{"id":220992,"date":"2025-05-29T05:24:50","date_gmt":"2025-05-29T05:24:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220992"},"modified":"2025-05-29T05:24:52","modified_gmt":"2025-05-29T05:24:52","slug":"draw-all-missing-reactants-and-or-products-in-the-appropriate-boxes-by-placing-atoms-on-the-canvas-and-connecting-them-with-bonds","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/draw-all-missing-reactants-and-or-products-in-the-appropriate-boxes-by-placing-atoms-on-the-canvas-and-connecting-them-with-bonds\/","title":{"rendered":"Draw all missing reactants and\/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds."},"content":{"rendered":"\n

Part A Draw a reasonable mechanism for the following reaction: OH CH3 H2O -CH3 Br Draw all missing reactants and\/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges needed. Electron-flow arrows should start on an atom or a bond and should end on an atom, bond, or location where a new bond should be created. Post-Lecture Assignment 6c mechanism Practice Problem: Nucleophilic Substitution Reactions + ? N CH, H2O) N s 1 :OH 11 CH,<\/p>\n\n\n\n

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

Assumed Reaction Setup<\/strong><\/h3>\n\n\n\n

From the parts you provided:<\/p>\n\n\n\n

    \n
  • One of the starting compounds is an alkyl bromide<\/strong> (\u2013CH\u2083\u2013Br).<\/li>\n\n\n\n
  • Water (H\u2082O) is the solvent\/nucleophile.<\/li>\n\n\n\n
  • The product has an \u2013OH (hydroxyl group) substituted where Br was.<\/li>\n\n\n\n
  • You mention SN1<\/strong>, which is unimolecular nucleophilic substitution.<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Complete Reaction Example<\/strong><\/h3>\n\n\n\n

    Let\u2019s assume the substrate is tert-butyl bromide<\/strong>:
    (CH\u2083)\u2083C\u2013Br<\/strong>
    With water as solvent\/nucleophile:<\/p>\n\n\n\n

    (CH\u2083)\u2083C\u2013Br + H\u2082O \u2192 (CH\u2083)\u2083C\u2013OH + HBr<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Mechanism: SN1 Pathway<\/strong><\/h3>\n\n\n\n

    Step 1: Formation of Carbocation (Rate-Determining Step)<\/strong><\/h4>\n\n\n\n
      \n
    • The C\u2013Br bond breaks heterolytically.<\/li>\n\n\n\n
    • Br\u207b leaves, forming a tertiary carbocation<\/strong>, which is stabilized by hyperconjugation and inductive effects from the three methyl groups.<\/li>\n<\/ul>\n\n\n\n

      (CH3)3C\u2013Br\u2192(CH3)3C++Br\u2212(CH\u2083)\u2083C\u2013Br \u2192 (CH\u2083)\u2083C\u207a + Br\u207b<\/p>\n\n\n\n

      Step 2: Nucleophilic Attack by Water<\/strong><\/h4>\n\n\n\n
        \n
      • Water (a weak nucleophile) attacks the electrophilic carbocation<\/strong>.<\/li>\n<\/ul>\n\n\n\n

        (CH3)3C++H2O\u2192(CH3)3C\u2013OH2+(CH\u2083)\u2083C\u207a + H\u2082O \u2192 (CH\u2083)\u2083C\u2013OH\u2082\u207a<\/p>\n\n\n\n

        Step 3: Deprotonation<\/strong><\/h4>\n\n\n\n
          \n
        • Another water molecule deprotonates the protonated alcohol to form the neutral alcohol<\/strong>.<\/li>\n<\/ul>\n\n\n\n

          (CH3)3C\u2013OH2++H2O\u2192(CH3)3C\u2013OH+H3O+(CH\u2083)\u2083C\u2013OH\u2082\u207a + H\u2082O \u2192 (CH\u2083)\u2083C\u2013OH + H\u2083O\u207a<\/p>\n\n\n\n

          \n\n\n\n

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

          This reaction follows an SN1 mechanism<\/strong>, which stands for “substitution nucleophilic unimolecular.” The reaction proceeds in three distinct steps. First, the leaving group<\/strong>, which is bromide (Br\u207b), dissociates from the carbon atom to form a carbocation<\/strong> intermediate. This is the rate-determining step<\/strong> and occurs slowly, especially if the carbon is not well stabilized. In this case, a tertiary carbocation<\/strong> forms, which is very stable due to hyperconjugation<\/strong> and the inductive effect<\/strong> of adjacent methyl groups. These effects help delocalize the positive charge, making the carbocation more stable and thus favoring the SN1 mechanism.<\/p>\n\n\n\n

          Once the carbocation is formed, the next step is nucleophilic attack<\/strong>. Water, a neutral molecule but with lone pairs on the oxygen, acts as a nucleophile<\/strong> and attacks the carbocation. Since water is a weak nucleophile<\/strong>, the reaction proceeds well only because the carbocation is highly reactive.<\/p>\n\n\n\n

          The product of this attack is a protonated alcohol<\/strong> (an oxonium ion), where the oxygen has three bonds and a positive charge. In the final step, a second molecule of water acts as a base<\/strong> and deprotonates this intermediate, yielding the neutral alcohol<\/strong> product and regenerating a hydronium ion (H\u2083O\u207a).<\/p>\n\n\n\n

          This reaction is favored in polar protic solvents<\/strong> like water that stabilize ions and can facilitate carbocation formation. Since the rate-determining step involves only the substrate, the reaction rate depends only on the concentration of the alkyl halide, not the nucleophile.<\/p>\n\n\n\n

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

          Part A Draw a reasonable mechanism for the following reaction: OH CH3 H2O -CH3 Br Draw all missing reactants and\/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges needed. Electron-flow arrows should start on an atom or a bond and should end on an atom, […]<\/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-220992","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220992","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=220992"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220992\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220992"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220992"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220992"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}