{"id":227902,"date":"2025-06-06T15:12:18","date_gmt":"2025-06-06T15:12:18","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=227902"},"modified":"2025-06-06T15:12:20","modified_gmt":"2025-06-06T15:12:20","slug":"select-the-appropriate-synthetic-sequence-to-accomplish-the-transformation-shown-b","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/06\/select-the-appropriate-synthetic-sequence-to-accomplish-the-transformation-shown-b\/","title":{"rendered":"Select the appropriate synthetic sequence to accomplish the transformation shown. B"},"content":{"rendered":"\n

22) Select the appropriate synthetic sequence to accomplish the transformation shown. Br<\/p>\n\n\n\n

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

The Correct Answer and Explanation is:<\/strong>1<\/a><\/sup><\/p>\n\n\n\n\n\n

Correct Answer: D) 1. KOtBu; 2. HBr, ROOR; 3. CH\u2083CH\u2082CCNa<\/strong><\/h3>\n\n\n\n
\n\n\n\n

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

The transformation shown is the conversion of a tert-butyl bromide on a cyclopentane ring<\/strong> into a cyclopentyl group with a side chain containing a terminal alkyne (triple bond)<\/strong>. Let\u2019s analyze this transformation step by step:<\/p>\n\n\n\n

\n\n\n\n

Step 1: Elimination Reaction<\/strong><\/h4>\n\n\n\n

Reagent: KOtBu (potassium tert-butoxide)<\/strong><\/p>\n\n\n\n

    \n
  • This strong, bulky base promotes E2 elimination<\/strong>, particularly favoring Hofmann product<\/strong> (less substituted alkene).<\/li>\n\n\n\n
  • From the tert-butyl bromide, KOtBu will abstract a \u03b2-proton and eliminate HBr, forming 1-cyclopentylethene (CH\u2082=CH\u2013cyclopentyl)<\/strong>.<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step 2: Anti-Markovnikov Hydrohalogenation<\/strong><\/h4>\n\n\n\n

    Reagents: HBr + ROOR (peroxides)<\/strong><\/p>\n\n\n\n

      \n
    • The alkene undergoes radical addition<\/strong> of HBr in the presence of ROOR<\/strong>, giving anti-Markovnikov addition<\/strong>.<\/li>\n\n\n\n
    • This results in the bromine adding to the terminal carbon<\/strong> of the double bond (CH\u2083\u2013CHBr\u2013cyclopentyl), forming a 1-bromoethyl-cyclopentane<\/strong>.<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 3: Nucleophilic Substitution with Sodium Acetylide<\/strong><\/h4>\n\n\n\n

      Reagent: CH\u2083CH\u2082C\u2261CNa (sodium propyne or sodium alkyne anion)<\/strong><\/p>\n\n\n\n

        \n
      • This is a strong nucleophile<\/strong> and will undergo SN2 substitution<\/strong> with the 1\u00b0 alkyl bromide made in step 2.<\/li>\n\n\n\n
      • The result is the substitution of the Br with CH\u2083CH\u2082C\u2261C\u2013 group (a terminal alkyne side chain).<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        Summary of the Transformation:<\/strong><\/h4>\n\n\n\n
          \n
        1. KOtBu<\/strong> \u2192 E2 elimination: forms alkene.<\/li>\n\n\n\n
        2. HBr\/ROOR<\/strong> \u2192 Anti-Markovnikov radical addition: adds Br to terminal carbon.<\/li>\n\n\n\n
        3. CH\u2083CH\u2082C\u2261CNa<\/strong> \u2192 SN2 attack: installs alkyne.<\/li>\n<\/ol>\n\n\n\n
          \n\n\n\n

          Why the Other Choices Are Incorrect:<\/strong><\/h4>\n\n\n\n
            \n
          • A<\/strong> involves Br\u2082\/h\u03bd (radical bromination) which doesn\u2019t fit the need.<\/li>\n\n\n\n
          • B & C<\/strong> introduce a Grignard or perform addition without radical conditions, not suitable here.<\/li>\n\n\n\n
          • E<\/strong> ends with H\u2083O\u207a (acidic workup), which doesn’t install an alkyne.<\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n

            Thus, choice D<\/strong> correctly sequences elimination \u2192 anti-Markovnikov addition \u2192 SN2 substitution<\/strong> to give the desired alkyne product.<\/p>\n\n\n\n

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

            22) Select the appropriate synthetic sequence to accomplish the transformation shown. Br The Correct Answer and Explanation is:1 Correct Answer: D) 1. KOtBu; 2. HBr, ROOR; 3. CH\u2083CH\u2082CCNa Explanation The transformation shown is the conversion of a tert-butyl bromide on a cyclopentane ring into a cyclopentyl group with a side chain containing a terminal alkyne […]<\/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-227902","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227902","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=227902"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227902\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227902"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227902"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227902"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}