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!
Answer:
The answer to your question is the letter A. 1
Explanation:
Unbalanced chemical reaction
Mg + O₂ ⇒ MgO
Reactants Elements Products
1 Magnesium 1
2 Oxygen 1
Balanced chemical reaction
2Mg + O₂ ⇒ 2MgO
Reactants Elements Products
2 Magnesium 2
2 Oxygen 2
Conclusion
The coefficients of the balanced equation are 2, 1, 2
1 is b 2 is a 3 is d 4 is a 5 is c
<h2>Heptene formed is -</h2><h2>
</h2>
Explanation:
The two possibilities when the peroxide is not present
+ HBr →
In presence peroxide,
≡
+ HBr →
- When peroxides are present in the reaction mixture, hydrogen bromide adds to the triple bond of heptane with regioselectivity.
- This reaction is opposite to that of Markovnikov's rule which says that when asymmetrical alkene reacts with a protic acid HX, then the hydrogen of an acid is attached to the carbon with more in number of hydrogen substituents, and the halide (X) group is attached to the carbon with more in number of substituents of alkyl.
- One mole of HBr adds to one mole of 1-heptane.
- The structure of heptene formed is -
