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

Options

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

Answer

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

### Step 1: Identify the Reactants
1,1-dichloroethane has the molecular formula C2H4Cl2. The structure can be represented as:
```
Cl
|
H - C - C - H
|
Cl
```

### Step 2: Understand the Reaction Conditions
Alcoholic KOH is a strong base that promotes elimination reactions, specifically E2 elimination. This means that it will facilitate the removal of a hydrogen atom and a leaving group (in this case, a chlorine atom) in a single concerted step.

### Step 3: Mechanism of the Reaction
In the E2 elimination mechanism:
- The base (KOH) abstracts a hydrogen atom from one of the carbon atoms adjacent to the carbon that has the leaving group (Cl).
- Simultaneously, the bond between the carbon and the chlorine atom breaks, resulting in the formation of a double bond between the two carbon atoms.

### Step 4: First Elimination Step
1. Remove one Cl and one H:
- From the structure, if we take the hydrogen from the carbon adjacent to the carbon with Cl, we can eliminate one Cl and one H.
- This results in the formation of a double bond:
```
H
|
H - C = C - H
|
Cl
```
This compound is vinyl chloride (C2H3Cl).

### Step 5: Second Elimination Step
2. Now, we can perform another elimination:
- The remaining Cl can be eliminated by the same mechanism. The base will abstract another hydrogen from the carbon adjacent to the remaining Cl, leading to the formation of a triple bond.
```
H
|
H - C ≡ C - H
```
This final product is ethyne (C2H2).

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

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