The stability of carbanions follows the order | vedclass

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(C) The stability of carbanions is primarily governed by the inductive effect ($+I$ effect) and resonance.$1$. Compound $IV$ is the most stable becaus

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$III < II < I < IV$

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(C) The stability of carbanions is primarily governed by the inductive effect ($+I$ effect) and resonance.$1$. Compound $IV$ is the most stable because the negative charge on the carbon is delocalized into the phenyl ring through resonance.$2$. Among the remaining alkyl carbanions $(I, II, III)$,stability decreases as the number of alkyl groups attached to the anionic carbon increases. This is because alkyl groups exert a $+I$ effect,which increases the electron density on the anionic carbon,thereby destabilizing it.$3$. Thus,the stability order is $1^{\circ} > 2^{\circ} > 3^{\circ}$.$4$. Comparing the given structures:- $I$ is a $1^{\circ}$ carbanion.- $II$ is a $2^{\circ}$ carbanion.- $III$ is a $3^{\circ}$ carbanion.- $IV$ is resonance-stabilized.Therefore,the increasing order of stability is $III < II < I < IV$.

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

Similar Questions

Identify the correct order of stability of the following carbocations:

In which of the following species is the central $C$-atom negatively charged?

Which of the following compounds forms the most stable carbocation: $(CH_3)_3C-Br$,$(C_6H_5)_3CBr$,$(C_6H_5)_2CHBr$,and $C_6H_5CH_2Br$?

The most stable carbonium ion among the following is:

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:

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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$