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

1. 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:

## marked as duplicate by Nilay Ghosh, Mithoron, Todd Minehardt, Waylander, M.A.R. ಠ_ಠMay 22 at 17:09

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.

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.

• Technically, each possibility has one more contributing structure containing $\ce{=\overset{+}{O}H}$. These structures require placing a positive charge on an electronegative atom but enable all atoms to have full valence shells. – Oscar Lanzi May 19 at 14:23

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 .

• You have two positions ortho to the hydroxyl group bit only one is listed as the major product. Why would you pick that one? See en.wikipedia.org/wiki/Naphthalene#Reactions_with_electrophiles and incorporate what that says into this answer. – Oscar Lanzi May 20 at 10:01
• Ok what about this, draw resonance structures on both sides(just to say: only one side that is drawn above it is permissible). – Vishesh Mangla May 27 at 6:15
• I would have posted an immage but it is not possible. – Vishesh Mangla May 27 at 6:15