Arrange the following carbanions in order of their decreasing stability.
A) `H_(3)C-C-=C^(-)`
B) `H-C-=C^(-)`
C) `H_(3)C-Cbar(H)_(2)`

Options

A. `A gt B gt C`
B. `B gt A gt C`
C. `C gt B gt A`
D. `C gt A gt B`

Answer

To determine the stability of the given carbanions, we will analyze each one based on their hybridization and the effect of substituents on the stability of the negative charge.

### Step 1: Identify the Carbanions
The carbanions given are:
A) `H3C-C≡C^(-)` (Carbanion A)
B) `H-C≡C^(-)` (Carbanion B)
C) `H3C-C^(-)(H)2` (Carbanion C)

### Step 2: Determine the Hybridization
1. **Carbanion A (`H3C-C≡C^(-)`)**:
- The carbon with the negative charge is involved in a triple bond (C≡C), which means it is sp hybridized.
- Hybridization: sp

2. **Carbanion B (`H-C≡C^(-)`)**:
- The carbon with the negative charge is also sp hybridized due to the triple bond.
- Hybridization: sp

3. **Carbanion C (`H3C-C^(-)(H)2`)**:
- The carbon with the negative charge is bonded to two hydrogens and one methyl group, making it sp3 hybridized.
- Hybridization: sp3

### Step 3: Analyze the Stability Based on Hybridization
- **Stability Order Based on Hybridization**:
- sp hybridized carbanions (A and B) are more stable than sp3 hybridized carbanions (C) because sp hybridized carbons have a higher s-character (50% s-character) compared to sp3 (25% s-character). Higher s-character means the negative charge is held closer to the nucleus, leading to greater stability.

### Step 4: Compare the Two sp Hybridized Carbanions
- Both carbanions A and B are sp hybridized. We need to consider the substituents:
- **Carbanion A (`H3C-C≡C^(-)`)** has a methyl group (which has a +I effect) that can destabilize the negative charge due to electron donation.
- **Carbanion B (`H-C≡C^(-)`)** has only hydrogen atoms, which do not have a +I effect and thus do not destabilize the negative charge.

### Step 5: Final Stability Order
Based on the analysis:
1. Carbanion B (`H-C≡C^(-)`) is the most stable (sp hybridized, no destabilizing substituents).
2. Carbanion A (`H3C-C≡C^(-)`) is less stable than B due to the +I effect of the methyl group.
3. Carbanion C (`H3C-C^(-)(H)2`) is the least stable due to being sp3 hybridized.

### Conclusion
The order of decreasing stability of the carbanions is:
**B > A > C**