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N'-[(1E)-1-(2-氟苯基)亚乙基]吡啶-3-碳酰肼

N'-[(1E)-1-(2-Fluoro-phen-yl)ethyl-idene]pyridine-3-carbohydrazide.

作者信息

Sreeja P B, Sithambaresan M, Aiswarya N, Kurup M R Prathapachandra

机构信息

Department of Chemistry, Christ University, Hosur Road, Bangalore 560 029, Karnataka, India.

Department of Chemistry, Faculty of Science, Eastern University, Sri Lanka, Chenkalady, Sri Lanka.

出版信息

Acta Crystallogr Sect E Struct Rep Online. 2014 Jan 8;70(Pt 2):o115. doi: 10.1107/S1600536813035009. eCollection 2014 Feb 1.

Abstract

The title compound, C14H12FN3O, adopts an E conformation with respect to the azomethine double bond whereas the N and methyl C atoms are in a Z conformation with respect to the same bond. The ketonic O and azomethine N atoms are cis to each other. The non-planar mol-ecule [the dihedral angle between the benzene rings is 7.44 (11)°] exists in an amido form with a C=O bond length of 1.221 (2) Å. In the crystal, a bifurcated N-H⋯(O,N) hydrogen bond is formed between the amide H atom and the keto O and imine N atoms of an adjacent mol-ecule, leading to the formation of chains propagating along the b-axis direction. Through a 180° rotation of the fluoro-phenyl ring, the F atom is disordered over two sites with an occupancy ratio of 0.632 (4):0.368 (4).

摘要

标题化合物C₁₄H₁₂FN₃O中,甲亚胺双键的构型为E型,而氮原子和甲基碳原子相对于同一键呈Z构型。酮羰基氧原子和甲亚胺氮原子彼此顺式排列。非平面分子(苯环之间的二面角为7.44 (11)°)以酰胺形式存在,C=O键长为1.221 (2) Å。在晶体中,酰胺氢原子与相邻分子的酮羰基氧原子和亚胺氮原子之间形成了一个分叉的N-H⋯(O,N)氢键,导致沿b轴方向形成链状结构。通过氟苯环的180°旋转,氟原子在两个位置上无序分布,占有率为0.632 (4):0.368 (4)。

https://cdn.ncbi.nlm.nih.gov/pmc/blobs/3d27/3998283/2a35f336da14/e-70-0o115-fig1.jpg

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