215 Exam-00-F - Chemistry 2 1 5 Final Examination Name...

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Unformatted text preview: Chemistry 2 1 5 Final Examination April 18, 2000 Name Please Print Dr. Edwin Vedejs (Sec 100) Dr. Masato Koreeda (Sec 300) (2 hr; 240 points) Signature Student ID # Please CHECK OFF your 216 Lab section. _101 Lecture ONLY _311 M 1-5 PM A624 CHEM SCHACH, ANDREW _110 M 1~5 PM A606 LAHSEN. AMY _312 M 1-5 PM A642 CHEM CAO, GANFENG _111 M 1—5 PM A612 CHEM CLARKE. NAGASH _330 T 1-5 PM A636 CHEM STOY, PATRICK _130 T 1-5 PM A606 CHEM CHEN. BIN _331 T 1-5 PM A642 CHEM SCHElDEMAN, MATTHEW _131 T 1-5 PM A612 CHEM CLAY. JULIA _332 T 1-5 PM A730 CHEM POON, STEVE _132 T 1-5 PM A618 CHEM SMALL, AARON _333 T 1.5 PM A736 CHEM TAN, LI _133 T 1-5 PM A624 CHEM ZHANG. BO _334 T 1-5 PM A742 CHEM ZHENG, NAN _134 T 1-5 PM A630 CHEM DENSMORE, CRYSTAL _350 w 12-4 pM A636 CHEM SCHACH, ANDREW _150 w 124 PM A606 CHEM LARSEN, AMY _351 w 124 PM A642 CHEM CA0, GANFENG _151 w 12-4 PM A612 CHEM TURPOFF, ANTHONY _352 w 1.5 pM A624 CHEM KOMAZIN, GLORIA _152 W 12- 4 PM A618 CHEM ZHANG. LIMING _353 W 1-5 PM A630 CHEM DENSMORE, CRYSTAL _170 TH 1-5 PM A606 CHEM CHEN. BIN _354 W 1-5 PM A742 CHEM HANN, CLAYTON _171 TH 1-5 PM A612 CHEM CLAY. JULIA _370 TH 1-5 PM A636 CHEM STOY, PATRICK ..172 TH 1-5 PM A618 CHEM TUHPOFF. ANTHONY _371 TH 15 PM A642 CHEM SCHEIDEMAN, MATTHEW _174 TH 1-5 PM A624 CHEM HANN. CLAYTON _372 TH 15 PM A730 CHEM POON, STEVE _190 F 1-5 PM A606 CHEM ZHANG. BO _373 TH 15 PM A736 CHEM TAN, LI _191 F 1-5 PM A612 CHEM CLARKE, NAGASH _374 TH 1-5 pM A742 CHEM ZHENG, NAN _301 Lecture ONLY _390 F 1-5 PM A618 CHEM ZHANG, LIMING _310 M 1-5 PM A618 CHEM SMALL, AARON __391 F 1-5 PM A624 CHEM KOMAZIN, GLORIA The exam has 14 pages in addition to this cover page. The last 4 pages include tables of pKa values for representative acids, electronegativity values for some elements, and bond dissociation energies for representative bonds, and a list of reagents. Individual point values are given in the corner of each answer space. GSI Initial PERIODIC CHART OF THE ELEMENTS 1 He .00280 5 6 7 B 10 L1 66‘ N 6 Ne 6.041 9.01218 10.81 12.011 160057153094 =": 9! 20.179 13 14 15 17 18 AI 51 p E Cl Ar 24. 305 808551-3376 35453 39.949 2‘ 926 31 32 33 35 . flflflfifieee 449559 47.990 509415 51.09654M-999 55.9947 590932 5999 63546 6539 9972 7259 49216 7996 70004 37 39 45 46 47 ‘5 49 51 52 53 Hell filmflflfiflflfifi I“ 954679 9762 639059 (98) 10107 02 10642 ca 11241 11492 11969 12175 12760 26.9045 13129 L357 5&71 227292 E73 75 75 7B 70 B1 83 Cs5 83156 mm. Hi Re Ir 77- Au Hg” TI Pb82 POIM Rn” 32006413733 29-1 IDES 179491909479 193.95 196207 190.2 192.22 19509 196-10050 204.393 2072 “(209) (:10) (222) B? 90403 104 105 107 PIWI HINIII (2:3) 2: .2 IDES (261) (262) (263) 58 59 B1 82 yfld 68 89 7O 71 EEEPEEHEEI EWWH 14012 140 14424 (145) 15036 59 ~ 1589 162.50 16726 69- t 17304 174967 ACTANIDE 11130 El M1dO1102 L103 ssmas Pa Cm“ EsW m1 ”W W.