A colourless liquid `A(b.p. 184^(@)C)` is sparingly soluble in warm water to which it gives feebly alkaline. On treating with `NaNO_(2)` and dil `HCl` in the cold solution, it yields a solution which reacts with an alkaline solution of `beta`-naphthol to give an orange yellow precipitate. Compound A is -

Options

A. `C_(6)H_(5)N_(2)Cl`
B. `C_(6)H_(5)NHNH_(2)`
C. `n - C_94)H_(9)NH_(2)`
D. `C_(6)H_(5)NH_(2)`

Answer

To determine the identity of compound A, we can follow these steps:

### Step 1: Analyze the properties of compound A
- Given that compound A is a colorless liquid with a boiling point of 184°C and is sparingly soluble in warm water, we can infer that it is likely an organic compound. The feebly alkaline nature suggests that it may contain an amine functional group.

**Hint:** Look for organic compounds that are colorless liquids with a boiling point around 184°C and have amine characteristics.

### Step 2: Consider the reaction with NaNO2 and dilute HCl
- The reaction of compound A with sodium nitrite (NaNO2) and dilute hydrochloric acid (HCl) indicates that it is likely an aromatic amine. This reaction typically leads to the formation of a diazonium salt.

**Hint:** Recall that aromatic amines react with NaNO2 and HCl to form diazonium salts.

### Step 3: Identify the diazonium salt formed
- When aniline (C6H5NH2) reacts with NaNO2 and HCl, it forms benzene diazonium chloride (C6H5N2Cl). This is a key intermediate in many organic reactions.

**Hint:** Consider the structure of aniline and how it transforms into a diazonium salt upon treatment with nitrous acid.

### Step 4: Reaction with beta-naphthol
- The formed diazonium salt can react with beta-naphthol (C10H7OH) in an alkaline medium to produce an orange-yellow precipitate, which is a characteristic azo dye.

**Hint:** Remember that diazonium salts can couple with phenolic compounds to form azo dyes, which are often brightly colored.

### Step 5: Conclusion
- Since the properties and reactions align with those of aniline, we conclude that compound A is aniline (C6H5NH2).

**Final Answer:** Compound A is aniline (C6H5NH2).