In reaction
`BrCH_(2)-CH_(2)Br+Zn to Y+ZnBr_(2)`.
Y is

Options

A. Ethyne
B. Ethene
C. Ethane
D. None

Explanation

To solve the reaction \( \text{BrCH}_2-\text{CH}_2\text{Br} + \text{Zn} \to Y + \text{ZnBr}_2 \), we need to analyze the reactants and the expected products.

### Step 1: Identify the Reactants
The reactant is 1,2-dibromoethane, which has the structure:
\[ \text{BrCH}_2-\text{CH}_2\text{Br} \]
This compound has two bromine atoms attached to the first and second carbon atoms of an ethane chain.

### Step 2: Understand the Role of Zinc
Zinc is a reducing agent in this reaction. It can facilitate the removal of bromine atoms from the dibromo compound. The reaction typically involves the elimination of bromine atoms, leading to the formation of a double bond.

### Step 3: Predict the Product
When zinc reacts with 1,2-dibromoethane, it removes the two bromine atoms, resulting in the formation of a double bond between the two carbon atoms. The product formed is ethene (ethylene), which has the structure:
\[ \text{CH}_2=\text{CH}_2 \]
Along with this, zinc bromide (\( \text{ZnBr}_2 \)) is produced as a byproduct.

### Step 4: Write the Balanced Reaction
The balanced reaction can be summarized as:
\[ \text{BrCH}_2-\text{CH}_2\text{Br} + \text{Zn} \to \text{CH}_2=\text{CH}_2 + \text{ZnBr}_2 \]

### Conclusion
The product \( Y \) in the reaction is ethene.

### Final Answer
Thus, \( Y \) is ethene (or ethylene).

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