In the reactions `HC-=CHoverset((1)NaNH_(2) // liq. NH_(3))underset((2)CH_(3)CH_(2)Br)rarrX`
`X overset((1)NaNH_(2) //liq. NH_(3))underset((2)CH_(3)CH_(2)Br)rarrY,X` and `Y` are `:`

Options

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

Answer

To solve the given problem, we will analyze the reactions step by step and identify the products X and Y formed in the reactions.

### Step 1: Identify the Reactants
The first reactant is acetylene (HC≡CH), and the second reactant is ethyl bromide (CH₃CH₂Br). The reagent used is sodium amide (NaNH₂) in liquid ammonia (NH₃).

### Step 2: Reaction of Acetylene with NaNH₂
When acetylene reacts with sodium amide in liquid ammonia, the acidic hydrogen from the terminal carbon of acetylene is removed by sodium amide. This results in the formation of a sodium acetylide intermediate.

**Reaction:**
\[ \text{HC≡CH} + \text{NaNH}_2 \rightarrow \text{HC≡C}^- \text{Na}^+ + \text{NH}_3 \]

### Step 3: Nucleophilic Substitution with Ethyl Bromide
The sodium acetylide (HC≡C⁻) acts as a nucleophile and attacks the ethyl bromide (CH₃CH₂Br). The bromine atom is displaced, resulting in the formation of product X.

**Reaction:**
\[ \text{HC≡C}^- \text{Na}^+ + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{HC≡C-CH}_2\text{CH}_3 + \text{NaBr} \]

### Step 4: Identify Product X
The product X formed from the above reaction is butyne (specifically 1-butyne, as the triple bond is at the terminal position).

**Product X:**
\[ \text{X} = \text{HC≡C-CH}_2\text{CH}_3 \quad (\text{1-butyne}) \]

### Step 5: Reaction of Product X with NaNH₂ Again
Now, product X (1-butyne) undergoes a similar reaction with sodium amide in liquid ammonia. The acidic hydrogen from the terminal carbon of 1-butyne is removed again, forming another sodium acetylide intermediate.

**Reaction:**
\[ \text{HC≡C-CH}_2\text{CH}_3 + \text{NaNH}_2 \rightarrow \text{HC≡C-CH}_2\text{C}^- \text{Na}^+ + \text{NH}_3 \]

### Step 6: Nucleophilic Substitution with Ethyl Bromide Again
The new sodium acetylide (HC≡C-CH₂⁻) then reacts with another molecule of ethyl bromide (CH₃CH₂Br), resulting in the formation of product Y.

**Reaction:**
\[ \text{HC≡C-CH}_2^- + \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{HC≡C-CH}_2\text{CH}_2\text{CH}_3 + \text{NaBr} \]

### Step 7: Identify Product Y
The product Y formed from this reaction is hexyne (specifically 3-hexyne, as the triple bond is between the third and fourth carbon).

**Product Y:**
\[ \text{Y} = \text{HC≡C-CH}_2\text{CH}_2\text{CH}_3 \quad (\text{3-hexine}) \]

### Final Answer
Thus, the products X and Y are:
- **X:** 1-butyne (HC≡C-CH₂CH₃)
- **Y:** 3-hexyne (HC≡C-CH₂CH₂CH₃)

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