X in the following reaction is
`Br_2+overset(CH_3-C-H)underset(H-C-CH_3)(||)overset("CCl_4")rarrX`

Options

A. (+)2,3-Dibromobutane
B. (-)2,3-Dibromobutane
C. `(pm)`2,3-Dibromobutane
D. meso-2,3-Dibromobutane

Explanation

To determine the product X in the reaction of bromine (Br₂) with a trans-alkene (specifically 2-butene) in carbon tetrachloride (CCl₄), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in the equation are Br₂ and a trans-alkene (specifically, 2-butene, which can be represented as CH₃-CH=CH-CH₃).

2. **Understand the Reaction Type**: The reaction between bromine and alkenes is an addition reaction. In this case, it is a halogen addition reaction where Br₂ adds across the double bond of the alkene.

3. **Mechanism of the Reaction**:
- The π electrons of the double bond in the alkene attack one of the bromine atoms, leading to the formation of a cyclic bromonium ion intermediate.
- The other bromine atom (Br) acts as a nucleophile and attacks the more substituted carbon of the bromonium ion from the opposite side (anti-addition).

4. **Formation of the Product**:
- The result of this anti-addition mechanism will yield 2,3-dibromobutane. Since the starting alkene is trans-2-butene, the addition of bromine will lead to a meso compound due to the symmetry of the product.

5. **Final Product Identification**: The final product X is identified as meso-2,3-dibromobutane.

### Conclusion:
Thus, the product X in the reaction is **meso-2,3-dibromobutane**.

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