Bromoethane on treatment with alcoholic KOH gives

Options

A. ethyl alcohol
B. butane
C. methane
D. ethylene

Answer

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

### Step 1: Identify the Reactants
The reactant in this case is bromoethane (C2H5Br) and the reagent is alcoholic KOH.

### Step 2: Understand the Reaction Type
Alcoholic KOH is known to act as a dehydrohalogenating agent. This means it will facilitate the elimination of hydrogen and bromine from bromoethane.

### Step 3: Elimination of HBr
In the reaction, a hydrogen atom (H) from one carbon and the bromine atom (Br) from the adjacent carbon are eliminated. This results in the formation of HBr as a byproduct.

### Step 4: Formation of Ethene
The elimination of H and Br leads to the formation of a double bond between the two carbon atoms, resulting in the formation of ethene (C2H4).

### Step 5: Write the Balanced Equation
The balanced equation for the reaction can be written as:
\[ \text{C}_2\text{H}_5\text{Br} + \text{KOH (alc)} \rightarrow \text{C}_2\text{H}_4 + \text{KBr} + \text{H}_2\text{O} \]

### Conclusion
Thus, the product of the reaction of bromoethane with alcoholic KOH is ethene.

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