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

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(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

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(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?

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