Boron tribromide (BBr3) cleaves ethers to give alkyl halides and alcohols. The reaction is thought to involve attack by a bromide ion on the Lewis acid–base adduct of the ether with BBr 3 (a strong Lewis acid). Propose a mechanism for the reaction of butyl methyl ether with BBr 3 to give (after hydrolysis) butan-1-ol and bromomethane.

Hello, everyone. Today, we have the following problem. The reagent boron tribromide is used to clear es producing alcohols and alkyl Halide after a subsequent hydrolysis reaction. Then we have a general reaction sequence shown here. This mechanism involves a bromide ion attacking a Lewis base adduct of the ether with our boron tribromide suggest a mechanism for the reaction of our ero ether with our bro boron tribromide and a subsequent hydrolysis reaction to producing alcohol propane one all and aroy ethane. So we have our starting reacting which is our ethyl propal E and with naming eats, we simply just need to draw the R group or the alky groups attached on either side of an oxygen. So we have an ethyl group, we have an ethel group and we have a purple group. No, this will first have to essentially react with our boron tetra tribromide first. So what this will look like because we will have the following molecules and we will have first a nucleic attack of the eater on the boron boron tribromide as the boron is able to act as a leis acid and accept those electrons. And in doing so, we have the following intermediate, we now our boron is negatively charged and her oxygen is positively charged. This is our adduct and this will stimulate the loss of one of the bromine ions which is favorable because bromide is a good leaving group as it is stable once it leaves the molecule. So once it leaves the molecule, we will have the following where we've stabilized the boron. But now we need to stabilize the oxygen. And so what we'll do is the bromide that just left will perform a nucleic attack on the carbon of that oxygen, which will essentially force the electrons between the carbon and the oxygen onto the subsequent oxygen to give us the following intermediate. And so now we have stabilized the oxygen and then we have our by product of our Brummel ethane. What occurs next is we will have our water and our hydrolysis reaction, our water will once again act on this Luis acid or the boron, which will nucleic at perform a nic attack on that molecule which gives us the following. And so as before our born was negatively charged and it is once more. So what will need to occur then is we will need to have an in intermolecular proton transfer using a pair of a long pair of electrons on this oxygen. You will deproteinate that oxygen. And upon doing that, we will return the electrons to that oxygen giving us this intermediate. And so one of our final steps, we'll be essentially losing this or group or this, either group as it is also stable on its own once it leaves the molecule. And so once it leaves the molecule, we will have formed two products. One being our prop pan, one all and the other and the other being our alcohol attached to that boron complex. And so with that, we have solved the problem overall, I hope it's 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
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
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

12. Alcohols, Ethers, Epoxides and Thiols

Ether Cleavage

Organic Chemistry

12. Alcohols, Ethers, Epoxides and Thiols

Ether Cleavage

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Problem 16

Textbook Question

Boron tribromide (BBr3) cleaves ethers to give alkyl halides and alcohols.

The reaction is thought to involve attack by a bromide ion on the Lewis acid–base adduct of the ether with BBr₃ (a strong Lewis acid). Propose a mechanism for the reaction of butyl methyl ether with BBr₃ to give (after hydrolysis) butan-1-ol and bromomethane.

Verified step by step guidance

Step 1: Recognize the role of BBr3 as a strong Lewis acid. It interacts with the ether oxygen atom, which has lone pairs of electrons, forming a Lewis acid-base adduct. This interaction increases the electrophilicity of the ether oxygen, making it more susceptible to nucleophilic attack.

Step 2: Propose the first step of the mechanism. The ether (butyl methyl ether, CH3-O-C4H9) reacts with BBr3 to form a Lewis acid-base adduct. This can be represented as CH3-O-C4H9 + BBr3 → CH3-O-BBr2-C4H9. The oxygen atom donates a lone pair to the boron atom, forming a coordinate covalent bond.

Step 3: Introduce the nucleophilic attack by bromide ion (Br⁻). The bromide ion attacks the carbon atom of the methyl group (CH3) in the ether, leading to cleavage of the C-O bond. This results in the formation of bromomethane (CH3Br) and a butoxyboron intermediate (C4H9-O-BBr2).

Step 4: Describe the hydrolysis step. The butoxyboron intermediate (C4H9-O-BBr2) undergoes hydrolysis in the presence of water. This reaction produces butan-1-ol (C4H9OH), boric acid (B(OH)3), and hydrogen bromide (HBr).

Step 5: Summarize the overall reaction. The mechanism involves the formation of a Lewis acid-base adduct, nucleophilic attack by bromide ion, cleavage of the ether bond, and subsequent hydrolysis to yield the final products: butan-1-ol (C4H9OH) and bromomethane (CH3Br).

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Key Concepts

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

Lewis Acids and Bases

Lewis acids are substances that can accept an electron pair, while Lewis bases are those that can donate an electron pair. In the context of the reaction with BBr3, the ether acts as a Lewis base, forming a complex with the Lewis acid BBr3. This interaction is crucial for the subsequent steps of the reaction, as it facilitates the cleavage of the ether bond.

Ethers and Their Cleavage

Ethers are organic compounds characterized by an oxygen atom bonded to two alkyl or aryl groups. The cleavage of ethers typically involves breaking the C-O bond, which can be achieved through various reagents, including strong Lewis acids like BBr3. Understanding the mechanism of ether cleavage is essential for predicting the products formed in the reaction.

Nucleophilic Substitution Reactions

Nucleophilic substitution reactions involve the replacement of a leaving group by a nucleophile. In the proposed mechanism, the bromide ion acts as a nucleophile that attacks the carbon atom bonded to the leaving group (the alkyl part of the ether). This step is critical for generating the alkyl halide and alcohol products after hydrolysis.

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