`CH_(3)-C-=C-CH_(3)underset(Pd-BaSO_(4))overset(H_(2))rarr`
(A) `underset(C Cl_(4))overset(Br_(2))rarr(B)`

Options

A. Meso
B. Racemic mixture
C. Diastereomers
D. All

Answer

To solve the given question step by step, we will analyze the reactions involving the alkyne and the reagents mentioned.

### Step 1: Identify the starting compound
The starting compound is 2-butyne, represented as CH₃-C≡C-CH₃. This is an alkyne with a triple bond between the second and third carbon atoms.

**Hint:** Recognize the structure of the alkyne and its functional groups.

### Step 2: Reaction with Rosenmund Reagent
When 2-butyne reacts with the Rosenmund reagent (Pd-BaSO₄ in the presence of H₂), the triple bond is reduced to a double bond. This reaction typically results in the formation of a cis-alkene due to the syn-addition of hydrogen across the double bond.

The product formed (let's call it compound A) is cis-2-butene (CH₃-CH=CH-CH₃).

**Hint:** Understand that the Rosenmund reduction converts triple bonds to double bonds and that it favors the formation of cis isomers.

### Step 3: Reaction with Br₂ in CCl₄
Next, compound A (cis-2-butene) undergoes a bromination reaction when treated with Br₂ in carbon tetrachloride (CCl₄). This is an electrophilic addition reaction where the double bond acts as a nucleophile and attacks the bromine molecule.

The bromine molecule (Br₂) will add across the double bond, resulting in the formation of a bromonium ion intermediate. The Br⁻ ion will then attack the more substituted carbon (or the other carbon) in an anti-addition manner.

**Hint:** Remember that bromination of alkenes involves the formation of a bromonium ion and leads to anti-addition of bromine atoms.

### Step 4: Formation of Products
The attack of Br⁻ can occur from either side of the bromonium ion, leading to two different products. These products will be enantiomers of each other due to their mirror-image relationship.

Thus, the final products formed (let's call them compound B) are the two enantiomers of 2,3-dibromobutane.

**Hint:** Recognize that the formation of enantiomers indicates that the product is a racemic mixture.

### Conclusion
The final product B consists of two enantiomers, which means it is a racemic mixture. Therefore, the answer to the question is that the final product B is racemic.

**Final Answer:** The final product B is a racemic mixture.