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线性分子周围π-空穴带的碱基最大占据率。

Maximal occupation by bases of π-hole bands surrounding linear molecules.

作者信息

Scheiner Steve

机构信息

Department of Chemistry and Biochemistry, Utah State University, Logan, Utah, USA.

出版信息

J Comput Chem. 2022 Feb 15;43(5):319-330. doi: 10.1002/jcc.26792. Epub 2021 Dec 3.

Abstract

Linear molecules such as CO contain a positive π-hole ring that surrounds C on the molecule's equator. Quantum calculations examine the question as to how many bases can simultaneously bind to this ring. Linear molecules examined are TO , where T = C, Si, Ge, Sn; bases are NCH and NH . CO engages in the weakest of the tetrel bonds, and can bind up to three NCH and two NH . Unlike σ-hole tetrel bonds, Si forms the strongest tetrel bonds, with interaction energies as high as 43 kcal/mol with NH . But like GeO , SiO can sustain only two bases in its equatorial ring. The π-hole ring of SnO can engage in up to four tetrel bonds with either NCH or NH , even though these bonds are weaker than those with GeO or SiO . As all of these complexes cast TO in the role of multiple electron acceptor, the resulting negative cooperativity makes each successive bond weaker than its predecessor as bases are added, as well as reducing the magnitude of the central molecule's π-hole.

摘要

诸如一氧化碳(CO)这样的线性分子在分子赤道面上含有一个围绕碳(C)的正π-空穴环。量子计算研究了有多少碱基能够同时与这个环结合的问题。所研究的线性分子是TO,其中T = C、Si、Ge、Sn;碱基是NCH和NH 。一氧化碳形成的是最弱的四元键,最多能结合三个NCH和两个NH 。与σ-空穴四元键不同,硅(Si)形成最强的四元键,与NH 的相互作用能高达43千卡/摩尔。但与GeO 一样,SiO 在其赤道环中只能容纳两个碱基。SnO的π-空穴环最多能与NCH或NH 形成四个四元键,尽管这些键比与GeO或SiO 形成的键弱。由于所有这些配合物都使TO充当多电子受体的角色,由此产生的负协同效应使得随着碱基的添加,每个相继的键都比前一个键弱,同时也减小了中心分子π-空穴的大小。

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