{"id":183257,"date":"2025-01-16T06:40:46","date_gmt":"2025-01-16T06:40:46","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183257"},"modified":"2025-01-16T06:40:48","modified_gmt":"2025-01-16T06:40:48","slug":"write-hond-line-structural-formalas-for-a-two-primary-alcohols","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/write-hond-line-structural-formalas-for-a-two-primary-alcohols\/","title":{"rendered":"Write hond-line structural formalas for (a) two primary alcohols"},"content":{"rendered":"\n

Write hond-line structural formalas for (a) two primary alcohols, (b) a secondary alcohol, and (c) a tertiary alcohol-all having the molecular formula CHO One way of naming alcohols is to name the alkyl group that is attached to the OH and add the word alabel Write bond-line formulas for (a) propyl alcohol and (b) isopropyl alcohol.<\/p>\n\n\n\n

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

Let’s break this down:<\/p>\n\n\n\n

(a) Two primary alcohols<\/h3>\n\n\n\n

A primary alcohol<\/strong> has the general structure where the -OH group is attached to a carbon atom that is only bonded to one other carbon (i.e., it’s a terminal carbon). For alcohols with the molecular formula C\u2083H\u2086O<\/strong> (CHO), the possible primary alcohols are:<\/p>\n\n\n\n

    \n
  1. 1-Propanol<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • Bond-line formula<\/strong>:<\/li>\n<\/ul>\n\n\n\n
         CH\u2083-CH\u2082-CH\u2082OH<\/code><\/pre>\n\n\n\n
      Here, the -OH group is attached to the terminal carbon (the last carbon), making it a primary alcohol.<\/p>\n\n\n\n
        \n
      1. 2-Propanol<\/strong> (though secondary, for completeness):<\/li>\n<\/ol>\n\n\n\n
          \n
        • Bond-line formula<\/strong>:<\/li>\n<\/ul>\n\n\n\n
             CH\u2083-CH(OH)-CH\u2083<\/code><\/pre>\n\n\n\n
          (b) A secondary alcohol<\/h3>\n\n\n\n
          A secondary alcohol<\/strong> has the -OH group attached to a carbon that is bonded to two other carbons. For a molecular formula of C\u2083H\u2086O, the only secondary alcohol is:<\/p>\n\n\n\n
            \n
          • 2-Propanol<\/strong> (Isopropanol):<\/li>\n\n\n\n
          • Bond-line formula<\/strong>:<\/li>\n<\/ul>\n\n\n\n
              CH\u2083-CH(OH)-CH\u2083<\/code><\/pre>\n\n\n\n
            The hydroxyl group is attached to the second carbon, which is bonded to two other carbons, making it a secondary alcohol.<\/p>\n\n\n\n
            (c) A tertiary alcohol<\/h3>\n\n\n\n
            A tertiary alcohol<\/strong> has the -OH group attached to a carbon that is bonded to three other carbons. However, with the molecular formula CHO<\/strong>, a tertiary alcohol is not possible because there aren’t enough carbons to attach to three other groups while also accommodating the -OH group.<\/p>\n\n\n\n
            (a) Propyl alcohol and (b) Isopropyl alcohol<\/h3>\n\n\n\n
            Now let’s look at the bond-line formulas:<\/p>\n\n\n\n
              \n
            • Propyl Alcohol (1-Propanol)<\/strong>:<\/li>\n\n\n\n
            • Bond-line formula<\/strong>:<\/li>\n<\/ul>\n\n\n\n
                CH\u2083-CH\u2082-CH\u2082OH<\/code><\/pre>\n\n\n\n
                \n
              • Isopropyl Alcohol (2-Propanol)<\/strong>:<\/li>\n\n\n\n
              • Bond-line formula<\/strong>:<\/li>\n<\/ul>\n\n\n\n
                  CH\u2083-CH(OH)-CH\u2083<\/code><\/pre>\n\n\n\n
                Explanation:<\/h3>\n\n\n\n
                  \n
                • Primary alcohols<\/strong>: These have the hydroxyl (-OH) group attached to a carbon that is connected to only one other carbon. In 1-propanol, the terminal carbon carries the -OH group, thus making it a primary alcohol.<\/li>\n\n\n\n
                • Secondary alcohols<\/strong>: The -OH group is attached to a carbon connected to two other carbons. In 2-propanol (isopropyl alcohol), the central carbon carries the -OH group and is attached to two other carbon atoms.<\/li>\n\n\n\n
                • Tertiary alcohols<\/strong>: The -OH group would be attached to a carbon that is connected to three other carbons. However, with a molecular formula of C\u2083H\u2086O, it isn’t possible to form a tertiary alcohol because of the limited number of carbon atoms.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"
                  Write hond-line structural formalas for (a) two primary alcohols, (b) a secondary alcohol, and (c) a tertiary alcohol-all having the molecular formula CHO One way of naming alcohols is to name the alkyl group that is attached to the OH and add the word alabel Write bond-line formulas for (a) propyl alcohol and (b) isopropyl […]<\/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-183257","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183257","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=183257"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183257\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183257"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183257"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183257"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}