{"id":202158,"date":"2025-03-17T08:29:35","date_gmt":"2025-03-17T08:29:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=202158"},"modified":"2025-03-17T08:29:38","modified_gmt":"2025-03-17T08:29:38","slug":"how-do-you-synthesize-cis-12-dibromocyclohexane-from-the-starting-product-of-1-methylcyclohexane","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/17\/how-do-you-synthesize-cis-12-dibromocyclohexane-from-the-starting-product-of-1-methylcyclohexane\/","title":{"rendered":"How do you synthesize cis-1,2-dibromocyclohexane from the starting product of 1-methylcyclohexane"},"content":{"rendered":"\n<p>How do you synthesize cis-1,2-dibromocyclohexane from the starting product of 1-methylcyclohexane<\/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>To synthesize <strong>cis-1,2-dibromocyclohexane<\/strong> from <strong>1-methylcyclohexane<\/strong>, follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Reaction Steps:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Allylic Bromination<\/strong> (Optional): Since 1-methylcyclohexane has a methyl group, it can undergo allylic bromination with <strong>NBS (N-Bromosuccinimide)<\/strong> and light (<strong>hv<\/strong>) to form <strong>1-bromomethylcyclohexane<\/strong>. However, this step is unnecessary if we proceed with direct bromination.<\/li>\n\n\n\n<li><strong>Hydrohalogenation to Form Cyclohexene<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Treat <strong>1-methylcyclohexane<\/strong> with <strong>concentrated sulfuric acid (H\u2082SO\u2084)<\/strong> or <strong>phosphoric acid (H\u2083PO\u2084)<\/strong> under heat.<\/li>\n\n\n\n<li>This causes <strong>E1 elimination<\/strong> of H\u2082O to form <strong>1-methylcyclohexene<\/strong> (a more substituted, stable alkene).<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Bromination of Alkene<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>React <strong>1-methylcyclohexene<\/strong> with <strong>Br\u2082<\/strong> in an inert solvent like <strong>CCl\u2084<\/strong> or CH\u2082Cl\u2082.<\/li>\n\n\n\n<li>This results in an <strong>anti-addition<\/strong> of bromine via a <strong>bromonium ion intermediate<\/strong>, forming <strong>trans-1-bromo-2-methylcyclohexane<\/strong>.<\/li>\n\n\n\n<li>However, we need the <strong>cis<\/strong> isomer, so this pathway isn&#8217;t ideal.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Catalytic Hydrogenation of Bromocyclohexene<\/strong> (Preferred Route for cis):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Instead of Br\u2082, use <strong>Br\u2082 in water (Br\u2082\/H\u2082O)<\/strong> to form <strong>1-bromo-2-hydroxycyclohexane<\/strong> (via anti-addition).<\/li>\n\n\n\n<li>Convert the alcohol (-OH) into a bromide using <strong>PBr\u2083<\/strong> or <strong>HBr<\/strong> to get <strong>cis-1,2-dibromocyclohexane<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Why This Works for cis?<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>halohydrin intermediate<\/strong> undergoes <strong>SN2 substitution<\/strong> with inversion, which compensates for the original anti-addition of Br\u2082, yielding the <strong>cis product<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Summary:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Convert <strong>1-methylcyclohexane<\/strong> \u2192 <strong>1-methylcyclohexene<\/strong> via <strong>acidic elimination<\/strong>.<\/li>\n\n\n\n<li>Brominate using <strong>Br\u2082\/H\u2082O<\/strong> to get <strong>1-bromo-2-hydroxycyclohexane<\/strong>.<\/li>\n\n\n\n<li>Convert <strong>-OH to Br<\/strong> using <strong>PBr\u2083<\/strong>, ensuring a final <strong>cis configuration<\/strong>.<\/li>\n<\/ol>\n\n\n\n<p>This pathway efficiently produces <strong>cis-1,2-dibromocyclohexane<\/strong> while avoiding the undesired trans product.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How do you synthesize cis-1,2-dibromocyclohexane from the starting product of 1-methylcyclohexane The correct answer and explanation is : To synthesize cis-1,2-dibromocyclohexane from 1-methylcyclohexane, follow these steps: Reaction Steps: Why This Works for cis? Summary: This pathway efficiently produces cis-1,2-dibromocyclohexane while avoiding the undesired trans product.<\/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-202158","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202158","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=202158"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202158\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202158"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202158"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202158"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}