CN1141309C - Preparation method of pyranoligosaccharide with 1-6 connection and 1,2-trans glycosidic band - Google Patents

Preparation method of pyranoligosaccharide with 1-6 connection and 1,2-trans glycosidic band Download PDF

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CN1141309C
CN1141309C CNB991251148A CN99125114A CN1141309C CN 1141309 C CN1141309 C CN 1141309C CN B991251148 A CNB991251148 A CN B991251148A CN 99125114 A CN99125114 A CN 99125114A CN 1141309 C CN1141309 C CN 1141309C
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孔繁祚
朱玉亮
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Research Center for Eco Environmental Sciences of CAS
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Abstract

本发明涉及1→6连接的、具有1,2反式糖苷键的六碳吡喃糖寡糖的合成的方法。用乙酰化的、1位为三氯乙酰亚胺酯的糖为糖基供体,用不保护的或部分保护的吡喃糖苷为糖基受体,在LEWIS酸催化下,可方便地制取双糖、三糖、四糖、五糖、六糖、七糖、八糖及大于八糖的寡糖。The present invention relates to a method for synthesizing 1→6 linked six-carbon pyranose oligosaccharides with 1,2 trans-glycosidic bonds. Using acetylated sugars with trichloroacetimidate at the 1-position as glycosyl donors and unprotected or partially protected pyranosides as glycosyl acceptors can be conveniently prepared under LEWIS acid catalysis Disaccharides, trisaccharides, tetrasaccharides, pentasaccharides, hexasaccharides, heptasaccharides, octasaccharides and oligosaccharides larger than octasaccharides.

Description

一种制备1→6连接的、具有1,2反式糖苷键吡喃寡糖的方法A method for preparing 1→6 linked pyran oligosaccharides with 1,2 trans-glycosidic bonds

本发明属于有生物活性的寡糖的制备技术领域,特别是涉及能用于合成1→6连接的、具有1,2反式糖苷键的吡喃式寡糖合成方法。The invention belongs to the technical field of preparation of biologically active oligosaccharides, and in particular relates to a synthesis method of pyran oligosaccharides which can be used to synthesize 1→6 linkages and have 1,2 trans glycosidic bonds.

1→6连接的、具有1,2反式糖苷键的吡喃式寡糖存在于很多天然产物中,如酵母细胞壁多糖含有α1→6连接的甘露寡糖片段,而植物的细胞壁的多糖含有β1→6连接的半乳寡糖片段,它们都具有1,2反式糖苷键的结构。在已报道的1→6连接的合成中,都是用保护—去保护的逐步合成策略。我们的研究组首次报道了通过原酸酯的中间体,用不保护的、或部分保护的糖基受体,能够选择性地得到1→6连接的、具有1,2反式糖苷键的双糖及三糖(见王为孔繁祚JOrg.Chem.63(1998)5744;王为 孔繁祚Angew.Chem.Int.Ed.38(1999)1247;孔繁祚 王为 中国发明专利971257788.4)。1→6-linked pyran oligosaccharides with 1,2 trans-glycosidic bonds exist in many natural products, such as yeast cell wall polysaccharides contain α1→6-linked mannan oligosaccharide fragments, and plant cell wall polysaccharides contain β1 →6-linked galactooligosaccharide fragments, all of which have a structure of 1,2 trans glycosidic bonds. In the reported synthesis of 1→6 linkage, the protection-deprotection stepwise synthesis strategy is used. Our group reported for the first time that a 1→6-linked, 1,2-trans glycosidic bonded double Sugar and trisaccharides (see Wang Wei Kong Fanzuo JOrg.Chem.63 (1998) 5744; Wang Wei Kong Fanzuo Angew.Chem.Int.Ed.38 (1999) 1247; Kong Fanzuo Wangwei China Invention Patent 971257788.4).

本发明的目的在于由活化的糖基供体与不保护的或部分保护的糖基受体直接偶连,不经过原酸酯的中间体,获得1→6连接的、具有1,2反式糖苷键的吡喃式寡糖的方法。The object of the present invention is to directly couple an activated glycosyl donor with an unprotected or partially protected glycosyl acceptor, without going through an orthoester intermediate, to obtain a 1→6-linked, 1,2 trans A method for glycosidically bonded oligopyranoses.

