FQE2006A澳大利亚化学竞赛试题

FINALSELECTIONEXAMINATIONforthe2007AUSTRALIANCHEMISTRYOLYMPIADTEAMPARTA2006Pleasenotethatthisanswerbookwillbephotocopiedwhenreturnedandthensplitsothatanswersaresenttotheappropriatemarkers.Forthisreasonitisextremelyimportantthatyouobserveinstructions6to8.InstructionstoStudent1.Youareallowed10minutestoreadthispaper,and3hourstocompletethequestions.2.Youarenotpermittedtorefertobooksornotesbutyoumayuseanonprogrammableelectroniccalculator.3.Allquestionstobeattempted.Aguidefortimeallocationissuppliedatthebeginningofeachquestion.4.Aperiodictablewithatomicmassesandthevaluesofsomephysicalconstantsareprovidedonthefollowingpage.Dataaresupplied,wherenecessary,witheachquestion.5.Answersmustprovideclearlylaidoutworkingandsufficientexplanationtoshowhowyoureachedyourconclusions.6.Answersmustbewrittenintheblankspaceprovidedimmediatelybeloweachquestion.Roughworkingmustbeonthebacksofpages.Onlymaterialpresentedintheanswerboxeswillbeassessed.7.EnsurethatyournameiswrittenintheappropriateplaceonALLofthepages(eventhoseyoumayhaveleftblank)inthisexaminationbooklet.8.Useonlyblackorbluepenforyourwrittenanswers,pencilorothercolouredpensarenotacceptable.SupervisorDeclarationIcertifythatthefinalselectionexaminationwascarriedoutunderstrictexaminationconditionsandthatnoimproperactionsoccurredduringtheexaminationperiod.NameofExamSupervisor:(pleaseprint)……………………………………………………………..……Signed:………………………………………………………………Date:……………………………Pleaseusetheenclosedpre-addressedExpressPostEnvelopetoreturntheExamination.(MrRWSwitzer,ASOChemistryprogram,POBox589,MudgeerabaQLD4213).EXAMINATIONSSHOULDBEEXPRESSPOSTEDONWEDNESDAY14thMARCHSOTHATTHEYARERECEIVEDBYFRIDAY17thMARCH2006.FinalPaper—PartAName:Page2of33CONSTANTSspeedoflightinvacuumc=2.998×108ms–1Planck’sconstanth=6.626×10–34Jselementarychargee=1.602×10–19Celectronmassme=9.109×10–31kgAvogadroconstantNA=6.022×1023mol–1FaradayconstantF=9.6485×104Cmol–1idealgasconstantR=8.3145JK–1mol–1Boltzmannconstantk=1.381×10–23JK–1atomicmassunitu=1.661×10-27kgpiπ=3.14159Ångström1Å=10–10mFinalPaper—PartAName:Page3of33Question1(36minutes)(a)Data:Ksp(Cd(OH)2)=7.20×10–15Ksp(Mg(OH)2)=5.61×10–12Ksp(Ca(OH)2)=5.02×10–6Thetablebelowsummariestheobservationsmadewhenonedropofamonoproticbasesolutionisaddedtoonedropofametalcationsolution.Allsolutionsareataconcentrationof0.100Mwithrespecttothereactingspeciesandtheprecipitatesformedaremetalhydroxides.Assumeeachdrophasthesamevolume.BaseMetalCationH2AsO3–CH3CO2–Cd2+PrecipitateformsPrecipitateformsMg2+PrecipitateformsNoprecipitateCa2+NoprecipitateNoprecipitate(i)Whichisthestrongerbase,H2AsO3–orCH3CO2–?(ii)GiventhatitcancausetheprecipitationofCd(OH)2,whatisthehighestpossiblevalueofpKbfortheethanoateion,CH3CO2–?Remembertotakeintoaccountthechangeinvolume.(iii)WhatisthelowestpossiblevalueofpKbfortheethanoateiongiventhatitcannotcausetheprecipitationofmagnesiumhydroxide?Morespaceisavailableonthenextpage.FinalPaper—PartAName:Page4of33(b)Onedropof0.200MAgNO3isaddedtoonedropof0.200MNH3toform[AgNH3]+and[Ag(NH3)2]+.ThefinalconcentrationsoffreeAg+andfreeNH3inthesolutionwerefoundtobe3.81×10–2Mand2.50×10–4Mrespectively.