What are X and Y respectively in the following reaction?
Z - product `overset(Y)larr` 2-Butyne `overset(x)rarr` E-product

Options

A. `Na//NH_(3(liq))andPd//BaSO_(4)+H_(2)`
B. `Ni//140^(@)CandPd//BaSO_(4)+H_(2)`
C. `Ni//140^(@)CandNa//NH_(3(liq))`
D. `Pd//BaSO_(4)+H_(2)andNa//NH_(3(aq))`

Answer

To determine the reagents X and Y in the reaction involving 2-butyne, we can analyze the transformation of 2-butyne into the respective products.

### Step-by-Step Solution:

1. **Identify the Structure of 2-Butyne**:
- 2-Butyne has the molecular formula C4H6 and the structure CH3-C≡C-CH3. It contains a triple bond between the second and third carbon atoms.

2. **Determine the Product E**:
- The product E is an alkene formed from 2-butyne. The description indicates that it is a trans-alkene (E configuration).
- To convert 2-butyne to a trans-alkene, we can use sodium (Na) in liquid ammonia (NH3). This reaction is known as the dissolving metal reduction, which leads to the formation of a trans-alkene.
- Therefore, **X = Sodium in liquid ammonia (Na/NH3)**.

3. **Determine the Product Z**:
- The product Z is another alkene, but it is a cis-alkene (Z configuration).
- To convert 2-butyne to a cis-alkene, we can use hydrogen gas (H2) in the presence of a catalyst like palladium (Pd) supported on barium sulfate (BaSO4). This method is known as the Lindlar catalyst, which selectively hydrogenates alkynes to cis-alkenes.
- Therefore, **Y = Palladium on barium sulfate (Pd/BaSO4)**.

4. **Summarize the Findings**:
- The reagents X and Y are:
- **X = Sodium in liquid ammonia (Na/NH3)**
- **Y = Palladium on barium sulfate (Pd/BaSO4)**

### Final Answer:
- X = Sodium in liquid ammonia (Na/NH3)
- Y = Palladium on barium sulfate (Pd/BaSO4)