Arrange the following carbanions in the decre | vedclass

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(C) The stability of a carbanion is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$.$IV$ $(p-CHO)$:

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$IV > I > II > V > III$

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(C) The stability of a carbanion is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$.$IV$ $(p-CHO)$: $-CHO$ is a strong $EWG$ ($-M$ effect), providing maximum stability.$I$ $(p-Br)$: $-Br$ is an $EWG$ ($-I$ effect), providing stability.$II$ $(C_6H_5-CH_2^-)$: Reference structure.$V$ $(p-CH_3)$: $-CH_3$ is an $EDG$ ($+I$ and hyperconjugation), decreasing stability.$III$ $(p-CH_3O)$: $-OCH_3$ is a strong $EDG$ ($+M$ effect), providing minimum stability.Thus, the decreasing order of stability is $IV > I > II > V > III$.

Arrange the following carbanions in the decreasing order of stability:
$I. \ p-Br-C_6H_4-CH_2^-$
$II. \ C_6H_5-CH_2^-$
$III. \ p-CH_3O-C_6H_4-CH_2^-$
$IV. \ p-CHO-C_6H_4-CH_2^-$
$V. \ p-CH_3-C_6H_4-CH_2^-$
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Similar Questions

The order of decreasing stability of the carbanions is:
$1.$ $(CH_3)_3C^-$
$2.$ $(CH_3)_2CH^-$
$3.$ $CH_3CH_2^-$
$4.$ $C_6H_5CH_2^-$

The stability of carbanions follows the order:
$I: CH_3CH_2CH_2\bar{C}H_2$
$II: CH_3\bar{C}HCH_2CH_3$
$III: (CH_3)_3\bar{C}$
$IV: CH_3\bar{C}(Ph)CH_2CH_3$

The most stable carbonium ion among the following is:

List the following carbocations in order of decreasing stability.

Which of the following carbocations is least stable?

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Arrange the following carbanions in the decreasing order of stability:$I. \ p-Br-C_6H_4-CH_2^-$$II. \ C_6H_5-CH_2^-$$III. \ p-CH_3O-C_6H_4-CH_2^-$$IV. \ p-CHO-C_6H_4-CH_2^-$$V. \ p-CH_3-C_6H_4-CH_2^-$Choose the correct answer from the options given below: