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WSU CHEM 345 - CHEM 348 Test 2 Key

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CHEM 348 Test 2 Key – Spring 2010 – 100 points possible 1) 2) base t-Butyl acetate enolate anion OO Imidazole p-Toluenesulfonate anion Phenoxide anion Conjugate acid pKa 25 t-Butyl acetate OO pKa 5 Imidazolium cation pKa <0 pKa 10 32 points possible acid iso-Butyl alcohol pKa 15 OH Cyclopentadiene pKa 15 HH Ethylbutanoate pKa 25 Diisopropyl amine pKa 35 NH Conjugate base iso-Butoxide anion O Cyclopentadienide anion Ethylbutanoate enolate anion Diisopropyl amide anion N3) The barrier to rotation for compound 1 about the indicated C‐C bond is about 65 kcal/mol. Compound 2 (called calicene) has not been synthesized, but computational methods agree the inter‐ring rotation should be about 32 kcal/mol. Using good molecular orbital arguments, explain using words and pictures why calicene (2) has much lower predicted barrier to rotation about the indicated C‐C bond. In the calicene system, an alternate Lewis structure is shown below where the inter‐ring double bond is broken in favor of a charge separated species. If you consider each ring independently and use Frost circles, you can justify this resonance form as significant because each ring attains aromaticity. 4) When 2‐cyclobutylpropan‐2‐ol is treated with a solution of HBr in Et2O, 2‐bromo‐1,1‐dimethylcyclopentane is the resulting product. Write the reaction and propose a mechanism using curved arrows notation in each step. 10 points each5) 28 points possible 6) Propose a synthesis of (E)‐5‐heptene‐2‐ol using precursors that contain fewer than four (< 4) carbon atoms. Use any necessary organic or inorganic reagents; any reagents or precursors used may contain more than four carbon atoms if fewer than four of the carbon atoms from that reagent or precursor remain in your final product. 10 points possible OHNaBH4MeOHOOCO2Eta) aq. NaOHb) aq. acid,ΔCO2EtOa) NaH, DMFBrb)OHPBr3,pyor HBr, Et2ONaBH4MeOHOHH3CCHONaO Haq. EtOH(Aldol Cond)(Hydride Red.)(Sn2 Rxn)(Acetoaceti c Ester S ynthesis)(Hydride Red.)There are many reasonable methods, t wo are shownMethod 1Method 2OHa)Mgob)c) H3O+HOBrPBr3Nao,NH3(l)BrHOa) LDAb)c) n eutralizeOH 7) (S)‐1‐phenyl‐4,5‐dihydroxypentan ‐1 ‐one can be converted into (1S, 4S)‐1‐phenyl‐2,7‐dioxabicyclo[2.2.1]heptane in one simple step. Deduce a mechanism for this reaction. 10 points possible PhOHOHOHcat. TsOHPhH, refluxDean-StarkOPhOHOOPhH==24S7PhHHOHOOSSSS1SFormationofanacetalPhOHOHOHHOPhOHHHOHBH+OPhHOHH2OOPhOHHBH+-H+


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