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.
1 answer:
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)
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Answer:
= 0.134;
= 0.866
The partial pressure of isopropanol = 34.04 Torr; The partial pressure of propanol = 5.26 Torr
Explanation:
For each of the solutions:
mole fraction of isopropanol () = 1 - mole fraction of propanol (
).
Given: mole fraction of propanol = 0.247. Thus, the mole fraction of isopropanol = 1 - 0.247 = 0.753.
Furthermole, the partial pressure of isopropanol = *vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr
The partial pressure of propanol = *vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr
Similarly,
In the vapor phase,
The mole fraction of propanol () =
Where, is the partial pressure of propanol and
is the partial pressure of isopropanol.
Therefore,
= 5.26/(34.04+5.16) = 0.134
= 1 - 0.134 = 0.866