{"id":199767,"date":"2025-03-12T15:26:40","date_gmt":"2025-03-12T15:26:40","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=199767"},"modified":"2025-03-12T15:26:42","modified_gmt":"2025-03-12T15:26:42","slug":"ch3ch2ch2br-na-cn-%e2%86%92","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/12\/ch3ch2ch2br-na-cn-%e2%86%92\/","title":{"rendered":"CH3CH2CH2Br + Na+ -CN \u2192"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">CH3CH2CH2Br + Na+ -CN \u2192<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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 class=\"wp-block-paragraph\">The given reaction:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">[<br>\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{Br} + \\text{Na}^+ -\\text{CN} \\rightarrow<br>]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Correct Answer:<\/strong><br>[<br>\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{CN} + \\text{NaBr}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Type of Reaction:<\/strong><br>This is a <strong>nucleophilic substitution reaction (S(_N)2 mechanism)<\/strong>. In this reaction, the bromopropane (CH(_3)CH(_2)CH(_2)Br) undergoes substitution with the cyanide ion (CN\u207b) from sodium cyanide (NaCN).<\/li>\n\n\n\n<li><strong>Mechanism of the Reaction:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>propyl bromide (CH\u2083CH\u2082CH\u2082Br)<\/strong> has a <strong>polar C-Br bond<\/strong>, where the <strong>carbon is partially positive<\/strong> due to the electronegativity of bromine.<\/li>\n\n\n\n<li>The <strong>cyanide ion (CN\u207b)<\/strong> is a strong nucleophile and attacks the <strong>electrophilic carbon<\/strong> from the <strong>backside<\/strong> in a <strong>one-step concerted mechanism<\/strong>.<\/li>\n\n\n\n<li>The <strong>Br\u207b leaves<\/strong>, and <strong>C-CN bond formation<\/strong> occurs simultaneously.<\/li>\n\n\n\n<li>This follows an <strong>S(_N)2 mechanism<\/strong> (bimolecular nucleophilic substitution), which occurs in <strong>one step with inversion of configuration<\/strong>.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Product Formation:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>propyl bromide (CH\u2083CH\u2082CH\u2082Br)<\/strong> is converted to <strong>propyl cyanide (CH\u2083CH\u2082CH\u2082CN)<\/strong>, also known as <strong>butanenitrile<\/strong>.<\/li>\n\n\n\n<li>Sodium bromide (NaBr) is formed as a by-product.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Why S(_N)2 Instead of S(_N)1?<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Primary alkyl halides<\/strong> (like CH\u2083CH\u2082CH\u2082Br) favor <strong>S(_N)2<\/strong> mechanisms because the primary carbon is <strong>less sterically hindered<\/strong>, making it easier for the nucleophile to attack.<\/li>\n\n\n\n<li>The <strong>S(_N)1 mechanism<\/strong>, which involves a carbocation intermediate, is less favorable here since a <strong>primary carbocation is highly unstable<\/strong>.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Importance of Reaction:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>This reaction is useful in organic synthesis for converting <strong>alkyl halides to nitriles<\/strong>, which can then be <strong>hydrolyzed to carboxylic acids<\/strong> or <strong>reduced to primary amines<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Thus, the final product is <strong>propyl cyanide (butanenitrile), CH\u2083CH\u2082CH\u2082CN<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>CH3CH2CH2Br + Na+ -CN \u2192 The correct answer and explanation is : The given reaction: [\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{Br} + \\text{Na}^+ -\\text{CN} \\rightarrow] Correct Answer:[\\text{CH}_3\\text{CH}_2\\text{CH}_2\\text{CN} + \\text{NaBr}] Explanation: Thus, the final product is propyl cyanide (butanenitrile), CH\u2083CH\u2082CH\u2082CN.<\/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 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