Ec—c benzene (L = 1.397 Å) = 534.0723 kj/mole
Ec—c graphite (L = 1.42 Å) = 503.3161 kj/mole
Being aware that the benzene has the three-electron bonds and also the interaction through the cycle, we can calculate the interaction through the cycle energy.
benzene on the basis of the three-electron bond, interaction through the cycle
(3)
from the equation we find L = 1.42757236 Å.
So, if the benzene molecule had a «clean» three-electron bond with a 1.5 multiplicity the c-c bond length would be L = 1.42757236 Å.
Now let us determine the energy of the «clean» three-electron bond with a 1.5 multiplicity knowing its length L = 1.42757236 Å:
Ec – c (L =1.42757236 Å) = 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)
(8)
n—the number of given value Y.
Let us calculate a from the equation
∑y = na + b∑ (1/x) + c∑ (1/x2), (9)
when n = 3.
Table 2. Calculation of ratios for relation E = f (L).
1/x1 = 0.64808814, x1 = 1.543, y1 = 347.9397
Σ (1/x2) = 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.
Finally, postulates of the three-electron bond theory (TBT) can be presented:
1) A chemical bond between two atoms may be established by means of three electrons with oppositely oriented spins (↑↓↑).
A • • • A (↑↓↑)
A • • • B (↑↓↑)
2) The electron shell of each atom in the stable molecule, ion, radical should have such a number of electrons which corresponds to the octet. A deviation from the octet results in an instability of a particle.
3) The state of the three-electron bond is determined by the octet rule.
4) The number of electrons participating in the chemical bond should be maximal and it’s then that the energy of the system will be minimal. Taking into consideration para 5 and 2.
5) In the course of establishing of the chemical bond electrons (their spins) are located in such a way that enables the interaction (attraction) to be maximal.
6) The aromatic bond is a three-electron bond in flat cyclic systems with a specific interaction of electrons through the cycle.
It is easy to show, that using three-electron bond one can explain paramagnetization and structure of oxygen molecule, structure of carboxylate anion, ozone, naphthalene and other organic and non-organic compounds. Let’s bring for the example structures of some compounds in terms of three-electron bond.
Naphthalene
Anthracene
Phenanthrene
Coronene
[18] -Annulene
It is interesting to note extreme symmetry of structures of naphthalene, anthracene, coronene and [18] -annulene, that is typical for the majority of aromatic compounds in general.
By the example of [18] -annulene it is possible to illustrate interaction through the cycle of central electrons of three-electron bonds. Interacting through the cycle, it shifts to the centre in the direction of inner atoms of hydrogen thus increasing electron density within the cycle and decreasing outside the cycle. And that’s why outside protons (12 Н) will give signals in the area of weaker field (reduction of electron density), and inner (6 Н) will give signals in the area of stronger field (increase of electron density). Thus this is observed in reality [13]. It also should be noted that inner protons bracing central electrons strengthen interaction through the cycle, and so stabilize aromatic system. But interaction through the cycle is decisive.
If aromatic system does not have inner protons, then outside protons will give signals in the area of weaker field (one of the features of aromatic compounds).
It is clear that in case of antiaromatic systems when there is no interaction (attraction) through the cycle, because central electrons have similar spins and push away, change in electron density in the centre of the cycle and outside the cycle will be reverse to aromatic systems.
Further we will continue demonstration of construction of organic and inorganic compounds.
Pyridine
Pyrimidine
Pyridazine
Pyrazine
1,3,5-Triazine
Quinoline
Isoquinoline
Indole
Purine
Furan, thiophene, pyrrole
Oxazole, thiazole, imidazole
Pyrazole
1,2,4-Triazole
1H-1,2,3-Triazole
Cyclopentadienyle anion
Carboxylate anion
Nitro compounds
Sulfonate anion
Organic acid amides and thioamides
Urea and thiourea
Guanidinium cation
Sodium malon ether
Sodium acetoacetic ether
Alyle cation
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