A hydrocarbon (A) on chlorinatio gives (B) ,which on reacting with alcoholic KOH changes into another hydrocarbon (C ) .The latter decolorizes Bayer's reagent and on ozonolysis forms formaldehyde only (A) is:

A hydrocarbon reacts with HI to give X which on reaction with aqueous KOH forms Y. Oxidation of Y gives 3-methyl, 2-butanone. The hydrocarbon is

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

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

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

A hydrocarbon C_6H_12 on oxonolysis gives only one product which does not reduce Fehling solution. The hydrocarbon is :

A neutral liquid A on treatment with Lucas reagent produces a compound B, which upon treatment with alcoholic KOH yields a compound C. The compound C decolourises bromine water and upon ozonolysis forms only methanal. Deduce the structures of A, B and C and explain the reactions.