“ u. (A221) (247) (2347) (2051) (252) (253) (N25?) (2L60) Name page 1 I. (24 points) For each problem below, draw the MAJOR product expected from the indicated reactants. 1_ CH3O “”2 1. NaN02/HCI ———> 2. CuCN CH3 (Synthesis 1999, 2045) H2CZCH‘COZCH3 COZCH3 —'—‘—’ quinidine (a complex tertiary (Tetrahedron Lett. amine that acts as 1999, 40, 7091) the catalyst) hint: the product contains all atoms of both carbonyl reactants H3c~ué / ‘CH OH ————» CH2=CH 2 + (CH3)3CO\C/O\C/0C(CH3)3 g ([5 (1 equivalent) (Tetrahedron: Asymmetry Show only the nitrogen-containing product 1 999, 10, 4653) (1 equivalent) 4. H‘IH C Bre CHSO CHsO’ \P(CGH5)3 @ OC/COZCH3 C4H9LI Cl \ _—, (Tetrahedron: Asymmetry Do not show the phosphorus-containing product 1999,10,3877) page 2 Name II. (24 points) Provide ALL necessary reagents for the indicated conversions; the conversions require at least two reagents or two procedures to complete the reaction to form the products as drawn; be sure to number individual steps if it is important to perform the procedures separately. 1. CH3 CH3 OH OH H30 H30 CH CH3 3 Q; OHS W” W Nae H C CH H3O CH3 0 3 3 O 09 (Tetrahedron Lett. 1999, 40, 9239) 3. o H3C H O H (Tetrahedron Lett. 1999, 40,6617) 0 4_ H OH CISH3 0 HC— (9 3H Glow/UN}: COZH H3N COZCH3 3 H cue (J. Org. Chem. 1999, 64, 9294) Name page 3 ill. (12 points) 2-Cyclopentenonone, below left, is a stable substance that is available commercially. ln contrast, cyclopentadienone (below, right) is highly reactive and cannot be isolated even at temperatures far below 0 °C. Provide a brief explanation for the difference in stability. You will need to draw and discuss at least one resonance structure to support your argument. resonance ex . lanation stable unstable IV. (16 points) Show intermediates in the following sequence: Li—CHCOZC(CH3)3 | CH0 O=P(OCH3)2 ————> CH3 0 (Syn/ett. 1999, 1919) a 80 C O CeH5{l3H(CH3) (C14H1802) . _ N-Br N CHZCSHS (CGH5C02)2 CO C(CH ) (5%) O 2 3 3 T 1 equivalent C5H5{|3H(CH3) Li‘N‘CHzCst, ‘——_ A nitrogen-containing enolate (don't show nitrogen- containing byproduct) page 4 Name V. (30 points) 1. A detailed structure of the tripeptide Lys-Met-Tyr is shown below. The pKa values for the conjugate acid form of the functional groups next to boxes are approximately 3.5, 9.0, and 10.5. (a) Select the best match of pKa and acidity and place the correct number in each box. (b) Draw the detailed structure expected at pH 1.0. (b) pH 1.0 structure (a) (6 pts; 3 answers) D NH2 pKa 10 (given (CH2)4 H ) HZNNNfiu 002H SCH3 Lys- -Met- Tyr 2. (6) The isoelectric points for the constituent amino acids of Lys-Met-Tyr are as follows: Lys: pl = 9.7 Met: pl = 5.8 Tyr: pl = 5.7 . Based on the above information, would you expect the isoelectric point (pl) of Lys—Met-Tyr to be closer to (circle best answer): 6 10 (9.7 + 5.8 + 5.7) 3 =7 3. Derivatives of Lys—Met—Tyr have been prepared by solid phase peptide synthesis. The