`H-C-=C-H+ 2NaNH_(2)rarr Aoverset("2 mole")underset(CH_(3)Cl)rarrB`
then 'B' is

Options

A. 1 - Butyne
B. 2 - Butyne
C. 2 - Pentyne
D. Propyne

Explanation

To solve the given question step by step, we will analyze the reactions taking place and determine the final product 'B'.

### Step 1: Identify the starting compound
The starting compound is **H-C≡C-H**, which is ethyne (acetylene).

### Step 2: Reaction with NaNH₂
When ethyne reacts with **2 moles of NaNH₂**, the sodium amide (NaNH₂) acts as a strong base. It abstracts a proton (H⁺) from the terminal carbon of ethyne, resulting in the formation of a carbanion (negatively charged carbon).

The reaction can be represented as:
\[ \text{H-C≡C-H} + \text{NaNH}_2 \rightarrow \text{H-C≡C}^- \text{Na}^+ + \text{NH}_3 \]

### Step 3: Formation of the carbanion
The carbanion formed is **H-C≡C⁻**. The negative charge on the carbon is stabilized due to its sp hybridization.

### Step 4: Reaction with CH₃Cl
Next, the carbanion reacts with **2 moles of CH₃Cl**. The carbanion acts as a nucleophile and attacks the electrophilic carbon in CH₃Cl, displacing the chloride ion (Cl⁻).

The reaction can be represented as:
\[ \text{H-C≡C}^- + \text{CH}_3\text{Cl} \rightarrow \text{H-C≡C-CH}_3 + \text{Cl}^- \]

### Step 5: Repeat the reaction
Since we have 2 moles of NaNH₂ and CH₃Cl, we can repeat the above reaction. The carbanion formed in the previous step can again react with another mole of CH₃Cl.

The second reaction is:
\[ \text{H-C≡C-CH}_3 + \text{CH}_3\text{Cl} \rightarrow \text{CH}_3\text{C≡C-CH}_3 + \text{Cl}^- \]

### Step 6: Final product
After both reactions, the final product 'B' is **CH₃C≡CCH₃**, which is **2-butyne**.

### Conclusion
Thus, the final product 'B' is **2-butyne**.

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