【作者】 孔治国；

【导师】 任爱民； 封继康；

【作者基本信息】 吉林大学 ， 物理化学， 2004， 硕士

【摘要】 近年来，与燃烧化学有关的自由基-分子反应研究日益受到关注，特别是自由电子在双键上的不饱和自由基与氧气的反应。该自由基最大的特点是能够与氧气形成共轭体系，使其初始的中间体具有相当的稳定性，利于反应进行。在双键自由基中，C3H5以其特别的稳定性和相当快的反应速率而被重视，但由于其体系较大，一直难以对其势能面作详尽理论计算。本论文即是利用量子化学计算方法对C3H5与O2的反应势能面进行了理论研究，并与C2H3等共轭体系进行了对比，期望得出C3H5这个反应的机理。主要内容概括如下：1、我们首先研究了反应入口问题，认为C*H=CH-CH3对O2的进攻方式就一种，对端基O的进攻，经计算，我们可以确定该进攻为无势垒进攻，O上的单电子直接与自由基上的自由电子形成共价键，生成初始化中间体m1的能量比反应物低68.93kCal/mol。因此在下面的讨论中，我们将主要讨论从m1开始的路径。2、从m1出发，我们得出了三个热力学允许的反应产物（即产物能量比反应物低），并找到了其与反应物连接的通道，其反应路径如下：Path1(I): R（CH3CH=C*H+O2）(m1（trans-CH3CH=CHOO）(m6（cis-CH3CH=CHOO）(P1（六元环中，C-H键断裂，同时C-O键断裂生成CH3CCH+O2H）。Path1(II): R（CH3CH=C*H+O2）(m1（trans-CH3CH=CHOO）(m6（cis-CH3CH=CHOO）(m9（C3H5中间C上的H发生1-3氢迁移转移<WP=54>到O上）(P1（C-O键断裂生成CH3CCH+O2H）。Path2(I): R（CH3CH=C*H+O2）(m1（trans-CH3CH=CHOO）(m2（形成COO三元环）(m3(C-C键断裂,同时形成C-O键,形成CH3CH-O-CHO) (P2（C-O键断裂生成CH3CHO+CHO）。Path2(II): R（CH3CH=C*H+O2）(m1(trans-CH3CH=CHOO）(m6（cis-CH3CH=CHOO）(m7（经双烯加成形成CCOO四元环）(m8(四元环中O-O键断裂形成CH3CH(O)CHO)(P2(C-C键断裂生成CH3CHO+CHO)。Path3(1)：R（CH3CH=C*H+O2）(m1（trans-CH3CH=CHOO）(m2（形成COO三元环）(m10(三元环中C上连接的H迁移到中间C上) (m11（三元环O-O键断裂）(P3(C-C键断裂生成CH3CH2+CO2)。Path3(1I)：R（CH3CH=C*H+O2）(m1（trans-CH3CH=CHOO）(m2（形成COO三元环）(m4（C-C bond,同时形成C-O键,生成CH3CH-O-CHO）(m5（-CHO上的H迁移到中间C上，形成CH3CH2OCO）( P3(O-C键断裂生成CH3CH2+CO2)。 以上路径中， Path3(1)中有一个能量较高的过渡态TSm2m10,它仅反应物低1.95Cal/mol，因此被认为在298.15K下，反应不容易进行，而在高温下存在一定竞争力。其它路径中的过渡态和中间体的相对能量均较低，都是可行的。特别是Path2(I) 全路径能量都较低，而且过程比较简单，是最可行的反应路径。 3.与C2H3+O2反应的对比作为同系物反应，我们将C3H5+O2的反应与C2H3+O2反应进行对比，发现两个反应的产物通道基本上是一致的，只是在C2H3体系中，C2H3体系的控速步骤是m1(TSm1m2，C3H5体系是m2(TSm2m3。还有就是C3H5+O2体系中所有中间体和过渡态的相对能量都比C2H3+O2低，这<WP=55>是由于共轭体系的扩大使体系的稳定性增加的缘故。我们分析共轭体系的增大引起稳定性增加的原因，认为：大的共轭体系可以提高环状的中间体和环状过渡态的稳定性，同时有利于加成过程中的π电子转移，从从而使C3H5体系的中间体和过渡态的相对能量比C2H3体系的普遍低许多，因而会使稳定性增加。并因此我们可以推断C3H5体系的反应速度要比C2H3体系的快很多。更多还原

