Name__Key__________________________ 215H W06-Exam No. 2 Page 2 I. (24 points) Pheromones are those chemicals that play a role in communication between individual animals or insects. Structure 1 is the pheromone of the Japanese peach moth and acts as a sexual attractant for the male. Devise the synthesis of racemic 1 starting from readily available 4-pentynol (2). You may use any monofunctional reagents of eight or fewer carbons. Show reagents/chemicals needed for each step, the product(s) after each step, and specify a solvent if this is important for reagent stability. Select methods that are likely to give the desired compound as the major product at each step. 2HO-CH2CH2CH2C C HCH3(CH2)7C CC(CH2)5CH3HHOHH(CH2)21 (racemate) 2HO-CH2CH2CH2C C HOp-TsOH(catalytic)O-CH2CH2CH2C C HO1. Na / NH3 (liq)2.IO-CH2CH2CH2C C (CH2)5CH3OHO-CH2CH2CH2C C (CH2)5CH3CH2(CH2)4CH3H-C-CH2CH2C C (CH2)5CH3OH3O+!PCC1. CH3(CH2)6-CH2MgBr2. aq NH4Cl1 (racemate)CH3(CH2)7-C-CH2CH2C C (CH2)5CH3OHHH2, Pd, BaSO4/quinoline II. (6 points) Treatment of the multi-functional group containing compound, 3, with NaBH4 in THF followed by the protection of the selectively produced alkoxide with TBSCl results in the formation of TBS ether 4 in 81% overall yield from 3 (Org. Lett. 2005, 7, 5593). Draw in the box below the structure of 4. OHOPhHOOCH3OCH3O31. NaBH4 in THF (room temperature)2. TBSCl, imidazole, DMFnote: Ph = phenyl; TBSCl = (tert-butyl)dimethylsilyl chlorideNHNimidazole:DMF: dimethylformamide (solvent)64OOPhHOOCH3OCH3OTBSName___Key________________________ 215H W06-Exam No. 2 Page 3 III (10 points) When trans-diol 5 was treated with one mol equivalent of lead tetraacetate, Pb[OC(=O)CH3]4, in the presence of a catalytic amount of trichloroacetic acid (pka~1.0), diketone 6 was produced together with lead diacetate and acetic acid (2 mol equivalents). In the box provided, draw the structure of the neutral intermediate and show the mechanism of its formation from trans-diol 5 and transformation of the intermediate X to diketone 6 through the use of the curved-arrow convention. You don’t need to balance each step. OHOHOOPb(O-C-CH3)4OCCl3C-OHOPb(O-C-CH3)2OHO-C-CH3O2++(catalytic)56 neutral intermediate X4CH3-C-OOPbCH3-COOH O-C-CCl3O2+OCOCH3CH3-C-OOPbCH3-COO2OCOCH3HOHOH5CH3-C-OOPb2OCOOOHHH3CO-C-CCl3OCH3-C-OOPb2OCOOOHH3COOHOCOH3COO6 IV. (12 points) Complete the following reaction scheme by providing in the boxes the structures of the corresponding products (J. Org. Chem. 2006, 71, 1390). Also, indicate for each of the compounds how many 13C NMR peaks are expected to be observed when run by the proton-decoupled method. HOHOOHOHHOOCH3(1 mol equiv)NaH (2.2 mol equiv)C CBrH(2.8 mol equiv)DMF0 °C to room temp.p-TsOH(catalytic)4 42Number of 13C peaks expected:2Number of 13C peaks expected:OCH3OOHOHOHOCH3OOOOH1016Name____Key________________________ 215H W06-Exam No. 2 Page 4 V. (8 points) Treatment of keto-diol 7 with a catalytic amount of p-toluenesulfonic acid (p-TsOH) produced a mixture of two stereoisomeric tricyclic ketals. Draw