# Can anyone please help me how do you determine kinetically the fastest product of sulphonation of $\beta$ napthol?

I was asked a question in a paper where I was asked to identify the kinetically fastest product of sulphonation of $$\beta$$ naphthol.

The answer that is provided was that the ortho position to the hydroxyl group is the position where the fastest attack would take place. Because when the carbo cation is formed, the most stable one initially.

But I did not find this explanation satisfactory because ultimately the whole molecule is in complete resonance and it should not matter whichever Ortho or para position the sulphonation happens because ultimately the positive charge would end up being stabilised by OH group. Please help me.

The solution argued that this is less kinetically fast as the initial carbocation is less stable. Also there are 2 Ortho positions, but it does not specify which is the faster one.

The solution given was that Oxy Tobias acid is fastest but based on above arguments , R acid is equally likely and Schaeffer's acid is more probable ...(as there is no steric factors

Also what would the thermodynamically most stable product be ??

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).