Name__key__________________________ 215 F07-Exam No. 2 Page 2 I. (15 points) For each of the following pairs of compounds, predict which compound is more acidic. Compare the two underlined Hs for each pair and circle the compound that is more acidic for each pair. C H 3 O 3 C HO H C l Cl l H a. b.(1) 3 O O H Cl O O H CH 32 O (2)a. b. C H O H H O (3) a. b. (4) (5) a. b. a. b. 3 3 H H H H 3 COOCH 3 H 3 COCH 3 OO OO HH HH 3 3 I. (14 points) (a) Show the structures of the two possible conjugate acids of methyl benzoate and explain, in several words, as to which of the two conjugate acids is more stable and would be expected to form preferentialy. Use drawings of pertinent resonance form(s) and several words to explain your answer. PhOCH 3 methyl benzoate 6 explantion: conjugate acids PhOCH 3 PhOCH 3 H H PhOCH 3 H PhOCH 3 H The positve in this conjugate acid structre is igly tabilzd by resnc. (b) The pKa value of the conjugate acid of DBU is considerably higher (~12) than that of a typical imine (estimated to be around 5-7). Explain this diference in pKa values using resonance structures of the conjugate acids of both DBU and an imine. N N NH 1,8-diazbicylo[5.4.0]- unec-7en (DBU) imine H H explantion: 4 13 N N N H H H N N H N N H The positve charge in the conjugate acid of DBU is tabilzd by rsoac. Therfor, this cnjugate acid h consieraly higer pKa (~12). Name__key__________________________ 215 F07-Exam No. 2 Page 3 II. (19 points) Treatment of quinic acid with a catalytic amount of p-toluenesulfonic acid (p-TsOH; pKa - 0.51) results in the formation of its lactone derivative shown below (Synlet 2007, 3219). Draw the structure of the expected uncharged, tetrahedral intermediate and propose in the box below a step-by-step, curved- arow reaction mechanism for this transformation from the hydroxy acid. You may use H-B and B- for the acid p-TsOH and its conjugate base, respectively. You do not need to balance each step. + p-TsOH H 2 O (i) Draw the mechanism through te formation f the uncharged, terahedral intermediate. uncharged, terahedral intrmdit (i) Draw the mechanism through te formation f the lactone product. lactone product HO OH HO OH OH HO OH HO O O HO H HO OH H HB HO H HO O O quinc aid 3 HO H HO OH HO H HO O H OH H H B HO H HO O H OH HB HO H HO O H O H H HO H HO O O H B IV. (7 points) Complete the following reaction scheme by drawing the correct structures in the boxes (Org. Let. 2007, 9, 4119). O OCH 3 LiN[Si(CH 3 )] 2 (1. mol equiv) THF, -78 °C 3 4 C 8 H 10 O 4 "enolate" H 3 COCN 1) 2) mild acidic workup with aq NH 4 Cl (1. ol equiv) * * Behaves imilarly as LDA. O OCH 3 Li O OCH 3 O H 3 CO +enatiomer Name__key__________________________ 215 F07-Exam No. 2 Page 4 V. (16 points) Draw the structure of the expected ?-hydroxy-aldehyde or aldol intermediate for the following aldol condensation reaction (J. Org. Chem. 2007, 72, 6493) and propose in the box below a step- by-step, curved-arow reaction mechanism for this transformation from the hydroxy acid. Use B- and B-H for the base catalyst and its conjugate acid, respectively. You do not need to balance each step. +H 2 O (i) Draw the mechanism through te formation f the !-hydroxy-aldehyde intermediate (i.e, an ldol). !-hydoxy-aldehyde intermeit (i) Draw the mechanism through te formation f the aldol condensation product. Aldol condensation pruct B 3 O O H O H O H O base catlyst O O H O H O H O + enatiomer O O H O H O OO H H HB O HO O H B O HO O VI. (9 points) Provide in the box below a step-by-step mechanism for the following reaction. Use B- and B-H for the base and the conjugate acid, respectively. Note that the base used here does not atack the ester C=O carbon. + O O CH 23 Cl O CH 23 H O K OC(H 3 ) (1.0 mol equiv) HOC(H 3 ) (solvent) 10 °C, 1.5 hr + enatiomer89% Cl O CH 23 H O B 9 Cl O CH 23 H O O O Cl O CH 23 O O CH 23 O Name__key__________________________ 215 F07-Exam No. 2 Page 5 VI. (21 points) Complete the following reaction schemes by providing in the boxes the structures of the corresponding products. (a) [J. Org. Chem. 2007, 72, 6270] + 5 C 7 H 8 N 2 O N NH O O OCH 23 CH 32 O OCH 23 O 185 °C, 5 h 4 H 2 O CH 23 LiOH, 2 O TF (solvent) 1. 2. H 3 O + 4 SOCl 2 PhNH 2 N N O O H N 4 (2 mol equivalents) K 2 CO 3 LiBr Br NH 2 NH 2 O N NH O O Cl N NH O O H N NH O O NHPh (b) [Org. Let. 2007, 9, 4259] + 4 H H 3 CO O O OO O N ! H 2 O C 14 H 14 O 5 (base) O O O CH 3 Name__key__________________________ 215 F07-Exam No. 1 Page 6 VII. (19 points) Complete the following reaction schemes by providing in the boxes the structures of the corresponding products. Make sure to indicate the stereochemistry where applicable. (1) [J. Am. Chem. Soc. 2006, 128, 6310; Tetrahedron Let. 2007, 48, 7567] 4 OCH 2 F 3 NH 3 (exces) CF 3 H 2 O (solvent) C 7 H 1 NO NH 2 O O O HO H O O Cl NaOCH 3 (ctlytic) HOCH 3 (solvent) 4 C 1 H 17 ClO 6 C 1013 ClO 5 (2) [Synthesi 207, 3219] OH O HO O O Cl CH 3 + H O O H 3 C O OCH 3 NaOCH 3 (ctlytic) HOCH 3 (solvent) 4 H 3 COCH 3 O (3) H HO O OCH 3 H O Ph + NaOCH 3 (ctlytic) HOCH 3 (solvent) 3 rflux H O Ph C 17 H 20 O C 17 H 20 O (4) [J. Org. Chem. 207, 2, 624] H O Ph (5) [J. Org. Chem. 207, 2, 649] OO HH 4 C 7 H 12 O+ enatiomer pieridne (mil bas) CH 2 l 2 rt, 1 hr OH H O Masato Koreeda Microsoft Word - 215 F07-E2-pp 2-6-key.doc
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