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) ->