The final major product of the following reaction is :
`CH_(3)-CH_(2)-C-=C-H overset((1)NaNH_(2)) underset((2)CH_(3)CH_(2)Br) to`

Options

A. `CH_(3)-CH-C-=C-CH_(2)-CH_(3)`
B. `CH_(3)-underset(CH_(2)CH_(3))underset(|)(CH)-C-=C-H`
C. `CH_(3)-CH_(2)- underset(NH_(2)) underset(|)(CH)-CH_(3)`
D. `H_(3)C-CH_(2)-C-=C-NH_(2)`

Explanation

To determine the final major product of the given reaction, we will follow these steps:

### Step 1: Identify the Reactants
The reactant provided is 1-butyne (CH₃-CH₂-C≡C-H) and the reagents are sodium amide (NaNH₂) and ethyl bromide (CH₃CH₂Br).

### Step 2: Role of Sodium Amide (NaNH₂)
Sodium amide is a strong base. It will abstract the acidic hydrogen from the terminal alkyne (the hydrogen attached to the sp-hybridized carbon). This results in the formation of a terminal alkyne anion.

**Reaction:**
\[ \text{CH}_3\text{-CH}_2\text{-C}\equiv\text{C-H} + \text{NaNH}_2 \rightarrow \text{CH}_3\text{-CH}_2\text{-C}\equiv\text{C}^- + \text{NH}_3 \]

### Step 3: Formation of the Anion
The anion formed is:
\[ \text{CH}_3\text{-CH}_2\text{-C}\equiv\text{C}^- \]

### Step 4: Nucleophilic Attack
The anion will act as a nucleophile and attack the electrophilic carbon in ethyl bromide (CH₃CH₂Br). The bromine atom is a good leaving group, which will leave as bromide ion (Br⁻).

**Nucleophilic Attack:**
\[ \text{CH}_3\text{-CH}_2\text{-C}\equiv\text{C}^- + \text{CH}_3\text{-CH}_2\text{Br} \rightarrow \text{CH}_3\text{-CH}_2\text{-C}\equiv\text{C-CH}_2\text{-CH}_3 + \text{Br}^- \]

### Step 5: Final Product
The final major product of this reaction is:
\[ \text{CH}_3\text{-CH}_2\text{-C}\equiv\text{C-CH}_2\text{-CH}_3 \]
This compound is 1-pentyne.

### Conclusion
Thus, the final major product of the reaction is 1-pentyne.

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