last step requires the coupling of a lysine fragment to Met-Tyr-OCHz-resin (where resin = cross-linked polystyrene beads). Show detailed (not abbreviated) structures for the reactants needed in the coupling reaction. Be sure to show necessary protecting groups; you may ignore the tyrosine phenolic OH group, and show resin without a detailed polymer structure. Analogy: Ingenito etal., J. Am. Chem. Soc. 1999, 121, 11369. A carboxyl-activated derivative of lysine that will A detailed structure of Met-Tyr—OCHz-resin acylate Met-Tyr-OCHz-resin with high efficiency. page 5 Name VI (20 points) 1. Draw the monocyclic starting material that was used to prepare the bicyclic ketone shown at right. 1. CZH5ONa 0020sz ——> 2. pH7 Tetrahedron Lett. 1999, 40, 8133 2. When the molecule shown below at left is heated with HCI/H20, both acetal groups are hydrolyzed. How would you perform a selective thioacetal hydrolysis? Show the reagent and the expected product. (Syn/ett. 1999, 1067) O S S 3. Deprotection of benzyl cis-4-tert-butylcyclohexanecarboxylate by hydrogenolysis using H2/Pd-C gave the expected product A. However, an attempt to accomplish the same result by treatment with NaOH followed by rapid neutralization to pH 7 using dilute acid gave two isomeric products A + B. Show the unstable intermediate and the product B. COZCH2C6H5 ‘l. (ct'd) 1. NaOH NaOH ——> 2. dil HCI tertC4H9 benzyl cis-4-tert—butyl- cyclohexanecarboxylate unstable intermediate. product B; does not (contains CH2C5H6) contain CHZC5H6 page 6 Name VII (50 points) For each problem below, draw the major product expected from the indicated reactants. (1) Cl (1 equivalent) —-—_—> OCH3 AlCl3 (1 equivalent) (J. Org. Chem. 2000, 65, 254) a (2) (1 equivalent) <><fi it C 0 ‘ii 0 (J. Org. Chem. 2000, 65, 2253) a 1. H30“: A 2. workup with NaOH C5H11NO (3) HO O (CH3)2C=O 1. NaBH4 H > -———> ———> H2304 2. NalO4 HO OH (catalytic) anomeric mixture 1. (C5H5)3P=C(CH3)2 (Org. Lett. 1999, 1, 2253) (2 equrvalents) 2. H2, 10% Pd/C page 7 Name 0 1. NaN3 (1 equivalent) 2 H2, 10% Pd C, 80020 (1 equivalent) (Tetrahedron Lett. 1991, 32 4503) hint: The product shows no CH3 signal in the 1H NMR spectrum and the IR stretching frequency for C=O in the starting compound remains unchanged. (Acc. Chem. Res. 1999, 32, 464) Note that these two produ-cts are diastereomers and are both end-o adducts as a result of steric constraints. Vll (continued) (4) (5) o§ ,OCH3 (CH3)38|O i 9 + ill 90 c (CH3)3S|O (l //C\ o OCH3 (J. Org. Chem. 2000, 65, 2257) page 8 Name VIII (10 points) The pKa values for the two acidic hydrogens of the enediol part of L-ascorbic acid (vitamin C) are 4.10 and 11.79. Which one of the two hydroxyl hydrogens at C-2 and C-3 is expected to be more acidic? H OH 2 2-OH ; 3-OH H OH 0 Circle one that applies. _ Explain briefly why the hydroxyl hydrogen selected above is more acidic than H 2 the other by using resonance structures of the acid. L-ascorbic acid IX (20 