`HC-=CHoverset("Excess")overset(NaNH_(2))rarrAoverset(C_(2)H_(5)Cl)rarrB` find the B.

Options

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

Answer

To solve the given problem, we need to follow the reaction steps involving the alkyne HC≡CH (ethyne) and the reagents NaNH₂ and C₂H₅Cl (ethyl chloride). Let's break it down step by step.

### Step 1: Reaction with NaNH₂
1. **Identify the acidic hydrogen**: The alkyne HC≡CH has acidic hydrogen atoms due to the sp-hybridized carbon atoms.
2. **Deprotonation**: When NaNH₂ (sodium amide) is added in excess, it acts as a strong base and abstracts the acidic hydrogen from the alkyne. This results in the formation of a carbanion.
\[ HC≡CH + NaNH₂ \rightarrow HC≡C^{-}Na^{+} + NH₃ \]
Here, we form the sodium acetylide (HC≡C⁻Na⁺).

### Step 2: Formation of A
3. **Formation of the carbanion**: Since NaNH₂ is in excess, it can deprotonate both hydrogen atoms from the alkyne, leading to the formation of a more stable carbanion:
\[ HC≡C^{-}Na^{+} + NaNH₂ \rightarrow HC≡C^{-}Na^{+} + NH₃ \]
This gives us the intermediate A, which is the sodium salt of the acetylide ion.

### Step 3: Reaction with C₂H₅Cl
4. **Nucleophilic substitution**: The carbanion (HC≡C⁻) is a strong nucleophile and will attack the electrophilic carbon in C₂H₅Cl (ethyl chloride). The chlorine atom is a good leaving group.
\[ HC≡C^{-} + C₂H₅Cl \rightarrow HC≡C-C₂H₅ + NaCl \]
Here, the nucleophile attacks the carbon attached to the chlorine, leading to the formation of a new carbon-carbon bond.

### Step 4: Formation of B
5. **Final product**: The final product B is butyne (C₄H₆), which can be represented as:
\[ HC≡C-C₂H₅ \]
This can also be written as:
\[ C₂H₅C≡C-H \]
Thus, the final product B is 1-butyne or 2-butyne depending on the orientation of the ethyl group.

### Conclusion
The final product B is:
\[ \text{B: } C₂H₅C≡C-H \text{ (1-butyne or 2-butyne)} \]

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