Bertozzi_Chem3A_Midterm2_1997

Bertozzi_Chem3A_Midterm2_1997 - 1[ED points Several pairs...

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Unformatted text preview: 1. [ED points] Several pairs of molecules are shown below. {ii} For each pair of molecules, 'ndioate wh h r ‘ tiome enantiorners diastereomers structural isomers identical enantiorners diastereorners structural isomers identical enanticmers diasterecmers structural isemers identical enanticmers diasterecrners structural iscmers identical erranticn'rers diasterecmers structural isomers identical 2. [1D points] Wig among saoh sst of structures below. a. + + b. GHSSH CH3IDH HaclflH rho/15H Hgo HES H2819 HEN giggle thg strongest Es among each set of structures below. I “H. 0 I1 {13 6:3 I}! HaCHCHE H H Hmi Mi Groin. CH i [55 points] Add the i ' ' start I r . Be sure to clearfy indicate sterecchemistry where relevant. The workup step is assumed where necessary. DO NOT include workup steps in any of the given boxes! a. 1. PBra 0” 2. Maria —I——h- optically active [circle one]? yes no I]. 1. i 2* E ———h— \i CHacH. o =c cpticalty active {circle one]? yes no 1. CHEMgBr 2. NaH 3. GHacl —h— NaOH 1. GHaLi 2. H2304, a 1 equiv NaCN D30 CH3 1:3 01 3-D 0’ 1. 1 equiv. GHalJ 2. 1 equiv. OH I HDMNHz 3 k. Br NaNHz Illh- Milt}! —I- "e Br C1UH1TOBr 4. [40 points] For each pair of reactions below.“ act' h t vr'l ta ta w '. WW '1 = 1;! I-_ no 4.-|;.l: 1". 1- .Nooreo'rtwiiibe given for a correct answer in the first part of the question with an inoonect reason in the second part NaSH,CH DH El NE —3-— N5” + NaBr D NB. NESEH, CH 30H N33 H 4- “Ear A. The (3—5 bond ls stronger than the G—Se bond B. H3? is better solvated than HS- C. H5? is more polarizahle than HS— A, I- is better solvated by protic solvents than by aprotic solvents B. 3N2 reactions proceed taster in polar sotvents than in nonpolar solvents C. I- is a weaker base than Cl- Ear H D U U0“ —n- + HBI‘ Bi' DH mo” “2° ”U + The nuciecphiie can attack trom either face oi the carbocaticn Carbocations are stabilized by resonance delocalization Inductive effects are distance-dependent 0' Nana “3 ':| _. + mm c: NaN i:i dfl _.3_... 6N3 + NaC] fi—Branohing reduces the rate of the 5:42 reaction Secondary substrates react more SIM}! than primaryr substrates in the ma reaction Primary carbocations are less stable than secondaryr carbocations A. G. it Has o if D karma —'— ”his + "are + CH30H + NeBr i—i" H GHgflNa —.. + CHaCiH + NaBr ”II-{E Inn"- The leaving group is involved in the rate-determining transition state Axial substituents engage in untavoraole steric interactions The base removes a proton anti to the leaving group if I:' o—i-oHa F36 0 «mo-{1 + H? _CH3 0 CFs E A. E C. Easioitv and nucleot'iuhilioityr correlate within a row of the periodic table Inductive efieots are distance dependent Weak nuoleophiles favor the 3M1 reaction 10 A. B. C. CHaflH Cl Br —... OCH; 4- HEr Br GHaDH OCHa D — + Hm The rate-determining step is unimdieeuiar Garbeeatiens are stabilized by hypereenjugatien Carbeeafiens are sp2 hybridized at the sits at positive charge H20. a El . _. + Weaker bases are better leaving groups The nusleephile is net invetved in the rate-determining step |Bhleride ion is better selveted than triflate ion 11 5. [3D points] PmfidEa-:.a =-.. w =.- Lt::' Ill-1 W {i e., e'lectren-puehing' ] Keep In mind that ea; Em should contain no more than fltree arrows a. .. PM Hscffikc’h * H202 —"' Hats” \ C”: + ”20 CH3 CH; 12 1. CHEI 2. CHaoH. a -—I- + GHasfiHa 13 E. [45 points] P i ' nthetic re a tram startin materi | to r d . Show your synthesis in a step-wise fashicn in the forwam direction. Be sure to shew each intermediate termed after each of your synthetic steps. Ycu dc not need tc shcw "rrcrkup' steps in your scheme -— they will he assumed. Ycu may use any additional crganic cr inorganic reagents in ycur scheme. {It may help to wcrk backward cn scratch paper betcre writing ycur answer in the fcnyeuj directicn in the space belcw.) 14 HO\/\/Br __________ a. HSWY ‘IE ...
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