The correct stability order of the following | vedclass

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(C) The stability of carbocations is determined by the electron-donating or electron-withdrawing nature of the substituents attached to the conjugated

Vedclass · Accepted Answer

$III > IV > I > II$

Vedclass · Answer

(C) The stability of carbocations is determined by the electron-donating or electron-withdrawing nature of the substituents attached to the conjugated system.$1$. In $III$ and $IV$, the groups $-NMe_2$ and $-OMe$ show a strong $+R$ (resonance) effect, which stabilizes the carbocation significantly.$2$. Since nitrogen is less electronegative than oxygen, the lone pair on $N$ is more easily donated, making $-NMe_2$ a stronger $+R$ group than $-OMe$. Thus, $III > IV$.$3$. In $I$, the $-CH_3$ group shows a weak $+I$ (inductive) effect and hyperconjugation, providing moderate stability.$4$. In $II$, the $-BMe_2$ group has an empty $p$-orbital and shows a $-R$ effect, which destabilizes the carbocation.$5$. Therefore, the correct stability order is $III > IV > I > II$.

The correct stability order of the following carbocations is:
$I$. $H_2C^+-CH=CH-CH_3$
$II$. $H_2C^+-CH=CH-BMe_2$
$III$. $H_2C^+-CH=CH-NMe_2$
$IV$. $H_2C^+-CH=CH-OMe$

Similar Questions

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

Which of the following is the most stable carbocation?

The shape of a carbonium ion (carbocation) is:

Most stable carbocation among the following is:

$CH_3-C(CH_3)_2-CH_2-CH_2^+ \to CH_3-C^+(CH_3)-CH(CH_3)-CH_3$
How many $H^-$ shifts are involved in the above rearrangement?

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The correct stability order of the following carbocations is:$I$. $H_2C^+-CH=CH-CH_3$$II$. $H_2C^+-CH=CH-BMe_2$$III$. $H_2C^+-CH=CH-NMe_2$$IV$. $H_2C^+-CH=CH-OMe$