3-Methylpent-2-ene on reaction with HBr in th | vedclass

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Question

(D) $3-$Methylpent$-2-$ene $(CH_3-CH=C(CH_3)-CH_2-CH_3)$ reacts with $HBr$ in the presence of peroxide via anti-Markovnikov addition to form $2-$bromo

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$4$

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(D) $3-$Methylpent$-2-$ene $(CH_3-CH=C(CH_3)-CH_2-CH_3)$ reacts with $HBr$ in the presence of peroxide via anti-Markovnikov addition to form $2-$bromo$-3-$methylpentane $(CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3)$.The product $CH_3-CH(Br)-CH(CH_3)-CH_2-CH_3$ contains two chiral centers at $C2$ and $C3$.Since the molecule is unsymmetrical,the number of stereoisomers is given by the formula $2^n$,where $n$ is the number of chiral centers.Here,$n = 2$,so the number of stereoisomers $= 2^2 = 4$.

$3-$Methylpent$-2-$ene on reaction with $HBr$ in the presence of peroxide forms an addition product. The number of possible stereoisomers for the product is:

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