本发明的目的是这样实现的:以酰化的、还原端活化的糖为糖基供体,以不保护的、或部分保护的糖苷为糖基受体,在LEWIS酸催化下进行偶连,得到保护的寡糖,按常规方法去除保护,即得到游离的寡糖。The purpose of the present invention is achieved in this way: with acylated, reducing end-activated sugars as glycosyl donors, unprotected or partially protected glycosides as glycosyl acceptors, coupling under LEWIS acid catalysis, The protected oligosaccharides are removed according to conventional methods to obtain free oligosaccharides.

本发明的合成方法在于:用乙酰化的、1位为活化基团的单糖为糖基供体,用不保护的单糖苷为糖基受体,所述的糖苷为烷基苷如甲基、乙基、烯丙基、烯丁基、苄基苷等,将它们溶于有机溶剂中,LEWIS酸催化作用下发生偶连,得到1→6连接的、具有1,2反式糖苷键的双糖,如下所示: The synthesis method of the present invention is: use the acetylated monosaccharide with the 1-position as the activation group as the glycosyl donor, and use the unprotected monosaccharide as the glycosyl acceptor, and the glycoside is an alkyl glycoside such as methyl , ethyl, allyl, enbutyl, benzyl glycosides, etc., they are dissolved in organic solvents, coupled under the catalysis of LEWIS acid to obtain 1→6 linked, with 1,2 trans glycosidic bonds Disaccharides, as follows:

X=OC(NH)CCl3 X=OC(NH) CCl3

将所得的双糖3按常规方法苯甲酰化或乙酰化,然后将还原端活化,即得到双糖供体,如下所示: The obtained disaccharide 3 is benzoylated or acetylated according to conventional methods, and then the reducing end is activated to obtain a disaccharide donor, as shown below:

R=酰基  X=OC(NH)CCl3 R=acyl X=OC(NH)CCl 3

将所得的双糖3按常规方法苯甲酰化,再选择性地去除乙酰基,即得到双糖受体,如下所示: Benzoylate the obtained disaccharide 3 according to the conventional method, and then selectively remove the acetyl group to obtain the disaccharide acceptor, as shown below:

按单糖供体1与单糖受体2的偶连条件,使1与双糖受体7偶连,或双糖供体5与单糖受体2偶连,都能得到三糖,如下所示:

Figure C9912511400062
According to the coupling conditions of monosaccharide donor 1 and monosaccharide acceptor 2, couple 1 to disaccharide acceptor 7, or couple disaccharide donor 5 to monosaccharide acceptor 2 to obtain a trisaccharide, as follows Shown:
Figure C9912511400062

X=OC(NH)CCl3 X=OC(NH) CCl3

按双糖3转化为双糖供体5的条件,三糖8或9可转化为三糖供体11,而按双糖3转化为双糖受体7的条件,三糖8或9可转化为三糖供体13,如下所示:

Figure C9912511400063
According to the conditions of converting disaccharide 3 into disaccharide donor 5, trisaccharide 8 or 9 can be converted into trisaccharide donor 11, while under the conditions of disaccharide 3 being converted into disaccharide acceptor 7, trisaccharide 8 or 9 can be converted is the trisaccharide donor 13, as follows:
Figure C9912511400063

R=苯甲酰基  X=OC(NH)CCl3 R=benzoyl X=OC(NH)CCl 3

按单糖供体1与单糖受体2偶连的条件,使双糖供体5与双糖受体7偶连,或单糖供体1与三糖受体13偶连,或三糖供体11与单糖受体2偶连,都能得到四糖,如下所示:

Figure C9912511400071
According to the coupling conditions of monosaccharide donor 1 and monosaccharide acceptor 2, disaccharide donor 5 is coupled with disaccharide acceptor 7, or monosaccharide donor 1 is coupled with trisaccharide acceptor 13, or trisaccharide Donor 11 is coupled with monosaccharide acceptor 2, both of which can obtain tetrasaccharides, as follows:
Figure C9912511400071

