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    Can someone please explain this reaction as I don't get why 3 moles of NaOH are used. I know the reaction between an ester and OH- ions will be sodium carboxylate but how do you know the CH3coo- part comes off?

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    This is not answering part a. I just want to know the reaction of aspirin and why it works like that. Thanks
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    (Original post by coconut64)
    Can someone please explain this reaction as I don't get why 3 moles of NaOH are used. I know the reaction between an ester and OH- ions will be sodium carboxylate but how do you know the CH3coo- part comes off?

    Thanks
    1) first equivalent NaOH deprotonates the benzoic acid

    2) 2nd equivalent hydrolyses the ester, forming the "carb acid" and alcohol

    3) but hang on, you still have strong base lying around, hence another eqvdeprotonates the new carb acid formed. you end up wth dicarboxylate.
 
 
 
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