Chemistry 9701 · AS & A Level · Carboxylic acids and derivatives
Carboxylic acids and derivatives — practice question
A reaction sequence based on propanoic acid is shown. $\text{CH}_3\text{CH}_2\text{CO}_2\text{H}$ undergoes reaction 1 to give $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$. In reaction 2, $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ is changed back into $\text{CH}_3\text{CH}_2\text{CO}_2\text{H}$. Propanoic acid reacts with $\text{CaCO}_3$ to produce three substances. Reaction 3 converts $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ into $\text{CH}_3\text{CO}_2\text{CH}_2\text{CH}_2\text{CH}_3$.
(a)[2]
Write an equation for reaction 1, with $[\text{H}]$ representing the reducing agent.
(b(i))[1]
What sort of reaction is reaction 2?
(b(ii))[2]
Suggest a suitable reagent and conditions for reaction 2.
(c)[2]
Write an equation for the reaction between propanoic acid and calcium carbonate, $\text{CaCO}_3$.
(d(i))[2]
Suggest a suitable reagent and conditions for reaction 3.
(d(ii))[1]
Which other product is formed in reaction 3?
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Balanced equation: $\mathrm{CH_3CH_2CO_2H + 4[H] \rightarrow CH_3CH_2CH_2OH + H_2O}$” …