【Abstract】 Recently, the studies on the radical reactions related to combustion chemistry continue to attract the extensive attention. Particularly, the reactions of the oxygen and the unsaturated radical which the free electron is on the double-bond are focused. There are so many characters of these radicals, and the most important one is that they can form some very steady conjugate systems with oxygen, which can make the initialized isomers more steady and make the reaction easier to continue. In these radicals, C3H5 is attended as their especial stability and the rapid rate in the reactions. Because the system is so large, there are few persons who had calculated its detailed potential energy surface. In this article, the theoretical investigations on the mechanisms for C3H5+O2 reaction is carried out based on the quantum chemical calculating methods. For gaining the whole mechanisms of C3H5 +O2, we compare this system with some others such as the system of C2H3. The following is the main results:At first, we study the entrance of the reaction, and only one attacking modes are found:allyl attacks the O atom in the O2. this attack is regarded as a process without energy barriers by our calculations and the article about the ethylene. Single electron at the O atom and free electron at <WP=57>the allyl are likely formed covalent bond, and produced a initialized isomer m1 whose energy is 68.93kCal/mol lower that the reactants. From abover, we can deduce that the most suitable entrance step is that allyl attacks O atom in the O2 and forms isomer m1. So, in following chapters, we will mainly discuss the path from beginning of isomer m1. 2. Starting from m1, we can gain three products which can produce in thermodynamics(their energies is lower than the reactants), and work out their produce channels which connect with reactants. The followings are these four path:Path1(I)： R（CH3CH=C*H+NO）(m1（trans-CH3CH=CHNO）(m6（cis- CH3CH=CHNO）(P1（C-H and C-O bonds in the six-member ring rupture and form CH3CCH +O2H）Path1(II)： R（CH3CH=C*H+NO）(m1（trans-CH3CH=CHNO）(m6（cis- CH3CH=CHNO）(m9(H at the 2-C in allyl thansfers N site by 1,3-Oshift)(P1（C-O bond ruptures and form CH3CCH +O2H）.Path2(I): R（CH3CH=C*H+O2）(m1（trans-CH3CH=CHOO）(m2（form COO three-member ring）(m3（C-C bond ruptures,at the same time forms C-O bond,then forms CH3CH-O-CHO) (P2（C-O bond ruptures and form CH3CHO+CHO）。Path2(II): R（CH3CH=C*H+O2）(m1(trans-CH3CH=CHOO）(m6（cis- CH3CH=CHOO）(m7（forms CCOO four-member ring by a diene addition reaction process）(m8(O-O bond in the four-member ring ruptures and forms CH3CH(O)CHO)(P2(C-C bond ruptures and form CH3CHO+CHO)。Path3(1)：R（CH3CH=C*H+O2）(m1（trans-CH3CH=CHOO）(m2（form COO three-member ring）(m10（H connects C at COO three –member ring site transfers C site ) (m11（O-O bond in the three-member ring ruptures）(P3(C-C bond ruptures and form CH3CH2+CO2)。 <WP=58>Path3(1I)：R（CH3CH=C*H+O2）(m1（trans-CH3CH=CHOO）(m2（form COO three-member ring）(m4（C-C bond ruptures and forms C-O bond at he same time ,then forms CH3CH-O-CHO）(m5（H connets C at CHO site transfer C site and form CH3CH2OCO）( P3(O-C bond ruptures and form CH3CH2+CO2)。Above the paths, There is only one transation TSm2m10 with high energy in Path3(1),so we can think that it might have some competitive produces at higher temperature, but the produces must be few. The energies of the other pathes are relatively low, so there are the feasible pathes in this system. Especiallly, The energies of the whole Path2(I) are relatively low and the course is simple, so it is the most feasible path in this system. 3.compared with C2H3+O2The path is similar to the main produce channel of C2H3. The only defference between these two system is that the step of m1( TSm1m2 is rate determinate step in system of C2H3, but it is m2(TSm2m3 in system of C3H5 as well as energies of transation and intermediates in system of C3H5 are lower than these in system of C2H3. We analy更多还原