Ethyl bromide on treatment with alcoholic KOH gives

Options

A. Ethylene
B. Ethanol
C. Acetic acid
D. Ethane

Explanation

To solve the question regarding the reaction of ethyl bromide with alcoholic KOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant**:
- The reactant is ethyl bromide, which has the chemical formula C2H5Br. Its structure can be represented as CH3-CH2-Br.

2. **Understand the Reagent**:
- The reagent is alcoholic KOH (potassium hydroxide in alcohol). This reagent is commonly used for elimination reactions.

3. **Determine the Type of Reaction**:
- The reaction between ethyl bromide and alcoholic KOH is a dehydrohalogenation reaction. This means that a hydrogen atom and a halogen atom (in this case, bromine) will be removed from the molecule.

4. **Mechanism of the Reaction**:
- The mechanism involved is an elimination reaction, specifically an E2 mechanism. In this process, the hydrogen atom is removed from the beta carbon (the carbon adjacent to the carbon bonded to the halogen), and the bromine atom is removed from the alpha carbon (the carbon that is directly bonded to the halogen).

5. **Identify the Carbons**:
- In ethyl bromide, the carbon bonded to bromine is the alpha carbon (C1), and the adjacent carbon is the beta carbon (C2).

6. **Elimination Process**:
- The hydrogen atom from the beta carbon (C2) is eliminated along with the bromine atom from the alpha carbon (C1). This results in the formation of a double bond between the two carbon atoms.

7. **Write the Product**:
- The product of this elimination reaction is ethylene (C2H4), which can also be represented as CH2=CH2.

### Final Answer:
The product formed when ethyl bromide is treated with alcoholic KOH is **ethylene (C2H4)**.

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