322af08_plq22_key

322af08_plq22_key - CHEMISTRY 322aL 3M7 M44 2008 W08...

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Unformatted text preview: CHEMISTRY 322aL{ 3M7! M44,- 2008 W08 Z [Orclctl‘cz 'fdv‘ sscoffms QUIZ " Fa ML Lab time @ “A 55.23” 23mg”). patch"k Thi§1test comprises this sheet I: told to limit the number of words in an answer. no so. and-ehree-numbered pages. Also, deduction it any part of an answer 1- W {our mmmm Te 5 will be - ai able from y TA at lab an'ffice w s A- 28-30, .-- it on initi.A he e -----> , ".11 - put out n the Study Row" by - p.m. Fri May tee s, nether i the S udy Roo~ or Dr Bllern': offi- 1 be hrown v t on M-. 23. d.‘ m is H May ,711-1:3 in he same vow . as were .1. previous exam: -.- Ni=l¢f 1. ('7) (a) (3) Regarding the prep of MethylPentenes from 4-methy1-2- pentanol and aq H2504: Use or 70% rather than 60% H 804 wit—(1”. MPs. State whether average carbocation itetime 1s anger or shorter in 70% acid, then explain only the gm], yield results, using <15 words. At most 1 point if any part of your answer is irrelevant to this question. a Wcaf {r (‘M :90“ \lagfir w: 70% ac; . 73M M/ y//// h: {A [Owl/t. Alc a. we. I‘M! 7% CC 1‘?) 4 (MI. Mp6 [ct/real] (b) (4) Two of the methylpentenes obtained were z-Me-l and 2-He-2. Draw the structures or a W and of a m alcohol either of which will react with aq I-IZSO‘II to give these two methylpentenes as almost the 3.91: products. 1 M <le $43 ' h; "O‘CVV‘C‘cM/wa,‘ - CUB‘ C " C “ma/J H '2 — i 2::- OH ‘1‘ - EEHEEEEh 2. (7) Note the gas chromatogram of the Mrs from EEEE 4-Me-2-pentanol ------- --> @ Egmwggwigfih (a) (2) Write IN above the peak associated with "’ the ma; mm component. (0‘, ac,“ M n‘.) EEEEE (b) (2) Peak area is taken as "ma-h". For pe 4, "m" 1" m-MMI- SA»- 6 m. .g‘, t (c) (3) Peaks ® and © are the MP3 of question . 1(b) . Without knowing their bp's, tell Egg-7:45;: which peak is which MP,- give your reason- :3... ing in < 12 words. Assume they come from a common intermediate. _ @a, 1’ 92-1—16? auva ‘3 [av JIM/c . Morretn Ml time-’- l 2 _ z (12 point this page) 1 l L 3. (23 In the last lab, you studied the Brz-catalyzed isomerization of I, gin—x-CR-CH—x (x = COOMe) to the trans isomer, III. The Br-containing radicals II—gauche and II-anti. general struc- ture Br-fH——-$Ho , are intermediates. 2: x (a)(2) Based on how one o: the rxn mixts exposed to light behaved, tell how one knows that light algae does not cause rxn. Answer only THIS question, using < 10 words. MI‘KZL [Ln/«4:4! uc‘fl‘ ’w i". am m [4,4111 (b)(6)(i)(3) Use of either too little or too much Br2 gives a poor yield of the desired product III. Explain in < 20 words. no (mfg 8n, (Va one! It“! c fl"[((m “£464 mica 0 M2 4 “a.” n We! 3%? (I «(14:48 (ii)(3) To save both time and materials. tell why it is better to star: with too little 8:2 than too much (<20 wds). (c)(4) 0.60 mmole Br; (Ml - 160) reacts with 2.16 g I (Ml - 144). All the 3:2 is consumed, while 2/3 of the I is converted to III; some I is left unreacted. (1) Calculate the man: of III produced, and (2) the initial male resin of I to Brz. flaww Page total a 14 pts: 11 for ques 3(d) (e), 3 for quea 4. (d) (a) On. can View I .. III as two steps: (1) I in DCMZL'III as supercooled liq. K " 15 the“ (2) Insacl “ IIIsolid' xx: 3 ‘23! = ' . (1) (5) In solution with 60.0 ml DCH, t e sin-T3! the mole fractions (“’41. of I and In . 