Meso`-` and racemic `-2,3-`dibromobutane on reaction with `Nal` in acetone give `:`

Options

A. `E-`But`-2-`ene and `Z-`But`-2-`ene, respectively
B. `Z-`But`-2-2`ene and `E-`But`-2-`ene, respectively
C. Both give `E-`But `-2-`ene
D. Both give `Z-`But`-2-`ene

Explanation

To solve the problem of what meso- and racemic-2,3-dibromobutane yield when reacted with NaI in acetone, we can follow these steps:

### Step 1: Understand the Structures
- **Meso-2,3-dibromobutane** has a plane of symmetry, making it achiral. The bromine atoms are on the same side (erythro configuration).
- **Racemic-2,3-dibromobutane** consists of two enantiomers (mirror images) where the bromines are on opposite sides (threo configuration).

### Step 2: Reaction Mechanism
- The reaction of 2,3-dibromobutane with NaI in acetone involves an **anti-elimination** mechanism. This means that the bromine atoms will be removed in such a way that they are eliminated from opposite sides of the molecule.

### Step 3: Analyze the Meso Form
- For **meso-2,3-dibromobutane**, when subjected to anti-elimination, the bromines will be removed from opposite sides, resulting in the formation of a **trans-alkene** (E form). The product will have the structure:
\[ \text{C} \equiv \text{C} \text{CH}_3 \text{CH}_3 \text{H} \text{H} \]

### Step 4: Analyze the Racemic Form
- For **racemic-2,3-dibromobutane**, the two enantiomers will also undergo anti-elimination. However, since they are enantiomers, the product will be a **cis-alkene** (Z form). The structure will be:
\[ \text{C} \equiv \text{C} \text{CH}_3 \text{H} \text{H} \text{CH}_3 \]

### Step 5: Conclusion
- Therefore, the meso-2,3-dibromobutane will yield a **trans-alkene (E form)**, and the racemic-2,3-dibromobutane will yield a **cis-alkene (Z form)**.

### Final Answer
- The products of the reactions are:
- Meso-2,3-dibromobutane → **E form**
- Racemic-2,3-dibromobutane → **Z form**