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(E)-2-甲基-3-(2-甲基-2-硝基乙烯基)-1H-吲哚和(E)-3-(2-甲基-2-硝基乙烯基)-2-苯基-1H-吲哚

(E)-2-Methyl-3-(2-methyl-2-nitrovinyl)-1H-indole and (E)-3-(2-methyl-2-nitrovinyl)-2-phenyl-1H-indole.

作者信息

Sonar Vijayakumar N, Parkin Sean, Crooks Peter A

机构信息

Department of Pharmaceutical Sciences, College of Pharmacy, University of Kentucky, Lexington KY 40536, USA.

出版信息

Acta Crystallogr C. 2005 Aug;61(Pt 8):o527-30. doi: 10.1107/S0108270105022298. Epub 2005 Jul 30.

DOI:10.1107/S0108270105022298
PMID:16082109
Abstract

In the title compounds, C12H12N2O2, (I), and C17H14N2O2, (II), respectively, the indole rings are planar and the vinyl groups lie out of the indole planes, making dihedral angles of 33.48 (5) and 41.31 (8) degrees , respectively. In (II), the dihedral angle between the phenyl and indole ring planes is 32.06 (6) degrees . In both molecules, the double bond connecting the methylnitrovinyl group and the indole nucleus adopts an E configuration. Notwithstanding the differences in space group [C2/c for (I) and P2(1)2(1)2(1) for (II)], the mode of packing of compounds (I) and (II) is determined by similar intermolecular N-H...O hydrogen-bonding interactions, forming chains that run parallel to [101] in (I) and [001] in (II).

摘要

在标题化合物C₁₂H₁₂N₂O₂(I)和C₁₇H₁₄N₂O₂(II)中,吲哚环是平面的,乙烯基位于吲哚平面之外,二面角分别为33.48(5)度和41.31(8)度。在(II)中,苯基和吲哚环平面之间的二面角为32.06(6)度。在两个分子中,连接甲基硝基乙烯基和吲哚核的双键采用E构型。尽管空间群不同[(I)为C2/c,(II)为P2(1)2(1)2(1)],但化合物(I)和(II)的堆积方式由类似的分子间N-H...O氢键相互作用决定,在(I)中形成平行于[101]的链,在(II)中形成平行于[001]的链。

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