Given the single-step reaction shown, draw the curved-arrow mechanism. Add curved arrows. Select Draw Rings More Erase H,C-CH2-0-CH3 + + : Br: Ho CH HC Br : 3 2 a
The Correct Answer and Explanation is:
To fully answer your question about drawing the curved-arrow mechanism for the reaction involving H₃C-CH₂-O-CH₃ (an ether), HBr (hydrobromic acid), and Br⁻, we’ll need to analyze what type of reaction is occurring and explain it thoroughly.
Correct Mechanism Overview:
The reaction shown is most likely an acidic cleavage of an ether by hydrobromic acid (HBr).
Starting reactants:
- H₃C-CH₂-O-CH₃: This is ethyl methyl ether.
- HBr: A strong acid that can protonate the ether oxygen.
Goal:
To cleave the ether bond and form ethanol (or ethyl bromide) and methyl bromide, depending on the pathway.
Step-by-Step Curved-Arrow Mechanism:
- Protonation of the ether oxygen:
The lone pair on the ether oxygen attacks the hydrogen from HBr, forming a protonated ether (R-O⁺H-R’) and leaving a Br⁻ ion.
Curved arrow: From the lone pair on oxygen to the hydrogen of HBr, and another from the H-Br bond to Br. - Cleavage of the C-O bond:
Once protonated, the oxygen has a positive charge and is more electrophilic.
The bond between CH₃-OH⁺ (or CH₂CH₃-OH⁺) breaks, and the electrons go to the oxygen, forming CH₃OH₂⁺ and leaving behind a CH₃⁺ carbocation or CH₂CH₃⁺ carbocation.
(More stable carbocation is formed — in this case, ethyl is more stable than methyl.) - Nucleophilic attack by Br⁻:
The Br⁻ ion attacks the carbocation formed, resulting in formation of ethyl bromide and methanol (or methyl bromide and ethanol, depending on the carbocation that forms).
This reaction is a classic example of acidic cleavage of ethers, which occurs in the presence of strong acids like HBr or HI. Ethers are generally unreactive, but under acidic conditions, they can undergo substitution reactions.
The mechanism begins with protonation of the ether oxygen by HBr. The oxygen atom has lone electron pairs, which make it a Lewis base. It donates a pair of electrons to a hydrogen ion (H⁺), forming an oxonium ion (a positively charged oxygen species). This step increases the polarity of the adjacent C-O bonds, making one of the alkyl groups more susceptible to leaving.
In the next step, the bond between the oxygen and one of the alkyl groups breaks, with the electrons going back to the oxygen. This results in the formation of a carbocation — a highly reactive intermediate. Between methyl and ethyl carbocations, the ethyl carbocation is more stable due to slight inductive effects and hyperconjugation.
Finally, the bromide ion (Br⁻), which was formed in the first step when HBr dissociated, acts as a nucleophile. It attacks the carbocation, forming a new C-Br bond and yielding ethyl bromide as the product. The other product is methanol, which results from the oxygen atom taking both electrons during bond cleavage.
This is a unimolecular substitution reaction (SN1) mechanism because it proceeds through a carbocation intermediate. This mechanism explains why ethers, although relatively inert, can be cleaved under acidic conditions, especially with strong nucleophiles like halides.
