1,2-dibromoethane reacts with alcoholic KOH to yields a product X. The hybridisation state of the Carbons present in X respectively , are

Options

A. sp,sp
B. `sp^3,sp^3`
C. `sp^3,sp^2`
D. `sp^3,sp^2`

Answer

To solve the problem, we need to analyze the reaction of 1,2-dibromoethane with alcoholic KOH and determine the hybridization states of the carbon atoms in the resulting product.

### Step-by-Step Solution:

1. **Identify the Reactant**:
The reactant is 1,2-dibromoethane, which has the following structure:
\[ \text{Br}-\text{C}-\text{C}-\text{Br} \]
Each carbon (C) is bonded to two hydrogen (H) atoms.

2. **Understand the Reaction Conditions**:
The presence of alcoholic KOH indicates that the reaction will undergo an elimination process rather than substitution. This is because alcoholic KOH favors the formation of alkenes or alkynes through elimination.

3. **Elimination Reaction**:
In this case, we can eliminate two hydrogen bromide (HBr) molecules from 1,2-dibromoethane. The elimination can occur in two ways:
- Removing H from one carbon and Br from the other carbon.
- Removing Br from one carbon and H from the other carbon.

This leads to the formation of an alkyne, specifically acetylene (ethyne).

4. **Product Formation**:
The product formed (X) is:
\[ \text{C} \equiv \text{C} \]
This is acetylene, which has a triple bond between the two carbon atoms.

5. **Determine Hybridization**:
In acetylene (C2H2), each carbon is involved in a triple bond with the other carbon. The hybridization of carbon in a triple bond is sp because:
- There are two sigma (σ) bonds and two pi (π) bonds formed by the overlap of p orbitals.
- The hybridization state is determined by the number of sigma bonds formed.

6. **Conclusion**:
Therefore, both carbon atoms in the product X (acetylene) are sp hybridized.

### Final Answer:
The hybridization state of the carbons present in X (acetylene) respectively is **sp** for both carbons.