Draw the products of the following reaction: the acid-catalyzed dehydration of 1-pentanol to 2-pentene.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n The acid-catalyzed dehydration of 1-pentanol to 2-pentene follows a typical E1 elimination mechanism. Here is the step-by-step explanation of the reaction:<\/p>\n\n\n\n The starting compound, 1-pentanol<\/strong> (CH\u2083CH\u2082CH\u2082CH\u2082OH), undergoes an acid-catalyzed dehydration (removal of a water molecule), resulting in the formation of 2-pentene<\/strong> (CH\u2083CH=CHCH\u2082CH\u2083). This reaction is promoted by the presence of an acidic catalyst, usually sulfuric acid (H\u2082SO\u2084).<\/p>\n\n\n\n The major product of this reaction is 2-pentene<\/strong>, which is an alkene formed by the elimination of water from 1-pentanol. Specifically, the product is 2-pentene<\/strong> in its cis<\/strong> or trans<\/strong> isomeric forms, with trans-2-pentene being the most predominant product due to its lower steric strain.<\/p>\n\n\n\n The acid-catalyzed dehydration of 1-pentanol to 2-pentene follows an E1 mechanism. The process involves the formation of a carbocation intermediate, which undergoes a rearrangement to form a more stable secondary carbocation at the 2-position. Finally, a proton is eliminated to form 2-pentene as the major product.<\/p>\n","protected":false},"excerpt":{"rendered":" Draw the products of the following reaction: the acid-catalyzed dehydration of 1-pentanol to 2-pentene. The Correct Answer and Explanation is: The acid-catalyzed dehydration of 1-pentanol to 2-pentene follows a typical E1 elimination mechanism. Here is the step-by-step explanation of the reaction: Reaction Overview: The starting compound, 1-pentanol (CH\u2083CH\u2082CH\u2082CH\u2082OH), undergoes an acid-catalyzed dehydration (removal of a […]<\/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-163543","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/163543","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=163543"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/163543\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=163543"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=163543"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=163543"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Reaction Overview:<\/h3>\n\n\n\n
Mechanism:<\/h3>\n\n\n\n
\n
The hydroxyl group (OH) of 1-pentanol is protonated by the acidic catalyst (H\u207a), turning it into a better leaving group (water, H\u2082O).
[
\\text{CH\u2083CH\u2082CH\u2082CH\u2082OH} + \\text{H\u207a} \\rightarrow \\text{CH\u2083CH\u2082CH\u2082CH\u2082OH\u2082\u207a}
]<\/li>\n\n\n\n
Once water is eliminated, a carbocation<\/strong> forms at the position where the hydroxyl group was originally attached (the 1-position). This carbocation is relatively unstable, so it undergoes a rearrangement<\/strong> to form a more stable carbocation. The 2-position carbocation is more stable than the 1-position because it is a secondary carbocation, whereas the original carbocation was primary.
[
\\text{CH\u2083CH\u2082CH\u2082CH\u2082OH\u2082\u207a} \\rightarrow \\text{CH\u2083CH\u2082CH\u2082\u207a} + \\text{H\u2082O}
]
This rearrangement results in the formation of a secondary carbocation at the 2-position.<\/li>\n\n\n\n
After the formation of the carbocation, a proton (H\u207a) is eliminated from the \u03b2-carbon (the 3-position) adjacent to the carbocation. This results in the formation of 2-pentene<\/strong>.
[
\\text{CH\u2083CH\u2082CH\u2082\u207a} \\rightarrow \\text{CH\u2083CH=CHCH\u2082CH\u2083}
]<\/li>\n<\/ol>\n\n\n\nProduct:<\/h3>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n