(IgnorethereactionofNH3withwaterforthisquestion).Assumeeachdrophasthesamevolume.(i)WriteanexpressionforthetotalAg(massbalance)intheresultingsolution.(ii)WriteanexpressionforthetotalNH3intheresultingsolution.(iii)UsingthedataprovidedandyouranswersabovecalculateK1,K2andthusKstabfortheformationofsilveramminecomplexes.FinalPaper—PartAName:Page5of33(c)Data:R=8.3145JK–1mol–1F=96485Cmol–1T=298KAsolutioncontainingionsofanunknownmetal,M2+,isaddeddropwisetoonedropofanacidifiedKMnO4solution.Intheoriginalsolutions[MnO4–]=[M2+].ThemetalionbecomesoxidizedtoMn+andMnO4–isreducedtoMn2+.AfterthreedropsareaddedthesolutionturnscolourlessindicatingalltheMnO4–hasbeenreduced.Assumeeachdrophasthesamevolume.(i)Givenitisaninteger(1,2,3…),whatisthechargeontheoxidizedmetalion,Mn+?(ii)Writehalfequationsandabalancedfullequationforthereaction.Thereductionpotentialoftheresultingsolutionisfoundtobe+0.17V.(iii)Whatisthestandardreductionpotentialforthehalf-reactionMn++(n-2)e–M2+?FinalPaper—PartAName:Page6of33Question2(16minutes)DATA:Gasconstant:R=8.3145Jmol–1K–1=8.3145kPaLmol–1K–1Pressure:760mmHg=1atm=1.013×105Pa(a)Ethanaldecomposesaccordingtothree-halvesorderkinetics,withanactivationenergyof199.2kJmol–1.Thetablebelowcompares(ataparticulartemperatureX)therateofreactiontoinitialconcentrationofethanal:P(CH3CHO)(mmHg)InitialRate(mmHgs–1)1.00×102-2.24×10–21.58×1024.47×10–22.51×1028.91×10–25.01×1022.51×10–1Verifythatthedatafollowsthree-halvesorderkinetics.(b)Itisfoundthat,attemperatureX,anincreaseintemperatureof40Kwouldincreasetherateofreactionbyafactorof4.98.CalculatethevalueofXfromthegivendata.X=FinalPaper—PartAName:Page7of33(c)Usingyouranswerinquestion2,calculatetherateconstantk(inSIunits,includingconcentrationunits)forthedecompositionattemperatureX.(Ifyoudidnotgetananswertoquestion2,assumeX=700K.)k=(d)Derivetheintegratedratelawforthisreaction.FinalPaper—PartAName:Page8of33(e)Hence,fillinthedataforthefollowingexperimentattemperatureX.Time(min)01.002.005.00P(CH3CHO)(mmHg)1.00×102FinalPaper—PartAName:Page9of33Question3(16minutes)(a)Drawamolecularorbitalenergydiagramforeachofthefollowing.Labeleachmolecularorbitalwithitsname,andshowelectronoccupanciesineachmolecule.(i)B2(ii)F2(iii)CNFinalPaper—PartAName:Page10of33(b)Completethefollowingtable:MoleculeLi2Be2B2C2N2O2F2Ne2HOMOσ!*2sBondorder0Paramagnetic?No(c)Drawthe90%inclusionsurfaceforeachmolecularorbitalofCN(oneexampleofeachdegeneratesetofpiorbitalswillsuffice).FinalPaper—PartAName:Page11of33(d)CNmoleculehasanabsorptionbandat9000cm–1(λ=1.11×10–6m).GiventhattheCN–iondoesnothaveanabsorptionbandinthisregionofthespectrum,suggestanelectronictransitionintheCNmoleculeresponsibleforthisabsorption.