{"id":217269,"date":"2025-05-21T11:51:12","date_gmt":"2025-05-21T11:51:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217269"},"modified":"2025-05-21T11:51:15","modified_gmt":"2025-05-21T11:51:15","slug":"convert-the-condensed-formula-hccchch32-into-its-expanded-structure","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/21\/convert-the-condensed-formula-hccchch32-into-its-expanded-structure\/","title":{"rendered":"Convert the condensed formula HCCCH(CH3)2 into its expanded structure."},"content":{"rendered":"\n<p>Convert the condensed formula HCCCH(CH3)2 into its expanded structure. Be sure to draw all hydrogen atoms. Draw the expanded structure. Select Draw Rings More Erase \/ 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>Let&#8217;s first understand the condensed formula <strong>HCCCH(CH3)2<\/strong> step-by-step, then expand it into its full structural formula with all hydrogens shown.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Analyze the condensed formula<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>HCCCH(CH3)2<\/strong><\/li>\n<\/ul>\n\n\n\n<p>This looks like an organic molecule made of carbons (C) and hydrogens (H). The formula can be broken down as follows:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>H-C\u2261C-C-H(CH3)2<\/strong><\/li>\n<\/ul>\n\n\n\n<p>But notice it\u2019s <strong>HCCCH(CH3)2<\/strong> without explicit bonds, so we interpret:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>H-C\u2261C-C-<\/strong> is the backbone (a chain with a triple bond)<\/li>\n\n\n\n<li>The <strong>C<\/strong> at the end is attached to <strong>(CH3)2<\/strong> which means two methyl groups (CH3) attached to that carbon.<\/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\">Step 2: Interpret the structure<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The first two carbons, H-C\u2261C-, form an acetylene (alkyne) part with a triple bond.<\/li>\n\n\n\n<li>The next carbon, C-, is attached to:\n<ul class=\"wp-block-list\">\n<li>One hydrogen (implied by the notation C-H)<\/li>\n\n\n\n<li>Two methyl groups (CH3)\u2082 attached to the same carbon.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>This carbon is a <strong>tertiary carbon<\/strong> because it bonds to three other carbons.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Draw the expanded structure<\/h3>\n\n\n\n<p>Let&#8217;s number the carbons:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>C1<\/strong> \u2014 bonded to one hydrogen and a triple bond to C2<\/li>\n\n\n\n<li><strong>C2<\/strong> \u2014 bonded to C1 with a triple bond and to C3 with a single bond<\/li>\n\n\n\n<li><strong>C3<\/strong> \u2014 bonded to C2, one hydrogen, and two methyl groups (CH3)<\/li>\n<\/ol>\n\n\n\n<p><strong>Explicit expanded structure:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C1: bonded to 1 H, triple bond to C2<\/li>\n\n\n\n<li>C2: triple bond to C1, single bond to C3, no hydrogens (because triple bond and 1 single bond already satisfy 4 bonds)<\/li>\n\n\n\n<li>C3: single bond to C2, bonded to 1 H, bonded to 2 methyl groups (each CH3)<\/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\">Step 4: Draw with all hydrogens shown<\/h3>\n\n\n\n<pre class=\"wp-block-code\"><code>     H\n     |\nH\u2014C\u2261C\u2014C\u2014H\n          |\n        CH3\n          |\n        CH3\n<\/code><\/pre>\n\n\n\n<p>More precisely, the third carbon is bonded to two methyl groups and one hydrogen:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C3 is connected to:\n<ul class=\"wp-block-list\">\n<li>One H<\/li>\n\n\n\n<li>Two CH3 groups<\/li>\n<\/ul>\n<\/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\">Full structural formula (with all hydrogens shown):<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C1: H\u2013C\u2261C\u2013 (triple bond)<\/li>\n\n\n\n<li>C2: \u2013C\u2013<\/li>\n\n\n\n<li>C3: \u2013C\u2013H<br>|<br>CH3<br>|<br>CH3<\/li>\n<\/ul>\n\n\n\n<p>or more visually:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>H\u2013C\u2261C\u2013C\u2013H\n       \/ \\\n     CH3 CH3\n<\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300 words):<\/h3>\n\n\n\n<p>The condensed formula <strong>HCCCH(CH3)2<\/strong> represents an alkyne with a substituent on the third carbon. Breaking down the formula, the first two carbons (H\u2013C\u2261C\u2013) form a triple bond typical of an alkyne group. The triple bond involves two \u03c0 bonds and one \u03c3 bond between carbon atoms C1 and C2. Carbon 1 has one hydrogen, completing its four bonds: one \u03c3 bond with hydrogen and a triple bond with C2.<\/p>\n\n\n\n<p>Carbon 2, which forms the other end of the triple bond, must have only one additional single bond to carbon 3 to complete its tetravalency, meaning it has no hydrogens bonded to it. This is consistent with the structure of alkynes where carbons in the triple bond tend to have fewer hydrogens.<\/p>\n\n\n\n<p>The third carbon (C3) is attached to the second carbon by a single bond. This carbon also binds to two methyl groups (CH3)\u2082 and one hydrogen atom. The presence of two methyl groups attached to C3 means it is a tertiary carbon (connected to three carbons).<\/p>\n\n\n\n<p>Each methyl group consists of one carbon bonded to three hydrogens, and all hydrogens are explicitly drawn to show the molecule&#8217;s complete structure.<\/p>\n\n\n\n<p>The overall structure shows a terminal alkyne with a branching at the third carbon, where the two methyl groups increase steric bulk and potentially affect the molecule\u2019s chemical behavior, like its reactivity or boiling point.<\/p>\n\n\n\n<p>By drawing all hydrogens, you accurately depict the molecule\u2019s saturation and hybridization. The hydrogens around C3 indicate that it is sp\u00b3 hybridized, whereas C1 and C2 in the triple bond are sp hybridized. This difference influences bond angles and molecular geometry \u2014 linear at the triple bond and tetrahedral at the branched carbon.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>If you&#8217;d like, I can generate a clean, drawn image of the expanded structure with all hydrogens explicitly shown. Would you want that?<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Convert the condensed formula HCCCH(CH3)2 into its expanded structure. Be sure to draw all hydrogen atoms. Draw the expanded structure. Select Draw Rings More Erase \/ H The Correct Answer and Explanation is: Let&#8217;s first understand the condensed formula HCCCH(CH3)2 step-by-step, then expand it into its full structural formula with all hydrogens shown. Step 1: [&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-217269","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217269","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=217269"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217269\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217269"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217269"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217269"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}