See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Carbanion Stability Analysis **Carbanion stability depends on how well the negative charge is stabilized.** The key factors are: 1. **Hybridization of the carbanion carbon:** $sp < sp^2 < sp^3$ (s-character stabilizes negative charge) 2. **Inductive effects:** Alkyl groups donate electron density; more alkyl groups → less stability 3. **Resonance stabilization:** Delocalization of negative charge increases stability ## Evaluation of Each Option: **(1) $HC \equiv \overset{\ominus}{C}$** — **sp hybridized carbanion** - Highest s-character (50%) → **maximum charge stabilization** - Triple bond geometry confines electrons close to nucleus - **Most stable carbanion** ✓ **(2) $H_2C=\overset{\ominus}{CH}$** — $sp^2$ hybridized carbanion - 33% s-character, less stabilized than option (1) - Allylic carbanion, but lacks full resonance delocalization here **(3) & (4) $sp^3$ hybridized carbanions** - Only 25% s-character - Destabilized by multiple alkyl groups (inductive donation) - Option (3): quaternary carbon → most electron-rich, least stable - Option (4): tertiary carbon → secondary to option (3) ## Answer: **(1)** is the most stable carbanion The **acetylide carbanion** ($HC \equiv \overset{\ominus}{C}^-$) has the highest s-character, which pulls electron density closest to the nucleus and stabilizes the negative charge most effectively.