322as05_plq2_2

322as05_plq2_2 - CHEMISTRY 322aL¢825€L 4 Cifihi-GGQA...

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Unformatted text preview: CHEMISTRY 322aL¢825€L 4 Cifihi-GGQA " #2. A pflcfi‘q («1‘ 39’ SECOND LAB QUIZ :50 ring 05 ~______________— é;:\ 1.(14) NAME 2. (6) Lab time 3. (8) T.A. 4.(10) TOTAL (38) This test comprises this sheet If told to use less than a certain number of words in and four numbered pages. an answer, D0 SO——deduction for excessive verbiage. ". — 1 — I ;z: 1: { f 7 1. (14) Dehydration of 4—methyl-2-pentanol, (CH3 CHZCH(OH)CH3, produces the following methylpentenes (MPs) without carbon skeleton rearrangement: 4—Me—1, 4—Me—2, 2—Me—1, and 2—Me-2. (a)(4) The differences in using 60% aqueous H2804 compared to 70% acid occur because average carbocation lifetime is longer in 70% acid. Complete the sentence below in < 15 wds, giving two consequences of (NOT reasons for) this fact--one concerns MP mixture composition, the other total MP yield. In the 70% acid one etse- ‘3 WWV 2451‘ MM Mp3) /0ul% 727996 yrpyfi y(b)(8) (i)(2) The girst ca bocat rearrz oement is 2° » 2°, standard molar AG, Aans 0. Considering m K, explain in <15 words why any substantial amount of rearranged MPs form at all. t;:" C“?W(L'f; ‘zo(§:a :t.() VVkfiQQDVL/G ;Q: 6(yy1014q,é2:— v “0 e” 4 fi‘w WC WMM‘ &C (\‘l ‘é \ G ? ‘Q’ZM,((/I, V'( . ’ CbL )Cen' (ii)(6) Only one of the MPs listed above MUST form via a 3° carbocation. Draw this MP and the 3° ion. Now explain in <12 words why th' MP is not the main produc rom the 3°.ion. C4 “6&— C 6 C143 LL~C~CELLCHL H"? ‘ "é-CL'Z 3 1 J 1W“; MUM“! [y/z/(e—LJJM (WM?) tfik 5’ (c)(2) All the methylpentenes of this lab boil at < 70°. Yet they distill out only slowly in the preparatize step even though the pot temperature was >90°. Explain this with <12 words. -2‘ p - 2.(6) Note the gas chromatogram of the commercial so ent, "Skelly B". . l i ' - l ' I u - ' : l - I ‘1 ' - I : I ' ' mmlmu-IIII-III “lull-- l ‘ I ' “In nm'mullnlml-I- “nmmIm‘m‘n-I-III-Im—IIIII“ . III- “II-Inmnmm'm-mflw-‘flu {mun-mumm- . I q$gmlmlnuiilnmmmmnnmywm Ill—mumm- . fimmmm-Immnm. V II ‘- ‘- . =_ mlummmnlmumllnll 5 ' - ' I-mmlIIflI-Illllnm III"..- I-IIIIII—Il-mnfll - ‘- mum—"ImmunHII-m III- n—III-HII-IIIIIII : . a i 3 III“..— lug-Imman III-- II...“ “mm-III“- LII-I “mammal-lull".- Innmllnmllmlnnnnlmllnminn mummmlllllIlII-I nmnmumunmlummlluln I IIIIHIIIIIIIIIIIIII—Il IIIII-IIIIII mflamwghqgn“. flfi_mgggafiggmnnmmun-u... w: IWéI' II I] 'I I "III- IIl-IIII “mum-mull -' ". E‘IL"! II IHI I‘Hln - III-ImIIIIl-IIIIIIIIIIII II...“ I n..- . Winn-umlmgmum-m-J I A III-Illn—num "Inn-III.“ nlumx‘“II’JI-x‘II-IIHIIIII-UIIIIIIIIIII IIIII-‘II-III- III-lull, " . yIquw Emu-[lull [HI-UllnmllIII-IIIIIII I...) .“B‘IIIII- MIN-lam,mull IHIIIIIII It; 'l-I-I-lmlnIIllnnI-II! III'JII’!I III..- ‘} A=:===IIIIIIIIIIII-l .‘alnllmIlnln‘nl-I'IIIII III-I _ V IImm-I'AI-Iu‘III'“In-“IINIIIIII “III-"Mllm- III-IIIIll-III III-l Ilulllfi III- lnnnmmlIIllliiilInml -' _:L‘WIIIIIII“III‘IIIII III-I’ III-Ill“.- IIIIIII-IIII I'll-“III’I-IIII’AIIII-Ik‘ iiiimlrnIlImIIu-lnlllllJun-r, III-Insul- LII-IIIIIIII IIIIIIIIIIIIIHIIIIIIIIIIInuilIIIIIIIIIIIIIIISIII.IIIKIIIIIIIIILSIII Ill-llllllllllllllfliIll-5....III-IIIITSIIIIIIIIIIIIIII III-I‘ll lull-IIIIIIIIIE- Immnmnlnnlmlnl ' III-IIIIIIII-"II-“Illlllumuml‘SEC-"mlflnIlIIIIIIIIIIIIIEIII-III- tPII-IIIIIIIIII-IIIIII‘II-IIII IInu-IIIB-“IIIWI[IinmnIIIII'mIIII- III-IIIIIIIIIIII-IIIIIIIIII l ‘0 I I.