R=酰基  X=OC(NH)CCl3 R=acyl X=OC(NH)CCl 3

按三糖8转化为三糖供体11及三糖受体13的条件,四糖14分别转化为四糖供体16及四糖受体18,如下所示:

Figure C9912511400072
According to the conditions under which trisaccharide 8 is converted into trisaccharide donor 11 and trisaccharide acceptor 13, tetrasaccharide 14 is converted into tetrasaccharide donor 16 and tetrasaccharide acceptor 18, respectively, as follows:
Figure C9912511400072

R=酰基  X=OC(NH)CCl3 R=acyl X=OC(NH)CCl 3

按单糖供体1与单糖受体2的偶连条件,使四糖供体16与四糖受体18偶连,得到八糖19,如下所示: According to the coupling conditions of monosaccharide donor 1 and monosaccharide acceptor 2, tetrasaccharide donor 16 was coupled with tetrasaccharide acceptor 18 to obtain octasaccharide 19, as follows:

R=苯甲酰基R = benzoyl

按常规方法去除保护,即得到游离的八糖。The protection is removed by conventional methods to obtain free octasaccharides.

用所制备的单、双、三、四糖供体与所制备的单、双、三、四糖受体以适当方式组合,可方便地制备出五糖、六糖、七糖、八糖、九糖、十糖、十一糖、十二糖、十四糖、十六糖。By combining the prepared mono-, di-, tri-, and tetra-saccharide donors with the prepared mono-, di-, tri-, and tetra-saccharide acceptors in an appropriate manner, pentasaccharides, hexasaccharides, heptasaccharides, octasaccharides, Nine sugars, ten sugars, undecyl sugars, twelve sugars, fourteen sugars, sixteen sugars.

所述的单糖为葡萄糖、半乳糖、甘露糖、塔罗糖、塔各糖、古勒糖、依德糖、阿罗糖。The monosaccharides are glucose, galactose, mannose, talose, tagose, gulose, edose, and allose.

所述的寡糖或由均一的单糖组成,或由上述的不同的单糖按任意组合方式而成,The oligosaccharides are either composed of uniform monosaccharides, or formed in any combination of the above-mentioned different monosaccharides,

所述的有机溶剂为酰胺、卤代烷、醚。Described organic solvent is amide, haloalkane, ether.

所述的LEWIS酸为银盐、三氟化硼、三甲基硅三氟甲磺酸酯。The LEWIS acid is silver salt, boron trifluoride, trimethylsilyl trifluoromethanesulfonate.

下面结合实施例对本发明进行详细的说明。 The present invention will be described in detail below in conjunction with the examples.

(1)双糖3的合成 (1) Synthesis of disaccharide 3

将单糖供体1 492毫克,1毫克分子(按Adv.Carbohydr.Chem.Biochem.50,1994,21的方法制备)单糖受体2242毫克,1.1毫克分子,溶于干燥的二甲基甲酰胺20毫升中,冷却至-40℃,在氮气保护下,加入TMSOTf 10微升,使反应物温度自然升至室温。4小时后加入三乙胺终止反应,按常规方法处理反应液,用硅胶柱层析法,以乙酸乙酯/石油醚为淋洗剂精制产物,得到3385毫克,产率70%。1492 milligrams of monosaccharide donors, 1 millimetre (prepared by the method of Adv. In 20 ml of amide, cool to -40 °C, under nitrogen protection, add 10 microliters of TMSOTf, and let the temperature of the reactant naturally rise to room temperature. After 4 hours, triethylamine was added to terminate the reaction, and the reaction solution was processed according to conventional methods, and the product was purified by silica gel column chromatography with ethyl acetate/petroleum ether as eluent to obtain 3385 mg with a yield of 70%.

