Alkyl cyanides undergo Stephen redyction to produce

Options

A. Aldehyde
B. Secondary amine
C. Primary amine
D. Amide

Answer

To solve the question regarding the product formed when alkyl cyanides undergo Stephan reduction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant**:
Alkyl cyanides have the general structure R-C≡N, where R is an alkyl group.

2. **Understand the Reagents**:
The Stephan reduction uses SnCl2 (tin(II) chloride) in the presence of HCl (hydrochloric acid).

3. **Mechanism of Reaction**:
- The first step involves the protonation of the nitrogen atom in the alkyl cyanide by HCl. The nitrogen has a lone pair of electrons and can accept a proton (H⁺), resulting in a positively charged nitrogen (R-C≡N-H⁺).
- This leads to the formation of a nitrilium ion (R-C≡N⁺).

4. **Nucleophilic Attack**:
- The next step involves the nucleophilic attack by SnCl2. The SnCl2 donates an electron pair to the positively charged carbon, converting the triple bond (C≡N) into a double bond (C=N) while releasing Cl⁻.
- This results in the formation of an imine intermediate (R-C=N-Cl).

5. **Hydrolysis**:
- The imine can then undergo hydrolysis when water is added. The water molecule attacks the carbon atom, leading to the formation of an aldehyde (R-C(=O)-H) and the release of ammonia (NH3) or ammonium ion (NH4⁺).

6. **Final Product**:
- The final product of the Stephan reduction of alkyl cyanides is therefore an aldehyde.

### Conclusion:
The product formed from the Stephan reduction of alkyl cyanides is an aldehyde.