Arrange the carbanions, `(CH_(3))_(3)bar(C),bar(C)Cl_(3),(CH_(3))_(2)bar(C)H,C_(6)H_(5)bar(C)H_(2)`, in order of their decreasing stability

Options

A. `(CH_(3))_(2)CH^(-) gt Cl_(3)C^(-) gt C_(6)H_(5)CH_(2)^(-) gt (CH_(3))_(3)C^(-)`
B. `Cl_(3)C^(-) gt C_(6)H_(5)CH_(2)^(-) gt (CH_(3))_(2)CH^(-) gt (CH_(3))_(3)C^(-)`
C. `(CH_(3))_3)C^(-) gt (CH_(3))_(2)CH^(-) gt C_(6)H_(5)CH_(2)^(-) gt Cl_(2)C^(-)`
D. `C_(6)H_(5)CH_(2)^(-) gt Cl_(3)C^(-) gt (CH_(3))_(3)C^(-) gt (CH_(3))_(2)CH^(-)`

Answer

B

Explanation

Due to -I-effect of the three Cl atoms, `Cl_(3)C^(-)` is the most stable. This is followed by `C_(6)H_(5)CH_(2)^(-)` which is stabilized by resonance. Out of `(CH_(3))_(3)C^(-)` and `(CH_(3))_(2)CH^(-), (CH_(3))_(2)CH^(-)` is more stable due to +I -effect of two rather than three `CH_(3)` groups. Thus option (b) is correct.