See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Stability of Carbanions **Key Concept:** Carbanion stability depends on the stabilization of the negative charge by adjacent groups. **Analysis of each option:** **(1) $CH_3\!-\!\dot{C}H_2$ (primary carbanion)** - Negative charge on $sp^3$ carbon with only one alkyl group - Minimal stabilization **(2) $CH_3\!-\!\dot{C}H\!-\!CH_3$ (secondary carbanion)** ✓ - Negative charge on $sp^3$ carbon with **two alkyl groups** - Greater inductive stabilization through electron-donating alkyl groups - **Most stable among options (1), (2), and (4)** **(3) $\dot{C}H_2\!-\!NO_2$ (carbanion with electron-withdrawing group)** - $NO_2$ is **highly electron-withdrawing** - Strongly **destabilizes** the negative charge - Makes this carbanion very unstable (least stable) **(4) $CH_3\!-\!\dot{C}H\!-\!C_2H_5$ (secondary carbanion)** - Similar to (2), but the second alkyl group is ethyl instead of methyl - Comparable stability to (2) **Conclusion:** The methyl carbanion ($\dot{C}H_3$, primary) is less stable than the secondary carbanion in option **(2)** because secondary carbanions benefit from **two electron-donating alkyl groups** providing greater stabilization through hyperconjugation and inductive effects. Option (3) is the worst due to the electron-withdrawing $NO_2$ group. **Answer: (2)** is correct.