Novel Synthesis and Crystal Structures of Two α, α′-bis-Substituted Benzylidene Cyclohexanones: 2,6-Bis-2-nitro(benzylidene)cyclohexanone and 2,6-Bis-4-methyl(benzylidene)cyclohexanone
作者:Ali Amoozadeh、Salman Rahmani、Grzegorz Dutkiewicz、Mehdi Salehi、Firouzeh Nemati、Maciej Kubicki
DOI:10.1007/s10870-011-0094-7
日期:2011.9
Two α,α′-bis-substituted benzylidene cycloalkanones have been synthesized in presence of SnCl4 and their crystal structures have been determined by means of X-ray diffraction. The bis(para-methyl) derivative, 2,6-bis-4-methyl(benzylidene)cyclohexanone 1 crystallizes in the orthorhombic space group Pbca with a = 9.413(2) Å, b = 10.787(2) Å, and c = 33.702(5) Å, while bis(ortho-nitro) derivative, 2,6-bis-2-nitro(benzylidene)cyclohexanone in monoclinic P21/n space group with a = 8.482(2) Å, b = 13.435(2) Å, c = 15.377(3) Å, and β = 92.96(2)°. In both compounds the olefinic bonds are in E-configuration, and the cyclohexyl rings adopt a sofa conformation. The phenyl rings are not coplanar with the planes of C=C–C(=O)–C=C fragments; the dihedral angles between these planes are 14.25(11) and 19.37(11)° in 1 and 60.50(6) and 63.26(6)° in 2. This twist might be regarded as the effect of the repulsive interactions between the hydrogen atoms from phenyl and cyclohexyl rings, and much larger values in 2 are certainly connected with the presence of nitro group in ortho-positions of the phenyl ring. It seems that, because of the lack of specific interactions the close packing requirements and the van der Waals forces are main factors determining the crystal packing. The phenyl rings are not coplanar with the planes of C=C–C(=O)–C=C fragments as the effect of the repulsive interactions between the hydrogen atoms from phenyl and cyclohexyl rings .
在氯化锡存在下合成了两种α,α′-双取代的亚苄基环己酮,并通过 X 射线衍射测定了它们的晶体结构。双(对甲)衍生物 2,6-双-4-甲基(亚苄基)环己酮 1 在正交空间群 Pbca 中结晶,a = 9.413(2) 埃,b = 10.787(2) 埃,c = 33。702(5) 埃,而双(正硝基)衍生物 2,6-双-2-硝基(亚苄基)环己酮在单斜 P21/n 空间群中,a = 8.482(2) 埃,b = 13.435(2) 埃,c = 15.377(3) 埃,β = 92.96(2)°。在这两种化合物中,烯烃键均呈 E 型构象,环己基环则呈sofa 型构象。苯基环与 C=C-C(=O)-C=C 片段的平面不共面;这些平面之间的二面角在 1 中分别为 14.25(11)°和 19.37(11)°,在 2 中分别为 60.50(6)°和 63.26(6)°。看来,由于缺乏特定的相互作用,紧密堆积的要求和范德华力是决定晶体堆积的主要因素。苯基环与 C=C-C(=O)-C=C 片段的平面不共面,这是苯基环和环己基环的氢原子之间排斥作用的结果。