Ochem B LabQuiz1

Ochem B LabQuiz1 - CHEMISTRY 322bL October 25, 2004 Fall...

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Unformatted text preview: CHEMISTRY 322bL October 25, 2004 Fall 2004 FIRST LAB QUIZ By Page 6;?” \.// NAME I 1.(12) 2. (9) Lab time __l_________— 3.(10) T.A. 4 (9) 5.(l4) 6. (6) This test comprises this sheet TOTAL (60) and six numbered pages. For questions about this quiz, see your TA first. TAs will have quizzes at office hours and make-up labs. For make—up labs, M eve convenes at 6:30, Tu morning at 9. -1— 1. (5) Recall the Grignard synthesis of 2-methyl-2-hexanol, CH3(CH2)3C(OH)(CH3)2, from n-BuBr and acetone, CH3COCH3. (a)(3) The prep of the Grignard reagent, R-MgX, is complicated by the coupling reaction (CR), the (further) reaction of R—MgX with R—Br. Explain in terms of relative rxn rates why the more dilute the rxn soln, the less coupling; use <20 wds. (b)(2) The intermediate bromomagnesium salt required protonation to obtain the product. Explain in <15 words why NH4C1 soln ‘ rather than plain water was used to get the alcohol. w J wha/ M§+L( J may? \ _, (Hm/Lg" ([9; MG”) WM W76 M VJ or 2. (7) Recal e Period 2 alkylation of anisole, Ph—OMe, by Ph3COH in HOAC with catalytic H2804 to give p—MeO—C6H4—CPh3. (a) 2) Put "rev(sible)" or "irr(eversible)" in each blank: The rxn a, “V works in HOAC because the Ph3C® rxn with HOAC is CQU , but JyLAgL, that with anisole is ( rwp . Rxn fails in EtOH because rxn \ “*4 of Ph3C+ with EtOH is I Y‘Y‘ (b)(5) Given the ratio Desired Prod to Side Prod is 10:1 at 20°, 1:1 at 70°, and varies exponentially with T change: Calculate the T to give DP:SP = 4:1. For full credit do NOT use log/expon L keys on your calculator but only that loglOZ = 0.3;i?/5 %'~45;a_2 85%SA 3. (5) For l-hexyne, n-C4H9CECH -——————9 2-hexanone, Q-C4H9-CO-CH3. (a)(3) An alkyl cation, AC, R—Eh—CH3, is more stable than is a vinyl cation, VC, RJELCHZ. Yet 13G for 1—hexene ——-; AC ~ 11G for 1-hexyne -—> VC. Explain in <20 words, considering only the difference in strength of the bonds broken on protonation. ossible NET MINUS score for answer beyond this scope. 1/, {Q7 «(K714 “4/ W 41K neon/c, V [M iflmm ;Z;L y’ ZRF7Z2‘ ,~;‘35‘.A2;&(76;a012 (b)(2) A student carries out the rxn until all the l—hexyne odor is gone. S/he tests a drop of the rxn mixt with 2,4—DNPH re— agent, getting only a trace of ppt. Tell why work—up of the rxn mixt is now pointless. Answer in <12 words. /” \ #49 (W norm MKKZL WW M/u/c’flfr We, 4. (4) For drying by distillation of dilute W solns in an OS: Given y = amt W, x = amt pot soln, and r = the vapor—to-liquid W conc ratio, the equation is ln(y/yo) = r[1n(x/xo)]. For r 10, calculate the fraction of the soln one must distill, fd ( 1 — (x/x0)), to distill 90% of the Water, i.e., to make (y/yo) = 0.1. For max credit, do not use the ln/log key on your calculator but only the value ln 10 = 2.3, and the approxima— tion that fd = -ln(x/x0) if/whenever |ln(x/x0)| < 0.3 452;) = :L-Myfl = ng/ _ 3 _ WW 5. (10) Recall the isatin/acetophenone aldol condensation: (a) (6) K ’k 64C 0411‘, ‘9’ >2 gal? //»¢rw\ 0/ (b) (4) U272) 7‘ " w l ), OW ) M W 0 R O CH3‘C-Ph l H Isatin Acetophenone MW = 147; mp = 193-195° MW = 120; g = 1.03 (1) Draw the structure of cpd I, the initial aldol product. (2) Calculate the MW of cpd I solely from its stoichiometric relationship to the starting materials; show calculation. (3) When the (nearly) colorless soln of cpd I is warmed with aq HCl, tell what visual change occurs and what it means regarding the change in electronic structure. 59 CN+ (:Liwcl'16l\ (;2'] (ad/(‘1:’ a; “2“47#-€?