Acetylene and ethylene react with alkaline `KMnO_(4)` to give

Options

A. Oxalic acid and formic acid
B. Acetic acid and ethylene glycol
C. Ethyl alcohol and ethylene glycol
D. None

Explanation

To solve the problem of what products are formed when acetylene and ethylene react with alkaline KMnO4, we can break down the reactions step by step.

### Step 1: Understanding the Reactants
- **Acetylene (C2H2)** has a triple bond between the two carbon atoms.
- **Ethylene (C2H4)** has a double bond between the two carbon atoms.
- **KMnO4** is a strong oxidizing agent.

### Step 2: Reaction of Acetylene with KMnO4
1. **Oxidation of Acetylene**:
- When acetylene reacts with KMnO4, the triple bond (C≡C) will undergo oxidation.
- The first step involves breaking one of the carbon-carbon bonds in the triple bond, leading to the formation of a double bond (C=C).
- The double bond will then react with KMnO4, leading to further oxidation.

2. **Formation of Oxalic Acid**:
- The oxidation process continues, and each carbon atom will be oxidized to form carboxylic acid groups (-COOH).
- The final product from the oxidation of acetylene will be **oxalic acid (C2H2O4)**.

### Step 3: Reaction of Ethylene with KMnO4
1. **Oxidation of Ethylene**:
- Ethylene has a double bond (C=C). When it reacts with KMnO4, the double bond will undergo syn addition.
- This means that the oxidizing agent will add hydroxyl groups (–OH) across the double bond.

2. **Formation of Formic Acid**:
- The double bond will break, and each carbon will form a hydroxyl group and be further oxidized to form carboxylic acid groups.
- The final product from the oxidation of ethylene will be **formic acid (HCOOH)**, and since there are two carbon atoms, two moles of formic acid will be produced.

### Final Products:
- **From Acetylene**: Oxalic acid (C2H2O4)
- **From Ethylene**: Two moles of formic acid (2 HCOOH)

### Summary of Products:
- Acetylene reacts with KMnO4 to give oxalic acid.
- Ethylene reacts with KMnO4 to give two moles of formic acid.