Draw the structure of the bromohydrin formed when (Z)-3-hexene reacts with Br2/H2O. Use the wedge/hash bond tools to indicate stereochemistry where it exists. If the reaction produces a racemic mixture, draw both stereoisomers. Separate multiple products using the sign from the drop-down menu.

(3R,4R)-4-bromohexan-3-ol

Explanation:

In this case, we have  reaction called halohydrin formation. This is a markovnikov reaction with anti configuration. Therefore the halogen in this case "Br" and the "OH" must have different configurations. Additionally, in this molecule both carbons have the same substitution, so the "OH" can go in any carbon.

Finally, in the product we will have chiral carbons, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)

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