C C H 2. O p-tsoh (catalytic) H 3 O +! O

Transcription

1 ame Key 15 W06-Exam o. Page I. ( 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 -pentynol (). 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. ( ) ( ) 7 ( ) 1 (racemate) - - p-ts - 1. a / 3 (liq). I ( ) 3 - ( ) ! - ( ) 5 3 P -- ( ) ( ) 6 - MgBr. aq l 3 ( ) 7 -- ( ) 5 3, Pd, BaS /quinoline 1 (racemate) II. (6 points) Treatment of the multi-functional group containing compound, 3, with ab in TF followed by the protection of the selectively produced alkoxide with TBSl results in the formation of TBS ether in 81% overall yield from 3 (rg. Lett. 005, 7, 553). Draw in the box below the structure of. 3 Ph 1. ab in TF (room temperature) 3. TBSl, imidazole, DMF note: Ph = phenyl; TBSl = (tert-butyl)dimethylsilyl chloride imidazole: DMF: dimethylformamide (solvent) 6 Ph TBS

2 ame Key 15 W06-Exam o. Page 3 III (10 points) When trans-diol 5 was treated with one mol equivalent of lead tetraacetate, Pb[(=) 3 ], in the presence of a catalytic amount of trichloroacetic acid (pk a ~1.0), diketone 6 was produced together with lead diacetate and acetic acid ( 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. Pb(-- 3 ) l Pb(-- 3 ) l Pb Pb Pb 3 --l Pb 3 neutral intermediate X IV. (1 points) omplete the following reaction scheme by providing in the boxes the structures of the corresponding products (J. rg. hem. 006, 71, 130). Also, indicate for each of the compounds how many 13 MR peaks are expected to be observed when run by the proton-decoupled method. 3 p-ts 3 a (. mol equiv) 3 Br (.8 mol equiv) DMF 0 to room temp. umber of 13 peaks expected: 10 umber of 13 peaks expected: 16

3 ame Key 15 W06-Exam o. Page V. (8 points) Treatment of keto-diol 7 with a catalytic amount of p-toluenesulfonic acid (p-ts) 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. 7 p-ts (catlytic) + + VI. (1 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! 3 a insect juvenile hormone 3 18 l () J. rg. hem. 006, 71, p-ts l

4 ame Key 15 W06-Exam o. Page 5 VII. (8 points) Treatment of triol 8 with excess trimethyl orthoacetate [ 3 ( 3 ) 3 ] in the presence of a catalytic amount of pyridinium p-toluenesulfonate (PPTS), a milder version of p-ts, cleanly produced a diastereomeric mixture of cyclic orthoester (8% yield) and mol equiv of 3. Subsequent treatment of this unstable product resulted in the formation of a new 5-membered cyclic ether 10 (1% yield) together with one mol equiv of 3 (J. rg. hem. 006, 71, 116) ( 3 ) 3 PPTS l room temp., 5 min BF 3 ( 3 ) l 10 min 10 ( ) IR: no peaks above 3100 cm cm -1 (strong) 13 MR:! ppm and 7 sp 3 13 peaks. + 3 (1) Provide in the box below a stepwise mechanism for the formation of cyclic orthoester from triol 8 by the use of the curve-arrow convention. You may use -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. 3 B B :B :B 16 () Draw in the box indicated the structure of compound 10 and show a stepwise mechanism for its formation from by the use of the curved-arrow convention. Use BF 3 to represent the Lewis acid and clearly indicate the stereochemical outcome for each step. You don t need to balance each step points for the mechanism 3 BF 3 BF F 3 B F 3 B ( )

5 VIII. (0 points) omplete each of the following reaction sequences as necessary. (1) [rg. Lett. 006, 8, 633]. ame Key 15 W06-Exam o. Page 6 TBS 3 3 S l ( 3 ) 3 8% TBS contains - S 3 K( 3 ) 3 % 37 3 TBS + KS 3 () [rg. Lett. 005, 7, 5163] l (gas) l 83% + enantiomer trans-isomer acceptable 1333 (3) [J. rg. hem. 006, 71, 166] Ph 3 S K( 3 ) 3 l 5% Ph + KS 3 () [Eur. J. rg. hem. 005, ] BrMg 1. aq l 10% aq S + 5% + enantiomer 6 1 ; IR: 300 cm -1 (strong and broad); no 13 peaks in the! ppm region.