Write stability order of following intermediates:
`(a)HC -= overset(Theta)C" "(b) CH_(3) - overset(Theta)CH - CH_(3)`
`(c) CH_(3) - C underset(CH_(3))underset(|)overset(CH_(3))overset(|)Theta `

Answer

To determine the stability order of the given intermediates, we need to analyze the hybridization and the inductive effects of the substituents present in each structure.

### Step-by-Step Solution:

1. **Identify the Structures**:
- (a) HC ≡ C⁻ (A): This is a carbon anion with a triple bond (sp hybridization).
- (b) CH₃ - C⁻H - CH₃ (B): This is a carbon anion with two methyl groups attached (sp³ hybridization).
- (c) CH₃ - C(CH₃)₂ (C): This is a carbon anion with three methyl groups attached (sp³ hybridization).

2. **Determine Hybridization**:
- For (a) HC ≡ C⁻: The carbon with the negative charge is sp hybridized.
- For (b) CH₃ - C⁻H - CH₃: The carbon with the negative charge is sp³ hybridized.
- For (c) CH₃ - C(CH₃)₂: The carbon with the negative charge is also sp³ hybridized.

3. **Stability Based on Hybridization**:
- The stability of carbanions is generally ordered as follows: sp > sp² > sp³. This means that the more s character in the hybrid orbital, the more stable the anion.
- Therefore, (a) HC ≡ C⁻ (sp) is more stable than (b) CH₃ - C⁻H - CH₃ (sp³) and (c) CH₃ - C(CH₃)₂ (sp³).

4. **Inductive Effects**:
- In (b), there are two methyl groups which provide a +I (inductive) effect, slightly destabilizing the negative charge.
- In (c), there are three methyl groups providing a stronger +I effect, making it less stable than (b).

5. **Final Stability Order**:
- Based on hybridization and inductive effects, the order of stability is:
1. (a) HC ≡ C⁻ (most stable)
2. (b) CH₃ - C⁻H - CH₃
3. (c) CH₃ - C(CH₃)₂ (least stable)

### Conclusion:
The stability order of the intermediates is:
**(a) > (b) > (c)**