1/7, i.e. xudu, :- 1/7. Calculate (1) mole M I+III in soln, and (2) mnol solid III obtained tron 3O muol I; note Keg above and assume no side products. 1‘ (I) Lgf‘ Wk(tffl’m‘/0(k ax, 3"=;—=X=(Ole @ < 0'1 M411 Hoffa OCM’ (may = .Ql—xl; a W 04 41% that, (0 Mme/4n): m" 9/“ Q 4‘ ;'(?O'/0)mu~o( éfl/I’é (ii) (3) Assuming IIIacl in in an ideal soln. calculate/deduce rum, I ; based on the data above. Give reasoning. ((17 x33“! 2 5!!" me“ hd‘c% {1% VII) fidfl =9 Km: 1+ (e) (“4533,174- ~n'r 1n lulu. The temperature dependence or 45'. Am ’1' -(20 cal/mole)/kelvin, Aug) -A[-R’I'(ln lg‘tlnn . Taking 'r in the term 31' (only) as constant. and RT - 600 cal/mole. calculateA'r. including 8.193. to change xxtln 13°“ 20 to 40. Note thatAlconstant-tufl - cw“)! - 10 %¢/KM = w aéw \\ I“ Kv‘lh k. Ar: ~60:o(o.6%}mv1 é, =Z4<§Lfl —20.5(cc~;1<)\ 3) One has a 20% ag soln (v/v)- of 1mm solute A. Calculate the 3913 traction one must evaporate to get a 50! aoln. You 4. ( 1- may assume a detinit 01 as 11 acre 1 1' answer. 1;} a.= emf/1,1,: m+' “9“; if; 33031373°E~wan a~ W) rain «'4‘, X = “‘ffi' "My {03 La. Fox $050 4: 9‘50“); 013‘: 3;... n: c 0‘1"?— ” o'rétfi N=¥Zafi¢awln =? “34, (507 ’7‘ ‘ M ' #0”? 56% -+’/ pjézl‘), 10. (2) A halt—saturated Org Solution contains 3.0 9 W per L. A certain Drying Agent system has relative humidity = 5.0%. A small excess of DA is added to the wet OS. Calculate the L A (+¢(§'J¢f0( 05 {W4 irh=m%‘ 0‘“- OJmC z: 3-017‘d/LOS Kg: Q30J%°5 [91‘ 6.0 x 11. (2) Six hundred grams of an OS contains 3.00 wt % W. For a DA of capacity 0.40 9 W per 9 DA, calculate how much DA is needed to C) jfitL/ dry the OS. For this calcn, assume all the W is removed. ($004373 0.) = C900 x W: lg 6.) A 0% @ J 605 a V‘a‘4vu'- ('gr\7 6;? “Emu @ a” ’ Wm :W— unextracted [FUR = 1/(1 + (z/n))n] for (a) use of B all at once, and (b) use of B so FUE is minimized. Take e3 = 20, (a) <2) ‘ (b) (2) I Q “M ' 4:C£ =3 -——--; {:Llcrh" u_yI/' (:(f'%=:)“ u 2 l | '6 2 "‘"' (0. OOLJ) '/ (0H3) ‘ m . 1 0.001+g-- ) (c)(2) As 2 increases, ' e FUE for (b) compa 1: to (a) (circle one)-- decreases, wh is worse. increases, hich 1s oet er. increases, which"; worse. 13. (2) I has mp = 100°, and L has mp = 130°. The most probabl‘ m- for a mixture of 90% J and 10% L is (circle one answer)-- 75° 99° 103° 115° 127° 135° ...
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This note was uploaded on 10/06/2009 for the course CHEM 322A taught by Professor Singer during the Spring '08 term at USC.

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322af08_plq22_key - CHEMISTRY 322aL 3M7 M44 2008 W08...

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