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2,2'-(二硫代二亚基)二苯甲酸 - N,N'-双 -(3 - 吡啶 - 甲基)乙二酰胺(1/1)

2,2'-(Disulfanediyl)dibenzoic acid-N,N'-bis-(3-pyridyl-meth-yl)ethane-diamide (1/1).

作者信息

Arman Hadi D, Miller Tyler, Poplaukhin Pavel, Tiekink Edward R T

出版信息

Acta Crystallogr Sect E Struct Rep Online. 2010 Sep 18;66(Pt 10):o2590-1. doi: 10.1107/S1600536810036494.

Abstract

The asymmetric unit of the title cocrystal, C(14)H(14)N(4)O(2)·C(14)H(10)O(4)S(2), comprises a twisted 2,2'-(disulfanediyl)dibenzoic acid mol-ecule [dihedral angle between the benzene rings = 76.35 (10)°] and a U-shaped N,N'-bis-(3-pyridyl-meth-yl)ethane-diamide mol-ecule with the pyridyl groups lying to the same side of the central diamide moiety [C-C-C-N = 113.8 (2) and -117.6 (2)°]. The latter aggregate into supra-molecular tapes propagating along the a axis via centrosymmetric eight-membered amide {⋯OCNH}(2) synthons. Intra-molecular N-H⋯O hydrogen bonds are observed. The 2,2'-(disulfanediyl)dibenzoic acid mol-ecules form carbox-yl-pyridine O-H⋯N hydrogen bonds, bridging a pyridine residue below the plane of the tape and one above the plane with two inter-vening N,N'-bis-(3-pyridyl-meth-yl)ethane-diamide mol-ecules. The supra-molecular chains are consolidated in the crystal packing by C-H⋯O contacts. An inter-molecular C-H⋯S inter-action also occurs.

摘要

标题共晶体C(14)H(14)N(4)O(2)·C(14)H(10)O(4)S(2)的不对称单元包含一个扭曲的2,2'-(二硫代二亚基)二苯甲酸分子[苯环之间的二面角 = 76.35 (10)°]和一个U形的N,N'-双(3-吡啶甲基)乙二酰胺分子,吡啶基团位于中央二酰胺部分的同一侧[C-C-C-N = 113.8 (2) 和 -117.6 (2)°]。后者通过中心对称的八元酰胺{⋯OCNH}(2) 合成子聚集形成沿a轴传播的超分子带。观察到分子内N-H⋯O氢键。2,2'-(二硫代二亚基)二苯甲酸分子形成羧基-吡啶O-H⋯N氢键,桥接带平面下方的一个吡啶残基和平面上方的一个吡啶残基,中间有两个N,N'-双(3-吡啶甲基)乙二酰胺分子。超分子链通过C-H⋯O接触在晶体堆积中巩固。还发生了分子间C-H⋯S相互作用。

https://cdn.ncbi.nlm.nih.gov/pmc/blobs/8073/2983345/87a3c73afa3f/e-66-o2590-fig1.jpg

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