`CH_(3)-C-=C-CH_(3)underset(Pd-BaSO_(4))overset(H_(2))rarr(A)`
`overset("cold. "KMnO_(4))rarr(B)`. Compound (B) is :

Options

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

Answer

To solve the given problem step by step, let's break down the reactions involved:

### Step 1: Identify the starting compound
The starting compound is 1-butyne, represented as \( CH_3-C \equiv C-CH_3 \).

### Step 2: Reaction with Hydrogen in the presence of Pd-BaSO4
The reaction of 1-butyne with hydrogen gas (\( H_2 \)) in the presence of palladium on barium sulfate (Pd-BaSO4) is known as the Rosenmund reduction. This reaction reduces the triple bond (alkyne) to a double bond (alkene) and specifically produces a cis-alkene.

Thus, the product \( A \) formed is **cis-2-butene**:
\[ CH_3-C \equiv C-CH_3 \xrightarrow{H_2, Pd-BaSO_4} \text{cis-}CH_3-CH=CH-CH_3 \]

### Step 3: Reaction with Cold KMnO4
Next, the compound \( A \) (cis-2-butene) reacts with cold potassium permanganate (\( KMnO_4 \)). This reagent is used for syn-dihydroxylation of alkenes, meaning it adds hydroxyl groups (\( OH \)) across the double bond in a syn fashion (both \( OH \) groups are added to the same side).

The reaction can be represented as:
\[ \text{cis-}CH_3-CH=CH-CH_3 \xrightarrow{cold \, KMnO_4} \text{meso-2,3-butanediol} \]

### Step 4: Identify the final product (B)
The final product \( B \) is **meso-2,3-butanediol**. This compound has a plane of symmetry, making it a meso compound.

### Conclusion
Thus, the compound \( B \) is **meso-2,3-butanediol**.

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