By the action of hydrobromic acid on ethyne we get -

Options

A. Methyl iodide
B. Ethyl iodide
C. Ethyledene bromide
D. 1, 2-Dibromo ethane

Answer

To determine the product formed by the action of hydrobromic acid (HBr) on ethyne (C2H2), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this reaction are ethyne (C2H2) and hydrobromic acid (HBr).

2. **Understand the Structure of Ethyne**:
- Ethyne has a triple bond between the two carbon atoms: H-C≡C-H.

3. **Initial Reaction with HBr**:
- When ethyne reacts with HBr, the hydrogen ion (H+) from HBr will add to one of the carbon atoms, while the bromide ion (Br-) will add to the other carbon atom.
- This results in the formation of a vinyl bromide (bromoethene) intermediate:
- H-C=C-Br (vinyl bromide).

4. **Further Reaction**:
- The vinyl bromide can further react with another molecule of HBr.
- The double bond in the vinyl bromide allows for the addition of another HBr molecule.
- The H+ will add to the carbon that does not already have a bromine atom, and the Br- will add to the carbon that already has a bromine atom.

5. **Final Product Formation**:
- After the second addition of HBr, the final product formed is 1,2-dibromoethane:
- C-H2Br-C-H2Br, which can be represented as Br-CH2-CHBr.

### Final Answer:
The product formed by the action of hydrobromic acid on ethyne is **1,2-dibromoethane**.

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