{"id":220042,"date":"2025-05-27T11:10:28","date_gmt":"2025-05-27T11:10:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220042"},"modified":"2025-05-27T11:10:30","modified_gmt":"2025-05-27T11:10:30","slug":"classify-each-reaction-as-homolysis-or-heterolysis","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/27\/classify-each-reaction-as-homolysis-or-heterolysis\/","title":{"rendered":"Classify each reaction as homolysis or heterolysis."},"content":{"rendered":"\n<p>Classify each reaction as homolysis or heterolysis. Heterolysis Homolysis Prev A. H3C-CH3 B. C. H3C- 3 of 17 CI -CH3 Br: Next CH3 CH3 H3C \u2753 :Br: CH3 1849 +<\/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 classify each reaction as <strong>homolysis<\/strong> or <strong>heterolysis<\/strong>, we need to understand how the bond is breaking:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Homolysis<\/strong>: The bond breaks evenly, and each atom takes one electron from the bond, forming <strong>two radicals<\/strong>.<\/li>\n\n\n\n<li><strong>Heterolysis<\/strong>: The bond breaks unevenly, with <strong>both electrons going to one atom<\/strong>, forming a <strong>cation and an anion<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>Let&#8217;s examine each case:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>A. H\u2083C\u2013CH\u2083 \u2192 H\u2083C\u2022 + \u2022CH\u2083<\/strong><\/h3>\n\n\n\n<p>This reaction breaks the bond between two carbon atoms in such a way that <strong>each carbon gets one electron<\/strong>, forming <strong>two radicals<\/strong>.<\/p>\n\n\n\n<p>\u2705 <strong>This is HOMOLYSIS.<\/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. H\u2083C\u2013Br \u2192 H\u2083C\u207a + :Br\u207b<\/strong><\/h3>\n\n\n\n<p>In this case, the bond between carbon and bromine breaks and <strong>both electrons go to Br<\/strong>, making it a bromide ion (:Br\u207b), while carbon becomes a <strong>carbocation (H\u2083C\u207a)<\/strong>.<\/p>\n\n\n\n<p>\u2705 <strong>This is HETEROLYSIS.<\/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. H\u2083C\u2013CH\u2083 \u2192 H\u2083C\u207a + :CH\u2083\u207b<\/strong><\/h3>\n\n\n\n<p>Here, the C\u2013C bond breaks and <strong>both electrons go to one carbon<\/strong>, making it a <strong>carbanion (:CH\u2083\u207b)<\/strong> and a <strong>carbocation (H\u2083C\u207a)<\/strong>.<\/p>\n\n\n\n<p>\u2705 <strong>This is HETEROLYSIS.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">Explanation<\/h2>\n\n\n\n<p>Bond dissociation reactions are crucial in understanding organic chemistry mechanisms. When a covalent bond breaks, the manner in which the electrons are distributed determines whether the process is <strong>homolytic<\/strong> (homolysis) or <strong>heterolytic<\/strong> (heterolysis).<\/p>\n\n\n\n<p>In <strong>homolysis<\/strong>, the shared pair of electrons in a bond is divided equally between the two atoms. This results in the formation of <strong>free radicals<\/strong>, which are species with an unpaired electron. Homolytic cleavage is common in non-polar bonds and under conditions like heat or UV light, which provide energy to break bonds symmetrically. For example, in reaction A (H\u2083C\u2013CH\u2083 \u2192 H\u2083C\u2022 + \u2022CH\u2083), the carbon-carbon bond breaks evenly, and each carbon retains one electron, forming two methyl radicals. This is typical of radical reactions such as halogenations.<\/p>\n\n\n\n<p>In <strong>heterolysis<\/strong>, both bonding electrons go to one of the atoms, creating a <strong>cation and an anion<\/strong>. This usually occurs in polar bonds where one atom is more electronegative. Reaction B (H\u2083C\u2013Br \u2192 H\u2083C\u207a + :Br\u207b) and reaction C (H\u2083C\u2013CH\u2083 \u2192 H\u2083C\u207a + :CH\u2083\u207b) are heterolytic cleavages. In B, the more electronegative bromine atom takes both electrons, resulting in a bromide ion and a methyl carbocation. In C, one carbon becomes positively charged while the other takes the electrons, forming a carbanion.<\/p>\n\n\n\n<p>Understanding these types of bond dissociation is essential for predicting reaction mechanisms, intermediates, and reactivity in organic chemistry.<\/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-banner5-52.jpeg\" alt=\"\" class=\"wp-image-220043\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Classify each reaction as homolysis or heterolysis. Heterolysis Homolysis Prev A. H3C-CH3 B. C. H3C- 3 of 17 CI -CH3 Br: Next CH3 CH3 H3C \u2753 :Br: CH3 1849 + The Correct Answer and Explanation is: To classify each reaction as homolysis or heterolysis, we need to understand how the bond is breaking: Let&#8217;s examine [&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-220042","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220042","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=220042"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220042\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220042"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220042"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220042"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}