(e)Be2hasabondorderofzero,yetitisobservedtoexist(withbonddissociationenergyof59kJmol–1,comparedwithLi2whichhasbondenergyof101kJmol–1).Explainthisobservationusingyourknowledgeofmolecularorbitaltheory.FinalPaper—PartAName:Page12of33Question4(24minutes)(a)DrawaLewisstructureforeachofthefollowingspecies.DeterminethemoleculargeometryofeachspeciesusingtheVSEPRmodel.Whereappropriate,alsodrawanycontributingstructures(resonancehybrids).Useformalchargetodecidewhichhybridisthemostlikelycontributor.(i)PCl3(ii)SF2(iii)OCN–(iv)IF4–(v)HPO32–FinalPaper—PartAName:Page13of33(b)ThemoleculeSOF4isafiveelectron-pairsystemwithamoleculargeometrybasedonatrigonalbipyramid.Indicateonthediagrambelowtheaxialandequatorialpositions.Discussanydifferencesinaxialandequatorialbondlengthsaswellasanydeviationsinbondanglesfrom180º,120ºand90º.IndicatethehybridstateofthecentralatominSOF4.FSFOFFSOF4:FSFOFF(c)FortheGroupVhydrides(EH3)whereEisnitrogen,phosphorus,arsenicorantimony,usetheVSEPRmodeltoexplainwhythebondangle(H–E–H)becomessmallerwhenmovingdownthegroup.FinalPaper—PartAName:Page14of33(d)(i)Aceticacid(CH3COOH)isaweakacidwhichpartiallydissociatesinawater-solution.Basedonthepartialdissociationofaceticacidwouldyouexpectthevan’tHofffactorforaceticacidinthissolutiontobelessthan1,equalto1orgreaterthan1?Explainyouranswer.(ii)Fromafreezingpointdepressionexperimentusinganon-polarsolvent,thevan’tHofffactorforaceticacidisfoundtobeverycloseto0.5.ExplainthisobservationusingLewisstructurestojustifyyouranswer.(e)A1.00molkg–1water-solutionofHFfreezesat–1.91ºC.Accordingtothesedata,whatisthepercentionisationofHFinthissolution?[Kfforwater=1.86ºCkgmol–1].FinalPaper—PartAName:Page15of33Question5(16minutes)Pleaseseepage19forstructuresofaminoacids.(a)Nucleicacidsarethebuildingblocksoflivingorganisms.ThemostcommonnucleicacidsareRNAandDNA.Theseareessentiallypolymersconstructedfromnucleotides.(i)Whatisanucleotide?(ii)ForthefollowingfourbasesofDNA,labelallH-donorsas“D”andH-acceptorsas“A”.NNNNHNH2NNHNNHONH2NHNNH2ONHNHOOH3CAdenineCytosineGuanineThymine(iii)Referringtothestructuresaboveandtothestructureofdeoxyribose,drawthemolecularstructureofdeoxyguanosinediphosphate.OHOOHOHDeoxyribose(iv)DrawthestructureoftheDNAstrandthatiscomplementaryto5’GCAGTGCAG3’.FinalPaper—PartAName:Page16of33(v)NametwoimportantfactorsthatmightdisruptbondinginDNA.(b)Theideaofisoelectricpoint(pI)underliesthetechniqueofgelelectrophoresis,whichisfrequentlyemployedintheseparationofaminoacidsandproteins.(i)WhatispI?(ii)CalculatethepIofglutamicacidgivenitspKas.pKa(α-COOH)=2.1pKa(sidechainCOOH)=4.3pKa(α-NH3+)=9.5FinalPaper—PartAName:Page17of33(iii)Drawthestructureofglutamicacid’szwitterionform.Edmandegradationisafrequentlyusedmethodforsequencingaminoacidsinpeptides.However,amajordrawbacktothistechniqueisthatthepeptidesbeingsequencedinthismannercannothavemorethan50to60residues.(iv)Whydoyouthinkthisisthecase?Acertainproteincomprisingalphahelicesandabetapleatedsheetshasundergonedegradationtoyieldseveralpeptideseachwithfewerthan50residues.Theshortestpeptide,whichwerefertoasA,hasundergonefurthercleavagebytrypsinandchymotrypsintoyieldthefollowingfragments:1.Ser-Glu-Pro-Arg2.Ala-Gln-Val3.His-Pro-Ile-Lys4.Gly-Phe(v)Whatlevelofproteinstructurereferstoalphahelicesandbetapleatedsheets?