The optically active alkene shown below was treated with HBr with and without pe

Question

The optically active alkene shown below was treated with HBr with and without peroxides. One of these reactions lead to the formation of two products, one of which is not optically active. The other reaction gives a single new product that is optically active. Which of the below statements best explains the observed results. The addition of HBr to the alkene in the absence of peroxide forms a carbocation intermediate that scrambles the original stereogenic center, giving two products. The addition of HBr to the alkene in the presence of peroxide forms the less-substituted bromide. The addition of HBr to the alkene in the absence of peroxide forms the less-substituted bromide. The addition of HBr to the alkene in the presence of peroxide follows Markovnikov?s rule, forming a chiral compound and a diastereomer that is meso. The addition of HBr to the alkene in the presence of peroxides brominates at either carbon of the double bond, giving a chiral product and a meso product. The product of addition in the absence of peroxides is chiral. The addition of HBr to the alkene in the absence of peroxide follows Markovnikov?s rule, forming a chiral compound and a diastereomer that is meso. The addition of HBr to the alkene in the presence of peroxide forms the less-substituted bromide.

Explanation / Answer

my answer is D

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