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Viefleur [7K]
3 years ago
15

Draw the structure of the bromohydrin formed when (Z)-3-hexene reacts with Br2/H2O. Use the wedge/hash bond tools to indicate st

ereochemistry where it exists. If the reaction produces a racemic mixture, draw both stereoisomers. Separate multiple products using the sign from the drop-down menu.

Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
5 0

Answer:

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

Explanation:

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

Finally, in the product we will have <u>chiral carbons</u>, 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)

I hope it helps!

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