摘要
通过比较C—O单键和CO双键的键能,预测醇的共轭碱很容易通过无键共振形成CO双键。由于C—C键能远小于C—H键能,所以甲基比氢原子更容易通过形成无键共振来稳定烷氧基负离子,由此导致气相中的酸性顺序为:(CH3)3COH>(CH3)2CHOH>CH3CH2OH>CH3OH。这个观点得到了量子化学计算的支持。
By comparing bond energies of C—O single bond and CO double bond,it can be predicted that the conjugate bases of alcohols are able to form CO double bond via no-bond resonance. Since the bond energy of C—C single bond is much lower than that of the C—H single bond,methyl group is easier to stabilize alkoxide ions via no-bond resonance compared with H atom,the acidity sequence of alcohols in vapor phase is:( CH3)3COH ( CH3)2CHOH CH3CH2 OH CH3 OH. This viewpoint has been supported by the result of quantum chemistry calculation.
出处
《大学化学》
CAS
2015年第1期68-71,共4页
University Chemistry
基金
福建省自然科学基金(No.2013J01361)
关键词
键能
无键共振
醇
酸性顺序
Bond energy
No-bond resonance
Alcohol
Acidity sequence