无序示例-氰基和乙酰基共占据无序(无序理解)

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无序示例-氰基和乙酰基共占据无序(无序理解)

案例来源：CCDC[1] 2414599. DOI[2]:10.5517/ccdc.csd.cc2m1l9l.

案例结构如图1所示（由ChemBioDraw[3]绘制）。

▲图1 案例结构

其晶体结构不对称ASU（asymmetric unit）如图2所示（由Olex2[4]呈现），包含一个完整的对乙酰基苯腈分子以及半个对乙酰基苯腈分子，后者中氧、甲基和氮做了无序处理，氧和甲基属于PART 1，氮属于PART 2，占有率均为0.5。

▲图2 案例结构ASU

进行对操作后，表观上看PART 1和PART 2结构如图3‒4所示，好像分别是对乙酰基苯腈和1,4-二乙酰苯混合物与对乙酰基苯腈和对苯二腈混合物。

▲图3 无序第一组分（PART 1）

▲图4 无序第二组分（PART 2）

实际上这是纯化合物对乙酰基苯腈的晶体结构，并不是混合物的晶体结构，按照图3‒4所示来理解是错误的，ASU中的半个对乙酰基苯腈穿过对称元素（反演中心），如图5所示（参阅视频“Mercury[5]基操-查看对称元素：https://www.bilibili.com/video/BV1U44y1U7ZF”）。

▲图5 对称元素查看

因为穿过对称元素，所以ASU中为半个对乙酰基苯腈分子，这就使得氰基和乙酰基成为共占据无序，而苯环本位碳重合，氰基碳和羰基碳重合，因此只需要对羰基氧、甲基和氰基氮做无序处理即可，占有率则由对称元素确定为0.5，所以上述无序应正确理解为如图6所示结构。

▲图6 案例结构无序正确理解

相关视频：

单晶结构解析练习396(数据还原-无序处理)：https://www.bilibili.com/video/BV197cjebESV

数据下载：

链接:https://pan.baidu.com/s/1nDqdPxoq5QCqiNsj_WS1AQ

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参考文献

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