322af09_lq2_key

322af09_lq2_key - CHEMISTRY 322a/325a December 2, 2009 FALL...

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Unformatted text preview: CHEMISTRY 322a/325a December 2, 2009 FALL 2009 SECOND LAB QUIZ BY 2398 NAMEM 1. (13) 2. (6) Lab time 3.(11) '1‘.A. 4. (10) TOTAL (40) L44: Qaf’i Iot'c’é‘a Schecade. ‘ SO“ 101 cull-Ma Charles Arden F (2330‘ [530 57L 0 {'h‘fimdlzft Abhimanyu Bhattacharya I)“ [46 1‘ 2‘30 Anna Dawsey n u -llgs-O Mike Kornoff Th ,4! {45 7.— 1:30 Vincent: Li T‘H I ‘ltfo am! MK {44 1~ 2.330 Shay Mallory 2-1330 dd Fn‘dft 1“” ’3". f‘fo (011.) Arjun Narayanan Eh (0.6 7‘30 “[0 74114 Janet Olsen fig (4A (lg) fit ?:3’0 Justin O'Neill FM mac 5%; [4r Prasanna Pullanikat W 1330 ~1L30 Andrey Rudenko F (9.130 a I I 30 Em Bm is Thu, Dec 10, 8:00 — 10:00 a.m., in the same room as all earlier exams. 1. (3) l'MQ.~L CoLé ~ vr. 616$ Mic/3 Dehydration of 4—methy1-2—pentanol, (CH3)2CHCH2CH(OH)CH3, gives (at least) the following methylpentenes (HPs): 3-Me-1, 4-Me-1, (Z)&(E)-4—Me-2, 2-Me-1, 2-Me-2, and (Z)&(E)-3-Me-2. 4-Me—2 is the main MP from either 60% or 70% acid. of 4-Me-2 to 2-Me—2 is larger from 60% acid. Tell why this means that the alkene mixture from 60% acid is farther from the equilibrium mixture (<15 words) Tam-t 03 MM ffeé/t 714mm Mo ~—9 {man 6M The ratio 1‘?) «zed «Ext. 2. (8) 3-Methy1-2—pentene(s) could come from either a secondary or a (a)(3) WM 51’ 1 «u. 3/« (c)(2) w (b)(3) tertiary carbocation, by removal of ne H+. Draw both ions, [1‘ l 3 C‘“: :ggwifgell CJ' <43 '5 ‘fi CW; ((565454)!ch Tell how the fact that one also gets 3-Me-1-pentene in this reaction means that at least some of the 3—Me-2(s) come from the secondary ion. Use drawings and <20 words. “3 Draw the structure of an al— (:6f3 llt" ) cohol which could dehy- l drate to the products of (b) without rearrangement. <:££7"<rt¥"<:‘ 5/4 H Total methylpentene yields in this prep are typically 20% — 60% based on the starting alcohol. Nearly all the alcohol reacts, but the ups contain very little other organics. Explain in <25 words; refer to hen the prep was carried out. 2: <5; #ACyM 4 (KM/Azabw m MC 1w» 6 {/w‘x ~40 «WroaaM/ Wm“ ()M/os («JP/74 ’T:£4L,C:é{g hi3 A7 2% 2.(6) Note the gas chromatogram of the commercial solvent, "Skelly B". _— I- ___-I_u—n—- ul- -. I —._m-I-_—mm.m - ._--I-II'-I_III-In r gmim———I-_I-III_ .F'! "_ -I._II mm“...- — “II_I_“--II in” "_Q _I'III-III III-II...“- flmm-_g—_ HM. 'l_-.-III-—- “II-I'- “_._IIIII-I II ‘ "-I-I-I'IIIIII'I-IIII- "II—.- II-“mi-. 1.?”II-IIll-II—II-Im- “nun-ulmnln .- Imm- ‘lI-IhI-I I 1—”. "I III-III..." flI-m-l—unu-nu . "I"- "nun—u Jun-lll-IIIII-I-l-I ' III—"I-III—IIIII-III #m- m_ l"_ ——IIIIII-- = I “In”... _. In." —“IIJI-II I I-I-‘II'IIIIIU . —III l—IIIl-Il- Ill-I --.‘Il-.lluI-IlllIInIlI-IIIIIIII Ill-lul-"IIII .- ‘VT‘III-l III- CII‘I-la It IIII-lIl'llll-IIIIIIII "Ill-r 1"..- u- IIIIILM' 1.....-- III-II I'll-II -1 Ill-IIII-I—III—III-II-l a I -I I. -0 nnmlnm-II-m- IP' J fln‘—4IIIIII “III-II III-III III-Ill II .7 I- . : Ill-IIIIIIIIIIIIIIIIIIIL—.lIl-Hullllhlmllnlemu-llulll‘m' All ‘ .IIII-IIIIIIIIII-I-nal—II—l-I -MIL' III-I'II-II-II‘. III-III Ill-lm-‘IIII III-IIIIIIIIII-IIIIII IIIIP-IIIIIJIII.‘IIIIIIIIIln—|III-IIInun-rallvlu III-“IIIIIIIIII-IIIII-llI-lz‘h‘I-hIIII-‘II-H Ill-Ile III-II InMJ-II“I I .III—I—I—IIII-IIIII—II'I J-J-.-I—I'I.‘-I—II-IIIIIII“Ira-III.“- - III-III" IIII-IIII-lI'm--III-IIII--emu-Inmnn'lllululnr.“III-II 1 III-IIIIIII lull-llIII-I'm-II-InllullIll—u-nII-unwunluaiurell-IIn-Is .m-ul-I-m-Im-Cfl-l’.III-I‘ll-I-Il-III-Inn-I-nI..."-I »--' ‘UIIIII-II. “jnuuum-III—I-n-II-u-I-u—nn-u-uu—nuu-n- -umm-------n_..