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Chapter 8: Problem 40

Treatment of 4 -penten-1-ol with aqueous \(\mathrm{Br}_{2}\) yields a cyclic bromo ether rather than the expected bromohydrin. Suggest a mechanism, using curved arrows to show electron movement.

Short Answer

Expert verified
The mechanism involves a bromonium ion intermediate followed by nucleophilic attack by the hydroxy group to form a cyclic bromo ether.

Step by step solution

01

Formation of Bromonium Ion

When 4-penten-1-ol contacts aqueous \(\mathrm{Br}_2\), the \(\mathrm{Br}_2\) molecule approaches the double bond. The \(\pi\)-electrons of the alkene interact with \(\mathrm{Br}_2\), leading to the formation of a bromonium ion. One bromine atom attaches to one of the carbons in the double bond, while the other gains a negative charge and leaves temporarily. The bridged bromonium ion results in a three-membered ring structure.
02

Nucleophilic Attack by Hydroxy Group

The oxygen of the hydroxy group in 4-penten-1-ol acts as a nucleophile and attacks the more substituted carbon of the bromonium ion, opening the three-membered ring. This forms a cyclic ether where the oxygen is now part of a new cyclic structure containing the bromine atom.
03

Stabilization of the Intermediate

The attack by the hydroxyl group stabilizes the intermediate. The oxygen now forms a stable cyclic bromo ether by bonding to the carbon that was part of the original \(\pi\)-bond.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bromonium Ion Formation
When we treat 4-penten-1-ol with aqueous bromine (\( \mathrm{Br}_2 \)), an interesting transformation begins with the formation of a bromonium ion. Here鈥檚 how it works:
  • The \( \pi \)-electrons of the alkene double bond move towards the \( \mathrm{Br}_2 \) molecule, causing a rearrangement of bonds.
  • One of the bromine atoms forms a bond with one of the carbons of the double bond. This leads the other bromine atom to temporarily gain a negative charge and leave as \( \mathrm{Br}^- \).
  • The result is a positively charged, three-membered ring structure known as a bromonium ion, where the bromine now shares bonds with both carbons from the original double bond.
This formation creates tension in the ring due to the three-membered structure, making it highly reactive and a prime target for subsequent reactions.
Nucleophilic Attack
Once the bromonium ion has formed, the reaction proceeds via a nucleophilic attack by the hydroxy group on the 4-penten-1-ol:
  • The oxygen atom in the hydroxy group is slightly negative and has a pair of non-bonding electrons, making it a perfect nucleophile.
  • This oxygen attacks the more substituted carbon of the bromonium ion, which is more electrophilic and accessible.
  • The attack helps break the strained three-membered bromonium ring, facilitating the reaction towards products.
The opening of the ring by the hydroxy group spells out a new step in the reaction, leading towards the formation of a cyclic structure that includes the oxygen atom as part of this newly formed product.
Cyclic Bromo Ether
The final step in this transformation involves the stabilization of the intermediate into a cyclic bromo ether:
  • The attack from the hydroxyl group onto the bromonium ion results in a ring closure, forming a five-membered cyclic ether structure.
  • This cyclic structure has both bromine and oxygen as part of the ring, creating what we refer to as a cyclic bromo ether.
  • This product is more stable than if the reaction progressed to form a bromohydrin, due to the favorable formation of a stable ring structure.
This entire sequence from bromonium ion formation, nucleophilic attack, to cyclic structure creation, demonstrates the elegant dance of atoms in organic reactions, making the final product a fascinating cyclic bromo ether.

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Most popular questions from this chapter

What product would you expect from the reaction of cyclopentene with NBS and water? Show the stereochemistry.

The cis and trans isomers of 2 -butene give different cyclopropane products in the Simmons-Smith reaction. Show the structures of both, and explain the difference.

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