Cold and dil. `KMnO_(4)` reacts with but-2-ene to form

Options

A. ethane-1, 2-diol
B. butane-1, 4-diol
C. butane-1, 3-diol
D. butane-2, 3-diol

Explanation

To solve the question of what product is formed when cold and dilute KMnO4 reacts with but-2-ene, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant**:
- The reactant is but-2-ene, which has the structure:
\[ \text{CH}_3-\text{C}=\text{C}-\text{CH}_3 \]
- This structure shows that there is a double bond between the second and third carbon atoms.

2. **Understand the Reaction Conditions**:
- The reaction is carried out with cold and dilute KMnO4. This reagent is known for its ability to perform syn-dihydroxylation of alkenes, which means it adds hydroxyl (OH) groups across the double bond.

3. **Reaction Mechanism**:
- When but-2-ene reacts with cold and dilute KMnO4, the double bond will be converted into a single bond, and two hydroxyl (OH) groups will be added to the carbon atoms that were involved in the double bond.
- The reaction can be represented as:
\[ \text{CH}_3-\text{C}=\text{C}-\text{CH}_3 + \text{KMnO}_4 \rightarrow \text{CH}_3-\text{C(OH)}-\text{C(OH)}-\text{CH}_3 \]

4. **Draw the Product Structure**:
- The product will have the structure:
\[ \text{CH}_3-\text{C(OH)}-\text{C(OH)}-\text{CH}_3 \]
- This indicates that both carbon atoms that were part of the double bond now have a hydroxyl group attached.

5. **Name the Product**:
- The product formed is a diol, specifically a vicinal diol because the hydroxyl groups are on adjacent carbons.
- The IUPAC name for this compound is 2,3-butanediol.

6. **Final Answer**:
- Therefore, the product formed when cold and dilute KMnO4 reacts with but-2-ene is **2,3-butanediol**.