When ethanol is treated with potassium bromide and concentrated `H_(2)SO_(4)`, ________ is produced.

Options

A. ethyl bromide
B. ethyl hydrogen sulphate
C. ethylene
D. ethane

Answer

To solve the question, "When ethanol is treated with potassium bromide and concentrated H₂SO₄, ________ is produced," we can follow these steps:

### Step 1: Identify the Reactants
- The reactants in this reaction are ethanol (C₂H₅OH), potassium bromide (KBr), and concentrated sulfuric acid (H₂SO₄).

### Step 2: Understand the Role of Concentrated H₂SO₄
- Concentrated sulfuric acid acts as a dehydrating agent and a catalyst in this reaction. It helps in converting the alcohol (ethanol) into a more reactive form by protonating the hydroxyl group (-OH), making it a better leaving group.

### Step 3: Protonation of Ethanol
- When ethanol is treated with concentrated H₂SO₄, the -OH group of ethanol gets protonated, forming an oxonium ion (C₂H₅OH₂⁺). This step increases the electrophilicity of the carbon atom bonded to the -OH group.

### Step 4: Formation of Carbocation
- The protonation leads to the departure of a water molecule (H₂O), resulting in the formation of a carbocation (C₂H₅⁺). This is a key step in the mechanism.

### Step 5: Nucleophilic Attack by Bromide Ion
- The bromide ion (Br⁻) from potassium bromide (KBr) acts as a nucleophile and attacks the carbocation. This forms a new carbon-bromine bond.

### Step 6: Final Product
- The final product of this reaction is ethyl bromide (C₂H₅Br). The overall reaction can be summarized as:
\[ \text{C₂H₅OH} + \text{KBr} + \text{H₂SO₄} \rightarrow \text{C₂H₅Br} + \text{KHSO₄} + \text{H₂O} \]

### Conclusion
- Therefore, when ethanol is treated with potassium bromide and concentrated sulfuric acid, ethyl bromide is produced.

### Final Answer
- The blank in the question can be filled with "ethyl bromide."

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