322bf05_plq2_2

322bf05_plq2_2 - CHEMISTRY 322bL W #2 «armies @7424)»...

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Unformatted text preview: CHEMISTRY 322bL W #2 «armies @7424)» Fq// or ,_ BY Bag: (A— NAME 1. (12) 2.(10) Lab time 3.(11) T.A. 4. (7) TOTAL (40) This test comprises this page and four numbered pages. A7th —NH2 (PNA), was made from aniline, 1. (8) p-Nitroaniline, p—OzN-CGH4 and p—OzN-C6H4-NH-Ac. Ph—NHZ, via acetanilide, Ph-NH-Ac, (a)(2) In this prep, the starting material was decolorized. Tell what behavior of a subsequent reaction could, in general, make am; .7 m?» “Mg/VJ «WWW "“4 "’9 601M”! (flm’é mole ratio = 50 in the rxn soln. If 1.1 mole (b)(2) The Water:PhNH2 2 converts 90% of it to Ph—NH-Ac, calculate ACZO per mole PhNH the minimum PhNH2:W reactivity ratio. Ll“? %fe W O-Ewgv 461/6 (“£0 moé oar/11 LJ (1-! ~04). {p “jab-2 5‘0 K 59:92- Z25? (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-OgN-C6H4-NH-AC under reflux; use <12 words. (ii) Tell why 6 M acid as fast as 12 m in this case; use 5 29 words. the structure of the intermediate resulting from at C—1 of the 2—naphthoxide anion. (1) Draw the for attack by E+ at 9:3. (2) Explain even though C-3 2. (4) Below is attack by 3* corresponding structure why substitution on 2—naphthol occurs at C-l, is less hindered. £2 CC) ¢g// ;¢;\ .2 H ED rm. 7 a. A 3 E. c4 :11] eff W? L 5 4- _- hum TM WM 73 [J PM (mush ’L‘CV rt”. .. 2 _ ’- - .151 Z ’ 0 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/(0.08'90Ac) . (1) (2) One-half of a GPA rxn mixt has aobsd = +10.0°. Assume the only OAC is oz-GPA, [ct] = +100°. Rearrange the formula, then calculate the mass of oz-GPA in this portion. 7; “0&5 1° 7: 30’“ flfiiggg “ 77570.73 1 3‘ (2) (2) Each half of the rxn mixture has 5.0 g total GPA. On addi- tion of cone H2804, the second half's ozobsd 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. gxg/If-T- 3/‘V—lef’3‘2‘fi17; /0" Fa. °%%=0.§f(k%) (3)(3) a—GPA has mp = 112°; the B anomer melts at 134°. A student got good quality 3, 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 3, 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. 153° I451 0"“ W‘s/7 ““Y’Wflvcflfl, Mmlo (Airfl‘ 5 “In” ée (1&1; [07..1/00. \ z W W WW am“ “70’: mm. (b) (3) Suppose one used B—glucose to make the GPAs. In coEarison 1 %L with use of a—glucose, as you did, the 9:925 fraction before adding H2804 would be (Cerle one choice in each set separated by /Is) / (nearly) the same / more. After adding H2804 the oz-GPA fraction would be less / red, 3% / more / >98%. 6M1. mum Cov-v -3- 321/ 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 I" twice. For each situation below, tell what the solid phase is 43% and what form(s) of glucose are in the solution. Solubility fflzLZ?— means the total glucose (wt %) in the satd soln at the moment. (“M smug/“Zamkyfilah’d’” i5 Mag/‘- .or (2) Within a few minutes, some re—precipitation, solid becomes larger chunks, solubility 29%. fi’Mk¥ 501% a] vL-A75{¢W6,' fJ/ACSIZQ'JZZ; “AR (0,” (3) Over the next ~40 min, much more solid dissolves, solubility 1,..gzz ‘ rises to 51%. Nature of solid and solubility cease changing. \ \ \ \ Ila/PX (A. ok‘kY/YQ/é( .(d/Pl 1/! $;fl‘e1l4tfén‘uu1 M‘E MQ (b)(3) The equilibrium mixture of a- and E-glucose is B:a = . . - Tell what the relationship is between the rate constant for a a S, kaag, and that for a a equilibrium mixture, k Give reasoning/calculation. \ Ion”! fof‘) UM fwlecs ofiqflt3m$gffiéi£~ “AbafiflynyuL :f? .44£L~«+47 ufl1‘r 12"j§;/4€f4«~‘/3 ’ZuL—afi -'-‘ MEX) ( /‘;‘ (1) Initially, much dissolves fast, undissolved solid powdery, aeeq mix‘ (c)(2) In general, in using logarithms with values having physiggl units, how does one avoid dealing with the problem of what, for example, the logarithm of moles/L, means? Use <20 wds. L0 ARM/“Ca gfifyxr [fl/0 “7 Mr A“. Mr )w W {633/ 4 match“? "/00 W74 “MM/L; ma, MW aflfifl), n w”? “WK C -4- 5. (7) This question deals with the osazone formation and reaction to give a triazole. CHQOH CHZOH ' Aldose Phenylosazdne (a)(4) Balance the above equation by putting a coefficient before the PhNHNH2 and writing the other product(s). Remember that one PhNHNH2 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 G—-———IM _—’ P k-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, é5W§§eh4,- assuming kl still > k2 and >> k_1. 1L3! M “(74? 1 M Mg ; ,é z womggom; rpm) pg [0401)] a AM “4147A, L‘MM m. M? bug‘MuMa/J 053er. [L830 1”“ cam ac [gals/om m ouemd 3?ij J (awe 46c g MM 23am) ()Vrzzi' ...
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This note was uploaded on 01/30/2009 for the course CHEM 322BL taught by Professor Singer during the Fall '07 term at USC.

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322bf05_plq2_2 - CHEMISTRY 322bL W #2 «armies @7424)»...

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