Winter 2008 blank-2 - l(44 points J00 2007 72 9181 A0 H NHB...

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Unformatted text preview: l (44 points) J00, 2007, 72, 9181 A0 H NHB | ‘Loal " OCH3 0:8:0 3) H3 is easin deprotonated with K2003. Using drawings and a few words. explain why Ha has such a low pKa. J. Am. Chem. Soc, 1999. 121, 6507 Alkaloids are of synthetic interest due to their antifungal and antibiotic activity. The following synthesis was used towards the synthesis of the tetracyciic core of two atkaloids. Complete the synthesis. b) OCHzCHa c) 2 CH3CH20H CH3CH20 0 (ko cat. H 80 2 4 d) ‘- 1) 101$ (reversible base) a CH30H20 0 $0 2)H30* N otV/£::j (protonation) / T + CH3CH20H 0 Nature Struct. Biol., 2001. 8. 858 e) The polysaccharide shown on the right is found in bacterial and parasitic cell walls. including pathogens such as Mycobacteriurn CH3 tuberculosis. Characterizeldescribe each of the glycosidic linkages. 0 OH 0 O OH " OH f) How many of the monomers are D- sugars? (circle one) Q) How many of the furanoses are or anomers? (circle one) 0 1 2 3 4 5 h) Is this pentasaccharide a reducing sugar? (circle one) yes or no ll (39 points) A) J00. 2008, 73, 661 compound B B) Ninhydrin is used to detect the presence of uncoupled amino acids during peptide synthesis. Ninhydrin reacts with a free u—amino group of the amino acid to form molecule E . O t Id 0 CH3 catacid O ca.aC| -——II- 0“ = H20 + m0 +H2”J\r°“ mm” OH O O 0 O ninhydrin molecule E The acid-catalyzed reaction between ninhydrin and the amino acid begins with the formation of an imine C. Show the stepwise mechanism. You may use HB as your acid and B‘ as its conjugate base. No stereochemistry needs to be shown. 0 0 CH3 H-B + HzNJWrOH —- o o Ill (43 points) A) lmine C (from the ninhydrin reaction on the previous page) undergoes a decarboxylation to form another imine intermediate D. In the presence of water. lmine D further reacts with water to yield molecule E and an aldehyde. Show the stepwise, acid cataiyzed mechanism for the transformation of lmine D to molecule E. You should use HB for your catalytic acid and B' for its conjugate base. 0 “CH3 0 H-B +H2 _b O i NHZ H CH3 0 molecuie E B) Molecule E then condenses with another equivalent of ninhydrin to form a symmetrical imine that has a deep purple color indicating the presence of the uncoupled amino acid. Draw the purple imine below. 0 o Gihw Ode - O 0 molecule E purple imine + H20 C) J00. 2007, 72, 9298 Fill in the reactants. Number separate steps if needed. IV (36 points) Peptide nucleic acids (PNA) are DNA mimetics in which the ribose phosphodiester backbone has been replaced by N-(2—aminoethyl)glycine units. DNA mimetics are of interest due to their potential as gene therapy agents. Base Base Base Base Q 4;)? Base = adenine. 0 Us!) 0 O 0 O cytosine. guanine or 2,277] go O_}LI',_O O_|g% thymine E{J\/N\/\HJK/N\/\n 0 0 9 DNA 9 PNA Complete the synthesis of two monomers incorported into a PNA dimer. A) Monomer A synthesis Monomer A a) O O b) BrCHZEOCHa, 1) NaOH (aq.) —h— HN I K2603 DAN 2) Hao* KHIOCH3 (protonatlon) O B) Monomer B synthesis MonomerB O HNJLONQ _.._ N/ —- ad 0 N Name the heterocyciic base: KWOH 0 d) Complete the reagents and products in the dimerization of Monomers A and B to form the PNA Dimer. Number separate steps if needed. e) H O JOL H 0 Base/\H’OH HQNNNJLOH o fi/VN‘JLOCH2CH3 O | D00 5 f) NH2 0 . bead N’j HN I 0% CAN 0 ._ oO Kfoo PNA Dimer Adenine binds to thymine via two hydrogen bonds. while cytosine binds to guanine via two hydrogen bonds. What two DNA bases do you expect this PNA Dimer to bind to? (Circle two) adenine, cytosine. guanine thymine V (40 points) (J. Am. Chem. Soc, 2008, 130, 2351) Ustiloxin D posesses OH \ O antimitotic activity and prevents cell division. Thus it shows promise __~‘ as an anti—cancer agent. Synthetically, it is derived from two modified HO amino acids (B-hydroxyisoleucine and N-methyl-B-hydroxytyrosine) OH hall” and two standard amino acids. Cleavage of an ether bond between 0 HN O O the B-hydroxyisoieucine and N-methyl-B-hydroxytyrosine in Ustiloxin _ 2': D results in the structure shown on the right. Ho" N HN\ H l— A) Draw the structures of the two modified amino acids. _ B) Draw the structures of the two standard amino acids and write their names in the bottom box. ll N-methyI-B-hydroxytyrosine B-hydroxyisoleucine C) (J. Am. Chem. Soc. 1996, 118, 495) Fill in the product of the following reaction. 0 OB" 1’ UAW” DMSO 0 2) (CH3CH2)3N (Swern) 5 D) (J. Am. Chem. Soc. 1996. 118, 495) lnositol and its phosphate esters play critical roles in both cellular structure and the regulation of cellular processes. Biosynthesis of lnositol begins with the enzyme myo- inositol 1-phosphate synthase which carries out an oxidation, enolization. intramolecular aldol reaction and reduction on the substrate glucose 6—phosphate. Complete the following scheme showing these four steps. No stereochemistry needs to be shown at newly formed stereocenters. “0 0 Ho 0 OH H OH H 7 open chain form of glucose 6-phosphate (with rotation around CS) enolization of C5 MAD+ oxidation of 05 aldol reaction reforms 6-membered ring reduction of 05 with NADH VI (38 points) A) Science, 2001, 294, 369 Part of the proposed mechanism for D-2-deoxyribose-S-phosphate aldolase is shown below. Provide the products of the first step and the arrows needed to complete the second step . product + arrows Lys 201 W LYS 2‘“ w Lys 201 SD-H w IL\ H H’fio H NIH H'O‘H _.. H, .H A 102_(0 HO H‘O‘H Sp ’N o9 H_N CH3 Asp 102 H B) (J. Am. Chem. Soc. 1997. 119, 4285-4291) Cysteine proteases have been implicated in diseases such as rheumatoid arthritis, muscular dystrophy and cancer metastasis; therefore, they are important targets in medicinal chemistry which has sought to develop inhibitors for this class of proteases. Complete the synthesis of the frame for a cysteine protease inhibitor. No stereochemistry needs to be shown. 1) LDA, THF, —78 °C 0 O CH3CH2CH20JKL /I/U\OCH2CH2CH3 O 2) H30+ (protonation) +CH2CH2CH3OH i l 0 I \ H N s s ..._ s s 2 ii CchHchzokf/j 0 o Ph {—1. H S S BocHN fl 0 0 | anhydrous BF30(CH2CH3)2 (Lewis Acid) 0 )L H20 Ho CF3 + SHCH2CH28H ...
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Winter 2008 blank-2 - l(44 points J00 2007 72 9181 A0 H NHB...

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