See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

**Stability of Carbanions** Carbanion stability is determined by the ability to stabilize the negative charge on carbon. Key factors: **1) Hybridization Effect (Primary Factor)** - $sp$ hybridized carbon (option A): 50% s-character - $sp^2$ hybridized carbon (option D): 33% s-character - $sp^3$ hybridized carbon (options B, C): 25% s-character s-orbitals are closer to the nucleus and lower in energy. **Higher s-character → more stable carbanion** because the negative charge is held closer to the nucleus. **2) Analysis of Each Option:** **(A) $HC \equiv C^{\ominus}$** - Triple bond = $sp$ hybridization at carbanion carbon - Highest s-character (50%) → **most stable** **(B) $C_6H_5^{\ominus}$ (Phenyl anion)** - $sp^2$ hybridized (aromatic carbon) - Cannot resonate (negative charge on aromatic ring is destabilizing) - Less stable than alkyne **(C) $(CH_3)_3C^{\ominus}$ (Tertiary carbanion)** - $sp^3$ hybridized - Electron-donating alkyl groups increase electron density (destabilizes) - Least stable **(D) $(CH_3)_2C=CH^{\ominus}$ (Allylic anion)** - $sp^2$ hybridized with some resonance stabilization - Less stable than acetylide anion **Answer: (A)** — The acetylide carbanion ($HC \equiv C^{\ominus}$) is most stable due to maximum s-character from $sp$ hybridization.