Arrange the carbanions,(CH_3)_3bar{C},bar{C}C | vedclass

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(C) The stability order is: $\bar{C}Cl_3 > C_6H_5\bar{C}H_2 > (CH_3)_2\bar{C}H > (CH_3)_3\bar{C}$.$1$. $\bar{C}Cl_3$: Most stable due to the strong $-

Vedclass · Accepted Answer

$\bar{C}Cl_3 > C_6H_5\bar{C}H_2 > (CH_3)_2\bar{C}H > (CH_3)_3\bar{C}$

Vedclass · Answer

(C) The stability order is: $\bar{C}Cl_3 > C_6H_5\bar{C}H_2 > (CH_3)_2\bar{C}H > (CH_3)_3\bar{C}$.$1$. $\bar{C}Cl_3$: Most stable due to the strong $-I$ effect of three $Cl$ atoms and the delocalization of the negative charge into the vacant $d$-orbitals of $Cl$.$2$. $C_6H_5\bar{C}H_2$: Stable due to resonance delocalization of the negative charge into the benzene ring.$3$. $(CH_3)_2\bar{C}H$: $2^{\circ}$ carbanion,stability decreases due to the electron-donating $+I$ effect of two $CH_3$ groups.$4$. $(CH_3)_3\bar{C}$: $3^{\circ}$ carbanion,least stable due to the destabilizing $+I$ effect of three $CH_3$ groups.

Arrange the carbanions,$(CH_3)_3\bar{C}$,$\bar{C}Cl_3$,$(CH_3)_2\bar{C}H$,$C_6H_5\bar{C}H_2$ in order of their decreasing stability.

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