{"id":181988,"date":"2025-01-13T07:18:10","date_gmt":"2025-01-13T07:18:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=181988"},"modified":"2025-01-13T07:18:12","modified_gmt":"2025-01-13T07:18:12","slug":"we-are-performing-a-reduction-of-3-nitroacetophenone-with-sodium-borohydride","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/we-are-performing-a-reduction-of-3-nitroacetophenone-with-sodium-borohydride\/","title":{"rendered":"We are performing a reduction of 3-nitroacetophenone with sodium borohydride"},"content":{"rendered":"\n<p>a) We are performing a reduction of 3-nitroacetophenone with sodium borohydride, write the reaction mechanism(where you show the movement of the electrons)<\/p>\n\n\n\n<p>b) When butanoic acid is treated with NaBH4, 1-butanol is NOT formed, but you can observe that something is happening. Which reaction can it be?<\/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<h3 class=\"wp-block-heading\">a) Reduction of 3-nitroacetophenone with Sodium Borohydride<\/h3>\n\n\n\n<p>The reduction mechanism involves the transfer of a hydride ion ((H^-)) from sodium borohydride ((NaBH_4)) to the carbonyl group of 3-nitroacetophenone, reducing the ketone to a secondary alcohol. The presence of the nitro group does not interfere with the reduction of the ketone under these mild conditions.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Steps of the Mechanism:<\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Hydride Attack on the Carbonyl Carbon<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A hydride ion ((H^-)) from (NaBH_4) attacks the electrophilic carbonyl carbon in 3-nitroacetophenone.<\/li>\n\n\n\n<li>The electrons in the carbon-oxygen (\\pi)-bond are pushed onto the oxygen atom, forming an alkoxide intermediate. <strong>Movement of electrons<\/strong>:<br>[<br>\\text{H:}^- + \\text{R-C(=O)-R&#8217;} \\rightarrow \\text{R-CHO(-)-R&#8217;}<br>]<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Protonation of the Alkoxide<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The alkoxide ion is protonated by water or an alcohol solvent, leading to the formation of the secondary alcohol, 3-nitro-1-phenylethanol. <strong>Final Product<\/strong>:<br>[<br>\\text{R-CHO(-)-R&#8217;} + \\text{H}_2\\text{O} \\rightarrow \\text{R-CH(OH)-R&#8217;} + \\text{OH}^-<br>]<\/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\">b) Reaction of Butanoic Acid with Sodium Borohydride<\/h3>\n\n\n\n<p>When butanoic acid ((CH_3CH_2CH_2COOH)) is treated with (NaBH_4), no reduction to 1-butanol occurs because (NaBH_4) is not strong enough to reduce carboxylic acids. Instead, a different reaction takes place: the <strong>acid-base reaction<\/strong> between the carboxylic acid and (NaBH_4).<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Explanation:<\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Reaction Details<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carboxylic acids are acidic and react with (NaBH_4), which is a mild base. This leads to the formation of the sodium salt of the acid and hydrogen gas ((H_2)).<br>[<br>CH_3CH_2CH_2COOH + NaBH_4 \\rightarrow CH_3CH_2CH_2COONa + H_2<br>]<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Why Reduction Does Not Occur<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sodium borohydride is selective and can reduce ketones and aldehydes but lacks the reducing power needed to break the stable (-COOH) bond in carboxylic acids. Stronger reducing agents like lithium aluminum hydride ((LiAlH_4)) are required to reduce carboxylic acids to alcohols.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Observable Evidence<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The reaction generates hydrogen gas, which is visible as effervescence. This confirms that a reaction is occurring, even though the carboxylic acid is not reduced to 1-butanol.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Summary:<\/h4>\n\n\n\n<p>The reaction between butanoic acid and (NaBH_4) is an acid-base reaction, resulting in the formation of sodium butanoate and hydrogen gas, not a reduction to alcohol. This highlights the selective nature of (NaBH_4).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>a) We are performing a reduction of 3-nitroacetophenone with sodium borohydride, write the reaction mechanism(where you show the movement of the electrons) b) When butanoic acid is treated with NaBH4, 1-butanol is NOT formed, but you can observe that something is happening. Which reaction can it be? The Correct Answer and Explanation is : a) [&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-181988","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181988","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=181988"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/181988\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=181988"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=181988"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=181988"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}