Fischer Esterification Lab Report

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For this experiment, a Fischer esterification reaction was observed. A Fischer esterification is a reaction that converts a carboxylic acid into an ester. Within the reaction, the hydroxyl portion, -OH, of the carboxylic acid is replaced by an -OR group. The byproduct is water which is also a nucleophile. Therefore, water can be added back into the compound and undergo hydrolysis on the newly formed ester which produces the starting carboxylic acid. To make sure the reversibility did not occur, the reaction mixture was heated to force the water to evaporate and therefore be removed from the overall reaction. The main idea of Fischer esterification is to form a carboxylic acid and make it a better electrophile under acidic conditions. This is …show more content…

Fischer was a well-known chemist who won the second Nobel Prize in chemistry. He is also notable for Fischer projections. His method was to run the reaction under acidic conditions. To do this, the oxygen was protonated to make it a good nucleophile. The carbon is then strong enough to react with the alcohol. This can then go on to make a good leaving group, water, which comes off. In the last step, a proton is removed. His method was ran under acidic conditions to make the electrophile a better electrophilic carbon. This reaction works very well.
An ester is derived from a carboxylic acid. The oxygen of the alcohol is the nucleophile and the carbonyl carbon of the carboxylic acid is the electrophile. The oxygen makes the carbon more electrophilic causing the alcohol to attack since the oxygen does not want to be positively charged. The solvent used was concentrated sulfuric acid which promoted the addition of the weak nucleophile, the alcohol of the carboxylic acid, to be replaced with an -OR group. A proton is gained back from the hydroxyl group of the carboxylic acid after being lost. After this, a proton shift occurs that gives up the leaving group of water which is the byproduct since it …show more content…

The first is the proton source. A certain one must be used. It is typically done with sulfuric acid, but hydrochloric acid can also be used. A strong base is not used so a strong acid is utilized. To make enough product, a strong acid is used to push the reactants into products to produce enough product. Phosphoric acid is an exception in that it can also be used but a strong acid is preferred to make the reaction go to completion. The second problem deals with equilibrium which is the most important. If the reaction was performed with an alcohol, methanol or ethanol, and a carboxylic acid, acetic acid, the equilibrium constant, Keq, would be 4. The Keq is the concentration of products divided by the concentration of reactants. In this case it would be ([ester][water])/([alcohol][acid]). The two products are ester and water and the two reactants are alcohol and carboxylic acid. The concentration of alcohol and ester are unknown, but they are equal to one another. The reactants are essentially the same and they should be in a 1:1 ratio to keep the concentrations equal to one another. In this reaction, a lot of waste is unwanted which this would produce excess waste. By producing a lot of waste, reactant, it increases the cost of waste disposal, so this method is not used. The main goal is to get the reactants to be equal to one another, so the reactants are (1-x). The

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