Identify X and Y in the following sequence
`C_(2)H_(5)Br overset(X) to "product" overset(Y) to C_(3)H_(7)NH_(2)`

Options

A. X=KCn , `Y=LiAlH_(4)`
B. X=KCn, `Y=H_(3)O^(+)`
C. `X=CH_(3)Cl , Y=AlCl_(3)/HCl`
D. `X=CH_(3)NH_(2) , Y=HNO_(2)`

Answer

To solve the problem, we need to identify the reagents X and Y in the reaction sequence starting from ethyl bromide (C₂H₅Br) and ending with propylamine (C₃H₇NH₂).

### Step-by-Step Solution:

1. **Identify the starting material and the final product**:
- Starting material: Ethyl bromide (C₂H₅Br)
- Final product: Propylamine (C₃H₇NH₂)

2. **Determine the change in carbon atoms**:
- Ethyl bromide has 2 carbon atoms.
- Propylamine has 3 carbon atoms.
- This indicates that one carbon atom is being added in the reaction.

3. **Identify the reagent X that adds one carbon atom**:
- To add a carbon atom to a molecule, we can use a cyanide ion (CN⁻) which can be introduced through potassium cyanide (KCN).
- The reaction of C₂H₅Br with KCN will yield ethyl cyanide (C₂H₅CN).

4. **Write the reaction for X**:
- C₂H₅Br + KCN → C₂H₅CN (ethyl cyanide)

5. **Identify the product after the first reaction**:
- The product after the reaction with KCN is ethyl cyanide (C₂H₅CN).

6. **Determine the reagent Y that converts ethyl cyanide to propylamine**:
- Ethyl cyanide (C₂H₅CN) can be reduced to propylamine (C₃H₇NH₂) using a reducing agent.
- A common reducing agent for this conversion is lithium aluminum hydride (LiAlH₄).

7. **Write the reaction for Y**:
- C₂H₅CN + LiAlH₄ → C₃H₇NH₂ (propylamine)

8. **Final identification of X and Y**:
- X = KCN (potassium cyanide)
- Y = LiAlH₄ (lithium aluminum hydride)

### Conclusion:
- The reagents are:
- X = KCN
- Y = LiAlH₄