Ethanol reacts with `SOCl_2` in pyridine to form

Options

A. ethyl chloride
B. acetyl chloride
C. chloroform
D. ethanoyl chloride

Explanation

To solve the problem of how ethanol reacts with thionyl chloride (SOCl2) in the presence of pyridine, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are ethanol (C2H5OH) and thionyl chloride (SOCl2). Pyridine is used as a solvent in this reaction.

2. **Understand the Reaction Type**:
- This is a substitution reaction where the hydroxyl (-OH) group of ethanol is replaced by a chlorine (Cl) atom. Thionyl chloride acts as a chlorinating agent.

3. **Mechanism of Reaction**:
- The -OH group of ethanol is protonated by the thionyl chloride, making it a better leaving group.
- The thionyl chloride then facilitates the substitution of the -OH group with a chlorine atom.

4. **Formation of the Product**:
- The product formed from this reaction is ethyl chloride (C2H5Cl). The reaction can be summarized as:
\[ \text{C2H5OH} + \text{SOCl2} \rightarrow \text{C2H5Cl} + \text{SO2} + \text{HCl} \]
- Here, ethyl chloride is formed along with sulfur dioxide (SO2) and hydrochloric acid (HCl) as by-products.

5. **Final Product**:
- The final product of the reaction is ethyl chloride (C2H5Cl).

### Conclusion:
Ethanol reacts with thionyl chloride in pyridine to form ethyl chloride.