Action of zinc on tetrabromoethane gives

Options

A. `CH_(3)-OH`
B. `CH-=CH`
C. `CH_(3)-CH_(3)`
D. `CH_(2)=CH_(2)`

Answer

To determine the product of the reaction between zinc and tetrabromoethane, we can follow these steps:

### Step 1: Understand the Structure of Tetrabromoethane
Tetrabromoethane has the formula C2H2Br4. It consists of two carbon atoms (C) and four bromine atoms (Br) attached to the carbon atoms. The structure can be represented as:
```
Br Br
| |
Br - C - C - Br
| |
H H
```

### Step 2: Identify the Action of Zinc
When zinc (Zn) is added to tetrabromoethane, it acts as a reducing agent. Zinc will react with the bromine atoms to form zinc bromide (ZnBr2). Each zinc atom can react with two bromine atoms.

### Step 3: Reaction with Bromine
Since tetrabromoethane has four bromine atoms, two zinc atoms will be required to react with all four bromine atoms:
- 2 Zn + 4 Br → 2 ZnBr2

### Step 4: Formation of the Product
After the reaction, the tetrabromoethane will lose the bromine atoms, and the two carbon atoms will form a triple bond between them due to the loss of the bromine atoms. The remaining structure will be:
```
H H
| |
C ≡ C
```
This structure represents ethyne (acetylene), which has the formula C2H2.

### Step 5: Conclusion
The final product of the reaction between zinc and tetrabromoethane is ethyne (C2H2).

### Final Answer:
The action of zinc on tetrabromoethane gives ethyne (C2H2).

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