The reaction of ethylene bromide with zinc yields:

Options

A. Alkane
B. Alkene
C. Alkyne
D. All of these

Answer

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

### Step 1: Identify the Reactants
Ethylene bromide, also known as 1,2-dibromoethane, has the molecular formula C2H4Br2. It consists of two carbon atoms (C) and two bromine atoms (Br).

### Step 2: Understand the Reaction Mechanism
When ethylene bromide reacts with zinc, the zinc acts as a reducing agent. The electronegative bromine atoms withdraw electron density from the carbon atoms, leading to the formation of a carbocation intermediate.

### Step 3: Formation of the Carbocation
As the bromine atoms are removed (one from each carbon), a carbocation is formed. The reaction can be summarized as follows:
- The bromine atoms leave as bromide ions (Br⁻).
- This results in the formation of a double bond between the two carbon atoms.

### Step 4: Formation of the Product
The product of this reaction is ethylene (C2H4), which is an alkene. The zinc bromide (ZnBr2) is also formed as a byproduct.

### Step 5: Write the Balanced Reaction
The overall reaction can be represented as:
\[ \text{C}_2\text{H}_4\text{Br}_2 + \text{Zn} \rightarrow \text{C}_2\text{H}_4 + \text{ZnBr}_2 \]

### Conclusion
Thus, the reaction of ethylene bromide with zinc yields ethylene (C2H4) and zinc bromide (ZnBr2).

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