Import Question JSON

$\text{The stability of carbanion is based on the stability of the negative charge in the carbon atom.}$ $\text{The more dispersed the negative charge, the more stable the carbanion.}$ $\text{All the carbanion is as follows:}$ $\text{C}_6\text{H}_5\text{CH}_2^-, \text{-CCl}_3, (\text{CH}_3)_3\text{C}^-, (\text{CH}_3)_2\text{HC}^-$ $\text{As the negative charge dispersed stability of the carbanion increases.}$ $\text{In -CCl}_3, \text{the negative charge is stabilized by the back bonding effect.}$ $\text{Carbon donates electrons in the vacant 3d orbital of C.}$ $\text{In C}_6\text{H}_5\text{CH}_2^-, \text{negative charge is stabilized by resonance effect.}$ $\text{But in the case of (CH}_3)_3\text{C}^- \text{and (CH}_3)_2\text{HC}^- \text{negative charge is destabilized because CH}_3 \text{group shows +I effect and destabilized the negative charge.}$ $\text{As number of CH}_3 \text{group increases +I effect also increases and carbanion became more unstable.}$ $(\text{CH}_3)_2\text{HC}^- \text{is more stable than (CH}_3)_3\text{C}^-.$ $\text{The stability order is as follows:}$ $\text{-CCl}_3 > \text{C}_6\text{H}_5\text{CH}_2^- > (\text{CH}_3)_2\text{HC}^- > (\text{CH}_3)_3\text{C}^-$ $\text{Thus, option 1 is the correct choice.}$