Consider the addition of HBr to 3,3-Dimethyl- | vedclass

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

Question

(D) The reaction proceeds via the following steps:$1$. Protonation of the alkene ($3,3$-Dimethyl-$1$-butene) by $HBr$ occurs to form a secondary carbo

Vedclass · Accepted Answer

Protonation of the alkene followed by a methyl shift and addition of bromide to the carbocation

Vedclass · Answer

(D) The reaction proceeds via the following steps:$1$. Protonation of the alkene ($3,3$-Dimethyl-$1$-butene) by $HBr$ occurs to form a secondary carbocation $(CH_3-C(CH_3)_2-CH^+-CH_3)$.$2$. This secondary carbocation undergoes a $1,2$-methyl shift to form a more stable tertiary carbocation $(CH_3-C^+(CH_3)-CH(CH_3)_2)$.$3$. Finally,the bromide ion $(Br^-)$ attacks the tertiary carbocation to yield the final product,$2$-bromo-$2,3$-dimethylbutane.

Consider the addition of $HBr$ to $3,3$-Dimethyl-$1$-butene shown below. What is the best mechanistic explanation for the formation of the observed product?

Similar Questions

The major product of the following reaction is $CH_3-CH(Br)-CH_2-CH_3 + \text{alc. } KOH \to $

Which of the following reagents is used to test for the unsaturation of an alkene?

The major product formed during the reaction of $1$-methylcyclopentene with $CH_3CO_3H$ is

Which of the following is the most stable alkene?

What is the product of the reaction of $CH_3CH=CH-C_6H_4-OH$ with $HBr$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module