Why does the -OH group in β-naphthol direct the incoming diazonium salt towards the ortho position? [duplicate]

Question

16. 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:

Answer 1

It's sort of a process of elimination. The electrophilic ion could potentially attack anywhere, but:

1. The hydroxyl group is activating, so the electrophile will prefer that ring. You are left with three open positions.

2. 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.

3. 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.

Answer 2

Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.

If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 .Both 1 and 2 are aromatic.

If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.

Therefore attack at alpha position is the major product.

UPDATE :

The resonance structures will have a structure in which all atoms possess complete valence electrons.

Reference: Pielhop, T.; Larrazábal, G. O.; Studer, M. H.; Brethauer, S.; Seidel, C.; Rudolf von Rohr, P. Lignin repolymerisation in spruce autohydrolysis pretreatment increases cellulase deactivation. Green Chem. 2015, 17 (6), 3521–3532 DOI: 10.1039/c4gc02381a.

Answer 3

I know it's some old jee stuff.Usually they ask for the most stable product. Think of the hydrogen bonding at ortho.This should stabilise the transition state.Draw the meisenheimer complex.It's low temperature therefore extra bonds provide extra stable transition state and the activation energy of reaction is lowered.It's kinetically favoured product as expected. Thermodynamics would check for steric factors and give para product. For kinetics/low temp look for any kind of stabilization which can lower energy of transition state .