Why does the -OH group in β-naphthol direct the incoming diazonium salt towards the ortho position? In the following reactions, the major product W is



I was doing this question which asks me the product of a diazotisation coupling reaction between diazonium chloride and β-naphthol. I know that the diazonium salt will act as an electrophile and compounds like phenol will direct it towards the para position, but I am not able to understand why the diazonium salt attaches itself to the ortho position of the -OH group in this compound.
The salt can also attach itself to the other benzene ring opposite to the -OH group. Also there are two ortho positions with respect to the -OH group. Can somebody please explain this to me?
This is the answer to the question:
It's sort of a process of elimination.  The electrophilic ion could potentially attack anywhere, but:


The hydroxyl group is activating, so the electrophile will prefer that ring.  You are left with three open positions.
One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing.  Hydroxyl follows that rule.
Finally -- electrophilic substitution in a naphthalene system is kinetically favored next to the other ring.  See here for example.


We have one position that's on the same ring as hydroxyl, ortho or para to that hydroxyl group, and next to the other ring.  That is the position shown in the book.