215 F07-F-pp 2-7-key with comments

215 F07-F-pp 2-7-key with comments - Name Ke I 215...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Name Ke I 215 F07—Final exam Page 2 I. (15 points) Using the pKa information given on page 12, match the following isoelectric point (pl) 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. 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. StrUCture at the iSOCICCUiC P0int? The predominant form of this tripeptide at pH 1 has a net charge of: (circle one) The predominant form of this tripeptide at pH 11 69 I has a net charge of: (circle one) H3 H “Egg WVDH ConflCa’t‘I‘i/f ,‘ at ya WW «Iv/mash? : -'/aY —2_ III. (20 points) Draw in the boxes below the full structures (including stereochemistry) of the products from each of the following reactions. (1) 1. o O H.) 6 N02 ’ O r" H H 4 O 2. HCI, H20,A (2) O H dimethyl— \© formamide (solvent) Name Key __ 215 F07—Final exam Page 3 IV. (20 points) Consider the esterification reaction (A to B) shown below. HCI o\ Ph/\n/OH + HOCH(CH3)2 = Ph/\n/ CH(CH3)2 + H20 0 O A B (1) (16 points) Which of the following would result in an increase in the amount of ester B formed? (2) If isotopically labeled H-1SOCH(CH3)2 (Cirde one) is used, the 18O label would end up in : V. (32 points) Treatment of L—phenylalanine with excess methanol in the presence of 1.5 mol equiv of gaseous HCl at room temperature produces the HCI salt of L—phenylalanine methyl ester in 95% yield. Provide in the box below a step—by-step mechanism, using the curved arrow convention, for the formation of this methyl ester HCl salt from L—phenylalanine. You may use H—B and B_ as a general acid and its conjugate base, respectively. 0 O H | 1. ' Ph/Ykoe C (gas, 5 mol equiv) Ph/YkOCHs 6) NH HOCH3 3 room teperature,5 hr ClgNHs 95% ( H—B L “0—H OH / Ph O_CH3*—~ Ph OH @NH3 H $77 of mTerMcd/b‘fes : 3,93%“ P M6011 pin/3172 5172/05: 2. P2304, Name Ke * 215 F07—Final exam Page 4 VI. (18 points) A levulinyl group[CH3C(=O)CH2CH2C(=O)—l is an extremely versatile hydroxyl-protecting group. A team of scientists at The Scripps 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. Natl. Acad. Sci., USA 2003, 100, 797]. :83 GAO o O <18) 0A0 0 O 0% L C 0013 D CCI3 /\o (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 NaBH4 is most convenient. In the reaction shown below, the levulinate ester of cyclohexanol, E, is treated with NaBH4 to afford cyclohexanol (F). The by—product (G) from this reaction has the molecular formula of C5H802. Provide in the box below the structure of this compound. c...+ ’3“. ,4 I (2) When a levulinate such as E is treated with hydrazine (NH2—NH2) in acetic acid, the levulinate—protected alcohol undergoes facile deprotection (see below). Draw in the box provided the structure of the by—product H. It might be easier if you solve # (3) before this question. 0 MO NHz—NH2 HO ————> + o CH3COOH E room temperature 5 min H(C5H8N20) (3) The reaction of levulinate E with hydrazine first produces its hydrazone derivative, which further undergoes an intramolecular reaction to produce H. Draw in the box below the structure of this hydrazone derivative of E. :0 Ts fur/more deaTmQM/IZ Man Name Kev 215 F07—Final exam Page 5 VII. (22 points) Provide in the box below a step—by—step mechanism for the following reaction using the curved arrow convention. Use B— and 8—H for the base and the conjugate acid, respectively. Make sure to show the mechanism for the catalytic cycle, i.e., regeneration of B’ [or ‘OC(CH3)3]. You do not need to rationalize the stereochemistry of the new chiral centers created [Tetrahedron Lett. 2007, 48, 4683]. Ph H ,H @e o o )s K OC(CH3)3 o/Lo o W Ph 0 (catalytiC) /\/K}\/u\ OCHon3 OCHzCHa tetrahydrofuran (solvent) . r' 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 H—B and B— for the acid and the conjugate base, respectively. 0 O p-TSOH (catalytic) H acetone (solvent) H0 66% 5’1? of‘l‘mfemcil‘da 1 3/03 64% "Ca Aw Mechcm , arr; {75(7): Name Kev 215 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) o o 09 1 )L 22 (a ' CI 0’ \Pha N(CH20H3)3 H3N H CHzOH 2. H30+ to pH ~1 Which amino acid is this? Use its 3-letter abbreviation as your answer. Ser answer (2) HBr/H3CC(=O)OH or H Br/FSCC(:O)OH (3) [J. Org. Chem. 2004, 69, 1038] 1. n-BuLi 1- LiCU(CH3)2 THF (solvent) THF (solvent) PhCH2\O/\/§ 2 O 2. aq NH4CI CI /U\OCH3 (4) [J. Org. Chem. 2007, 72, 9541 J o OCH3 ACO& HBr (gas) ———> ACO OAC CHZCI2 OAC H C AC: 3 (acetyl) O (solvent) a-anomer (320}1220ll w WMS c4454 5 MIC-Ef:fi_2‘ Name Ke I 215 F07—Final exam Page 7 X. (34 points) Complete the following transformations as indicated by the data provided. Clearly indicate the stereochemistry where applicable. Sequential experimental steps must be numbered. (1) 1 . S/j O H THF (solvent) : Li /P\ aq (2) [Tetrahedron 2007, 63, 5855] C9H18082 o 2 > N 1. KH H _ THF (solvent) / (2 mol equw) 2. PhCHzBr CH2 ——> \ O o (3) {J. Org. Chem. 2007, 72, 5943] C12H17NO3 H OCH3 H OH H C + H0 MS 3 O + 2HOCH HO toluene,A Hi3C O 3 H OCH3 H 9 anlgo4 Q 0 0K. (4) [Tetrahedron Lett. 2007, 48, 5623] Out; 1, H OH 0 H . 5 2 LDA (1 mol equw) PhCHZO/WLO I THF 2. CHZCHC" (solvent) Ph aq NH4CI -30 °C diastereomeric mixture at a new chiral center (Show that stereochemistry with My) HacXOCH3 H30 p—TsOH ...
View Full Document

Page1 / 6

215 F07-F-pp 2-7-key with comments - Name Ke I 215...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online