Stephen reduction `(SnCl_(2)//HCl)` converts cyanides to

Options

A. Aldehydes
B. Ketones
C. Amines
D. Acids

Answer

To solve the question regarding the conversion of cyanides using Stephen reduction `(SnCl2/HCl)`, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The Stephen reduction is a specific type of reduction reaction that involves the use of tin(II) chloride (SnCl2) in the presence of hydrochloric acid (HCl). It is primarily used to reduce nitriles (or cyanides) to aldehydes.

**Hint**: Recall that reduction reactions often involve the gain of electrons or the addition of hydrogen.

2. **Identify the Reactants**: The reactant in this case is a cyanide, which can be represented as RCN, where R is an alkyl group.

**Hint**: Remember that cyanides are compounds containing the cyano group (-CN).

3. **Reaction Mechanism**: The mechanism of the Stephen reduction involves the following steps:
- The cyanide (RCN) reacts with HCl to form a salt.
- The tin(II) chloride (SnCl2) then reduces the cyanide through a single electron transfer mechanism.
- This reduction leads to the formation of an intermediate that is then hydrolyzed.

**Hint**: Focus on the role of SnCl2 as a reducing agent and how it interacts with the cyanide.

4. **Formation of Aldehyde**: After the hydrolysis of the reduced intermediate, the final product obtained is an aldehyde (RCHO).

**Hint**: Aldehydes are characterized by the presence of the carbonyl group (C=O) at the end of the carbon chain.

5. **Conclusion**: Therefore, the Stephen reduction converts cyanides to aldehydes.

**Final Answer**: The correct answer to the question is option 1: aldehydes.