If we take naphthalene then we can react it with electrophiles. For example if we treat it with nitric acid under mild conditions.
Now the nitration reaction does not appear possible to reverse, but the two nitronapthelenes are different in terms of thermodynamics. In the case of the 1-nitronaphthalene there is a steric repulsion between the hydrogen on the 8 position and the nitro group. In the case of the 2-nitronaphthalene there is no strong steric repulsion between the hydrogens on the 1 and 3 positions and the nitrogroup.
There is a reason why the less thermodynamically favored product will form, this is a kinetic reason. If we react naphthalene with an electrophile at the 1 position we go via a cationic intermediate. Two of the five lower energy resonance forms keep one of the two rings aromatic. This will lower their energy.
By now you should be learning that the more low energy resonance forms you can draw the lower the energy the ion has, also if you can draw resonance forms which are extra stable then the energy of the combination of the resonance forms will be lower. Lower energy means more stable. If we repeat the intellectual process for the reaction at the two position on naphthalene then we get a less stable cationic intermediate.
So what happens when naphthalene reacts is that the kinetic product (product which is formed more quickly) is higher in energy than the product whose formation is slower. If the formation of the product where the naphthalene has reacted at the one position (alpha isomer) can be reversed then if we stew the reaction mixture for a long time then we will get the beta isomer as our product.
While nitration is very irreversible the reactions of sulfuric acid and things like tert-butyl cations are reversible. Since 1870 (Merz, Chem. Ber., 1870, 3, 196) it has been known that naphthalene-1-sulfonic acid when heated at 130 oC in sulfuric acid will transform into naphthalene-2-sulfonic acid.
Another example is the reaction of tert-butyl chloride with benzene using aluminium chloride as the cataylst. If we do the reaction then we will have a Friedal-Crafts alkylation reaction. The first stage will be the reaction to form tert-butyl benzene.
Then a second molecule of tert-butyl chloride reacts to form 1,4-di-tert-butyl benzene.
Then things get a bit complex, one of the tert-butyl groups will come off the benzene ring and then we can react the tert-butyl cation at the meta site to form the 1,3-di tert-butyl benzene.
This can then react with more of the tert-butyl chloride to form the final product, I suspect that only one product will form. The reason is that the steric effects of the tert-butyl groups will oppose the loss of the protons from the cationic intermidates.
The steric energy of 1,2,3-tri-tert-butylbenzene is 75.3531 kcal per mole, 1,2,4–tri-tert-butylbenzene is 38.2984 kcal per mole and 1,3,5–tri-tert-butylbenzene is 14.3038 kcal per mole. On the other hand the steric energy for 1,3-di-tert-butylbenzene is 9.5103 kcal per mole. The steric energy of the tert-butyl cation is only 3.6054 kcal per mole. I think that a steric argument explains why the 1,3,5-product is a major product from the reaction of benzene with tert-butyl chloride.
Organic chemistry teaching 