{"id":218895,"date":"2025-05-24T16:06:21","date_gmt":"2025-05-24T16:06:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218895"},"modified":"2025-05-24T16:06:23","modified_gmt":"2025-05-24T16:06:23","slug":"draw-all-the-structural-isomers-for-the-molecular-formula-c3h7br","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/24\/draw-all-the-structural-isomers-for-the-molecular-formula-c3h7br\/","title":{"rendered":"Draw all the structural isomers for the molecular formula C3H7Br."},"content":{"rendered":"\n<p>Draw all the structural isomers for the molecular formula C3H7Br.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-2-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The molecular formula <strong>C\u2083H\u2087Br<\/strong> represents <strong>bromopropane<\/strong>, a compound that contains three carbon atoms, seven hydrogen atoms, and one bromine atom. There are <strong>two structural isomers<\/strong> for C\u2083H\u2087Br:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Structural Isomers of C\u2083H\u2087Br<\/strong>:<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>1. 1-Bromopropane (n-propyl bromide)<\/strong><\/h4>\n\n\n\n<p><strong>Structure:<\/strong><br>Br\u2013CH\u2082\u2013CH\u2082\u2013CH\u2083<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In this structure, the bromine atom is attached to the <strong>first carbon<\/strong> of a straight-chain propane molecule.<\/li>\n\n\n\n<li>This is the <strong>normal (n-) isomer<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>2. 2-Bromopropane (isopropyl bromide)<\/strong><\/h4>\n\n\n\n<p><strong>Structure:<\/strong><br>CH\u2083\u2013CH(Br)\u2013CH\u2083<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Here, the bromine atom is attached to the <strong>second (middle) carbon<\/strong> in the chain.<\/li>\n\n\n\n<li>This results in a <strong>branched structure<\/strong>, also referred to as the <strong>iso<\/strong> form.<\/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>Explanation <\/strong><\/h3>\n\n\n\n<p>The molecular formula <strong>C\u2083H\u2087Br<\/strong> consists of three carbon atoms, seven hydrogen atoms, and one bromine atom. This formula corresponds to <strong>haloalkanes<\/strong>, also known as <strong>alkyl halides<\/strong>, specifically the <strong>brominated derivatives of propane (C\u2083H\u2088)<\/strong>. When one hydrogen atom in propane is substituted by a bromine atom, the result is a <strong>monobromopropane<\/strong> isomer.<\/p>\n\n\n\n<p>To determine the number of <strong>structural isomers<\/strong>, we analyze all the unique ways a bromine atom can be bonded to the carbon skeleton of propane.<\/p>\n\n\n\n<p>Propane (C\u2083H\u2088) has a straight chain of three carbon atoms. The bromine can be attached either to the <strong>end carbon (C-1)<\/strong> or the <strong>middle carbon (C-2)<\/strong>. Attaching Br to either of the terminal carbons gives <strong>1-bromopropane<\/strong>. However, because of molecular symmetry, attaching Br to the first or third carbon yields the same compound. Attaching the bromine to the middle carbon gives <strong>2-bromopropane<\/strong>.<\/p>\n\n\n\n<p>There are <strong>no other arrangements<\/strong> of the carbon skeleton with only three carbon atoms that would result in a different structure. Unlike longer carbon chains, three-carbon chains do not allow for more complex branching.<\/p>\n\n\n\n<p>Therefore, only <strong>two structural isomers<\/strong> exist:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>1-bromopropane<\/strong>, a straight-chain compound with Br on the end.<\/li>\n\n\n\n<li><strong>2-bromopropane<\/strong>, a compound where Br is attached to the central carbon, creating a branched appearance.<\/li>\n<\/ol>\n\n\n\n<p>These isomers differ in their <strong>physical and chemical properties<\/strong>, such as boiling point, reactivity, and steric effects, due to the position of the bromine atom within the molecule.<\/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-378.png\" alt=\"\" class=\"wp-image-218896\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n","protected":false},"excerpt":{"rendered":"<p>Draw all the structural isomers for the molecular formula C3H7Br. The Correct Answer and Explanation is: The molecular formula C\u2083H\u2087Br represents bromopropane, a compound that contains three carbon atoms, seven hydrogen atoms, and one bromine atom. There are two structural isomers for C\u2083H\u2087Br: Structural Isomers of C\u2083H\u2087Br: 1. 1-Bromopropane (n-propyl bromide) Structure:Br\u2013CH\u2082\u2013CH\u2082\u2013CH\u2083 2. 2-Bromopropane (isopropyl [&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-218895","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218895","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=218895"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218895\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218895"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218895"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218895"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}