{"id":237401,"date":"2025-06-17T08:06:22","date_gmt":"2025-06-17T08:06:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=237401"},"modified":"2025-06-17T08:06:24","modified_gmt":"2025-06-17T08:06:24","slug":"starting-with-propanol-chchchoh-show-how-would-you-synthesize-the-following-substances","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/17\/starting-with-propanol-chchchoh-show-how-would-you-synthesize-the-following-substances\/","title":{"rendered":"Starting with Propanol (CHCHCHOH), show how would you synthesize the following substances"},"content":{"rendered":"\n
    \n
  1. (a) Starting with Propanol (CH
    CH
    CH
    OH), show how would you synthesize the following substances? Use equations only, showing structures of reactants, reagents and all intermediate products. (i) 3-Hexanol {CH
    CH
    CH
    (OH)CHCH
    CH
    } (ii) Propanoic acid (CH
    CH
    COOH) (iii) 2-Pentene (CH
    CH
    CH=CHCH
    ) (iv) Ethyl propyl ether (CH
    CH
    CH
    OCH
    CH
    ) (b) Give the structure the major product(s) formed when 2-Butanol (CH
    CH
    CH
    CH
    OH) reacts with the following reagents: (i) KMnO
    \/H
    O
    (ii) SOCl
    (iii) Conc. H
    SO
    \/heat (iv) PCC\/CH
    Cl<\/li>\n<\/ol>\n\n\n\n
    \"\"<\/figure>\n\n\n\n

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

    (a) Synthesis from Propanol<\/h3>\n\n\n\n

    (i) 3-Hexanol<\/h4>\n\n\n\n
      \n
    1. Convert propanol to propyl bromide using PBr\u2083: CH\u2083CH\u2082CH\u2082OH + PBr\u2083 \u2192 CH\u2083CH\u2082CH\u2082Br<\/li>\n\n\n\n
    2. Perform a Grignard reaction with ethylene oxide to obtain 3-pentanol: CH\u2083CH\u2082CH\u2082MgBr + C\u2082H\u2084O \u2192 CH\u2083CH\u2082CH\u2082CH(OH)CH\u2083<\/li>\n\n\n\n
    3. Extend the carbon chain using another Grignard reagent to yield 3-hexanol: CH\u2083CH\u2082CH\u2082CH(OH)CH\u2083 + CH\u2083CH\u2082MgBr \u2192 CH\u2083CH\u2082CH\u2082CH(OH)CH\u2082CH\u2083<\/li>\n<\/ol>\n\n\n\n

      (ii) Propanoic Acid<\/h4>\n\n\n\n
        \n
      1. Oxidize propanol using a strong oxidizing agent like KMnO\u2084 or CrO\u2083: CH\u2083CH\u2082CH\u2082OH \u2192 CH\u2083CH\u2082COOH<\/li>\n<\/ol>\n\n\n\n

        (iii) 2-Pentene<\/h4>\n\n\n\n
          \n
        1. Convert propanol to propyl bromide using PBr\u2083: CH\u2083CH\u2082CH\u2082OH + PBr\u2083 \u2192 CH\u2083CH\u2082CH\u2082Br<\/li>\n\n\n\n
        2. React with sodium ethoxide to form 2-pentene via elimination: CH\u2083CH\u2082CH\u2082Br + C\u2082H\u2085ONa \u2192 CH\u2083CH\u2082CH=CHCH\u2083<\/li>\n<\/ol>\n\n\n\n

          (iv) Ethyl Propyl Ether<\/h4>\n\n\n\n
            \n
          1. Convert propanol to propyl bromide using PBr\u2083: CH\u2083CH\u2082CH\u2082OH + PBr\u2083 \u2192 CH\u2083CH\u2082CH\u2082Br<\/li>\n\n\n\n
          2. Perform Williamson ether synthesis using sodium ethoxide: CH\u2083CH\u2082CH\u2082Br + C\u2082H\u2085ONa \u2192 CH\u2083CH\u2082CH\u2082OCH\u2082CH\u2083<\/li>\n<\/ol>\n\n\n\n

            (b) Reactions of 2-Butanol<\/h3>\n\n\n\n

            (i) KMnO\u2084 \/ H\u2083O\u207a<\/h4>\n\n\n\n

            Strong oxidation forms butanone and eventually butanoic acid: CH\u2083CH\u2082CH(OH)CH\u2083 \u2192 CH\u2083CH\u2082COCH\u2083 \u2192 CH\u2083CH\u2082COOH<\/p>\n\n\n\n

            (ii) SOCl\u2082<\/h4>\n\n\n\n

            Forms 2-chlorobutane via substitution: CH\u2083CH\u2082CH(OH)CH\u2083 + SOCl\u2082 \u2192 CH\u2083CH\u2082CHClCH\u2083<\/p>\n\n\n\n

            (iii) Concentrated H\u2082SO\u2084 \/ Heat<\/h4>\n\n\n\n

            Under acidic dehydration, 2-butene is formed: CH\u2083CH\u2082CH(OH)CH\u2083 \u2192 CH\u2083CH=CHCH\u2083<\/p>\n\n\n\n

            (iv) PCC \/ CH\u2082Cl\u2082<\/h4>\n\n\n\n

            Mild oxidation leads to butanone: CH\u2083CH\u2082CH(OH)CH\u2083 \u2192 CH\u2083CH\u2082COCH\u2083<\/p>\n\n\n\n

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

            Each transformation is guided by standard organic chemistry principles. The oxidation of propanol yields propanoic acid, whereas conversion to a Grignard reagent allows chain extension to 3-hexanol. Williamson ether synthesis efficiently forms ethyl propyl ether. For 2-butanol, oxidation and dehydration reactions play key roles in obtaining ketones, carboxylic acids, and alkenes. Understanding these fundamental reactions provides insight into broader organic synthesis strategies.<\/p>\n\n\n\n

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

            The Correct Answer and Explanation is: (a) Synthesis from Propanol (i) 3-Hexanol (ii) Propanoic Acid (iii) 2-Pentene (iv) Ethyl Propyl Ether (b) Reactions of 2-Butanol (i) KMnO\u2084 \/ H\u2083O\u207a Strong oxidation forms butanone and eventually butanoic acid: CH\u2083CH\u2082CH(OH)CH\u2083 \u2192 CH\u2083CH\u2082COCH\u2083 \u2192 CH\u2083CH\u2082COOH (ii) SOCl\u2082 Forms 2-chlorobutane via substitution: CH\u2083CH\u2082CH(OH)CH\u2083 + SOCl\u2082 \u2192 CH\u2083CH\u2082CHClCH\u2083 (iii) Concentrated […]<\/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-237401","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237401","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=237401"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/237401\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=237401"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=237401"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=237401"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}