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_(2)-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 Reactant
The reactant is butyne, specifically 1-butyne, which can be represented as:
\[ \text{CH}_3\text{CH}_2\text{C} \equiv \text{CH} \]

### Step 2: Reaction with Sodium Amide (NaNH₂)
Sodium amide (NaNH₂) is a strong base. When it reacts with 1-butyne, it abstracts the acidic hydrogen from the terminal alkyne (the hydrogen attached to the carbon of the triple bond). This results in the formation of a carbon anion:
\[ \text{CH}_3\text{CH}_2\text{C} \equiv \text{C}^- \]
Here, the negative charge is stabilized by the sodium ion (Na⁺).

### Step 3: Nucleophilic Attack on Ethyl Bromide (CH₃CH₂Br)
Next, we introduce ethyl bromide (CH₃CH₂Br) to the reaction. The carbon anion formed in the previous step acts as a nucleophile and attacks the carbon atom in ethyl bromide that is bonded to the bromine atom. This leads to the displacement of bromide ion (Br⁻) and the formation of a new carbon-carbon bond:
\[ \text{CH}_3\text{CH}_2\text{C} \equiv \text{C}^- + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{C} \equiv \text{C-CH}_2\text{CH}_3 + \text{NaBr} \]

### Step 4: Final Product Formation
The product formed from this reaction is an internal alkyne with six carbon atoms:
\[ \text{CH}_3\text{CH}_2\text{C} \equiv \text{C-CH}_2\text{CH}_3 \]
This compound is known as hex-3-yne.

### Conclusion
The final major product of the reaction is:
\[ \text{Hex-3-yne} \]