é7&£27" 0 409112577 l4” 2 /“67/’/w {3750/ ME 4AM 73? fag/‘9'} W W (ffflm. Exactly 50 mmole isatin gives 10.0 g crude cpd I; 3.0 g of this yields 12 mmole pure cpd II. (1) Calculate the yield of cpd II based on isatin, as a percentage of theory. (2) For the calculation to be correct, what criterion must the sam— pling method from the crude cpd I meet? Use <10 wds. Ixrg 5‘0 Ma $ UM fCI/‘Mfllma Mflm/ {hf ct Scum/IQ. -4- 6. (9) Recall the methyl benzoate prep as actually done: 2 ~ 7 PhCOOH + 16 g HOMe —————-—————e—— PhCOOMe + HOH MW = 122 MW = 32 MW = 98 MW = 136 bp = 66° bp = 199° (a)(5) If one takes 4.88 g PhCOOH, obtains 24 mmol ester, and recov— ers 1/7 of the starting acid, calculate-— ‘ (1) the percentage of benzoic acid converted to product. ‘tXK @ WM! 62-41%: "/Lt. ;t$6 mm/ 04L ' 4% d& can): 32... 1w“. 0- the yield of ester as a percentage of theory. (9 . 6. 60% K /~/1 @fl W “"7 = {70% (b)(2) In this prep and usually, even a large excess of methanol can be separated from the product ester befigre distillation. Tell how this was done, in < 15 words. cam. raft «m baa/fir. CED’?4(3 (2) H (C)(2) After separating the aq Na2C03—extracted ester/acid DCM soln, it usually still contains some dispersed aq phase (tiny drop— lets). These should be removed with Na2804 before distilla— tion. Tell why this is done even though the distillation efficiently chases water well before the pot T reaches 60°. QM C Urn/033 my mm an. M and WW/ 472w 7. (14) PhCHO was converted to Dilantin via benzoin and benzil. (a) (5) To dimerize PhCHO "head—to—head" to benzoin, Ph—CO-CH(OH)—Ph, the enamine species below attacks a (second) Ph-CHO. Show electron movement and the resulting next intermediate ONLY. OH (0 O’H (90 (b) (6) Summarize the copper—catalyzed benzoin oxidation by filling in the blanks: Compared to 2-propanol, benzoin is easily oxi- . . . . ‘57-:0 dized because its carbinol g, CLI——OH, is 01 to a L— ‘T——‘— (A/ In two one—electron steps, the organic material, benzil, “0 A -—>) Id , is produced as (write struc - (an CA any) 15 re uced to Cu+. In (re)oxidizing 2 Cu+, NH4NO3 is reduced to NH4N02, which decomposes to gas and water. An intermediate is the brown gas H101). , hich appears after the benzoin oxidation is . Cm CW) 0”") (c) (3) Note the structure of Dilantin® below. Vigorous hydrolysis cleaves all amide/imide bonds. Write the structure 0 resulting principal organic product as i_t exists .a_1; 6' ‘ (The other products of the reaction are CO3= and NH3. H G C/ \ CGH/s' C=0 AW “W NHL c / 0/ \N\ (.3) 6 The striking visual property of Glucose Phenylosazone (Period 4B) is due to its chromophore, an a-diimine unit, -N=$-$=N— The a-diketone, benzil (from Period 6), has the same property to a lesser degree. (a) Name the common property in < 5 words, and (b) draw the structure of benzil and circle ONLY the relevant group of atoms. (a) Wat“? 00/” par—p M (B 0 U) 9. (3) Recall the glucose pentaacetate (GPA) reaction mixture. The reaction is summarized in the equation below: Heat other C6H7O(OH)5‘ + excess ACZO ——-——————9 C6H7O(OAC)5 + product Cat NaOAc Mn 44%“ ,4) s AM (F 504%. Neither glucose nor NaOAc dissolves in Ac20 at room tempera- ture. Explain why thejfi dissolves during the heating period. State specifical y whether the dissolution is mostly because (1) the NaOAc is soluble in hot Aczo, or (2) the nature of the medium changes, or (3) the NaOAc reacts to give a soluble product. This is not to be your ENTIRE answer. Wow OW A W [0% (40/4: , fut wavmflzof WH< VI) Q/ué/e. ...
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Ochem B LabQuiz1 - CHEMISTRY 322bL October 25, 2004 Fall...

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