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 when alcoholic KOH is treated with 1,1-dichloroethane, we can follow these steps:

### Step 1: Identify the structure of 1,1-dichloroethane
1,1-dichloroethane has the molecular formula C2H4Cl2. The structure can be represented as:
```
Cl
|
CH3-C-CH2
|
Cl
```
Here, the carbon atom bonded to both chlorine atoms is the alpha carbon, and the adjacent carbon is the beta carbon.

### Step 2: Understand the reaction with alcoholic KOH
Alcoholic KOH is a strong base and promotes elimination reactions. In this case, it will facilitate a beta elimination reaction, where a hydrogen atom from the beta carbon and a chlorine atom from the alpha carbon are removed, resulting in the formation of a double bond.

### Step 3: Perform the first elimination
From the structure, we can remove one Cl atom and one H atom:
- Remove one Cl from the alpha carbon (C with Cl)
- Remove one H from the beta carbon (C adjacent to the alpha carbon)

The reaction can be summarized as:
```
1,1-dichloroethane → CH2=CH2 + HCl
```
This results in the formation of ethylene (CH2=CH2).

### Step 4: Perform the second elimination
If we treat the product (ethylene) further with alcoholic KOH, we can perform another elimination reaction. This time, we will remove another hydrogen atom from the beta carbon and a chlorine atom from the alpha carbon (if we consider that we can form a terminal alkyne):
```
CH2=CH2 → C≡C + HCl
```
This results in the formation of ethyne (C≡C).

### Final Product
The final product after treating 1,1-dichloroethane with alcoholic KOH is ethyne (C2H2).

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

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