Which among the following depicts the correct order of acidity ?

Options

A. `CH_(2) = CH_(2) gt CH_(3) - CH= CH_(2) gt CH_(3) -C-=CH`
B. `CH-=CHgtCH_(3) - C-= CHgtCH_(2) =CH_(2) gt`
C. `CH-= CH gt CH_(2) =CH_(2) gt CH_(3) -C -= CH`
D. `CH_(3) -CH_(3)gt CH_(2) = CH_(2) gt CH_(3) -C -=CH`

Answer

To determine the correct order of acidity among the given compounds, we will analyze the hybridization of the carbon atoms in each compound and their corresponding S character. The acidity of a compound is influenced by the hybridization of the carbon atom bonded to the hydrogen atom that is being removed (protonated).

### Step-by-Step Solution:

1. **Understanding Hybridization and S Character**:
- Acidity increases with increasing S character of the hybridized orbital.
- The order of S character is as follows:
- sp (50% S character) > sp2 (33% S character) > sp3 (25% S character).

2. **Analyzing Each Compound**:
- **CH2=CH2 (Ethene)**: The carbon atoms are sp2 hybridized (33% S character).
- **CH3-CH=CH2 (Propene)**: The carbon with the double bond is sp2 hybridized (33% S character), while the CH3 group is sp3 hybridized (25% S character).
- **CH≡CH (Ethyne)**: The carbon atoms are sp hybridized (50% S character).
- **CH3-C≡CH (Propyne)**: The carbon with the triple bond is sp hybridized (50% S character), while the CH3 group is sp3 hybridized (25% S character).

3. **Comparing Acidity**:
- **CH≡CH (Ethyne)** has the highest acidity due to its sp hybridization (50% S character).
- **CH3-C≡CH (Propyne)** is less acidic than ethyne because of the +I effect from the CH3 group, which donates electrons.
- **CH2=CH2 (Ethene)** and **CH3-CH=CH2 (Propene)** are less acidic than both ethyne and propyne due to their sp2 hybridization (33% S character).

4. **Correct Order of Acidity**:
- Based on the analysis, the correct order of acidity is:
- CH≡CH > CH3-C≡CH > CH2=CH2 > CH3-CH=CH2

5. **Final Conclusion**:
- The second option provided in the question correctly depicts the order of acidity.