{"id":222737,"date":"2025-05-31T17:14:34","date_gmt":"2025-05-31T17:14:34","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222737"},"modified":"2025-05-31T17:14:36","modified_gmt":"2025-05-31T17:14:36","slug":"h-h-cl-c-cl-c-h-h-hutch-section-13-11","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/h-h-cl-c-cl-c-h-h-hutch-section-13-11\/","title":{"rendered":"H H Cl C Cl C H H Hutch Section 13.11"},"content":{"rendered":"\n<p>H H Cl C Cl C H H Hutch Section 13.11<\/p>\n\n\n\n<p>Give the multiplicity of the following signals. A) H<br>B) H<br>C) H<br>12 Kay Saudbex<br>H H Cl C Cl C H H Hutch Section 13.11<\/p>\n\n\n\n<p>Give the multiplicity of the following signals. A) H<br>B) H<br>C) H<br>12 Kay Saudbex<\/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 answer the question on the <strong>multiplicity of signals<\/strong> for hydrogens in the given molecule, let&#8217;s first interpret the structure from your text. It seems to be referring to <strong>1,2-dichloropropane (CH\u2083\u2013CHCl\u2013CH\u2082Cl)<\/strong> based on the sequence of atoms and placement of Cl and H atoms.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Structure of 1,2-Dichloropropane (CH\u2083\u2013CHCl\u2013CH\u2082Cl):<\/strong><\/h3>\n\n\n\n<pre class=\"wp-block-code\"><code>   H   Cl       Cl   H\n   |   |         |   |\nH\u2013C\u2013C\u2013C\u2013H \u2192 CH\u2083\u2013CHCl\u2013CH\u2082Cl\n   |   |         |   |\n   H   H         H   H\n<\/code><\/pre>\n\n\n\n<p>Label the hydrogens for analysis:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>A) H<\/strong> on the methyl group (CH\u2083\u2013)<\/li>\n\n\n\n<li><strong>B) H<\/strong> on the central carbon (\u2013CHCl\u2013)<\/li>\n\n\n\n<li><strong>C) H<\/strong> on the CH\u2082Cl group<\/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>A) Methyl group protons (CH\u2083\u2013)<\/strong><\/h3>\n\n\n\n<p>These hydrogens are adjacent to the CHCl carbon, which has <strong>1 hydrogen<\/strong> (H_B).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Rule:<\/strong> n + 1 where n = number of neighboring nonequivalent protons.<\/li>\n\n\n\n<li><strong>n = 1 \u2192 multiplicity = 2 \u2192 doublet<\/strong><\/li>\n<\/ul>\n\n\n\n<p>\u2705 <strong>Multiplicity: Doublet<\/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>B) Methine proton (\u2013CHCl\u2013)<\/strong><\/h3>\n\n\n\n<p>This hydrogen is adjacent to:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One CH\u2083 group (3 hydrogens)<\/li>\n\n\n\n<li>One CH\u2082 group (2 hydrogens)<\/li>\n\n\n\n<li><strong>n = 3 + 2 = 5 \u2192 multiplicity = 6 \u2192 sextet<\/strong><\/li>\n<\/ul>\n\n\n\n<p>\u2705 <strong>Multiplicity: Sextet<\/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>C) CH\u2082Cl protons<\/strong><\/h3>\n\n\n\n<p>These hydrogens are adjacent to the CHCl carbon (1 hydrogen)<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>n = 1 \u2192 multiplicity = 2 \u2192 doublet<\/strong><\/li>\n<\/ul>\n\n\n\n<p>\u2705 <strong>Multiplicity: Doublet<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h2>\n\n\n\n<p>In proton nuclear magnetic resonance (\u00b9H NMR) spectroscopy, <strong>multiplicity<\/strong> refers to how a hydrogen atom\u2019s signal splits due to neighboring non-equivalent hydrogens, a result of <strong>spin-spin coupling<\/strong>. The <strong>n + 1 rule<\/strong> determines this: a hydrogen with <em>n<\/em> equivalent neighbors will show a signal split into <em>n + 1<\/em> peaks.<\/p>\n\n\n\n<p>For 1,2-dichloropropane, we examine the molecule\u2019s environment:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>A) Methyl protons (CH\u2083\u2013)<\/strong>: These three equivalent hydrogens are next to a carbon (CHCl) bearing one hydrogen. That single neighboring proton splits the CH\u2083 signal into a <strong>doublet<\/strong> (1 + 1 = 2).<\/li>\n\n\n\n<li><strong>B) Methine proton (\u2013CHCl\u2013)<\/strong>: This proton is between a CH\u2083 group (3 H) and a CH\u2082Cl group (2 H). These five neighboring hydrogens split the signal into <strong>six peaks<\/strong>, a <strong>sextet<\/strong> (5 + 1 = 6).<\/li>\n\n\n\n<li><strong>C) CH\u2082Cl protons<\/strong>: These two equivalent protons are adjacent to the CHCl carbon with one proton. Thus, the signal is split into a <strong>doublet<\/strong> (1 + 1 = 2).<\/li>\n<\/ul>\n\n\n\n<p>It\u2019s important to assume that coupling occurs only through <strong>three bonds<\/strong>, and that the protons are not magnetically equivalent to the one being observed. Additionally, electronegative atoms like chlorine withdraw electron density but do not split neighboring hydrogen signals significantly, as Cl has no proton for coupling.<\/p>\n\n\n\n<p>Thus, multiplicities are based on direct proton-proton couplings, yielding:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A) Doublet<\/li>\n\n\n\n<li>B) Sextet<\/li>\n\n\n\n<li>C) Doublet<\/li>\n<\/ul>\n\n\n\n<p>This analysis provides insight into molecular structure and local chemical environments in NMR spectroscopy.<\/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-banner5-176.jpeg\" alt=\"\" class=\"wp-image-222738\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>H H Cl C Cl C H H Hutch Section 13.11 Give the multiplicity of the following signals. A) HB) HC) H12 Kay SaudbexH H Cl C Cl C H H Hutch Section 13.11 Give the multiplicity of the following signals. A) HB) HC) H12 Kay Saudbex The Correct Answer and Explanation is: To 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-222737","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222737","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=222737"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222737\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222737"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222737"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222737"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}