Compound 'A' on chlorination gives compound 'B'. 'B' reacts with alc. KOH to give gas 'C' which decolorizes Baeyer reagent. Ozonolysis of compound 'C' gives only HCHO compound. Compound 'A' is:

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