(2)双糖供体5的制备 (2) Preparation of disaccharide donor 5

将双糖3 550毫克,1毫克分子,溶于15毫升吡啶中,室温、搅拌下加入苯甲酰氯1.3毫升,用TLC监测反应,反应完成后,按常规方法处理反应液,定量得到双糖4。将4溶于10毫升二氯甲烷中,加入二氯化钯100毫克,在室温、搅拌下进行反应,用TLC监测。反应完成后,用常规方法处理反应液,产物用硅胶柱层析法精制,得到1位为游离羟基的苯甲酰化双糖744毫克,产率90%。将此双糖826毫克,1毫克分子溶于15毫升干燥二氯甲烷中,加入三氯乙睛0.2毫升,2毫克分子,DBU 27微升,在室温、搅拌下进行反应,用TLC监测。反应完成后,用常规方法处理反应液,产物用硅胶柱层析法精制,得到双糖供体5 920毫克,产率95%Dissolve 550 mg of disaccharide 3, 1 milligram molecule, in 15 milliliters of pyridine, add 1.3 milliliters of benzoyl chloride under stirring at room temperature, monitor the reaction with TLC, after the reaction is completed, process the reaction solution according to a conventional method, and quantitatively obtain disaccharide 4 . 4 was dissolved in 10 ml of dichloromethane, 100 mg of palladium dichloride was added, and the reaction was carried out under stirring at room temperature, monitored by TLC. After the reaction was completed, the reaction solution was processed by conventional methods, and the product was purified by silica gel column chromatography to obtain 744 mg of benzoylated disaccharide with a free hydroxyl group at the first position, with a yield of 90%. Dissolve 826 mg of this disaccharide, 1 millimole, in 15 milliliters of dry dichloromethane, add 0.2 milliliter of trichloroacetonitrile, 2 millimoles, and 27 microliters of DBU, react at room temperature under stirring, and monitor with TLC. After the reaction was completed, the reaction solution was processed by a conventional method, and the product was purified by silica gel column chromatography to obtain 920 mg of disaccharide donor 5 with a yield of 95%.

(3)双糖受体7的合成

Figure C9912511400083
(3) Synthesis of disaccharide receptor 7
Figure C9912511400083

将双糖3 554毫克,1毫克分子,溶于15毫升吡啶中,室温、搅拌下加入苯甲酰氯1.3毫升,用TLC监测反应,反应完成后,按常规方法处理反应液,定量得到双糖4。将4溶于15毫升无水甲醇中,加入乙酰氯0.75毫升,用TLC监测。反应完成后,用常规方法处理反应液,产物用硅胶柱层析法精制,得到双糖受体7 625毫克,产率90%。Dissolve 554 mg of disaccharide 3, 1 milligram molecule, in 15 milliliters of pyridine, add 1.3 milliliters of benzoyl chloride under stirring at room temperature, monitor the reaction with TLC, after the reaction is completed, process the reaction solution according to a conventional method, and quantitatively obtain disaccharide 4 . Dissolve 4 in 15 ml of anhydrous methanol, add 0.75 ml of acetyl chloride, and monitor with TLC. After the reaction was completed, the reaction solution was processed by conventional methods, and the product was purified by silica gel column chromatography to obtain 7625 mg of disaccharide acceptor with a yield of 90%.

(4)三糖8及9的合成

Figure C9912511400091
(4) Synthesis of trisaccharides 8 and 9
Figure C9912511400091

按合成双糖3的基本相同的条件,由1 541毫克,1.1毫克分子与7 694毫克,1毫克分子(1+2)得到三糖8 720毫克,产率70%。或由5 1066毫克,1.1毫克分子与2 220毫克,1毫克分子(2+1)得到三糖9 720毫克,产率70%。According to the same conditions for the synthesis of disaccharide 3, 8 720 mg of trisaccharide was obtained from 1 541 mg, 1.1 mg molecule and 7 694 mg, 1 mg molecule (1+2), with a yield of 70%. Or from 5 1066 mg, 1.1 mg molecule and 2 220 mg, 1 mg molecule (2+1) to obtain 9 720 mg of trisaccharide, yield 70%.

