{"id":223316,"date":"2025-06-01T12:54:25","date_gmt":"2025-06-01T12:54:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223316"},"modified":"2025-06-01T12:54:28","modified_gmt":"2025-06-01T12:54:28","slug":"what-would-the-skeletal-structure-of-this-be","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/01\/what-would-the-skeletal-structure-of-this-be\/","title":{"rendered":"What would the skeletal structure of this be"},"content":{"rendered":"\n<p>What would the skeletal structure of this be? Molecular formula consistent with mass spec: C3H6Br2 1H<\/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\"><strong>Correct Skeletal Structure:<\/strong><\/h3>\n\n\n\n<p>The molecular formula <strong>C\u2083H\u2086Br\u2082<\/strong> corresponds to <strong>1,2-dibromopropane<\/strong> (or alternatively 1,3-dibromopropane, but the NMR data can distinguish them \u2014 see below).<\/p>\n\n\n\n<p>A likely <strong>skeletal structure<\/strong> for C\u2083H\u2086Br\u2082 is:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>   Br        Br\n   |         |\nCH3\u2013CH\u2013CH2   or   Br\u2013CH2\u2013CH2\u2013CH2\u2013Br\n<\/code><\/pre>\n\n\n\n<p>However, based on <strong>\u00b9H NMR<\/strong> (which was mentioned), the <strong>correct structure<\/strong> is:<\/p>\n\n\n\n<p><strong>1,2-dibromopropane<\/strong>:<br><strong>CH\u2083\u2013CHBr\u2013CH\u2082Br<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>The molecular formula <strong>C\u2083H\u2086Br\u2082<\/strong> tells us the molecule contains three carbon atoms, six hydrogens, and two bromine atoms. First, calculate the degree of unsaturation (index of hydrogen deficiency, or IHD):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Saturated hydrocarbon formula: C\u2099H\u2082\u2099\u208a\u2082 = C\u2083H\u2088<\/li>\n\n\n\n<li>Actual hydrogen count = 6 \u2192 missing 2 hydrogens<\/li>\n\n\n\n<li>Each degree of unsaturation = 2 hydrogens \u2192 IHD = 1<\/li>\n<\/ul>\n\n\n\n<p>But each <strong>halogen (Br)<\/strong> is counted like a hydrogen for IHD purposes. So:<br>Adjusted hydrogen count = 6 + 2 (for 2 Br) = 8<br>Now compare to C\u2083H\u2088 \u2192 No hydrogen deficiency \u2192 <strong>IHD = 0<\/strong><br>\u2705 Therefore, <strong>no rings or double bonds<\/strong> are present.<\/p>\n\n\n\n<p>We now consider isomers with this formula:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>1,2-dibromopropane<\/strong>: CH\u2083\u2013CHBr\u2013CH\u2082Br<\/li>\n\n\n\n<li><strong>1,3-dibromopropane<\/strong>: BrCH\u2082\u2013CH\u2082\u2013CH\u2082Br<\/li>\n\n\n\n<li><strong>2,2-dibromopropane<\/strong>: (CH\u2083)\u2082CBr\u2082 \u2192 Not possible with only 6 H<\/li>\n<\/ol>\n\n\n\n<p><strong>\u00b9H NMR Evidence (mentioned in your prompt):<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In <strong>1,2-dibromopropane<\/strong>, we expect:\n<ul class=\"wp-block-list\">\n<li>A doublet for CH\u2083 (next to CH)<\/li>\n\n\n\n<li>A multiplet for CH (next to CH\u2083 and CH\u2082Br)<\/li>\n\n\n\n<li>A triplet or complex pattern for CH\u2082Br<br>\u2192 This gives <strong>three types of protons<\/strong> \u2192 matches <strong>1,2-dibromopropane<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>In contrast, <strong>1,3-dibromopropane<\/strong> would give only <strong>two<\/strong> types of protons (CH\u2082Br and central CH\u2082), due to symmetry.<\/p>\n\n\n\n<p>So the <strong>correct structure<\/strong> consistent with <strong>C\u2083H\u2086Br\u2082<\/strong> and <strong>\u00b9H NMR<\/strong> is:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p><strong>CH\u2083\u2013CHBr\u2013CH\u2082Br (1,2-dibromopropane)<\/strong><\/p>\n<\/blockquote>\n\n\n\n<p>This molecule has <strong>three unique proton environments<\/strong>, no unsaturation, and matches both molecular formula and NMR.<\/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-banner5-13.jpeg\" alt=\"\" class=\"wp-image-223317\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What would the skeletal structure of this be? Molecular formula consistent with mass spec: C3H6Br2 1H The Correct Answer and Explanation is: Correct Skeletal Structure: The molecular formula C\u2083H\u2086Br\u2082 corresponds to 1,2-dibromopropane (or alternatively 1,3-dibromopropane, but the NMR data can distinguish them \u2014 see below). A likely skeletal structure for C\u2083H\u2086Br\u2082 is: However, based on [&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-223316","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223316","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=223316"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223316\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223316"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223316"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}