CH_3-C(CH_3)=CH_2 xrightarrow[R_2O_2]{HBr} X | vedclass

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(A) $1$. The reaction of $CH_3-C(CH_3)=CH_2$ with $HBr$ in the presence of peroxide $(R_2O_2)$ follows the Anti-Markovnikov addition (Kharasch effect)

Vedclass · Accepted Answer

$X = CH_3-CH(CH_3)-CH_2-Br, Y = CH_3-CH(CH_3)-CH_2-OCH_3$

Vedclass · Answer

(A) $1$. The reaction of $CH_3-C(CH_3)=CH_2$ with $HBr$ in the presence of peroxide $(R_2O_2)$ follows the Anti-Markovnikov addition (Kharasch effect). The $Br$ atom attaches to the less substituted carbon,yielding $X = CH_3-CH(CH_3)-CH_2-Br$ (isobutyl bromide).$2$. The reaction of the primary alkyl halide $(X)$ with sodium methoxide $(CH_3ONa)$ proceeds via an $S_N2$ mechanism to form an ether,$Y = CH_3-CH(CH_3)-CH_2-OCH_3$ (methyl isobutyl ether).

$CH_3-C(CH_3)=CH_2$ $\xrightarrow[R_2O_2]{HBr} X$ $\xrightarrow{CH_3ONa} Y$; $X$ and $Y$ are:

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