322as09_plq2_2_key

322as09_plq2_2_key - CHEMISTRY 322aL ‘ W MSG-G—...

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Unformatted text preview: CHEMISTRY 322aL ‘ W MSG-G— pracl‘f M SECOND W QUIZ .300? NAME 1; Lab time T.A. , 4. (11% / TOTAL ,"‘(40) 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. labs and fic ri). If u in ~ your quiszill e 0:00 a mi: i the s ' ’Ee go to SGM 12 . \ _l_ 1. (l2) Dehydration of 4-methyl~2-pentanol, (CH3)2CHCH2CH(OH)CH3, produces the following methylpentenes (MPs) without carhgn aggletgn rearrangement: 4‘Me-l, 4—Me—2, 2—Me—1, and 2—Me-2. (a)(3) Whether 60% and 70% H2304 was used affected the MP isomer {agig and the total yield. Circle one choice in each pair separated by / below: ,l~ In the 70% acid one gets a /(Eé$§§2>/ smaller fraction of dfizlcjt/ rearranged products. This occ 3 because the 70‘ acid is o-sic, and therefore average carbocation life— / shorter. rwrw (b)(3) Students who got the highest ratio of Z‘Me—Z to 4—Me-2 also got the Lowest total MP yield. Correlate these data, refer— ring also to which acid they used. Use < 25 words. 669 Wf‘ Yea/Vt, flroagwifl’l War! /¢‘7[¢1£<‘me4/. €¢gqfidcafi Mo was? «)6 M4 ((‘W‘W)7‘o (w. 3% Long law/cf c «4’ tion. Identify this MP, draw the 3° ion, nd the why this MP is a ming; product from this ion. C ‘3 /(::> /@ ;_M{-1 M Chg-'0 aha/@613. QUZL {‘14}! at'V-flmmfiy AEfiBMe—Lwambrtj m1» gamma 14%ij (d)(2) Total methylpentene yields ranged from 20% - 70%. All the starting alcohol was consumed, yet even MP samples not re— distilled had little organic side product(s). Explain; refer to how the prep was done and the nature of the side products. Rm T’ wow» .4. /00°Cgfm 56452 147$) W ((1, W Lu A04, “WOW (CY, Cw~ ) f [(9 5/35 /ow Vpr J l000=9 U “(77¢ oath/{5. time is 2- Mfié 2. (6) Note the gas chromatogram of the commercial solvent, "Skelly B". l , l l _ ’ I." “u —4 u ' 7 _” mum—u -_u_l-Vu.-I- '- — ‘ I—-_- __II-I— ’4'" g n— -15 -- In." _. h - — .- --""._."--....."==........--.._--_..."'=.....-"_=--_..""==...==_..--"".. * " " " =3.- - —l-I_-_-I-_I--_- -- "I". I “Il—_“-—_.II__- - "I!Il-u ‘ ——-IIII-III- ~——--.—-—-- “1:...- A‘T‘ III-I IIII-I:==III"IIIIII glnufiII-fiPJIiI—Il- nu- ——ImI-- II uu_lIIIIII--IIIII—IIVLI-IIIII v “I. _ III _m-—llIl—IIII~IlI-IIIIIII ' -ngw‘n-=III III-I'--..‘II| IIIIIHIIIIII—IIIII-I InI-IEA A - II II lull-Imlu-Il-I-III-I-Im CNN-’1.- II‘ .II IKIL'JI--. I. .‘1‘ III—- lI-I_III-Iml-'l I - 'I I l “hme—EIII—m- Il" ZJ-K'AIIIII' "III-n III... -I IIII I. I! ll . UIHI|“’AII ‘ III-III...- I.-I-I.l I‘l-mI-M‘IIIII’II-II-I-I III—LLDEIII-I-nq- I.- III III-- III-In.- II. III- .II "II-l‘.--' III-III.“ 1-..IIIII .Il"‘ ' s I I III-IIIIIIm-IIIIIIIIE I ‘I-h—IIl‘I-AI-uI—Ill III-"It‘l-IO'J-‘II ERA-I- mlnI—m-MI-I—II' ~-.-I—Ik'l_-I—IIIII ‘ II '1; ll LE!“ , III-mur-In-I-III- I'd-I—I-IIIIII III-IIII-II-u-Il‘ II... I“ U I-- I ll‘ ‘ V l I... III-- I. III...- IIIII '1... II..- IIIII III—I...- III-I “VI- III-IIIIII I IIII. .. I...lu-ll-m-l-lllll-ZIIIIIII-III..-IIIIIIIIIIIIIIIIIR‘I-ll! .' Ill-E EIIII‘. 7 II-Il-Il-I-I-I ---—-------- “III-InlII-ll III-IIIIIIIIII-m222:—-—;_—._ I—Il-IIIII-I-Iml I-I-I-I —-----I “In. _----_I --—------ an. 3- 'IIIIIMI 'l lull-u..- Iii-Innu- III-IIIIII-IIIIIIIIll-II...u Inigo-Inn-nunuunuunuuunIII-III-IIIIII-Inuuu-u- " MM (a) (2) Which peak is associated with the most volatile component? Number .1 (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 R). Then calculate the area in "squarelets" (use h and Xb to the nearest 56 squarelet edge). Also state how one determines Kb using <10 words. i E :5 i I E E i E I I E I I I I I i I i - ! E! ii 55 :a E! I: 55 .