In the reaction
`H-C-=CH underset((2)CH_3CH_2Br)overset((1)NaNH_2//liq.NH_3)toXunderset((2)CH_3CH_2Br)overset((1)NaNH_2//liq.NH_3)toY`
X and Y are

Options

A. X=1-Butyne : Y=3-Hexyne
B. X=2-Butyne : Y=3-Hexyne
C. X=2-Butyne : Y=2-Hexyne
D. X=1-Butyne : Y=2-Hexyne

Explanation

To solve the given problem, we will analyze the reaction step by step.

### Step 1: Identify the starting compound
The starting compound is an alkyne, specifically ethyne (acetylene), which has the structure:
\[ H-C \equiv C-H \]

### Step 2: First reaction with NaNH₂ in liquid ammonia
When ethyne reacts with sodium amide (NaNH₂) in liquid ammonia, it acts as a strong base and deprotonates one of the hydrogen atoms. This results in the formation of an acetylide anion:
\[ H-C \equiv C^- \]

### Step 3: Nucleophilic attack on ethyl bromide
The acetylide anion then acts as a nucleophile and attacks ethyl bromide (CH₃CH₂Br). The bromine atom is displaced, leading to the formation of 1-butyne:
\[ H-C \equiv C-CH₂CH₃ \]
This compound is 1-butyne.

### Step 4: Second reaction with NaNH₂ in liquid ammonia
Next, we take the product from the first reaction (1-butyne) and treat it again with sodium amide in liquid ammonia. The strong base will again deprotonate the terminal hydrogen of 1-butyne, forming another acetylide anion:
\[ H-C \equiv C-CH₂CH₃ \rightarrow C \equiv C-CH₂CH₃^- \]

### Step 5: Nucleophilic attack on ethyl bromide again
The newly formed acetylide anion will again act as a nucleophile and attack another molecule of ethyl bromide (CH₃CH₂Br). The result of this reaction will be the formation of 3-hexane:
\[ C \equiv C-CH₂CH₃ + CH₃CH₂Br \rightarrow C \equiv C-CH₂-CH₂-CH₃ \]
This compound is 3-hexane.

### Summary of products
- **X** is 1-butyne (C₄H₆)
- **Y** is 3-hexane (C₆H₁₄)

### Final Answer
- **X = 1-butyne**
- **Y = 3-hexane**