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

Arrange the following carbocations in the correct order of stability:

Which of the following is the most stable carbanion?

The correct stability order for the following species is:
$(I)$ $o-CN-C_6H_4-CH_2^-$
$(II)$ $m-CN-C_6H_4-CH_2^-$
$(III)$ $p-CN-C_6H_4-CH_2^-$
$(IV)$ $C_6H_5-CH_2^-$
(Note: The image shows the structures of these carbanions.)

In which pair is the second ion more stable than the first?

In which of the following reactions does a rearrangement take place with a change in the carbon skeleton?

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