Answer: A. Work backwards from ethyl ethanoate, \(\text{CH}_3\text{COOC}_2\text{H}_5\). It forms by esterification of ethanol with ethanoic acid, so Y and Z are these two compounds.
Y is oxidised by acidified \(\text{KMnO}_4\) to give Z. Primary alcohols oxidise to carboxylic acids, so Y = ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) and Z = ethanoic acid (\(\text{CH}_3\text{COOH}\)).
X reacts with steam to form Y (ethanol). The standard hydration of an alkene gives an alcohol: \(\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}\). So X = ethene, \(\text{C}_2\text{H}_4\).
The empirical formula of \(\text{C}_2\text{H}_4\) (divide by 2) is \(\text{CH}_2\) — option A.
Every O Level Pure Chemistry TYS MCQ solution video, plus full Sec 3 / Sec 4 topic recordings.
Find out more