322bsu09_plq3_key

322bsu09_plq3_key - CHEMISTRY 322bLf§ifitflf~...

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Unformatted text preview: CHEMISTRY 322bLf§ifitflf~ dqnfiflniflhpiulfii- -9?RiNG~3499 I90 [area-(79L act)? Sign LAB QUIZ - J'QMM-er‘ -————--—""'—'f"--—-———--—‘—‘““‘“'—~fl.u~nu--.--_-—"--""-—----p NAME . ______ Lab time T.A. TOTAL (35) This test comprises this page, four numbered pages, and a graph. ab Quizzes will oe - ailable from D-s as usual. ~on a‘d Tue -nly TAB who do not sually have a y - fice hour or ' 1 ill hold sp- ial of.ice hours iFri Apr 30/May 1. If -u initial her- —----> , our - iz will bf put in t e ~tudy Roo- by Mon, M». 4, 5 p ALL - izzes rot picked - willfce thrown out an ." 26. ‘1‘ [021/0 1. (10) n-Nitroaniline, p-OzN-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 1! words, tell what one should do before adding Ac20/NaOAc. ‘J/L era/L ‘ o 6M4 Yd/W M475 afla JC/ (b)(3) Recall that the acetylation of Ph-NH2 was done by mixing Aczo into an aq soln of Ph-NH§$>hefgze "free-basing" it. Given CD /Zénv that the half-life (tx) of Aczo in water is 240 sec, calculate ‘ 6! how much time you have to add base so that only 10% of the Aczo has hydrolyzed. Useful formula: tm/tn a (1n m)/(ln n), 014au~4LdL where m anqh are the fractions remaining. //«dzw /07{3 Mcé fi 70% Hwak‘w, : 0.14‘2. =;> C9043 : M‘ORC k 0"” “flows t 0.4’0 5. {am} = {3&7} (to, (e ’,.a”97;)(3) Using <15 words explain why hydrolyzing n-nitroacetanilide with refluxing 6 M aq HCl, bp ~ 108°, is superior to using either weaker or stronger aq HCl. Considering both conven- ience and speed of reaction. 6 M 17‘C/ h Mar édk/iv qawfvyi—L 2'? Matt r (Mk «fa/é) M Ad (d)(2) 2—Naphthol, ClQH7-OH, has pKa ~ 10. Its anion is the main reactive form in diazonium coupling at pH >6. Explain why even if Ph-N§:’were not deactivated at high pH, it would pointless to operate at pH >11. Refer to the 2-N anion frac- tion at pH 2 11. Recall loglo({A‘]/[HA]) a pH - pKa. flit—NW3”) [4‘] 2/0, m. >?o"/o ' -2- 2. (2) Regarding oeazone formation: When glucose monophenylhydrazone (GMPH) with the indicated 15N- labeled N is treated with excess unlabeled Ph-NH—NHZ, most of the 15N is found in the NH3: CHalsN-NHPh CHaN-NHPh | + 2 HzN-NH-pn -——.. + lSNH3 + Pr:an + H20 CH-OH ' _ CH-N-NHPh If unlabeled GMPH is treated With excess primary amino labeled H 15N-NHPh, C-l only C-2 only both C-1 and C-2 neither 3. (6) ' Dissolution Step 1 Step 2 k k k dissoln 1 2 gposolid ——__._, GPOin soln .L;""" IN ~.____, 1: (é/JOJ’a/n 0K; «40 Cg‘WL 04017) (b)(49 (i) As described above, Referring to 5 given in (a), would this change Explain your answer in <20 words. 3M (0, if ad (04*? /0'§o/d 5/0“): #31 10~5§/a( m flaretfire; filmafiw m5? K 19):? 57‘4/ X meéd. / z=7 nt cetates (GPAs). (1)(2) A portion of a GPA rxn mixt has aobsd a +10.0°. Assume the only OAC is a-GPA, [a] a +100°. Rearrange the formula, then calculate the mass of a—GPA in this portion. ddéf/ ‘_ a j 04‘ ‘: [x3 .- --\r/000 g :l/JJ‘Z: (2)(2) On addition of cone H2304 to the above por ion, aobsd climbs Given that this is a 90:10 mixture of azfi GPA ano- mers, calculate the total GPA mass and then the fraction of d-GPA hefigxg acidification. Full credit only for a minimum path calculation including use of the result of (1 ) above. glob-6W4 now = (-L$3x%:qfiff (b)(3) Suppose one used B-glucose to make the GPAs. In comparison with use of a-glucose, as you did, the MBA firminn before adding H2804 would be (circle one choice in each by /'s) less / (nearly) the same / set separated Interpret the following: “Anhydrous a—D-Glucose, [a] a +53°, water solubility at 25° a 52.7 wt %f Be specific about the solid phase and define the nature of the glucose in solution. 