Review. Benzene on the basis of the three-electron bond. Theory of three-electron bond in the four works with brief comments (review). 2016.. Volodymyr Bezverkhniy
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493.3097 kj/mole
Taking into account that the benzene c-c bond energy with a 1.658 multiplicity is equal to Ec-c benzene = 534.0723 kj/mole, the difference will make:
ΔE = 534.0723 kj/mole – 493.3097 kj/mole = 40.7626 kj/mole.
40.7626 kj/mole is the energy of interaction through the cycle per one c-c bond. Therefore, the energy of interaction through the cycle will be two times higher:
E1 = 40.7626 kj/mole ∙ 2 = 81.5252 kj/mole (19.472 kcal/mole)
It is clear that the three interactions through the cycle present precisely the working benzene delocalization energy which is:
E = 3E1 = 3 ∙ 81.5252 kj/mole = 244.5756 kj/mole (58.416 kcal/mole)
benzene on the basis of the three-electron bond, delocalization energy
It is also possible to calculate the benzene molecule energy gain in comparison with the curved cyclohexatriene (let us assume that energy of C-H bonds in these molecules is similar). For this we calculate the sum of energies of single and double c-c bonds in cyclohexatriene:
E2 = 3Ec—c +3Ec═c = 2890.286 kj/mole
The energy of six benzene c-c bonds with a 1.658 multiplicity is equal to:
E3 = 6 · 534.0723 kj/mole = 3204.434 kj/mole
Therefore, the gain energy of benzene compared to cyclohexatriene will amount to:
E = E3 – E2 = 3204.434 kj/mole – 2890.286 kj/mole = 314.148 kj/mole (75.033 kcal/mole).
2.2. Experimental
Let’s show more detailed calculation of ratios for our mathematical relations. Let’s consider relation Multiplicity = f (L) and E = f (L) for С-С bonds, where multiplicity is multiplicity of bond, L – length of bond in Å, Е – energy of bond in kj/mole.
As initial points for the given bonds we will use ethane, ethene and acetylene. For the length of bonds let us take the findings [7]:
bond lengths in ethane, ethylene and acetylene
As usual, the С-С bond multiplicity in ethane, ethylene and acetylene is taken for 1, 2, 3. For the energy of bonds let us take the findings [7, p. 116]:
energies of bonds in ethane, ethylene and acetylene
If we have two variants and we received the set of points and we marked them on the plane in the rectangular system of coordinates and if the present points describe the line equation y = ax + b that for choose the coefficients a and b with the least medium-quadratic deflection from the experimental points, it is needed to calculate the coefficients a and b by the formulas:
(4)
(5)
n-the number of given values x or y.
If we want to know how big is the derivative, it is necessary to state the value of agreement between calculated and evaluated values y characterized by the quantity:
(6)
The proximity of r2 to one means that our linear regression coordinates well with experimental points.
Let us find by the method of selection the function y = a + b/x + c/x2 describing the dependence multiplicity = f (L) and E = f (L) in best way, in general this function describes this dependence for any chemical bonds.
Let us make some transformations for the function y = a + b/x + c/x2, we accept
X = 1/x,
than we’ll receive: Y = b1 + cX, that is the simple line equality, than
(7)
(8)
n—the number of given value Y.
Let us find a from the equality:
∑y = na + b∑ (1/x) + c∑ (1/x2), (9)
when n = 3.
Let us find now multiplicity = f (L) for C─C, C═C, C≡C.
Table 1. Calculation of ratios for relation Multiplicity = f (L).
1/x1 = 0.64808814, x1 = 1.543, y1 = 1
Σ (1/x2) = 1.66729469, Σ (1/x) = 2.22534781 when n = 3
c = 11.28562201,
b = – 5.67787529,
a = – 0.06040343
Let us find from the equation:
Multiplicity C—C (ethane) = 1.
Multiplicity C═C (ethylene) = 2.
Multiplicity C≡C (acetylene) = 3.
Multiplicity C—C (graphite) (L = 1.42 Å) = 1.538 ≈ 1.54.
Multiplicity C—C (benzene) (L = 1.397 Å) = 1.658.
As we can see the multiplicity C—C of benzene bond is 1.658 it is near the bond order of 1.667 calculated by the method MO [8, p. 48].
It should be noted that the а, b, с coefficients for this y = a + b/x + c/x² function in case of using three pairs of points (х1, у1), (х2, у2) and (х3, у3) are defined explicitly; actually, they (the coefficients) are assigned to these points. In that way we find these coefficients for working further with the equation. For making certain that this dependence y = a + b/x + c/x² describes well the Multiplicity = f (L) and E = f (L) functions it will take only to perform correlation for four or more points. For example, for the dependence Multiplicity = f (L) for C-C bonds we should add a fourth point (Lc—c = 1.397 Å, Multiplicity = 1.667) and obtain an equation with r² = 0.9923 and the coefficients а = – 0.55031721, b = – 4.31859233, с = 10.35465915.
As it is difficult, due to objective reason, to define four or more points for the Multiplicity = f (L) and E = f (L) equations for a separate bond type, we will find the а, b, с coefficients using three points (as a rule they are the data for single, double and triple bonds). The dependences obtained in such a way give good results as regards the bond multiplicity and energies.
We’ll find the dependence E = f (L) for the C—C bonds
b1 = b + c/x1, Y = b1 + cX
As usual:
(7)