Let us work backwards from the final product to identify compound 'A':

1. Ozonolysis of compound 'C' gives only methanal ($\text{HCHO}$), which indicates that 'C' is a symmetrical alkene with a double bond at the terminal carbon atoms. Therefore, 'C' must be ethene ($\text{CH}_2=\text{CH}_2$) .
2. Ethene ('C') is a gas that decolourises Baeyer's reagent (cold, dilute aqueous $\text{KMnO}_4$), confirming the presence of a double bond .
3. Compound 'C' (ethene) is formed by the reaction of compound 'B' with alcoholic KOH. This is a dehydrohalogenation ($\beta$-elimination) reaction. Thus, 'B' must be an alkyl halide, specifically chloroethane ($\text{C}_2\text{H}_5\text{Cl}$) .
4. Compound 'B' (chloroethane) is formed by the chlorination of compound 'A'. Therefore, 'A' must be the corresponding alkane, which is ethane ($\text{C}_2\text{H}_6$) .

The series of reactions is: $\text{C}_2\text{H}_6 \xrightarrow{\text{Cl}_2/h\nu} \text{C}_2\text{H}_5\text{Cl} \xrightarrow{\text{alc. KOH}} \text{C}_2\text{H}_4 \xrightarrow{\text{O}_3, \text{Zn}/\text{H}_2\text{O}} 2\text{HCHO}$