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Alcohols and Esters: Question 8

Syllabus 18.2

Structured AS 8 marks

Methyl propanoate, CH3CH2COOCH3\text{CH}_3\text{CH}_2\text{COOCH}_3, can be hydrolysed under acidic or alkaline conditions.

(a) Name the alcohol and the carboxylic acid from which methyl propanoate could be formed by esterification. [2]

(b) Write an equation, using structural formulas, for the acid hydrolysis of methyl propanoate using dilute sulfuric acid under reflux, and state whether this reaction goes to completion. [3]

(c) Write an equation, using structural formulas, for the alkaline hydrolysis of methyl propanoate using aqueous sodium hydroxide, and name the two organic products formed. [3]

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Worked solution

Part (a): Identifying the alcohol and acid

Methyl propanoate, CH3CH2COOCH3\text{CH}_3\text{CH}_2\text{COOCH}_3, is named from its acid part (“propanoate”, from propanoic acid, CH3CH2COOH\text{CH}_3\text{CH}_2\text{COOH}) and its alcohol part (“methyl”, from methanol, CH3OH\text{CH}_3\text{OH}). It could be formed by esterifying these two together, catalysed by an acid:

CH3CH2COOH+CH3OH⇌CH3CH2COOCH3+H2O\text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{OH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{H}_2\text{O}

Part (b): Acid hydrolysis

Refluxing methyl propanoate with dilute sulfuric acid simply reverses the esterification reaction, breaking the ester back down into the carboxylic acid and the alcohol:

CH3CH2COOCH3+H2O⇌CH3CH2COOH+CH3OH\text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{OH}

Because acid hydrolysis is governed by the same equilibrium as esterification, this reaction is reversible and does not go to completion. Refluxing produces an equilibrium mixture of ester, water, acid and alcohol rather than complete conversion.

Part (c): Alkaline hydrolysis

Refluxing methyl propanoate with aqueous sodium hydroxide hydrolyses the ester, but the hydroxide ion reacts to form the sodium salt of the carboxylic acid rather than the free acid itself:

CH3CH2COOCH3+NaOH→CH3CH2COONa+CH3OH\text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{CH}_3\text{OH}

The two organic products are sodium propanoate, CH3CH2COONa\text{CH}_3\text{CH}_2\text{COONa}, and methanol, CH3OH\text{CH}_3\text{OH}.

Final answers

  • (a) Methanol, CH3OH\text{CH}_3\text{OH}, and propanoic acid, CH3CH2COOH\text{CH}_3\text{CH}_2\text{COOH}.
  • (b) CH3CH2COOCH3+H2O⇌CH3CH2COOH+CH3OH\text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{OH}; reversible, does not go to completion.
  • (c) CH3CH2COOCH3+NaOH→CH3CH2COONa+CH3OH\text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{CH}_3\text{OH}; products are sodium propanoate and methanol.