A compound is treated with `NaNH_(2)` to give sodium salt. Identify the compound-

Options

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

Answer

To solve the problem of identifying the compound that reacts with NaNH₂ to give a sodium salt, we can follow these steps:

### Step 1: Understand the Reactivity of NaNH₂
NaNH₂ (sodium amide) is a strong base and is known to deprotonate acidic hydrogens. It will react with compounds that have sufficiently acidic hydrogen atoms.

**Hint:** Look for compounds with acidic protons that can be removed by a strong base.

### Step 2: Identify Potential Compounds
Common compounds that can react with NaNH₂ include:
- Alkynes (like ethyne, C₂H₂)
- Alcohols
- Carboxylic acids
- Ammonium salts

Among these, alkynes are known to have acidic protons due to the sp-hybridized carbon atoms.

**Hint:** Focus on compounds that contain sp-hybridized carbons, as they have more acidic hydrogens.

### Step 3: Analyze Ethyne (C₂H₂)
Ethyne (C₂H₂) is an alkyne that has the structure H-C≡C-H. The hydrogen atoms attached to the terminal carbon are relatively acidic compared to those in alkanes or alkenes.

**Hint:** Consider the structure of ethyne and its acidic hydrogen atoms.

### Step 4: Reaction with NaNH₂
When ethyne (C₂H₂) is treated with NaNH₂, the strong base will deprotonate one of the hydrogen atoms, leading to the formation of sodium ethynide (HC≡C⁻Na⁺) and hydrogen gas (H₂).

The reaction can be represented as:
\[ \text{C}_2\text{H}_2 + \text{NaNH}_2 \rightarrow \text{HC}\equiv\text{C}^- \text{Na}^+ + \frac{1}{2} \text{H}_2 \]

**Hint:** Write down the balanced chemical equation to visualize the reaction.

### Step 5: Conclusion
Since ethyne is the compound that reacts with NaNH₂ to form a sodium salt (sodium ethynide), we can conclude that the compound in question is C₂H₂.

**Final Answer:** The compound is C₂H₂ (ethyne).