Partial reduction of Cyanides with `SnCl_2` / HCl gives X which on hydrolysis form Y . What is X and Y ?

Explanation

To solve the problem, we will follow a systematic approach to understand the partial reduction of cyanides and the subsequent hydrolysis.

### Step-by-Step Solution:

1. **Identify the Starting Material:**
- The starting material is a cyanide, which can be represented as RCN, where R is an alkyl or aryl group.

2. **Partial Reduction of Cyanides:**
- The question states that cyanides undergo partial reduction using `SnCl2` in the presence of `HCl`. This reaction typically leads to the formation of an aldimine.
- The reaction can be represented as:
\[ RCN \xrightarrow{SnCl_2/HCl} RCH=NH \]
- Here, RCH=NH is the aldimine formed, which we will denote as X.

3. **Identify Compound X:**
- From the above reaction, we can conclude that:
- **X = Aldimine (RCH=NH)**

4. **Hydrolysis of Aldimine:**
- The next step involves the hydrolysis of the aldimine (X). Hydrolysis of an aldimine typically yields an aldehyde and ammonia.
- The hydrolysis reaction can be represented as:
\[ RCH=NH + H_2O \rightarrow RCHO + NH_3 \]
- In this reaction, RCHO is the aldehyde formed.

5. **Identify Compound Y:**
- From the hydrolysis reaction, we can conclude that:
- **Y = Aldehyde (RCHO)**

### Final Answer:
- **X = Aldimine (RCH=NH)**
- **Y = Aldehyde (RCHO)**