The position para to the OH group in 2-napthol is "blocked" by the junction with the second benzene ring. If you had 1-napthol, it would react at the para position. In a lot of cases, the regiochemical outcome (ortho vs para) of aromatic substitution reactions is dictated by sterics; lower steric requirement in the transition state = lower activation energy barrier = higher rate of product formation (than a more sterically hindered product). In cases where both ortho- and para- products are possible, there should be a distribution of products (which relates to their relative rates of formation). Thus, the aniline will react preferentially at the para position, because that's less sterically hindered.
Hope that helps
|