points) Draw a Fischer projection formula for each of the following two compounds and assign its absolute stereochemistry using the D/L-designation. (1) (53H OCH3 (2) HZOH HOCHW /H H O H i 8 HH H OH /N\ 0 H3O CH3 HO OH H OH stereochemical stereochemical designation: designation: D ; L D ; L Circle one that Circle one that applies. applies. a Draw in an open-chain aldehyde form. CHZOH D—glyceraldehyde page 9 Name X (18 points) A disaccharide is cleaved by a B-glycosidase, an enzyme that specifically hydrolyzes a B-glycosidic linkage. When this disaccharide is treated with excess dimethyl sulfate [(CHS)ZSO4]/NaOH/H20 and then hydrolyzed under aqueous acidic conditions at around 90 °C, two monosaccharide products are formed: 2,3,4-tri-O-methy|-D-mannopyranose (1) and 2,3,4,6-tetra-O—methyl-D-galactopyranose (2). H OH B—D—mannopyranose B—D—galactopyranose 1. Draw in the box below the structure of the disaccharide. Each of the monosaccharide units should be drawn in the Haworth projection formula and the stereochemistry of the anomeric hydroxyl group in the B-form. 2. Draw in the boxes given below the structures of the two methylated monosaccharide products 1 and 2 in their most stable chair conformers. Assume the stereochemistry at the anomeirc hydroxyl group to be in the B-form. For 1: For 2: page 10 Name Xi (16 points) 1. Provide in the boxes below the structures of the monomer that is required to prepare each of the following two polymers. (1) radical polymerization ‘— ' ’ _ CF20F20F2CF2CF2CF2 — “Teflon" monomer (2) N N N anionic ? g g polymerization —CH2-?—CH2-$—CH2—?— F0 so i=0 OCH3 OCH3 OCH3 monomer "Super Glue" 2. Treatment of 3,3-dimethyloxetane with a Lewis acid such as BF3 provides a polymer. Draw in the box below the structure of a segment of the polymer consisting of three monomer units. H3 CH3 _________> O 3,3-dimethyl- oxetane a Page 11 E m H B e: @H e e: 90 e..\ d E ”on ..C\ ”NIH G \C/ G .6 H G H s T I \ @\ x S l \ _ . l H 9 G .\ . x a A 01C 1 H _ OIC O 2 01C 1 1 C C H .C G G B /. olN 2 . / _ H / 3 CICIC E 6 HIN __ x \ /H t G .0 ./ H 3 C H _ C / CHC HICI s . O H 3 C H C H e w e a: c, m m a H H H. /H A/ v "r m. H H \ m C no! 2 2 5 O 7 a O O O 5 . H H l/ H \ H H H H H .m ”on H x x H H _ H c m o c\ H, \ s/ \NIH LH 0/ ,c, a e a a 2 .;H ,C\ H/ H6» H "_n A H 2 [I - la In I .1 m /o.. .. mic/H m oIC/ .n/ H. onc\ H _ c H H N/ ._L_ cnc\ H G HLTH a o u 0” .l. \ . . .. N/ H3 H, H C, Km c Hr H H /H / H [m e ”0 c c m c \ e e; m E m ) e B m G: H G G e E B "on .. G H .. H 3 T n ..0|.rlo 6:. 61. o\ \H ®.\o. .\0 e o.” /N\ e e w o A e .. l_l .I. .C / 0 CNN .OHC l x H _ O O | \ H _ G k . . .. 2 / / x F OIC / HC . \ \ OIC @ N x C U m .m. w H o a. c _ “ouc onc /c =. H H CH m IWl nm. a C / /3 "/0. OHC\ /H C /N\ / — C H H. / e _ O C um: H C C 4 unnu. _ _ _ _ 4. 0. 3 4. 4. 8 5. 2. 1 4 O 4 6 9 9. 9 O ) 1 d 3 a .H H H H O u H l O H M w .. \H @H 65" \0 .