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Computational study of selectivity in the [PtIICl4]2−-catalysed arylation of arenes by diaryliodonium reagents: arene activation at PtIV centres
Dalton Transactions ( IF 3.5 ) Pub Date : 2017-10-23 00:00:00 , DOI: 10.1039/c7dt03602g Allan J. Canty 1, 2, 3, 4 , Alireza Ariafard 1, 2, 3, 4, 5
Dalton Transactions ( IF 3.5 ) Pub Date : 2017-10-23 00:00:00 , DOI: 10.1039/c7dt03602g Allan J. Canty 1, 2, 3, 4 , Alireza Ariafard 1, 2, 3, 4, 5
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The mechanism for the [PtIICl4]2−-catalysed reaction of Ph2IIII(TFA) (TFA = trifluoroacetate) with naphthalene (NapH) to give β-phenylation over α-phenylation of naphthalene has been examined by computational methods. In support of the mechanism proposed by Sanford based on experimental evidence, reaction commences by oxidative phenyl transfer from [Ph2I]+ to [PtIICl4]2−, giving trans-[PhPtIVCl4(TFA)]2−. Transformation to cis-[PhPtIVCl3(TFA)]− leads to reaction with NapH at the β-position in an inner-sphere Concerted Metalation Deprotonation (CMD) manner to give cis-[Ph(Nap)PtIVCl3]− and trifluoroacetic acid. Reductive elimination yields β-PhNap, and coordination of chloride regenerates [PtIICl4]2− for subsequent catalytic cycles. The selectivity for β-phenylation over α-phenylation is attributable to steric factors in the CMD PtIV transition state containing a higher coordination number than that occurring for related reactions in PdII catalysis that gives α-phenylation.
中文翻译：
二芳基碘鎓试剂在[Pt II Cl 4 ] 2-催化芳烃芳基化中选择性的计算研究：Pt IV中心的芳烃活化
通过计算方法研究了[Pt II Cl 4 ] 2-催化Ph 2 I III（TFA）（TFA =三氟乙酸盐）与萘（NapH）生成β-苯基化反应而不是萘的α-苯基化反应的机理。。为了支持桑福德基于实验证据提出的机理，反应通过从[Ph 2 I] +到[Pt II Cl 4 ] 2-的氧化苯基转移而开始，得到反式-[PhPt IV Cl 4（TFA）] 2-。转化为顺式-[PhPt IVCl 3（TFA）] -以内球协同金属化去质子化（CMD）的方式与NapH处的NapH反应，生成顺式-[Ph（Nap）Pt IV Cl 3 ] -和三氟乙酸。还原消除产生β-PhNap，氯化物的配位体在随后的催化循环中再生[Pt II Cl 4 ] 2-。β-苯基化相对于α-苯基化的选择性归因于CMD Pt IV过渡态中的空间因素，该立体因素具有比在产生α-苯基化的Pd II催化中相关反应更高的配位数。
更新日期：2017-11-15
中文翻译：
二芳基碘鎓试剂在[Pt II Cl 4 ] 2-催化芳烃芳基化中选择性的计算研究：Pt IV中心的芳烃活化
通过计算方法研究了[Pt II Cl 4 ] 2-催化Ph 2 I III（TFA）（TFA =三氟乙酸盐）与萘（NapH）生成β-苯基化反应而不是萘的α-苯基化反应的机理。。为了支持桑福德基于实验证据提出的机理，反应通过从[Ph 2 I] +到[Pt II Cl 4 ] 2-的氧化苯基转移而开始，得到反式-[PhPt IV Cl 4（TFA）] 2-。转化为顺式-[PhPt IVCl 3（TFA）] -以内球协同金属化去质子化（CMD）的方式与NapH处的NapH反应，生成顺式-[Ph（Nap）Pt IV Cl 3 ] -和三氟乙酸。还原消除产生β-PhNap，氯化物的配位体在随后的催化循环中再生[Pt II Cl 4 ] 2-。β-苯基化相对于α-苯基化的选择性归因于CMD Pt IV过渡态中的空间因素，该立体因素具有比在产生α-苯基化的Pd II催化中相关反应更高的配位数。
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