{"id":185812,"date":"2025-01-23T07:48:45","date_gmt":"2025-01-23T07:48:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185812"},"modified":"2025-01-23T07:48:47","modified_gmt":"2025-01-23T07:48:47","slug":"naoh","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/23\/naoh\/","title":{"rendered":"NaOH"},"content":{"rendered":"\n<ol class=\"wp-block-list\">\n<li>NaOH 2. CICH3 G) 1. NaOEt 2. PhCH2Br 3. H30+, Heat OEt EtO<\/li>\n<\/ol>\n\n\n\n<p>Show the result and mechanism for these 2 reactions:<\/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>Let&#8217;s go through each reaction step by step.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Reaction 1: NaOH<\/h3>\n\n\n\n<p>This reaction involves sodium hydroxide (NaOH), a strong base, reacting with a substrate. However, you haven&#8217;t specified the substrate, so I&#8217;ll assume we&#8217;re dealing with a common reaction mechanism such as the <strong>dehydrohalogenation<\/strong> of an alkyl halide, or the <strong>ester hydrolysis<\/strong>.<\/p>\n\n\n\n<p>Let&#8217;s assume we&#8217;re working with an <strong>alkyl halide<\/strong> (R-X), and the reaction is:<\/p>\n\n\n\n<p><strong>R-X + NaOH \u2192 R-OH + NaX<\/strong><\/p>\n\n\n\n<p>This reaction involves the following mechanism:<\/p>\n\n\n\n<p><strong>Step 1: Nucleophilic Substitution<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The hydroxide ion (OH\u207b) from NaOH acts as a nucleophile. It attacks the electrophilic carbon atom in the alkyl halide (R-X), which is attached to a halogen (Cl, Br, or I).<\/li>\n\n\n\n<li>The nucleophile displaces the halide ion, leaving behind the alcohol (R-OH).<\/li>\n<\/ul>\n\n\n\n<p>The reaction proceeds via <strong>SN2 (Substitution Nucleophilic Bimolecular)<\/strong> if the carbon is primary or secondary (typically). In the case of tertiary carbons, the mechanism may proceed via <strong>SN1 (Substitution Nucleophilic Unimolecular)<\/strong>, where the leaving group departs first, forming a carbocation, which is then attacked by the hydroxide ion.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Reaction 2: NaOEt, PhCH2Br, H3O+, Heat, OEt<\/h3>\n\n\n\n<p>This sequence involves several steps:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Step 1: Nucleophilic Substitution<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>NaOEt<\/strong> (sodium ethoxide) acts as a strong nucleophile and attacks the benzyl bromide (PhCH2Br).<\/li>\n\n\n\n<li>The ethoxide ion (EtO\u207b) displaces the bromide ion (Br\u207b), leading to the formation of <strong>PhCH2OEt<\/strong> (benzyl ethyl ether).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Step 2: Acid Work-Up<\/strong>\n<ul class=\"wp-block-list\">\n<li>After this nucleophilic substitution, the reaction is treated with <strong>H\u2083O\u207a<\/strong> (acidic work-up), which will protonate the ether and help in its final preparation.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Step 3: Heating<\/strong>\n<ul class=\"wp-block-list\">\n<li>The reaction mixture is heated to promote the ether formation. Heat is often used to encourage substitution reactions and eliminate any byproducts.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>In this sequence, the key mechanism is <strong>SN2<\/strong> because the benzyl carbon (PhCH2) is primary, making it favorable for backside attack by the ethoxide ion.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>NaOH<\/strong> typically leads to nucleophilic substitution or ester hydrolysis.<\/li>\n\n\n\n<li><strong>NaOEt, PhCH2Br, H3O+, Heat, OEt<\/strong> involves an <strong>SN2 reaction<\/strong>, where the ethoxide ion displaces the bromide from the benzyl group to form a benzyl ethyl ether.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Show the result and mechanism for these 2 reactions: The Correct Answer and Explanation is : Let&#8217;s go through each reaction step by step. Reaction 1: NaOH This reaction involves sodium hydroxide (NaOH), a strong base, reacting with a substrate. However, you haven&#8217;t specified the substrate, so I&#8217;ll assume we&#8217;re dealing with a common reaction [&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-185812","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185812","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=185812"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185812\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185812"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185812"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185812"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}