Compound 'A' on chlorination gives compound 'B', compound 'B' reacts with alc. KOH gives gas 'C', which decolourizes Baeyer's reagent and ozonolysis of compound 'C' gives only HCHO compound. Then compound 'A' is:

C_{2} H_{6}

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C_{2} H_{4}

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C_{4} H_{10}

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C_{2} H_{5} C l

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Solution

The correct option is A $C_{2} H_{6}$
The compound A is ethylene $C_{2} H_{6}$ .
It on chlorination gives compound B which is ethyl chloride.

$C_{2} H_{6} A c h l o r i n a t i o n - ------ \to C_{2} H_{5} C l B$

On dehydrohalogenation with alc. KOH, ethyl chloride gives ethylene (gas C).
$C_{2} H_{5} C l B \to C H_{2} = C H_{2} C$

Ethylene decolourizes Baeyer reagent.
$C_{2} H_{4} + H_{2} O + [O] \to C H_{2} O H - C H_{2} O H$

Ozonolysis of ethylene gives formaldehyde.
$C H_{2} = C H_{2} O z o n o l y s i s - ----- \to 2 H C H O$ .

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