AS Chemistry 9701: Forming and Hydrolysing Propyl Ethanoate

Propyl ethanoate, $\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3$ ( $M_r = 102$ ), can be made from propan-1-ol and ethanoic acid, and can also be hydrolysed back to its starting materials.

(a) Write an equation, using structural formulas, for the formation of propyl ethanoate from propan-1-ol and ethanoic acid, and name this type of reaction. [3]

(b) State the role of the concentrated sulfuric acid used in this reaction. [1]

(c) A $5.10\ \text{g}$ sample of propyl ethanoate is refluxed with an excess of aqueous sodium hydroxide until hydrolysis is complete. Calculate the maximum mass of sodium ethanoate, $\text{CH}_3\text{COONa}$ ( $M_r = 82$ ), formed. [3]

(d) Explain, in terms of the position of equilibrium, why hydrolysis of propyl ethanoate with aqueous sodium hydroxide goes to completion, whereas hydrolysis with dilute aqueous acid does not. [2]

Part (a): Writing the esterification equation

Propan-1-ol and ethanoic acid condense together, with loss of a water molecule, to form propyl ethanoate:

$\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}$

This is an esterification reaction (a type of condensation reaction), and, because it is catalysed by an acid rather than driven to completion, it reaches a position of equilibrium rather than going fully to products, hence the $\rightleftharpoons$ sign.

Part (b): Role of concentrated sulfuric acid

The concentrated sulfuric acid acts as a catalyst: it speeds up the rate at which the reaction reaches equilibrium, without itself being used up or appearing in the overall equation.

Part (c): Calculating the mass of sodium ethanoate

First, convert the mass of ester to moles using its $M_r$ : $n(\text{propyl ethanoate}) = \frac{5.10}{102} = 0.0500\ \text{mol}$

Alkaline hydrolysis of an ester is a $1:1$ reaction with hydroxide ions, and produces one mole of the sodium salt of the acid per mole of ester: $\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$

So $n(\text{CH}_3\text{COONa}) = n(\text{ester}) = 0.0500\ \text{mol}$ .

Converting back to a mass using $M_r(\text{CH}_3\text{COONa}) = 82$ : $m = 0.0500 \times 82 = 4.10\ \text{g}$

(Check: since sodium ethanoate has a smaller $M_r$ than propyl ethanoate, 82 vs 102, the product mass should be somewhat less than the starting mass. $4.10\ \text{g} < 5.10\ \text{g}$ , which is consistent.)

Part (d): Why alkaline hydrolysis goes to completion

Acid-catalysed hydrolysis is simply the reverse of esterification: it is governed by the same equilibrium, so refluxing an ester with dilute aqueous acid only ever produces an equilibrium mixture of ester, water, acid and alcohol. It does not go to completion.

Alkaline hydrolysis is different because the hydroxide ion reacts with the ester to directly form a carboxylate ion (here, the ethanoate ion) rather than the neutral carboxylic acid. The carboxylate ion is stabilised (its negative charge is delocalised) and has no tendency to react back with the alcohol to re-form the ester. Because the products are effectively removed from the equilibrium, and $\text{OH}^-$ is continually consumed, the reaction is driven to completion in one direction.

Final answers

(a) $\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}$ ; esterification.
(b) Catalyst, increases the rate of reaching equilibrium without being consumed.
(c) $4.10\ \text{g}$ of sodium ethanoate.
(d) Alkaline hydrolysis forms an unreactive, stabilised carboxylate ion so the reaction goes to completion; acid hydrolysis is the reverse of a reversible esterification equilibrium, so it does not.