Nerolin : williamson ether synthesis

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In the acid-base equilibrium involving KOH and naphthol, where does the equilibrium lie?

2-naphthol pka= 9.5
H₂O  pka= 15

2-naphthol + OH^- ----->  naphtholate + H₂O

The equilibrium should be shifted towards the products.

( I thought  NaOH should be  added in  large excess so I had anyway also some OH- anions  that competes with naphtholate. )
 must  KOH/NaOH be added in stochiometric quantity? So it all reacts with the naphthol....


thanks

If you have OH- in the solution you will get some hydrolysis of the ethyl iodide, its not neccesary to have excess hydroxide.

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Naphtolate ion is much softer nucleophile then OH- and reacts faster.

Is it much more softer than OH^- because  its negative charge is delocalized on the two rings...is it correct?

But how can I say that CH₂CH₃I is a soft electrophile?
Because the C-I  carbon has only a small partial positive charge?

I've never used the hard-soft theory in order to predict the competition between two nucleophiles in a Sn₂ reaction...usually I say that stronger nucleophile is the species that will attack the substrate. (In in this case was OH- the stronger nucleophile)

Thanks

Are you working from a particular procedure that specifies what the ratio of KaOH to naphthol is?

In the acid-base equilibrium involving KOH and naphthol, where does the equilibrium lie?

The naphtolate is more polarizable just as the ethyl iodide, that makes them softer. You can compare, a ester carbonyl is a hard electrophile and it will not be attacked by naphtolate but by hydroxide.
https://en.wikipedia.org/wiki/HSAB_theory

Naphtolate ion is much softer nucleophile then OH- and reacts faster.