215F07-E3-key - Name K(Z i 215 F07-Exam No 3 Page 2 I(29...

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Unformatted text preview: Name K (Z i 215 F07-Exam No. 3 Page 2 I. (29 points) Streptomycetes are soil—dwelling, filamentous, Gram—positive saprophytic bacteria. They are responsible for over 50% of the known microbial metabolites, including many antibiotics used in human and veterinary medicine. In an article published earlier this year, a group of scientists from Czech Republic reported the isolation of a number of extracellular carbohydrate metabolites from Streptomyces coelicolor A3(2) [J. Nat. Prod. 2007, 70, 768]. The structures of some of these carbohydrates are shown below. HOHgC HOHgC o H CHQOH H H 0% H Ho CH20H Ho Ho 0 0 HH H O H HO H OH0 OH H 1 CH20H 3 Answer the following questions about the structures of these carbohydrates 1 — 3 and their hydrolysis products. (1) For each of these three carbohydrates, answer if it is a reducing sugar. Circle yes if it is a reducing sugar and, no if not a reducing sugar. For 1: For 2: For 3: (2) How many of the sugar units of trisaccharide 1 (3) IS the sugar unit 0f glycoside 2 D sugar? are D sugars? Circle one that applies. CirCIC one that applies. 01 2 ©. (4) Is the glycosidic bond in glycoside 2 or or [3? (5) Is the stereochemistry of the anomeric OH in Circle one that applies. furanose 3 or or [3? Circle one that applies. @5- @3- (6) Draw in the box below a Fischer projection . . . . formula for the carbohydrate product, obtainable (7) Draw 1n the box below a Flscher prolectlon upon acid hydrolysis of glycoside 2, in an formula for furanose 3 in an open—chain form open-chain form. Name [LC Z 215 F07-Exam No. 3 Page 3 II. (17 points) Cyclopentanoid 5 shown below has been proposed to be biosynthesized from the C—4 oxidized D—glucose 4 via the “ene-diol” intermediate under acid-catalyzed conditions (Tetrahedron Lett. 2002, 43, 451). Provide in the small box below a structure for the ene—diol and propose in the box below a step—by-step, curved—arrow reaction mechanism for this transformation from 4 to 5. Use +B—H and B: for the acid catalyst and its conjugate base, respectively. You do not need to explain the stereochemistry at the newly created chiral centers in 5. HO, H OH III. (8 points) Provide in the boxes below the structures of the reaction products. anomeric mixture 0 - ’ l *7 W 0—4 6/ L 9:70”, ‘ CH3 0 CH3 0 o V O O y \F O (S (/l fuygoms’rel’aCle) H30 H30 C14HzoO9 p—TsOH or 1. NaOCH3 BF3-(O(CngC3)H3)2 H SC 0 SPh (catalytic) cata ytic Q HSCOH PhSH O O O O VCH3 (solvent) Y \F O 2. NaH H30 H30 (excess) PhCHgBr B-anomer; C13H2207S (excess) THF + H3CC(=O)OH (solvent) Name K9 14’ 215 F07-Exam N0. 3 Page 4 IV (8 points) Treatment of keto—diol 6 with p—TSOH results in the highly diastereoselective formation of spiroketal 7 (Angew. Chem. Int. Ed. 2007, 46, 7664). Draw the stereostructure of the major product 7 by completing the remaining structure in the box. H300 0 H300 O O HO p-TsOl-l @O W (catalytic) @O H + H20 H300 a OH $015313? “300 E 6 The term used to rationalize the fact that the above . H stereoisomer is the more stable diastereomer is: Anomeric effect V. (16 points) Methyl a—D—glucofuranoside can readily be converted into methyl a—D—glucopyranoside under acidic conditions. Draw in the box below a step—by—step, curved—arrow mechanism for this reaction. Use +BH and B—H for the acid catalyst and its conjugate base, respectively. 0 O HO methyl a-D-glucopyranoside OCH3 Name _____/<_€7[______ 215 FO7—Exam No. 3 Page 5 VI (14 points) Draw in the box below a step—by-step, curved-arrow mechanism for the reaction shown below including the aqueous acid workup step (Synthesis 2007, 3185). Use B— and B—H for the base ('OCHZCH3) and its conjugate acid, respectively. J (1 mol equiv) HSCCHQO O H3CCHZOMN HOCHzCHs J H H “01“”) O l N + HOCH20H3 O 2. H30+ workup H‘O (protonation) ‘ > ) ' ,. VII. (12 points) Complete the following reaction schemes by providing in the boxes the structures of the corresponding products for each of the following reaction schemes. (1) [Synthesis 2007, 3185] 1. LiAlH4 (excess) 0% H30+ THF (solvent) Ht A 2. aqueous O O workup ll V1 H O (0/2 C7H11N02 £02sz + HOCHQCH3 + C02 (2) [Tetrahedron Lett. 2007, 48, 4761] Li 0 \\ . <3]}E!)\N,OCH3 + OSI(CHZCH3)3 H30+ workup (5H3 THF (solvent) + H2N+ (CH3)OCH3 Name K62 ' 215 F07-Exam o. 3 Page 6 VIII. (16 points) Complete the following reaction schemes by providing in the boxes the structures of the corresponding products. (1) ‘P ““‘J\OCH 1. NaOCH3 (1 mol equiv) 3 benzene OCH3 —__——_> 2. HCl/H30+ H3 C O (protonation) + HOCH3 + 002 (2) [see: Org. Lett. 2007, 9, 1879] lithium enolate o —/_/CU‘U ow dj THF (solvent) m —78 °c + enantiomer + Cu(CHQCHgCH2CH3) BrCHQPh + LiBr ...
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