June 19, 2019, 09:12:23 AM
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Topic: RCOOR +NaOH --> ???  (Read 247 times)

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Offline xshadow

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RCOOR +NaOH --> ???
« on: June 13, 2019, 04:49:38 AM »
Hi, studying  org. Chem 1 my textbook says that an ester can react with a base like NaOH giving a saponification reaction , i.e ""RCOOH"" -->  RCOO- + H+

Later I studied the  acidity of α C-H  directly bonded to C=O

Now an ester like  ketons or an aldehyde  has an "acdic Hydrogen on α carbon" .

So can we have the deprotonation and the formation of an "enolate ester"??
Then an ""aldolic addition"" in order to give an ""ester-hemiacetal"" like this(?):




I suppose that the first reaction is the major one because in the second one we have the formation of an hemiacetal,not a very stable species

But can the second one theoretically occours at a very very low percentage??

Thanks!!

Offline AWK

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Re: RCOOR +NaOH --> ???
« Reply #1 on: June 13, 2019, 05:03:37 AM »
2. Claisen condensation needs a stronger base.
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Offline xshadow

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Re: RCOOR +NaOH --> ???
« Reply #2 on: June 13, 2019, 05:59:46 AM »
2. Claisen condensation needs a stronger base.

NaOH ≈  OH-

Now RO- is used in aldolic addition...and OH- is a base similar to RO- because pka H2O≈ pka ROH ...so why isn't enough?!

Thanks

Offline AWK

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Re: RCOOR +NaOH --> ???
« Reply #3 on: June 13, 2019, 07:20:53 AM »
Some reactions require an absolutely anhydrous environment, and then the thinking of incorporating the ionic product of water into consideration is useless.
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Offline clarkstill

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Re: RCOOR +NaOH --> ???
« Reply #4 on: June 13, 2019, 07:54:32 AM »
Does a Claisen condensation need anhydrous conditions? I didn't think so. I think this is just a case of relative rates - the hydrolysis is fast and irreversible (since once the acid is formed it will be rapidly and irreversibly deprotonated). For the Claisen condensation to occur the enolate needs to form - pKa of alcohol/water is ~15 and esters are about 25, so only 1 in 1010 molecules in solution will be able to participate in the Claisen, so the overall rate is much slower since the effective concentration is so low.

I wouldnt be surprised if you got trace amounts of claisen by-products in cases where ester hydrolysis is slow. Also, to the OP, the hemi-acetal isn't stable, but will convert to a ketone.

Online rolnor

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Re: RCOOR +NaOH --> ???
« Reply #5 on: June 13, 2019, 09:09:20 AM »
If you have malonate ester you can have Claisen in water?

Offline hollytara

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Re: RCOOR +NaOH --> ???
« Reply #6 on: June 13, 2019, 08:57:03 PM »
In Claisen, we usually use alkoxide that matches the ester: ethoxide for ethyl esters, methoxide for methyl esters....

So if there is nucleophilic substitution at carbonyl, the product is the same as the reactant - exchange one ethyl group for the other. 

Offline David Tan

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Re: RCOOR +NaOH --> ???
« Reply #7 on: June 14, 2019, 04:15:14 PM »
Saponification is essentially an alkaline hydrolysis of an ester. For saponification, the hydroxide is acting as a nucleophile instead of a base. Although there is an alpha proton next to the carbonyl carbon of the ester and could potentially be deprotonated by the hydroxide, it won't happen. That's because if an enolate is formed, the negative charge on the deprotonated carbon would get delocalised into the adjacent C=O moiety which would compete with the delocalisation of electrons on the oxygen atom of OR. Hence, the enolate would not be stabilised due to fellow competition. However if we have a ketone or aldehyde, the alpha proton can be deprotonated because there is no competition for delocalisation and the enolate formed would be stabilised. Hope it helps. 
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