What final product will form when alcoholic KOH is treated with 1,1- dirbomoethane ?

Options

A. Ethene
B. Ethane -1, 2 - diol
C. Acetaldehyde
D. Ethyne

Answer

To solve the problem of what final product forms when alcoholic KOH is treated with 1,1-dibromoethane, we can follow these steps:

### Step 1: Understand the structure of 1,1-dibromoethane
1,1-dibromoethane has the following structure:
- It is derived from ethane (C2H6) where both bromine atoms are attached to the first carbon atom.
- The structure can be represented as:
\[ \text{Br} - \text{C}(\text{Br}) - \text{C}(\text{H}) - \text{H} \]
or more simply as:
\[ \text{Br} - \text{C}(\text{Br}) - \text{C}(\text{H}) - \text{H} \]

### Step 2: Reaction with alcoholic KOH
Alcoholic KOH acts as a strong base and will promote an elimination reaction (dehydrohalogenation). In this case, it will remove a hydrogen atom and a bromine atom from the molecule.

### Step 3: Identify the elimination process
- The elimination will occur at the β-position (the carbon adjacent to the carbon with the leaving group).
- The base (KOH) will abstract a hydrogen atom from the β-carbon, leading to the formation of a double bond between the two carbon atoms.

### Step 4: First elimination step
- The first elimination will yield an alkene (specifically, ethylene):
\[ \text{Br} - \text{C}(\text{Br}) - \text{C}(\text{H}) \xrightarrow{\text{KOH}} \text{C} = \text{C} + \text{HBr} \]
This results in the formation of:
\[ \text{C} = \text{C}(\text{H}) - \text{H} \]

### Step 5: Second elimination step
- Since alcoholic KOH is present in excess, a second elimination can occur. The base will again abstract a hydrogen from the β-carbon of the alkene formed in the first step.
- This will lead to the formation of a triple bond (alkyne):
\[ \text{C} = \text{C}(\text{H}) \xrightarrow{\text{KOH}} \text{C} \equiv \text{C} + \text{HBr} \]

### Step 6: Final product
The final product after both eliminations is:
\[ \text{C} \equiv \text{C} \text{ (ethyne)} \]

### Conclusion
The final product formed when 1,1-dibromoethane is treated with alcoholic KOH is **ethyne (C2H2)**.