{"id":238323,"date":"2025-06-18T04:32:02","date_gmt":"2025-06-18T04:32:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=238323"},"modified":"2025-06-18T04:32:30","modified_gmt":"2025-06-18T04:32:30","slug":"compound-a-is-an-l-aldopentose","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/18\/compound-a-is-an-l-aldopentose\/","title":{"rendered":"Compound A is an L-aldopentose"},"content":{"rendered":"\n

Compound A is an L-aldopentose. Oxidation of A with HNO3 yields an optically active aldaric acid B. Wohl degradation of A yields an L-aldotetrose C. Reduction of C with NaBH4 produces an optically active alditol D. Using this information, help draw structures A, B, C, and D.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Correct Structures:<\/h3>\n\n\n\n
    \n
  1. Compound A (L-aldopentose):<\/strong> L-arabinose<\/li>\n\n\n\n
  2. Compound B (aldaric acid):<\/strong> L-arabinaric acid<\/li>\n\n\n\n
  3. Compound C (aldotetrose from Wohl degradation):<\/strong> L-erythrose<\/li>\n\n\n\n
  4. Compound D (alditol from reduction):<\/strong> L-erythritol<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Explanation <\/h3>\n\n\n\n

    The problem begins with Compound A<\/strong>, which is described as an L-aldopentose<\/strong>. Aldopentoses are five-carbon sugars with an aldehyde group at C1 and hydroxyl groups on the remaining carbons. The term “L” refers to the stereochemistry relative to L-glyceraldehyde.<\/p>\n\n\n\n

    Among L-aldopentoses, we consider L-arabinose<\/strong> and L-xylose<\/strong>. However, further information helps narrow the identity.<\/p>\n\n\n\n

    Next, oxidation of A with nitric acid (HNO\u2083)<\/strong> forms Compound B<\/strong>, an aldaric acid<\/strong>. Nitric acid oxidizes both the aldehyde (C1) and the terminal alcohol (C5) to carboxylic acids. Importantly, Compound B is optically active<\/strong>. This rules out L-xylose, which would yield meso-xarabaric acid<\/strong>, a non-optically active (achiral) compound upon full oxidation. Therefore, Compound A must be L-arabinose<\/strong>, which forms optically active L-arabinaric acid<\/strong>.<\/p>\n\n\n\n

    Then, Wohl degradation<\/strong> removes one carbon from the aldehyde end (C1) of Compound A to give a four-carbon aldotetrose (Compound C)<\/strong>. The product of this step from L-arabinose is L-erythrose<\/strong>, because it preserves the stereochemistry of carbons 2 through 4 of the original molecule.<\/p>\n\n\n\n

    Reduction of Compound C (L-erythrose)<\/strong> with NaBH\u2084<\/strong> reduces the aldehyde group at C1 to a primary alcohol, yielding an alditol<\/strong> (Compound D). This reduction product is L-erythritol<\/strong>, which is optically active and retains chirality at both central carbon atoms (C2 and C3).<\/p>\n\n\n\n

    Each step in the sequence preserves or transforms the stereochemistry in a predictable way:<\/p>\n\n\n\n

      \n
    • Oxidation retains chirality at all centers.<\/li>\n\n\n\n
    • Wohl degradation removes the top carbon but retains the configuration of the remaining carbons.<\/li>\n\n\n\n
    • Reduction with NaBH\u2084 introduces no new stereocenters, only converts an aldehyde to an alcohol.<\/li>\n<\/ul>\n\n\n\n

      Therefore, the logical structures are:<\/p>\n\n\n\n

        \n
      • A: L-arabinose<\/strong><\/li>\n\n\n\n
      • B: L-arabinaric acid<\/strong><\/li>\n\n\n\n
      • C: L-erythrose<\/strong><\/li>\n\n\n\n
      • D: L-erythritol<\/strong><\/li>\n<\/ul>\n\n\n\n
        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Compound A is an L-aldopentose. Oxidation of A with HNO3 yields an optically active aldaric acid B. Wohl degradation of A yields an L-aldotetrose C. Reduction of C with NaBH4 produces an optically active alditol D. Using this information, help draw structures A, B, C, and D. The Correct Answer and Explanation is: Correct Structures: […]<\/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-238323","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238323","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=238323"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/238323\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=238323"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=238323"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=238323"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}