322af07_plq2_1_key

322af07_plq2_1_key - CHEMISTRY 322aL/325aL I émcfib W...

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Unformatted text preview: CHEMISTRY 322aL/325aL I émcfib. W ‘9)? SECOND LAB QUIZ- BY Pigs-2‘ MAL Lab time T.A. This test comprises this sheet If told to use less than a certain number of words in and four numbered pages. an answer, DO SO-—deduction for excessive verbiage . PLEASE m; Graded qu'zzes gila be available from TAs Thu and b“ C ’ Fri, W at (check-out) labs and office hours ("special office hours" for TAs with no lab or regular office hours Thu or Fri). IE . . . a 1 . gaiLl—be_pue—ia—ehe—Seaéy—RoomT—SGM—lOST—atrir1rfin., “ . J; . .33 i i . 3 m; M is Thu, Dec(v\< 8:00 - 10:00 a.m.,—a'en—e-he—same -roeae—as‘ali—prevéeus_axames 47 £= It 1. (12) Dehydration of 4-methyl-2-pentanol, (CH3 CHCHZCH(OH)CH3, produces the following methylpentenes (MPs) without carbon m rearrangement: 4-Me-1, 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 lishorter) (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 W, for the required carbocation rearrangements by circling one choice: large negative small negative large positive small positive RMmch ‘ «4 «Mg. <15") (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 form from this ion. 3 C I @6643‘tc-3CI/p CM. Chi, @(bla-c=C[/-C({LC[7(3 @ (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. {M fl 4’MQ7‘fiy/fi'mfm/ L’— \ GARCéZ—cé/rgéchZ: M I y“, 4L (0%? «10%— -2- pziaé 2. (6) Note the gas chromatogram of the commercial sol "Skelly B". I y ' :l - u 1- i :I ‘ ; II...IIIIII..IIIIIIIIII - IHIIIII ' ' III-Ifi-I%=IIIIIIIIII“IIIII II—IIIIIIUIIHIIIIIIIIIIIII I..qu 1 III—II. III-IIIIIIIIHIIIIIIIIIIIIIIIIVAWK-YII ‘IIIIIIIIIIIIIIiIIIIIIII :II IIIIIIIIHIIIII. III-“IIIIIIIIII- living-4- ...IIIIIIIIIIIIIIIII ' III-III... IIIIIIII—IIIIIIII AIIHHCII IIIIIIIIIIIIIIIII... _ .I III... II... ' "II-“IIIIIIIIIIETAIIIIIIIIIIIIIIIIIIIIHII I 3 ' .IIIIIIIIIIII'IIIIIIIIIIIIIIII IIIIIII. “II-II... ‘ 1 ' :30; i m- In‘l-IIIII-I IIIIIII—IIIIII—I. “III-IIIImIIIIIII IIIHHIIIIIIIII. I... II... III-IIIIIIIIIIIIIIIII III-IIIIIIIIIIIII... IImIIIIIIIIIIIIIIIIIIIIIIIIIIII IIIII===II...III-IIIIIIIIIIIIIIIIIIIIIIIIIII—III—IIIIIIIIIIIIIIIIIIIIIIIIIIII II... II....I—IIIIIIIIIn-IIIIIIIIIIIIImnull—IIIIIIIIIIIIIIIIIIIIIII. ‘ III-IIIIIIIIIIIII-IIIII. III—IIInnIIIIIIIIIIUIII—II—IIIIIIIIIIIIIIIIII ‘ IIIIIIIIIIIIIIIIIIIIIII III... lag-III..."IIIIIII..I_IIIIIIIIIIIIIIII ‘r IIIIIIIIIIIIIIIIIIIIIII III“... III. IIHIIIIIHIIIIIIIIIIIIIIIIIIII'I'IIIII IIIIIIIIIIIIIIIIIIIIIII I. IIIIIIIIIIIIIIIIIIII'IIIIIIIIIIIIIIIIIIII III-III IIIIIIIIIIIIIIIIIIIIIIIIII. “II-IIIIII-IIIIIII IIIIIIIIIIIIIIIIIIII y"II-III III-IIIIIIIIIIIIIIIIIIIIIII WEIIIIIIIIIIIIIIIIIIII IIIIIIIII I TI... IIIIIIIIIIIIIIIIIIIIII II... 0 IIIIIIIIIIIIIIII—II IIIIIIIIIIIIIIILVIIII II==IIIII III-III... .II III-22.3.. I.’.‘I.a1...IIHIII.IIU IIIIIIIIIIIIIIII-‘IIIII I... emu-ulnflw UIINIIIIMII IIIIIUIIIIIIIIIIIIIIIIIIIIIII'G‘MI‘IIIIAIIII I" III-III Ink'JIIIIIII .) A. Ill-Illl‘41'lIIIIIlI—IIIIIIIII-IIIIIIII'J l 1‘ II .IIVIIIII ILII'II (III-“IIIIIIIIIIIIIIIIU IFDIIIL‘mIIIIIIIIIII—IIIIIII III. I. .7 .II'IIII .MIIIIIII-III...II...IIIIK'I'AIIII'E'IIIIIIHIIIIIIIIIIIIIIII. .III‘VAII'AIIIII :II..IIIIIIIIIII......=:IIIIIIIIIII"IE “II-"IIIIII‘III IIIIIIIIIIIIIII“ 'AIIIIII -l.....II.II.III.IIIIII IIII’III..."IIIuVIIIIIIIIIIII—“IIIIIIIIIIIInU b." 'AIIIIII. IIIIIIIIIIIIIIIIIIIIIIIIIIIZ I ‘I—VAIIIIIIIHAII’EIIIIIII‘IIIIIIIIIIII'AII I I NIIIII'AL‘ IIIIIIIIIIIIIIIIIIIIIIIIII'. ' VIII-IRWIYAIII‘IIIIII‘III II'AIII IIIQIIII l.