Arrange the following carbocations in decreas | vedclass

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(D) The stability of carbocations is determined by factors like resonance and inductive effect.$I$ is an ethyl carbocation $(CH_3-CH_2^+)$,which is a

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$III > II > I$

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(D) The stability of carbocations is determined by factors like resonance and inductive effect.$I$ is an ethyl carbocation $(CH_3-CH_2^+)$,which is a primary $(1^{\circ})$ carbocation stabilized only by the inductive effect of one methyl group.$II$ is an isopropyl carbocation $((CH_3)_2CH^+)$,which is a secondary $(2^{\circ})$ carbocation stabilized by the inductive effect and hyperconjugation of two methyl groups.$III$ is a benzyl carbocation $(Ph-CH_2^+)$,which is highly stabilized by resonance with the phenyl ring.Comparing their stabilities: Resonance stabilization in $III$ makes it the most stable. Between $I$ and $II$,the secondary carbocation $II$ is more stable than the primary carbocation $I$ due to more hyperconjugation and inductive effect.Therefore,the decreasing order of stability is $III > II > I$.

Arrange the following carbocations in decreasing order of stability:
$I: CH_3-CH_2^+$
$II: (CH_3)_2CH^+$
$III: Ph-CH_2^+$

Similar Questions

The incorrect statement about carbene $(CH_2)$ is:

The order of stability of the following carbocations is:
$I$. $CH_2 = CH - CH_2^+$
$II$. $CH_3 - CH_2 - CH_2^+$
$III$. $C_6H_5CH_2^+$

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:

Which of the following is the most stable carbocation?

The correct order of stability is:

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Arrange the following carbocations in decreasing order of stability:$I: CH_3-CH_2^+$$II: (CH_3)_2CH^+$$III: Ph-CH_2^+$