{"id":183608,"date":"2025-01-16T17:45:04","date_gmt":"2025-01-16T17:45:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183608"},"modified":"2025-01-16T17:45:06","modified_gmt":"2025-01-16T17:45:06","slug":"hydrolysis-of-2-bromo-2-methylpropane-tert-butyl-bromide-yields-2-methylpropan-2-ol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/hydrolysis-of-2-bromo-2-methylpropane-tert-butyl-bromide-yields-2-methylpropan-2-ol\/","title":{"rendered":"Hydrolysis of 2-bromo-2-methylpropane (tert-butyl bromide) yields 2-methylpropan-2-ol"},"content":{"rendered":"\n

Hydrolysis of 2-bromo-2-methylpropane (tert-butyl bromide) yields 2-methylpropan-2-ol. (CH), CBr + 2H\u2082O \u2192 (\u0441\u043d.), \u0441\u043e\u043d + \u043d\u043e + Br Give the SN1 mechanism. Draw structures including electrons and charges and add curved arrows. Details count.<\/p>\n\n\n\n

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

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

The hydrolysis of 2-bromo-2-methylpropane<\/strong> (tert-butyl bromide) follows an SN1<\/strong> mechanism. In this mechanism, the reaction proceeds via a two-step process: first, the leaving group departs to form a carbocation, and then the nucleophile (water in this case) attacks the carbocation to form the final product.<\/p>\n\n\n\n

Here\u2019s a step-by-step breakdown of the SN1 mechanism<\/strong> for the hydrolysis of tert-butyl bromide:<\/p>\n\n\n\n

Step 1: Formation of the carbocation<\/strong><\/h3>\n\n\n\n
    \n
  1. 2-bromo-2-methylpropane (tert-butyl bromide)<\/strong> contains a tertiary carbon attached to the bromine atom. The bromine is a good leaving group, so it can easily dissociate from the carbon atom.<\/li>\n\n\n\n
  2. As bromine leaves, a tertiary carbocation<\/strong> is formed. This carbocation is highly stable due to the inductive and hyperconjugation effects of the three methyl groups attached to the central carbon.<\/li>\n<\/ol>\n\n\n\n
      \n
    • The leaving group is Br\u207b<\/strong>.<\/li>\n\n\n\n
    • The intermediate formed is a tert-butyl cation (C\u2084H\u2089\u207a)<\/strong>. This step is the rate-determining step in the SN1 reaction.<\/li>\n<\/ul>\n\n\n\n

      Step 2: Nucleophilic attack by water<\/strong><\/h3>\n\n\n\n
        \n
      1. After the carbocation is formed, the water molecule, acting as a nucleophile, attacks the carbocation. Since water is polar, it readily donates a lone pair of electrons to the electrophilic carbocation.<\/li>\n\n\n\n
      2. The oxygen atom of the water molecule bonds to the carbocation, forming 2-methylpropan-2-ol<\/strong> (tert-butyl alcohol).<\/li>\n\n\n\n
      3. Finally, water acts as a base and deprotonates the oxonium ion formed, yielding the neutral alcohol.<\/li>\n<\/ol>\n\n\n\n
          \n
        • The final product is 2-methylpropan-2-ol<\/strong> (tert-butyl alcohol).<\/li>\n<\/ul>\n\n\n\n

          Overall Reaction:<\/h3>\n\n\n\n

          [
          \\text{C\u2084H\u2089Br} + \\text{2H\u2082O} \\rightarrow \\text{C\u2084H\u2089OH} + \\text{HBr}
          ]<\/p>\n\n\n\n

          Curved Arrows:<\/h3>\n\n\n\n
            \n
          • One curved arrow shows the departure of Br\u207b<\/strong> from the carbon atom, leading to the formation of the carbocation.<\/li>\n\n\n\n
          • Another curved arrow shows the nucleophilic attack by water<\/strong> (H\u2082O), with the lone pair on oxygen attacking the carbocation.<\/li>\n<\/ul>\n\n\n\n

            In conclusion, the SN1<\/strong> mechanism is characterized by the formation of a stable carbocation intermediate. The reaction rate depends on the stability of the carbocation, which is why a tertiary carbocation like in tert-butyl bromide favors this mechanism. The reaction is also sensitive to the nature of the solvent, with polar protic solvents like water facilitating the carbocation formation and nucleophilic attack.<\/p>\n\n\n\n

            I cannot display images directly here, but you can refer to the provided link for a detailed visual representation of the mechanism, where curved arrows and structures are shown clearly.<\/p>\n\n\n\n

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

            Hydrolysis of 2-bromo-2-methylpropane (tert-butyl bromide) yields 2-methylpropan-2-ol. (CH), CBr + 2H\u2082O \u2192 (\u0441\u043d.), \u0441\u043e\u043d + \u043d\u043e + Br Give the SN1 mechanism. Draw structures including electrons and charges and add curved arrows. Details count. The Correct Answer and Explanation is : The hydrolysis of 2-bromo-2-methylpropane (tert-butyl bromide) follows an SN1 mechanism. In this mechanism, the […]<\/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-183608","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183608","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=183608"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183608\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183608"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183608"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183608"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}