-ply-I‘mWIS-IIIfl-IIBIIIIIIIIIII- 1 ' - 0 (a)(2) Which peak is associated with the highest boiling component? Number ___fi;____ (b)(4) Recall that one estimates the areas of GC peaks by assuming they are triangular and calculating “Kb-h". Using the base- line indicated, mark peak 2 to show how this is done: Draw ' i the height and fib. Then calculate the area in "squarelets" (use h and %b to the nearest % squarelet edge). Also etete how one determines %b using <10 words. -3- fl the bromzne-catalyzed 3. (8) This ques and 4 and 5 on p 4 deal with isomerization of I, gis—X-CH=CH—X (X = COOMe) to the trans isomer, III. The Br—containing species, II—gauche and II— anti, different conformations of X-gH-QH—X, are intermediates. Br (a)(2) What was the evidence that the reaction is not caused by light along? Limit your answer to this Wu and <10 words. 7% Ml My? 0% 6 V1, ‘ (b)(4) Consider a mechanism for I a III in which a propagation step is II-anti + I e III + II—gaughe, 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 ' — ' species. (2) Sum the steps, showing that they do add up to the overall reaction. (c) (2) The by-product dibromide formed is almost all mesg rather thaQ‘ ragemig, although both I and III are present and react toathe /\ dibromides. This means bromine addition here is-—circle one choice below (spec = stereospecific, sel = stereoselective)—- sel but not spec not sel nor spec spec and sel spec but not sel A7 2:50 4. (6) Consider I a III as occurring in two steps: (1) I in soln 9 III as supercooled liq (scl) with AG = 0; then (2) IIISG;L —> IIIsolid, i.e., some III crystallizes, withAGxtln < 0. One defines thln by ln thln = ’(AGxtln/RT)' (a)(2) If IIISC1 forms an ideal soln with DCM, state what thln = 20 means quantitatively (one sentence of <15 words). @hkfln“ 12:a xzads‘mm) 0,! Xm(cZQ/0§=/wglfw{¢%‘ (b)(4) Assume that mmolsolute = 100-(Xsolute), e.g., XIII = 0:02 corresponds to 2 mmol III in soln. Calculate mmol solid III (i::> ' from 20 mmol I if a rxn mixt is run to equilibrium at the T-— 46(eéir (i) where XIII = 0.07; afterward cooling lowers XIII to 0.02. “A 1“ )(EH: :.(9.0‘7 “it "1LL46/ 3:? x:1: 5'C9.CT7. jafqu, Amy-L 7Mwl£ FZ—Mmiarbaw‘l 2:9 ; (ii) where XIII = 0.02 (with no further cooling). I MQJMpo/ (AI/(6“ QX; >407, ‘ mflflg 5. (2) 0.60 mmol Br2 (MW = 160) and 2.16 g I (MW = 144) react com— pletely, giving only III and product(s) from adding Br2 to I. [16; L¢n¢or- Calculate the mole ratio of III to Br—containing product(s). = 2.16 ; l W6”: ‘WM" 641‘ 0.6 m A7501 :9 fiat;aé0 : fT+mmolEEg> “(a ma; 253— :VML; 6. (2) The W solubility in an Organic Solvent is 200 mg/100 g soln. A certain Drying Agent forms only a 4-hydrate; the relative humidity for DA/DA-4W is 20%. One adds DA to 200 g W—satd OS, (:)’L{/ which also has 1.0 9 water droplets in it. At equilibrium, the average composition of all phases containing DA is DA-2W. Calculate the total mass of W, dissolved and as droplets, 6’ remaining, i.e., the total amount of W not bound by the DA. :ZT/Z ail L \ 2 6M9 ‘ MW M /%f L4 K ‘1 WW“ wag—ah ...
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322as05_plq2_2 - CHEMISTRY 322aL¢825€L 4 Cifihi-GGQA...

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