(5)三糖供体11的制备

Figure C9912511400092
(5) Preparation of trisaccharide donor 11
Figure C9912511400092

将三糖8 1028毫克,1毫克分子溶于15毫升干燥吡啶,加入苯甲酰氯0.39毫升,3.3毫克分子,反应在室温、搅拌下进行,用TLC监测。反应完成后,用常规方法处理反应液,定量得到三糖10,将10溶于15毫升二氯甲烷中,加入二氯化钯120毫克,在室温、搅拌下进行反应,用TLC监测。反应完成后,用常规方法处理反应液,产物用硅胶柱层析法精制,得到1位为游离羟基的三糖1170毫克,产率90%。将此三糖1300毫克,1毫克分子溶于15毫升干燥二氯甲烷中,加入三氯乙睛0.2毫升,2毫克分子,DBU 27微升,在室温、搅拌下进行反应,用TLC监测。反应完成后,用常规方法处理反应液,粗产物可直接用于下一步反应。Dissolve 1028 mg, 1 mg molecule of trisaccharide 8 in 15 ml of dry pyridine, add 0.39 ml, 3.3 mg molecule of benzoyl chloride, and react at room temperature under stirring, monitored by TLC. After the reaction was completed, the reaction liquid was processed by conventional methods to quantitatively obtain trisaccharide 10. Dissolve 10 in 15 ml of dichloromethane, add 120 mg of palladium dichloride, react at room temperature under stirring, and monitor with TLC. After the reaction was completed, the reaction solution was processed by conventional methods, and the product was purified by silica gel column chromatography to obtain 1170 mg of trisaccharides with a free hydroxyl group in the first position, with a yield of 90%. Dissolve 1300 mg of this trisaccharide, 1 millimole, in 15 ml of dry dichloromethane, add 0.2 ml of trichloroacetate, 2 millimole, and 27 microliters of DBU, react at room temperature with stirring, and monitor with TLC. After the reaction is completed, the reaction solution is processed by a conventional method, and the crude product can be directly used in the next reaction.

(6)三糖受体13的制备 (6) Preparation of trisaccharide receptor 13

按3转化7的方法,由1028毫克,1毫克分子的三糖8经由10而得到三糖受体13,两步合并的产率为90%。According to the method of converting 3 into 7, trisaccharide acceptor 13 was obtained from 1028 mg, 1 mg molecule of trisaccharide 8 via 10, and the combined yield of the two steps was 90%.

(7)四糖的制备

Figure C9912511400101
(7) Preparation of tetrasaccharides
Figure C9912511400101

按单糖供体1与单糖受体2偶连的条件,进行双糖供体5与双糖受体7的偶连,得到四糖14 1501毫克,产率70%。从NMR谱图上可看出,在δ5.17,5.05,4.82,4.73有四个H-1,在δ2.11,2.03,1.98,1.94四个CH3CO。According to the coupling conditions of monosaccharide donor 1 and monosaccharide acceptor 2, the coupling of disaccharide donor 5 and disaccharide acceptor 7 was carried out to obtain 1501 mg of tetrasaccharide 14 with a yield of 70%. It can be seen from the NMR spectrum that there are four H-1 at δ5.17, 5.05, 4.82, and 4.73, and four CH 3 CO at δ2.11, 2.03, 1.98, and 1.94.

(8)四糖供体16的制备

Figure C9912511400102
(8) Preparation of tetrasaccharide donor 16
Figure C9912511400102

按三糖8转化为三糖供体11的条件,由四糖14 1502毫克,1毫克分子得到四糖供体16 1729毫克,产率90%。According to the conditions of converting trisaccharide 8 into trisaccharide donor 11, 1 milligram of tetrasaccharide 14 1502 mg was used to obtain tetrasaccharide donor 16 1729 mg with a yield of 90%.

(9)四糖受体18的制备

Figure C9912511400103
(9) Preparation of Tetrasaccharide Acceptor 18
Figure C9912511400103

按三糖8转化为三糖受体13的条件,由四糖14 1502毫克,1毫克分子得到四糖受体18 1478毫克,产率90%。According to the conditions of converting trisaccharide 8 into trisaccharide acceptor 13, 1 milligram of tetrasaccharide 14 1502 mg was used to obtain tetrasaccharide acceptor 18 1478 mg with a yield of 90%.

