Taking into account the stability of various | vedclass

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

Question

(A) The reaction of cyclopropyl-ethene with $HBr$ initiates with the protonation of the double bond to form a cyclopropylmethyl carbocation. This carb

Vedclass · Accepted Answer

Vedclass · Answer

(A) The reaction of cyclopropyl-ethene with $HBr$ initiates with the protonation of the double bond to form a cyclopropylmethyl carbocation. This carbocation is stabilized by $\sigma$-resonance (bent bonds of the cyclopropane ring). However,it can undergo ring expansion to relieve ring strain and form a more stable carbocation. The cyclopropylmethyl carbocation can rearrange to a cyclobutyl carbocation. The cyclobutyl carbocation is then attacked by the nucleophile $Br^-$ to form cyclobutyl bromide. Therefore,the pathway involving ring expansion to a cyclobutyl system is the most likely outcome.

Taking into account the stability of various carbocations and the rules governing mechanisms of carbocation rearrangements,which reaction is most likely to occur during the given reaction?

Similar Questions

The compound $(X)$ will be:

Which of the following carbocations is most stable?
$A) (CH_3)_2CH-CH^+-CH_3$
$B) (CH_3)_2C^+-CH_2CH_3$

The most stable carbonium ion is:

Define carbocation,carbanion,and free radical.

In which of the following compounds,the $C-Cl$ bond ionisation shall give the most stable carbonium ion?

Vedclass Test Series

Exam Paper Generator

Online Exam Module