Aldoximes on reduction with `LiAlH_(4)` or `Na//C_(2)H_(5)OH` give

Answer

To solve the question regarding the reduction of aldoximes with lithium aluminium hydride (LiAlH₄) or sodium in ethanol (Na/C₂H₅OH), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Aldoximes**:
- Aldoximes are compounds with the general structure RCH=NOH, where R is an alkyl or aryl group. They are derived from aldehydes.

2. **Reduction Agents**:
- The reduction agents mentioned in the question are lithium aluminium hydride (LiAlH₄) and sodium in ethanol (Na/C₂H₅OH). Both are known to reduce functional groups.

3. **Reaction Mechanism**:
- When aldoximes are treated with LiAlH₄ or Na/C₂H₅OH, the double bond between carbon and nitrogen (C=N) in the aldoxime is reduced.
- The reduction process involves the addition of hydrogen atoms to the nitrogen and carbon atoms.

4. **Formation of Primary Amines**:
- The reduction of aldoximes leads to the formation of primary amines. Specifically, the product of the reduction will be RCH₂NH₂.
- This indicates that the carbon atom of the aldoxime becomes a methylene group (–CH₂–) and the nitrogen atom is converted to an amine group (–NH₂).

5. **Final Products**:
- Therefore, the final products of the reduction of aldoximes with LiAlH₄ or Na/C₂H₅OH are primary amines. The general reaction can be summarized as:
- RCH=NOH + [H] → RCH₂NH₂ (Primary Amine)

### Conclusion:
When aldoximes are reduced with lithium aluminium hydride (LiAlH₄) or sodium in ethanol (Na/C₂H₅OH), they yield primary amines.