{"id":220970,"date":"2025-05-29T05:06:42","date_gmt":"2025-05-29T05:06:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220970"},"modified":"2025-05-29T05:06:44","modified_gmt":"2025-05-29T05:06:44","slug":"study-the-nmr-spectrum-of-1-bromobutane-c4h9br","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/study-the-nmr-spectrum-of-1-bromobutane-c4h9br\/","title":{"rendered":"Study the NMR spectrum of 1-bromobutane (C4H9Br)"},"content":{"rendered":"\n<p>Study the NMR spectrum of 1-bromobutane (C4H9Br) (Figure 20).<\/p>\n\n\n\n<p>1H-NMR (300 MHz): \u03b4 0.92 (3H, t, J = 6 Hz, \u2013Me), 1.45 (2H, h, J = 6Hz, CH2), 1.85 (2H, q, J = 6Hz, CH2), 3.41 (2H, t, J = 6Hz, CH2Br) ppm.<\/p>\n\n\n\n<p>13C-NMR (75.43 MHz): \u03b4 13, 22, 31, 53 ppm.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-504.png\" alt=\"\" class=\"wp-image-220971\"\/><\/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>The NMR spectra provided\u2014both \u00b9H-NMR and \u00b9\u00b3C-NMR\u2014are consistent with the structure of <strong>1-bromobutane (CH\u2083CH\u2082CH\u2082CH\u2082Br)<\/strong>. Let\u2019s analyze each in detail to support the structure.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>\u00b9H-NMR (300 MHz)<\/strong><\/h3>\n\n\n\n<p>Chemical shifts (\u03b4 in ppm) and splitting patterns:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>0.92 ppm (3H, triplet, J = 6 Hz)<\/strong>\n<ul class=\"wp-block-list\">\n<li>This is a methyl group (\u2013CH\u2083) attached to a CH\u2082 group.<\/li>\n\n\n\n<li>The triplet arises due to coupling with the two neighboring protons on the adjacent \u2013CH\u2082\u2013.<\/li>\n\n\n\n<li>This corresponds to the terminal methyl group in the chain.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>1.45 ppm (2H, multiplet \u2013 appears as a hextet, J = 6 Hz)<\/strong>\n<ul class=\"wp-block-list\">\n<li>A \u2013CH\u2082\u2013 group next to both a CH\u2082 and a CH\u2083.<\/li>\n\n\n\n<li>The multiplicity suggests coupling with five equivalent protons (two on one side, three on the other).<\/li>\n\n\n\n<li>This is the penultimate \u2013CH\u2082\u2013 next to the terminal methyl.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>1.85 ppm (2H, quartet, J = 6 Hz)<\/strong>\n<ul class=\"wp-block-list\">\n<li>A \u2013CH\u2082\u2013 group adjacent to the CH\u2082Br group.<\/li>\n\n\n\n<li>Coupling likely comes from the neighboring CH\u2082 (middle of chain).<\/li>\n\n\n\n<li>Shifted downfield slightly due to the proximity of the electronegative Br (via the next carbon).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>3.41 ppm (2H, triplet, J = 6 Hz)<\/strong>\n<ul class=\"wp-block-list\">\n<li>This is the \u2013CH\u2082Br group.<\/li>\n\n\n\n<li>The deshielded shift (downfield) is due to the electronegative bromine.<\/li>\n\n\n\n<li>Triplet due to coupling with the adjacent CH\u2082.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>\u00b9\u00b3C-NMR (75.43 MHz)<\/strong><\/h3>\n\n\n\n<p>Chemical shifts:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>13 ppm<\/strong> \u2013 methyl carbon (\u2013CH\u2083)<\/li>\n\n\n\n<li><strong>22 ppm<\/strong> \u2013 CH\u2082 next to CH\u2083<\/li>\n\n\n\n<li><strong>31 ppm<\/strong> \u2013 middle CH\u2082<\/li>\n\n\n\n<li><strong>53 ppm<\/strong> \u2013 CH\u2082 bonded to Br (most deshielded)<\/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\"><strong>Conclusion (Correct Answer):<\/strong><\/h3>\n\n\n\n<p>The \u00b9H-NMR and \u00b9\u00b3C-NMR spectra fully support the structure of <strong>1-bromobutane<\/strong>. Each hydrogen and carbon environment is distinct and corresponds with expected chemical shifts and splitting due to neighboring protons. The NMR data confirms the linear structure of 1-bromobutane with a terminal bromine on a four-carbon chain.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner4-89.jpeg\" alt=\"\" class=\"wp-image-220972\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Study the NMR spectrum of 1-bromobutane (C4H9Br) (Figure 20). 1H-NMR (300 MHz): \u03b4 0.92 (3H, t, J = 6 Hz, \u2013Me), 1.45 (2H, h, J = 6Hz, CH2), 1.85 (2H, q, J = 6Hz, CH2), 3.41 (2H, t, J = 6Hz, CH2Br) ppm. 13C-NMR (75.43 MHz): \u03b4 13, 22, 31, 53 ppm. The Correct Answer [&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-220970","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220970","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=220970"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220970\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220970"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220970"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220970"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}