322af02_lq2_key

322af02_lq2_key - CHEMISTRY 322aL/325aL December 4, 2002...

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Unformatted text preview: CHEMISTRY 322aL/325aL December 4, 2002 FALL 2002 SECOND LAB QUIZ By Bags NAME___.____££:Jé;___________ l.(l4) 2. (6) Lab time 3.(12) T.A. 4. (8) TOTAL (40) This test comprises this sheet If told to use less than a I certain number of words in and four numbered pages. an answer, D0 SO—-deduction for excessive verbiage. PL E 3913; Graded quizzes will be available from TAs Thu and Fri, Dec 5 and 6, at (check-out) labs and office hours ("special" office hours for TAs with no lab or regular office hours Thu or Fri). If you initial here — — — — ——> , your quiz will be put in the Study Room, SGM 102, at 5 p.m., Fri, Dec 6. All quizzes will be thrown out Jan 31. FINAL EXAM is Thu, Dec 12, 2:00 — 4:00 p.m. in the same rooms as all the previous exams. —1- flji:/(_]L 1. (14) Dehydration of 4—methyl—2-pentanol, (CH3)CHCH2CH(OH)CH3, produced the following methylpentenes (MP5) without eerben egeleeen rearrangement: 4—Me—1, 4—Me-2, 2—Me—l, and 2-Me—2. (a)(4) The differences in using 70% aqueous H2804 compared to 60% acid occur because average carbocation lifetime is shorter in 60% acid. Complete the sentence below in < 20 wds, giving two consequences of (NOT reasons for) this fact--one concerns total methylpentene yield, the second ieemer reLiQLsL. In the 60% acid one gets—- A. . M yfeW , bqf/«M Lug Narmwjz/ (b)(6) Two of the MPs listed above can arise from either of two (non— primary) carbocations. Choose one of these MPs, write its structure, and the structures of both possible immediate precursor carbocations. On the carbocation structures, draw an arrow to the carbon from which H+ must be lost. r bit/Md ZPMC— Z (if WA A M @ éwcwwwvcy; ‘thgw'cb/tcb‘fi ' 0‘” C” (4/3)» (c)(4) All the methylpentenes of this lab boil at < 70°. Yet they distill out only slowly in the preparative step even though the pot temperature was ~95°. Explain this with <12 words, and draw enly the slow step common to the mechanism for forma— tion of all the methylpentenes, showing electron movement. Mp5 have. '60 ____ 9; mm mm - WW (wow )fiwm7~ {WW MVAOC/d/MZGH})L(4‘(£4L.€/{-Cb/j éco/A ' Mité 2.(6) Note the gas chromatogram of the commercial solvent, "Skelly B". ,.‘. !.. ~ :s.: .-1' - I; ' l l‘:: V {PI-IIIIIIIIIIIIIII .- |'-' u.IlIiI :-= lulll===:==::==IlI-IIII-llllllgm,our:inn-Inlull-z.null-llI-III I III-Ill- a! III-l III-IIIIIII-IliluuuuHamlin-nagnuv‘nnull-Ill.-Ill-lla-Illulul annulus-llulull IIIIIIIIIIIIIIi-uu lIulNI' III-IIIIIIIIIIIIIIII _ Ill-Il-IlIlIlllI-fiifiiillInpc‘=:= 4+e%-====: I'll-II... I _' IIIIIIIIIIIIIIIIIIII +; Eggs" wwsemmmrws 4.? III-I:=:==IIIII====: f%5fiiiiiillllI====:: i=lui==Ill-l:===:=====III-Eliii- III-IIIIIIIIIIIIIIIIll-lIII-IIIIIIIIIIIIIIIII flm—m'lIIIIIIInI-III I m%nmmnmm—mmu—m_u “III-l VIII-mm- I . - mmmmmmmm— ' III.- IIII'JN’IZ'III’A‘I-l-II-m—n- I‘ll-III“...— mI-I-ufl l_——IIIZ‘INI Eli! III-III.“ mmmllm-l fIflA—IIIUE'Am mill-In“..- III-III-IIIIIIIl-l-I- III-III- III I II’ III" I}! III-IIIII-IIIIIIII-IIIII-I ‘ "Emma-mu.“ I “II-ml" lull-1‘ Imnm-lmummu III—IIIII-IIIIII-IIIIIIII “III—— Ill-Ill III-hum Sill-“III...” ImIIIIII-IIIIIIIIImI III."- mm In... mull-III..."- "um-mm... II..- - III-ll" “II-III“ [III-III.“an- ' ' "II-III-IIII-IIII “Influx-HII-Illfllnlmlnmlllull-mam...“ ‘ : “WEWIfllnlm "m‘ln.I‘1IlIII-Il-IIIIIIl-l-I-IIHHI III-I'Z.m‘-IIIII- Inllfllw All-ll.- lIl-EI. l-TJI'IIIIIllmII-IIII-I—V "I III-I'll UNI III-Il- ' I I lxunnunnllllllnm Ifzfl-H'JI-I-ll'lll-Iml'mI-LWE III-l .- If ,III-II- ' nil-III..- Infill-IIIIIIK'VAIII-HIII-lIlmmIlmIm-m‘luIl‘VJI-m-I " I.--“--.-. III-III.- lull—IIII'mIII-II'IIIIII-m.II-IIIIIIIIIIII-‘ .’--IIIIII-IIIII-IIIII III-.- II-I'mn'AIIIflII-l lmnmmunnnllllllnll. III-l.