Arrange the following carbocations in decreas | vedclass

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Question

(D) The stability of carbocations is determined by resonance and inductive effects.In structure $B$,the positive charge is on the carbon atom adjacent

Vedclass · Accepted Answer

$B > A > C$

Vedclass · Answer

(D) The stability of carbocations is determined by resonance and inductive effects.In structure $B$,the positive charge is on the carbon atom adjacent to the oxygen atom. The lone pair on the oxygen atom can donate electron density through resonance,which significantly stabilizes the carbocation.In structure $C$,the positive charge is on the carbon atom at the $\beta$-position relative to the oxygen atom. The oxygen atom exerts an electron-withdrawing inductive effect ($-I$ effect),which destabilizes the carbocation.In structure $A$,the carbocation is a simple secondary carbocation without any adjacent heteroatom effects.Therefore,the stability order is $B > A > C$.

Arrange the following carbocations in decreasing order of stability.

Similar Questions

Choose the correct statement regarding the formation of carbocations $A$ and $B$ given:
$CH_3-CH_2-CH=CH_2 + HBr \to CH_3-CH_2-CH_2-CH_2^+ + Br^-$ $(A)$
$CH_3-CH_2-CH=CH_2 + HBr \to CH_3-CH_2-CH^+-CH_3 + Br^-$ $(B)$

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

Due to the presence of an unpaired electron,free radicals are

Which one of the carbanions is the least stable?

The shape of a carbonium ion (carbocation) is:

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