Question

In which of the following pairs of carbanion the first one is more stable than second.

• A.
• B. $\mathrm { H } _ { 2 } \mathrm { C } = \stackrel { \ominus } { \mathrm { C } } \mathrm { H }$
• C.
• D. $\mathrm { H } _ { 3 } \mathrm { C } - \stackrel { \ominus } { \mathrm { C } } \mathrm { H } _ { 2 }$

Answer 1

Answer: (B)

$\mathrm { H } _ { 2 } \mathrm { C } = \stackrel { \ominus } { \mathrm { C } } \mathrm { H }$

Answer 2

(due to p $\pi$ -d $\pi$ back bonding).

(stability of anion, sp > sp2 > sp3)

$\left( \mathrm { CH } _ { 3 } \right) _ { 3 } \stackrel { \ominus } { \mathrm { C } } < \mathrm { H } _ { 3 } \mathrm { C } - \stackrel { \ominus } { \mathrm { C } } \mathrm { H } _ { 3 }$ (stability order of carbanion inversely related to +I)