首页> 外文期刊>Acta Crystallographica, Section B. Structural Science >Crystal engineering using bisphenols and trisphenols. Complexes with hexamethylenetetramine (HMTA): Strings, multiple helices and chains-of-rings in the crystal structures of the adducts of HMTA with 4,4'-thiodiphenol(1/1), 4,4'-sulfonyldiphenol(1/1)
【24h】

Crystal engineering using bisphenols and trisphenols. Complexes with hexamethylenetetramine (HMTA): Strings, multiple helices and chains-of-rings in the crystal structures of the adducts of HMTA with 4,4'-thiodiphenol(1/1), 4,4'-sulfonyldiphenol(1/1)

机译:使用双酚和三酚的晶体工程。与六亚甲基四胺(HMTA)的配合物:HMTA与4,4'-硫代二酚(1/1),4,4'-磺酰基二酚(1/1)加合物的晶体结构中的线,多重螺旋和环链

获取原文
获取原文并翻译 | 示例
           

摘要

The 4,4/-bisphenols (1), X(C6H4OH)(2) [a, X = nil; b, X = O; c, X = S; d, X = SO2; e, X = CO; f, X = CH2; g, X = CMe2; h, X = C(CF3)(2)], when co-crystallized from alcoholic solutions with hexamethylenetetramine, (CH2)(6)N-4 (HMTA), form 1:1 adducts (4a)-(h). 4,4'-Thiodiphenol-hexamethylenetetramine (1/1) (4c), C12H10O2S.C6H12N4, and 4,4'-sulfonyldiphenol-hexamethylenetetramine (1/1), (4d), C12H10O4S.C6H12N4, are orthorhombic, Pmn2(1), (4c) a = 15.029 (2), b = 9.7954 (8), c = 5.9817 (11)Angstrom and (4d) a = 14.779 (2), b = 10.2558 (15), c = 5.9817 (8) Angstrom, with Z = 2, and the structures consist of zigzag chains comprising strings of alternating bisphenol and HMTA units, each lying across mirror planes and linked by O-H ... N hydrogen bonds. In addition, both (4c) and (4d) exhibit C-H ...pi(arene) hydrogen bonds with one CH2 group of the HMTA unit acting as a donor to two different arene rings; (4d) also exhibits multiple C-H ... O=S hydrogen bonds with three C-H bonds in each HMTA unit acting as donors towards a single sulfone O atom. 4,4'-Isopropylidenediphenol-hexamethylenetetramine (1/1), (4g), C15H16O2.C6H12N4, is monoclinic, C2/c, a = 25.093(6), b = 7.1742(13), c = 23.612(7)Angstrom, beta = 110.42(2)degrees, with Z = 8, and again the structure is built from chains of alternating bisphenol and HMTA units linked by O-H ... N hydrogen bonds, but these now form double helices around twofold rotation axes; the double helices are themselves Linked into sheets by C-H ... O hydrogen bonds. The trisphenol (2), CH3C(C6H4OH)(3), forms three adducts (5a)-(c) with HMTA, having trisphenol:HMTA ratios of 1:2 (5a), 2:3 (5b) and 1:1 (5c). 