{"id":227150,"date":"2025-06-06T02:35:49","date_gmt":"2025-06-06T02:35:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=227150"},"modified":"2025-06-06T02:35:51","modified_gmt":"2025-06-06T02:35:51","slug":"tools-for-bonds-atoms-and-advanced-template-toolbars","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/06\/tools-for-bonds-atoms-and-advanced-template-toolbars\/","title":{"rendered":"Tools (for bonds), Atoms, and Advanced Template toolbars"},"content":{"rendered":"\n<p>Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default.<br>(E)-1,2-dibromo-3-isopropyl-2-hexene Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default.<\/p>\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<h3 class=\"wp-block-heading\">Correct Answer:<\/h3>\n\n\n\n<p>To <strong>draw (E)-1,2-dibromo-3-isopropyl-2-hexene<\/strong>, follow these steps using your molecular drawing tool (like ChemDraw or MarvinSketch), using the <strong>Tools<\/strong>, <strong>Atoms<\/strong>, and <strong>Advanced Template<\/strong> toolbars:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step-by-Step Drawing Instructions:<\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Draw the Carbon Backbone (Hexene):<\/strong>\n<ul class=\"wp-block-list\">\n<li>Start by creating a 6-carbon straight chain (<strong>hexane<\/strong>).<\/li>\n\n\n\n<li>Add a <strong>double bond between carbon 2 and carbon 3<\/strong> (\u2192 this makes it <strong>2-hexene<\/strong>).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Add Substituents:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Place a <strong>bromine (Br) atom<\/strong> on <strong>carbon 1<\/strong>.<\/li>\n\n\n\n<li>Place another <strong>Br atom on carbon 2<\/strong>.<\/li>\n\n\n\n<li>Place an <strong>isopropyl group (\u2013CH(CH\u2083)\u2082)<\/strong> on <strong>carbon 3<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Assign (E)-configuration:<\/strong>\n<ul class=\"wp-block-list\">\n<li>On the double bond at C2=C3, you have:\n<ul class=\"wp-block-list\">\n<li>On <strong>C2<\/strong>: Br (high priority) and C1 (chain).<\/li>\n\n\n\n<li>On <strong>C3<\/strong>: isopropyl (high priority) and C4 (chain).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The <strong>(E)-configuration<\/strong> means the <strong>higher priority groups (Br and isopropyl)<\/strong> must be on <strong>opposite sides<\/strong> of the double bond.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>So, when drawing the molecule, make sure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>Br on carbon 2<\/strong> is on one side of the double bond.<\/li>\n\n\n\n<li>The <strong>isopropyl group on carbon 3<\/strong> is on the <strong>opposite side<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p><strong>(E)-1,2-dibromo-3-isopropyl-2-hexene<\/strong> is a substituted alkene, and understanding its structure requires knowledge of IUPAC nomenclature, stereochemistry (E\/Z), and functional groups.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Parent Chain and Numbering:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The longest continuous chain with the double bond has <strong>six carbon atoms<\/strong>, hence <strong>hexene<\/strong>.<\/li>\n\n\n\n<li>The double bond begins at carbon 2, so we name it <strong>2-hexene<\/strong>.<\/li>\n\n\n\n<li>The chain is numbered from the end <strong>closest to the double bond<\/strong>, ensuring substituents have the lowest possible numbers.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Substituents:<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Two bromine atoms<\/strong> are attached to carbon atoms 1 and 2: thus, <strong>1,2-dibromo<\/strong>.<\/li>\n\n\n\n<li>An <strong>isopropyl group (\u2013CH(CH\u2083)\u2082)<\/strong> is attached to carbon 3: hence, <strong>3-isopropyl<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Stereochemistry \u2013 E\/Z Notation:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The double bond between C2 and C3 allows for <strong>cis-trans (E\/Z) isomerism<\/strong>.<\/li>\n\n\n\n<li>To determine (E) or (Z), we apply the <strong>Cahn-Ingold-Prelog priority rules<\/strong>:\n<ul class=\"wp-block-list\">\n<li>On C2: Br > C1<\/li>\n\n\n\n<li>On C3: isopropyl > C4<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The <strong>(E)-isomer<\/strong> has the higher-priority groups on <strong>opposite sides<\/strong> of the double bond.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>This molecule\u2019s accurate drawing requires careful placement of atoms and substituents to reflect the correct connectivity and 3D arrangement, particularly for the <strong>(E)-stereochemistry<\/strong>, which impacts physical and chemical properties significantly.<\/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-banner4-504.jpeg\" alt=\"\" class=\"wp-image-227151\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default.(E)-1,2-dibromo-3-isopropyl-2-hexene Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default. The Correct Answer and Explanation is: Correct Answer: To draw (E)-1,2-dibromo-3-isopropyl-2-hexene, follow these steps [&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-227150","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227150","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=227150"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227150\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227150"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227150"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227150"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}