322as07_plq2_2_key - CHEMISTRY 322aL$335aL...

Unformatted text preview: ' CHEMISTRY 322aL$335aL ~November-297-EQOG —£%é§-i066- \ w mum. 73mm W" 07 Y ""'-\ K NAME 1.(12) Lab time 53,23“! 7m Hm”). TOTAL This test comprises this sheet If told to use ess than a certain number of words in and four numbered pages. an answer, D0 SO--deduction for excessive verbiage. Graded quizzes will be available from TAs and ri, Nov 30 and Dec 1, at (check-o labs and ours“ for TAs with no lab or ' hours Thu or Fri). If you initial —————— > , your quiz y Room, SGM 102, at 5 p.m., c 1. All quizzes 1 be thrown out Jan 26. EXAM is Thu, Dec 7, 8:00 — 10:00 a.m., rooms as all previous exams. -1- érl’L 1. (12) Dehydration of 4—methyl—2-pentanol, (CHEZHCH2CH(OH)CH3, produces the following methylpentenes (MPs) without m m rearrangement: 4—Me-l, 4—Me-2, 2—Me-1, and 2-Me-2. (a) (4) The 60% and 70% H2804 give different W results. Circle one choice in each pair separated by / below: In the 60% acid the total MP yield is lower /- and one gets a larger /. fraction of rearranged products. This is because the 60% acid is -/ less basic, and therefore average carbocation lifetime is longer @ (b) (6) (i) (2) Only two of the MPs can come directly from the initial carbocation. Yet substantial amounts of other MPs formed rapidly at < 100°. Tell what this implies about the magni— tude of A , the free energy of mm, for the required carbocation rearrangements by circling one choice: large negative small negative large positiv)e small positive RM {uck 79km? M K lead (ii) (4) The only P listed above which must form via a 3° carbocation is 2- Me- 1. Draw the 3° ion and the min product which foCrm from this ion. ClT/3 (PC1713 CC—SCfi/LCl/Lcl‘f [@CHJ-C‘?‘ Cljl'CIC7LLC137l (9 3" MM» WV-‘J’CA’W 7') (c) (2) Methylpentene samples re—distilled on a steam bath generally left a small pot residue of material, not dimers and higher polymers from methylpentenes. Name or draw this material. A -2- pzifié 2.(6) Note the gas chromatogram of the commercial sol "Skelly B". :HII- . "marl? ' “i #I‘E' A = I * —III — I l 1 ' IIIIIIIIIIIIIIIII ! II—In-II-‘IINIIIIIIIIIIIII II”- 2 IIII—III ' : IIIIIIII.“II"IIIIIIIIIIVAW£1IIIIIII“IIIIIIIIIIIIIIIII i - IIIIIIIIIIIII—II-II ' —“'_§' “In. 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' . -IlulIlAI—IIIIII—II IIIIIIIIIIIIIIIL‘III " ' “Egg:- :- IIIIg III-"mun 31:;1IIIII"IIIIIII‘I'IIIIIIIIIIIIII’ R‘IIIIIII 4 - l I; ' n III-lammIIIIIIII-IIIEIIDIIIHI‘; 'lIIIIIII—IIIIIIIIIIIIIIIIIII'IIUIIIIIVIIIII Ill-II III-IIIIIIIIIIIIIIIIIIIII IDHIIWJIIIII'IIII 'IIIII III. I. IlII'IIII ‘ MIIIIIIII IIIIIIIIIIIIIIK-E’AIIIIEIIIIIIHIIIII _ IIIIIIII IIIL‘KJII 'AIIIII ' AIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIMIIIII’lI-IIIIIIIIIIII—IIIIIII-‘II'AIIIIII _ ~ IIIIIIIIIIIIIIIIIIIIII IIII’III-IIIIIIGIII'IIII‘Im—“IIIIIIIIIIII-’40 LI~ 'AIIIIIII lIIIIIIIIIIIIIIIIIIIIIIIIIIZ ‘I—VAIIIII‘IUAIIIIII—III‘IIIIIIIIIII'AII W w-NIIIIIII ' IIIIIIIIIIIIIIIIIIIIIIIIII v—IIIIKWImI—IIIIIII II' III 9"! III-MI IIIIIIIII IIIIIIIIIIIII'JIIIIIIIIIIIIIIIIIIIIIIIIIIIIN-I "5:2: 22:; III. l-l _ :IIIII IIII IIIIIIIIIIIII'AIIIIIIIIIIIIIIIIIIIIIIIIISIFIIGIIIIIIIKIIII-I III. INII‘ 'lIl III-"IIIIIIIIII'AIIIIIII-IIIIIIIIIIIIIIII’AIIIK III-‘IIIII’IIIIIIII llll II-‘l II| III-IIIIIIIIIIIIl—I-IIIIIIIIIIIIIIIIIIIIIIIIIIIIAIIIIIImI-n———..-——- ——_. __--.:~u IIIIIIIIII‘I-IIII-IIII-IIII-III Ill-Ill..- —-I——II Ill-Ill: I-I -———- ml “— III “I: III.