trans-But`-2-`ene reacts with cold dilute `KMnO_(4)` solution to yield

Options

A. butane`-2,3-`dione
B. `(+-)`butan`-2-`ol
C. meso-butane`2-,3-`diol
D. `(+-)`butane`-2,3-`diol

Explanation

To solve the question regarding the reaction of trans-But-2-ene with cold dilute KMnO₄, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of trans-But-2-ene**:
- trans-But-2-ene has the following structure:
```
CH3-CH=CH-CH3
```
- The double bond is between the second and third carbon atoms, and the two methyl groups (CH3) are on opposite sides of the double bond.

2. **Understand the Reaction with Cold Dilute KMnO₄**:
- Cold dilute KMnO₄ is known to perform syn-dihydroxylation, which means it adds two hydroxyl (OH) groups across the double bond in a syn fashion (both groups add to the same side).

3. **Mechanism of the Reaction**:
- When trans-But-2-ene reacts with cold dilute KMnO₄, the double bond is broken, and two hydroxyl groups are added to the carbons involved in the double bond.
- The reaction proceeds through the formation of a cyclic manganate ester intermediate.

4. **Product Formation**:
- Since the addition is syn, both OH groups will be added to the same side of the double bond.
- The resulting product will have the following structure:
```
CH3-CHOH-CHOH-CH3
```
- This compound is 2,3-butanediol.

5. **Stereochemistry**:
- The product formed will be a racemic mixture because the addition of OH groups can occur from either side of the double bond, leading to two enantiomers (R and S forms of 2,3-butanediol).

6. **Final Answer**:
- The product of the reaction of trans-But-2-ene with cold dilute KMnO₄ is a racemic mixture of 2,3-butanediol.

### Summary:
The reaction of trans-But-2-ene with cold dilute KMnO₄ yields a racemic mixture of 2,3-butanediol.