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by: Columbus Torphy


Marketplace > Iowa State University > Chemistry > CHEM 537 > PHYS ORGANIC CHEM I
Columbus Torphy
GPA 3.63


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Class Notes
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This 7 page Class Notes was uploaded by Columbus Torphy on Sunday September 27, 2015. The Class Notes belongs to CHEM 537 at Iowa State University taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/214549/chem-537-iowa-state-university in Chemistry at Iowa State University.




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Date Created: 09/27/15
CHEM 537 SECOND EXAM NOVEMBER 14 2007 0 NAME ANSWER KEY SHOW ALL YOUR WORK PARTIAL CREDIT WILL BE AWARDED THE EXAM 1S 7 PAGES NO NOTES OR BOOKS ARE ALLOWEDII YOU MAY USE A SIMPLE CALCULATOR AND MOLECULAR MODELS R 1987 CALKMOL h PLANCK S CONSTANT 6626 x 103934 m2 kg 5 kb BOLTZMANN S CONSTANT 138 x 103923 m2 kg 5392 K391 High 98 Low 51 Average 78 1 6 points Rank the following seven compounds from most acidic to least acidic para nitrobenzoic acid acetic acid tert butanol methanol phenol HI HCl HIgtHClgt para nitrobenzoic acid gtacetic acidgtphenolgtmethanolgt tert butanol 2a 2 points Are the following reactions stereoselective Yes X No 2b 2 points Is the bromination stereospecific Yes X No Br Br Br2 H C02H H02C H H020 COZH racemlc Maleic acid H cozH H020 H Br Br Br CO H g 2 Bra H020 H HO C gt meso 2 H COZH Fumaric acid Br See problem 622A of Anslyn and Dougherty s text 3 1 point Can diastereotopic groups be differentiated by chiral reagents Xyes See page 319 of Anslyn and Dougherty s text no Pl FA gF GR ADP INFORMATION ON THF R ACK OF Tng PAGF l CHEM 537 SECOND EXAM NOVEMBER 14 2007 NAME ANSWER KEY 4 Consider the following compounds Osswald PWurthner F J Am Chem Soc 2007 ASAP 2430 X Y R 1 3 MeOPhO 3MeOPhO 26IF r2Ph 2 F F 261Pr2Ph 3 Br H 26IPr2Ph 4 1 Cl Ph 5 Br Br 4a 4 points What is the stereochemical relationship between the two compounds as drawn atropisomers 1 point for vaguer term conformationalrotational isomers or to mention that they are enantiomers 4b 3 points What stereochemical descriptorsnomenclature are used to describe the above compounds M and P helical descriptors 4c 4 points The authors made the above compounds 15 in order to discover compounds in which the above interconversion barrier is greater than 22 kcalmol Why would this be desirable The compounds would then not be able to interconvert at ambient temperature and individual isomers could be isolated and studied 4d 5 points The halflife of the conversion of compound 4 from the lefthand side of the reaction to the right hand side at 303 K was 255 minutes What is the reaction rate per second ln2half life rate 069315300 sec 45 X 10395 per second Pl FAQF GRAD lNFOPhATTON 0N THF RAf K 0F Tl UQ PAGF CHEM 537 SECONDEXAM Nowmm 14 2007 NAME ANSWER KEY 4e 5 pmn39s The eehveheheh wampuurd 5 going fmm me le hard me thhe when m the nghthand me was studmd hyehEyhhe plot 5 se enhe Eynngequannnvahd m y at why hm Yes because he memememen ehsy shes eh a band when em Lheefehe ehe wuqu expect a ehe ehep pmcess 4f 4 palms The mmxssxmcuef cmmkappals D znsetas are mehe Eynng equa un Hewevex these lessamhzrs muk Iappam he u 5 Why he he valxd 7 The pmbabzhtynf39le uehsmehscaue mumsfurm mm m2 enhe 011121 suemesemex us an an 4e 5 palms seed eh the shave hhfemheh em gweh the Eynng eqlaunn below what he the expemed AG Dfanuvaunn Ad In ImaJmul enhe