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322bf05_lq2_key

322bf05_lq2_key - CHEMISTRY 322bL FALL 2005 SECOND LAB QUIZ...

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Unformatted text preview: CHEMISTRY 322bL November 28, 2005 FALL 2005 SECOND LAB QUIZ BY Rage NAME I ; E i l.(13)______ 2 (7) Lab time 3.(ll) T.A. 4. (9) This test comprises this page and TOTAL(40) four numbered pages. Limit your answers as directed in the questions. Deduction for irrelevant information even if correct. EQEE; FINAL EXAM is Mon, Dec 12, 8 - 10 a.m. in SGM 124, the regular lecture room. TAs will have quizzes at check-out and office hours. Tuesday morning lab check-out starts at 9:30. -1- flJ£$I3 (13) Recall that aniline, Ph-NHZ, was converted to p-nitroaniline l. (PNA) via acetanilide, Ph-NH-Ac, and p—OzN-C6H4—NH-Ac. (a)(3) Ph—NH2 reacts directly with HNO3/I-IZSO4 giving a poor yield of total nitroanilines. Name the side reaction and correlate it with how the Ph—NHZ you used looked; use <15 wds total. “/Zi) flu Oxl‘c/u4‘ovx; H1003 ‘m, Ora/«21w; 776$ Dal/i [0 W (- (4' @ M VZZ (é flan/(1‘7 «2% . (b)(3) One made Ph—NH—Ac from Ac 0 in If one starts with mole ratios of 50:1 H20:Ph—NH2, 1:1 Ph-NH2:Ac20, and gets Ph-NH-Ac in 80% yield, tell why the Ph—NH2:H20 reactivity ratio must e con31;i:r 1y 2 203;, a ‘ ¢N aCYlfihm @ f0 kYM : Ire—ac NINA yoc/ oat/MUM, :prv = MO.)6¢ owns!» w‘cvem Q ”a Na L’fidfdz/j 9'7 W‘ WILLC W446} riL (c)(3) When n-OzN-C6H4-NH3 -———> PNA using ammonia as the‘éLse, 4(- some nucleophilic aryl substitution occurs. Yet PNA is ob- tained in ~100% yield. Explain with structural drawings and <15 words detailed mechanism unnecessary and irrelevant. 1,1,3”: #5 (HOT 0x «41(4) mcl aM/l (“4,74, /’ (‘NJ fl HgN‘vdd n-fllu'k. (79 #101, A/Eéj ”~; I lb)ézg 67M;7{ 2. (4) B-Naphthol (2-H), C10H7—OH, has pKa ~ 10. Its rxn rate with aq Ar—N2(D is 1000K greater at pH 9 than at 6. (a) Based on this, tell what the reactive £93m of 2-H is, explaining in <15 wds. (b) Explain why the rxn rate increases <10% as pH goes from 11 a 12. Recall that pH — pKa = log10([A']/[HA]) @ C4) “1% am moan, M (MWWM ( Jo Arm [4%(‘013’ :7 (MM/Y Aka/73m ‘7‘!) With}? sérwl‘ C? (A) M [ml-“”2 It‘ll/4’1") MW»! 770% Maw], 44‘ w-«n, imp/14 MM M/Ma 4' pée7 3. (18) In Pds 6G, 9, and 10, you performed some glucose chemistry: OH on 7 pts on E; . 2 a“ 0 this page . vie/fit. “0%” H0 HOOH HO HO (equfloflll) (axial) a-D—(+)—Glucopynnose B-HH-G‘Wm (a)(4) An equilibrated glucose solution contains 96 g B— and 54 g a-anomer. If this solution is slowly evaporated at 65° until lflfl 9 total glucgse has crystallized, give (a) the composition of the pregipitate, and (b) the composition of the 50 g glu- cose remaining in soln. Explain your answers, showing calcu- 1ation(s) as necessary. The three crystalline forms of glu- cose in equilibrium with solns are anhydrous a (metastable <55°, stable 55°—98°), a-hydrate (stable <55°), and S (stable >98°). Note that a-anomer in eglntign is neither anhydrous nor hydrate(d), it is just aeangmer. (a (a) Pace/off”? oi [Maelfl @% finghm. £1193 erfd/c he” L4,“; (1046(9me of (IX; (an 6 Am a»: K pé‘) (1) (Lax/Jam M (W Wfioxrfi‘m MafA/inuumif (:7 323g! ”(7,4 (b)(3) Two students, A and B, measure the solubility of a—glucose monohydrate (the hydrate) at 25°. Each shakes a large excess of finely powdered hydrate with water. A then immediately filters the mixture; B filters it after it has stood >1 hour. After evaporation of the filtrates, A finds the solution was 30% total glucose; B finds 50%. Assume each is correct. Referring to which component's cone in the soln determines the hydrate's solubility, explain the difference in the results. Use <20 words total; some data in part (a) may be useful. #76006? W‘é 0M :65. _% .. 3 .. ,, (3. contd; 11 pts this page) /J i " // (c)(4) In the Glucose PentaAcetate prep, a—glucose, Aczo, and gate; lytig NaQAg gave mostly B-GPA. (1) Explain this in terms of glucose anomerization and acetylation rates (<15 words). (2) If NaOAc is replaced by H2304, the facts about the rates of (1) are the same, but the main product is a-GPA. Explain in <15 words. (game, lcuwé fl / @ m Naoflc ccf JLW “K 3 (d)(4) A glucose solution contains 1/3 a-anomer of [a] = +111° and 2/3 E-anomer of [a] = +21°. (1) Calculate [almixt° (2) Calculate this mixture's observed rotation if one has 5.0 g total glucose in 12 mL aq soln, and 10 mL soln has path length = 0.8 dm. FORMULA: [a] = aobsd/{(g/mL) -1(dm)}; full credit only for the simplest possible calculation. @3 (I) ($35“sz Gmc JOE/OK (39“ 8>Cok36 = {fa/1°) r %(110)=(37 [email protected] cuff 2 s/° 9 9.5.. : 21>}.t l‘ «(65:41. hug: ( ax‘oé’) ‘5’ (IL/4 My 3m'°.0$’ 5km“ U¥e)(3) The rate constant k for a first-order rxn A a B, is the slope of the (linear) semi-log plot of [A] (or of some property = (constant)[A]) vs time. If the rxn is reversible, one must plot [A] t)'[A]final to get a linear plot. (1) For what 2:99:33 oes the slope of the line then give the rate con— stant, and what does one need to know to calculate kAaB from the slope? Answer THIS QUESTION ONLY, in <12 words. {QM a «é fl A4 goal. MpK/‘w‘ Ulwé 141M Wflfl ¢67 mtxfi was? 4. (9) Recall the Glucose -> Phenlesazone -> 1,2,3-Triazole sequence. ‘1‘ ‘? (i=0 (|:=NNHC6Hs CHOH C=NNHC6HS I ($30K), + CGHSNHNH1 —> (anion), c1110“ (6 XCess) CH20H Aldose Phenylomzone (a) (4) Circle the best answer in each set of choices: prepares an osazone from an a—hydroxyaldehyde, PhNH-NH has two roles. One role is as a(n) araphile / nucleophile / electrophile / stereofuge / radical source / chiral auxiliary. When one 2 £1 Since a ECH-OH e >C=N—, one PhN‘HNH2 must get cyclized / —Q¢C / dehydrated / / oxidized / derearranged. Vvo +.\ Z Dissolution Step 1 Step 2 [QT I k dissoln k1 k2 . ___.____.> . -———" GPOsolid V GPOin soln ‘—-——k I" _—_"' P ' -1 (b) (3) As you did the triazole prep, the figmtign of the deep red Cu(II)-containing Interflediate was fast compared to its Leas; gm giving (P) roduct, i.e., the latter step was rate determin- ing overall. But if one starts with a much lower CuSO4 conc, the gygrall rate slows. Explain in < 20 words. 9 w WA {cum} I 1126/53 {drag 41;E£M3.flw «$4 Wm // ,CM aflj oJW woéaw/ W W m [Cot-Ma, (c) (2) In osazones, the :C=N—bonds of the oz—diimine group, —N=C—C= —, are fairly readily cleaved. In 1,2,3—triazoles, these bonds are much more resistant to attack. Explain this, referring to the character of the ring; use < 7 words. £1”? 54.. qwamaxll‘c‘COSQeomQ a3 nofj ...
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