Structure at the isoelectric point: The predominant form of this tripeptide at ph 1 has a net charge of: (circle one)

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1 ame Key 21 F07-Final exam Page 2 I. (1 points) Using the pka information given on page 12, match the following isoelectric point (pi) values (3.08, 6.00, and 10.76) to their corresponding amino acids by writing the correct isoelectric point value in the box beside each amino acid. Arg pi = Glu pi = 3.08 Val pi = II. (18 points) Draw the full structure (including stereochemistry) for the tripeptide Cys-Pro-Arg (all L amino acids) in the predominant form it would exist at its isoelectric point. Use the Fischer projections. tructure at the isoelectric point: The predominant form of this tripeptide at p 1 has a net charge of: (circle one) The predominant form of this tripeptide at p 11 has a net charge of: (circle one) III. (20 points) Draw in the boxes below the full structures (including stereochemistry) of the products from each of the following reactions. (1) 2 F 3 (C 2 C 3 ) 3 2. Cl, 2,! Cl (2) itro-containing product 6 Cl salt dimethylformamide (solvent) 2 C 2 C 1 10

2 ame Key 21 F07-Final exam Page 3 IV. (20 points) Consider the esterification reaction (A to ) shown below. Cl C(C 3 ) 2 C(C3 ) 2 A (1) (16 points) Which of the following would result in an increase in the amount of ester formed? 2 If the following is done: Yield of would increase (circle one) (2) Excess C(C 3 ) 2 is added. Yes o ac(c 3 ) 2 is used in place of C(C 3 ) 2 Yes o C(C 3 ) 2 is removed during the reaction Yes o Dilute aqueous Cl is used instead of concentrated Cl Yes o If isotopically labeled - 18 C(C 3 ) 2 (circle one) is used, the 18 label would end up in : V. (32 points) Treatment of L-phenylalanine with excess methanol in the presence of mol equiv of gaseous Cl at room temperature produces the Cl salt of L-phenylalanine methyl ester in 9% yield. Provide in the box below a step-by-step mechanism, using the curved arrow convention, for the formation of this methyl ester Cl salt from L-phenylalanine. You may use - and - as a general acid and its conjugate base, respectively. 2 3 Cl (gas, mol equiv) C 3 C 3 room teperature, hr 3 9% Cl 3 3 C 3 3 C 3 3 C 3 3 C 3 3 C 3 3 C

3 ame Key 21 F07-Final exam Page VI. (18 points) A levulinyl group[c 3 C(=)C 2 C 2 C(=)-] is an extremely versatile hydroxyl-protecting group. A team of scientists at The cripps Research Institute in La Jolla, CA, reported in 2003 that the levulinate group (indicated by a rectangular box) in differentially protected disaccharide C can be deprotected selectively to provide alcohol D in quantitative yield [Proc. atl. Acad. ci., UA 2003, 100, 797]. CCl 3 C D CCl 3 (1) The methods used to achieve this selective deprotection take advantage of the difference in electrophilicity between aldehyde/ketone and ester carbonyl groups. Among these methods known, the use of a is most convenient. In the reaction shown below, the levulinate ester of cyclohexanol, E, is treated with a to afford cyclohexanol (F). The by-product (G) from this reaction has the molecular formula of C 8 2. Provide in the box below the structure of this compound. E a C 3 0 C G (C 8 2 ) (2) When a levulinate such as E is treated with hydrazine ( 2-2 ) in acetic acid, the levulinate-protected alcohol undergoes facile deprotection (see below). Draw in the box provided the structure of the by-product. It might be easier if you solve # (3) before this question. E 2-2 C 3 C room temperature min F F (C 8 2 ) (3) The reaction of levulinate E with hydrazine first produces its hydrazone derivative, which further undergoes an intramolecular reaction to produce. Draw in the box below the structure of this hydrazone derivative of E

4 ame Key 21 F07-Final exam Page VII. (22 points) Provide in the box below a step-by-step mechanism for the following reaction using the curved arrow convention. Use - and - for the base and the conjugate acid, respectively. Make sure to show the mechanism for the catalytic cycle, i.e., regeneration of - [or C(C 3 ) 3 ]. You do not need to rationalize the stereochemistry of the new chiral centers created [Tetrahedron Lett. 2007, 8, 683]. C 2 C 3 K C(C 3 ) 3 (catalytic) tetrahydrofuran (solvent) 69% C 2 C 3 : C 2 C 3 C 2 C 3 C 2 C 3 : C 2 C 3 C 2 C 3 22 VIII. (22 points) Provide in the box below a step-by-step mechanism for the following acid-catalyzed reaction using the curved arrow convention. Use - and - for the acid and the conjugate base, respectively. p-ts (catalytic) acetone (solvent) 66% 22

5 ame Key 21 F07-Final exam Page 6 IX. (39 points) Complete the following transformations as indicated by the data provided. Clearly indicate the stereochemistry where applicable. (1) (2) 3 C 2 Which amino acid is this? Use its 3-letter abbreviation as your answer. er Cl answer C 3 to p ~1, (C 2 C 3 ) 3 C 2 2 C2 2, Pd/C 3 with added DCC r/ 3 CC(=) or r/f 3 CC(=) (3) [J. rg. Chem. 200, 69, 1038] C 2 n-uli TF (solvent) 2. Cl C 3 () [J. rg. Chem. 2007, 72, 91] C 3 3 LiCu(C 3 ) 2 TF (solvent) 2. aq Cl C (Z-isomer) C 3 Ac Ac Ac = 3C r (gas) C 3 Ac (acetyl) Ac C 2 Cl 2 (solvent) C 3 Ac Ac Ac r Cs 2!-anomer C C 3 Ac Ac Ac Csr

6 ame Key 21 F07-Final exam Page 7 X. (3 points) Complete the following transformations as indicated by the data provided. Clearly indicate the stereochemistry where applicable. equential experimental steps must be numbered. (1) Li TF (solvent) 2. aq Cl (2) [Tetrahedron 2007, 63, 8] C C 2 (2 mol equiv) K TF (solvent) 2. C 2 r (3) [J. rg. Chem. 2007, 72, 93] C Kr 3 C 3 C C 3 C 3 C 3 C 3 C 3 p-ts 3 C toluene,! 3 C C 3 2 C 3 C 8 18 () [Tetrahedron Lett. 2007, 8, 623] C 2 C 2 C 3 2 LDA TF (solvent) -30 C dianion diastereomers C 2 C 3 (1 mol equiv) 2. aq Cl C 2 C 3 diastereomeric mixture at a new chiral center (show that stereochemistry with ) 3 C C 3 2 C 3 C 2C 3 C 2C 3 3 C p-ts C 3