CH_3-CH=CH-CH_3 xrightarrow[R_2O_2, Delta]{HB | vedclass

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(A) The reaction proceeds via a free radical mechanism.In the first step,the bromine radical $(Br^{\bullet})$ attacks the $sp^2$-hybridized carbon ato

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Racemic

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(A) The reaction proceeds via a free radical mechanism.In the first step,the bromine radical $(Br^{\bullet})$ attacks the $sp^2$-hybridized carbon atom of the symmetric alkene $CH_3-CH=CH-CH_3$.This attack results in the formation of a planar radical intermediate,$CH_3-\dot{C}H-CH(Br)-CH_3$.In the subsequent step,the hydrogen atom from $HBr$ can attack either face of the planar radical with equal probability.This leads to the formation of both enantiomers of $2-bromobutane$ in equal amounts,resulting in a racemic mixture.

$CH_3-CH=CH-CH_3 \xrightarrow[R_2O_2, \Delta]{HBr} \text{ (Anti-Markownikoff's addition)}$
Comment on the optical activity of the products.

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$CH_3-CH=CH-CH_3 \xrightarrow[R_2O_2, \Delta]{HBr} \text{ (Anti-Markownikoff's addition)}$Comment on the optical activity of the products.