but if the rings are due to temperature particles and,
\(f=\cfrac{c}{\lambda}\)
given,
\(139*\cfrac { \sqrt { 3 } }{ 2 } \quad \le \quad { a_{ \psi } }\quad \le \quad 139\quad \, pm\)
and
\(2\pi a_{\psi}=n\lambda\)
with
\(n=3\)
\(f=\cfrac{n.c}{2\pi a_{\psi}}\)
What is the significance of \(f\)?
If a reagent with an needed functional group also has a temperature particle layer (just below the outer \(p^+\), \(e^-\) layer) at this frequency, does reaction occurs more readily? Is the functional group accepted more readily?
For a lone particle, \(f\), because of
\(E=h.f\)
\(E=\cfrac{n.hc}{2\pi a_{\psi}}\)
is the energy of \(\psi\) that constitute the particle. \(f_{ph}=f\) here, is then the energy of each of the two tori on the benzene.
How to manipulate \(f_{ph}\)? Once again, it is a specific temperature. At this temperature, all collisions between the reagents and benzene molecules result in an exchange of radicals, hopefully between a \(H\) from the benzene ring and the functional group from the reagent. The difference in energy levels between the temperature \(\psi\) clouds on the benzene and the reagent with the needed functional group is the hindrance against chemical reactions.
As the functional group reagent is often smaller than benzene (low \(\lambda\), high \(f\), high \(E\)), it is the energy of temperature particle clouds on benzene that has to increase to facilitate a substitution. Conversely, the reagent that provides the functional group is lowered in temperature and made to collide with benzene.
Heat up benzene and spray it at the reagent which remains at a lower temperature.
If \(f_{ph}\) can be raised by increasing the temperature of benzene, then at the specific temperature when \(E=h.f_{ph}\) is equal to the energy level of the temperature clouds on the reagent, the energy barrier is surmounted and reactions proceeds.
But what charge is the benzene \(\psi\) torus? Under normal conditions, benzene rejects all reagents and is chemically stable. So, the benzene torus is positive and repulses positive temperature clouds on other possible reagent including other benzene rings.
Collisions bring the reactants close enough for \(\psi\) to interact as waves and be attractive, but that still requires compatible energy levels as \(\psi\) clouds merge and subsequently separate.
Which bring us to endothermic and exothermic reactions....because electrical charges and energy levels do not change temperature charges and energy levels. Electric and temperature are two orthogonal types of fields that do not affect each other.
Have a nice day...
