CBSE Class 10 Science 2017 · Paper 31/1/1/C · Question 21

(a) (i) Oxidation (ethanol to ethanoic acid): ${\text{CH}}_3{\text{CH}}_2\text{OH}+2[\text{O}]\stackrel{{\text{alkaline KMnO}}_4\text{, heat}}{\to }{\text{CH}}_3\text{COOH}+{\text{H}}_2\text{O}$. (ii) Addition (ethene to ethane): ${\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\stackrel{\text{Ni catalyst}}{\to }{\text{CH}}_3-{\text{CH}}_3$. (iii) Substitution (methane to chloromethane): ${\text{CH}}_4+{\text{Cl}}_2\stackrel{\text{sunlight}}{\to }{\text{CH}}_3\text{Cl}+\text{HCl}$. (b) Ethanol, ${\text{C}}_2{\text{H}}_5\text{OH}$ (ethyl alcohol), is the second member of the homologous series of alcohols, with the functional group $-\text{OH}$. When it is heated at 443 K with excess concentrated sulphuric acid it is dehydrated to ethene: ${\text{CH}}_3{\text{CH}}_2\text{OH}\underset{443\text{K}}{\overset{{\text{hot conc. H}}_2{\text{SO}}_4}{\to }}{\text{CH}}_2={\text{CH}}_2+{\text{H}}_2\text{O}$. The concentrated sulphuric acid acts as a dehydrating agent: it removes water from ethanol.