How does the first step in the reaction of propene with Br 2 differ from the first step in the reaction of propene with HBr?

Hello, everyone. Today, we have the following problem, consider how cyclops reacts with hydrochloric acid and how cyclops reacts with chlorine. What makes the first steps step of these two reactions different? So we will label the reaction with hydrochloric acid as one and the reaction with chlorine two. So for reaction one, we have our cyclopse where we draw that alkene at any carbon because it's not specified. And in its reaction with hydrochloric acid, we have the alkene from a nucleic attack on that hydrogen expelling the chlorine. And then of course, this will lead to the following carbo cion intermediate. Now we're only focused on the first step and our second reaction, we have Cyclin once again, but this time, it's reacting with molecular chlorine. And in this reaction, we will have a concerted mechanism where the, where in the alkene attacks one of the chlorine, the same chlorine that it attacks will also attack the other carbon of that alkene. And the chlorine will be dispelled, which will essentially form a bridged intermediate known as a chlorinum ion. And so then, of course, the subsequent reaction will involve an anti addition of two chlorine across this alkene. But with the first step, we can note that the reaction one just involves a simple nucleic attack. However, the second reaction involves a bridged intermediate ion that is formed. And the first step react reaction results in a carbon formation. And so with that, we have solved the problem overall, I hope it helped. And until next time.

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1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
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31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

10. Addition Reactions

Halogenation

Organic Chemistry

10. Addition Reactions

Halogenation

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4:28 minutes

Problem 21a

Textbook Question

How does the first step in the reaction of propene with Br₂ differ from the first step in the reaction of propene with HBr?

Verified step by step guidance

The first step in the reaction of propene with Br₂ involves the formation of a bromonium ion intermediate. This occurs through an electrophilic addition mechanism where the π-electrons of the double bond in propene attack one of the bromine atoms in Br₂, leading to the formation of a cyclic bromonium ion and a bromide ion (Br⁻).

In contrast, the first step in the reaction of propene with HBr involves the formation of a carbocation intermediate. The π-electrons of the double bond in propene attack the hydrogen atom in HBr, resulting in the addition of a proton (H⁺) to one of the carbon atoms in the double bond, leaving behind a carbocation on the other carbon atom.

The key difference lies in the intermediates formed: the reaction with Br₂ forms a bromonium ion, which is a three-membered cyclic structure, while the reaction with HBr forms a carbocation, which is a planar, positively charged species.

The driving force for the reaction with Br₂ is the polarization of the Br-Br bond due to the electron-rich double bond in propene, whereas the reaction with HBr is driven by the electrophilic nature of the proton (H⁺) in HBr.

Understanding these differences is crucial because the nature of the intermediate (bromonium ion vs. carbocation) determines the subsequent steps and the stereochemistry of the final product in each reaction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electrophilic Addition Reactions

Electrophilic addition reactions are a fundamental type of reaction in organic chemistry where an electrophile reacts with a nucleophile, resulting in the formation of a more saturated product. In the case of alkenes like propene, the double bond acts as a nucleophile, attacking the electrophile, which can be a halogen or a hydrogen halide. Understanding this concept is crucial for analyzing how different electrophiles interact with alkenes.

Mechanism of Bromination

The bromination of alkenes involves the formation of a cyclic bromonium ion intermediate, which is a key step in the reaction mechanism. This intermediate is formed when Br2 approaches the double bond, leading to a three-membered ring structure. The subsequent attack by a bromide ion results in the formation of a vicinal dibromide. This mechanism differs from that of hydrogen halides, where no such cyclic intermediate is formed.

Hydrogen Halide Addition Mechanism

The addition of hydrogen halides, such as HBr, to alkenes follows a different mechanism compared to halogen addition. In this case, the hydrogen atom from HBr acts as the electrophile, leading to the formation of a carbocation intermediate. The stability of this carbocation influences the regioselectivity of the reaction, which is not a factor in the bromination mechanism. Understanding these differences is essential for predicting the products of these reactions.

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