{"id":186664,"date":"2025-01-26T05:40:28","date_gmt":"2025-01-26T05:40:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186664"},"modified":"2025-01-26T05:40:30","modified_gmt":"2025-01-26T05:40:30","slug":"calculate-the-index-of-hydrogen-deficiency-degrees-of-unsaturation-for-the-following-molecule-with-molecular-formula-c5h10o2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/26\/calculate-the-index-of-hydrogen-deficiency-degrees-of-unsaturation-for-the-following-molecule-with-molecular-formula-c5h10o2\/","title":{"rendered":"Calculate the index of hydrogen deficiency (degrees of unsaturation) for the following molecule with molecular formula C5H10O2"},"content":{"rendered":"\n<p>Calculate the index of hydrogen deficiency (degrees of unsaturation) for the following molecule with molecular formula C5H10O2, and propose a structure for this compound consistent with the following IR, 1H NMR, and 13C NMR.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-472-1024x541.png\" alt=\"\" class=\"wp-image-186665\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-473-1024x384.png\" alt=\"\" class=\"wp-image-186666\"\/><\/figure>\n\n\n\n<p>Part 1 (1 point) The index of hydrogen deficiency for this molecule is: Part 2 (1 point) Propose a structure for this compound consistent with the above spectra.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p><strong>Part 1: Index of Hydrogen Deficiency (IHD) Calculation<\/strong><\/p>\n\n\n\n<p>The Index of Hydrogen Deficiency (IHD), also known as degrees of unsaturation, indicates the number of rings and \u03c0 bonds (double bonds or triple bonds) in a compound. It can be calculated using the formula:<\/p>\n\n\n\n<p>[ \\text{IHD} = \\frac{2C + 2 &#8211; H}{2} ]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C<\/strong> = Number of carbon atoms<\/li>\n\n\n\n<li><strong>H<\/strong> = Number of hydrogen atoms<\/li>\n<\/ul>\n\n\n\n<p>For the molecular formula C\u2085H\u2081\u2080O\u2082:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C<\/strong> = 5<\/li>\n\n\n\n<li><strong>H<\/strong> = 10<\/li>\n<\/ul>\n\n\n\n<p>Applying the formula:<\/p>\n\n\n\n<p>[ \\text{IHD} = \\frac{2(5) + 2 &#8211; 10}{2} = \\frac{10 + 2 &#8211; 10}{2} = \\frac{2}{2} = 1 ]<\/p>\n\n\n\n<p>Therefore, the IHD for C\u2085H\u2081\u2080O\u2082 is <strong>1<\/strong>, indicating one degree of unsaturation, which could be a double bond or a ring.<\/p>\n\n\n\n<p><strong>Part 2: Structure Proposal Based on Spectral Data<\/strong><\/p>\n\n\n\n<p>Given the molecular formula C\u2085H\u2081\u2080O\u2082 and the provided IR and NMR spectra, the compound is likely <strong>ethyl acetate (CH\u2083COOCH\u2082CH\u2083)<\/strong>.<\/p>\n\n\n\n<p><strong>IR Spectrum Analysis:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A strong absorption around 1735 cm\u207b\u00b9 suggests a carbonyl group (C=O), typical of esters.<\/li>\n<\/ul>\n\n\n\n<p><strong>\u00b9H NMR Spectrum Analysis:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A doublet at \u03b4 1.3 ppm (6H) indicates two methyl groups (CH\u2083) adjacent to a CH\u2082 group.<\/li>\n\n\n\n<li>A singlet at \u03b4 2.0 ppm (3H) corresponds to a methyl group (CH\u2083) attached to a carbonyl group (C=O).<\/li>\n\n\n\n<li>A septet at \u03b4 5.0 ppm (1H) suggests a CH group adjacent to a CH\u2082 group, characteristic of an ethyl group (CH\u2082CH\u2083) attached to an ester group.<\/li>\n<\/ul>\n\n\n\n<p><strong>\u00b9\u00b3C NMR Spectrum Analysis:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Signals at \u03b4 14.0 ppm and \u03b4 60.0 ppm correspond to methyl (CH\u2083) and methylene (CH\u2082) carbons, respectively.<\/li>\n\n\n\n<li>A signal at \u03b4 170.0 ppm indicates a carbonyl carbon (C=O).<\/li>\n<\/ul>\n\n\n\n<p>These spectral features align with the structure of ethyl acetate, confirming its identity.<\/p>\n\n\n\n<p>For a visual explanation, you might find this video helpful:<\/p>\n\n\n\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=mtELSzGlAoo&amp;utm_source=chatgpt.com\" target=\"_blank\" rel=\"noopener\">1H NMR of C5H10O2 (pentatonic acid, valeric acid)<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the index of hydrogen deficiency (degrees of unsaturation) for the following molecule with molecular formula C5H10O2, and propose a structure for this compound consistent with the following IR, 1H NMR, and 13C NMR. Part 1 (1 point) The index of hydrogen deficiency for this molecule is: Part 2 (1 point) Propose a structure for [&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-186664","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186664","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=186664"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186664\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186664"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186664"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186664"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}