Choose the correct statement regarding the fo | vedclass

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

Question

(A) In the electrophilic addition of $HBr$ to but$-1-$ene,the proton $(H^+)$ attacks the double bond to form a carbocation intermediate.Carbocation $A

Vedclass · Accepted Answer

Carbocation $B$ is more stable and formed relatively at a faster rate.

Vedclass · Answer

(A) In the electrophilic addition of $HBr$ to but$-1-$ene,the proton $(H^+)$ attacks the double bond to form a carbocation intermediate.Carbocation $A$ is a primary $(1^\circ)$ carbocation $(CH_3-CH_2-CH_2-CH_2^+)$,while carbocation $B$ is a secondary $(2^\circ)$ carbocation $(CH_3-CH_2-CH^+-CH_3)$.Secondary carbocations are more stable than primary carbocations due to the inductive effect and hyperconjugation.Because carbocation $B$ is more stable,its formation has a lower activation energy,meaning it is formed at a faster rate.

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

Similar Questions

What is the minimum energy required for the homolytic bond fission of the $C-H$ bond in the following molecule?

Rank the transition states that occur during the following reaction steps in order of increasing stability (least $\to$ most stable):
$1.$ $CH_3-O^+H_2 \to CH_3^+ + H_2O$
$2.$ $(CH_3)_3C-O^+H_2 \to (CH_3)_3C^+ + H_2O$
$3.$ $(CH_3)_2CH-O^+H_2 \to (CH_3)_2CH^+ + H_2O$

Which of the following species $(A)$,$(B)$,$(C)$,and $(D)$ contain an $sp^2$ hybridized carbon atom? $(A) CH_3^+$,$(B) CH_3^-$,$(C) \dot{C}H_3$,$(D) :CH_2$

The correct stability order for the following carbocations is:

Which among the following carbocations has the maximum stability?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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^-$ $(B)$