To fully answer your question about drawing the curved-arrow mechanism for the reaction involving H\u2083C-CH\u2082-O-CH\u2083<\/strong> (an ether), HBr<\/strong> (hydrobromic acid), and Br\u207b<\/strong>, we\u2019ll need to analyze what type of reaction is occurring and explain it thoroughly.<\/p>\n\n\n\n The reaction shown is most likely an acidic cleavage of an ether<\/strong> by hydrobromic acid (HBr)<\/strong>.<\/p>\n\n\n\n Starting reactants<\/strong>:<\/p>\n\n\n\n Goal<\/strong>: This reaction is a classic example of acidic cleavage of ethers<\/strong>, 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.<\/p>\n\n\n\n The mechanism begins with protonation<\/strong> 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\u207a), forming an oxonium ion<\/strong> (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.<\/p>\n\n\n\n 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<\/strong> \u2014 a highly reactive intermediate. Between methyl and ethyl carbocations, the ethyl carbocation is more stable due to slight inductive effects and hyperconjugation.<\/p>\n\n\n\n Finally, the bromide ion (Br\u207b)<\/strong>, which was formed in the first step when HBr dissociated, acts as a nucleophile<\/strong>. It attacks the carbocation, forming a new C-Br bond and yielding ethyl bromide<\/strong> as the product. The other product is methanol<\/strong>, which results from the oxygen atom taking both electrons during bond cleavage.<\/p>\n\n\n\n This is a unimolecular substitution reaction (SN1)<\/strong> 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.<\/p>\n\n\n\n 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 H\u2083C-CH\u2082-O-CH\u2083 (an ether), HBr (hydrobromic […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-218017","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218017","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=218017"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218017\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218017"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218017"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218017"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nCorrect Mechanism Overview<\/strong>:<\/h3>\n\n\n\n
\n
To cleave the ether bond and form ethanol (or ethyl bromide)<\/strong> and methyl bromide<\/strong>, depending on the pathway.<\/p>\n\n\n\n
\n\n\n\nStep-by-Step Curved-Arrow Mechanism<\/strong>:<\/h3>\n\n\n\n
\n
The lone pair on the ether oxygen<\/strong> attacks the hydrogen from HBr<\/strong>, forming a protonated ether<\/strong> (R-O\u207aH-R’) and leaving a Br\u207b<\/strong> ion.
Curved arrow<\/strong>: From the lone pair on oxygen to the hydrogen of HBr, and another from the H-Br bond to Br.<\/li>\n\n\n\n
Once protonated, the oxygen has a positive charge and is more electrophilic.
The bond between CH\u2083-OH\u207a<\/strong> (or CH\u2082CH\u2083-OH\u207a) breaks, and the electrons go to the oxygen, forming CH\u2083OH\u2082\u207a<\/strong> and leaving behind a CH\u2083\u207a<\/strong> carbocation or CH\u2082CH\u2083\u207a<\/strong> carbocation.
(More stable carbocation is formed \u2014 in this case, ethyl is more stable than methyl.)<\/li>\n\n\n\n
The Br\u207b ion<\/strong> attacks the carbocation<\/strong> formed, resulting in formation of ethyl bromide<\/strong> and methanol<\/strong> (or methyl bromide and ethanol, depending on the carbocation that forms).<\/li>\n<\/ol>\n\n\n\n
\n\n\n\n<\/h3>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-119.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"