CBSE Class 10 Science 2020 · Paper 31/1/3 · Question 17

(a) (i) Ethanol, ${\text{CH}}_3{\text{CH}}_2\text{OH}$: $$\begin{array}{ccccccc}& & \text{H}& & \text{H}& & \\ & & |& & |& & \\ \text{H}& -& \text{C}& -& \text{C}& -& \text{O}-\text{H}\\ & & |& & |& & \\ & & \text{H}& & \text{H}& & \end{array}$$ (ii) Ethanoic acid, ${\text{CH}}_3\text{COOH}$: $$\begin{array}{ccccccc}& & \text{H}& & \text{O}& & \\ & & |& & \Vert & & \\ \text{H}& -& \text{C}& -& \text{C}& -& \text{O}-\text{H}\\ & & |& & & & \\ & & \text{H}& & & & \end{array}$$ (b) When ethanol changes into ethanoic acid, oxygen is added to it (and two hydrogen atoms are removed), so the conversion is an oxidation. The oxidising agent used is alkaline potassium permanganate, ${\text{KMnO}}_4$ (or acidified potassium dichromate, ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$).