322bs09_plq2_1_key

322bs09_plq2_1_key - CHEMISTRY 322bL/325bL...

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Unformatted text preview: CHEMISTRY 322bL/325bL —Ap¥é§-%OT‘€€09 —SPH$NG—3008 m EY Lab t ime‘ T.A. This test comprises this page and four numbered pages. .1. o. 'zzea w' Ts who u '3' availabl ’ TAB as usual. We r I f , . A .« i . g . v i . ‘ 1' r : ri will ho d a a1 offlcle hourfzs Th/F , ay 1/2. If}! 9' - ' I l '7 ’ ' : ou initia here 2 ‘ , your/ Quiz will be“ put I _ v7 . ,LMon, r’May 5 . e trirown Gui: . ,1 rooms as te . l earlier xamg”. “\ i I/ _ x55 3, \D [m 2': / ~30a/o. W “ M 2 * IO 1. (7) Aniline, Ph-NHZ, was converted to n~nitroaniline, p—OzN—C6H4- NH2 (PNA), via acetanilide. Ph—NH~Ac, and n—OZN-C6H4~NH—Ac. (a)(5) For the rxn of Ph~NH2 with Aczo in dilute aq soln: (i)(2) The aq acidic Ph-NHZ soln was treated with decolorizing char- coal. For this to be useful, tell what any side rxn(s) in the acetylation procedure must not do. Use < 7 words. ~ (,0/ [Md] prootqaz, [tn Mac] a {C pr 5. (ii)(3) Acetanilide is best made from aq Ph—NH3C>by rapidly mixing in Aczo and then quickly adding aq NaOAc. Suppose you added 3.5 mL Aczo but forgot to get the NaOAc in advance: The hydroly— sis half—life of Aczo at 25° is about 4 minutes. If you can add the NaOAc soln in 2 min, how much Aczo should you now add? Make an estimate, or better do a calculation. ~( A ((0,, 1,; < ware, 6L. 2: —— «70 /r\. (NIL Mica/fill] (b)(2) In one lab section, a bottle marked "Nitrating Agent" failed completely. Acetanilide dissolved in the rxn mixture and gave solid on pouring it into ice water, but on acidic hydrolysis, no orange—red color appeared. Identify the solid, giving your reasoning in <20 words. [how’th (05% 03 2. (3) For a general arene diazonium ion, Ar-NASK (a) Draw a reso- nance structure not involving the ring which correlates with the ion's coupling with 2-naphthol. (b In <15 words, explain how the fact at one can prepare Ar-NZ in aq soln shows that it is only a weak electrophile. .— .. €70 @ Q N“N® wot—fl W He?“ Utfl ("L 0K 0041‘)» [iryvvz .- éfl’NEé’ Mi“) 3. (9) This question deals with glucosazone (glucose phenylosazone, GPO), and its cyclization to a 1,2,3-triazole. (a)(3) GPO is a his(phenylhydrazone), but its formation requires 3 PhNHNH2 per glucose. One PhNHNHZ (PH) reacts with the C~2 ::CH~OH of glucose so that C-2 can later react with another PH to form a phenylhydrazone. Write a balanced equation, no; a mechanism, showing gnlx the 2° alcohol/PH reaction. \;H .4. \ .. gob, «- MUH‘U/{V ’C_..0 + {hwy}? (b)(3) The overall scheme for GPO -——- triazole is shown below: / Dissolution Step 1 Step 2 kdissoln k1 GPOsolid -—————-——. opoin soln WT“ in ~—-——-——-.. 9 -l The reaction was done in aq i-PrOH with CuSO4 and aq H2804 catalysis. Students who were impatient, heating stronly enough to get vigorous boiling, often got slower reaction. Using < 25 words, explain this. Consider that under more vigorous conditions, all processes should go faster, other things being the same. "-—' O‘H‘Jfl. W W figf (m: (Lima «#47 {VITO/1‘ Mela Mott dome; m +4.)» .2 mg 52.6“ promo «(0604/ X 551014 A! M Mag? f M4”) (c)(3) Decolorizing the reaction soln changed it from muddy greenish to a cleaner green. Yet the triazole was often stained tan to medium brown. This is due to air oxidation to red-brown tarry material of a reaction by-product, used in an earlier lab. Iden ' ' subst e and tell why it was (still) in the soln from which the product crystallized. Use < 20 words. flush-u. [oh—M4... ofiaf (mam hrs «‘1‘ Queen [:4 Mfflhflhfl‘yeé ML: «4. £‘/h‘飓——9MI~, n“. -3- 4. (5) The three solid forms of glucose and the temp range over which each is in stable e ilibrium with .u .- soln are as follows: (1) anhydrous a (55° — §§55, a~hydrate < 55° and (3) e 1 (anhydrous) B (> 98°). Assum- t a - or um at 23° the solution is 50 wt 1: Lana; glucose with W W Exactly 100 g B—glucose is dissolved completely in water. equilibrium, there is 160 9 aqueous solution plus solid(s). Describe the equilibrium system at 23°, giving reasoning and/or calculation(s): Tell how many grams of each component are in the 160 9 sq solution, and state the component(s) in, and mass of, each solid phase. The a-hydrate is 10/11 glu— cose, e.g., 10 g anhydrous gives 11 g hydrate. At +9.0°. After treatment with conc H2804. a this is the rotation of the 90:10 a:B GPA that a-GPA is only optically active component, and that total GPA is conserved. Calculate the GPA isomeric composition hgfigzg reaction with 32804. L; 4/ GM %6; W #6495 $064 : we. c be“; 16% = .35 11"€L 3‘; ‘ ircpis - flakyln‘flt MM 08 flaw «tofu [AIM/ya ~ f5 W (lewd L444 a4 Nrfi/mlon / wk?“ «MAYJ jMwA'" 6. (7) For the glucose anomers, the equilibrium ratio is 5:3 B:a. The rate laws have the same o"- . eaag _ - switching a and 3 gives the equation for Bad. Below the k's are labeled km and k5, respectively. (a)(3) Show that the rate ggnstant ratio, kazkg, must a 5:3; recall at equilibrium the rate for a a 3 must = the rate for B a a. ’1‘ =j/a [dawn] ¢ «4 Cfi'wml m‘ I; - [k4 Mm] (b)(4) The slope of the plot of (adjusted) rotation vs time is = {:3 kaaeq(mixt). Justify in <20 words that ka = (5/3)kageq. Combine this with the result of (a) to show that ka + k3 = k a»eq(mixt)’ :7 QM iwlecs ck raSwwlec/f K) g on Chad.“ 3 awe/€04 f £7 Mt‘xé, 0 me 9'); “£8 = : {.497 - 7r (2 %‘€*4‘7, 1‘? [A “6/3 °Hfifi=£7 At 40°, the solubility of B—D-glucose in an a solution (a:3 = 3:5) is 76%. Calculate its solubility in a solution containing only the B anomer u ,Q C\ 5\ .f f“\ 6\ ,0 “F- SE ...
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This note was uploaded on 10/12/2009 for the course CHEM 322AL taught by Professor Jung during the Spring '07 term at USC.

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322bs09_plq2_1_key - CHEMISTRY 322bL/325bL...

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