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Two stereoisomers are obtained from the reaction of HBr with (S)-4-bromo-1-pentene. One of the stereoisomers is optically active, and the other is not. Draw the structure of the optically active stereoisomer.

User UncleO
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Answer:

See explanation

Step-by-step explanation:

In this case, we have an addition reaction. Additionally, this is a marknovnikov addition, therefore the "Br" atom would be added in the most substituted carbon (in this case carbon a). And we are going to have 2 enantiomers (2S,4S)-2,4-dibromopentane and (2R,4S)-2,4-dibromopentane. In the case of (2R,4S)-2,4-dibromopentane we will have a symmetry plane (a point in the molecule in which we can divide the molecule into two equal parts). When this happens we will have a mesocompound and we will not have optical activity.

See figure 1

I hope it helps!

Two stereoisomers are obtained from the reaction of HBr with (S)-4-bromo-1-pentene-example-1
User Lee Theobald
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