{"id":240957,"date":"2025-07-03T13:52:25","date_gmt":"2025-07-03T13:52:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=240957"},"modified":"2025-07-03T13:52:27","modified_gmt":"2025-07-03T13:52:27","slug":"draw-the-structure-of-a-compound-of-molecular-formula-c4h8o-that-has-a-signal-in-its-13c-nmr-spectrum-at-160-ppm","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/03\/draw-the-structure-of-a-compound-of-molecular-formula-c4h8o-that-has-a-signal-in-its-13c-nmr-spectrum-at-160-ppm\/","title":{"rendered":"Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectrum at > 160 ppm."},"content":{"rendered":"\n<p>Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectrum at > 160 ppm. Then draw the structure of an isomer of molecular formula C4H8O that has all of its 13C NMR signals at &lt; 160 ppm.<\/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<p>To draw the structures and provide explanations for the given compounds, let\u2019s break down the molecular formula C4H8O into its possible structural configurations, focusing on their 13C NMR spectra.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>Compound with a 13C NMR signal at &gt;160 ppm:<\/strong><\/h3>\n\n\n\n<p>For a compound with molecular formula C4H8O and a signal at <strong>&gt;160 ppm<\/strong> in the 13C NMR spectrum, we are looking for a structure where a carbonyl group (C=O) is present. The carbonyl carbon will appear at <strong>&gt;160 ppm<\/strong> because it is highly deshielded, typically in the range of <strong>160-220 ppm<\/strong>. A typical compound of this type would be an <strong>\u03b1,\u03b2-unsaturated carbonyl compound<\/strong> like <strong>but-3-en-2-one<\/strong> (methyl vinyl ketone).<\/p>\n\n\n\n<p><strong>Structure of But-3-en-2-one (C4H8O):<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">iniCopyEdit<code>    CH2=CH-CO-CH3\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>carbonyl group (C=O)<\/strong> at position 2 is the deshielded carbon responsible for the 13C NMR signal at <strong>>160 ppm<\/strong> (around 200 ppm).<\/li>\n\n\n\n<li>The <strong>C=C<\/strong> bond is also conjugated with the carbonyl group, which slightly shifts the carbonyl signal downfield.<\/li>\n<\/ul>\n\n\n\n<p>In the 13C NMR spectrum, the signal from the <strong>carbonyl carbon<\/strong> will be observed at <strong>&gt;160 ppm<\/strong>, and other carbons will appear at lower ppm ranges.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Isomer with all 13C NMR signals at &lt;160 ppm:<\/strong><\/h3>\n\n\n\n<p>For the second compound, we need a structure with <strong>no carbonyl group<\/strong>, as carbonyl carbons give signals above 160 ppm. A <strong>saturated alcohol or ether<\/strong> could be a potential isomer. One possible isomer is <strong>butanol<\/strong> (butan-1-ol), a saturated alcohol.<\/p>\n\n\n\n<p><strong>Structure of Butanol (C4H8O):<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">markdownCopyEdit<code>    CH3-CH2-CH2-OH\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In this structure, all carbons are attached to hydrogens and a hydroxyl group (-OH).<\/li>\n\n\n\n<li>The <strong>13C NMR signals<\/strong> will be in the range of <strong>0-70 ppm<\/strong>, as none of the carbons are deshielded by a carbonyl group or other highly electronegative groups.<\/li>\n\n\n\n<li>The <strong>C-O<\/strong> signal will be around <strong>60-70 ppm<\/strong>, which is typical for carbons attached to an oxygen atom in alcohols.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of NMR Signals:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>But-3-en-2-one<\/strong> (\u03b1,\u03b2-unsaturated ketone): The signal at <strong>>160 ppm<\/strong> corresponds to the carbonyl carbon.<\/li>\n\n\n\n<li><strong>Butanol<\/strong> (saturated alcohol): All signals are below <strong>160 ppm<\/strong>, with no carbonyl group contributing a high-shielding signal.<\/li>\n<\/ul>\n\n\n\n<p>By understanding how functional groups like carbonyls influence the chemical shifts in the NMR spectrum, we can accurately predict and explain the different chemical environments in these isomers of C4H8O.<\/p>\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-banner8-60.jpeg\" alt=\"\" class=\"wp-image-240958\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectrum at > 160 ppm. Then draw the structure of an isomer of molecular formula C4H8O that has all of its 13C NMR signals at &lt; 160 ppm. The Correct Answer and Explanation is: To draw the [&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-240957","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240957","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=240957"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/240957\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=240957"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=240957"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=240957"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}