- III-IIIIIII-II-II-II-IIIZ ‘I-VA-Illl‘lfl 'AI-I-II-II-l‘I-II-I-IIII'AII-ldlll-I- “III-III..." I... “nil-“'1 [mm-m": "An-IanI- IIIII-Il-IIII-lel-mIIII-ImII-mnmn‘n-Il IVA-III-In‘II-I Ill-II-III. I-IIl-III-III'A-IIII-I-I-II-IIII!“III-III-Illln‘I-IlIII-KI-IIIIIII‘V-I- ' “WIMI-m-Imflfll-lIII-.I-Ill-uu Au-I-lll-IFII§‘I lmflI-IIII-IINUI nnmnl-mmmllllluI-in—fimiln-III “ml-unnu—G=Illllllulllh; lIllllllIlIIlIlI-IllI-l-I—l-I III-mn- I...“ u.- l-Illll-ll lln-m-I III-I...- _ null-mm 'IIIII-IIII-I-IIIIII'tI-I-IIIII III-III.- IDII-IIIIIII III-III-II-IIIII-l . “mu-“mumnlnnnunllun mIII-I- III-III-IIIIIIIIIIII-I-IIII I] s - . FrumumlI-II-m-llm-II-III-I- . 6 , 56444.1(‘14-6 (a)(2) Which peak is associated with the lowest volatility component? Number z’f (b)(4) Recall that one estimates the areas of GC peaks by assuming they are triangular and calculating "Kb-h“. Without actual getting the area, mark peak a to show hOW this is done; draw the baseline, the height and Mb. Also state how one geter— miggs Mb; use < 15 words. 3 , W16 603174 hfl/Z 3. (20 pts, 12 on this page) Iq<7 ii 2; {’L' In the last experiment you studied the bromine-catalyzed isomerization of I, gig—X—CH=CH—X (X = COOMe) to the trans isomer, III. The Br—containing species, II-gauche and II— anti, different conformations of X-CH-—§fl—X,are intermediates. l Br (a)(2) What was the evidence that the bromide anion, Br", is not involved? Limit your answer to this question and <20 words. 0W 6n m/a mg #7qu war/42¢. Mr M mgr/W “47%” «MW. (b)(6) Consider a mechanism for I a III in which a propagation step is II-anti + I e III + II—gauche, i.e., a mechanism in which II—anti does not expel Br- but "hands" it directly to I. (1) Write the other propagation step(s) of this mechanism and circle the h i — r i species. (2) Write a chain;termin; aging step which does not involve a radical-radical reaction but which does irreversibly remove bromine. (c)(4) g;g—X-CH=CH-X (I) is a liquid,and solid by—product dibromide is much less soluble in common solvents than is the desired product, trans—X—CH=CH—X (III). To get a reasonable yield of easily purified III, is it better to use too little Br2 and get 70% III, 5% dibromide, and 25% unreacted I; or to use enough Br2 to cause all the I to react and get 80% III and 20% dibromide? EXolain our answer in <20 wds. f/I’L‘ “Mutt fl flm +Ww MCI/LXCLWM. -4- 3. (continued, 8 points on this page) ‘ i. '- I g (d)(4) Consider the isomerization of I 9 III as taking place in two steps: (1) I in soln a III in soln at the same conc, then (2) most III crystallizes. N fl . (i) (2) Given AG for step (2) , AGxtln = —830 cal, calculate the . effective "equilibrium constant" for the crystallization, thln. Use ZLG = -RT ln thln; take RT M "’ 0 -AG. +23 k:)cf%ht= (Z_ ,ZZFF: :: £2: 6CM9 :: ‘f:<:) (23(7iggi) 600 cal. (ii)(2) Given that Ke for step (1) = 7.5, calculate the equilibrium q . I . constant, Koverall, for I in soln e III as SOlld. J’oa‘a. (HAW WV! w (QM (fa/:50) Fm , Kouech 7’ ((1) :7,(K4_0 =_ 30.5 (e)(4) Suppose 0.60 mmol Br2 (MW = 160) and 2.16 g of I (MW = 144) react completely to give only III and the product of adding Br2 to I. (1) Calculate the yield of III as a percentage of theory (for this calculation, assume the Br2 is solely cata— lytic), and (2) the mass yield of Br—containing product. 1 60 (0 WW/ f 2 /@MW/ . 4/ 13/th @ _V(’Z‘Zf/L€«A MM? E? > [CO ~ (9.66 : (Wk I‘M/r ‘Z7éiéi @ ('17 fléO Mml 4(‘AWMFAX Mu) '2 want/go: 309‘ M = 30+W7/w, x a.éww/=( xgzgéél ...
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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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322af02_lq2_key - CHEMISTRY 322aL/325aL December 4, 2002...

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