{"id":215257,"date":"2025-05-16T15:43:54","date_gmt":"2025-05-16T15:43:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=215257"},"modified":"2025-05-16T15:43:56","modified_gmt":"2025-05-16T15:43:56","slug":"draw-a-six-carbon-alkyne-that-can-exist-as-diastereomers-8","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/16\/draw-a-six-carbon-alkyne-that-can-exist-as-diastereomers-8\/","title":{"rendered":"Draw a six-carbon alkyne that can exist as diastereomers."},"content":{"rendered":"\n<p>Draw a six-carbon alkyne that can exist as diastereomers. You do not need to specify wedge or dashed bonds. Select Draw Rings More Erase \/ MICH 5 3 2 o<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Certainly! Let&#8217;s first understand the problem clearly:<\/p>\n\n\n\n<p><strong>Task:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Draw a six-carbon alkyne (C6)<\/li>\n\n\n\n<li>It must be able to exist as <strong>diastereomers<\/strong> (meaning it must have at least two stereocenters, or chiral centers, but not be mirror images of each other\u2014i.e., not enantiomers)<\/li>\n\n\n\n<li>No need for wedge or dashed bonds (just the skeletal structure)<\/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 1: Understand Diastereomers in Alkynes<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Alkyne<\/strong> means a carbon-carbon triple bond (C\u2261C).<\/li>\n\n\n\n<li>To have diastereomers, the molecule must have <strong>at least two stereocenters<\/strong> (chiral carbons).<\/li>\n\n\n\n<li>Stereocenters are <strong>tetrahedral carbons<\/strong> attached to four different groups.<\/li>\n\n\n\n<li>Alkynes have sp hybridized carbons at the triple bond, so these carbons <strong>cannot be stereocenters<\/strong> (because they are linear, with only two substituents).<\/li>\n\n\n\n<li>So the chiral centers must be <strong>away from the alkyne<\/strong>.<\/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: Design the molecule<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Six carbons total, one alkyne somewhere in the chain.<\/li>\n\n\n\n<li>Add two chiral centers <strong>not adjacent to the alkyne<\/strong> but in the chain so that the molecule can have diastereomers.<\/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\">Example:<\/h3>\n\n\n\n<p>Draw the structure for <strong>4-hexyne with chiral centers at C2 and C3<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon numbering from left to right: C1 &#8211; C2 &#8211; C3 &#8211; C4 &#8211; C5 &#8211; C6<\/li>\n\n\n\n<li>Alkyne triple bond between C4 and C5 (C4\u2261C5)<\/li>\n\n\n\n<li>Chiral centers at C2 and C3, each bonded to four different groups.<\/li>\n<\/ul>\n\n\n\n<p>Let&#8217;s define groups:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C1: CH3-<\/li>\n\n\n\n<li>C2: chiral center (with substituents H, CH3, C1-alkyl, and rest of chain)<\/li>\n\n\n\n<li>C3: chiral center<\/li>\n\n\n\n<li>C4: sp carbon, triple bond to C5<\/li>\n\n\n\n<li>C5: sp carbon, triple bond to C4<\/li>\n\n\n\n<li>C6: terminal methyl or substituent.<\/li>\n<\/ul>\n\n\n\n<p>For instance:<\/p>\n\n\n\n<p>CH3-CH<em>R-CH<\/em>S-C\u2261C-CH3<\/p>\n\n\n\n<p>Where CH<em>R and CH<\/em>S are chiral centers at C2 and C3.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Draw it simply:<\/h3>\n\n\n\n<p>CH3-CH-CH-C\u2261C-CH3<br>| |<br>X Y<\/p>\n\n\n\n<p>(X and Y represent substituents to ensure chirality)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Why can it have diastereomers?<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>With two stereocenters (C2 and C3), there are 2^2 = 4 stereoisomers possible.<\/li>\n\n\n\n<li>These include pairs of enantiomers and pairs of diastereomers.<\/li>\n\n\n\n<li>Diastereomers differ in configuration at <strong>one<\/strong> stereocenter but not both.<\/li>\n\n\n\n<li>Because alkyne does not affect stereochemistry directly, the chiral centers generate diastereomers.<\/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\">Explanation in detail (300+ words):<\/h3>\n\n\n\n<p>An alkyne is a hydrocarbon containing a carbon-carbon triple bond. The carbons in the triple bond are sp hybridized and linear, so they cannot be stereocenters because they are bonded to only two other atoms, unlike sp3 carbons which have four different substituents.<\/p>\n\n\n\n<p>To have diastereomers, a molecule must have at least two stereocenters \u2014 carbons with four different groups attached \u2014 and these stereocenters must be arranged so that some stereoisomers are non-mirror images of each other.<\/p>\n\n\n\n<p>For a six-carbon alkyne, a practical way to achieve this is to place the alkyne functional group somewhere in the carbon chain and have two chiral centers elsewhere in the molecule.<\/p>\n\n\n\n<p>Take 4-hexyne as an example: this molecule has six carbons with a triple bond between carbons 4 and 5. If carbons 2 and 3 are sp3 hybridized carbons each bonded to four different groups (such as H, CH3, and different alkyl chains), then each can be a stereocenter.<\/p>\n\n\n\n<p>Having two stereocenters means the molecule can exist as four stereoisomers (2^2 = 4). Among these stereoisomers, pairs of enantiomers (mirror images) and pairs of diastereomers (non-mirror image stereoisomers) exist. The diastereomers differ at one stereocenter but not the other, resulting in different physical and chemical properties.<\/p>\n\n\n\n<p>This structural design is consistent with the requirement of a six-carbon alkyne capable of existing as diastereomers.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw a six-carbon alkyne that can exist as diastereomers. You do not need to specify wedge or dashed bonds. Select Draw Rings More Erase \/ MICH 5 3 2 o The Correct Answer and Explanation is: Certainly! Let&#8217;s first understand the problem clearly: Task: Step 1: Understand Diastereomers in Alkynes Step 2: Design the molecule [&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-215257","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215257","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=215257"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215257\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=215257"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=215257"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=215257"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}