Methanoic acid is heated with conc. `H_(2)SO_(4)`, to form

Options

A. `CO`
B. `CO_(2)`
C. `CH_(4)`
D. `(COOH)_(2)`

Answer

To solve the question regarding the reaction of methanoic acid (formic acid, HCOOH) when heated with concentrated sulfuric acid (H₂SO₄), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are methanoic acid (HCOOH) and concentrated sulfuric acid (H₂SO₄).

**Hint:** Remember that concentrated sulfuric acid acts as a dehydrating agent in organic reactions.

### Step 2: Understand the Role of Concentrated H₂SO₄
Concentrated sulfuric acid will act as a dehydrating agent, meaning it will facilitate the removal of water from methanoic acid during the reaction.

**Hint:** Dehydrating agents typically promote the elimination of water from organic compounds.

### Step 3: Mechanism of the Reaction
1. The concentrated H₂SO₄ protonates the carbonyl oxygen of methanoic acid, forming a positively charged intermediate.
2. This intermediate undergoes dehydration (loss of water), leading to the formation of a carbonyl compound.

**Hint:** Look for the formation of a carbocation or resonance-stabilized intermediate during the reaction.

### Step 4: Formation of Carbon Monoxide
As the reaction proceeds, the unstable intermediate can lose a hydrogen ion (H⁺) to form carbon monoxide (CO). The final product of this reaction is carbon monoxide.

**Hint:** The stability of the products is crucial; unstable intermediates often lead to the formation of gases like CO.

### Step 5: Conclusion
The main product formed when methanoic acid is heated with concentrated sulfuric acid is carbon monoxide (CO).

**Final Answer:** CO (Carbon Monoxide)

### Summary of Steps:
1. Identify reactants: Methanoic acid (HCOOH) and concentrated H₂SO₄.
2. Recognize the role of concentrated H₂SO₄ as a dehydrating agent.
3. Understand the mechanism involving protonation and dehydration.
4. Conclude that carbon monoxide (CO) is formed as the main product.