The left hand isomer is clearly the kinetically favoured product and the right hand isomer is the thermodynamically favoured product. However, I have no idea why. Can you explain?
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1 Answer
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In contrast to many other electrophilic aromatic substitution reactions, aromatic sulfonation is reversible, in other words it is an equilibrium. If you use a large excess of $\ce{SO3}$ you push the reaction to the sulfonation side; if you heat the sulfonated product in the absence of $\ce{SO3}$ it will de-sulfonate. This makes sulfonation a nice method to 1) control where a second substituent will be introduced or 2) serve as a protecting group (block second substituent introduction) - and then when you are done, you simply de-sulfonate.
Now to your question
If we look at the intermediates leading to the 2 isomers we see that we can draw 2 resonance structures that preserve aromaticity in one ring for the transition state leading to naphthalene-1-sulfonic acid, but only one resonance structure where the other aromatic ring remains intact for the 2-isomer (draw the resonance structures to convince yourself).
Therefore the transition state leading to the 1-isomer will be lower in energy and naphthalene-1-sulfonic acid will be formed fastest - it is the kinetically favored product.
However, naphthalene-2-sulfonic acid is a few kcal/mol more stable than its 1-isomer. This is due to an adverse 1,8 steric interaction present in the 1-isomer.
Because the sulfonation reaction is reversible, if we run it long enough at high temperature (favoring reversibility), eventually we will wind up with the thermodynamically preferred (more stable) product, the 2-isomer.