(vi)Wheredotrypsinandchymotrypsincleaverespectively?FinalPaper—PartAName:Page18of33(vii)DetermineallpossibleaminoacidsequencesforpeptideAusingtheaboveinformation.(Youmayuseabbreviations.)(viii)Drawthestructureofpeptidefragment4fromabove,labellingtheNandCterminusaswellasthepeptidebond.What’sspecialaboutthepeptidebond?FinalPaper—PartAName:Page19of33FinalPaper—PartAName:Page20of33Question6(20minutes)(a)Namethefollowingcoordinationcompounds:(i)K2[CuCl4](ii)NH3PdSCNSCNNH3(iii)[Cr(ONO)2(en)2]NO2(b)Foreachofthefollowingcomplexions,drawalabelledd-orbitalsplittingdiagram,andhencecalculatethecrystalfieldstabilisationenergyandspin-onlymagneticmoment.(i)[Ti(NH3)6]3+FinalPaper—PartAName:Page21of33(ii)[NiBr4]2–(iii)[FeF6]3–(iv)[Co(CN)6]4–FinalPaper—PartAName:Page22of33(c)Sketchallstereoisomersofthecomplexionsinthefollowingcompounds:(i)[Cr(SCN)3(H2NCH2CH2NHCH2CH2NH2)](ii)[Co(meso-2,3-diaminobutane)2Cl2]Br(Donotconsiderstereoisomersoftheligands;thereisnoneedtoconsidercomplexesof(R,R)-2,3-diaminobutane,forexample.)FinalPaper—PartAName:Page23of33(d)Nameaformofstructuralisomerismwhichisexhibitedbyoneofthecompoundsin(c).Giveanexampleofthissortofisomeroftherelevantcompound.FinalPaper—PartAName:Page24of33Question7(20minutes)Dataforallparts:R=8.3145JK–1mol–1;0°C=273.15K(a)Glycogenphosphorylasecatalysesthereaction(glycogen)n+Pi(glycogen)n–1+glucose-1-phosphateΔG0=+3.05kJmol–1.Inmusclecellsat310K,theconcentrationsofthesespeciesareasfollows:Concentration(10–3molL–1)Pi10glucose-1-phosphate0.003Doesglycogenphosphorylasecausethesynthesisorthedegradationofglycogeninmuscle?Explain.FinalPaper—PartAName:Page25of33(b)WhitephosphorusburnsreadilyinairwithabrilliantwhiteflametogiveP4O10.Usethefollowingdatatoestimatethetemperatureoftheflameinsurroundingsat298K.Takeairtobea1:4mixtureofoxygenandnitrogengases.Assumethatallspeciesentertheflameatambienttemperature,298K,andleavetheflameattheflametemperature.(Enthalpyvaluesareat298K;youmayassumeheatcapacitiesareindependentoftemperature.)ΔfH0(kJmol–1)Cp(JK–1mol–1)P(white,s)24.7O2(g)29.4P4O10(s)–2984212N2(g)29.1FinalPaper—PartAName:Page26of33(c)(i)Galliumisunusualamongmetalsforitsverylownormalmeltingpoint(i.e.,at1.00bar)of29.78°C.Athighpressures,thisdropsstillfurther;forinstance,at100.00barthemeltingpointdropsto29.58°C.EstimateΔfusU0,ΔfusH0andΔfusS0forgalliumatitsnormalmeltingpoint,giventhedensitiesbelow.