{"id":218500,"date":"2025-05-23T21:34:14","date_gmt":"2025-05-23T21:34:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218500"},"modified":"2025-05-23T21:34:16","modified_gmt":"2025-05-23T21:34:16","slug":"determine-the-intermediates-and-the-product-for-the-following-reaction-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/determine-the-intermediates-and-the-product-for-the-following-reaction-2\/","title":{"rendered":"Determine the intermediates and the product for the following reaction."},"content":{"rendered":"\n

Determine the intermediates and the product for the following reaction. (11 (CH)Cul Br2 CHyCH2OH LiBr LizCO,, DMF 2) H2O CHC [11 (CH,) Cui 21 H2O KOC(CHjh<\/p>\n\n\n\n

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

The reaction you are describing appears to be a two-step transformation<\/strong> involving an organocuprate reagent, a brominated substrate<\/strong>, and a nucleophilic substitution<\/strong> or reductive work-up<\/strong>. Based on your description, here is the cleaned-up version of the reactants and conditions (interpreted as best as possible from the formatting):<\/p>\n\n\n\n

\n\n\n\n

Step 1:<\/strong><\/p>\n\n\n\n

    \n
  • Reagents:<\/strong> (CH\u2083)\u2082CuLi (Gilman reagent), Br\u2082CHCH\u2082OH, LiBr, Li\u2082CO\u2083, DMF<\/li>\n\n\n\n
  • Conditions:<\/strong> Organocuprate substitution in DMF<\/li>\n<\/ul>\n\n\n\n

    Step 2:<\/strong><\/p>\n\n\n\n

      \n
    • Reagents:<\/strong> H\u2082O, CH\u2083CN (possibly), KOC(CH\u2083)\u2083 (a strong non-nucleophilic base, potassium tert-butoxide)<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step-by-step Explanation:<\/strong><\/h3>\n\n\n\n

      Step 1 \u2013 Nucleophilic Substitution with Gilman Reagent:<\/strong><\/h4>\n\n\n\n

      (CH\u2083)\u2082CuLi is a Gilman reagent<\/strong>, a type of organocuprate that can undergo nucleophilic substitution<\/strong> with alkyl halides, particularly primary halides like BrCH\u2082CH\u2082OH (2-bromoethanol).
      The organocuprate replaces the bromine<\/strong> atom with a methyl<\/strong> group: BrCH2CH2OH+(CH3)2CuLi\u2192CH\u2083CH2CH2OH\\text{BrCH}_2\\text{CH}_2\\text{OH} + (CH\u2083)\u2082CuLi \\rightarrow \\text{CH\u2083CH}_2\\text{CH}_2\\text{OH}<\/p>\n\n\n\n

      So the intermediate<\/strong> is propanol<\/strong> (CH\u2083CH\u2082CH\u2082OH).<\/p>\n\n\n\n

      Step 2 \u2013 Dehydration or Elimination Reaction:<\/strong><\/h4>\n\n\n\n

      In the second step, treatment with a strong base<\/strong> like potassium tert-butoxide (KOtBu) in a protic solvent leads to elimination of water<\/strong> (E2 elimination), forming an alkene: CH\u2083CH2CH2OH\u2192KOtBu \/ HeatCH2=CHCH3+H2O\\text{CH\u2083CH}_2\\text{CH}_2\\text{OH} \\xrightarrow{\\text{KOtBu \/ Heat}} \\text{CH}_2=\\text{CHCH}_3 + H\u2082O<\/p>\n\n\n\n

      So the final product<\/strong> is propene (CH\u2082=CHCH\u2083)<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      Summary:<\/strong><\/h3>\n\n\n\n
        \n
      • Intermediate:<\/strong> Propanol (CH\u2083CH\u2082CH\u2082OH)<\/li>\n\n\n\n
      • Final Product:<\/strong> Propene (CH\u2082=CHCH\u2083)<\/li>\n<\/ul>\n\n\n\n

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

        This reaction sequence involves a two-step synthetic transformation commonly encountered in organic chemistry involving organocuprate reagents and elimination reactions<\/strong>.<\/p>\n\n\n\n

        In the first step, the organocuprate reagent, dimethylcuprate [(CH\u2083)\u2082CuLi]<\/strong>, reacts with 2-bromoethanol (BrCH\u2082CH\u2082OH)<\/strong>. Organocuprates are known for their ability to perform nucleophilic substitutions<\/strong> on primary alkyl halides<\/strong>, displacing the halide with an alkyl group. In this case, one of the methyl groups from the Gilman reagent replaces the bromide atom on the substrate, yielding propanol (CH\u2083CH\u2082CH\u2082OH)<\/strong> as the intermediate. The reaction is conducted in polar aprotic solvents like DMF, with LiBr and Li\u2082CO\u2083 aiding in halide displacement and neutralization.<\/p>\n\n\n\n

        In the second step, this intermediate undergoes elimination. Potassium tert-butoxide (KOtBu)<\/strong> is a strong, non-nucleophilic base that favors E2 elimination<\/strong>. Upon heating, it abstracts a \u03b2-hydrogen from propanol, promoting elimination of water to form a double bond<\/strong>. The product of this elimination is propene (CH\u2082=CHCH\u2083)<\/strong>. This step illustrates a common method for converting alcohols into alkenes through base-induced dehydration.<\/p>\n\n\n\n

        Thus, the reaction sequence efficiently converts a brominated alcohol precursor into an alkene via nucleophilic substitution followed by elimination. The key transformations include SN2 displacement by an organocuprate and E2 elimination under basic conditions.<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Determine the intermediates and the product for the following reaction. (11 (CH)Cul Br2 CHyCH2OH LiBr LizCO,, DMF 2) H2O CHC [11 (CH,) Cui 21 H2O KOC(CHjh The Correct Answer and Explanation is: The reaction you are describing appears to be a two-step transformation involving an organocuprate reagent, a brominated substrate, and a nucleophilic substitution or […]<\/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-218500","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218500","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=218500"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218500\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218500"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218500"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218500"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}