Practice Lab Quiz 3 Key

Practice Lab Quiz 3 Key - CHEMISTRY 322aL...

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Unformatted text preview: CHEMISTRY 322aL 1Efi£§£=£fi;=ififii=? iflflflfl!3:2¢fl$= Wm QUIZ " 5-me 800% m__L<__E_Y___ Lab time T.A. TOT This test comprises this sheet and three numbered pages. Tea . ~‘ 1 be available from your TA at i- »- - office hours Apr 27 T‘nghfhh ‘ _ ’//// is Wed, May 6, 11-1 in SGH 123, as were . ‘ .;. earlier ams and quizzes in this course. -1- $4? ‘1': [:5 1. (13) For the methylpentene (MP) prep from 4- e hyl-z-pentanol, (CH3)2CHCH2CH(OH)CH3 (SM). and 60% or 70% aq 32304: (a)(5) Even at 90°, the water solubility of the SM is low. room T, the negation mixture was initially one phase. But at (1) Explain, using a chemical equation and <15 words. (ii) As MPs distill ouE, an 5rganIc Layer appears in the pot although the pot T ~ 90° and [H SO41 has increased. Explain in <15 words; refer to what is in this new organic layer. @“7 ‘I‘W€'2/‘,0MJZ0/M/ 0': “$790) Frame/{Q [7‘0 W~goC4M< «$809131 £\O(.7Lf- {fgflj ' A3141. b/Jto @LCC) £3)! 704/003“ch kcjlym) Mu) («/ch (Cu, W 604ml?) 0M4 W0h€W€ (cf ¢, (b)(4) As the GC analysis showed, some of the MP5 must come from rearranged carbocations. actually formed. is larger from the stronger acid. as among c rbocations; use <25 words total. ’ {a} 4.1,: $6»ch (Act-l; (czfghcac—c/Mb’: , 3 fi’w g)” ((+JCH:c~c£/v6h[3 r 1!” (9/461, ~1 «.4 our; M U‘C) LL Anny)». «x81, «AMI/647a 64x4on AMYM 0‘: W =? M 15:0 a PU h Wwigmwwfl 2. (4) Note the gas chromatogram of a MP mixture: The area of a peak is estimated as 'Hb-h'. Tell how Mb is found, using <10 words, then draw the baseline and mark Mb on the arrowed peak. nib" a. 5% 7‘» it /<::::) «nd°£11\ QXL hw(((:’/\1J:7£¢?f'~’3>qa (jqwldg (\Wa {WW 661491?“ @ Ti ll ‘1 ll IIIIIIIIII Illllll III \_—_——/ ‘ (i) Draw structures of two such MPs (ii) Explain why the rearranged MP traction Refer to competing process- J II!“ "t t:— ' Iiifiil! "Will"... h 9 iii! .‘aiiliiili‘l' ilfflli . ll 54" lllllill‘lmn I ""Iiili ll "E! grit Iilll I III I u “‘1 n —. a n l (13 points this page) - 2 - Z 2 3. (23) In the last experiment, you studied the Brz-catalyzed isomeri~ zation of I, gig—x-CH-CH-X (X - COOMe) to the trans isomer, III. The Br-containing radicals II—gauche and II-anti, gener- al structure Br-fH—-—$HO, are intermediates. x X (a)(3) Brz in solution rapidly attacks 2-butene danidnidaiix. Explain why this mode of addition does nan occur with I. Refer to the key difference in the groups attached to the alkene C's in the two compounds; use <15 words. *COOM‘J w fflmjlr e~ afi‘fl/V‘awiuf) MW. e"; /%4 m.— m/W' «c /‘ ‘rm‘c HM], (b)(4) Brz additidn to I, is a two-step aide reaction. Its second step is a termination step of the radical chain with respect to the I a III isomerization. Draw the missing structures for 2. L. the balanced reaction below (disregard stereochemistry): ‘dc JflfYW(C '2? 6r ‘CH-CH-/+»Bz~?r ———, ?? flymCMAq/flr + Br- IUM mt I )2 U I 1 show-— S" x X x A¢UC . (c)(6) The "successful" reaction mixture contained initially 100 mg Br , MW a 160, and 2.2 g I, MW a 144. (i)(2) If all the Br 2 , . 2 forms a solid additidn product (AP) with I, calculate the mass of the AP. (ii)(2) If the conversion of I a adiid III is 75%, calculate the purity of the solid III, as mass %, assuming all the AP also precipitates. (iii)(2) Tell what one should do after filtering off the solid to obtain pure III; use <10 wds. (5) MW ,4,0 2 100 we 2c ($25!” = How (o./q ) 1/95 (a) 4,: 2.23 M75“ .- zryét 221: ' W 2 (if 1 fluffy, W ‘23 W { 7.7%) may flwysM/wac m M ow growl ..3- 0 3. (continued, 10 points on this page) /<;7 Z (d) (8) One can view I 4 III as two steps: (1) I in DCM::III as supercooled liq, Keq a 1; then (2) IIIscl a IIIsolid. (i)(4) At equilibrium in solution with 48.0 mmol DCM, the sum of the mole fractions of I and III - 1/5, i.e. X(I+III) - 0.20. Calculate (1) mmols I+III in soln, and (2) mmol solid III obtained from 30 mmol I. Q ‘ maulocmqfi a (I) ‘65-;- —-O.'L 737’ Kallmuw/l 9r W(€:rfl?_0.,b ( l ;;3 wxwtol C 1:\fJZZj @ Ofil C(‘g mmo @ (’1) MM! 504% E1: 2 30‘- /5L) Mme] 5"{8 (Iv-4,71 M Gm 394.4910) (data!) “a (e)(6) At 102°(375 K), the mp of III, (19 tln a 0 and thus thln s 1. Taking R a 2.0 cal/mole-K, and 43 x 20 cal-(AI), one can use 435 a -R(375 +15T)1n thln to calculate T for a given thln. In (c Hg I 2.0% t (i)(3) For thln - , calcu a the T (Tsatd) at which an ideal solution in DCM is just saturated in III. 910-137“: -1(37§+AT)1‘0g x-mso— «LL/Mb?“ ‘9 AT: -6%é K =5» FM =<J75—é}€é)K = 3/0.? K = 31+ 0c (ii)(1) In <10 words, tell what thln means. Kxfl” ‘2 m (0!“ buy WI: fivfig firm. (iii)(2) Tell what the maximum possible yield of 19:31 III is as a fraction of I if the reaction is run at 2T sumed) then cooled. Explain in <20 words. jél (:nSYgfiéfl‘ [2525d[}p2? ob (V’Yfl h9L53T¢Zfi:2 (Z 1 (AA: {XVI/W/ M? W"! W 57% “WI EL/i‘~/fl/£“SO%~ [fowl 53””me £vm] satd (all Brz con- ...
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This note was uploaded on 07/22/2009 for the course CHEM 322AL taught by Professor Jung during the Summer '07 term at USC.

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Practice Lab Quiz 3 Key - CHEMISTRY 322aL...

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