問題詳情

四、下列九個類黃酮化合物(1 至 9)係分離自菲律賓楠(Machilus philippinensis),其分離方法/步驟如下:The EtOH extract (160 g) of the dried leaves of M. philippinensis (774 g) was partitionedinto fractions soluble in CH2Cl2 (38.30 g), EtOAc (10.51 g), n-BuOH (31.02 g) and H2O(74.12 g) by a liquid-liquid partitioning process. Part of the EtOAc-soluble fraction (6.03 gout of 10.51 g) was fractionated on a Sephadex LH-20 column (815 × 35 mm, MeOH) togive six fractions (E1~6). An aliquot of fraction E2 (45.0 mg out of 500.7 mg) was separatedby a semi-preparative RP-18 HPLC column (Phenomenex® Prodigy ODS-3, 250 × 10 mm,5 μm), each run 5.0 mg, delivered by 16% MeCN in H2O for 30 min, to 25% MeCN in 40 minby a linear gradient mode, and then MeCN for 15 min, with a flow rate of 2.5 mL/min anddetection at 300 nm. After nine runs, the fractions containing pure compounds were evaporatedunder reduced pressure to give 3 (2.9 mg, tR = 21.39 min), 4 (10.0 mg, tR = 22.61 min), 6(7.3 mg, tR = 30.86 min), 7 (4.8 mg, tR = 36.36 min), and 9 (1.9 mg, tR = 42.33 min),respectively. An aliquot of fraction E3 (273.8 mg out of 1.20 g) was separated on a Lobar(low-pressure) RP-18 column (LiChroprep RP-18, size B, 310 × 25 mm; 40-63 μm, Merck),delivered by a stepwise gradient of MeOH―H2O from 30:70 to 100:0, to give sixsubfractions (E3-1~6). Fractions E3-3 (24.7 mg) and -5 (2.0 mg) were pure 5 and 8,respectively. An aliquot of fraction E6 (201.0 mg out of 756.0 mg) was separated on the sameLobar RP-18 column, delivered by a stepwise gradient of MeOH―H2O from 35:65 to 85:15,to give five subfractions (E6-1~5). Fractions E6-2 (14.4 mg) and -4 (2.3 mg) were pure 1 and2, respectively.


【題組】⑴以上圖右之方式表達化合物 1 至 9 之分離流程。(10 分)

參考答案

答案:D
難度:簡單0.885878
統計:A(110),B(28),C(81),D(1700),E(0) #
個人:尚未作答書單:辦公處所之面積規劃區分表

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