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I'm following this tutorial for synthesizing esters. I don't understand why adding water to the separation funnel doesn't push the equilibrium reaction point back to the left, or why draining it doesn't push the reaction to the right, given the tutorial's explanation of Le Chatelier's Principle:

When the reaction reaches equilibrium there is still a large amount of reactants left in the mixture resulting in a poor yield of the ester.

The yield of ester can be improved by increasing the concentration of one of the reactants (either the alcohol or the carboxylic acid). By Le Chatelier's Principle an excess of one reactant will drive the reaction to the right, increasing the production of ester, and therefore increasing the yield of ester. In our experiment we are using an excess of carboxylic aicd (acetic acid which is also known as ethanoic acid).

From the balanced chemical equation, you will also appreciate that the presence of water in the reaction mixture will drive the equilibrium to the left, favouring the formation of reactants rather than ester.

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    $\begingroup$ It is a question of temperature. If water is added at the temperature of the reaction, the back reaction is favored, and the ester will be hydrolyzed. But if water is added to the ester at room temperature, the reaction will be so slow that no visible reaction will happen during the time of the operation. $\endgroup$ – Maurice Oct 21 '20 at 18:56
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    $\begingroup$ You are preparing ethyl acetate at elevated temperature in the presence of H2SO4 as a catalyst. The extraction in the presence of water is at room temperature. Huge difference in rate and contact time. BTW the tutorial states that acetic acid is in excess. The volumes of ethanol and acetic acid say otherwise. Ethanol (~0.25 mol); acetic acid (~0.16 mol). Ethyl acetate is partially soluble in water (8.3g/100mL). You should saturate the aqueous layer with salt. $\endgroup$ – user55119 Oct 21 '20 at 19:03
  • $\begingroup$ @user55119, why saturate the water with salt? Thanks $\endgroup$ – CL22 Oct 21 '20 at 20:06
  • $\begingroup$ You have 2 layers, ethyl acetate and water. But the former is partially soluble in the latter. By adding NaCl and forming a brine solution, the ethyl acetate becomes less soluble in the aqueous phase there by increasing the yield of the ester. $\endgroup$ – user55119 Oct 21 '20 at 23:08
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Draining the aqueous phase cannot change anything. If the phases are in equillibrium before (both internally and with each other), they stay that way if they are separated.

Now if you add water to the aqueous phase, the reaction in the unpolar phase (which is saturated with water, as the phases are in equillibrium) is in the first moment not affected.

However the acid, alcohol and ester have different solubilities in the aqueous phase. That means the concentrations in the unpolar phase will change a bit, and so does the reaction equillibrium.

But: This is all irrelevant, because the reaction happens under reflux. While you are standing there with the cooled mixture in the separation funnel, the reaction rate is much lower, you will never notice a change.

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