Chemistry 5070 · O Level · Carboxylic acids

Carboxylic acids — practice question

Propanoic acid may be written using the formula $\text{CH}_3\text{CH}_2\text{COOH}$.
(a(i))[2]

When propanoic acid reacts with methanol, $\text{CH}_3\text{OH}$, an ester is produced. Name the ester formed and draw its displayed formula.

(a(ii))[2]

Propanoic acid is a weak acid. $\text{CH}_3\text{CH}_2\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_3\text{O}^+$ Explain how this equation shows that: • $\text{CH}_3\text{CH}_2\text{COOH}$ is an acid by referring to proton transfer • $\text{CH}_3\text{CH}_2\text{COOH}$ is a weak acid.

(a(iii))[2]

Propanoic acid reacts with magnesium. Name both products of this reaction.

(a(iv))[2]

Magnesium is a solid at room temperature. Describe the motion and separation of the particles in a solid.

(b(i))[3]

Fig. 7.1 presents the simplified structures of two molecules that react to make a polyamide. Complete Fig. 7.2 to show the structure of two repeat units of this polyamide. Show every atom and every bond in the linkages.

(b(ii))[2]

Polyamides are polymers. State what is meant by the term polymer.

(b(iii))[1]

Polyamides are condensation polymers. State one way in which condensation polymerisation differs from addition polymerisation.

Worked solution & mark scheme

This 14-mark question has a full step-by-step worked solution and mark scheme. One marking point: methyl propanoate

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