Question

The correct order of stability for the following carbanions is: \[ \text{CH}_2=\text{CH}^- ,\quad \text{CH}_3-\text{CH}_2^- ,\quad \text{CH}\equiv\text{C}^- \]

• A. \(\text{CH}=\text{C}^->\text{CH}_2=\text{CH}^->\text{CH}_3-\text{CH}_2^-\)
• B. \(\text{CH}_3-\text{CH}_2^->\text{CH}_2=\text{CH}^->\text{CH}\equiv\text{C}^-\)
• C. \(\text{CH}_2=\text{CH}^->\text{CH}\equiv\text{C}^->\text{CH}_3-\text{CH}_2^-\)
• D. \(\text{CH}\equiv\text{C}^->\text{CH}_3-\text{CH}_2^->\text{CH}_2=\text{CH}^-\)

Hint:

For carbanions, stability increases with increasing \(s\)-character of the hybrid orbital holding the negative charge.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The stability of a carbanion (a species with a negative charge on carbon) depends on the electronegativity of the carbon atom holding the charge. Electronegativity is directly proportional to the s-character of the hybrid orbital.
Step 2: Key Formula or Approach:
Stability of carbanion \(\propto\) Electronegativity of carbon \(\propto\) % s-character.
Step 3: Detailed Explanation:
1. \(\text{CH} \equiv \text{C}^-\): The carbon is \(sp\) hybridized. s-character = \(50%\). High electronegativity allows better stabilization of the negative charge.
2. \(\text{CH}_2 = \text{CH}^-\): The carbon is \(sp^2\) hybridized. s-character \(\approx 33.3%\). Intermediate stabilization.
3. \(\text{CH}_3 - \text{CH}_2^-\): The carbon is \(sp^3\) hybridized. s-character = \(25%\). Low electronegativity makes it the least stable.
Additionally, in \(\text{CH}_3 - \text{CH}_2^-\), the \(+I\) effect of the methyl group further destabilizes the carbanion by increasing electron density on the already negative carbon.
Thus, the stability order is: \(sp>sp^2>sp^3\).
Step 4: Final Answer:
The correct stability order is \(\text{CH} \equiv \text{C}^->\text{CH}_2 = \text{CH}^->\text{CH}_3 - \text{CH}_2^-\).