A hydrocarbon A on chlorination gives B which on heating with alcoholic potassium hydroxide changes into another hydrocarbon C. The latter decolourises Baeyer's reagent and on ozonolysis forms formaldehyde only. A is

Given 2 O /H O Cl alc./KOH 3 2 2 A B C CH O Hydrocarbon → → → Since hydrocarbon C give only CH₂O on ozonolysis, C should be CH₂ = CH₂ hence going backward A should be ethane. Thus the reactions are Cl /hv 2 alc. CH CH CH CH Cl 3 3 3 2 KOH (A) B → → O /H O 3 2 CH CH HCHO 2 2 (D) →