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选择氧化催化剂通常为多组分复合氧化物.一般认为,高价过渡金属的端末双键氧(M=O)是烷烃活化的中心,而非金属端氧(NM=O)与烷烃活化无关.但近期的理论研究发现,复合氧化物中非金属端氧也可能参与烷烃活化.本文采用密度泛函方法(B3LYP)对比V=O和P=O的脱氢活性,并深入揭示二者的差异.H脱除反应可以视为是质子偶联电子传递的过程.对于V/P复合氧化物,V5+充当电子的受体,而V=O和P=O均可接受质子.由于P=O具有更强的质子化能力,导致PO–H键能比VO–H有利6–10 kcal/mol.对于烷烃活化,V=O和P=O脱氢的能垒均可与反应焓变很好地关联,但二者线性回归的截距相差6.2 kcal/mol,说明在相同的焓驱动下,P=O脱氢需要克服更高的能垒.根据Marcus模型,反应的能垒不仅取决去反应焓变,还与内部重组能有关.计算表明,在脱氢过程中,P=O需克服的重组能为128–140 kcal/mol,比V=O过程高出21–23 kcal/mol.这很好地解释了前面的计算结果.应该指出的是,除了反应热力学驱动和重组能外,在势能曲线相交处的电子耦合作用(?HAB?)亦对能量有一定的影响.丁烷选择氧化制顺酐可能经过2-丁烯,丁二烯,2,5-二氢呋喃和丁烯酸内酯等一系列中间体,共有8个H原子在反应过程中需要脱除.对于丁烷的脱氢,P=O的能垒仅比V=O低1.3 kcal/mol,说明初始反应时二者是竞争的.但对于2-丁烯和2,5-二氢呋喃,二者活化能的差距增加为6–7 kcal/mol,说明这时P=O脱氢将占主导.而对丁烯酸内酯活化,二者活化能的差异又缩小到2.5 kcal/mol,表明V=O又具有一定的竞争力.事实上,这种能垒的差异与端氧的亲核性密切相关.P=O更具亲核性,因此有利于被更具酸性的C–H键进攻.根据Evens的估计,烷烃C–H键的p Ka为50左右,而烯丙基性C–H为43.这就很好地解释了为什么2-丁烯和2,5-二氢呋喃更容易和P=O发生反应,而丁烷脱氢二者差异不大的原因.这些理论研究可以加深我们对复合氧化物催化剂上活性位点的认识,并为催化剂的理性设计提供理论支撑.
The selective oxidation catalyst is usually a multi-component composite oxide.It is generally believed that the terminal transition double bond oxygen (M = O) of a high-valent transition metal is the center of alkane activation whereas the non-metal terminal oxygen (NM = O) Recent theoretical studies have found that non-metallic end-oxygen in the composite oxide may also participate in the activation of alkanes.In this paper, the density functional theory (B3LYP) is used to compare the dehydrogenation activity of V = O and P = O, and to reveal the difference between the two. H removal reaction can be considered as the process of proton-coupled electron transfer. For V / P complex oxide, V5 + acts as an acceptor for electrons, whereas both V = O and P = O accept protons. Since P = O has more The strong protonation capacity leads to a PO-H bond that is 6-10 kcal / mol more favorable than VO-H. The energy barriers for V = O and P = O dehydrogenation for alkane activation are well correlated with the enthalpy of reaction , But the intercept of the two linear regression is 6.2 kcal / mol, which shows that under the same enthalpy driving, P = O dehydrogenation needs to overcome higher energy barrier.According to the Marcus model, the energy barrier of the reaction depends not only on the reaction enthalpy , Which is also related to internal recombination.The calculation shows that the recombination energy to be overcome by P = O during dehydrogenation is 128-140 kcal / mol, which is 21-23 kcal / m higher than that of V = O This well explains the previous results. It should be noted that in addition to the thermodynamics of driving and recombination energy, the electron coupling at the intersection of potential curves (? HAB?) also has some effect on energy. Selective oxidation of alkane to maleic anhydride may take place through a series of intermediates such as 2-butene, butadiene, 2,5-dihydrofuran and butenolide, and a total of 8 H atoms need to be removed during the reaction. Butane dehydrogenation, the energy barrier of P = O is only 1.3 kcal / mol lower than that of V = O, indicating that the initial reaction is competitive, but for 2-butene and 2,5-dihydrofuran, both The difference of activation energy increased to 6-7 kcal / mol, indicating that P = O dehydrogenation will dominate at this time, while the activation of butenolide, the difference between the two activation energy reduced to 2.5 kcal / mol, indicating that V In fact, the difference of this energy barrier is closely related to the nucleophilicity of terminal oxygen, and P = O is more nucleophilic and thus favored by more acidic C-H bonds . According to Evens’s estimation, the p Ka of an alkane C-H bond is around 50, whereas the allylic C-H is 43. This explains very well why 2-butene and 2,5-dihydrofuran are more Easily react with P = O while butane dehydrogenates The difference between the two is not the reason.These theoretical studies can deepen our understanding of the active sites on the composite oxide catalyst and provide theoretical support for the rational design of the catalyst.