In the reaction `CH_3-C-=C-Hoverset((i) NaNH_2//NH_3(l))rarr(A)overset(CH_3-CH_2-CH_2-Br)rarr(B)` The product B is

Options

A. `CH_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`

Answer

To solve the given reaction step-by-step, we need to analyze the reactants and the reagents involved.

### Step 1: Identify the starting material
The starting material is propyne, which is represented as:
\[ \text{CH}_3-\text{C} \equiv \text{C}-\text{H} \]

### Step 2: Reaction with Sodium Amide (NaNH₂)
When propyne reacts with sodium amide (NaNH₂) in liquid ammonia (NH₃), it acts as a strong base. The sodium amide deprotonates the terminal hydrogen of propyne, leading to the formation of the sodium salt of propyne:
\[ \text{CH}_3-\text{C} \equiv \text{C}^- \text{Na}^+ \]
This is product A.

### Step 3: Nucleophilic Substitution with Alkyl Bromide
Next, we add an alkyl bromide, specifically butyl bromide (CH₃-CH₂-CH₂-Br), to product A. The negatively charged carbon (the terminal carbon of the alkyne) acts as a nucleophile and attacks the electrophilic carbon in the butyl bromide:
\[ \text{CH}_3-\text{C} \equiv \text{C}^- + \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{Br} \rightarrow \text{CH}_3-\text{C} \equiv \text{C}-\text{C} \text{H}_2\text{CH}_2\text{CH}_3 + \text{NaBr} \]

### Step 4: Final Product
The final product B, after the nucleophilic substitution, is:
\[ \text{CH}_3-\text{C} \equiv \text{C}-\text{C}-\text{H}_2-\text{CH}_2-\text{CH}_3 \]
This compound is a hexane derivative, specifically hex-2-yne.

### Conclusion
Thus, the product B is:
\[ \text{Hex-2-yne} \]

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