in the boxes provided below the structures of these products. Make sure to clearly indicate their stereostructures. OHOH744+ + H2Op-TsOH(catlytic)OHOHOHOHOH VI. (12 points) Provide the structures of the expected major products for each of the following reactions. Make sure to clearly indicate their stereochemistry and, where applicable, label isotopes. (1) My work from the last century! CH3ONaCH3OHH218OHClO4(catalytic)44insect juvenile hormone0 °COCH3HOOHOCH3OHOHOCH3OOHOCH318OH (2) J. Org. Chem. 2006, 71, 1139. OHOSiSiOCH2CH3OC21H34O4Si2p-TsOH(catalytic)CH2Cl2C21H34O4Si24SiSiOCH2CH3OOHOName____Key________________________ 215H W06-Exam No. 2 Page 5 VII. (28 points) Treatment of triol 8 with excess trimethyl orthoacetate [H3CC(OCH3)3] in the presence of a catalytic amount of pyridinium p-toluenesulfonate (PPTS), a milder version of p-TsOH, cleanly produced a diastereomeric mixture of cyclic orthoester 9 (98% yield) and 2 mol equiv of CH3OH. Subsequent treatment of this unstable product 9 resulted in the formation of a new 5-membered cyclic ether 10 (91% yield) together with one mol equiv of CH3OH (J. Org. Chem. 2006, 71, 1416). H3C OHHOOHOOH3COHH3C OCH3H3CC(OCH3)3HH(C8H14O3)8PPTS(catalytic)+ 2 CH3OHCH2Cl2room temp., 5 min910BF3•O(CH2CH3)2(catalytic)CH2Cl210 minIR: no peaks above 3100 cm-1 1739 cm-1 (strong)13C NMR: ! 171.1 ppm and 7 sp3 13C peaks.+ CH3OH (1) Provide in the box below a stepwise mechanism for the formation of cyclic orthoester 9 from triol 8 by the use of the curve-arrow convention. You may use H-B to represent PPTS. Make sure to indicate the stereochemistry for each of the intermediates when applicable. Please also note that the entire reaction is reversible. You don’t need to balance each step. CH3COCH3OCH3OCH3916H BCH3COCH3OCH3OCH3HCH3COCH3OCH3H3C OHHOOH8CH3C OCH3H3C OHOOHOCH3H:BCH3C OCH3H3C OHOOHOCH3H BCH3C OCH3H3COHOOHOCH3HCH3C OCH3H3COHOOHHOOH3COHH3C OCH3HHH:B (2) Draw in the box indicated the structure of compound 10 and show a stepwise mechanism for its formation from 9 by the use of the curved-arrow convention. Use BF3 to represent the Lewis acid and clearly indicate the stereochemical outcome for each step. You don’t need to balance each step. OOH3COHH3C OCH3HH948 points for the mechanism(C8H14O3)10BF3OOH3COHH3COCH3HHBF3OOH3COHH3CHHOH3CH3CHOHHOF3B OCH3F3B OCH3HOH3CH3CHOHOName____Key________________________ 215H W06-Exam No. 2 Page 6 VIII. (20 points) Complete each of the following reaction sequences as necessary. (1) [Org. Lett. 2006, 8, 633]. 4OHNOTBSOHOCH3SO2Cl(1 mol equiv)(CH3CH2)3N98%contains N-H99%+ KOSO2CH3C22H39NO4SiC22H37NO3SiKOC(CH3)3(1 mol equiv)4ONOTBSOHOSO2CH3NOTBSOO (2) [Org. Lett. 2005, 7, 5163] NOOOHHCl (gas)(catalytic)CH2Cl283%4C13H23NO3C13H23NO3+ enantiomerNOOOHHtrans-isomeracceptable (3) [J. Org. Chem. 2006, 71, 1696] CH2Cl295%4OOPhCH2OOHKOC(CH3)3(1 mol equiv)SO2CH3+ KOSO2CH3OOPhCH2O (4) [Eur. J. Org. Chem. 2005, 4929] BrMgOOO10% aq H2SO412. aq NH4Cl54%4C6H12O2; IR: 3400 cm-1 (strong and broad); no 13C peaks in the ! 110-220 ppm region.+