{"id":241758,"date":"2025-07-04T03:29:31","date_gmt":"2025-07-04T03:29:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=241758"},"modified":"2025-07-04T03:29:33","modified_gmt":"2025-07-04T03:29:33","slug":"draw-the-structural-formula-for-the-b-ketoester-formed-by-the-claisen-condensation-of-ethyl-butanoate-with-the-following-ester","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/04\/draw-the-structural-formula-for-the-b-ketoester-formed-by-the-claisen-condensation-of-ethyl-butanoate-with-the-following-ester\/","title":{"rendered":"Draw the structural formula for the B-ketoester formed by the Claisen condensation of ethyl butanoate with the following ester"},"content":{"rendered":"\n<p>Draw the structural formula for the B-ketoester formed by the Claisen condensation of ethyl butanoate with the following ester. Assume a 1:1 stoichiometry. You do not have to consider stereochemistry. If more than one product is possible, only draw the major product. Do not draw organic or inorganic by-products<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Correct Answer:<\/strong>Generated code<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>O    O\n      ||   ||\nCH3CH2CH-C-C-O-CH2CH3\n      |\n      C=O\n      |\n      O\n      |\n      <\/code><\/pre>\n\n\n\n<p><em>Note: This is a 2D representation. A more common way to draw this structure is:<\/em>Generated code<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>O=C-OCH2CH3\n      |\nCH3CH2CH-C(=O)-OCH2CH3<\/code><\/pre>\n\n\n\n<p><em>Which can be written linearly as CH3CH2CH(COOCH2CH3)C(=O)COOCH2CH3<\/em><\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>The reaction described is a crossed Claisen condensation between ethyl butanoate and diethyl oxalate. The Claisen condensation is a carbon-carbon bond-forming reaction that occurs between two ester molecules in the presence of a strong base, typically an alkoxide.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Identifying the Nucleophile and Electrophile:<\/strong>\u00a0A successful crossed Claisen condensation requires one ester that can form an enolate (the nucleophile) and one that cannot. Ethyl butanoate (CH\u2083CH\u2082CH\u2082COOEt) has two acidic \u03b1-protons on the carbon adjacent to the carbonyl group. In the presence of a base like sodium ethoxide (NaOEt), it will be deprotonated to form a nucleophilic enolate. Diethyl oxalate (EtOOC-COOEt), on the other hand, has no \u03b1-protons and therefore cannot form an enolate. It will act as the electrophile. This selectivity ensures that the crossed-condensation product is the major product, minimizing the self-condensation of ethyl butanoate.<\/li>\n\n\n\n<li><strong>Reaction Mechanism:<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Enolate Formation:<\/strong>\u00a0The ethoxide base removes an \u03b1-proton from ethyl butanoate.<\/li>\n\n\n\n<li><strong>Nucleophilic Attack:<\/strong>\u00a0The resulting enolate of ethyl butanoate attacks one of the highly electrophilic carbonyl carbons of diethyl oxalate.<\/li>\n\n\n\n<li><strong>Elimination:<\/strong>\u00a0A tetrahedral intermediate is formed, which then collapses by eliminating the ethoxide group (-OEt) from the oxalate moiety.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Product Formation:<\/strong>\u00a0This sequence of steps results in the formation of a new carbon-carbon bond between the \u03b1-carbon of the original ethyl butanoate and the carbonyl carbon of the diethyl oxalate. The final product is a \u03b2-ketoester. In this specific case, the product is diethyl 2-ethyl-3-oxobutanedioate. The structure contains the four-carbon backbone from butanoate, which is now substituted at its \u03b1-position with an ethoxycarbonyl group originating from the diethyl oxalate.<\/li>\n<\/ol>\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-banner5-178.jpeg\" alt=\"\" class=\"wp-image-241775\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the structural formula for the B-ketoester formed by the Claisen condensation of ethyl butanoate with the following ester. Assume a 1:1 stoichiometry. You do not have to consider stereochemistry. If more than one product is possible, only draw the major product. Do not draw organic or inorganic by-products The Correct Answer and Explanation is: [&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-241758","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241758","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=241758"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/241758\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=241758"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=241758"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=241758"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}