- IIIIIIIIIII IIIIIIIIIIIIII’AIIIIII'IIIIIIIIIII'AIIII IIIIIIK‘III. II' 22:: :22. III! IHI\ IIIIIIIIIII IIIIII..IIIII'AIIIIIIIIIIIIIIIIIVAIIIII"VIIIIIIIBIIIIIIIKIXIIII III. I“. "I .uImIIIIIIIIII-IIIII'SIIIIIIIIIIIIIIIIII‘II-Il ‘I-I-I-D_‘III--V.Il III-I I... IK‘I AI. “nun-I'm.III-III...IlllIllllllllllIlIlAIllI-l-u-Iilll-u—‘_-uI--—- ——-- ---_:- III-“IIII'IIIIII-Ill I-III—lll III-II...- m .—____- mm". ____. _--- I... “y “- IIIII'IIIIIll-IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII III-IIIIIIIIIIIIII IIII IIIIII. IIIIIIIIIII-IIIIIIIIIIIIIIII‘III-III... IIIIIIIIIIICIHIIIIII'IIIIII III.‘ XIII III II'~ III-IIIIIII IIIIIIIIII‘IIIIIII IIIIIIIII-IIIIIIIIIIEII'IIIIIIIIJIIIIII .II'JIII IIIKII' ' I....IIIIIIIIIIIIIIIIIIIIII. | III-IIIIIIIIIIIIIIIIIIIIIIIIII IIIIIII .IIIIIAI @41th ~ or 1 (a) (2) Which peak is associated with the most volatile component? of km- Number 1- {l‘u/ (b) (4) Recall that one estimates the areas of GC peaks by assuming they are triangular and calculating “lab-h". Mark peak 4 to show how this is done: Draw a proper baseline, the height and 39.). Then calculate the area in "squarelets" (use h and 39b to the nearest 3,4 W) . Also state how one Wings )9.) using <10 words. ‘ Q “45"0 :2 {96, WA =7, M=+L \ Mi“? 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-cH-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-ganghe, i.e., a mechanism in which II-anLi does not expel Br- but "hands" it directly to I. 1) Write the other propagation step(s) of this mechanism and circle the chain;gar:ying species. (2) Sum the steps, showing that they do add up to the overall reaction. A . <;<ji~. c(vc/2 / ‘ x—i 1 F 13w *{or new mf‘ =4‘5 / (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 solid Br—contain- ing by-products form. Tell why one might choose to use less Br2 while leaving more unreacted I; use <15 words. 9410/ Lead 5'7, Alan gf-‘m' M (fl , :§ wriyun‘fi‘afiw 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 ratio of III to Br-containing product(s). 40% 4 10,0 «Ml = 6.0 mmol -4- #7 £740 4. (4) Consider I 9 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 Simplic1ty, assume mmolsolute = 100-(Xsolute), e.g., XIII = 0.02 means 2 mmol III in sgln. Calculate mmol solid 111 from 20 mmol I if rxn is run to equilibrium at the T—- (a) where XIII = 0.03. ‘5 K; :. 0.03 QM «in!!! 5 (b) where XIII = 0.06; after rxn is oxen, mixture is cooled to T @ which XIII to 0.02. n, W era? a: 1an 1% may :7 3mm! ZI/flfi. 43?“. Why? an...) ‘ l - = [2 WM 5. (6) Not:.<'e«t'-.‘Kc'e)(data:(4v l till-'0’ L g g Reaction AG 11593.1). .6 I (1) n—Bu-Br —---> n-Bu+ + Br' + 178 (2) n-Bu+ + I“ ----> n-Bu-I — 171 Sum: n-Bu-Br + I‘ ----> n-Bu-I + Br' +7 (a (3) Set up the equilibrium empression for the Sum, and calculate Keq (2 sig figs), = e(—-AG/RT); RT = 0.66 kcal/mole. 1 K7: Evin-1:35“? 2 6—7:;6 : L's-.10-:- fldJ I It? (b)(3) In acetone sat ated in both N, and NaBr, take [I'] - 2.0 M and [Br'] = 3.0-1=‘7 M. Pl ‘ these values into the above Ke expression, calcula - [n-z I]/[n-BuBr], and state what this ratio means about the (c. value for the overall rxn, n-Bu-Br + NaI --—-> n-Bu-I NaBr in acetone satd in both salts. I (:4 HQ $3.040", ‘ L 04‘ fi [‘07 [*wWJ" 1.0 — (“hid LL\ 4 V‘ (IL: Nqfir 3- r. (q ‘ m «m: l m W7”) ‘ K? ...
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This note was uploaded on 02/13/2008 for the course CHEM 322AL taught by Professor Jung during the Fall '07 term at USC.

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322af07_plq2_1_key - CHEMISTRY 322aL/325aL I émcfib W...

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