{"id":188434,"date":"2025-02-07T03:34:02","date_gmt":"2025-02-07T03:34:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188434"},"modified":"2025-02-07T03:34:04","modified_gmt":"2025-02-07T03:34:04","slug":"construct-an-efficient-three-step-synthesis-of-12-epoxycyclopentane-from-bromocyclopentane-by-dragging-the-appropriate-formulas-into-the-bins","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/07\/construct-an-efficient-three-step-synthesis-of-12-epoxycyclopentane-from-bromocyclopentane-by-dragging-the-appropriate-formulas-into-the-bins\/","title":{"rendered":"Construct an efficient three-step synthesis of 1,2-epoxycyclopentane from bromocyclopentane by dragging the appropriate formulas into the bins"},"content":{"rendered":"\n<p>Construct an efficient three-step synthesis of 1,2-epoxycyclopentane from bromocyclopentane by dragging the appropriate formulas into the bins. Note that each bin will hold only one item, and not all of the given reagents or structures will be used.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-113.png\" alt=\"\" class=\"wp-image-188435\"\/><\/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>Unfortunately, I don&#8217;t have access to the image you linked, so I can&#8217;t directly view it. However, I can help guide you through a general synthetic route for converting bromocyclopentane to 1,2-epoxycyclopentane. Let me explain a three-step synthesis based on common reagents and reaction strategies.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Nucleophilic Substitution (SN2) with a Strong Nucleophile<\/h3>\n\n\n\n<p>The starting material, bromocyclopentane, contains a bromine atom attached to a carbon, making it a good substrate for a nucleophilic substitution reaction. The goal in this first step is to substitute the bromine atom with a hydroxyl group (-OH) through an SN2 reaction. This can be achieved by treating bromocyclopentane with a strong nucleophile like sodium hydroxide (NaOH) or potassium hydroxide (KOH).<\/p>\n\n\n\n<p><strong>Reaction:<\/strong><br>Bromocyclopentane + NaOH \u2192 Cyclopentanol<\/p>\n\n\n\n<p>In this reaction, the hydroxide ion (OH-) attacks the electrophilic carbon attached to the bromine, resulting in the displacement of the bromine atom and the formation of cyclopentanol.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Formation of an Epoxide (Cyclopentanone Intermediate)<\/h3>\n\n\n\n<p>In the second step, cyclopentanol undergoes an oxidation reaction to form cyclopentanone (a ketone). This can be accomplished using an oxidizing agent like PCC (Pyridinium chlorochromate) or other mild oxidizers. The alcohol group of cyclopentanol is oxidized to a carbonyl group, yielding cyclopentanone.<\/p>\n\n\n\n<p><strong>Reaction:<\/strong><br>Cyclopentanol + PCC \u2192 Cyclopentanone<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Epoxidation of Cyclopentanone (or Cyclopentene)<\/h3>\n\n\n\n<p>The final step involves the formation of an epoxide. If cyclopentanol was directly transformed into a double bond (via dehydrogenation or a different mechanism), the next step would be the epoxidation of the resulting alkene. The addition of a peracid (like m-CPBA) to the cyclopentene would result in the formation of the epoxide ring.<\/p>\n\n\n\n<p><strong>Reaction:<\/strong><br>Cyclopentene + m-CPBA \u2192 1,2-Epoxycyclopentane<\/p>\n\n\n\n<p>The peracid (m-CPBA) adds oxygen across the double bond, resulting in the formation of the epoxide group (an oxygen bridge between carbon 1 and carbon 2).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Steps:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Nucleophilic substitution<\/strong>: Bromocyclopentane + NaOH \u2192 Cyclopentanol<\/li>\n\n\n\n<li><strong>Oxidation<\/strong>: Cyclopentanol + PCC \u2192 Cyclopentanone<\/li>\n\n\n\n<li><strong>Epoxidation<\/strong>: Cyclopentene + m-CPBA \u2192 1,2-Epoxycyclopentane<\/li>\n<\/ol>\n\n\n\n<p>This sequence is efficient because it involves common reactions like nucleophilic substitution, oxidation, and epoxidation, which are widely used in organic synthesis.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Construct an efficient three-step synthesis of 1,2-epoxycyclopentane from bromocyclopentane by dragging the appropriate formulas into the bins. Note that each bin will hold only one item, and not all of the given reagents or structures will be used. The Correct Answer and Explanation is : Unfortunately, I don&#8217;t have access to the image you linked, [&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-188434","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188434","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=188434"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188434\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188434"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188434"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188434"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}