The hydrocarbon 'a' is ethane.

Ethane on chlorination, gives 1-chloroethane, which reacts with alcoholic KOH to give a dehydro-halogenation product (Ethene). The ethene formed from this undergoes ozonolysis to break the double bond and form two molecules of formaldehyde.

'a' - Ethane
'b' - 1-chloroethane
'c' - Ethene

Given information:
- Hydrocarbon A
- On chlorination gives B
- B on heating with alcoholic potassium hydroxide changes into C
- C decolourises Baeyer's reagent
- On ozonolysis, C forms formaldehyde only

To find: Hydrocarbon A

Solution:

Step 1: Chlorination of hydrocarbon A

- Hydrocarbon A is a saturated hydrocarbon since it undergoes chlorination.
- The reaction of A with chlorine is a substitution reaction in which one or more hydrogen atoms are replaced by chlorine atoms.
- The product B is a chlorinated hydrocarbon of A.

Step 2: Heating of B with alcoholic potassium hydroxide

- Alcoholic potassium hydroxide is a strong base that can cause elimination reactions in organic compounds.
- When B is heated with alcoholic potassium hydroxide, an elimination reaction takes place, and a double bond is formed between two carbon atoms.
- The product C is an unsaturated hydrocarbon.

Step 3: Baeyer's reagent test

- Baeyer's reagent is an alkaline solution of potassium permanganate. It is used to test for the presence of unsaturation in organic compounds.
- Unsaturated hydrocarbons react with Baeyer's reagent and decolourise it.
- Since C decolourises Baeyer's reagent, it must be an unsaturated hydrocarbon.

Step 4: Ozonolysis of C

- Ozonolysis is a chemical reaction in which ozone (O3) reacts with an unsaturated compound to form two or more products.
- The products of ozonolysis can be used to identify the structure of the unsaturated compound.
- When C is subjected to ozonolysis, it forms formaldehyde only.
- This indicates that C is an alkene with a double bond between two carbon atoms.

Conclusion:

- Hydrocarbon A is a saturated hydrocarbon that undergoes chlorination to form a chlorinated hydrocarbon B.
- On heating with alcoholic potassium hydroxide, B forms an unsaturated hydrocarbon C with a double bond between two carbon atoms.
- Since C decolourises Baeyer's reagent and forms formaldehyde only on ozonolysis, it must be an alkene.
- Therefore, hydrocarbon A must be ethane, which is the only saturated hydrocarbon among the given options.

You have to track the reaction products reversely like it is given that on ozonolysis of C it gives formaldehyde only so C must be ethene. and it is given that B on treatment with alcoholic KOH gives ethene so it means that B must be chloroethane which on elimination reaction gives ethene and so A must be ethane which on chlorination gives B chloroethane. so ans is (a) ethane