{"id":228121,"date":"2025-06-07T03:30:52","date_gmt":"2025-06-07T03:30:52","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=228121"},"modified":"2025-06-07T03:30:54","modified_gmt":"2025-06-07T03:30:54","slug":"hgoac2-etnh2-2-nabh4-ccl4-br2-chcl3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/07\/hgoac2-etnh2-2-nabh4-ccl4-br2-chcl3\/","title":{"rendered":"Hg(OAc)2, EtNH2 2) NaBH4, CCl4, Br2, CHCl3"},"content":{"rendered":"\n

Hg(OAc)2, EtNH2 2) NaBH4, CCl4, Br2, CHCl3<\/p>\n\n\n\n

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

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

e.<\/strong>
Reactant: 2-methyl-2-butene
Reagents:<\/p>\n\n\n\n

    \n
  1. Hg(OAc)\u2082, EtNH\u2082<\/strong><\/li>\n\n\n\n
  2. NaBH\u2084<\/strong><\/li>\n<\/ol>\n\n\n\n

    This is an example of oxymercuration\u2013reduction<\/strong>, but instead of water, an amine (EtNH\u2082)<\/strong> is used as the nucleophile. This results in amino-mercuration<\/strong>, where the nucleophile adds to the more substituted carbon of the alkene, following Markovnikov’s rule.<\/p>\n\n\n\n

    Product:<\/strong> The ethylamino group (EtNH) adds to the more substituted carbon (tertiary), and hydrogen goes to the less substituted carbon (secondary).
    So the product is:
    2-ethylamino-2-methylbutane<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    h.<\/strong>
    Reactant: 1,2-dimethylbenzene (o-xylene)
    Reagent: Br\u2082 in CCl\u2084<\/strong><\/p>\n\n\n\n

    This is a radical bromination<\/strong> at the benzylic position<\/strong>, not an electrophilic aromatic substitution, because CCl\u2084 is a non-polar solvent favoring radical conditions. Since both benzylic positions (methyl groups) are equivalent, bromination occurs at one of them.<\/p>\n\n\n\n

    Product:<\/strong> Benzyl bromide<\/strong> derivative \u2014 one of the methyl groups becomes a \u2013CH\u2082Br<\/strong><\/p>\n\n\n\n

    So the product is:
    1-bromomethyl-2-methylbenzene<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    l.<\/strong>
    Reactant: 1,2-dihydronaphthalene
    Reagent: Br\u2082 in CHCl\u2083<\/strong><\/p>\n\n\n\n

    In alkenes<\/strong>, Br\u2082 in non-nucleophilic solvents<\/strong> (like CHCl\u2083) leads to anti-addition<\/strong> of bromine across the double bond, forming a vicinal dibromide<\/strong>.<\/p>\n\n\n\n

    Product:<\/strong> Bromines add anti to each other across the former double bond.<\/p>\n\n\n\n

    So the product is:
    trans-1,2-dibromo-1,2-dihydronaphthalene<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Summary of Products:<\/h3>\n\n\n\n
      \n
    • e.<\/strong> 2-ethylamino-2-methylbutane<\/li>\n\n\n\n
    • h.<\/strong> 1-bromomethyl-2-methylbenzene<\/li>\n\n\n\n
    • l.<\/strong> trans-1,2-dibromo-1,2-dihydronaphthalene<\/li>\n<\/ul>\n\n\n\n

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

      Each reaction represents a different classical organic transformation. In part e<\/strong>, the reaction uses oxymercuration\u2013reduction to achieve Markovnikov-selective addition<\/strong> of an amine nucleophile across an alkene. The mercury complex adds to the less substituted carbon while the nucleophile (ethylamine) adds to the more substituted center. Subsequent reduction with NaBH\u2084 removes the mercury, leading to a stable amine-containing product.<\/p>\n\n\n\n

      In part h<\/strong>, benzylic bromination<\/strong> is shown. Under radical conditions (Br\u2082 in CCl\u2084), bromine atoms preferentially abstract hydrogen from benzylic positions due to the resonance stabilization of the resulting radical. This leads to selective monobromination of one methyl group on the aromatic ring, resulting in a benzylic bromide. This type of reaction is valuable for installing good leaving groups on aromatic rings for further substitution or elimination.<\/p>\n\n\n\n

      In part l<\/strong>, halogenation of an alkene<\/strong> is demonstrated. Br\u2082 in CHCl\u2083 adds across the alkene in a stereospecific anti<\/strong> manner via a bromonium ion intermediate. The product is a vicinal dibromide<\/strong>, with the two bromine atoms on opposite faces of the former double bond. This transformation is useful in organic synthesis to introduce halogen atoms that can later undergo substitution or elimination reactions.<\/p>\n\n\n\n

      These reactions showcase the versatility of alkenes and aromatic compounds in organic synthesis, demonstrating how different reagents and conditions direct selective functionalization.<\/p>\n\n\n\n

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

      Hg(OAc)2, EtNH2 2) NaBH4, CCl4, Br2, CHCl3 The Correct Answer and Explanation is: e.Reactant: 2-methyl-2-buteneReagents: This is an example of oxymercuration\u2013reduction, but instead of water, an amine (EtNH\u2082) is used as the nucleophile. This results in amino-mercuration, where the nucleophile adds to the more substituted carbon of the alkene, following Markovnikov’s rule. Product: The ethylamino […]<\/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-228121","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228121","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=228121"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228121\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=228121"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=228121"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=228121"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}