\0. @. .H. H \C H S .OHSHO 1.. o l. HID \ .. \ II o OHC HINIH .0. O OHC H O D We. _ _ . J». ../ HJO/ OHN/ OIC/ 4.. H _ .mU \ . \ /C\ N enflu H _ .|.. . . 2 . / \C . HC .OHC . | x C C m .m. H H m H 0 fl H cx . / .. / .ouc\ /H E H «H CH A n "m. e a _ H. o x; _ H _ page 12 Table 1.2 Electronegativity Values for Some Elements 1 2 13 14 15 16 17 H Table 2.4 Average Bond Energies [kJ/mol (above) and kcal/mol (be/owl] Single Bonds C=O (ketones) 179 page 13 The Reagent List Shown below is a list of key reagents (not always the whole recipe) which may be useful for solving questions on the final exam. Reagent classification or specialized use Chapter 13 NaBH4 nucleophilic hydride reduction LiAlH4 nucleophilic hydride reduction RMgX nucleophilic carbon RLi nucleophilic carbon Raney Ni desulfurization HSCHZCHZSH thiacetal/thioketal formation 4-CH3C6H4503H (TSOH) organic-soluble acid CSHSNH+ ClCrO3' oxidant ClC(O)C(O)Cl/H3CS(O)CH3 then (CH3CH2)3N oxidant HZNNH2 hydrazone formation HZNOH oxime formation BF3°O(CH2CH3)2 Lewis acid source (CH3)3C(CH3)ZSiCl (TBDMS-Cl) protection H39 9H3 H30-(I3-$i-Cl H30 CH3 (CHSCHZCHZCHZLNT‘ nucleophilic F source dihydropyran (CngO) protection fl 0 Chapter 14 AgZO, NaOH, H20 oxidant Chapter 15 (C4H9)2A1H (DIBAL) Lewis-acidic reduction hydride source Chapter 17 KH base LiN [CH(CH3)2]2 strong base Chapter 19 R2CuLi and RzMgLi conjugate addition nucleophilic substitution P(C6 5)3 phosphorous nucleophile RR'C' - P“(C6H5)3 Wittig reagent RC(O)CHR'P(O)(OC2H5)2 modified Wittig (Homer-Emmons) reagent 1,3—dithiane a carbonyl equivalent SVS HgClle-‘IZO thioacetal/thioketal hydrolysis page 14 Chapter 20 Chapter 21 Chapter 22 Chapter 24 Chapter 25 Historical From Chem 210 source of X+ source of N02+ XZ/FeX3 (X = Br, c1) HZSO4/HNO3 or NaNO3 RC(O)X/A1C13 (x = Br, Cl) sto4/503 source of acylium ion N—bromosuccinimide (NBS) benzoyl peroxide radical initiator azobis(isobutyronitrile) [AIBN] radical initiator 9+3 9+3 H30—(IJ—N2N—CIJ—CH3 ON ON (CH3CH2CH2CH2)3$nH source of (CH3CH2CH2CH2)3Sn- and H- SnCIZ/HCl reductant NaN 041130711120, 0 °C CuX (X = Br, Cl, CN) H3PO2 (hypophosphorous acid) reductant NaIO4 oxidant C6H1,N=C=NC5Hll (dicyclohexylcarbodiimide) N-hydroxysuccinimide ROC(=O)N3; RO(=O)C1 (szenzyl or tert—butyl) [tert—butyl-O-C(=O)—]ZO CH3 0 0 “39,043 H 3C\ l H 3C/C\O O O/C\ sulfonation radical bromination nitrosation of amines Sandmeyer reaction RCOOH activation additive for RCOOH activation N-protection N—protection HzNNHC(O)NH2 For 216 labs; not for 215 exams HZNNHC6H3(N02)2 For 216 labs; not for 215 exams OsO4 oxidation KMnO4 oxidation peroxyacid (e. g., 3-chloroperoxybenzoic acid) oxirane formation 03 then (CH3)ZS or Zn ozonolysis NaNH2 base NaH base K+ ‘OC(CH3)3 bulky base H2/Pd hydrogenation H2/Pd/CaCO3 hydrogenation BH3 or 9—BBN then HZOZ/NaOH hydroboration PBr3 e.g., R-OH -> R-Br SOCl2 e.g., R-OH -> R-Cl 4-CH3CGI-LSOZC1 (TsCl) CH3SOZC1 (MsCl) tosylate formation mesylate formation ...
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