In the reaction :
`CH_(3)-C-=C-Hoverset((i)NaNH_(2)//NH_(3)(I))rarr(A)overset(CH_(3)-CH_(2)-CH_(2)-Br)rarr (B)`
the product (B) is :

Options

A. `H_(3)-C-=C-CH_(2)-CH_(2)-CH_(3)`
B. `CH_(3)-CH=CH_(2)`
C. `CH_(3)-CH_(2)-C-=C-CH_(2)-CH_(3)`
D. `CH_(3)-CH=C=CH-CH_(2)-CH_(3)`

Explanation

To solve the given reaction step by step, we will analyze the transformation from the starting material to product B.

### Step 1: Identify the Starting Material
The starting material is `CH3-C≡C-H`. This is an alkyne known as propyne.

### Step 2: Reaction with NaNH2/NH3
The first step involves treating propyne with sodium amide (NaNH2) in ammonia (NH3). Sodium amide acts as a strong base and will deprotonate the terminal hydrogen of the alkyne.

- **Reaction**:
\[ CH3-C≡C-H + NaNH2 \rightarrow CH3-C≡C^{-}Na^{+} + NH3 \]

Here, the terminal hydrogen (H) is removed, resulting in a negatively charged alkyne (acetylide ion) and the release of ammonia gas.

### Step 3: Formation of Product A
The product A is the acetylide ion, which can be represented as:
\[ A: CH3-C≡C^{-}Na^{+} \]

### Step 4: Reaction with Alkyl Halide
Next, the acetylide ion (A) reacts with an alkyl halide, specifically `CH3-CH2-CH2-Br` (1-bromopropane). The acetylide ion acts as a nucleophile and attacks the electrophilic carbon in the alkyl halide, leading to the substitution of the bromine atom.

- **Reaction**:
\[ CH3-C≡C^{-} + CH3-CH2-CH2-Br \rightarrow CH3-C≡C-CH2-CH2-CH3 + Br^{-} \]

### Step 5: Formation of Product B
After the nucleophilic attack, the product B is formed, which is:
\[ B: CH3-C≡C-CH2-CH2-CH3 \]

This product is an alkyne with a longer carbon chain.

### Final Answer
The product (B) is `CH3-C≡C-CH2-CH2-CH3`.

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