“Inn-IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII III-I I‘ll-I‘ll! IIIIIIIIIIIIIIIIIII‘IIIIIIII(III-IIIIII'l SIIIIIIHIIIIIIIIIII'IIIIIIIIII lIII IIIIIF ' IPIIIIIIIIII IIIIIIIIIIIIIIIIII . IIIIIIIIIIIIIVIIIIIIIIIII-IJIIIIII II III. IIIIIII?‘ ' IIIIIIIIIIIIIIIIIIIIIIIIII . IIIIIIIIIIIIUIIIIIIII'AIIIIIIIIIIIIIIIIIII IIIIIIII E: ' Awmhc 9L 1 (a)(2) Which peak is associated with the most volatile component? u er L km- Nmb {I‘M/ (b)(4) Recall that one estimates the areas of GC peaks by assuming they are triangular and calculating "Kb-h". Mark peak 4 to show how this is done: Draw a proper baseline, the height and Kb. Then calculate the area in "squarelets" (use h and Kb to the nearest % sguezelet.eggs). Also state how one determines Kb using <10 words. ‘ Q to (jawed/1‘ a4 74"” A9 wen/Mng‘tée ”$44” firm : 56 “V50 swam/45 [ta/f/Jfflf‘ “Mo New? =9 WH/uénffim—ilf‘ '3' fiJZ’? 3. (9) This ques and 4 on p 4 deal with the bromine—catalyzed isomer— ization of I, gig—X—CH=CH-x (X = COOMe) to the trans isomer, III. The Br—containing species, II—gauche and II—anti, dif— ferent conformations of x-gH-—$H—X, are intermediates. Br (a)(4) Consider a mechanism for I a III in which a propagation step is II-anti + I a III + II—gaughe, i.e , a mechanism in which 11- anti does not expel Br- but "hands" it directly to I. 1) Write the other propagation step(s) of this mechanism and circle the chain_garxying species. (2) Sum the steps, showing that they do add up to the overall reaction. _ <;<:L~. C'VC/k. 6W VL/ov- Jew M1“ = 3 (b)(2) In this prep, liquid dimethyl maleate (I) is converted to solid dimethyl fumarate (III). Maximum conversion of I a III requires enough Br2 that about 12 % by mass sglid Br—contain— ing by-products form. Tell why one might snags: to use less Br2 while leaving more unreacted I; use <15 words. 6 MM 8'7, Alma lef-ém' W'( (M :fi wr‘yanfl‘mfin {ZZ- c)(3) Exactly 80 mg Br2 (MW = 160) reacts with 10.0 mmol I (MW = 144). The yield of III is 60% based on I; all the Br2 reacts by addition to I. Calculate the mass LQLLQ of III to Br- containing product(s). 60% 4 {010 MM/ = 6.0 Mma/ : 6.0 MM] )0 LP“?— : 36+ “" (0364;? -4- /5l 2210 4. (4) Consider I a III as occurring in two steps. (1) I in soln a III as supercooled liq (scl) with.AG = 0; then (2) IIIscl a IIIsolid'_i;e" some III crystallizes, with AGxtln < 0. For simpliCity, assume mm°lsolute = 100'(xsolute)' e.g., XIII = 0.02 means 2 mmol III in soln. Calculate mmol solid III from 20 mmol I if rxn is run to equilibrium at the T—— (a) where XIII = 0.03. a {1: :. 0.03 J“! M‘i% QM wCWlk $ (b) where XIII = 0.06; after rxn is Qxfifi, mixture is cooled to T @ :whijl XIII to 0.2:. w: 10;!” 1% =7 2mm! Urn/:15" 45‘“ mm mm! DIM/r 6 wl MOI n (l n~Bu-Br —--—> g—Bu+ + Br‘ (2) n-Bu+ + I‘ -———> n-Bu-I Sum: n—Bu =r + I‘ -———> n—Bu-I + Br' +7 (a)(3) Set up the e- ilibrium ergression for th: um, and calculate Keq (2 sig fig , = e(—-AG/RT); RT = 0 6 kcal/mole. K . _ 7 _ Kfi= For-5&4: m e 656 = 2-5.40 f (b)(3) In acetone saturate- in both NaI -nd NaBr, take [I'] = 2,0 M and [Br'] = 3.0-1!' M. Plug thes- values into the above Ke expression, cal"late [n—BuI]/[n-BuB and state what this q ratio means -ybut the Keq value for the ‘verall rxn, n—Bu—Br + NaI -—-—- n—Bu—I + NaBr in acetone satd .. both salts. C :59 1?] F319 '10 £1 2- 5,“) -f a #190? "at RM. 351 “ 1.0 [“55" ' ...
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