{"id":179398,"date":"2024-12-30T19:42:36","date_gmt":"2024-12-30T19:42:36","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=179398"},"modified":"2024-12-30T19:42:38","modified_gmt":"2024-12-30T19:42:38","slug":"draw-a-lewis-structure-for-each-compound-include-all-nonbonding-pairs-of-electrons","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/12\/30\/draw-a-lewis-structure-for-each-compound-include-all-nonbonding-pairs-of-electrons\/","title":{"rendered":"Draw a Lewis structure for each compound. Include all nonbonding pairs of electrons"},"content":{"rendered":"\n<p>Draw a Lewis structure for each compound. Include all nonbonding pairs of electrons.<\/p>\n\n\n\n<p>(a) CH3COCH2CHCHCOOH<\/p>\n\n\n\n<p>(b) NCCH2COCH2CHO<\/p>\n\n\n\n<p>(c) CH2CHCH(OH)CH2CO2H<\/p>\n\n\n\n<p>(d) CH2CHC(CH3)CHCOOCH3<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s start by drawing the Lewis structures for each compound. We will also identify and include all nonbonding pairs of electrons (lone pairs) in the drawings. Here\u2019s the step-by-step approach for each molecule.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(a) <strong>CH\u2083COCH\u2082CH=CHCOOH (Methyl vinyl ketone)<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count the atoms and arrange them:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The compound has a total of 10 carbon (C) atoms, 12 hydrogen (H) atoms, and 2 oxygen (O) atoms.<\/li>\n\n\n\n<li>The structure contains a ketone group (C=O), a carboxylic acid group (COOH), a double bond between carbons (C=C), and single bonds between the other atoms.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Constructing the skeleton:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Start with the central C-C backbone: <strong>CH\u2083-C-O-CH\u2082-CH=CH-COOH<\/strong>.<\/li>\n\n\n\n<li>The first part (CH\u2083-C=O) is an acetyl group.<\/li>\n\n\n\n<li>The second part (CH\u2082-CH=CH) is the vinyl group.<\/li>\n\n\n\n<li>Finally, add the carboxyl group (-COOH).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Assigning bonds and lone pairs:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The O in the carboxyl group (COOH) has two lone pairs. The C=O bond in the ketone has no lone pairs on the O.<\/li>\n\n\n\n<li>Each C-H bond has two electrons in the form of shared pairs, and each C=C bond has four electrons.<\/li>\n\n\n\n<li>The remaining lone pairs are on the oxygen atoms.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(b) <strong>NCCH\u2082COCH\u2082CHO (2-Cyanopropionyl aldehyde)<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count the atoms and arrange them:<\/strong>\n<ul class=\"wp-block-list\">\n<li>This compound contains 7 carbon (C), 9 hydrogen (H), 2 oxygen (O), and 1 nitrogen (N) atom.<\/li>\n\n\n\n<li>The structure involves a nitrile group (-C\u2261N), an aldehyde group (-CHO), and a carbonyl group (-C=O).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Constructing the skeleton:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Start with the chain: <strong>N\u2261C-CH\u2082-C=O-CH\u2082-CHO<\/strong>.<\/li>\n\n\n\n<li>Place the nitrile group at the beginning, then a -CH\u2082 group, followed by a carbonyl group, and another -CH\u2082 group, finishing with an aldehyde group.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Assigning bonds and lone pairs:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The nitrile group has a triple bond between C and N with one lone pair on the N.<\/li>\n\n\n\n<li>The aldehyde and carbonyl oxygens have lone pairs, as typical for carbonyl groups.<\/li>\n\n\n\n<li>Hydrogens are attached to C atoms, completing each C\u2019s valence.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(c) <strong>CH\u2082=CH-CH(OH)-CH\u2082CO\u2082H (3-Hydroxybutenoic acid)<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count the atoms and arrange them:<\/strong>\n<ul class=\"wp-block-list\">\n<li>This compound has 6 carbon (C), 8 hydrogen (H), 3 oxygen (O) atoms.<\/li>\n\n\n\n<li>It contains a hydroxyl group (-OH), a carboxyl group (-COOH), and a double bond (C=C).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Constructing the skeleton:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The backbone is <strong>CH\u2082=CH-CH(OH)-CH\u2082CO\u2082H<\/strong>.<\/li>\n\n\n\n<li>Place the hydroxyl group (-OH) on the third carbon, and the carboxyl group (-COOH) on the last carbon.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Assigning bonds and lone pairs:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The carboxyl group has lone pairs on the oxygen atoms.<\/li>\n\n\n\n<li>The hydroxyl group also has lone pairs on the oxygen.<\/li>\n\n\n\n<li>The C=C double bond and single C-H bonds fill the other valence.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(d) <strong>CH\u2082=CH-C(CH\u2083)-CHCOOCH\u2083 (Methyl acrylate)<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count the atoms and arrange them:<\/strong>\n<ul class=\"wp-block-list\">\n<li>This compound contains 8 carbon (C), 12 hydrogen (H), 2 oxygen (O) atoms.<\/li>\n\n\n\n<li>The structure includes an alkene (C=C), a methyl group (-C(CH\u2083)), and an ester group (-COOCH\u2083).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Constructing the skeleton:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The backbone is <strong>CH\u2082=CH-C(CH\u2083)-CHCOOCH\u2083<\/strong>.<\/li>\n\n\n\n<li>Attach the methyl group to the central carbon, the ester group (-COOCH\u2083) to the terminal carbon, and the C=C bond in the middle.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Assigning bonds and lone pairs:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The ester group (-COOCH\u2083) contains lone pairs on the oxygen atoms.<\/li>\n\n\n\n<li>The other carbons form bonds as needed to satisfy valency.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of the Lewis Structures<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>(a)<\/strong> Involves an acetyl group (CH\u2083-C=O), a vinyl group (CH\u2082=CH), and a carboxyl group (-COOH), with lone pairs on oxygen atoms.<\/li>\n\n\n\n<li><strong>(b)<\/strong> Contains a nitrile group (C\u2261N), aldehyde group (-CHO), and a carbonyl group (C=O) with lone pairs on oxygen atoms.<\/li>\n\n\n\n<li><strong>(c)<\/strong> Includes a hydroxyl group (-OH), a carboxyl group (-COOH), and a C=C bond.<\/li>\n\n\n\n<li><strong>(d)<\/strong> Features an alkene (C=C), a tertiary carbon (C(CH\u2083)\u2082), and an ester group (COOCH\u2083), with lone pairs on oxygen atoms.<\/li>\n<\/ul>\n\n\n\n<p>The lone pairs are placed around oxygen atoms in carbonyl and hydroxyl groups, and other nonbonding pairs are assumed to complete the octet for each atom.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw a Lewis structure for each compound. Include all nonbonding pairs of electrons. (a) CH3COCH2CHCHCOOH (b) NCCH2COCH2CHO (c) CH2CHCH(OH)CH2CO2H (d) CH2CHC(CH3)CHCOOCH3 The correct answer and explanation is: Let&#8217;s start by drawing the Lewis structures for each compound. We will also identify and include all nonbonding pairs of electrons (lone pairs) in the drawings. Here\u2019s 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-179398","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/179398","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=179398"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/179398\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=179398"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=179398"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=179398"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}