Review. Benzene on the basis of the three-electron bond. Theory of three-electron bond in the four works with brief comments (review). 2016. - страница 8
1/x>1 = 0.64808814, x>1 = 1.543, y>1 = 1
Σ (1/x>2) = 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 (х>1, у>1), (х>2, у>2) and (х>3, у>3) 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
b>1 = b + c/x>1, Y = b>1 + cX
As usual:
(7)
(8)
n—the number of given value Y.
Let us calculate a from the equation
∑y = na + b∑ (1/x) + c∑ (1/x>2), (9)
when n = 3.
Table 2. Calculation of ratios for relation E = f (L).
1/x>1 = 0.64808814, x>1 = 1.543, y>1 = 347.9397
Σ (1/x>2) = 1.66729469, Σ (1/x) = 2.22534781 when n = 3
c = – 1699.18638789,
b = 5065.62912191,
a = – 2221.34518418
(2)
Let us calculate from the equation:
Ec—c (ethane) = 347.9397 kj/mole
Ec═c (ethylene) = 615.4890 kj/mole
Ec≡c (acetylene) = 812.2780 kj/mole.
2.3. Conclusion
As we can see, three-electron bond enables to explain aromaticity, find delocalization energy, understand aromatic bond’s specificity. Aromatic bond in benzene molecule is simultaneous interaction of three pairs of central electrons with opposite spins through the cycle. But whereas central electrons are the part of three-electron bond, then it is practically interaction of six three-electron bonds between themselves, that is expressed in three interactions through cycle plus six three-electron bonds. We shouldn’t forget in this system about important role of six atom nucleuses, around which aromatic system is formed. Properties of nucleuses especially their charge will influence on properties of aromatic system.