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    What conditions favour and thereby catalyse esterification (the formation of an ester from a carboxylic acid and alcohol), as opposed to favouring and catalysing de-esterification (the transformation of an ester into a carboxylic acid and an alcohol)? I know H+ ions are involved as a catalyst (speeding up both directions), so clearly these are not the main issue - so what would you add to favour the esterification, and what would you add to favour the 'decomposition'?
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    (Original post by Big-Daddy)
    What conditions favour and thereby catalyse esterification (the formation of an ester from a carboxylic acid and alcohol), as opposed to favouring and catalysing de-esterification (the transformation of an ester into a carboxylic acid and an alcohol)? I know H+ ions are involved as a catalyst (speeding up both directions), so clearly these are not the main issue - so what would you add to favour the esterification, and what would you add to favour the 'decomposition'?
    Concentrated acids are also dehydrating agents and help to absorb the water as it is formed pulling the equilibrium to the side of the ester.

    Aqueous systems help the equilibrium to be established in the reverse direction by providing an excess of the water needed for the hydrolysis.
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    (Original post by Big-Daddy)
    What conditions favour and thereby catalyse esterification (the formation of an ester from a carboxylic acid and alcohol), as opposed to favouring and catalysing de-esterification (the transformation of an ester into a carboxylic acid and an alcohol)? I know H+ ions are involved as a catalyst (speeding up both directions), so clearly these are not the main issue - so what would you add to favour the esterification, and what would you add to favour the 'decomposition'?
    Dehydrating conditions.... You need to shift the equilibrium by removing water from the reaction. If you carry the reaction out in toluene, and assuming the alcohol used for the esterification is sufficiently soluble in toluene, you can use a dean-stark apparatus to remove the water as it is generated.
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    (Original post by charco)
    Concentrated acids are also dehydrating agents and help to absorb the water as it is formed pulling the equilibrium to the side of the ester.

    Aqueous systems help the equilibrium to be established in the reverse direction by providing an excess of the water needed for the hydrolysis.
    (Original post by JMaydom)
    Dehydrating conditions.... You need to shift the equilibrium by removing water from the reaction. If you carry the reaction out in toluene, and assuming the alcohol used for the esterification is sufficiently soluble in toluene, you can use a dean-stark apparatus to remove the water as it is generated.
    I see, so H+ is a catalyst for both directions (as I said originally), but also helps shift the equilibrium to the right (ester formation) as it lowers the concentration of water (you'd need concentrated acid to have any significant impact, though). However, this H+ catalyst is also needed to ensure the reaction reaches equilibrium reasonably quickly (and if the H+ concentration were too low, it would take much longer to reach equilibrium) - so if you want to convert the ester to alcohol and carboxylic acid, what conditions would you use? Wouldn't excess of water lower the H+ concentration and make the reaction too slow to reach equilibrium?
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    (Original post by JMaydom)
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    (Original post by charco)
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    Any help? (See my post directly above)
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    (Original post by Big-Daddy)
    I see, so H+ is a catalyst for both directions (as I said originally), but also helps shift the equilibrium to the right (ester formation) as it lowers the concentration of water (you'd need concentrated acid to have any significant impact, though). However, this H+ catalyst is also needed to ensure the reaction reaches equilibrium reasonably quickly (and if the H+ concentration were too low, it would take much longer to reach equilibrium) - so if you want to convert the ester to alcohol and carboxylic acid, what conditions would you use? Wouldn't excess of water lower the H+ concentration and make the reaction too slow to reach equilibrium?
    If something is a catalyst it will speed the rate at which a system reaches equilibrium. They are one and the same, not two separate points.
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    (Original post by JMaydom)
    If something is a catalyst it will speed the rate at which a system reaches equilibrium. They are one and the same, not two separate points.
    I know, I said that in my own post. What I am asking is:

    I can see that increasing H+ concentration will act to shift the equilibrium to the right and encourage ester formation as it lowers the concentration of water (so the system will act to produce more water, and in turn more of the ester).

    How would you encourage an ester (in the presence of H2O) to revert back (by hydrolysis?) to an alcohol and carboxylic acid? Would you add a lot of water, and a lot of H+ (the latter being to keep the rate high, as a catalyst; the former being to push the equilibrium to the left)?
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    (Original post by Big-Daddy)
    I see, so H+ is a catalyst for both directions (as I said originally), but also helps shift the equilibrium to the right (ester formation) as it lowers the concentration of water (you'd need concentrated acid to have any significant impact, though). However, this H+ catalyst is also needed to ensure the reaction reaches equilibrium reasonably quickly (and if the H+ concentration were too low, it would take much longer to reach equilibrium) - so if you want to convert the ester to alcohol and carboxylic acid, what conditions would you use? Wouldn't excess of water lower the H+ concentration and make the reaction too slow to reach equilibrium?
    To add to JM's post (above), it doesn't matter from which direction you approach the equilibrium, the catalyst provides an altrernative route with a lower activation energy. It speeds up the attainment of equilibrium.

    The other 'conditions' are to try to improve the yield. Strictly speaking this does not affect the equilbrium position, but rather by using concentrated sulphuric acid you remove water from the equilibrium causing the system to respond by making more and, at the same time, more ester.

    Hydrolysis of esters is favoured by excess water and slow attainment, hence heat speeds up the attainment of equilibrium. But you are not going to get any extra yield of alcohol/acid unless you arrange for the alcohol to be distilled off as formed (not usually carried out), or for the acid to be neutralised as formed, hence the use of sodium hydroxide as the hydrolysing reagent.

    So for preparation of ester: alcohol + acid (concentrated acid & heat)

    For hydrolysis of ester: ester + excess water (strong NaOH & heat)
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    (Original post by charco)
    To add to JM's post (above), it doesn't matter from which direction you approach the equilibrium, the catalyst provides an altrernative route with a lower activation energy. It speeds up the attainment of equilibrium.

    The other 'conditions' are to try to improve the yield. Strictly speaking this does not affect the equilbrium position, but rather by using concentrated sulphuric acid you remove water from the equilibrium causing the system to respond by making more and, at the same time, more ester.

    Hydrolysis of esters is favoured by excess water and slow attainment, hence heat speeds up the attainment of equilibrium. But you are not going to get any extra yield of alcohol/acid unless you arrange for the alcohol to be distilled off as formed (not usually carried out), or for the acid to be neutralised as formed, hence the use of sodium hydroxide as the hydrolysing reagent.

    So for preparation of ester: alcohol + acid (concentrated acid & heat)

    For hydrolysis of ester: ester + excess water (strong NaOH & heat)
    So OH- will also function as a catalyst (just like H+) for the esterification reaction? But you would use concentrated H+ to favour production of water and thus production of the ester, whereas you would use water and OH- when hydrolysing the ester into the constituent carboxylic acid and alcohol, correct?
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    (Original post by Big-Daddy)
    So OH- will also function as a catalyst (just like H+) for the esterification reaction? But you would use concentrated H+ to favour production of water and thus production of the ester, whereas you would use water and OH- when hydrolysing the ester into the constituent carboxylic acid and alcohol, correct?
    If by saying "to favour the production of water" you mean to absorb the water as it is formed pulling the reaction to the side of the ester, then yes.
 
 
 
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