322bf07_plq2_2_key

322bf07_plq2_2_key - CHEMISTRY 322bLF$§5bb...

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Unformatted text preview: CHEMISTRY 322bLF$§5bb cJfinfii-ifiT—QOOV —&PRTM3fiflH¥F- #52 prqclll‘a— 73,414 fir SECOND LAB QUIZ ~ g” 07 NAME K Lab time BYEEQE T.A. TOT L (40) This test comprises this page, four numbered pages, and a graph. Lab On” 5 will be available from TAs as usual. Mo e only TAs who do n rually have any off“ our Th or Fri will hold special office hour ri Apr 26/27. If you initial here ----- > , your quiz 1 be put in the Study Room by , Apr 30, 5 p.m. ALL quizzes not p up w' e thrown out on Mon May 21. -1— /7£=/0 1. (10) n—Nitroaniline. p~02N-C6H4-NH2 (PNA), was made from aniline, Ph—NHZ, via acetanilide, Ph-NH-Ac, and n-OzN—C6H4—NH-Ac. (a)(2) On adding red—brown aniline to aq HCl, one gets a pale yellow soln and some dark gum sticking to the flask walls. In < 10 0 4/52, words, tell what one should do before adding Aczo/NaOAc Lflg/érgfiwom (0/:4 (,Q-fp ovum/1n afi/MZ (b)(3) Initial PhNHz: Water mole ratio (mr) = 0.020. Adding one Aczo per PhNH2 gives 75% Ph-NH-Ac, 25% hydrolysis; final PhNH2:W mr = 0.005. Taking the effective average mt as the geometric mean, i.e., the square root of (mrinit)(mrfinal)' calculate the ACZO reactivity ratio PhNszw. gffgcfiv-L aVW My- =W = 0on (at 3 2w WA Marl?“ “ “70‘“!er .-, N = mm :[ 5’00! C Vania) (c)(3) To the extent that direct nitration of Ph—NH2 with HNO3/HZSO4 gives a mononitroaniline, it gives the wrong isomer for this prep. Explain ONLY this, writing a reaction and <15 words. @qm um +— [,1de M14843 @ (3&3 (MIL/y) M?’ 045466" (d)(2) 2-Naphthol, CloH7-OH, has pKa ~ 10. Its anion is the main reactive form in diazonium coupling ion at pH > 6. Explain why the speed of this reaction is essentially the fiamg at pH 12 and pH = 13. Refer to the 2-N anion fraction at the two pH's. Recall loglo([A']/[HA]) = pH — pKa. -2- [$1 2. (6) This question deals with osazone formation. Key steps involve the decomposition of a MonoPhenylHydrazone to an IminoKetone: CH—N—N—Ph __ C/ 5b H—C—OH . \\C__ /__ ow avrwj [1- ”H Mafflfl “”4930“ , 1" (a)(3) Draw the enehydrazino tautomer of the MPH (no mech required for this part); this species has a 6- membered ring inglnding the hydroxyl H hydro en- -bonded to the N attached to the Ph. (b)(3) Now show a simultaneous movement of 3 electron pairs to give h K; h 1 h h d . Ia t e I if :;4: so“:d:\7: er pr? ’u:2,14;; ’K7I1b(/ 3. (4) For the triazole prep in W/i-PrOH, G1ucosePhenlesazone dis- solved and formed a deep red complex with cuso4. The color faded, and product was isolated after work-up. Dissolution Step 1 Stop 2 I"dis-c.9111 k1. . k2 GPO-011d GPO“ I01!) ‘—-—I" '_""’ P -1 (a)(2) Given that the cu complex is an intermediate, tell which step, 1, or 2, was rate—determining. Explain in < 15 words. 561,09. 0% (m m y“ [fill “(Mm/«ll, a4 (1 almjm C 6 @4332} W“ (b)(2) In plain water, the rxn temp is higher. Product yield is good, but the overall reaction is much slower, and the red color never appears although CuSO4 is still catalytic. k1 is still > k2, and >> k_1. Explain in < 25 words. ”(M 64 Ml: (Ml//%'0U{fl// wéc mm Le 7 my arse.) M MMAHW LE)“ th/é€ N‘”0 4. (10) This question deals with the Glucose PentaAcetates (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/(0.08-gOAC). (1)(2) One—half of a GPA rxn mixt, Sampl, has aobsd = +12.0°. AsSume the only OAC is a—GPA, [a] = +100°. Rearrange the formula, then calculate the mass of a-GPA in this portion. = OKOA‘J : (2.0 (740% “3.0.0:: WWW? {mow (2)(2) The other half of the rxn mixture, Sampz, has 5.0 g natal GPA. On addition of conc H2804, the SampZ's aobsd rapidly climbs to +35°. 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. (3)(3) a-GPA has mp = 112°; the B anomer melts at 134°. A student got good quality 8, mp ~ 132°, from Sampl above. SampZ gave Prod2, of mp = 106° - 109°. To test whether #2 was slightly impure a or grossly impure 8, she took a mixture mg of B and Prod2 and got mp = 79° - 85°. Tell what this means about Prod2, giving your reasoning in <25 words. b)(3) Suppose one used 8- -glucose to make the GPAs. In comparison 1 with use of a-glucose, as you did, the MBA fraction before adding H2804 would be (circle one choice in each set separated 2 g, by /'s) / (nearly) the same / more. r After adding H2804/ the a- -GPA fraction would be less / A {’4 (nearly) the same >98% 0 . -4- 5.(5)(a)(3) At 25° the water solubility of a—glucose monohydrate (AH) with only a-anomer in soln is 30 (wt)%, i.e., 30 g a-glucose per 70 9 water. Assuming this ratio is unchanged with B anomer also in soln, calculate AH's solubility in a mutarotat- ed soln, anomeric composition 5:3 3:0. Hint: Recall how solubility is defined for this system, calculate the mass of B-glucose in equilibrium with 30 g a anomer and note that it affects both total glucose and total solution masses. fluff- 6 m‘ azur‘fléfl'm WVfl 346A «Q. NIH: 305% 3rd?) OK‘W’UI‘ 30 {-50 72.482 : {livfl 70*g0kabjguw “’0 flaw”, M 3,. J (00 4-51) (b)(2) If the above result is approximately the measured value, tell what this means about how each anomer affects the saturation gonggntnatign of the other. Use < 12 words. a «(a M M N 1 Cf KM [afiwwflxdamm fiaifigfl 6. (5) Note the attached graph for the mutarotation of a-glucose. (a)(2) Calculate the specific rotation of a-glucose, stating what l;r'”':ETdatum you are using from the graph. Take [aleq = 52°. SpeCific rotation, [a], = aobsd/mass-0.08, for this. oL‘Ka‘l CvLJ (30 C67 afllnloyla/lltvn 1‘1! Uwfid ctr/5'2 “it; \% 0 "" ° 5 0] = lbs/50.08 = 53.75' C ‘f {‘vamw = -=["3v4~a(~m (b)(3) Suppose all the times on the graph had the same additive error, e.g., all were plotted 3 min later. (i) Would this affect the calculation of (a), just above? (ii) Would it affect the evaluation of the rate constant? Explain each 1% answer briefly. W I (2} YM) i=0 «ITUQWL {1,4de 2% m) N0; 5/. 1%. wfi‘rcfé/ m. 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