{"id":262252,"date":"2025-07-20T13:35:06","date_gmt":"2025-07-20T13:35:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=262252"},"modified":"2025-07-20T13:35:09","modified_gmt":"2025-07-20T13:35:09","slug":"when-2-methyl-13-butadiene-reacts-in-the-presence-of-hbr-the-major-product-is-1-bromo-3-methyl-2-butene","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/20\/when-2-methyl-13-butadiene-reacts-in-the-presence-of-hbr-the-major-product-is-1-bromo-3-methyl-2-butene\/","title":{"rendered":"When 2-methyl-1,3-butadiene reacts in the presence of HBr the major product is 1-bromo-3- methyl-2-butene."},"content":{"rendered":"\n<p>When 2-methyl-1,3-butadiene reacts in the presence of HBr the major product is 1-bromo-3- methyl-2-butene. Provide a complete mechanism to show the formation of the product and explain why this product is major. (Hint: consider the reactive intermediates that are possible) 2-methyl-1,3-butadiene HBr Br 1-bromo-3-methyl-2-butene ?? ???? &#8211; ???? +Br B<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Of course. Here is the complete mechanism and explanation for the reaction.<\/p>\n\n\n\n<p>The reaction of 2-methyl-1,3-butadiene with HBr is an electrophilic addition to a conjugated diene. The formation of 1-bromo-3-methyl-2-butene as the major product is governed by the formation of the most stable carbocation intermediate and the thermodynamic stability of the final product.<\/p>\n\n\n\n<p><strong>Complete Mechanism:<\/strong><\/p>\n\n\n\n<p><strong>Step 1: Protonation to form a resonance-stabilized allylic carbocation.<\/strong><br>The reaction begins when one of the pi bonds of the diene acts as a nucleophile, attacking the electrophilic proton of HBr. Protonation occurs at the C1 carbon because this generates the most stable possible carbocation: a tertiary allylic carbocation. This intermediate is highly stabilized by both the inductive effect of the methyl group and by resonance.<\/p>\n\n\n\n<p><strong>Step 2: Nucleophilic attack by bromide ion.<\/strong><br>The positive charge on the allylic carbocation is delocalized across both the C2 and C4 positions, as shown in the resonance structures. The bromide ion (Br\u207b), acting as a nucleophile, can attack either of these electrophilic carbons.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>1,2-Addition (Kinetic Product):<\/strong>\u00a0If the bromide attacks C2, it forms the 1,2-adduct, 4-bromo-3-methyl-1-butene. This product often forms faster because the C2 carbon bears more of the positive charge in the resonance hybrid (it&#8217;s a tertiary vs. a primary carbocation).<\/li>\n\n\n\n<li><strong>1,4-Addition (Thermodynamic Product):<\/strong>\u00a0If the bromide attacks C4, it forms the 1,4-adduct, 1-bromo-3-methyl-2-butene.<\/li>\n<\/ul>\n\n\n\n<p><strong>Explanation for Why 1-bromo-3-methyl-2-butene is the Major Product:<\/strong><\/p>\n\n\n\n<p>The major product is determined by thermodynamic stability. We must compare the stability of the two possible products. The 1,4-adduct, 1-bromo-3-methyl-2-butene, has a double bond that is&nbsp;<strong>trisubstituted<\/strong>&nbsp;(it is bonded to three other carbon atoms). In contrast, the 1,2-adduct, 4-bromo-3-methyl-1-butene, has a double bond that is only&nbsp;<strong>monosubstituted<\/strong>.<\/p>\n\n\n\n<p>According to Zaitsev&#8217;s rule, more highly substituted alkenes are more stable due to hyperconjugation. Because the 1,4-adduct contains a more substituted and therefore more stable double bond, it is the thermodynamically favored product. Under conditions that allow the reaction to reach equilibrium, the more stable product will predominate, making 1-bromo-3-methyl-2-butene the major product of this reaction.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1224.jpeg\" alt=\"\" class=\"wp-image-262269\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>When 2-methyl-1,3-butadiene reacts in the presence of HBr the major product is 1-bromo-3- methyl-2-butene. Provide a complete mechanism to show the formation of the product and explain why this product is major. (Hint: consider the reactive intermediates that are possible) 2-methyl-1,3-butadiene HBr Br 1-bromo-3-methyl-2-butene ?? ???? &#8211; ???? +Br B The Correct Answer and Explanation [&hellip;]<\/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-262252","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/262252","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=262252"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/262252\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=262252"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=262252"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=262252"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}