Arrange the following carbanions in the decre | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$IV > I > II > V > III$

Vedclass · Answer

(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:
$I. \ p-Br-C_6H_4-CH_2^-$
$II. \ C_6H_5-CH_2^-$
$III. \ p-CH_3O-C_6H_4-CH_2^-$
$IV. \ p-CHO-C_6H_4-CH_2^-$
$V. \ p-CH_3-C_6H_4-CH_2^-$
Choose the correct answer from the options given below:

Similar Questions

Which free radical from the following is least stable?

The correct order of stability for the following carbanions is:
$CH_2=CH^{-}, CH_3-CH_2^{-}, CH\equiv C^{-}$

Based upon an understanding of product stability,predict the product formed when the following dianion reacts with one equivalent of acid $(H^+)$.

Which of the following is the most stable carbanion?

Rank the following radicals in order of increasing stability (least $ < < < $ most).

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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: