下述构型的正确命名是:(图见问题补充)A. (1)R (2)R (3)S B.(1)R (2)S (3)S C.(1)S (2)S (3)S D.(1)R (2)R (3)R
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![下述构型的正确命名是:(图见问题补充)A. (1)R (2)R (3)S B.(1)R (2)S (3)S C.(1)S (2)S (3)S D.(1)R (2)R (3)R](/uploads/image/z/9972221-5-1.jpg?t=%E4%B8%8B%E8%BF%B0%E6%9E%84%E5%9E%8B%E7%9A%84%E6%AD%A3%E7%A1%AE%E5%91%BD%E5%90%8D%E6%98%AF%3A%EF%BC%88%E5%9B%BE%E8%A7%81%E9%97%AE%E9%A2%98%E8%A1%A5%E5%85%85%EF%BC%89A.++%281%29R+%282%29R+%283%29S+++++++++++++++++++++++++++++++++++++++++++++++B.%281%29R+%282%29S+%283%29S++++++++++++++++++C.%281%29S+%282%29S+%283%29S+++++++++++++++++++++++++++++++++++++++++++++++D.%281%29R+%282%29R+%283%29R)
下述构型的正确命名是:(图见问题补充)A. (1)R (2)R (3)S B.(1)R (2)S (3)S C.(1)S (2)S (3)S D.(1)R (2)R (3)R
下述构型的正确命名是:(图见问题补充)
A.
(1)R (2)R
(3)S
B.(1)R (2)S
(3)S
C.(1)S (2)S
(3)S
D.(1)R (2)R
(3)R
下述构型的正确命名是:(图见问题补充)A. (1)R (2)R (3)S B.(1)R (2)S (3)S C.(1)S (2)S (3)S D.(1)R (2)R (3)R
答案选C,三个手性碳的构型都是S构型.
由费歇尔投影式判断手性碳R、S构型的方法:按照基团排列的先后顺序选择一个顺序在最后的基团(以1为例,就是-H),将它排除再外.另外三个基团也按照基团排列的先后顺序,给出一个编号,如图:
再根据这个编号,确定从小到大的旋转方向(如图为顺时针,不包括-H);同时由于未编号基团-H处于水平位置上,所以所得构型翻转(如果在竖直方向上则不翻转),所以上图翻转后为逆时针旋转,手性碳为S构型.
第二个结构式中-H不编号,编号基团中-Br:①,-CH2OH:②,-CH3:③.顺时针,-H处于水平位置,需要翻转.所以为逆时针,S构型.
第三个结构式中-CH2NH2不编号,编号基团中-Br:①,-Cl:②,-CH2NHCH3:③.逆时针,-CH2NH2处于竖直位置,不需要翻转,所以是S构型.
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