首页> 外文OA文献 >2,2′-[4,10-Bis(carb­oxy­meth­yl)-4,10-diaza-1,7-diazo­niacyclo­dodecane-1,7-di­yl]diacetate dihydrate
【2h】

2,2′-[4,10-Bis(carb­oxy­meth­yl)-4,10-diaza-1,7-diazo­niacyclo­dodecane-1,7-di­yl]diacetate dihydrate

机译:二乙酸2,2'-[4,10-双(咔­氧基­甲基)-4,10-二氮杂-1,7-二重氮环环十二烷-1,7-二­基]二乙酸酯

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摘要

In the title compound, C16H28N4O8·2H2O, the 12-membered macrocycle has twofold crystallographic symmetry and the asymmetric unit comprises one half-mol­ecule. The four carbox­yl/carboxyl­ate groups reside on the same side of the macrocycle. The mol­ecule is a double zwitterion with two of the carb­oxy­lic acid H atoms transferred to the two N atoms on the opposite sides of the macrocycle, resulting in both N atoms having positive charges and leaving the two resulting carboxyl­ate groups with negative charges. The two remaining carb­oxy­lic acid groups and the carboxyl­ate groups form O—H⋯O hydrogen bonds with the crystal water mol­ecules. The H atoms bound to the N atoms within the macrocyle are engaged in two equivalent hydrogen bonds with the adjacent N atoms.
机译:在标题化合物C16H28N4O8·2H2O中,十二元大环具有双重晶体对称性,不对称单元包含一个半分子。四个羧基/羧酸根基团位于大环的同一侧。该分子是双两性离子,其中两个羰基羟基酸H原子转移至大环相反侧的两个N原子,导致两个N原子均带正电荷,而使两个羧酸根基团带负电荷。剩下的两个氧羰基羧酸基团和羧酸根基团与结晶水分子形成O-H = O氢键。大环内与N原子键合的H原子与相邻的N原子以两个等效的氢键结合。

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