Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic $11.1$ — Formulae, functional groups and terminology (Parts $\mathrm{(a)}$, $\mathrm{(c)}$, $\mathrm{(d)}$)
• Topic $11.6$ — Alcohols (Part $\mathrm{(b)}$)
• Topic $11.8$ — Polymers (Part $\mathrm{(e)}$)
(a)(i) $C_{n}H_{2n+1}OH$
(a)(ii) Same functional group (they also share similar chemical properties and display a trend in physical properties).
(b)(i)
Method 1 (glucose): Temp = $25\text{-}35^{\circ}\text{C}$. Conditions: yeast (catalyst), absence of oxygen.
Method 2 (ethene): Temp = $300^{\circ}\text{C}$. Conditions: acid catalyst, $60\text{ atm}$ pressure.
(b)(ii) $C_{6}H_{12}O_{6} \rightarrow 2C_{2}H_{5}OH + 2CO_{2}$
(b)(iii) $C_{2}H_{4} + H_{2}O \rightarrow C_{2}H_{5}OH$
(c)(i) $C_{4}H_{10}O_{2}$
(c)(ii) $2$
Explanation: Butane-1,4-diol has two hydroxyl ($-OH$) groups, so it requires two moles of a monoprotic acid (ethanoic acid) to fully esterify both ends.
(d)(i) $C_{2}H_{3}O_{2}$
Explanation: The molecular formula of butanedioic acid is $C_{4}H_{6}O_{4}$. Dividing by the greatest common divisor ($2$) yields the simplest ratio $C_{2}H_{3}O_{2}$.
(d)(ii) Hydrogen
Explanation: Sodium is a reactive metal that displaces hydrogen from carboxylic acids.
(e)(i) A diagram showing a polyester sequence is required. It should show the ester linkage explicitly drawn as $-O-C(=O)-$ alternating with the carbon backbone blocks from the diol and dicarboxylic acid. To show two repeat units, there should be three fully displayed inter-block ester links with the correct orientation and continuation bonds at each end.
(e)(ii) Condensation (polymerisation)
Explanation: The reaction between a diol and a dicarboxylic acid releases a small molecule (water) as the ester linkages form, which is the defining characteristic of condensation polymerisation.
