When methyl alcohol is heated to `140^(@)C` with conc. `H_(2)SO_(4)`, the product formed is :

Options

A. Ethyl sulphate
B. Dimethyl ether
C. Ethane
D. Ethanoyl sulphate

Explanation

To solve the question, we need to understand the reaction that occurs when methyl alcohol (methanol, CH₃OH) is heated with concentrated sulfuric acid (H₂SO₄) at 140°C. Here’s a step-by-step breakdown of the process:

### Step 1: Understand the Reactants
- We start with methyl alcohol (methanol), which has the chemical formula CH₃OH.
- We are also using concentrated sulfuric acid (H₂SO₄) as a catalyst.

### Step 2: Protonation of Methyl Alcohol
- When methanol is heated with concentrated H₂SO₄, the alcohol gets protonated.
- The oxygen atom in methanol has lone pairs and can accept a proton (H⁺) from sulfuric acid.
- This leads to the formation of an oxonium ion (CH₃OH₂⁺), where the oxygen carries a positive charge.

### Step 3: Nucleophilic Attack
- Another molecule of methanol (CH₃OH) can now act as a nucleophile.
- The lone pair of electrons on the oxygen of the second methanol molecule attacks the positively charged oxonium ion.
- This results in the formation of a bond between the two methanol molecules.

### Step 4: Elimination of Water
- During this reaction, a water molecule (H₂O) is eliminated.
- The reaction results in the formation of dimethyl ether (CH₃OCH₃).

### Step 5: Conclusion
- Therefore, when methyl alcohol is heated to 140°C with concentrated H₂SO₄, the product formed is dimethyl ether (CH₃OCH₃).

### Final Answer
The product formed is **dimethyl ether**.

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