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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: xshadow on March 28, 2021, 07:24:42 AM

Title: Nerolin : williamson ether synthesis
Post by: xshadow on March 28, 2021, 07:24:42 AM

HI!!

I have a doubt about the reaction of:

 β-naphthol  + CH₃CH₂I  ----->  nerolin


The reaction is done with KOH + EtOH (or MeOH).


In the first step the base KOH deprotones the b-naphthol so I get a better nucleophile, a charged one
Then the naphtholate gives a Sn₂ reaction with  iodoethane to give the desired product

NOW my question is why cant't OH^- (from KOH)  compete with the naphtolate giving a Sn2 reaction, with the formation  ethanol as side product?
OH- is also a strong nucleophile and I have lots of OH- ions in the mixture.


Can it happen actually?
Is it perhaps better using a base like NaH that is a strong base but a worst nucleophile  (H- ha actually a strong steric hindrance cause solvatation)?
Thanks

Title: Re: Nerolin : williamson ether synthesis
Post by: Babcock_Hall on March 28, 2021, 11:06:16 AM

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

Title: Re: Nerolin : williamson ether synthesis
Post by: xshadow on March 28, 2021, 11:35:05 AM

Quote from: Babcock_Hall on March 28, 2021, 11:06:16 AM

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

Title: Re: Nerolin : williamson ether synthesis
Post by: rolnor on March 28, 2021, 12:09:22 PM

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

Title: Re: Nerolin : williamson ether synthesis
Post by: xshadow on March 28, 2021, 02:32:44 PM

Quote from: rolnor on March 28, 2021, 12:09:22 PM

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

Title: Re: Nerolin : williamson ether synthesis
Post by: rolnor on March 28, 2021, 02:57:19 PM

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

Title: Re: Nerolin : williamson ether synthesis
Post by: Babcock_Hall on March 29, 2021, 09:27:00 AM

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

Title: Re: Nerolin : williamson ether synthesis
Post by: rolnor on March 29, 2021, 01:48:10 PM

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