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This cheat sheet covers how esters form from carboxylic acids and alcohols, and how esters break down by hydrolysis. Students need it because esterification and ester hydrolysis combine organic reaction patterns, equilibrium ideas, acid-base conditions, and stoichiometry. Worked examples help connect reaction schemes to mole ratios, limiting reactants, percent yield, and product prediction.

It is designed as a quick reference for solving exam-style organic chemistry problems.

Key Facts

  • Fischer esterification follows the general reaction RCOOH+ROHRCOOR+H2O\mathrm{RCOOH + R'OH \rightleftharpoons RCOOR' + H_2O} under acid catalysis.
  • An acid catalyst such as H2SO4\mathrm{H_2SO_4} increases the reaction rate but is not consumed in the overall esterification reaction.
  • Fischer esterification is reversible, so removing H2O\mathrm{H_2O} or using excess alcohol shifts equilibrium toward ester formation by Le Châtelier's principle.
  • Acid hydrolysis of an ester follows RCOOR+H2ORCOOH+ROH\mathrm{RCOOR' + H_2O \rightleftharpoons RCOOH + R'OH} and is the reverse of Fischer esterification.
  • Base hydrolysis, or saponification, follows RCOOR+OHRCOO+ROH\mathrm{RCOOR' + OH^- \rightarrow RCOO^- + R'OH} and is usually effectively irreversible because a carboxylate ion forms.
  • For esterification stoichiometry, the mole ratio between carboxylic acid and alcohol is usually 1:11:1, so n=mMn=\frac{m}{M} is used to find the limiting reactant.
  • Percent yield is calculated with % yield=actual yieldtheoretical yield×100%\%\text{ yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100\%.
  • The ester name is built from the alcohol part first as an alkyl group, followed by the carboxylic acid part ending in -oate\text{-oate}.

Vocabulary

Ester
An ester is an organic compound containing the functional group RCOOR\mathrm{RCOOR'}.
Fischer esterification
Fischer esterification is the acid-catalyzed reaction of a carboxylic acid with an alcohol to form an ester and water.
Hydrolysis
Hydrolysis is a reaction in which water breaks a bond, such as converting an ester into a carboxylic acid and an alcohol under acidic conditions.
Saponification
Saponification is base hydrolysis of an ester that produces a carboxylate salt and an alcohol.
Equilibrium
Equilibrium is the state in a reversible reaction where the forward and reverse reaction rates are equal.
Limiting reactant
The limiting reactant is the reactant that is used up first and determines the maximum amount of product that can form.

Common Mistakes to Avoid

  • Forgetting that Fischer esterification is reversible is wrong because the ester yield depends on equilibrium, not just on the starting mole amounts.
  • Using NaOH\mathrm{NaOH} and writing a carboxylic acid product is wrong because base hydrolysis forms RCOO\mathrm{RCOO^-} first, not neutral RCOOH\mathrm{RCOOH}.
  • Naming the ester from the acid first is wrong because ester names place the alcohol-derived alkyl group first and the acid-derived -oate\text{-oate} part second.
  • Treating the acid catalyst as a reactant in stoichiometry is wrong because the catalyst speeds the reaction but does not set the theoretical yield.
  • Ignoring the 1:11:1 mole ratio in simple esterification is wrong because equal reaction coefficients mean moles, not grams, must be compared.

Practice Questions

  1. 1 Ethanoic acid reacts with ethanol to form ethyl ethanoate and water: CH3COOH+CH3CH2OHCH3COOCH2CH3+H2O\mathrm{CH_3COOH + CH_3CH_2OH \rightleftharpoons CH_3COOCH_2CH_3 + H_2O}. If 0.250mol0.250\,\mathrm{mol} of ethanoic acid reacts with excess ethanol, what is the theoretical amount of ester in moles?
  2. 2 A student forms 6.60g6.60\,\mathrm{g} of ethyl ethanoate, C4H8O2\mathrm{C_4H_8O_2}, from a theoretical yield of 8.80g8.80\,\mathrm{g}. Calculate the percent yield using % yield=actual yieldtheoretical yield×100%\%\text{ yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100\%.
  3. 3 Methyl propanoate reacts with aqueous NaOH\mathrm{NaOH}. Write the organic products of CH3CH2COOCH3+OH\mathrm{CH_3CH_2COOCH_3 + OH^- \rightarrow} after base hydrolysis.
  4. 4 Explain why adding excess alcohol can increase ester yield in Fischer esterification but does not change the identity of the ester formed.