A compound 'A' formula of `C_(3)H_(6)Cl_(2)` on reaction with alkali can give 'B' of formula `C_(3)H_(6)O` or 'C' of formula `C_(3)H_(4). 'B'` on oxidation gave a compound of the formula `C_(3)H_(6)O_(2).'C'` with dilute `H_(2)SO_(4)` containing `Hg^(2+)` ion gave 'D' of formula `C_(3)H_(6)O` which with bromine and alkali gave the sodium salt of `C_(2)H_(4)O_(2)`. Then 'A' is

Options

A. `CH_(3)CH_(2)CHCl_(2)`
B. `CH_(3)C Cl_(2)CH_(3)`
C. `CH_(2)CICH_(2)CH_(2)Cl`
D. `CH_(3)CHCICH_(2)Cl`

Answer

To solve the problem, we need to analyze the given information step by step and identify the compound 'A' based on its reactions and products.

### Step 1: Identify Compound 'A'
The compound 'A' has the formula \( C_3H_6Cl_2 \). This indicates that it is a dichloro compound with three carbon atoms. The possible structure could be 1,2-dichloropropane or 1,3-dichloropropane.

### Step 2: Reaction with Alkali
When compound 'A' reacts with alkali, it can give either compound 'B' (with the formula \( C_3H_6O \)) or compound 'C' (with the formula \( C_3H_4 \)).

- If we consider 1,2-dichloropropane, it can undergo elimination to form propene (\( C_3H_6 \)) or can react with alkali to form an alcohol (\( C_3H_6O \)).
- If we consider 1,3-dichloropropane, it can also lead to similar products.

### Step 3: Analyze Compound 'B'
Compound 'B' has the formula \( C_3H_6O \). This suggests that it could be an alcohol or an aldehyde. If we assume it is propanal (an aldehyde), it can be oxidized to form a carboxylic acid.

- On oxidation, compound 'B' gives \( C_3H_6O_2 \), which is consistent with propanoic acid.

### Step 4: Analyze Compound 'C'
Compound 'C' has the formula \( C_3H_4 \). This indicates that it is likely an alkyne, specifically propyne (\( CH_3C \equiv CH \)).

- When propyne reacts with dilute \( H_2SO_4 \) in the presence of \( Hg^{2+} \), it undergoes hydration to form an alcohol, which corresponds to compound 'D' with the formula \( C_3H_6O \).

### Step 5: Analyze Compound 'D'
Compound 'D' is formed from compound 'C' and has the formula \( C_3H_6O \). This can be propan-2-ol or propanal.

- When compound 'D' reacts with bromine and alkali, it forms the sodium salt of \( C_2H_4O_2 \), which is sodium acetate.

### Conclusion
Based on the analysis, the structure of compound 'A' that fits all the reactions and products is **1,2-dichloropropane**.

Thus, the answer to the question is:
**Compound 'A' is 1,2-dichloropropane.**