See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Most Stable Carbanion Analysis **Step 1: Understand carbanion stability factors** Carbanions (negatively charged carbon) are stabilized by: - Electron-withdrawing groups (inductive effect) - Delocalization of negative charge (resonance) - Hybridization ($sp > sp^2 > sp^3$) **Step 2: Evaluate each option** **(A) $CH_3^{\ominus}$** — Methyl carbanion: No stabilization, highly unstable **(B) $CH_2=CH-CH_2^{\ominus}$** — Allyl carbanion: Has resonance stabilization but negative charge on $sp^2$ carbon **(C) Benzyl carbanion ($C_6H_5-CH_2^{\ominus}$):** Negative charge on $sp^3$ carbon, but resonance delocalizes charge into aromatic ring through conjugation — highly stable **(D) $NO_2$-substituted benzyl carbanion:** Same benzyl structure BUT with strong electron-withdrawing $NO_2$ group on the benzene ring **Step 3: Why option D is most stable** The $NO_2$ group is highly electron-withdrawing (strong -I effect), which: - Stabilizes the aromatic ring structure - Further polarizes the $\pi$ electrons toward the ring - Enhances resonance stabilization of the negative charge - Increases overall carbanion stability through both inductive and resonance effects **Answer: (D)** combines benzyl carbanion resonance stabilization with the additional electron-withdrawing effect of $NO_2$, making it the most stable.