During debromination of meso-2,3-dibromobutane, with `Zn` dust`//CH_(3)COOh` the major compound formed is-

Options

A. n-butane
B. 1-butane
C. cis-2-butane
D. trans-2-butane

Answer

To solve the question regarding the debromination of meso-2,3-dibromobutane with zinc dust in the presence of acetic acid (CH₃COOH), we can follow these steps:

### Step 1: Understand the Structure of Meso-2,3-Dibromobutane
Meso-2,3-dibromobutane has the following structure:

```
Br Br
| |
CH₃-C-CH-CH₃
| |
H H
```

This compound has two bromine atoms attached to the second and third carbon atoms of a butane chain.

### Step 2: Identify the Reaction Conditions
The reaction is carried out using zinc dust (Zn) and acetic acid (CH₃COOH). Zinc is a reducing agent that will facilitate the removal of bromine atoms (debromination) from the molecule.

### Step 3: Mechanism of Debromination
During debromination, the bromine atoms will be removed, and a double bond will be formed between the second and third carbon atoms. The reaction will result in the formation of an alkene.

### Step 4: Determine the Product
Since meso-2,3-dibromobutane is a symmetrical molecule, the removal of the two bromine atoms can lead to the formation of a double bond in two possible configurations: cis or trans. However, the presence of zinc dust and acetic acid favors the formation of the cis product.

The major product formed will be:

```
H H
| |
CH₃-C=C-CH₃
```

This structure corresponds to **cis-2-butene**.

### Step 5: Conclusion
The major compound formed during the debromination of meso-2,3-dibromobutane with zinc dust and acetic acid is **cis-2-butene**.

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