Arrange the following carbocations in the cor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The stability order is $II < I < IV < III$.$1$. Structure $III$ (methoxycyclohexyl cation) is the most stable due to the $+M$ effect of the oxygen

Vedclass · Accepted Answer

$II < I < IV < III$

Vedclass · Answer

(A) The stability order is $II $1$. Structure $III$ (methoxycyclohexyl cation) is the most stable due to the $+M$ effect of the oxygen atom,which donates electron density to the positively charged carbon.$2$. Structure $IV$ (ethylcyclohexyl cation) is more stable than $I$ (methylcyclohexyl cation) because the ethyl group provides more hyperconjugation and inductive stabilization than the methyl group.$3$. Structure $II$ is a bridgehead carbocation. According to Bredt's rule,bridgehead carbocations in small bicyclic systems are highly unstable because they cannot achieve the required $sp^2$ planar geometry due to severe ring strain.

Arrange the following carbocations in the correct order of stability:

Similar Questions

The species $A$,$B$,$C$,$D$ formed in the following bond cleavages respectively are:

What is the decreasing order of stability (most stable $\to$ least stable) of the following carbocations?

Which one of the carbanions is the least stable?

Assertion : Carbanions like ammonia have pyramidal shape.
Reason : The carbon atom carrying negative charge has an octet of electrons.

Which of the following is the most stable free radical?

Vedclass Test Series

Exam Paper Generator

Online Exam Module