Identify X and Y in the following
`H_(2)underset(Br)underset(|)(C)-underset(Br)underset(|)(C)H_(2)+KOH overset("alcohol")to X overset(NaNH_(2))to Y`

Options

A. `X-CH_(3)CHBr,Y-CH_(2)=CH_(2)`
B. `X-CH_(2)OH-CH_(2)OH,Y-CH_(2)=CH_(2)`
C. `X-CH_(2)CHBr,Y-CH-=CH`
D. `X-CH-=CBr,Y-CH-=CH`

Explanation

To solve the problem, we need to identify the compounds X and Y in the given reaction sequence. The reaction involves a compound with two bromine substituents and a potassium hydroxide (KOH) treatment, followed by a reaction with sodium amide (NaNH₂).

### Step 1: Identify the Starting Compound
The starting compound is represented as:
```
H₂C(Br)C(Br)H₂
```
This indicates a dibrominated alkane. The structure suggests it is 2,2-dibromopropane (since there are two bromine atoms on the same carbon).

### Step 2: Reaction with KOH
When 2,2-dibromopropane reacts with KOH in an alcoholic medium, it undergoes elimination to form an alkene. The elimination reaction will lead to the formation of propene (C₃H₆).

So, the reaction can be summarized as:
```
2,2-dibromopropane + KOH (alcohol) → propene (X)
```

### Step 3: Reaction with NaNH₂
Next, propene (X) reacts with sodium amide (NaNH₂). Sodium amide is a strong base and can deprotonate the alkene to form an alkyne. In this case, propene can be converted to propyne (C₃H₄).

So, the reaction can be summarized as:
```
propene (X) + NaNH₂ → propyne (Y)
```

### Final Identification
- X = Propene (C₃H₆)
- Y = Propyne (C₃H₄)

### Summary of the Solution
1. The starting compound is identified as 2,2-dibromopropane.
2. Upon treatment with KOH, it forms propene (X).
3. Propene then reacts with NaNH₂ to form propyne (Y).