{"id":187890,"date":"2025-02-06T07:55:44","date_gmt":"2025-02-06T07:55:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187890"},"modified":"2025-02-06T07:55:46","modified_gmt":"2025-02-06T07:55:46","slug":"what-is-the-predicted-major-product-of-the-reaction-shown","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/what-is-the-predicted-major-product-of-the-reaction-shown\/","title":{"rendered":"What is the predicted major product of the reaction shown"},"content":{"rendered":"\n

What is the predicted major product of the reaction shown? 1. NaOH\/Br2 2. HO* ?? Br “. Br OH Br Br II III OBT 0 OH dan OH IV V What is the predicted major product of the reaction shown? NaOH, H2O butanal + 2,2-dimethylpentanal ?? 0 ?? 0 ??? H H H OH 0 | II III IV OH V OH What reactants would give the aldol condensation product shown? H H HO H H III ??? . H H H IV V<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

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

he reaction sequence you’ve provided involves two distinct processes:<\/p>\n\n\n\n

    \n
  1. Treatment with NaOH and Br\u2082<\/strong>: his suggests a halogenation reaction, specifically the haloform reaction<\/strong>, which typically occurs with methyl ketones.<\/li>\n\n\n\n
  2. Treatment with NaOH and H\u2082O<\/strong>: his indicates an aldol condensation<\/strong>, a reaction between carbonyl compounds in the presence of a base.
    iven the reactants\u2014butanal<\/strong> and 2,2-dimethylpentanal<\/strong>\u2014let’s analyze each step:
    Step 1: NaOH and Br\u2082 Treatment<\/strong><\/li>\n<\/ol>\n\n\n\n

    he haloform reaction involves the halogenation of methyl ketones at the alpha position, leading to the formation of a carboxylate and a haloform (e.g., chloroform, bromoform).owever, in this case, neither butanal nor 2,2-dimethylpentanal are methyl ketones; they are aldehydes.ldehydes can undergo halogenation at the alpha position in the presence of halogens and base, leading to alpha-halo aldehydes.herefore, under these conditions, both aldehydes are likely to form alpha-bromo derivatives:<\/p>\n\n\n\n

      \n
    • Butanal<\/strong>: H\u2083-CH\u2082-CH\u2082-CHO \u2192 CH\u2083-CH\u2082-CH(Br)-CHO<\/li>\n\n\n\n
    • 2,2-Dimethylpentanal<\/strong>: CH\u2083)\u2083C-CH\u2082-CH\u2082-CHO \u2192 (CH\u2083)\u2083C-CH\u2082-CH(Br)-CHO
      Step 2: NaOH and H\u2082O Treatment (Aldol Condensation)<\/strong><\/li>\n<\/ul>\n\n\n\n

      n the aldol condensation, an enolate ion formed from one carbonyl compound attacks the carbonyl carbon of another.iven the two aldehydes, the reaction can proceed as follows:<\/p>\n\n\n\n

        \n
      1. Enolate Formation<\/strong>: aOH deprotonates the alpha position of one of the aldehydes, forming an enolate ion.<\/li>\n\n\n\n
      2. Nucleophilic Addition<\/strong>: he enolate attacks the carbonyl carbon of the other aldehyde, leading to the formation of a \u03b2-hydroxy aldehyde (aldol product).<\/li>\n\n\n\n
      3. Dehydration<\/strong>: nder the basic conditions, the \u03b2-hydroxy aldehyde can lose a water molecule to form an \u03b1,\u03b2-unsaturated aldehyde.
        onsidering the structures:<\/li>\n<\/ol>\n\n\n\n
          \n
        • Butanal Enolate<\/strong>: H\u2083-CH\u2082-CH=CH-O\u207b<\/li>\n\n\n\n
        • 2,2-Dimethylpentanal Enolate<\/strong>: CH\u2083)\u2083C-CH\u2082-CH=CH-O\u207b
          he enolate from butanal is less hindered and more likely to form.t can attack the carbonyl carbon of 2,2-dimethylpentanal, leading to the following aldol product before dehydration:
          H\u2083-CH\u2082-CH(OH)-CH\u2082-CH\u2082-C(CH\u2083)\u2083
          pon dehydration, the major product would be:
          H\u2083-CH\u2082-CH=CH-CH\u2082-C(CH\u2083)\u2083
          his compound is an \u03b1,\u03b2-unsaturated aldehyde resulting from the cross-aldol condensation of butanal and 2,2-dimethylpentanal.
          Conclusion<\/strong><\/li>\n<\/ul>\n\n\n\n

          he predicted major product of the reaction sequence is 3,3-dimethyl-5-heptenal<\/strong>, formed through an aldol condensation between butanal and 2,2-dimethylpentanal.
          or a visual explanation of the aldol condensation mechanism, you might find this video helpful:
          \ue200video\ue202Aldol Addition Reactions, Intramolecular Aldol Condensation\ue202turn0search9\ue201<\/p>\n","protected":false},"excerpt":{"rendered":"

          What is the predicted major product of the reaction shown? 1. NaOH\/Br2 2. HO* ?? Br “. Br OH Br Br II III OBT 0 OH dan OH IV V What is the predicted major product of the reaction shown? NaOH, H2O butanal + 2,2-dimethylpentanal ?? 0 ?? 0 ??? H H H OH 0 […]<\/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-187890","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187890","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=187890"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187890\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187890"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187890"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187890"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}