Answer:
The structures are shown in the figure.
Explanation:
The primary hydrogens are those which are attached to primary carbon.
Primary carbons are the carbons which are attached to only one carbon.
Primary carbons is bonded to three hydrogens.
In order to draw such structure we will draw structures which will have carbon with three hydrogens or no hydrogens (quaternary)
The structures are shown in the figure with clear marking.
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!
