Which of the following is the most stable car | vedclass

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(C) The stability of a carbanion is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$ due to the indu

Vedclass · Accepted Answer

$p-O_2N-C_6H_4-CH_2^-$

Vedclass · Answer

(C) The stability of a carbanion is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$ due to the inductive and resonance effects.$1.$ $CH_3^-$ is a simple alkyl carbanion.$2.$ $CH_3CH_2^-$ is less stable than $CH_3^-$ due to the $+I$ effect of the methyl group.$3.$ $p-O_2N-C_6H_4-CH_2^-$ contains a nitro group $(-NO_2)$,which is a strong electron-withdrawing group ($-I$ and $-M$ effect). This group effectively disperses the negative charge on the carbon atom,making it the most stable among the given options.$4.$ $p-CH_3-C_6H_4-CH_2^-$ contains a methyl group $(-CH_3)$,which is an electron-donating group ($+I$ and hyperconjugation),thus destabilizing the carbanion.Therefore,the correct option is $C$.

Which of the following is the most stable carbanion?

Similar Questions

The correct order of the relative stability of the following carbanions is:
$I$. $HC \equiv C^{-}$
$II$. $CH_2=CH-CH_2^{-}$
$III$. $H_2C=CH^{-}$
$IV$. $(C_6H_5)_2CH^{-}$

What are free radicals,carbanions,and carbocations? How are they formed?

Consider the following compounds:
$(I)$ $(CH_3)_3C-CH^{\bullet}-C_6H_5$
$(II)$ $(C_6H_5)_3C^{\bullet}$
$(III)$ $2$-methylbicyclo$[2.2.1]$hept$-2-$yl radical (as shown in the image)
Hyperconjugation occurs in:

Find the correct stability order for the following carbocations:

The total number of electrons around the carbon atom of a methyl free radical is:

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