`CH-=Choverset("Excess "NaNH_(2))toAoverset(C_(2)H_(5)Cl)toB`, find the B.

Options

A. `C_(2)H_(5)-C-=C-C_(2)H_(5)`
B. `HC-=C-C_(2)H_(5)`
C. `H_(2)C=CH-C_(2)H_(5)`
D. `CiHC=CHC_(2)H_(5)`

Answer

To solve the question, we need to follow the steps of the reaction involving the alkyne and the reagents provided. Let's break it down step by step.

### Step 1: Identify the starting material
The starting material is an alkyne, specifically ethyne (acetylene), which can be represented as:
\[ \text{CH} \equiv \text{CH} \]

### Step 2: React with sodium amide (NaNH₂)
When ethyne reacts with excess sodium amide (NaNH₂), it acts as a strong base and abstracts a proton (H⁺) from the terminal carbon of the alkyne. This results in the formation of a carbanion (C⁻) at that carbon:
\[ \text{CH} \equiv \text{C} + \text{NaNH}_2 \rightarrow \text{C} \equiv \text{C}^- + \text{NH}_3 \]

Since excess NaNH₂ is used, the carbanion can be formed again by deprotonating the terminal carbon:
\[ \text{C} \equiv \text{C}^- + \text{NaNH}_2 \rightarrow \text{C} \equiv \text{C}^{2-} + \text{NH}_3 \]

### Step 3: Formation of the nucleophile
Now, we have a di-anionic species (C≡C²⁻), which can act as a nucleophile.

### Step 4: React with ethyl chloride (C₂H₅Cl)
Next, we react this di-anionic species with ethyl chloride (C₂H₅Cl). The nucleophile (C≡C²⁻) will attack the electrophilic carbon in ethyl chloride, leading to the substitution of the chloride ion (Cl⁻):
\[ \text{C} \equiv \text{C}^{2-} + \text{C}_2\text{H}_5\text{Cl} \rightarrow \text{C} \equiv \text{C} - \text{C}_2\text{H}_5 + \text{Cl}^- \]

### Step 5: Identify the products
After the reaction, we have two products:
- **A**: The intermediate formed after the first reaction with NaNH₂, which is the alkyne with a carbanion (C≡C²⁻).
- **B**: The final product after the reaction with ethyl chloride, which is 1-butyne:
\[ \text{B} = \text{C}_2\text{H}_5\text{C} \equiv \text{C} \]

### Final Answer
- **A**: C≡C²⁻ (the di-anionic species)
- **B**: 1-butyne (C₂H₅C≡C)