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:

Options

A. `C_(2)H_(6)`
B. `C_(2)H_(4)`
C. `C_(4)H_(8)`
D. `C_(2)H_(5)Cl`

Answer

To determine the compound 'A', we will analyze the information given step by step.

### Step 1: Identify Compound 'A'
The problem states that compound 'A' undergoes chlorination to yield compound 'B'. Chlorination typically occurs with alkanes, resulting in haloalkanes.

Assuming 'A' is an alkane, we can start with the simplest alkane, ethane (C2H6).

### Step 2: Chlorination of Compound 'A'
When ethane (C2H6) is chlorinated (in the presence of UV light), it forms chloroethane (C2H5Cl) as compound 'B':
\[ \text{C2H6} + \text{Cl2} \xrightarrow{h\nu} \text{C2H5Cl} + \text{HCl} \]

### Step 3: Reaction of Compound 'B' with Alcoholic KOH
Next, compound 'B' (chloroethane) reacts with alcoholic KOH. This reaction leads to elimination, forming ethene (C2H4), which is a gas:
\[ \text{C2H5Cl} + \text{KOH (alc)} \rightarrow \text{C2H4} + \text{KCl} + \text{H2O} \]

### Step 4: Gas 'C' Decolorizing Baeyer Reagent
The gas 'C' is ethene (C2H4), which can decolorize Baeyer’s reagent (KMnO4) due to its ability to undergo oxidation to form ethylene glycol. The reaction is:
\[ \text{C2H4} + \text{KMnO4} \rightarrow \text{C2H6O2} + \text{MnO2} \]
This decolorizes the purple color of KMnO4, forming a brown precipitate (MnO2).

### Step 5: Ozonolysis of Compound 'C'
Finally, ozonolysis of ethene (C2H4) yields formaldehyde (HCHO):
\[ \text{C2H4} + \text{O3} \rightarrow 2 \text{HCHO} \]
This confirms that compound 'C' gives only HCHO upon ozonolysis.

### Conclusion
Based on the steps above, the compound 'A' is ethane (C2H6).

### Final Answer:
**Compound 'A' is C2H6 (ethane).**

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