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Topic: Nerolin : williamson ether synthesis  (Read 356 times)

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Offline xshadow

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Nerolin : williamson ether synthesis
« 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
« Last Edit: March 28, 2021, 08:35:27 AM by xshadow »

Offline Babcock_Hall

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Re: Nerolin : williamson ether synthesis
« Reply #1 on: March 28, 2021, 11:06:16 AM »
In the acid-base equilibrium involving KOH and naphthol, where does the equilibrium lie?

Offline xshadow

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Re: Nerolin : williamson ether synthesis
« Reply #2 on: March 28, 2021, 11:35:05 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

Offline rolnor

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Re: Nerolin : williamson ether synthesis
« Reply #3 on: March 28, 2021, 12:09:22 PM »
Naphtolate ion is much softer nucleophile then OH- and reacts faster.

Offline xshadow

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Re: Nerolin : williamson ether synthesis
« Reply #4 on: March 28, 2021, 02:32:44 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

Offline rolnor

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Re: Nerolin : williamson ether synthesis
« Reply #5 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

Offline Babcock_Hall

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Re: Nerolin : williamson ether synthesis
« Reply #6 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?

Offline rolnor

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Re: Nerolin : williamson ether synthesis
« Reply #7 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.

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