`LiAlH_4` / ether reduces methyl cyanide into

Options

A. methylamine
B. ethylamine
C. methylimine
D. ethylimine

Explanation

To solve the question of how `LiAlH4` (lithium aluminium hydride) in ether reduces methyl cyanide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is methyl cyanide, which has the structure CH3-C≡N (a methyl group attached to a cyano group).
- The reducing agent is `LiAlH4`, which is known for its strong reducing properties.

2. **Understand the Reaction Mechanism**:
- `LiAlH4` dissociates into Li⁺ and AlH₄⁻ ions. The AlH₄⁻ ion is the active reducing species.
- The cyano group (C≡N) will be attacked by the hydride ion (H⁻) from AlH₄⁻.

3. **First Hydride Transfer**:
- The hydride ion (H⁻) attacks the carbon of the cyano group (C≡N), resulting in the formation of an imine salt.
- The reaction can be represented as:
\[ CH3-C≡N + H⁻ \rightarrow CH3-C=NH \]
- This intermediate is known as an imine salt (specifically, a methyl imine salt).

4. **Second Hydride Transfer**:
- The imine salt formed in the previous step is still reactive and will undergo further reduction.
- Another hydride ion from `LiAlH4` will attack the nitrogen atom of the imine salt, leading to the formation of an amine.
- The reaction can be represented as:
\[ CH3-C=NH + H⁻ \rightarrow CH3-CH2-NH2 \]
- This results in the formation of ethylamine (C2H5NH2).

5. **Final Product**:
- The final product of the reduction of methyl cyanide by `LiAlH4` is ethylamine (C2H5NH2).

### Conclusion:
The reduction of methyl cyanide by `LiAlH4` in ether yields **ethylamine** (C2H5NH2).