{"id":189864,"date":"2025-02-10T15:07:41","date_gmt":"2025-02-10T15:07:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189864"},"modified":"2025-02-10T15:07:43","modified_gmt":"2025-02-10T15:07:43","slug":"acid-catalyzed-dehydration-of-2-butanol-yields-2-butene-as-the-major-product","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/10\/acid-catalyzed-dehydration-of-2-butanol-yields-2-butene-as-the-major-product\/","title":{"rendered":"Acid-catalyzed dehydration of 2-butanol yields 2-butene as the major product"},"content":{"rendered":"\n<p>Acid-catalyzed dehydration of 2-butanol yields 2-butene as the major product. Draw curved arrows to show the movement of electrons in this step of the reaction mechanism.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-243-1024x106.png\" alt=\"\" class=\"wp-image-189865\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-244-1024x152.png\" alt=\"\" class=\"wp-image-189866\"\/><\/figure>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>\ue200i\ue202turn0image0\ue202turn0image1\ue202turn0image3\ue202turn0image7\ue201The acid-catalyzed dehydration of 2-butanol to form 2-butene proceeds through a three-step mechanism:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Protonation of the Alcohol<\/strong>: The hydroxyl group (-OH) of 2-butanol is protonated by the acid catalyst, converting it into a better leaving group, water (H\u2082O).<\/li>\n\n\n\n<li><strong>Loss of Water<\/strong>: The protonated alcohol undergoes a loss of water, resulting in the formation of a carbocation intermediate at the second carbon (C\u2082).<\/li>\n\n\n\n<li><strong>Deprotonation to Form the Alkene<\/strong>: The carbocation formed at C\u2082 is deprotonated by a base (often the conjugate base of the acid catalyst), leading to the formation of 2-butene.<\/li>\n<\/ol>\n\n\n\n<p>This mechanism is classified as an E1 (unimolecular elimination) reaction, where the rate-determining step is the formation of the carbocation intermediate.<\/p>\n\n\n\n<p><strong>Explanation<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Step 1: Protonation of the Alcohol<\/strong>: The lone pair of electrons on the oxygen atom of the hydroxyl group attacks a proton (H\u207a) from the acid catalyst. This protonation increases the electrophilicity of the oxygen, making it a better leaving group.<\/li>\n\n\n\n<li><strong>Step 2: Loss of Water<\/strong>: The protonated alcohol is now a good leaving group. It departs as water (H\u2082O), leaving behind a carbocation intermediate at the second carbon (C\u2082).<\/li>\n\n\n\n<li><strong>Step 3: Deprotonation to Form the Alkene<\/strong>: The carbocation at C\u2082 is unstable and seeks to stabilize by losing a proton (H\u207a) from an adjacent carbon (C\u2083). This deprotonation results in the formation of a double bond between C\u2082 and C\u2083, yielding 2-butene as the major product.<\/li>\n<\/ul>\n\n\n\n<p>The formation of the carbocation intermediate is the rate-determining step in this mechanism. The stability of the carbocation intermediate is crucial; a more stable carbocation will form more readily, leading to a faster reaction. In the case of 2-butanol, the carbocation formed at C\u2082 is relatively stable due to hyperconjugation and inductive effects from the adjacent methyl group.<\/p>\n\n\n\n<p>The E1 mechanism is favored under acidic conditions, especially with secondary alcohols like 2-butanol, because the protonation of the alcohol and the subsequent loss of water are both facilitated by the acidic environment. Additionally, the formation of a stable carbocation intermediate is more favorable in this context.<\/p>\n\n\n\n<p>In summary, the acid-catalyzed dehydration of 2-butanol to 2-butene involves the protonation of the alcohol, loss of water to form a carbocation intermediate, and deprotonation to yield the alkene product. This E1 mechanism is characterized by the formation of a carbocation intermediate and is favored under acidic conditions.<\/p>\n\n\n\n<p>For a visual explanation of this mechanism, you might find the following video helpful:<\/p>\n\n\n\n<p>\ue200video\ue202Acid-Catalyzed Dehydration Mechanism\ue202turn0search3\ue201<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Acid-catalyzed dehydration of 2-butanol yields 2-butene as the major product. Draw curved arrows to show the movement of electrons in this step of the reaction mechanism. The Correct Answer and Explanation is : \ue200i\ue202turn0image0\ue202turn0image1\ue202turn0image3\ue202turn0image7\ue201The acid-catalyzed dehydration of 2-butanol to form 2-butene proceeds through a three-step mechanism: This mechanism is classified as an E1 (unimolecular elimination) [&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-189864","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189864","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=189864"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189864\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189864"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189864"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}