{"id":229588,"date":"2025-06-08T09:52:33","date_gmt":"2025-06-08T09:52:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=229588"},"modified":"2025-06-08T09:52:36","modified_gmt":"2025-06-08T09:52:36","slug":"what-is-the-likely-product-of-the-sn2-reaction-shown","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/08\/what-is-the-likely-product-of-the-sn2-reaction-shown\/","title":{"rendered":"What is the likely product of the SN2 reaction shown"},"content":{"rendered":"\n

What is the likely product of the SN2 reaction shown? OH HCI OH IV 0 0 II I 0 IV None of these OHz<\/p>\n\n\n\n

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

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

The question asks for the likely product of the SN2 reaction<\/strong> shown, where butan-2-ol<\/strong> reacts with HCl<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Step-by-step Reaction Mechanism<\/strong><\/h3>\n\n\n\n

The reaction given is:<\/p>\n\n\n\n

    \n
  • Starting material: CH3CH(OH)CH2CH3<\/strong> (butan-2-ol)<\/li>\n\n\n\n
  • Reagent: HCl<\/strong><\/li>\n<\/ul>\n\n\n\n

    This is not a typical SN2 reaction with a strong nucleophile and a good leaving group. However, here’s how it proceeds mechanistically:<\/p>\n\n\n\n

      \n
    1. Protonation of the alcohol<\/strong>: CH3CH(OH)CH2CH3+HCl\u2192CH3CH(OH2+)CH2CH3+Cl\u2212CH3CH(OH)CH2CH3 + HCl \\rightarrow CH3CH(OH_2^+)CH2CH3 + Cl^-CH3CH(OH)CH2CH3+HCl\u2192CH3CH(OH2+\u200b)CH2CH3+Cl\u2212 The hydroxyl group is protonated to form a better leaving group \u2014 water.<\/li>\n\n\n\n
    2. Loss of water (leaving group leaves)<\/strong>: CH3CH+CH2CH3+H2OCH3CH^+CH2CH3 + H2OCH3CH+CH2CH3+H2O This forms a secondary carbocation<\/strong>. However, since SN2 mechanisms do not involve carbocation intermediates<\/strong>, this step is more typical of SN1<\/strong>, not SN2.<\/li>\n\n\n\n
    3. Nucleophilic attack<\/strong>:
      Chloride ion (Cl\u207b) attacks the carbocation.<\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      SN2 vs. SN1 Consideration<\/strong><\/h3>\n\n\n\n
        \n
      • SN2 reactions<\/strong> are one-step<\/strong> and involve backside attack<\/strong>.<\/li>\n\n\n\n
      • SN2 is favored on primary carbons<\/strong>, not secondary<\/strong> or tertiary<\/strong>.<\/li>\n\n\n\n
      • Butan-2-ol is a secondary alcohol<\/strong>, and the intermediate is a secondary carbocation<\/strong>.<\/li>\n\n\n\n
      • Hence, this reaction is more SN1-like<\/strong>, despite the question mentioning SN2.<\/li>\n<\/ul>\n\n\n\n

        However, assuming we treat this as an SN2<\/strong> (as asked), and HCl is present, the best scenario is the chloride displacing the OH group<\/strong>, forming: CH3CH(Cl)CH2CH3(2-chlorobutane)CH3CH(Cl)CH2CH3 \\quad \\text{(2-chlorobutane)}CH3CH(Cl)CH2CH3(2-chlorobutane)<\/p>\n\n\n\n

        \n\n\n\n

        Correct Product Identification<\/strong><\/h3>\n\n\n\n

        Let’s now analyze the options:<\/p>\n\n\n\n

          \n
        • I<\/strong> = 1-chlorobutane \u2192 Wrong position.<\/li>\n\n\n\n
        • II<\/strong> = 2-butanol with protonated water (intermediate) \u2192 Not a final product.<\/li>\n\n\n\n
        • III<\/strong> = 2-chlorobutane<\/strong> \u2192 Correct<\/strong>.<\/li>\n\n\n\n
        • IV<\/strong> = 1-butanol \u2192 Wrong alcohol.<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          \u2705 Correct Answer: III<\/strong><\/h3>\n\n\n\n
          \n\n\n\n

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

          The reaction shown involves a secondary alcohol (butan-2-ol) reacting with hydrochloric acid (HCl). While the question labels this as an SN2 reaction, the actual reaction conditions and intermediates suggest a mechanism closer to SN1. In this case, HCl serves two purposes: it protonates the hydroxyl (-OH) group of the alcohol, making it a good leaving group (water), and it supplies a nucleophile (Cl\u207b).<\/p>\n\n\n\n

          The first step is the protonation of the alcohol, converting the -OH into -OH\u2082\u207a, a much better leaving group. This is followed by the departure of water, forming a secondary carbocation (on the second carbon of butane). Normally, SN2 reactions do not form carbocations; they proceed in one concerted step. However, since this is a secondary carbon, it is not ideally suited for SN2 either. Still, if we assume a concerted substitution as the question suggests, the Cl\u207b can directly displace the leaving group, forming the substitution product.<\/p>\n\n\n\n

          The most likely substitution product is 2-chlorobutane<\/strong>, where the hydroxyl group is replaced by a chloride at the second carbon. This corresponds to structure III<\/strong> among the answer choices. The other options represent incorrect positions for substitution (like option I, which is 1-chlorobutane), intermediates (option II), or unchanged alcohols (option IV).<\/p>\n\n\n\n

          Thus, despite the mechanistic ambiguity, option III<\/strong> correctly represents the substitution product expected under acidic conditions with HCl and a secondary alcohol.<\/p>\n\n\n\n

          \n\n\n\n

          Correct answer: III<\/strong>.<\/p>\n\n\n\n

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

          What is the likely product of the SN2 reaction shown? OH HCI OH IV 0 0 II I 0 IV None of these OHz The Correct Answer and Explanation is: The question asks for the likely product of the SN2 reaction shown, where butan-2-ol reacts with HCl. Step-by-step Reaction Mechanism The reaction given is: This […]<\/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-229588","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229588","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=229588"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229588\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229588"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229588"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229588"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}