`C_2H_5OH+SOCl_2overset("Pyridine")rarrx+y+z` in this reaction x, y and z respectively are

Options

A. `C_2H_4CI_2, SO_2,HCI`
B. `C_2H_5CI,SO_2,HCI`
C. `C_2H_5Cl, SOCl`
D. `C_2H_4,SO_2,Cl_2`

Explanation

To solve the reaction of ethanol (C₂H₅OH) with thionyl chloride (SOCl₂) in the presence of pyridine, we can break down the process step by step.

### Step 1: Understanding the Reactants
- Ethanol (C₂H₅OH) is an alcohol.
- Thionyl chloride (SOCl₂) is a reagent commonly used to convert alcohols into alkyl chlorides.
- Pyridine acts as a base in this reaction, helping to facilitate the conversion.

### Step 2: Reaction Mechanism
1. **Proton Abstraction**: Pyridine abstracts a proton (H⁺) from the hydroxyl group (-OH) of ethanol. This forms an ethoxide ion (C₂H₅O⁻) and a pyridinium ion (C₅H₅NH⁺).

\[ C_2H_5OH + Pyridine \rightarrow C_2H_5O^- + C_5H_5NH^+ \]

2. **Formation of Alkyl Chloride**: The ethoxide ion then reacts with thionyl chloride. The sulfur atom in SOCl₂ is electrophilic, and it will react with the ethoxide ion to form ethyl chloride (C₂H₅Cl).

\[ C_2H_5O^- + SOCl_2 \rightarrow C_2H_5Cl + SO_2 + HCl \]

### Step 3: Products of the Reaction
From the above steps, we can identify the products:
- **X**: Ethyl chloride (C₂H₅Cl)
- **Y**: Sulfur dioxide (SO₂)
- **Z**: Hydrogen chloride (HCl)

### Final Answer
Thus, in the reaction \( C_2H_5OH + SOCl_2 \overset{\text{Pyridine}}{\rightarrow} x + y + z \), the products are:
- **X** = C₂H₅Cl (ethyl chloride)
- **Y** = SO₂ (sulfur dioxide)
- **Z** = HCl (hydrochloric acid)