An alkene X (C_4H_8) on reaction with HBr gav | vedclass

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(A) $1$. The alkene $X$ is $2$-methylpropene $(CH_3-C(CH_3)=CH_2)$.$2$. Reaction of $2$-methylpropene with $HBr$ follows Markovnikov's rule to give $Y

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$2-$methylpropene,tert-butyl bromide,tert-butylbenzene

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(A) $1$. The alkene $X$ is $2$-methylpropene $(CH_3-C(CH_3)=CH_2)$.$2$. Reaction of $2$-methylpropene with $HBr$ follows Markovnikov's rule to give $Y$,which is $tert$-butyl bromide $(CH_3-C(Br)(CH_3)-CH_3)$.$3$. Reaction of $tert$-butyl bromide with benzene in the presence of anhydrous $AlCl_3$ (Friedel-Crafts alkylation) gives $Z$,which is $tert$-butylbenzene $(C_6H_5-C(CH_3)_3)$.$4$. $tert$-butylbenzene has no benzylic hydrogen atoms,so it is resistant to oxidation by $KMnO_4-KOH$.

An alkene $X$ $(C_4H_8)$ on reaction with $HBr$ gave $Y$ $(C_4H_9Br)$. Reaction of $Y$ with benzene in the presence of anhydrous $AlCl_3$ gave $Z$,which is resistant to oxidation with $KMnO_4-KOH$. What are $X, Y, Z$ respectively?

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