322bs05_plq2_2

322bs05_plq2_2 - Lab time This test comprises this page and...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lab time This test comprises this page and four numbered pages. 1 (9) Recall that aniline, Ph—NHZ, was converted to p—nitroaniline (PNA) via acetanilide, Ph-NH—Ac, and p-OZN—C6H4-NH—Ac. (a)(2) The main menonitroaniline from reaction of Ph—NH2 with conc HZSO4/HNO3 is the meta isomer. Considering the very fast reaction between Ph—NH2 and strong acid, explain this using a chemical equation and < 15 words. M—MLA. + W * 49% . (b)(2) To get clean Ph—NH—Ac from aq Ph—NH3 , swirled, and then quickly added aq NaOAc. between additions, the reaction fails. Explain in <20 words. Answer THIS question only-—possible NET negative otherwise. flax) Mdrvlfaed r—Q f/flflC. #oyh w«‘[{ M?!’ :03; (c)(5) Usually acetanilides, Ac-NH—Ar's, can be cleanly cleaved to Ac-O and Ar-NH2 by aq‘DbH. But when p—nitroacetanilide is one added Aczo, v If one takes 31 min boiled with 10 M aq NaOH, the amine part gives largely a sharp smelling gas and a phenoxide anion. (1) Draw the structures of these two products, and (2) name the reaction from the standpoint of the aromatic ring; three letters suffice. A) (AC/[u AF/"C (I) NHJ «M G) 1 NHL 2—Naphthol's pKa 10. A 0.01 M soln at pH 9 couples with Ar—NZC> about as fast as does a 0.1 M soln at pH 8. (1) Explain (<20 wds); consider the Reactive §pecies. (2) A 0.001 M soln at pH = 10 reacts half as fast. Does this mean there is a different RS at this pH? 07 TL'V k: s 2. (5) Explain in <15 words. 0J4 \cVaZI’ C74 #6 M4 ‘50, 1“} 04 ‘m 0.0 /fl p7, £4,115 qul‘d‘t 3. (9) - 2 - This question deals with the cyclization of glucosazone (glucose ghenlesazone) to a 1,2,3-triazole. (a)(3) Note the structure of GPO, below. Draw the structure of the triazole to which it cyclizes. OK to write the tetrahydroxy- butyl group (circled) as just R in your drawing. CH=NNHC6H5 =NNHC6H, (1) (2) Gposolid Dissolution Step 1 Step 2 k - k k dissoln 1 2 —_._——’ ._________, spoin soln k In .______, p -1 (3) Given Step 1 Keq >> 1, and >1 equivalent of Cu(II) per GPO: Based on your observations, would increasing the [Cu(II)] speed up the overall reaction? Explain in <20 wds. (,{utcléelY} fM ML Wf 1 6463:) M 6400‘ Max Kiev“, 1, 771,1 [FM] Mf' VRCWé/ fill/pm {Cu(II)} (3) After the reaction was complete, treatment with decolorizing charcoal (DC) and hot filtration caused a visible change. Tell specifically what the change was and how it indicated that certain impurities were removed. Use < 20 words. M (M) (0/44 P—t’ /(:7/of%l [44% New (m arm/yeZZr/w) §2z1~V7r Sllfgg /0V7¢1a%c7{£? $004 =7 0am mWU£4/. -3- 1 4. (5) The three crystalline forms of glucose and the temperature range over which each is in stable equilibrium with a saturat- ed solution are as follows: (1) anhydrous a (55° - 98°), (2) a—hydrate (< 55°), and (3) (anhydrous) B (> 98°). Assume that the equilibrium com-osition in solution is 63:37 Bza, and that the £3..l glucose in so u o' s 0 wt % at 23'. A system contains 140 9 total glucose (anhydrous basis) and enough water to make any solid hydrate(s) plus 200 9 aqueous solution. Describe the ultimately stable system at 23°, giving reasoning/calculation(s). Tell how many grams of each component are in the 200 g aq solution, and state the compo- n-u in and mass of each Solid chase. The a-hydrate is 10/11 glucose, i.e., 10 g anhydrous gives 0 'vo . e. 5.(9)(a)(6) B—Glucose pentaacetate (B-GPA) is the majority anomer from reaction of a—glucose with acetic anhydride and NaOAc, al- though the a-GPA is 90% at equilibrium. Tell what this means (1) about how NaOAc interacts with a-glucose, (2) concerning how fast Aczo reacts with the components resulting from (1), and (3) about the equilibration of the anomeric GPAs from (2). | .Wll, '4' fde 2 5.(b)(3) A 0-mL filter i GPA reaction mixture stands 8 it has an observed rotation, cm high in the polarimeter tube; aobsd = +16°. After adding 2 mL conc H2804, aobsd rises to ° The spec1f1c rotation of a—GPA = +100°; take that of W W“? Assume the final GPA mixture is 90% a and 10% B and that total GPA is conserved. Calculate the B—GPA fraction before adding the acid. You may find it useful to calculate what aobsd would be for 100% a-GPA. Deduction for including unnecessary data in your calculation, even if final answer is correct. f? $04) ace/At Am r1444 3 NH? [00%, W W WILVL: rma. E—GPA to be 0°“ we ¢va( —f~ I é; 8' (€?'<3:/¢¢ 6. (2) If cpd F‘s rxn rate, d[F]/dt, = (constant)-[F], what function of [F] varies linearly with elapsed t(ime)? Answer in <5 wds. Careful! The function's values must be pure numbers. EIF3, .. r at 1% [- ./ ‘If’c "I +° 22-2 C 7. (3) A solution contains a set of mutarotating diastereomers, Phred with [a] = +70°, and Pseudophred with [a] = —30°. Their equilibrium mixture has [a] = +5.0°. Calculate the fraction or percentage of each in the equilibrium mixture. Show your work and/or give reasoning. X [9/6 K t {W 3? ['K‘”’3(Mc ...
View Full Document

This note was uploaded on 10/12/2009 for the course CHEM 322B at USC.

Page1 / 5

322bs05_plq2_2 - Lab time This test comprises this page and...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online