What are the products obtained when ethyl alcohol `(C_(2)H_(5)OH)` reacts with thionyl chloride?

Options

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

Answer

To determine the products obtained when ethyl alcohol (C₂H₅OH) reacts with thionyl chloride (SOCl₂), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are ethyl alcohol (C₂H₅OH) and thionyl chloride (SOCl₂).

### Step 2: Understand the Reaction Mechanism
Thionyl chloride is a reagent that converts alcohols into alkyl chlorides. In this reaction, the hydroxyl group (-OH) of ethyl alcohol will be replaced by a chlorine atom (Cl).

### Step 3: Formation of the Intermediate
When ethyl alcohol reacts with thionyl chloride, the -OH group of the alcohol will first react with SOCl₂. This leads to the formation of an intermediate where the oxygen atom of the alcohol forms a bond with sulfur, creating a sulfonate ester. The reaction can be represented as follows:

\[ \text{C}_2\text{H}_5\text{OH} + \text{SOCl}_2 \rightarrow \text{C}_2\text{H}_5\text{OSOCl} + \text{HCl} \]

### Step 4: Elimination of the Leaving Group
In the next step, the sulfonate ester (C₂H₅OSOCl) will undergo elimination. The chlorine atom from the sulfonate ester will leave, and a proton will be removed from the ethyl group, leading to the formation of ethyl chloride (C₂H₅Cl) and the byproducts sulfur dioxide (SO₂) and hydrochloric acid (HCl):

\[ \text{C}_2\text{H}_5\text{OSOCl} \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{SO}_2 + \text{HCl} \]

### Step 5: Write the Final Products
The final products of the reaction are:
1. Ethyl chloride (C₂H₅Cl)
2. Sulfur dioxide (SO₂)
3. Hydrochloric acid (HCl)

### Final Answer
The products obtained when ethyl alcohol reacts with thionyl chloride are:
- Ethyl chloride (C₂H₅Cl)
- Sulfur dioxide (SO₂)
- Hydrochloric acid (HCl)