《2019南京大学结构化学双语课件CH6LEC2.ppt》由会员分享，可在线阅读，更多相关《2019南京大学结构化学双语课件CH6LEC2.ppt（40页珍藏版）》请在三一文库上搜索。
1、6.2 More about Energy Levels and Molecular Orbitals,The BOUND and SUM of energy levels （能级的 “和” 与 “界”） Alternant Hydrocarbons （交替烃）,Bound of the Energy Levels,Theorem,The p energy levels of the conjugated hydrocarbons are bounded, satisfying |x|3,|x| 2,|x| 3,|x| 3,|x| 2,PROOF：,One atom connects with
2、 no more than two other atoms,Secular Equation of Hexatriene,Each line contains no more than three non-zero terms,Suppose, there exists an eigenstate, with the largest contribution from the ith orbital. Then the ith line reads,Here, the kth and lth atom are connected to ith atom.,更复杂的体系,The is one e
3、igenstate corresponding to x = 2,The is one eigenstate corresponding to x = 3,Alternant Hydrocarbons（交替烃）,参与共轭的原子可以被交替染色的共轭碳氢分子,Definition,*,*,*,*,*,*,*,*,*,*,*,*,*,?,?,Properties of alternant Hydrocarbons,For alternant hydrocarbons, the p energy levels +x and x occurs in pairs (x). For the paired M
4、Os, the atomic orbital coefficients for labeled atoms (*) are the same, while for unlabeled atoms, the only differences are that they are opposite in sign.,Theorem,Proof,For some alternant hydrocarbon molecule, suppose there exists one energy level x. For each labeled atom i,( r,s,t stand for three
5、unlabeled atoms connected to atom i, but it could be less than three ),For each unlabeled atom j,( l,m,n stand for three labeled atoms connected to atom i, but it could be less than three ),Obviously,Unlabeled,Unlabeled,Which means, wavefunction with ci, -cj as AO coefficients are also eigenstate, t
6、he corresponding eigenvalue(energy level) is x.,交替烃的非键轨道,For A.H.s with odd-number of atoms, the AOs from the unlabeled atoms do not have contributions to the non-bonding MO.,While the summation of the coefficients from all the atoms connect to one unlabeled atoms is zero,交替烃的非键轨道举例,a,-a,-a,a,-a,a,0
7、,0,0,0,0,0,0,a,-a,a,-2a,0,0,0,a,-a,a,0,0,6.3 Charge Densities and Bond Orders,Charge Densities Bond Orders Free Valences 自由价 Molecular Graph 分子图,1. Charge Densities,Butadiene,x1=1.618,x2=0.618,x3=-0.618,x4=-1.618,Charge Densities,Energy level 1, x1=1.618, is occupied by two electrons, more probably
8、appear on atom two and three.,Probability of finding 1 electron (from MO y1) on atom 1 is calculated by,Probability of finding 1 electron (from MO y2) on atom 1 is calculated by,Charge Densities,Since y1 and y2 are the only two MOs occupied by electrons, the probability of finding 1 electron (from a
9、ll of the MOs) on atom 1 is calculated by,This could also be regarded as the p electron charge density of atom 1.,Charge Densities,In general, the p electron charge density of atom r could be calculated by the following formula,Number of electrons Occupying orbital m,Coefficient of the rth AO in mth
10、 MO.,2. Bond Orders,Bond Orders,For connected atoms, the p-bond orders are defined as,This parameter reflects the strength of the corresponding p-bond, therefore can be correlated to the bond lengths,Bond Orders,Butadiene,Therefore, bond 1-2 is much stronger than bond 3-4, butadiene is often denoted
