반응 #1951035

반응 방정식

MeOH CHCl3

3-Formyl-2,5,6-trichloro-1-[3,5-di-O-(p-toluoyl)-2-deoxy-β-D-ribofuranosyl]indole

sodium methoxide

→

4.114

수율 69.0%

5,6-Dichloro-2-methoxy-3-formyl-1-(2-deoxy-β-D-ribofuranosyl)indole

수율 69.0%

용매

MeOH

MeOH

반응 조건

후처리

1. 1
   기타The solvent was then removed under vacuum
2. 2
   추출extracted with EtOAc (2×100 mL)
3. 3
   세척The combined organic extracts were washed with brine (25 mL)
4. 4
   건조dried over MgSO4
5. 5
   여과filtered
6. 6
   기타evaporated
7. 7
   기타to yield a damp solid
8. 8
   workup.ADDITIONFractions containing product
9. 9
   기타evaporated
10. 10
    기타to yield a white solid
11. 11
    기타The solid was recrystallized
12. 12
    온도from warm MeOH

실험 절차

3-Formyl-2,5,6-trichloro-1-[3,5-di-O-(p-toluoyl)-2-deoxy-β-D-ribofuranosyl]indole (4.113, 198 mg, 0.33 mmol) was suspended in dry MeOH (10 mL) to which was added sodium methoxide (75 mg, 1.4 mmol). The suspension was stirred at room temperature for 16 h, after which time the solution clarified. The solvent was then removed under vacuum, and the residue was suspended in water (50 mL), and extracted with EtOAc (2×100 mL). The combined organic extracts were washed with brine (25 mL), dried over MgSO4, filtered and evaporated to yield a damp solid. The crude material was dissolved in MeOH (1 mL) and subjected to column chromatography (40×350 mm) on silica gel with 10% MeOH/CHCl3. Fractions containing product were pooled and evaporated to yield a white solid. The solid was recrystallized from warm MeOH to yield 82 mg (68%) of 4.114 as a white crystalline solid: mp 186-187° C.; Rf0.3 (10% MeOH/CHCl3); 1H-NMR (500 MHz, DMSO-d6): δ 10.21 (s, 1H), 8.29 (s, 1H), 8.22 (s, 1H), 6.32 (dd, 1H), 5.37 (b, 1H, D2O exch.), 5.15 (b, 1H, D2O exch.), 4.43 (s, 3H), 4.41 (m, 1H), 3.83 (d, 1H), 3.68 (m, 2H), 2.53 (m, 1H), 2.11 (m, 1H). 13C-NMR (125 MHz, DMSO-d6): δ 183.22, 159.16, 129.40, 125.04, 124.73, 124.66, 121.11, 114.24, 100.02, 87.29, 82.99, 70.07, 64.16, 61.03, 38.43. HRMS (EI) m/z calcd. for C15H15Cl2NO5 359.0327, found 359.0320. Anal calcd for C15H15Cl2NO5.½H2O: C, 48.80; H, 4.37; N, 3.79. Found: C, 48.48; H, 4.61; N, 3.61.

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