The total number of alkyl bromides (including | vedclass

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(C) The reaction of $3,3-dimethylbut-1-ene$ with $HBr$ proceeds via the formation of a carbocation intermediate. Initially, the protonation of the dou

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$3$

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(C) The reaction of $3,3-dimethylbut-1-ene$ with $HBr$ proceeds via the formation of a carbocation intermediate. Initially, the protonation of the double bond yields a secondary carbocation, which undergoes a $1,2-methyl$ shift to form a more stable tertiary carbocation. $1$. The direct addition of $Br^-$ to the secondary carbocation gives $2-bromo-3,3-dimethylbutane$. This molecule has a chiral center, so it exists as a pair of enantiomers ($R$ and $S$). $2$. The addition of $Br^-$ to the rearranged tertiary carbocation gives $2-bromo-2,3-dimethylbutane$, which is achiral. Thus, the total number of alkyl bromides formed is $2$ (enantiomers) $+ 1$ (rearranged product) $= 3$.

The total number of alkyl bromides (including stereoisomers) formed in the reaction $(CH_3)_3C-CH=CH_2 + HBr \rightarrow$ will be

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