Arrange the following carbanions in order of their decreasing stability,
(A) `H_(3)C-C-=C^(-)` (B )`H-C-=C^(-)` (c) `H_(3)C-CH_(2)^(-)`

Options

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

Answer

To arrange the given carbanions in order of their decreasing stability, we need to analyze the hybridization and the effects of substituents on the stability of the negative charge. The carbanions provided are:

(A) `H₃C-C≡C⁻`
(B) `H-C≡C⁻`
(C) `H₃C-CH₂⁻`

### Step 1: Determine the Hybridization of Each Carbanion

1. **Carbanion A (`H₃C-C≡C⁻`)**:
- The carbon with the negative charge is involved in a triple bond (C≡C).
- It has 1 sigma bond (C-C) and 2 pi bonds (C≡C), plus the negative charge.
- **Steric Number** = 1 (sigma bond) + 1 (negative charge) = 2.
- **Hybridization** = sp.

2. **Carbanion B (`H-C≡C⁻`)**:
- The carbon with the negative charge is also involved in a triple bond.
- It has 1 sigma bond (C-C) and 2 pi bonds (C≡C), plus the negative charge.
- **Steric Number** = 1 (sigma bond) + 1 (negative charge) = 2.
- **Hybridization** = sp.

3. **Carbanion C (`H₃C-CH₂⁻`)**:
- The carbon with the negative charge is attached to two hydrogens and one carbon.
- It has 1 sigma bond (C-CH₂) and 1 negative charge.
- **Steric Number** = 1 (sigma bond) + 1 (negative charge) = 2.
- **Hybridization** = sp³.

### Step 2: Analyze the Stability Based on Hybridization

- **Stability and Hybridization**:
- sp hybridized carbanions (A and B) have 50% s character, which means the negative charge is held closer to the nucleus, making them more stable.
- sp³ hybridized carbanions (C) have 25% s character, making them less stable due to the negative charge being held further from the nucleus.

### Step 3: Consider the Inductive Effects of Substituents

- **Carbanion A (`H₃C-C≡C⁻`)**:
- The presence of the methyl group (CH₃) provides a +I (inductive) effect, which can destabilize the negative charge slightly.

- **Carbanion B (`H-C≡C⁻`)**:
- There are no alkyl groups providing a +I effect, making it more stable than A.

- **Carbanion C (`H₃C-CH₂⁻`)**:
- The negative charge is on a carbon that is sp³ hybridized, which is less stable than the sp hybridized carbanions. The +I effect from the CH₃ group also destabilizes the negative charge.

### Step 4: Conclusion on Stability Order

Based on the hybridization and the inductive effects, we can arrange the carbanions in order of their decreasing stability:

1. **B (`H-C≡C⁻`)** - Most stable due to sp hybridization and no destabilizing +I effect.
2. **A (`H₃C-C≡C⁻`)** - Less stable than B due to the +I effect from the CH₃ group.
3. **C (`H₃C-CH₂⁻`)** - Least stable due to sp³ hybridization and the presence of the +I effect.

### Final Answer:
**Order of decreasing stability**: B > A > C

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