77v. hejck 71?) 213% m 9519) Lidfl/‘é/‘fl/Zlé/ 50/14 (A; fia/flérfM Nt‘fl (cf/f/ 04‘Ayd’Vl/é7. (a)(3) Calculate the specific rotation of a-glucose, stating what datum you are using from the graph. Take [a]e = 52°. Specific rotation, [a], a 12.5 (aobsd/mass), fgr this. the rate constant. (iii) When you became aware of the error, could you correct the rate constant value without re—Elotting the data? Explain briefly. @ M0} ML 6:0 “5 W‘? (61%. 0 ((1) X150) AKC Zr [email protected] (Q U“) yea. J/afl a jwz‘ Ad? ‘ Wc‘f‘ VOLLIIQ $ «é % Q‘C’L (mace/«M CCC) ZMC I‘ll/U)J W mo?’ Mime]. ,_,‘4.9 so” _— NJ _ h: ‘l J I i __._H—.__-_._._.__._..——._.-—..__.—._.__.__.__ __._..__.._.._..___..__._._..___. __.____._.._.._.__._—_.__,.. - —________._.___,__H__-_._______.._H_-_.____.___._._—._____ _. -._.—___—__.__._.__.___—___..——~.._——__._-_F.__._____—..______ __.. ._ _..___..___ -—-._._..._.______.__.___.__—_..__..._._.__.__._—._._ __.“_._..__..._. __._ ____._._.. _____u._.__.._.__——._—____.—.._.—_.__.___.__._._—.__.___ __ __._. __—.___.._...__—.___—__.—_________—-___._.._..-—___.___.__.___ ._ _._._... _____-—_.._._._.y______.—__..____—__._.-.._._—___.—_.__ _. .. .. ._ _—_..__-——_..—..__..___——.___—___‘___.._.____—__.______—__ ; ... __._..__._—.___—-—_.___.___—__—__._.—._._____—___.____.____. __.. .- m .___—._._____.____-—._—__——_._——__._——___...__. “GD- _u—a—__«___.—._.______._—___—_._.—_._.__-—¢___—.____ _— -.—.._. __._._—-___.——___-_._...-—__~——._—.._..-—-____.__———._._ H... __.- —__—_——__—___—q_______—__—_.—__—_. __._. __._... “__._—__._fi—fl—q_—__._—___-____——___—___ -_.._. —._...._..__ __.—-—_._-—___.--__..—.___._.-—_.__.___——.._—.__.__ hid—g __Iu-h“ —-—fi_'———_fl--————n-.fil—Q———Ii-—---—u—u—n—~_I———— _ “__._. _———_—._.__..—____———._—_.__-__m—.__—n_ ‘ "__._... -._.._-._ u—__—__.—_.__...._—__..-_.__._—._-.__._—_._—._ _. __._—__._ -——_._—____—__..—_.____..__—_.._——__.._—____—___-__. ‘ n q __._-..__...._.__ “__._—hm__h___.fi_...____.__._ H--. -—.....—.._ __._—__«__.—.-—._——._-.-—___-——_~—__—___.—— .__..._—-.._ ..._—u..._...... __._ —__..—...__-—__—..._..._—___-_.._-_._.._..—.——_._ __._—__._ ....-—.._.-. _-—._.---—_-———_-_-m...——..__-—_._-.—_.-—._——.__ __._--.— ...__...._.._ __.__-._—u...—__.-—q._u—..._ __._—__._...— -.____.._— u. -n___ q.—_..—-_——__———_——__-—__—_——.—_—— ——_-——.———_._—m..--—._. -—-_—._—-—_—__-——_.——_ _—. um..-“ _—-_.-——_———_ —-.m--_——_—_* -1. .——._——__.. -—_-——-——.-——-_. “nu-n— _--—._———_—__ __._-__.u —~_—.—_—. _——__——_._—_—— __._—__._.“— ' _u-un—wi ——_u_—- __._— hw-flh—n—_—-_-—**H .1.- u-_--.-.——- __._ ----—_--——__ .u. —-- __-— __- ii I $106 ,fube It. cm to (0 ml. ’0) _ - _ ( ' 2E mark ‘0‘ ’41 ' _E 8 H H! l l __-_-__._...— _—'_... -—-____._..——._-.-.-__.. —...-_—.._.......___ I l ! ! —_——-- --— u——__fi___i——- h—H--*m_-fi_‘—-fi-“__—-~—a-IH—_fl- -——.—u_—--.—-_.u——. -——--_—-————_u_-u——-u—- ‘h-n-“h— q-p-u——— -—-——--————--————-—— —--————u---—a—.——--— ——u—_-———-——u—-—.—--————-——— ‘4’ mmmh_—__—— __._— .- — _.u_——..—..._——_¢— u. 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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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322bsu09_plq3_key - CHEMISTRY 322bLf§ifitflf~...

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