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在稀碱中形成的醛固酮异构体的手性。

Chirality of isomers of aldosterone formed in dilute alkali.

作者信息

Duax W L, Griffin J F

机构信息

Medical Foundation of Buffalo, Inc., New York 14203.

出版信息

Chirality. 1991;3(1):71-5. doi: 10.1002/chir.530030112.

DOI:10.1002/chir.530030112
PMID:2039684
Abstract

In the presence of dilute alkali at room temperature aldosterone undergoes rearrangement to form 11 beta,18:18,21-diepoxy-20,21-dihydroxy-4-pregnen-3-one (V). X-Ray crystallographic analysis demonstrates that isomers of both 18R, 20S, 21S and 18R, 20S, 21R configuration are formed rather than the 18R, 20R, 21R isomer postulated on the basis of examination of 1H-NMR spectra. The spectra appears to be consistent with the major component of the mixture. The 20S configuration observed is in agreement with the chirality assigned to the degradation product obtained when the same alkaline solution of aldosterone is subjected to reflux. The crystals of (V) are monoclinic P2(1), Z = 4 with a = 20.891(2), b = 6.3848(5), c = 16.067(2)A, beta = 122.09 degrees(1) with two molecules in the asymmetric unit. Molecule A has the 20S,21S configuration and the molecules in the second site are a mixture of the 20S, 21S and 20S, 21R configuration in the ratio of 3:2.

摘要

在室温下,醛固酮在稀碱存在下会发生重排,形成11β,18:18,21-二环氧-20,21-二羟基-4-孕烯-3-酮(V)。X射线晶体学分析表明,形成的是18R, 20S, 21S和18R, 20S, 21R构型的异构体,而非基于1H-NMR光谱检查推测的18R, 20R, 21R异构体。该光谱似乎与混合物的主要成分一致。观察到的20S构型与醛固酮的相同碱性溶液回流时得到的降解产物所确定的手性一致。(V)的晶体为单斜晶系P2(1),Z = 4,a = 20.891(2),b = 6.3848(5),c = 16.067(2)Å,β = 122.09°(1),不对称单元中有两个分子。分子A具有20S,21S构型,第二个位点的分子是20S, 21S和20S, 21R构型的混合物,比例为3:2。

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