11、 as,Bond Order vs. Bond Length,共轭C-C键键长反比于其键级,3. Free Valence,描述共轭分子中，不同原子的反应活性差异,Equal Charge Density， Unequal Reactivity,Free Valence,Definition,Fr=nmax-nr,nr : 原子 r与相邻原子间的共轭键p键级之和。 nmax : 碳原子所能取的最大 nr值。代表共轭分子中参与共轭的碳原子的最大成键能力。 Fr : 自由价，代表第r个原子的剩余成键能力，即反应活性。,6.3 Charge Densities and Bond Orders,Fre
12、e Valence,nmax is calculated from the model compound,(CH2)3C,*,n*=1.732,三亚甲基甲烷 (Trimethylenemethane, TMM),nmax=1.732,For butadiene,F1 = F4 = 1.73-0.89 =0.84 F2 = F3 = 1.73-0.89-0.45 =0.39,1,4-addition is preferred to 1,2- or 2,3-additions,a,b,a,b,g,FaFb,Fg FaFb,Anthracene,Naphthalene,4. Molecular Gr
13、aph,0.84,0.89,0.45,0.39,0.52,0.55,0.73,0.80,0.10,0.45,0.40,6.4 Aromaticity,Aromaticity, Anti-aromaticity, 4n+2 rule Resonance Energy （共振能） The Eight-parameter Scheme The Five-parameter Scheme,1. Aromaticity, anti-aromaticity and 4n+2 rule,Cyclic conjugated hydrocarbons are, either, Aromatic Having 4
14、n+2 pi electrons, extra stabilities is originated from the p-electron delocalization,Anti-aromatic Having 4n pi electrons, destabled due to the p-electron delocalization,2. Resonance Energy, a Criteria of Aromaticity,RE = Ep ER,ER,?, 0 Aromatic, benzene, etc. = 0 non-aromatic, acyclic hydrocarbons 0
15、 anti-aromatic, cyclobutadiene, etc.,In unit of b,以一种大小相等的非芳香性共轭分子作为参考结构， 计算共振能-共轭所导致的稳定化能。,参考结构,早期，参考结构被定义为定域的双键 (Ethene)。 例如，苯分子的参考结构为三个孤立的双键。,Ep = 2(a+2b)+4(a+b)=6a+8b ER = 3*2*(a+b) =6a+6b,RE = 2b,2,1,-1,-2,1,-1,Resonance Energy of Some Typical Molecules,* Incorrect prediction,3. The Eight-param
16、eter Scheme,以八个参考结构代替一个参考结构,(23),(22),(22),(21),(20),(12),(11),(10),LJ Shaad and BA Hess, Jr. J. Am. Chem. Soc., 93,305 (1971).,The Eight-parameter scheme,上述权重值是通过对大量非芳香性分子的Ep进行最小二乘拟合得到的。,并对每一种参考结构对于能量的贡献赋予一定的权重。,The Eight-parameter scheme,Total reference energy of the molecule can be calculated by
17、counting the reference structure energies of each structure type.,ER= 3E22+3E12= 3*(2.0699+0.4660) =7.61,RE = 8-7.61 = 0.39 (b),The Eight-parameter scheme,分子共振能的计算与 Kekul 结构的画法有关, 分子有多种Kekul 结构时，RE 是各种 Kekul结构共振能的平均值.,ER= 4E22+E20 +2E12+ 4E11 = 13.128,ER= 3E22+2E21 +3E12+ 2E11 + E10 =13.1225,ER= 3E2
18、2+2E21 +3E12+ 2E11 + E10 =13.1225, = (13.128+2*13.1135)/3= 13.124 Ep = 13.68 RE = 13.68-13.124 = 0.56 (b),The Eight-parameter scheme,Define: REPE = RE/N,Resonance Energies per Electron,For Benzene REPE = 0.39/6 =0.065 For Naphthalene REPE = 0.55/10 = 0.055,The Eight-parameter scheme,4. Five-paramete
19、r Scheme,J-T-H five-parameter Scheme,Yuansheng Jiang (江元生） Au-Chin Tang （唐敖庆） Roald Hoffmann,Problem Set: 7，8，10，14,Charge Densities,For neutral A.Hs, the p electron charges on all of the carbon atoms are equal to 1.,Theorem,Proof,For A.H.s, the energy levels occurs in pairs (x). The molecular orbitals of each pair should have the property:,Charge Densities,Proof,The probability of finding the rth AO in all of the MOs is equal to 1,m: the MO index for bonding MOs m: the MO index for anti-bonding MOs,