Question

In which of the following 2nd anion is more stable than first?

• A. $O_2N-\overset{e}{CH_2}$ and $F-\overset{e}{CH_2}$
• B. $\overset{e}{CF_3}$ and $CCl_3$
• C. $F_3C-\overset{e}{CH_2}$ and $Cl_3C-\overset{e}{CH_2}$
• D. $CH_3-\underset{||}{C}- \overset{e}{CH_2}$ and $H_2N-\overset{e}{CH_2}$

Answer 1

Answer: (B)

(b)

Answer 2

The stability of carbanions is enhanced by electron-withdrawing groups.

(a) $O_2N-\overset{e}{CH_2}$ is stabilized by strong resonance (-R) and inductive (-I) effects of the $NO_2$ group. $F-\overset{e}{CH_2}$ is stabilized only by the inductive (-I) effect of F. Resonance stabilization is much stronger than inductive stabilization. So, the first anion is more stable.

(b) $\overset{e}{CF_3}$ vs $\overset{e}{CCl_3}$. While F has a stronger -I effect than Cl, Cl (a third-period element) can stabilize the carbanion by accepting electron density into its vacant 3d orbitals (d-orbital resonance/negative hyperconjugation). F (a second-period element) does not have accessible d-orbitals. This d-orbital stabilization in $\overset{e}{CCl_3}$ is a more significant factor than the stronger -I effect in $\overset{e}{CF_3}$. Therefore, $\overset{e}{CCl_3}$ is more stable than $\overset{e}{CF_3}$.

(c) $F_3C-\overset{e}{CH_2}$ vs $Cl_3C-\overset{e}{CH_2}$. The $CF_3$ group is a stronger electron-withdrawing group by induction (-I) than the $CCl_3$ group. Thus, $F_3C-\overset{e}{CH_2}$ is more stable.

(d) $CH_3-\underset{||}{C}- \overset{e}{CH_2}$ is strongly stabilized by resonance with the carbonyl group, delocalizing the negative charge onto oxygen. $H_2N-\overset{e}{CH_2}$ is destabilized by the lone pair of nitrogen (positive resonance effect) and only weakly stabilized by the inductive effect of nitrogen. So, the first anion is more stable.