322bsu05_plq3

322bsu05_plq3 - CHEMISTRY 322bL/325bL SPRING 2005 rac / B...

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Unformatted text preview: CHEMISTRY 322bL/325bL SPRING 2005 rac / B ‘<: (Er_\r’ Y Eagfi NAME ,l.(12) 2.(10) Lab time 3.(11) T.A.____________________________ 4 (7) TOTAL (40) This test comprises this page and four numbered pages. a1. Mo and Tue L.- Quizzes will oe av.ilab1e from T“ as u 0 1y TAs who do not usually have an offic- hour T or Fr 8/29. f you ill hold spec'al off'ce hours Th/FAi Apr initial here — ~--> , you. quiz ill be p t in t e Study Room by Mon, M.y 2, 5 p.m. ALL - izzes not nicked up will be th own on on Mon May 3. EXAM i~ Wednesday ay 4, : - 10 a.:., in the F17. save rooms as all othe |OTE: exams t is semezter. 2311/1 1. (8) p—Nitroaniline, p—OZN—C6H4~NH2 (PNA), was made from aniline, Ph-NHZ, via acetanilide, Ph-NH-Ac, and n-OzN-C6H4-NH-Ac. (a)(2) In this prep, the starting material was decolorized. Tell what behavior of a subsequent reaction could, in general, make 9'”er “42/644 ' RWW "‘4 ‘4’ “We”! ‘Wfi' (b)(2) The WaterzPhNH2 mole ratio = 50 in the rxn soln. If 1.1 mole Ac20 per mole PhNH2 converts 90% of it to Ph-NH-Ac, calculate the minimum PhNH2:W reactivity ratio. L/f We W 0.2% 400 0 mg: maé wf fit 147 (1.1‘04). fp with“ fox 5.733233? (c)(4) (i) State the safet and ease of handlin advantage of using ~6 m aq HCl rather than "concentrated" (~12 M) to hydrolyze p-OzN—C6H4—NH-Ac under reflux; use <12 words. (ii) Tell why 6 M acid works as fast as 12 M in this case; use < 20 words. 2) [70 N07” “Ha «AMJL 71-1/0 K “(WjWKh/WJ (CC) 3%)6 7:2 h«‘\7[&(dt71:5)__'m% “M fit [0mm W, 2. (4) Below is the structure of the intermediate resulting from attack by E+ at C-1 of the 2-naphthoxide anion. (1) Draw the corresponding structure for attack by E+ at 2;}. (2) Explain why substitution on 2-naphthol occurs at C-1, even though C-3 is less hindered. 2 0 / /" .2 H ED é» (find. I mum V‘M WM =§ m r-‘u rm...“ 5:; for NW). -2- ’ AZ 2' a 3. (10) This question deals with the Glucose Pe cetates (GPAs). (a)(7) For a mixture containing only one Optically Active Compound (OAC), using your 10 mL grad cyl as a polarimeter tube, the following formula applies: [a] = aobsd/(O.O8-gOAC). (1)(2) One-half of a GPA rxn mixt has aobsd = +10.0°. Assume the only OAC is a-GPA, [a] = +100°. Rearrange the formula, then calculate the mass of a-GPA in this portion. 3% W 1 W“! i (2)(2) Each half of the rxn mixture has 5.0 9 total GPA. On addi- tion of conc H2804, the second half's aobsd rapidly climbs to +34°. Calculate the fraction, or %, of a-GPA now in this portion. Full credit only for a minimum path calculation including use of the result of (1) above. ng/lf=- lfixl-ngthi‘C; Fm. “M = 0-35‘(8§%) (3)(3) a—GPA has mp = 112°; the B anomer melts at 134°. A student got good quality 8, mp ~ 132°, from portion (1) above. The second half gave Prod #2, of mp = 107 - 110°. To test whether #2 was slightly impure a or grossly impure B, she took a mixture mp of B and #2 and got mp = 82 - 87°. Tell what this means about #2, giving your reasoning in <25 words. (b)(3) Suppose one used B—glucose to make the GPAs. In comparison 1 fl with use of a—glucose, as you did, the mfiEA fraction before N do adding H2804 would be (circle one choice in each set separated by /'s) / (nearly) the same / more. i:2i? After adding H2804 the a—GPA fraction would be less / 6H1. " (nearly) the same / more / >98%. -3- 4. (11) This question deals with the solubility and mutarotatory behavior of glucose. (a)(6) When finely powdered anhydrous a—glucose is mixed with water at room temp, it changes solid form and reverses solubility 2 L1 twice. For each situation below, tell what the solid phase is 6hr and what form(s) of glucose are in the solution. solubility /mx¢z?r means the total glucose (wt %) in the satd soln at the moment. (1) Initially, much dissolves fast, undissolved solid powdery, “(fl solubility 37%. {W So/flzulowkyz 9L; J’th i5 .or (2) Within a few minutes, some re—precipitation, solid becomes q/ larger chunks, solubility 29%. 501% a 9L- 1174 Wé 2 waft/0,3; [L ‘ sad” (3) Over the next ~40 min, much more solid dissolves, solubility ;_:.s=: ‘ rises to 51%. Nature of solid and solubility cease changing. \ \ \ \ fo/(‘X (4 oK‘kYJVW/é( 9/91 1/3 Jug oil [bn‘un’ wu‘g Mg (b)(3) The equilibrium mixture of a— and B-glucose is Bza = . . - Tell what the relationship is between the rate constant for a a B, kaefi, and that for a e equilibrium mixture, k Give reasoning/calculation. WWW? pm! a“; MW“; “1:2 fwlees $96,3mokg15A. "L73%V““”" :5? ,A4£L-—+df'ubq‘x~ ‘2 ’Efr’4éfdvflfl/3 (M AL,” 2 fafwv WK) aeeq mix- (c)(2) In general, in using logarithms with values having phyfiigg; units, how does one avoid dealing with the problem of what, for example, the logarithm of moles/L, means? Use <20 wds. hq~¥ q \PMLC‘WOO‘ "/00 CON—(th WM/L/ W /007(5§_7.:Q), {0 (MM? -4- Affiz7 5. (7) This question deals with the osazone formation and reaction to give a triazole. 1‘? ‘f=° fHOH (CI-10H),l + I 5 CHZOH CHZOH Aldose Phenylosam'ne (a)(4) Balance the above equation by putting a coefficient before the PhNHNH2 and writing the other product(s). Remember that one PhNI-INH2 acts as an oxidizing agent. (b)(3) As done in lab, GPO dissolved and rapidly formed a deep red intermediate with CuSO4 (IM). This then decomposed over 25 — 40 min to give the product (P): Dissolution Step 1 Step 2 k - k k dissoln 1 2 ———-—’ GPOsolid '*—_"—' " GPOin soln <hE-—-——-IM """"" P -l The overall reaction rate was not sensitive to the Cu(II) conc. If this conc were sufficiently low, the overall rate could become proportional to it. Explain in < 25 words, é; yea4,- assuming k1 still > k2 and >> k_l. 3 1 W ré/é :JZ‘ Zép0h~(¢q3£(qfflf); rpm) 5:; [cacao] a low WA, EMM #1:” IL- MIL 601411“?me afikrvafl‘ jterB fl elem 5k [rM34/0w« 1a (Neurch %% J (awk— ACC ‘f Duh ox) mfi. ...
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This note was uploaded on 03/30/2010 for the course CHEM 322 B taught by Professor Singer during the Spring '09 term at USC.

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322bsu05_plq3 - CHEMISTRY 322bL/325bL SPRING 2005 rac / B...

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