322bs05_plq2_1

322bs05_plq2_1 - CHEMISTRY 322bL/325bL 10rd (AV-e Mflér...

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Unformatted text preview: CHEMISTRY 322bL/325bL 10rd (AV-e Mflér SECONDLABQUIZ " \9’ 0r . BY Eflgfi NAME l. (12)._____ 2. (9) Lab time 3. (10) T.A. 4.(10) TOTAL (41) This test comprises this page and four numbered pages. Lab Quizzes will be available from TAs as usual. Mon and Tue TAs who do not have scheduled lab or any office hour Th or Fri will hold special office hours Th/Fri.W NOTE: FINAL EXAM is Wed, Mayq’, 8 — 10 a.m.v,1-rrr-m xams - an e a @122 . WW. -1-- 1. (9) Aniline, Ph-NHZ, was converted to p-nitroaniline, p—OzN-C6H4— NHZ (PNA), via acetanilide, Ph-NH—Ac, and p-OzN-C6H4-NH-Ac. (a)(3) Suppose on reaction of a 2 mole % aq Ph-NH2 soln (rest essen- tially Water) with 1.0 mole AC20 per mole Ph—NHZ, one isolates a 75% yield of Ph-NH-Ac. Showing your work, calculate the PhNH2:W minimum reactivity ratio toward Ac20. 75% YZJ 4 MAW/lg =7 2:22 vw wiflt W, N: gov?) x. 1% = [157050) (b)(3) Acetanilide was in fact made from aq Ph-NH§:>by adding Ac20 and aq NaOAc, in that order. In <20 words, explain why this gives a cleaner product than does adding NaOAc first. Answer only THIS QUESTION; possible NET negative for irrelevancy(s). MW W ~ _ CWQ m‘Ulzlt—a “£71459? Jazz/gm)ch AIL flag. «WW (c)(3) For Ph—NH-Ac » Q—OzN-C6H4_NH_ACI fiQLQLQQLQLXZQQ amide ydroly_ . sis is an undesired side reaction. But excess H2804 slows the side rxn; explain with a chemical equation and < 12 words; /4< Vfivoq LAW”? 0/1“ 440+ *Hw f”? #So‘eaHvlgda 2. (3) The coupling rate of 2—naphthol with an arenediazonium ion, Ar-NAEZ increases rapidly with pH from 7 to 10, rises more slowly from 10 to 12, and then decreases at pH 2 13. Explain ONLY the high pH rate decrease, using simple chemical e (s) and < 15 word (9 pts this page) a 3. (12) This question deals with glucosazone (glucose phenylosazone, GPO), and its cyclization to a 1,2,3-triazole. (a)(4) A key step in osazone formation is the internal redox decompo— sition of glucose monophenylhydrazone (MPH). Redraw the MPH tautomer below as a six—membered ring including an H-bond, then show its decomposition by a simultaneous three—electron— pair movement; show the products also. H _H 5’ I . .I- I . //~/" / .CH—Ny—N—céHs —— c: ( 70kg —- <\ +HLN 4% II- C-‘-’-Q—H ’ ' 0 v Vow—H“ 0 (‘3‘ arrive): IMJ saw 6M Awfao C64), (b)(5) The overall scheme for GPO -—-——) triazole is shown below: Dissolution Step 1 Step 2 k - k k dissoln 1 l 2 GPOsolid "’ GPOin soln d$——————-IM " P k—l As done in lab, GPO dissolved rapidly, giving an intensely red Cu(II)-GPO complex (IM), which then disappeared over 20 — 30 (:::> min to give Product. Based onizhese data, identify the rate- determining-step; explain in < words. K m 2 .2 ME. 52M WW“ “4 tank/[W WC I‘KC’W M W 0Q¢W {/ow/r. In plain water at higher T, but with same reactant amounts, k1 and k2 are both greater than above (and k1 still >> k_l). But complete reaction takes 5 — 6 hours, and the red color never appears. Explain in < 15 words. {fewer 0\KV‘4[/ MU ttfflwwjé [07,“ @ lad/4% W ~49 N” ()1 Mo é Jada/L, mmz‘ EJ- m 79 m MK W[£K =3 its form mg Jo .r/ow, 3. (concl) 2 (c) (3) Write the structure of the bis(phenylhydrazone) which would cyclize to the triazole below (aniline is the other product). I R ’ ’ N N :N I : 7' NH I'VI/ u \ | (411 m c M /04« Ph 4. (14 pts, 7 on this page). Other glucose work you did involved its solubility, its mutarotation, and its pentaacetates (GPAs) . ('31! , OH | ‘ . CH: 0 CH2 0 “0% I 110% on HO 1 HO OH H0 H0 (equatorial) (axial) a-Da(+)-Glucopyranos¢ B-o-(+ )—G|ucopyranose (a) (7) The water solubilities at 25° of the three solid forms of glucose in an WM m are as follows (weight per weight of solution): oz—anhydrous 62% (metastable with respect to its hydrate below 55°); ot—hydrate 51%; B—anhydrous (only solid 8 form) 72%. Exactly 70 g finely powdered B is fig shaken with 30 g water. Tell what form(s) of glucose is/are /‘ in soln, and as solid(s) if applicable, at various times: @ (1) After one minute, some solid remains undissolved. sob-4 a, 6 ; folk Qua/ij 6‘ (2) After an hour, no solid is‘ present. Q jolt/k (21’ wk fl MFXfW. (3) After a week, a little < 40 g solid has precipi ed; the system has stopped changing. , / 62,0 '3in 10““Jgkyd qC/Df “(v/FA W q ~ \ l.‘ \ a , J g of; {VA}. .(A’JJA J A A M( . I — 4 - 4. (concl, 7 pts this page) 2 +1 0 , (b)(4) A solution contains a—GPA, MW = 390, [a] = as its only optically active component. Sixty percent of rxn mixt soln ube 0.80 dm to the 10 mL mark. ield f om the wholo reaction mix- a—glucose. Recalling that one started with solid a- explain why one got only about a 4:1 Bza GPA ratio. Use <15 words; ZERO unless answer this, and ONLY this, question. I Some .QQQW (Kw/WV) aCy/afe/ 6% {Md/m {WM adv £636qu first—order reaction from the 5. (3) To get a rate constant for a the logarithm of slope of concentration vs time, one can plot 22%;}?C) any property that is proportional to the concentration, i.e., conc = (some constant)-(va1ue of the property). In fact, one need not even know what the constant is! Explain why not in 7%k_ 699%? "<20 words, giving a numerical example. at? W/ LOO MW m/w‘vue ...
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This note was uploaded on 10/12/2009 for the course CHEM 322B at USC.

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322bs05_plq2_1 - CHEMISTRY 322bL/325bL 10rd (AV-e Mflér...

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