{"id":226293,"date":"2025-06-05T05:34:54","date_gmt":"2025-06-05T05:34:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226293"},"modified":"2025-06-05T05:34:56","modified_gmt":"2025-06-05T05:34:56","slug":"draw-the-two-major-organic-products-of-the-reaction-shown","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/draw-the-two-major-organic-products-of-the-reaction-shown\/","title":{"rendered":"Draw the two major organic products of the reaction shown"},"content":{"rendered":"\n<p>Draw the two major organic products of the reaction shown: Select Draw OH conc HBr<\/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>Thank you for the image. The compound shown is an <strong>allylic alcohol<\/strong>:<br><strong>3-buten-2-ol<\/strong> derivative with methyl substitutions \u2014 specifically, <strong>3-methylbut-2-en-1-ol<\/strong>.<\/p>\n\n\n\n<p>You&#8217;re reacting this with <strong>concentrated HBr<\/strong>, which will convert the \u2013OH into a good leaving group (H\u2082O), enabling <strong>carbocation formation<\/strong>, and then substitution by Br\u207b.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udd2c Step-by-Step Mechanism:<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 1: Protonation of the alcohol<\/strong><\/h4>\n\n\n\n<p>The \u2013OH group is protonated by HBr \u2192 becomes H\u2082O, a good leaving group.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 2: Formation of carbocation<\/strong><\/h4>\n\n\n\n<p>Loss of water leads to a <strong>primary carbocation<\/strong> at the terminal carbon \u2014 but this is <strong>allylic<\/strong>, so <strong>resonance stabilization<\/strong> allows <strong>delocalization<\/strong> of the positive charge over two positions:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>   CH2=CH\u2013CH(+)-CH3   \u21cc   CH2(+)-CH=CH\u2013CH3\n<\/code><\/pre>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Step 3: Attack by Br\u207b<\/strong><\/h4>\n\n\n\n<p>The bromide ion (Br\u207b) can attack at either resonance-stabilized carbocation center:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>The <strong>primary<\/strong> carbon (original position)<\/li>\n\n\n\n<li>The <strong>tertiary<\/strong> carbon after resonance<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Major Organic Products:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>1-Bromo-3-methylbut-2-ene<\/strong><br>(Br attaches to the original \u2013OH position)<\/li>\n\n\n\n<li><strong>3-Bromo-3-methylbut-1-ene<\/strong><br>(Br attaches to the tertiary carbon via resonance)<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83e\uddea Structures:<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Product 1:<\/h4>\n\n\n\n<p><strong>CH\u2082=CH\u2013C(CH\u2083)\u2082\u2013Br<\/strong><br>(Allylic substitution at the terminal CH\u2082)<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Product 2:<\/h4>\n\n\n\n<p><strong>CH\u2083\u2013C(Br)=CH\u2013CH\u2083<\/strong><br>(Br attaches at the tertiary carbon, double bond shifts)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"> Explanation:<\/h3>\n\n\n\n<p>When 3-methylbut-2-en-1-ol reacts with concentrated HBr, the alcohol undergoes substitution by a bromine atom. This transformation proceeds through a two-step mechanism: first, protonation of the hydroxyl group by HBr forms water, a good leaving group. Once water departs, a carbocation intermediate is formed. In this case, the resulting carbocation is <strong>allylic<\/strong>, meaning the positive charge is adjacent to a double bond. This allows for <strong>resonance stabilization<\/strong>, where the positive charge can delocalize across two carbon atoms.<\/p>\n\n\n\n<p>Because of this resonance, the Br\u207b nucleophile has <strong>two electrophilic centers<\/strong> to attack: the carbon where the OH originally was (a primary position), and the more substituted tertiary carbon formed via resonance. Both of these pathways lead to <strong>major organic products<\/strong>, because both are resonance contributors of a stable allylic carbocation.<\/p>\n\n\n\n<p>The first product results from bromide attacking the <strong>primary position<\/strong>, giving <strong>1-bromo-3-methylbut-2-ene<\/strong>. This is considered major because it follows direct substitution. The second product forms when bromide attacks the <strong>tertiary resonance form<\/strong>, leading to <strong>3-bromo-3-methylbut-1-ene<\/strong>. This product is also major due to the <strong>stability of the tertiary carbocation<\/strong>.<\/p>\n\n\n\n<p>The reaction showcases two fundamental concepts in organic chemistry: <strong>carbocation resonance stabilization<\/strong> and <strong>nucleophilic substitution at allylic positions<\/strong>. These allow for multiple favorable pathways, yielding more than one major product. Such outcomes are especially common with allylic systems where resonance and substitution can compete or cooperate<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner7-212.jpeg\" alt=\"\" class=\"wp-image-226296\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the two major organic products of the reaction shown: Select Draw OH conc HBr The Correct Answer and Explanation is: Thank you for the image. The compound shown is an allylic alcohol:3-buten-2-ol derivative with methyl substitutions \u2014 specifically, 3-methylbut-2-en-1-ol. You&#8217;re reacting this with concentrated HBr, which will convert the \u2013OH into a good leaving [&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-226293","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226293","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=226293"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226293\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=226293"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=226293"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=226293"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}