- E5 -: a a; i ii a: fljf‘fl 3. (14), 11 pts on this page, 3 pts on p 4. This question deals with the bromine—catalyzed isomerization of I, cis-X—CH=CH-X (X = COOMe) to the trans isomer, III. The Br-containing species, II—gauche and II-anti, formations of X-CH—-$H‘X, are intermediates. Br different con- (a)(3) Last fall, three students simultaneously took their Br — containing rxn mixts to the east SGM patio. All held their flasks at shoulder height facing the sun. Student #1 was wearing a white shirt, #2 a red sweatshirt, and #3 a black sweater. The order in which they got reaction—-fading of Br2 color and pptn of solid—-was #1 first, then #2, then #3. Explain this result in mechanistic terms, using <20 words. W¢ #759,! ref/«c “2769 NW! MIKE ‘ out“ Mr ,flm W. :7 ,2 94 was W (b)(8) Consider I » III as occurring in twg steps: (1) I in soln a III as supercooled liq (scl) withng = 0, i.e. K 1; then (2) III8C1 » IIIsolid, with Gxtln < 0, i e.. thln > 1- It At 102°(375 K), the mp of III, [Ldgtln = o and thus thln = 1. Taking R = 2.0 cal/mole-K, and (3G = 20 cal-(A30, one can use 1&5 = —R(31§,QT)ln thln to calculate T for a given thln. (i)(4) One reacts a soln having XI = 0.20 at a T at which III’s solu- bility > XI = 0.10 (recall that in soln, X = X ). Calcu- late Tsatd or III; thln = l/(mole frac solubility . +lo Ar: - 1C37r+mjw =—/717-%A1” l 3 131‘ 3 “ ’2C).5: k: 2%? 7:;fal : “3C)S-Ikr (:l3:LO<€) :i/ (ii)(4) Assuming complete rxn before cooling, I » III in soln and therefore <50% sol Calculate the fraction of I a I at the T at which 2.0 mmol I the above gives 50% of id III on cooling. solid III if one reacts 20 mmol II stays in solution. 2.0 rum/fl vs Fd/l, §4go ROMA: «A (alt: 4f 919((L'é ? /6 W/m/é) :9 we Sdt‘r/flM r : 0.80 (@022) -4- fl 3. (continued) C7£¥:: (c)(3) Exactly 120 mg Brz (MW = 160) is reacted with 15.0 mmol I (MW 2 144). All the Brz reacts, giving dibromide(s) from Br2 addition to I. The yield of solid III is 70%. Calculate the mole ratio of solid III dibromide(s). mama 6r» : w : 0‘7S“mm/ % ’— rwg‘o lKUmwlflL’m 70% :. [(7-4 “MI/“4 2 /4/~ 4 ote the data be ow. "m-n" in the left column means "a mix ure of CaClz'm H20 and CaClz'n H20 in equilibrium". CaC12 hydrate mixture Relative humidity (rh), % O- 1.25 1—2 4.2 4 — 6 21 I” 6—saturateo aq soln 30 ,' / (a)(3) The solubility ~f W in an Organic Solution is 5.0 g/L. One has 100 mL of W-=1td OS, which also has 2.0 9 W argyleta. On J1 i/éL adding 4 — 6 g CaC ., it forms a solid of average formula 6 CaC12-3W. Calculat he final mass of W in the 100 mL OS. 8‘9 x 1“ ‘ H (M M/e “Al/WVva Ova/‘40! 0.70 Am 7W0 0,3 )c 0.1+ = (2) The OS is separated 539% the as id. A new low capacity, only ) §: 7 0.020 gw/gDA, but Veh‘ Bfficien Drying Agent is added. “a t Calculate how much VEDA is needed 0 dry the CaClZ-treated \ soln above, assuming it removes all the water. iv m _ M C1Z JQ.07O A ' .‘ Vérpfi “’09 ‘7 “a 04 O.0L0 (fl/am, " (c)(3) Leslie he Lab Loser (L3) decided to save tim; by using enough VEDA /0 remove all the W completely in one ste- Calculate gV A needed just to rgmove the W droplets from ‘_e original and tell why L lost nearly all his/her O 2.0 W “ ‘ l [00 WM~ CJM/ Q‘IOVé mffi rid/h. ...
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This note was uploaded on 10/12/2009 for the course CHEM 322AL taught by Professor Jung during the Spring '07 term at USC.

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322as09_plq2_2_key - CHEMISTRY 322aL ‘ W MSG-G—...

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