{"id":228027,"date":"2025-06-06T18:03:06","date_gmt":"2025-06-06T18:03:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=228027"},"modified":"2025-06-06T18:03:08","modified_gmt":"2025-06-06T18:03:08","slug":"curved-arrows-are-used-to-illustrate-the-flow-of-electrons","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/06\/curved-arrows-are-used-to-illustrate-the-flow-of-electrons\/","title":{"rendered":"Curved arrows are used to illustrate the flow of electrons"},"content":{"rendered":"\n<p>Curved arrows are used to illustrate the flow of electrons. Follow the curved arrows and draw the products of the following reaction. Include all lone pairs and charges as appropriate. Ignore inorganic byproducts.<\/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<h3 class=\"wp-block-heading\"><strong>Correct Product:<\/strong><\/h3>\n\n\n\n<p>The product of this reaction is an <strong>amide<\/strong>, formed by the substitution of the <strong>chloride<\/strong> group (Cl\u207b) on the carbonyl carbon with an <strong>amine group<\/strong> (from the nitrogen nucleophile). The lone pair of electrons on the nitrogen attacks the electrophilic carbonyl carbon, forming a tetrahedral intermediate. Subsequently, the leaving group (Cl\u207b) departs, and the carbonyl is restored.<\/p>\n\n\n\n<p>The correct product will be:<\/p>\n\n\n\n<p><strong>Structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A carbon double-bonded to oxygen (carbonyl).<\/li>\n\n\n\n<li>Single bonded to the nitrogen nucleophile (now part of the amide).<\/li>\n\n\n\n<li>The nitrogen is bonded to a methyl group and a hydrogen.<\/li>\n\n\n\n<li>The Cl atom leaves as Cl\u207b.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>This reaction is a classic example of <strong>nucleophilic acyl substitution<\/strong>, where a nucleophile (in this case, a nitrogen atom from a methylamine) attacks a carbonyl carbon of an acyl compound (an acyl chloride).<\/p>\n\n\n\n<p>The carbonyl carbon in the acyl chloride is electrophilic due to the strong electron-withdrawing effect of both the oxygen and the chlorine. This makes it highly susceptible to attack by nucleophiles. The curved arrow from the lone pair on the nitrogen to the carbonyl carbon represents the <strong>nucleophilic attack<\/strong>, which momentarily breaks the \u03c0 bond of the C=O, shifting electrons onto the oxygen and forming a <strong>tetrahedral intermediate<\/strong>.<\/p>\n\n\n\n<p>A second curved arrow shows electrons collapsing back from the negatively charged oxygen to reform the carbonyl double bond. As this happens, the bond between the carbon and the chlorine is broken, and Cl\u207b is expelled as the leaving group.<\/p>\n\n\n\n<p>This results in the formation of an <strong>amide<\/strong>, where the nitrogen group is now attached to the carbonyl carbon, and chlorine has been replaced.<\/p>\n\n\n\n<p>Important to note:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The lone pair on nitrogen initiates the reaction.<\/li>\n\n\n\n<li>The intermediate is not stable and rapidly collapses to eject the chloride.<\/li>\n\n\n\n<li>Chloride is a good leaving group, which makes this reaction efficient.<\/li>\n<\/ul>\n\n\n\n<p>Properly drawing the product includes:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Showing the new C\u2013N bond.<\/li>\n\n\n\n<li>Including lone pairs on oxygen and nitrogen.<\/li>\n\n\n\n<li>Indicating the formal charge (if any), though in this case the neutral amide is formed.<\/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-banner4-571.jpeg\" alt=\"\" class=\"wp-image-228029\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Curved arrows are used to illustrate the flow of electrons. Follow the curved arrows and draw the products of the following reaction. Include all lone pairs and charges as appropriate. Ignore inorganic byproducts. The Correct Answer and Explanation is: Correct Product: The product of this reaction is an amide, formed by the substitution of 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-228027","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228027","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=228027"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228027\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=228027"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=228027"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=228027"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}