When `1,1,2,2-`tetrabomopropane is heated with zinc powder in alcohol, which is formed `:`

Options

A. propyne
B. propene
C. propane
D. propadiene

Explanation

To solve the problem of what is formed when `1,1,2,2-tetrabromopropane` is heated with zinc powder in alcohol, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant**: The reactant is `1,1,2,2-tetrabromopropane`, which has the structure:
\[ \text{CH}_3\text{CBr}_2\text{CHBr}_2 \]

2. **Understand the Reaction Conditions**: The reactant is heated with zinc powder in alcohol. Zinc is a reducing agent and will facilitate the removal of bromine atoms from the alkyl halide.

3. **Dehalogenation Reaction**: When `1,1,2,2-tetrabromopropane` is treated with zinc, a dehalogenation reaction occurs. This means that the bromine atoms will be removed from the molecule.

4. **Count the Bromine Atoms**: The molecule has 4 bromine atoms (2 from each carbon). Therefore, when zinc removes these bromine atoms, it will lead to the formation of a carbon-carbon triple bond.

5. **Formation of Alkyne**: The removal of 4 bromine atoms will result in the formation of a triple bond between the two central carbon atoms. The structure will change from:
\[ \text{CH}_3\text{CBr}_2\text{CHBr}_2 \rightarrow \text{CH}_3\text{C} \equiv \text{C} \text{H} \]
This product is known as **propine** (or propyne).

6. **Final Product**: The final product of the reaction is propine, which can be represented as:
\[ \text{C}_3\text{H}_4 \quad \text{(or CH}_3\text{C} \equiv \text{C} \text{H)} \]

### Conclusion:
When `1,1,2,2-tetrabromopropane` is heated with zinc powder in alcohol, the product formed is **propine**.