ρ(gcm–3)Ga(s)5.904Ga(l)6.095FinalPaper—PartAName:Page27of33(ii)Thefullreasonwhygalliummeltsatareducedtemperatureunderhighpressureinvolvesboththefirstandsecondlawsofthermodynamics.Fillintheblankswithappropriatewords,numbersorsymbols,accordingtothekeybelow,tocompletethefollowingexplanationofthisphenomenon.A=athermodynamicquantity(e.g.,ΔfG,H,T,…)B=“greater”or“less”C=anumberormathematicalformula(e.g.,2.0J,U+PV,…)D=aworddescribingathermodynamicprocess(e.g.,“isothermal”,“spontaneous”)Because<0,wisathighpressurethanatatmosphericpressure.Sinceisconstant,itfollowsthatqmustbethanatatmosphericpressure(althoughitisstillthanzero).Nowingeneral,ΔSsurr=.However,atthemeltingpoint,ΔSuniversemustbeequaltobecauseatthistemperaturemeltingis;soΔSsurr=,whichisindependentofpressure.ThusthemeltingtemperatureTfus=,whichissmallerathighpressurethanatatmosphericpressure.DCCCCBBABAFinalPaper—PartAName:Page28of33OrganicSection(32minutes)Question8(8minutes)(S)-1-bromo-1,2-diphenylethanemayundergoeliminationviaanE2mechanism.BrKOtBu(a)Providethestructuresofthepossibleeliminationproducts.(b)Giveareasonwhypotassiumtert-butoxide(KOtBu)isusedinpreferencetosodiummethoxide(NaOMe)toeffecttheelimination.(c)Provideamechanismforthereaction.Youneednotincludetransitionstates.FinalPaper—PartAName:Page29of33(d)Predictthemajorproductformedbythereaction,usingappropriatediagramstorationaliseyourchoice.Question9(7minutes)Considerthefollowingreactionscheme:BrcompoundA(anylide)1.PPh3!,CH3CN(solvent)2.NaOH(a)Classifyanddrawthemostprobablemechanismforthefirstreactionstep.Giveabriefreasonforyourchoiceofreactionmechanism.FinalPaper—PartAName:Page30of33(b)GivethestructureoftheylideproductA.Reactionof3-methylbut-2-en-1-olwithhydrochloricacidaffordedcompoundB,whichwassubsequentlytreatedwithpyridiniumchlorochromatetoformC.Finally,CreactedwithcompoundAabovetogiveD.(c)GivethestructuresofcompoundsB–D,clearlyshowinganystereochemicalfeatures.FinalPaper—PartAName:Page31of33Question10(8minutes)(a)Explain,usingappropriatediagrams,whymonobrominationofnitrosobenzeneoccursinthemetapositioninpreferencetotheparaposition.ONnitrosobenzene(b)Proposeasynthesisform-chloroanilinestartingfrombenzene.FinalPaper—PartAName:Page32of33Question11(9minutes)TrihexyphenidylisadrugusedtotreatthesymptomsofParkinson’sdisease.Synthesisoftrihexyphenidylmayproceedasfollows.PiperidinereactswithformaldehydetoformcompoundE.NHOCH2+CH2N+piperidineformaldehydecompoundE(a)IdentifytheelectrophilicsiteincompoundEthatismostsusceptibletonucleophilicattack.CH2N+TreatmentofaceticacidwithSOCl2resultsintheformationofcompoundF,whichsubsequentlyreactswithbenzeneinthepresenceofAlCl3togiveG.CompoundGproducesapaleyellowprecipitatewhentreatedwithI2/NaOH.TreatmentofGwithLDAfollowedbysubsequentreactionwithEyieldsH(C14H19NO),atertiaryamine.CompoundHgivesacolouredprecipitateuponadditionof2,4-dinitrophenylhydrazine.(b)ProvidethestructuresofcompoundsF–H.FinalPaper—PartAName:Page33of33TreatmentofcyclohexanolwithPBr3producescompoundI,whichreactswithMgindryethertoformcompoundJ.(c)ProvidethestructuresofcompoundsIandJ.(d)WhymusttheconversionfromItoJbeperformedinadrysolvent?Finally,compoundJreactswithcompoundH,leadingtotheformationoftrihexylphenidylfollowingaqueousworkup.Trihexyphenidyldoesnotdecolourisedilutepotassiumpermanganatesolution.(e)Providethestructureoftrihexylphenidyl.