Propyne and propene can be distinguished by
`conc. H_2SO_4`
`Br_2` in `"CCl"_4`
dil. `KMnO_4`
`AgNO_3` in ammonia

Options

A. `conc. H_2SO_4`
B. `Br_2` in `"CCl"_4`
C. dil. `KMnO_4`
D. `AgNO_3` in ammonia

Answer

To distinguish between propyne (C₃H₄) and propene (C₃H₆), we can analyze each of the given reagents step by step:

### Step 1: Reaction with Concentrated H₂SO₄
- **Propyne (C₃H₄)**: When propyne reacts with concentrated sulfuric acid (H₂SO₄) in the presence of mercuric sulfate (HgSO₄), it undergoes hydration to form a ketone (propanone).

\[ \text{C₃H₄} + H₂SO₄ \xrightarrow{HgSO₄} \text{CH₃C(=O)CH₃} \quad (\text{Propanone}) \]

- **Propene (C₃H₆)**: When propene reacts with concentrated sulfuric acid, it forms an alkyl hydrogen sulfate, which can further hydrolyze to form an alcohol (propan-2-ol).

\[ \text{C₃H₆} + H₂SO₄ \rightarrow \text{CH₃CH(OSO₃H)CH₃} \quad (\text{Alkyl hydrogen sulfate}) \]

**Conclusion**: Both reactions lead to different products, but there is no observable physical change, so they cannot be distinguished by this method.

### Step 2: Reaction with Br₂ in CCl₄
- **Propyne (C₃H₄)**: When propyne reacts with bromine (Br₂) in carbon tetrachloride (CCl₄), it adds across the triple bond to form a dibromide.

\[ \text{C₃H₄} + 2 \text{Br₂} \rightarrow \text{C₃H₄Br₂} \]

- **Propene (C₃H₆)**: When propene reacts with bromine in CCl₄, it also adds across the double bond to form a dibromide.

\[ \text{C₃H₆} + \text{Br₂} \rightarrow \text{C₃H₄Br₂} \]

**Conclusion**: Both compounds react with bromine and decolorize the bromine solution, so they cannot be distinguished by this method.

### Step 3: Reaction with Dilute KMnO₄
- **Propyne (C₃H₄)**: When propyne reacts with dilute potassium permanganate (KMnO₄), it undergoes oxidative cleavage to form carboxylic acids.

\[ \text{C₃H₄} + \text{KMnO₄} \rightarrow \text{CH₃COO⁻} + \text{HCOO⁻} \]

- **Propene (C₃H₆)**: When propene reacts with dilute KMnO₄, it also undergoes oxidation to form an alcohol.

\[ \text{C₃H₆} + \text{KMnO₄} \rightarrow \text{CH₃CH(OH)CH₃} \]

**Conclusion**: Both propyne and propene react with dilute KMnO₄ and decolorize the purple color, so they cannot be distinguished by this method.

### Step 4: Reaction with AgNO₃ in Ammonia
- **Propyne (C₃H₄)**: When propyne reacts with silver nitrate (AgNO₃) in ammonia, it forms a white precipitate of silver acetylide.

\[ \text{C₃H₄} + \text{AgNO₃} \rightarrow \text{AgC≡C} + \text{other products} \]

- **Propene (C₃H₆)**: When propene reacts with AgNO₃ in ammonia, no reaction occurs because propene does not have an acidic hydrogen to react with.

**Conclusion**: Propyne forms a white precipitate, while propene does not react. Therefore, this method can successfully distinguish between propyne and propene.

### Final Answer
The correct answer is **AgNO₃ in ammonia** as it can distinguish between propyne and propene.

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