1,1,1-Tris(4-hydroxyphenyl)ethane-hexamethylenetetramine (1/2) (5a), C20H18O3 (C6H12N4)2, is orthorhombic, P2(1)2(1)2(1), a = 6.9928(10), b = 14.0949(15), c = 30.999(4)Angstrom, with Z = 4, and the trisphenol units and half the HMTA units form a triple helix around a 2(1) axis, in which each strand consists of alternating phenol and HMTA units, linked as usual by O-H ... N hydrogen bonds. The remaining HMTA units, which are external to the triple helix, are connected to it by O-H ... N hydrogen bonds and are formed into externally buttressing stacks. The triol (3), 1,3,5-C6H3(OH)(3), forms a 2:3 adduct (6) with HMTA. 1,3,5-Trihydroxybenzene-hexamethylenetetramine (2/3), (6), C6H6O3.(C6H12N4)(1.5), is monoclinic, C2/c, a = 23.598 (2), b = 7.136 (2), c = 19.445 (3) Angstrom, beta = 96.822(11)degrees, with Z = 8, and the dominant structural motif consists of centrosymmetric rings containing two molecules each of (3) and HMTA, connected by O-H ... N hydrogen bonds; these rings are themselves linked into a chain-of-rings by further HMTA units lying on twofold rotation axes. The hydrogen-bonding patterns are codified using the graph-set approach.
机译:4,4 /双酚(1),X(C6H4OH)(2)[a,X = n; b,X = O; c,X = S; d,X = SO 2; e,X = CO; f,X = CH 2; g,X = CMe 2; h,X = C(CF3)(2)],当从乙醇溶液与六亚甲基四胺(CH2)(6)N-4(HMTA)共结晶时,会形成1:1加合物(4a)-(h)。 4,4'-噻二酚-六亚甲基四胺(1/1)(4c),C12H10O2S.C6H12N4和4,4'-磺酰基二酚-六亚甲基四胺(1/1),(4d),C12H10O4S.C6H12N4是正交的,Pmn2(1 ),(4c)a = 15.029(2),b = 9.7954(8),c = 5.9817(11)埃和(4d)a = 14.779(2),b = 10.2558(15),c = 5.9817(8) Z = 2的埃,并且该结构由之字形链组成,该之字形链包括交替的双酚和HMTA单元的串,每个单元横穿镜面并通过OH ... N氢键连接。另外,(4c)和(4d)都显示出C-H ... pi(芳烃)氢键,其中HMTA单元的一个CH 2基团充当两个不同的芳烃环的供体; (4d)还表现出多个C-H ... O = S氢键,每个HMTA单元中的三个C-H键充当单个砜O原子的供体。 4,4'-异丙基二烯二酚-六亚甲基四胺(1/1),(4g),C15H16O2.C6H12N4,是单斜晶的,C2 / c,a = 25.093(6),b = 7.1742(13),c = 23.612(7)埃β= 110.42(2)度,Z = 8,并且该结构再次由通过OH ... N氢键连接的双酚和HMTA交替单元链构成,但现在它们围绕着两个旋转轴形成了双螺旋;双螺旋本身通过C-H ... O氢键连接成片。三酚(2)CH3C(C6H4OH)(3)与HMTA形成三个加合物(5a)-(c),三酚:HMTA比为1:2(5a),2:3(5b)和1:1 (5c)。 1,1,1-三(4-羟基苯基)乙烷-六亚甲基四胺(1/2)(5a),C20H18O3(C6H12N4)2,是正交晶,P2(1)2(1)2(1),a = 6.9928( 10),b = 14.0949(15),c = 30.999(4)埃,Z = 4,三酚单元和一半HMTA单元围绕2(1)轴形成三重螺旋,其中每条链由苯酚和HMTA单元交替排列,通常通过OH ... N氢键连接。在三重螺旋外部的其余HMTA单元通过O-H ... N氢键与之连接,并形成外部支撑堆叠。三醇(3)1,3,5-C6H3(OH)(3)与HMTA形成2:3加合物(6)。 1,3,5-三羟基苯-六​​亚甲基四胺(2/3),(6),C6H6O3。(C6H12N4)(1.5)是单斜晶的,C2 / c,a = 23.598(2),b = 7.136(2),c = 19.445(3)埃,β= 96.822(11)度,Z = 8,并且主要结构基序由中心对称环组成,该中心对称环包含通过(OH)... N氢键连接的两个分子(3)和HMTA;这些环本身通过位于双向旋转轴上的其他HMTA单元链接成环链。氢键模式使用图集方法进行编码。

著录项

相似文献

  • 外文文献
  • 中文文献
  • 专利
获取原文

客服邮箱:kefu@zhangqiaokeyan.com

京公网安备:11010802029741号 ICP备案号:京ICP备15016152号-6 六维联合信息科技 (北京) 有限公司©版权所有
  • 客服微信

  • 服务号