{"id":233515,"date":"2025-06-13T07:28:24","date_gmt":"2025-06-13T07:28:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233515"},"modified":"2025-06-13T07:28:27","modified_gmt":"2025-06-13T07:28:27","slug":"a-ketone-may-react-with-a-nucleophilic-hydride-ion-source-e-g-lialh-or-h-and-subsequently-a-proton-source-e-g-h%e2%82%82o-or-h-to-give-a-secondary-alcohol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/a-ketone-may-react-with-a-nucleophilic-hydride-ion-source-e-g-lialh-or-h-and-subsequently-a-proton-source-e-g-h%e2%82%82o-or-h-to-give-a-secondary-alcohol\/","title":{"rendered":"A ketone may react with a nucleophilic hydride ion source (e.g., LiAlH or &#8220;H&#8221;) and, subsequently, a proton source (e.g., H\u2082O or &#8220;H+&#8221;) to give a secondary alcohol."},"content":{"rendered":"\n<p>A ketone may react with a nucleophilic hydride ion source (e.g., LiAlH or &#8220;H&#8221;) and, subsequently, a proton source (e.g., H\u2082O or &#8220;H+&#8221;) to give a secondary alcohol. Given the labeled ketone, determine the atom that is attacked by the hydride and proton. Which atom is attacked by the hydride ion source H in the first step? 1. H- 2. H+ Which atom is protonated in the second step of the reaction by H+?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>In the nucleophilic reduction of a <strong>ketone<\/strong> to a <strong>secondary alcohol<\/strong>, the <strong>reaction proceeds in two main steps<\/strong> involving a <strong>hydride ion source<\/strong> (like LiAlH\u2084 or NaBH\u2084) and then a <strong>proton source<\/strong> (like H\u2082O or H\u2083O\u207a). Understanding which atoms are involved in each step is crucial to predicting the mechanism and outcome.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Correct Answers:<\/strong><\/h3>\n\n\n\n<p><strong>Step 1 \u2013 Nucleophilic attack by H\u207b:<\/strong><br><strong>\u2192 The hydride ion (H\u207b) attacks the carbon atom of the carbonyl group (C=O).<\/strong><\/p>\n\n\n\n<p><strong>Step 2 \u2013 Protonation by H\u207a:<\/strong><br><strong>\u2192 The negatively charged oxygen atom (O\u207b) formed after hydride attack is protonated.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong> Explanation (Textbook Style):<\/strong><\/h3>\n\n\n\n<p>Ketones contain a carbonyl group (C=O), where the carbon atom is electrophilic due to the electron-withdrawing nature of the double-bonded oxygen. This makes the carbon susceptible to attack by nucleophiles. When a nucleophilic hydride ion source like lithium aluminum hydride (LiAlH\u2084) or sodium borohydride (NaBH\u2084) is used, the hydride ion (H\u207b) donates a pair of electrons to the electrophilic carbon of the carbonyl group.<\/p>\n\n\n\n<p>In the <strong>first step<\/strong>, the hydride ion acts as a nucleophile and attacks the <strong>carbon atom<\/strong> of the C=O group. This breaks the \u03c0 bond of the carbonyl group, resulting in the formation of a <strong>tetrahedral alkoxide intermediate<\/strong>. The oxygen now bears a negative charge (O\u207b), as it has accepted both bonding electrons from the broken double bond.<\/p>\n\n\n\n<p>In the <strong>second step<\/strong>, the reaction mixture is treated with a <strong>proton source<\/strong>, such as water or dilute acid. This provides a proton (H\u207a), which is <strong>accepted by the negatively charged oxygen atom (O\u207b)<\/strong> of the alkoxide intermediate. This step neutralizes the charge and yields a <strong>secondary alcohol<\/strong> as the final product.<\/p>\n\n\n\n<p>Thus, the overall transformation involves <strong>nucleophilic attack at the carbon<\/strong> atom by H\u207b followed by <strong>electrophilic protonation of the oxygen<\/strong> atom by H\u207a. This two-step process is a common method for reducing ketones into alcohols in organic synthesis.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Summary:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Hydride attacks:<\/strong> Carbon (C) of the carbonyl group.<\/li>\n\n\n\n<li><strong>Protonation occurs at:<\/strong> Oxygen (O\u207b) of the intermediate alkoxide.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-259.jpeg\" alt=\"\" class=\"wp-image-233516\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A ketone may react with a nucleophilic hydride ion source (e.g., LiAlH or &#8220;H&#8221;) and, subsequently, a proton source (e.g., H\u2082O or &#8220;H+&#8221;) to give a secondary alcohol. Given the labeled ketone, determine the atom that is attacked by the hydride and proton. Which atom is attacked by the hydride ion source H in the [&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-233515","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233515","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=233515"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233515\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233515"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233515"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233515"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}