{"id":223858,"date":"2025-06-02T12:27:41","date_gmt":"2025-06-02T12:27:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223858"},"modified":"2025-06-02T12:27:43","modified_gmt":"2025-06-02T12:27:43","slug":"organic-chemistry-smartwork5-for-organic-chemistry-07-this-is-a-molecule-drawing-question","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/organic-chemistry-smartwork5-for-organic-chemistry-07-this-is-a-molecule-drawing-question\/","title":{"rendered":"Organic Chemistry Smartwork5 for Organic Chemistry 07 This is a Molecule Drawing question"},"content":{"rendered":"\n<p>Organic Chemistry Smartwork5 for Organic Chemistry 07 This is a Molecule Drawing question \/ It is WOF Question 16 (1 point): Draw the structure of (E)-1-chlorobut-1-ene. 5th attempt Feedback<\/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\/image-24.png\" alt=\"\" class=\"wp-image-223859\"\/><\/figure>\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><strong>Correct structure (line formula)<\/strong><br>Use the double-bond shown horizontally so you can place the ligands in their correct stereochemical positions:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>      Cl           H\n       \\         \/\n        C1 == C2\n        \/         \\\n       H         CH2-CH3\n<\/code><\/pre>\n\n\n\n<p>\u2014 the chlorine on C1 and the ethyl group (-CH2-CH3) on C2 lie on <em>opposite<\/em> sides of the C1=C2 \u03c0-bond, satisfying the <strong>(E)<\/strong> (\u201centgegen\u201d, German for \u201copposite\u201d) configuration. A compact, machine-readable form is <strong>Cl\/C=C\/CC<\/strong>, the E-configured SMILES.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Why this is the right drawing<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Identify the parent skeleton<\/strong><br><em>But-1-ene<\/em> is a four-carbon chain (but-) with a double bond starting at C-1 (-1-ene). Write the backbone as<br><code>C1=C2-C3-C4<\/code>.<\/li>\n\n\n\n<li><strong>Place the substituent<\/strong><br>The name carries the prefix <strong>1-chloro-<\/strong>, meaning a chlorine atom replaces one of the hydrogens on C-1, the same carbon already involved in the double bond. Temporarily, therefore, C-1 is attached to Cl, H and C-2; C-2 is attached to H, C-1 and C-3.<\/li>\n\n\n\n<li><strong>Assign CIP priorities<\/strong><br>For alkene stereochemistry we compare the two substituents attached to each vinylic carbon.\n<ul class=\"wp-block-list\">\n<li>At <strong>C-1<\/strong> the higher atomic-number atom is Cl (Z = 17) versus H (Z = 1), so <strong>Cl<\/strong> gets priority 1, <strong>H<\/strong> priority 2.<\/li>\n\n\n\n<li>At <strong>C-2<\/strong> we compare the first atoms of each substituent: a carbon of an ethyl group (Z = 6) versus H (Z = 1). Thus <strong>CH2-CH3<\/strong> is priority 1, <strong>H<\/strong> is priority 2.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Determine E or Z<\/strong><br>If the two priority-1 groups appear on the <em>same<\/em> side of the double bond, the isomer is <strong>Z<\/strong> (\u201czusammen\u201d, together). If they lie on <em>opposite<\/em> sides, the isomer is <strong>E<\/strong>. The name specifies <strong>(E)<\/strong>, so your sketch must place the Cl and the ethyl group across from one another. A convenient way is to draw the double bond horizontally, Cl above C-1 and CH2CH3 below C-2 (or vice-versa); the hydrogens automatically occupy the remaining, opposite positions.<\/li>\n\n\n\n<li><strong>Check valencies<\/strong><br>Each sp\u00b2 carbon in the alkene must have four bonds in total (counting the double bond as two). The completed structure you just drew satisfies this, confirming chemical correctness.<\/li>\n<\/ol>\n\n\n\n<p>Because only one stereogenic \u03c0-bond exists and its configuration is explicitly defined as <strong>E<\/strong>, the molecule you have drawn\u2014Cl\/C=C\/CC\u2014is the single, correct answer expected by Smartwork.<\/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-banner7-72.jpeg\" alt=\"\" class=\"wp-image-223860\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Organic Chemistry Smartwork5 for Organic Chemistry 07 This is a Molecule Drawing question \/ It is WOF Question 16 (1 point): Draw the structure of (E)-1-chlorobut-1-ene. 5th attempt Feedback The Correct Answer and Explanation is: Correct structure (line formula)Use the double-bond shown horizontally so you can place the ligands in their correct stereochemical positions: \u2014 [&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-223858","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223858","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=223858"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223858\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223858"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223858"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223858"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}