322bf07_plq2_1_key

322bf07_plq2_1_key - CHEMISTRY 322bL anmmtfimké77‘2306...

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Unformatted text preview: CHEMISTRY 322bL anmmtfimké77‘2306 —PA££—2096 1H Pracfm gr SECONDLABQUIZ‘ 07 m KEY Lab time BY Eigfi This test comprises this page and four numbered pagesJ plus junta“. will be available at lab check-out a hours; Tu a.m. lab meets will be available at SGM your quiz wil 9 Fri Dec 1, quizzes initial here 2, l unclaimed tests will be thrown out Jan 1 I Fri a.m., , 2007. put out in the study room, (1 FINAL EXAM is Mon, Dec 8 - 10 a.m. in the lecture room, SGM 124, as were all previous exams. (a)(2) To get clean Ph-NH-Ac from aq Ph-NH}? (6) Recall that aniline, Ph—NHZ, was converte -1- £:/0 to p-nitroaniline (PNA) via acetanilide, Ph-NH-Ac, and p-OZN-C6H4—NH—Ac. I0 one added Ac20, swirled, and then added aq NaOAc. Reversing the order of addition leaves some unreacted Ph—NHZ. Explain in <20 words. Answer THIS question only--possible NET negative otherwise. fqu/ NaO/itc {(1rny (I'M 6? (Cato [eh]. {Wk Gyro gig-Ice? flL‘Nfi/L’OM 2‘4)“ Mtg mf‘ Mcf. (b)(4) Give two reasons why the hydrolysis of p-nitroacetanilide was done using refluxing 20% aq HCl, bp ~ 108°. One reason re— lates to ease of handling; for the other tell why the rate is g9; slower than with "conc"(38%) HCl. Use <40 words total. 2 ,9 mm a w» M «WW \ 6 kt “Xi gag} ,4“ mm ,o/ 3 1 TM 4/1 /x “W 4“ cant, MJ (4) 2-Naphthol (HA)'s pKa = 10. Assume its anion, A", reacts 105x faster than HA with Ar—Néa. From pH — pKa = loglo([H1/[HA]), the rxn rate falls 10X per pH unit from 9 a 6. (a) Calculate the pH (pH=) at which amt prod from #3,: amt prod from HA. (b) Explain why from (pH: —1) to pH = O, the rxn rate stays nearly constant; use <15 words. - 2 — {{Jf f: :LI’CD 3. (10) This question deals with the formation of glucosazone (GPO) and its cyclization to a 1,2,3—triazole. (a)(4) Balance the chemical equation below. Include correct coeffi— cients for all species, including three gins: Products. c=o C=NNHC6H5 H» , #11011 J (:I=NNHC6H5 f“ [9 A N [71 1/ +' N 3 f. 14’; ($HOH)H + QHSNHNH2—> ((|:H0H)n .6 M W t‘ pilsggzzm a) Wu; (b)(6) The overall scheme for GPO —————’ triazole is shown below: ' Dissolution Step 1 Step 2 k - k k d(issoln) 1 2 GPO - —————> GPO- —-—" m ____., p solid in aq i_PrOH in soln d—k 1 (1) (3) Given the usual observations, namely that solid GPO dissolves rapidly with simultaneous formation of the deep red IM, tell what this means about the rate-determining step and the rela— tive values of kd, k1, and k2. Use <20 words. £05 2: 57‘ L CIM [447‘ #Je (7m! jg .L,/, (2) (3) A few students got slower reaction with slower GPO dissolu- tion. All had heated more strongly, boiling away much of the i-PrOH. Explain these results in <25 words, referring to which step(s) are slowed and why. CJW‘ law ("Iota/1‘ 6'00“; I W4 1w “ML, m. 14 #6 W2 -3- 4. (5) The three crystalline forms of glucose and the temperature range over which each is in stable equilibrium with a saturat— ed aqueous solution are as follows: (1) anhydrous a (55° - 98°), (2) a—hydrate (< 55°, MW = 198, i.e , 10 9 glucose corresponds to 11 g hydrate), and (3) (anhydrous) B (> 98°). An equilibrated solution contains 48 g B- and 27 g a—anomer. The solution is slowly evaporated at 30° until 25 9 total glucose remains in solution. Explaining your answers, give (a) the comp sition and amount of each solid that crystal- lizes, and ( ) the amount of each anomer in solution. 5.(6)(a)(3) A 10—mL filtered portion of the GPA reaction mixture stands 8 cm high in the polarimeter tube; it has an observed rotation, aobsd = +13°. .After adding 2 mL conc H280 , aobsd rises to +45°. The speCific rotation of a-GPA = +10 °; take that of B-GPA = 0°. Assume the final GPA mixture is 90% a and 10% 8. Calculate the B-GPA fraction before adding the acid. Full credit only for minimum path calculation. :2 +130 :7 %=~0.26 (26%)=i7‘f% oV" W030 W,” 8 / (b)(3) The literature mp's for the GPAs are 134° for 3; 112° fOr a. A student gets 130° - 132° for his/her 3, 103° - 111° for a, and a mixt mp of equal amounts = 116° - 120°. Assuming the B is >95% pure, interpret and explain these data; use <15 words. Sue“... «Aaf 6%.; “#va up, gamma A :2. «c 7 '“ /7€=7 6. (9) The mutarotation of glucose is a first-order reaction, i e., the reaction rate is proportional to the conc of the starting material: —d[A]/dt = k[A], where t is the elapsed time and k is the rate constant. The solution to this differential equation is ln([A]/[A]O) = -kt; note that the - sign means k is positive since [A] is decreasing. In a first-order proc- ess, the rate of relative change is constant, e.g., if 20% reacts in one minute the 20% of what remains reacts in the next minute. Specifically, the half—time (or half-life), t”, the amount of time for [A]now a %[A]now, is often reported. (a)(3) Refer to, but do not detach, the following page, which is a semi—log plot of optical rotation data vs time for a—glucose. Determine tx, including units, from the time interval between two values on the vertical axis. Mark the plot and/or state your reasoning. 15y» =- /3’7’0 YMQ .A‘M (b)(2) From the basic equation, ln(%) = -kt%. Use this and your answer to (a) to get the rate constant, k; include its units. 1,4 AU/flaaé‘l? €01) =3 wig. :-o.1 _ 0‘37} a<EvIQTM¢CVQ ‘i’;:E:;’~ N (c)(2) The rate constant gotten from the plot is k for a—anomer » equilibrium mixture. Given that the eq mixt anomeric ratio is B:a = 5:3, calculate kaag from (b). ICMLQ e7 lM/(‘Lf v3 '72 fl £$_.g> gjfigvml.Kf-: (d)(2) Note that the first data point, the one at 7 min/18°, does not "fit". It was in fact ignored in drawing the line. Rational— ize its position, given that time was counted from the begin; fling of the dissolving of a—glucose in water; use <15 words. flof 4,6? 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322bf07_plq2_1_key - CHEMISTRY 322bL anmmtfimké77‘2306...

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