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

Concept: The stability of a carbanion depends mainly on:
Hybridization of the negatively charged carbon
Electronegativity (more \(s\)-character stabilizes negative charge)
Resonance effects Greater the \(s\)-character of the orbital holding the negative charge, greater is the stability.
Step 1: Analyze hybridization of each carbanion
\( \text{CH}_3-\text{CH}_2^- \): Carbon is \(sp^3\)-hybridized (25% \(s\)-character)
\( \text{CH}_2=\text{CH}^- \): Carbon is \(sp^2\)-hybridized (33% \(s\)-character)
\( \text{CH}\equiv\text{C}^- \): Carbon is \(sp\)-hybridized (50% \(s\)-character)
Step 2: Apply stability rule Higher \(s\)-character \(\Rightarrow\) greater electronegativity \(\Rightarrow\) better stabilization of negative charge. Thus: \[ sp>sp^2>sp^3 \]
Step 3: Write the correct order \[ \text{CH}\equiv\text{C}^->\text{CH}_2=\text{CH}^->\text{CH}_3-\text{CH}_2^- \] Final Answer: \[ \boxed{\text{Option (A)}} \]