Alkyl halide undergoes sequence of reaction to form primary amine. Identify X and Y in the following sequence
`C_2H_5Broverset(X) rarr"Product " overset(Y)rarrC_3H_7NH_2`

Options

A. `X=KCN,Y=H_3O^+`
B. `X=KCN,Y=LiAlH_4`
C. `X=CH_3Cl,Y=AlCl_3//HCl`
D. `X=CH_3NH_2,Y=HNO_2`

Answer

To solve the problem, we need to identify the reagents X and Y in the reaction sequence that converts bromoethane (C2H5Br) into propanamine (C3H7NH2).

### Step-by-Step Solution:

1. **Identify the Starting Material**:
The starting material is bromoethane (C2H5Br).

2. **Determine the Product**:
The final product is propanamine (C3H7NH2). This indicates that there is an increase in the carbon chain length from 2 carbons (in bromoethane) to 3 carbons (in propanamine).

3. **Identify Reagent X**:
To increase the carbon chain from 2 to 3, we can use a nucleophile that adds a carbon group. A common reagent for this purpose is potassium cyanide (KCN).
- **Reaction**: C2H5Br + KCN → C2H5CN (ethanenitrile)
This reaction involves an SN2 mechanism where the cyanide ion (CN-) attacks the bromoethane, displacing the bromine atom and forming ethanenitrile.

4. **Identify Reagent Y**:
The next step is to convert the nitrile (C2H5CN) into a primary amine (C3H7NH2). A suitable reagent for this reduction is lithium aluminum hydride (LiAlH4).
- **Reaction**: C2H5CN + LiAlH4 → C2H5CH2NH2 (propanamine)
This reaction reduces the nitrile group to a primary amine.

5. **Conclusion**:
Therefore, we have identified:
- X = KCN
- Y = LiAlH4

### Final Answer:
- X = KCN
- Y = LiAlH4