When ethyl alcohol `(C_(2) H_(5) OH)` reacts with thionyl chloride in the presence of pyridine the product obtained is .

Options

A. `CH_(3) CH_(2) Cl+ HCI`
B. `C_(2) H_(5) Cl+ HCI+SO_(2)`
C. `CH_(3) CH_(2) Cl +H_(2) O + SO_(2)`
D. `CH_(3) CH_(2) Cl + Cl_(2) + SO_(2)`

Answer

To determine the product obtained when ethyl alcohol (C₂H₅OH) reacts with thionyl chloride (SOCl₂) in the presence of pyridine, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are ethyl alcohol (C₂H₅OH) and thionyl chloride (SOCl₂). Pyridine is used as a base to facilitate the reaction.

### Step 2: Understand the Reaction Mechanism
In this reaction, the hydroxyl group (-OH) of ethyl alcohol is substituted by a chlorine atom (Cl). Thionyl chloride acts as a chlorinating agent.

### Step 3: Reaction Process
1. The oxygen atom in the alcohol (C₂H₅OH) is more electronegative and will attract electrons from the sulfur atom in SOCl₂. This leads to the formation of a positively charged sulfur intermediate.
2. The negatively charged chloride ion (Cl⁻) from SOCl₂ will then attack the carbon atom bonded to the hydroxyl group, leading to the substitution of the -OH group with a Cl atom.

### Step 4: Write the Product
The product formed from this reaction is ethyl chloride (C₂H₅Cl). Along with this, sulfur dioxide (SO₂) and hydrogen chloride (HCl) are also produced as by-products.

### Step 5: Final Products
Thus, the final products of the reaction are:
- Ethyl chloride (C₂H₅Cl)
- Sulfur dioxide (SO₂)
- Hydrogen chloride (HCl)

### Conclusion
The product obtained when ethyl alcohol reacts with thionyl chloride in the presence of pyridine is ethyl chloride (C₂H₅Cl).

---