(10)八糖的制备

Figure C9912511400104
(10) Preparation of octasaccharides
Figure C9912511400104

按单糖供体1与单糖受体2偶连的条件,进行四糖供体16与四糖受体18的偶连,得到八糖19 3358毫克,产率70%。从NMR谱图上可看出,在δ5.33-4.61有八个H-1,在δ2.09,2.03,1.97,1.89有四个CH3CO。According to the coupling conditions of monosaccharide donor 1 and monosaccharide acceptor 2, the coupling of tetrasaccharide donor 16 and tetrasaccharide acceptor 18 was carried out to obtain 3358 mg of octasaccharide 19 with a yield of 70%. It can be seen from the NMR spectrum that there are eight H-1 at δ5.33-4.61, and four CH 3 CO at δ2.09, 2.03, 1.97, and 1.89.

Claims (5)

1.一种制备1→6连接的、具有1,2反式糖苷键吡喃寡糖的方法,其特征在于:1. A method for preparing 1 → 6 connection, with 1,2 trans-glycosidic bond pyran oligosaccharides, characterized in that: (1)用乙酰化的、1位为活化基团的单糖为糖基供体,用不保护的单糖苷糖基受体,将它们溶于有机溶剂中,LEWIS酸催化作用下发生偶连,得到1→6连接的、具有1,2反式糖苷键的双糖,如下所示:
Figure C9912511400021
(1) Use acetylated monosaccharides with 1-position as the activating group as glycosyl donors, and use unprotected monoglycoside glycosyl acceptors, dissolve them in organic solvents, and couple under the action of LEWIS acid catalysis , resulting in a 1→6 linked disaccharide with 1,2 trans glycosidic linkages, as shown below:
Figure C9912511400021
X=OC(NH)CCl3 X=OC(NH) CCl3 (2)将所得的双糖3按常规方法苯甲酰化或乙酰化,然后将还原端活化,即得到双糖供体,如下所示:
Figure C9912511400022
(2) Benzoylate or acetylate the obtained disaccharide 3 according to a conventional method, and then activate the reducing end to obtain a disaccharide donor, as follows:
Figure C9912511400022
R=酰基  X=OC(NH)CCl3 R=acyl X=OC(NH)CCl 3 (3)将所得的双糖3按常规方法苯甲酰化,再选择性地去除乙酰基,即得到双糖受体,如下所示:
Figure C9912511400023
(3) Benzoylate the obtained disaccharide 3 according to a conventional method, and then selectively remove the acetyl group to obtain a disaccharide acceptor, as shown below:
Figure C9912511400023
(4)按单糖供体1与单糖受体2的偶连条件,使1与双糖受体7偶连,或双糖供体5与单糖受体2偶连,都能得到三糖,如下所示:
Figure C9912511400024
(4) According to the coupling conditions of monosaccharide donor 1 and monosaccharide acceptor 2, 1 is coupled with disaccharide acceptor 7, or disaccharide donor 5 is coupled with monosaccharide acceptor 2, and three sugar, as follows:
Figure C9912511400024
R=酰基 X=OC(NH)CCl3 R=acyl X=OC(NH)CCl 3 (5)按双糖3转化为双糖供体5的条件,三糖8或9可转化为三糖供体11,而按双糖3转化为双糖受体7的条件,三糖8或9可转化为三糖供体13,如下所示:
Figure C9912511400031
(5) According to the condition that disaccharide 3 is converted into disaccharide donor 5, trisaccharide 8 or 9 can be converted into trisaccharide donor 11, and according to the condition that disaccharide 3 is converted into disaccharide acceptor 7, trisaccharide 8 or 9 can be converted to the trisaccharide donor 13 as follows:
Figure C9912511400031
