Can anyone please help me how do you determine kinetically the fastest product of sulphonation of $\beta$ napthol?
Electrophile can substitute 2-naphthol ring on 3 possible positions: 1-position through intermediate I, 6-position through intermediate II, and 3-position through intermediate III (See following diagram). Among these three intermediates, intermediate I is the most stable (You must visit Mechanism of formation of 2-naphthol red dye to get excellent explanation why substitution at 1-position is the preferable one over 3-position, although both are ortho to the phenolic $\ce{OH}$ group). Thus, the kinetically fastest product of sulphonation of 2-naphthol is 2-hydroxynaphthalene-1-sulfonic acid (Oxy Tobias acid). There will be no 3-hydroxynaphthalene-2-sulfonic acid (Ref.1).



The best evidence for this preference has been displayed during specific reaction using Claisen rearrangement (See above diagram in the box). After first rearrangement free phenolic $\ce{OH}$ group was allylated again and subjected to heat. For our surprise, the rearragement is proceeded to give 100% 1,1-diallyl derivative as shown. No 1,3-diallyl-2-naphthol was detected (according to NMR evidence).

Based on Ref.1, some 2-hydroxynaphthalene-6-sulfonic acid (Schaeffer's acid) has also been produced under kinetic control. Yet, longer heating time allowed to give disubstituted product (2-hydroxynaphthalene-1,6-disulfonic acid) preferably, thus thermodynamic product cannot be determined by these evidence.

Reference:


Paolo Beltrame, Giorgio Bottaccio, Paolo Carniti, Giuseppina Felicioli, “Sulfonation of 2-naphthol by sulfuric acid: rate measurements and kinetic model,” Ind. Eng. Chem. Res. 1992, 31(3), 787-791 (https://doi.org/10.1021/ie00003a022).