---I.-m- lulu-III‘l-‘IIII-I u-II— . I.- —-'.I---- l...__. l"...... II-I-II-I -- . i... m... l.....“.. I... I! III. III. III-IIIII‘IIIII‘II III-mI-u-I-IIII'II III-- Ill-IIIIII-“IIIIII‘IIIIIHIII' lull-“IF ‘IIIIII’III “IIIIIIIlI-IIIIIIIL .IIII [I lI-II-IIII lIIIIIlIII—lllflli Mll- I'IIII IJIII IIII-L I... lllflul-lIlII'JIII I I-I'Inl-II-II-IIIl-I'IIII Ill—ll.- I-ll-III-II-“II- III-III—Ilnlll-IIIII ‘ (z:;)_,ap‘9‘ufl‘Llilfl4. ’i (a)(2) Which peak is associated with the most volatile component? Number lJE. _ (h)(4) Recall that one estimates the areas of GC peaks by assuming they are triangular and calculating 'Mb-h'. Mark peak(§)to show how this is done: Draw a proper baseline, the height and Kb. Then calculate the area in "squarelets' (use h and Mb to the nearest x squarelet edge). Also shin: how one determines ’(f tgwusing ‘10 2mg, ) t- M; Co‘ny V.) 30 ‘5er ‘1“)de Silas "n/Lb = 7 sclrzé «(9&4 $7 ' mm x: {fit {we 09% @M “arm W -3- 3. (21 pts total. 11 pts on this page) Z In the last experiment, Period 10, you studied the bromine— catalyzed isomerization of I, cia-x-CH-CH-X (x = COOMe) to the trans isomer, III. The bromine-containing species, II-gauche and II-anti, are intermediates. (a)(3) For reaction mixtures with Brz and exposed to light, the color fades as the product precipitates. Yet the product isolated i/QL, does not contain Br! Explain, using <20 words; refer to mole ratio of I to Brz. “‘—’—“““‘ V‘ (A q %ZC/ O ‘4 ml CY ‘ Ina/i mé‘o 136m 7: 20-291 _. 6 - t \ v SMbyfiMAgrm (b)(5) Using the labels above, I, II-g, etc., write a mechanism for I a III, with Br- as the chain carrier. §um tHE'sEeps and state how this shows that Br. is catalytic for this process. V'UT9QE filtgirvj ’ZT‘ {_. £?r~. -———4€3 LZZ::~5F\ £13 :2" 14¢ 51L mt[7 ,MMHV_ ‘ W53.” Swat r EE+@ (c) ) I is a liquid; III is a solid. The by-product dibromide(s). DBs, are solids. Method A uses a small amount of Br2 and gets 70% I converted to III, 5% DBs, and 20% unreacted I. Method 3 uses much more Brz, giving 85% III. 15% DBs, and no I. To get a good yield of high purigz III easily by recrystallization, which method is better? Justi y your answer in < 20 words. _ 4 _ 3. (concl, 10 pts on this page) i 2 (e)(6) Suppose the I a III Kequilib in salutign, = 1, and that the main driving force for the reaction is the crystallization (xtln) of III. This amounts to viewing the reaction as two steps: I a supercooled liquid III (IIIBcl) a IIIcrYs. (1H4) At 102°(375 K), the mp of III, 45ml = o and thus thln‘: 1 Taking R a 2.0 cal/mole-K, and Li a 20 cal-(am), use AXG = -R(375 +£3T)ln thln to calculate T for thln = 25. (ii)(2) Using < 12 words, tell what "thln = 25" means. /@ 20137“ = 1075 *ATJc/"Lé‘; ‘ . 3.7.2 CC) = — z+z+ ~4.~HFAT —.-= 41': -?/.3 K :9 (2 2%17 l< (’4 10.7 *3) @ <56 £6 WM flu; solubility Xfwh (We) '3 5%. Wk. (f)(4) The Brz-catalyzed IaIII conversion is run at T1, where XIII a 0.050, until all Brz is gone. Then the rxn mixt is cooled to T2, Where a 0.010. A second reaction run, using the same amounts of solvent, I, and Brz as above is done entirely at the T where XIII = 0.020. 6w. 1 = 6%1 Mb? 2c; = 2cm mm, £3673“. (IMO/316i X]: mac" {132 10.010, K13+Dj a 0.060 G34 fit-(4% K;=~ Kmfio‘owm ’ :71 mars“ Sch 3 Mi {of ...
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This note was uploaded on 03/04/2010 for the course CHEM 322AL taught by Professor Jung during the Fall '07 term at USC.

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322af09_lq2_key - CHEMISTRY 322a/325a December 2, 2009 FALL...

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