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 determine the final product formed when 1,1-dibromoethane is treated with alcoholic KOH, we can follow these steps:

### Step 1: Identify the Reactants
1,1-Dibromoethane (C2H4Br2) is a compound with two bromine atoms attached to the first carbon atom of the ethane chain. Alcoholic KOH is a strong base that can facilitate elimination reactions.

**Hint:** Look for the structure of 1,1-dibromoethane to visualize the bromine atoms' positions.

### Step 2: Understand the Reaction Type
The reaction between 1,1-dibromoethane and alcoholic KOH is a dehydrohalogenation reaction. This means that the reaction will involve the elimination of hydrogen halides (HBr in this case).

**Hint:** Remember that dehydrohalogenation typically leads to the formation of alkenes or alkynes.

### Step 3: Mechanism of the Reaction
1. The alcoholic KOH will act on the 1,1-dibromoethane. The potassium (K) will bond with one of the bromine (Br) atoms, leading to the elimination of KBr.
2. The hydroxide ion (OH-) will abstract a hydrogen atom from the adjacent carbon, resulting in the formation of a double bond.

**Hint:** Focus on how the elimination of Br and H occurs to form a double bond.

### Step 4: First Elimination
After the first elimination, we will have:
- CH2=CHBr (vinyl bromide)

**Hint:** Keep track of the changes in the molecular structure after each step.

### Step 5: Second Elimination
Now, the vinyl bromide (CH2=CHBr) will undergo a similar reaction:
1. KOH will again react, with K bonding to Br and OH- taking H from the carbon adjacent to the Br.
2. This results in the formation of a triple bond.

The final product after the second elimination will be:
- C≡C (ethyne or acetylene)

**Hint:** Consider the stability of the formed product and the type of bonds created during the reaction.

### Step 6: Identify Side Products
Along with the main product (ethyne), the side products of the reaction will be:
- 2 KBr (from the two eliminations)
- 2 water molecules (from the hydroxide ions abstracting hydrogen)

**Hint:** Always account for the by-products formed during the reaction.

### Conclusion
The final product formed when 1,1-dibromoethane is treated with alcoholic KOH is ethyne (acetylene). The side products are KBr and water.

**Final Answer:** Ethyne (C2H2) is the main product formed.