`Ph-CH_2-C-=CHunderset(Y)overset(X)hArrPh-C-=C-CH_3`.
The reagents X and Y respectively are

Options

A. Lindlar catalyst , `NaNH_2`
B. `NaNH_2` and alc . KOH
C. Pt catalyst , Wilkison's catalyst
D. Alc. KOH and `NaNH_2`

Answer

To solve the question, we need to identify the reagents X and Y that facilitate the transformation of the given organic compounds.

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- The starting compound is a terminal alkyne: Ph-CH₂-C≡CH (where Ph represents a phenyl group).
- The product is an internal alkyne: Ph-C≡C-CH₃.

2. **Determine Reagent Y:**
- To convert a terminal alkyne to an internal alkyne, we typically need a reagent that can abstract a proton and form a carbon anion.
- In this case, the reagent Y is **sodium amide (NaNH₂)**. Sodium amide will abstract the proton from the terminal alkyne, resulting in the formation of a carbon anion, which can then undergo tautomerization to form the internal alkyne.

3. **Determine Reagent X:**
- To convert the internal alkyne back to a terminal alkyne, we need a reagent that can add a proton to the carbon anion formed from the internal alkyne.
- The reagent X is **alcoholic KOH (potassium hydroxide in alcohol)**. This reagent will abstract a proton from the terminal alkyne, leading to the formation of the internal alkyne.

4. **Final Answer:**
- Therefore, the reagents X and Y are:
- X = Alcoholic KOH
- Y = Sodium Amide (NaNH₂)

### Summary:
- The reagents X and Y respectively are:
- X = Alcoholic KOH
- Y = Sodium Amide (NaNH₂)