Consider the following carbanions
`(i) CH_(3)- overset(Ө)CH_(2)" "(ii) CH_(2) = overset(Ө)CH" "(iii) CH -= overset(Ө)C `
Correct order of stability of these carbanions in decreasing order is

Options

A. `1 gt 2 gt 3`
B. `2 gt 1 gt 3`
C. `3 gt 2 gt 1`
D. `3 gt 1 gt 2`

Answer

To determine the correct order of stability of the given carbanions, we will analyze the hybridization and s-character of each carbanion:

1. **Identify the Hybridization of Each Carbanion:**
- **(i) CH₃-Ø-CH₂**: The carbon atom with the carbanion has four bonds (three from CH₃ and one from CH₂) and is **sp³ hybridized**.
- **(ii) CH₂=Ø-CH**: The carbon atom with the carbanion has three bonds (two from CH₂ and one from CH) and is **sp² hybridized**.
- **(iii) CH≡Ø-C**: The carbon atom with the carbanion has two bonds (one from CH and one from the triple bond) and is **sp hybridized**.

2. **Determine the s-character of Each Hybridization:**
- **sp³ hybridization** has **25% s-character**.
- **sp² hybridization** has **33% s-character**.
- **sp hybridization** has **50% s-character**.

3. **Stability Analysis:**
- Carbanions with higher s-character are more stable because the electrons are held closer to the nucleus. Therefore, the stability order based on the s-character is as follows:
- **(iii) CH≡Ø-C** (sp hybridized, 50% s-character) - Most stable
- **(ii) CH₂=Ø-CH** (sp² hybridized, 33% s-character) - Moderately stable
- **(i) CH₃-Ø-CH₂** (sp³ hybridized, 25% s-character) - Least stable

4. **Final Order of Stability:**
- The correct order of stability of the carbanions in decreasing order is:
- **(iii) > (ii) > (i)**

Thus, the final answer is: **(iii) > (ii) > (i)**.