R=苯甲酰基 X=OC(NH)CCl3 R=benzoyl X=OC(NH)CCl 3 (6)按单糖供体1与单糖受体2偶连的条件,使双糖供体5与双糖受体7偶连,或单糖供体1与三糖受体13偶连,或三糖供体11与单糖受体2偶连,都能得到四糖,如下所示: (6) According to the coupling conditions of monosaccharide donor 1 and monosaccharide acceptor 2, disaccharide donor 5 is coupled with disaccharide acceptor 7, or monosaccharide donor 1 is coupled with trisaccharide acceptor 13, Or trisaccharide donor 11 is coupled with monosaccharide acceptor 2 to obtain tetrasaccharides, as follows: R=酰基  X=OC(NH)CCl3 R=acyl X=OC(NH)CCl 3 (7)按三糖8转化为三糖供体11及三糖受体13的条件,四糖14分别转化为四糖供体16及四糖受体18,如下所示: (7) According to the conditions under which trisaccharide 8 is converted into trisaccharide donor 11 and trisaccharide acceptor 13, tetrasaccharide 14 is converted into tetrasaccharide donor 16 and tetrasaccharide acceptor 18 respectively, as follows: R=酰基  X=OC(NH)CCl3 R=acyl X=OC(NH)CCl 3 (8)按单糖供体1与单糖受体2的偶连条件,使四糖供体16与四糖受体18偶连,得到八糖19,如下所示: (8) According to the coupling conditions of monosaccharide donor 1 and monosaccharide acceptor 2, tetrasaccharide donor 16 is coupled with tetrasaccharide acceptor 18 to obtain octasaccharide 19, as follows: R=苯甲酰基R = benzoyl 按常规方法去除保护,即得到游离的八糖,Remove the protection by conventional methods to obtain free octasaccharides, (9)用所制备的单、双、三、四糖供体与所制备的单、双、三、四糖受体以适当方式组合,可制备出五糖、六糖、七糖、八糖、九糖、十糖、十一糖、十二糖、十四糖、十六糖。(9) By combining the prepared mono-, di-, tri-, and tetra-saccharide donors with the prepared mono-, di-, tri-, and tetra-saccharide acceptors in an appropriate manner, pentasaccharides, hexasaccharides, heptasaccharides, and octasaccharides can be prepared , nine sugars, ten sugars, eleven sugars, twelve sugars, fourteen sugars, sixteen sugars.
2.如权利要求1所述的一种制备1→6连接的、具有1,2反式糖苷键吡喃寡糖的方法,其特征在于:所述的单糖为葡萄糖、半乳糖、甘露糖、塔罗糖、塔各糖、古勒糖、依德糖、阿罗糖。2. a kind of preparation 1 → 6 as claimed in claim 1 is connected, have 1, the method for 2 trans-glycosidic bond pyran oligosaccharides, it is characterized in that: described monosaccharide is glucose, galactose, mannose , talose, tallose, gule sugar, edose, allose. 3.如权利要求1所述的一种制备1→6连接的、具有1,2反式糖苷键吡喃寡糖的方法,其特征在于:所述的寡糖或由均一的单糖组成,或由权利要求2中所述的不同的单糖按任意组合方式而成。3. A method for preparing 1 → 6 linked pyran oligosaccharides with 1,2 trans-glycosidic bonds as claimed in claim 1, characterized in that: the oligosaccharides are composed of uniform monosaccharides, Or form by any combination of different monosaccharides described in claim 2. 4.如权利要求1所述的一种制备1→6连接的、具有1,2反式糖苷键吡喃寡糖的方法,其特征在于:所述的有机溶剂为酰胺、卤代烷、醚。4 . A method for preparing 1 → 6 linked oligopyranose with 1,2 trans glycosidic bonds as claimed in claim 1 , characterized in that: the organic solvent is amides, alkyl halides, and ethers. 5.如权利要求1所述的一种制备1→6连接的、具有1,2反式糖苷键吡喃寡糖的方法,其特征在于:所述的LEWIS酸为银盐、三氟化硼、三甲基硅三氟甲磺酸酯。5. a kind of preparation 1 → 6 as claimed in claim 1 connects, has the method for 1,2 trans-glycosidic bond pyran oligosaccharides, it is characterized in that: described LEWIS acid is silver salt, boron trifluoride , Trimethylsilyl trifluoromethanesulfonate.
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