorbic acid reduced form of vitamin C O HO OH H O O O oxidized form of vitamin C Purpose of the problem Revision of enols and an assessment of stability by conjugation Suggested solution There can be two keto forms with one carbonyl group and two keto or ester forms with two carbonyl groups O HO OH H OH HO O O HO OH H O HO O O HO OH H OH O O O HO OH H OH O OH Two forms have greater conjugation than the other two and the favoured form preserves the ester rather than a ketone and so has extra conjugation Solutions for Chapter 42 Organic chemistry of life 511 O HO OH H OH HO O O HO OH H OH HO O O HO OH H OH HO O O HO OH H OH HO O PROBLEM 8 The amino acid cyanoalanine is found in leguminous plants Lathyrus spp but not in proteins It is made in the plant from cysteine and cyanide by a twostep process catalysed by pyridoxal phosphate Suggest a mechanism We suggest you use the shorthand form of pyridoxal phosphate shown here CO2H NH2 CO2H NH2 pyridoxal CN N H CHO OH Me O P O O OH N H CHO CN SH pyridoxal phosphate shorthand Purpose of the problem Exploration of a new reaction in pyridoxal chemistry using pyridoxal itself rather than pyridoxamine Suggested solution The reaction starts with the formation of the usual imineenamine equilibrium but what looks like an SN2 displacement of SH by CN turns out to be an elimination followed by a conjugate addition Any attempt at an SN2 displacement would simply remove the proton from the SH group Notice that the pyridoxal is regenerated PROBLEM 9 Assign each of these natural products to a general class such as amino acid metabolite terpene polyketide explaining what makes you choose that class Then assign them to a more specific part of the class such as pyrrolidine alkaloid grandisol polyzonimime serotonin scytalone pelletierine Purpose of the problem Practice at the recognition needed to classify natural products Suggested solution Grandisol and polyzonimine have ten carbon atoms each with branched chains having methyl groups at the branchpoints They are terpenes and specifically monoterpenes You might also have said that polyzonimine is an alkaloid as it has a basic nitrogen Serotonin is an amino acid metabolite derived from tryptophan Scytalone has the characteristic unbranched chain and alternate oxygen atoms of a polyketide an aromatic pentaketide in fact Pelletierine is an alkaloid specifically a piperidine alkaloid They are also an insect pheromone grandisol a defence substance polyzonimine an important human metabolite serotonin a fungal metabolite scytalone and a toxic compound from hemlock pelletierine Solutions for Chapter 42 Organic chemistry of life 513 PROBLEM 10 The piperidine alkaloid pelletierine mentioned in problem 9 is made in nature from the amino acid lysine by pyridoxal chemistry Fill in the details from this outline H2N CO2H NH2 H lysine pyridoxal RNH2 is pyridoxamine N H NHR N H CoAS O O N H CO2H O N H O pelletierine Purpose of the problem A more thorough exploration of the biosynthesis of one group of alkaloids Suggested solution The first stage produces the usual pyridoxal imineenamine compound and decarboxylation gives a compound that can cyclize and give the cyclic iminium salt by loss of pyridoxamine H2N CO2H NH2 H RCHO pyridoxal H2N CO2H N H R CO2 NH2 N R H Enz N H NHR H Enz N H RNH2 Now the enol of acetyl CoA adds to the iminium salt to complete the skeleton of the piperidine alkaloids Hydrolysis and decarboxylation gives pelletierine PROBLEM 11 Aromatic polyketides are typically biosynthesized from linear ketoacids with a carboxylic acid terminus Suggest what polyketide starting material might be the precursor of orsellinic acid and how the cyclization might occur polyketide precursor orsellinic acid Purpose of the problem More detail on polyketide folding Suggested solution Looking at this problem as if it were a chemical synthesis we could disconnect orsellinic acid by aldol style chemistry But how are we to go further Those cis alkenes and alcohols are a problem This is easily resolved as the alkenes are enols and we need to replace them by the corresponding ketones We discover a linear polyketide derived from an acetate starter and three malonyl CoA units The only CC bond that needs to be made is the one See p 1162 of the textbook Solutions for Chapter 42 Organic chemistry of life 515 that closes the sixmembered ring Enolization then gives aromatic orsellinic acid PROBLEM 12 Chemists like to make model compounds to see whether their ideas about mechanisms in nature can be reproduced in simple organic compounds Natures reducing agent is NADPH and unlike NaBH4 it reduces stereopecifically p 1150 of the textbook A model for a proposed mechanism uses a much simpler molecule with a close resemblance to NADH Acylation and treatment with MgII causes stereospecific reduction of the remote ketone Suggest a mechanism for this stereochemical control How would you release the reduced product N OH Ph Me H Cl Ph O N O Ph Me H Ph O O Mg2 N O Ph Ph O OH H Purpose of the problem An example of a model compound to support mechanistic suggestions Suggested solution The ketone is too far away from the chiral centre for there to be any interaction across space The idea was that the side chain would bend backwards so that the benzene ring would sit on top of the pyridine ring and that this could happen with NADH too N O Ph Me H Ph O O Mg2 N Ph Me H O O Ph O Mg2 N Ph O Ph O OH H product 516 Solutions Manual to accompany Organic Chemistry This is a difficult problem but examination of the proposed mechanism should show you that binding to the magnesium holds the side chain over the pyridine ring Enzymatic reactions often use binding to metals to hold substrates in position Of course in this example the substrate is covalently bound to the reagent but simple ester exchange with MeO in MeOH releases it PROBLEM 13 Both humulene a flavouring substance in beer and caryophylene a component of the flavour of cloves are made in nature from farnesyl pyrophosphate Suggest detailed pathways How do the enzymes control which product will be formed humulene H H caryophyllene OPP farnesyl pyrophosphate Purpose of the problem Some serious terpene biosynthesis for you to unravel Suggested solution Judging from the number of carbon atoms 15 and the pattern of their methyl groups these closely related compounds are clearly sequiterpenes They can both be derived from the same intermediate by cyclization of farnesyl pyrophosphate without the need to isomerize an alkene The elevenmembered ring in humulene can accommodate three Ealkenes OPP humulene H Caryophyllene needs a second cyclization to give a fourmembered ring the stereochemistry is already there in the way that the molecule foldsand a proton must be lost The enzymes control the processes so that the starting material is held in the right shape and more subtly to make the wrong PROBLEM 14 This experiment aims to imitate the biosynthesis of terpenes A mixture of products results Draw a mechanism for the reaction To what extent is it biomimetic and what can the natural system do better Purpose of the problem Reminder of the weaknesses inherent in and the reassurance possible from biomimetic experiments Suggested solution The relatively weak leaving group acetate is lost from the allylic acetate with Lewis acid catalysis to give a stable allyl cation This couples with the other isopentenyl acetate in a way very similar to the natural process However what happens to the resulting cation is not well controlled Loss of each of the three marked protons gives a different product In the enzymatic reaction loss of the proton would probably be concerted with CC bond formation as a basic group such as an imidazole of histidine or a carboxylate anion would be in the right position to remove one of the protons selectively These experiments still give us confidence that the rather remarkable reactions proposed for the biosynthesis are feasible M Julia et al J Chem Res 1978 268 269