eehveheheh Bf compuurd e from the lemand m me hehehehame enhe equenehezana K7 meme values eh me exam esasxml mugexuxwgwsm axxxunm hee x muame AG e 1 927 eeJthemma 1o mu 4 x m H e 23 heeymex 4h 4 palms when yeuexpeeuhe Ahhemue atmva unenexgym he campalable In me AG eeseu eea alum27 Why at why hem Eecause Lb xhexmhwxemh us a uni sEp prmess Lb mmmscupm and mmmscupm me cms39an39s mu he deanal pugs om mmnoNon mkovms PAGE 1 x CHEM 537 SECOND EXAM 0 NOVEMBER 14 2007 0 NAME ANSWER KEY 5a 3 points Draw a plot of the starting concentration of a reactant B versus the observed rate that shows saturation in the reactant at higher concentrations of B Blo 5b 4 points Based on your graph above why would one want to run a reaction at high initial concentrations of the reactant B At high concentrations of B the observed rate becomes zero order in B and therefore the rate law becomes simplified pl FASF GRADE TN39FOPMATTON ONTH39F RACK OF THIS PAGE l 6c 3 points CHEM 537 SECOND EXAM NOVEMBER 14 2007 NAME ANSWER KEY he of l 6 39 39 n a t hiosynthesis of vitamin B J Am Chem Soc 2007 129 48604861 An analysis oft 39 T h 39 39 Com anson he V L 4 displacement of henzyl chloiide derivaives 9 and to the ieadion catalyzed by TPV nthase h 39 l Vmus up m M I the mndel 1 enzymaic reactions circles 39 and the 8 a Q O 7 B E t m aw U 1 as a his 6 6a 7 points 39 L value tell you about the mechanism of the ieadion A r bar L L m up 6b 4 points How are sigma yaluescales used for Hammett plots de ned henzoic acid See page 445 of Anslyn and Dougherty39s text for fuithet discussion U reaaion mechanism A change of mechanism occuis as suhstituents are changed PLEA SF mum lNFnRMATlnN 0N quotVHF RA CK nFTlils DA m l CHEM 537 SECOND EXAM I NOVEMBER 14 2007 I NAME ANSWER KEY 7 The following reaction has a kinetic isotope effect of we 54 Lafmnce Gorelsky Fagnou J Am Chem S05 2007 ASAP won 13 mowu Clof PCy3HBF4 5 mm 0 8r H Ease Mame 3L 1 Mesumene 135 7a 1 point Is am a normal x or inverse isowpe e 2 ffect 7b 1 point Is mas aprimaryx or secondary isotope effect 70 4 points What does due isotope effect indicate about due reaction mechanism n A r 7d Spoints 39 H quot L L A 39 39 39 u uh 1 39 u 1uhm1w 8 8 points The three pigs for citric acid are 31 47 and 54 Explain why the hm pk HOZC Y wZH HO COZH See problem 515 of Anslyn and Dougherty s next P FASF GRADF VNFORMA39HON ON THF RANK OF THVS PAGF i CHEM 537 SECOND EXAM NOVEMBER 14 2007 0 NAME ANSWER KEY 9 2 points What is the Curtin Hammett Principle The Curtin Hammett Principle states that the ratio of products is determined by the relative heights of the highest energy barriers leading to the different products and is not significantly influenced by the relative energies of any isomers conformers or intermediates prior to the highest energy transition states 10 2 points A common method for intermediate identification is trapping of the intermediate with an added reagent To be effective what key quality must a trapping reagent possess A faster reaction time with the proposed reactive intermediate to compete with the standard reaction pathway 11 9 points For each structure below identify whether the two ethyl groups are homotopic H enantiotopic E diastereotopic D or constitutionally heterotopic CH on the time scale where ring conversion is very slow Et Et Mm aw Erma D D H See the discussion for the related problem 632 of Anslyn and Dougherty s text Pl FAR GRAD lNF RhATION ON THE RAFK OF THIS PAGE


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