6.1.3 The allyl system. The normalization relationship can be introduced to make the eqs. solvable,c+c+c = 1. There are two practical approaches to find the coefficients.Firstmethod:usetheeguations towrite all of the coefficientsinterms of one of them.Fromeq. [A] we have: -V2c1 +C2 = 0hence C2 = V2c1.C2 - V2c3 = 0 [C]Now use eq. [C]:Substitute C2 = V2c1 → V2c1 - V2c3 = 0 → C3 = C1now insert these values into the normalization condition, and hence find cci+c2+c =1→ ci+(V2c1)2+ci = 1→4ci= 1→ C1= 1/2with Ei = α + V2βC2 = V2/2, C3 = 1/2; →1 = (Φ1+V2Φ2 +Φ3)/2
6.1.3 The allyl system • The normalization relationship can be introduced to make the eqs. solvable, 𝒄𝟏 𝟐 + 𝒄𝟐 𝟐 + 𝒄𝟑 𝟐 = 𝟏 • There are two practical approaches to find the coefficients. • First method: use the equations to write all of the coefficients in terms of one of them. From eq. [A] we have: − 𝟐𝒄𝟏 + 𝒄𝟐 = 𝟎 hence 𝒄𝟐 = 𝟐𝒄𝟏. Now use eq. [C]: 𝒄𝟐 − 𝟐𝒄𝟑 = 𝟎 [𝑪 Substitute 𝒄𝟐 = 𝟐𝒄𝟏 𝟐𝒄𝟏 − 𝟐𝒄𝟑 = 𝟎 𝒄𝟑 = 𝒄𝟏 now insert these values into the normalization condition, and hence find c1 : 𝒄𝟏 → 𝒄𝟏= 𝟏/𝟐 𝟐 + 𝒄𝟐 𝟐 + 𝒄𝟑 𝟐 = 𝟏 𝒄𝟏 𝟐 + ( 𝟐𝒄𝟏) 𝟐+𝒄𝟏 𝟐 = 𝟏 𝟒𝒄𝟏 𝟐 = 𝟏 𝒄𝟐 = 𝟐/𝟐 , 𝒄𝟑 = 𝟏/𝟐; 𝝍𝟏 = (𝜱𝟏+ 𝟐𝜱𝟐 + 𝜱𝟑)/𝟐 with 𝐸1 = 𝛼 + 2𝛽
6.1.3 The allyl system. The second method: as we know the relationship between the coefficients, we might justset one of them to have the value 1 and work out the rest, then normalize at the end.. Let us set c,= 1; from [A] we have:-V2c1 + C2 = 0, put C1 = 1 , giving C2 = V2? Now we use this value for c, in [C]:C2 - V2c3 = 0, giving C3 = 1C2= V2 C3 =1.ThecoefficientsarethereforeC1 = 1ci + c + c = V1 + 2 + 1 = 2: Now normalize the coefficients:C2 = V2/2.Thenormalizedcoefficientsare:C1 = 1/2C3 = 1/2Which method do you recommend?
• The second method: as we know the relationship between the coefficients, we might just set one of them to have the value 1 and work out the rest, then normalize at the end. • Let us set c1= 1; from [A] we have: − 𝟐𝒄𝟏 + 𝒄𝟐 = 𝟎, put 𝒄𝟏 = 1 , giving 𝒄𝟐 = 𝟐 6.1.3 The allyl system • Now we use this value for c2 in [C]: 𝒄𝟐 − 𝟐𝒄𝟑 = 𝟎, giving 𝒄𝟑 = 1 • The coefficients are therefore: 𝒄𝟏 = 𝟏 𝒄𝟐 = 𝟐 𝒄𝟑 = 𝟏 • Now normalize the coefficients: 𝒄𝟏 𝟐 + 𝒄𝟐 𝟐 + 𝒄𝟑 𝟐 = 𝟏 + 𝟐 + 𝟏 = 𝟐 • The normalized coefficients are: 𝒄𝟏 = 𝟏/𝟐 𝒄𝟐 = 𝟐/𝟐 𝒄𝟑 = 𝟏/𝟐 Which method do you recommend?
一6.1.3 The allyl systemThefinalresultsaresummarizedinthetableE,=α-V2βWMOwavefunctionMOnumberenergy1E=+=++3E,=αL2E2 = αEx.273E=-2=-3(please recall that we once employed a graphical method toHE,=α + V2βTdeal withthe -MOs ofthis molecule inthefirst semester!HHHHHAllylanion（4元e）Allyl cation (2元e)
• The final results are summarized in the table. 6.1.3 The allyl system Ex.27 Allyl cation (2e) Allyl anion (4e) 3 2 1 E1= + 𝟐 E3= 𝟐 E2= (please recall that we once employed a graphical method to deal with the -MOs of this molecule in the first semester! )
小测验I.H,+可能具有直线形和正三角形两种结构，休克尔近似下,可令α=α，βH-H=β分别写出两种结构的久期行列式II.折叠苯C.H的结构图右图所示，其元由C原子p,轨道组成，休克尔近似下有αc=α, βc-c=βN1.快速写出其元分子轨道的久期方程(矩阵形式）2.运用对称性，推导其元分子轨道的能量及组成形式25
小测验 I. H3 + 可能具有直线形和正三角形两种结构, 休克尔近似 下,可令H = , H-H=, 分别写出两种结构的久期行列式; II. 折叠苯C6H6的结构图右图所示, 其𝜋6 6由C原子pz轨道组成, 休克尔近似下有C = , C-C=. 1. 快速写出其分子轨道的久期方程(矩阵形式); 2. 运用对称性, 推导其分子轨道的能量及组成形式
6.1.4 1,3-Butadiene.TheTsystemofbutadienecomprises.Thesecularequationsare:four p orbitals in a row?0β0α-EHHC10ββα-EC2= 00ββα-EC300βα-EC4HH: To solve this problem, we would first need to find the determinant of the 4 × 4 matrix,set itto zeroand then solvethe resultingquarticinE: This already sounds like very hard work and although in this case it might just bepossibletodothis by hand: Luckily, symmetry comes to our aid and reduces this problem to something very mucheasier
6.1.4 1,3-Butadiene • The π system of butadiene comprises four p orbitals in a row: • The secular equations are: • To solve this problem, we would first need to find the determinant of the 4 × 4 matrix, set it to zero and then solve the resulting quartic in E. • This already sounds like very hard work and although in this case it might just be possible to do this by hand. • Luckily, symmetry comes to our aid and reduces this problem to something very much easier