class problem set1

1 1. A student carries out the transformations shown below to elucidate the stereochemistry of compound A , an intermediate in the synthesis of a complex natural product. Using the spectra of compound B , knowledge of the chemical sequence, and the NOE data supplied, perform the following operations: a. Assign all of the resonances in the spectrum b. Determine as many coupling constants as possible, and report your data in ACS format c. Deduce the relative configuration at as many positions as possible in products A and B . O HO Me HO OH Me Me OEt A B O I 2 , NaHCO 3 I CO 2 Et 1) Ac 2 O, py 2) Bu 3 SnH AIBN,  O AcO Me CO 2 Et Decoupling the signal at 1.20 ppm reduces the signal at 2.81 to a doublet with J = 6.3 Hz Decoupling the signal at 0.90 ppm reduces the signal at 2.60 to a dd with J = 5.4, 4.5 Hz 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 ppm 4.35 ppm 2149.190 2151.881 2153.698 2156.381 2158.190 2160.880 2174.323 2179.714 2180.337 2185.723 3.80 3.85 ppm 1886.654 1890.976 1896.954 1901.275 1914.531 1917.201 1924.829 1927.500 2.75 2.80 2.85 ppm 1391.927 1398.588 1404.951 1411.437 1418.120 2.60 2.65 ppm 1284.094 1289.273 1291.373 1294.143 1296.440 1298.668 1301.383 1303.612 1305.944 1308.674 1310.811 1315.987

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3 Proton  NOE A 4.36 D (7%), G (2%), F (7%), I (6 %) B 4.31 D (2 %), E (5 %), G (5 %), H (6 %), K (3 %) C 3.99 F (4 %), I (3 %), J ( 9%) D 3.84 A (6 %), B (2 %), E (9 %) E 3.79 B (5 %), D (9 %), G (3 %) F 2.81 A (7 %), C (4 %), I (9 %) G 2.60 A (2 %), B (5 %), E (3 %), K (6 %) H 1.98 B ( 5 %) I 1.16 A (5 %), F (9 %) J 1.01 C (8 %) K 0.90 A (8 %), B (2.5 %), G (7 %)

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5 2. A student continues the same synthesis, and carries out the transformations shown below to elucidate the stereochemistry of intermediate X . Using the spectral data for compound Y , and knowledge of the chemical sequence, deduce the missing stereochemistry in product X a. Assign all of the resonances in the spectrum b. Determine as many coupling constants as possible, and report your data in ACS format c. Deduce the stereochemistry at as many positions as possible in product X (you will need the results of question 11 for this). O MeO Me OAc Me Me 2 BBr Br MeO Me OH Me OAc 1) HSnBu 3 2) NaOH, MeOH 3) PCHO, H + MeO Me O Me O Ph X Y 1 H NMR (500 MHz, C 6 D 6 )  A : 7.32-7.09 (m, 5H) B : 5.40 (s, 1H) C : 3.99 (dd, J = 13.1, 11.0 Hz, 1H) D : 3.86 (dd, J = 13.1, 4.6 Hz, 1H) E : 3.46 (quint, J = 6.6 Hz, 1H) F : 3.29 (dd, J = 10.1, 3.7 Hz, 1H) G : 3.14 (s, 3H) H : 2.20-2.14 (m, 1H) I : 2.15-2.07 (m, 1H) J : 1.65 (d, J = 6.5 Hz, 3H) K : 1.20 (d, J = 6.3 Hz, 3H) L : 1.12 (d, J = 6.7 Hz, 3H). Decoupling the signal at 1.20 ppm ( K ) simplifies the signal at 2.15-2.07 ( I ) (1077.30 – 1032.7 Hz) to a multiplet from 1067.85 to 1042.15 Hz. Hydrogens match sum of all JS ↓ L is F must be on ~ same Face due H2 -o to seeing Each other 25 . 1 G ] Balso sees C not D one i ! I ph B - C E / E - T - - & O - O ↑ + r i = 4 YS I C Age CH3 [ ⊥ t z ju L Cl one F # k & " /" D 2 E - · S B c I is on Save Face of D While Diff . than