{"id":229580,"date":"2025-06-08T09:46:20","date_gmt":"2025-06-08T09:46:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=229580"},"modified":"2025-06-08T09:46:23","modified_gmt":"2025-06-08T09:46:23","slug":"draw-the-fischer-projections-of-the-four-aldotetroses","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/08\/draw-the-fischer-projections-of-the-four-aldotetroses\/","title":{"rendered":"Draw the Fischer projections of the four aldotetroses."},"content":{"rendered":"\n<p>Draw the Fischer projections of the four aldotetroses. Draw the D-sugar on the left and its L-isomer directly to the right of it. Be sure you select the appropriate hydroxy group so that the carbon bonded to the oxygen is connected to it. Draw D-sugar Draw sugar CH2OH CH2OH<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/image-287.png\" alt=\"\" class=\"wp-image-229581\"\/><\/figure>\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>The question asks you to draw the Fischer projections of the four aldotetroses \u2014 specifically, the D- and L-isomers of one aldotetrose. Aldotetroses are sugars with four carbon atoms and an aldehyde group at the top. There are four stereoisomers (two pairs of D- and L-enantiomers), but this question focuses on just one pair.<\/p>\n\n\n\n<p>From the image, we see the Fischer projections for <strong>D-erythrose<\/strong> on the left and <strong>L-erythrose<\/strong> on the right.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Correct Structures:<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">D-erythrose (D-sugar 1):<\/h4>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>CHO\n |\nH\u2013C\u2013OH\n |\nH\u2013C\u2013OH\n |\nCH2OH\n<\/code><\/pre>\n\n\n\n<h4 class=\"wp-block-heading\">L-erythrose (L-sugar 1):<\/h4>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>CHO\n |\nHO\u2013C\u2013H\n |\nHO\u2013C\u2013H\n |\nCH2OH\n<\/code><\/pre>\n\n\n\n<p>These are correctly drawn in the image. The distinguishing feature between D- and L- isomers lies in the orientation of the hydroxyl group on the <strong>highest-numbered chiral carbon<\/strong> (carbon 3 in aldotetroses). In D-sugars, the hydroxyl on the bottom chiral center points to the <strong>right<\/strong>, and in L-sugars, it points to the <strong>left<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>Aldotetroses are monosaccharides composed of four carbon atoms and an aldehyde functional group. The general structure includes an aldehyde at carbon 1 and hydroxyl groups at the other chiral centers. They have two chiral centers (C2 and C3), meaning they can form 2\u00b2 = 4 stereoisomers. These stereoisomers are divided into two sets of enantiomers (mirror images): erythrose and threose, each with D- and L-forms.<\/p>\n\n\n\n<p>In a Fischer projection:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The most oxidized carbon (the aldehyde, CHO) is placed at the top.<\/li>\n\n\n\n<li>Horizontal lines represent bonds coming <strong>out of the plane<\/strong> toward the viewer.<\/li>\n\n\n\n<li>Vertical lines represent bonds going <strong>into the plane<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>For D-aldoses, the hydroxyl group on the <strong>last chiral center<\/strong> (farthest from the aldehyde) is on the <strong>right<\/strong>.<br>For L-aldoses, it&#8217;s on the <strong>left<\/strong>.<\/p>\n\n\n\n<p>The image correctly displays <strong>D-erythrose<\/strong> and its mirror image <strong>L-erythrose<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Both have hydroxyl groups on the same side for C2 and C3 (right for D, left for L).<\/li>\n\n\n\n<li>This defines them as the \u201cerythro\u201d pair, where the -OH groups are on the same side.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the image is accurate for one of the four aldotetroses. The other two (threose and its mirror) would have the hydroxyls on opposite sides at the chiral centers.<\/p>\n\n\n\n<p>This fundamental understanding is crucial in stereochemistry and carbohydrate biochemistry, where the configuration determines biochemical reactivity and biological function.<\/p>\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-banner6-142.jpeg\" alt=\"\" class=\"wp-image-229582\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Fischer projections of the four aldotetroses. Draw the D-sugar on the left and its L-isomer directly to the right of it. Be sure you select the appropriate hydroxy group so that the carbon bonded to the oxygen is connected to it. Draw D-sugar Draw sugar CH2OH CH2OH The Correct Answer and Explanation is: [&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-229580","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229580","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=229580"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229580\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229580"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229580"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229580"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}