{"id":187706,"date":"2025-02-06T04:49:00","date_gmt":"2025-02-06T04:49:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187706"},"modified":"2025-02-06T04:49:03","modified_gmt":"2025-02-06T04:49:03","slug":"bromobenzene-ml-density-of-bromobenzene-g-ml","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/06\/bromobenzene-ml-density-of-bromobenzene-g-ml\/","title":{"rendered":"Bromobenzene (ml) density of bromobenzene (g\/mL)"},"content":{"rendered":"\n

Bromobenzene (ml) density of bromobenzene (g\/mL) Conc HNO3 (mL) Conc H2SO4 (mL) filter paper (g) filter paper + 4-NO2-Br-Benzene actual yield of 4-NO2-Br-Benzene (g) actual yield of 2-NO2-Br-Benzene (ml) 2-NO2BrBenz-density (g\/mL) Theoretical yield (both isomers combined) Percent yield (both isomers combined) Ratio of ortho product Ratio of para product 4.5 1.5 4 4 0.098 3.933 3.832 1.4 1.719 ? ? ? ?<\/p>\n\n\n\n

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

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

To determine the theoretical yield, percent yield, and product ratio of the nitration reaction of bromobenzene, let\u2019s go through the calculations step by step.<\/p>\n\n\n\n

Step 1: Determine the Moles of Bromobenzene<\/strong><\/h3>\n\n\n\n
    \n
  • Given volume of bromobenzene = 4.5 mL<\/strong><\/li>\n\n\n\n
  • Given density of bromobenzene = 1.5 g\/mL<\/strong><\/li>\n\n\n\n
  • Molar mass of bromobenzene (C\u2086H\u2085Br) = 157.01 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n

    [
    \\text{Mass of bromobenzene} = \\text{Volume} \\times \\text{Density} = 4.5 \\times 1.5 = 6.75 \\text{ g}
    ]<\/p>\n\n\n\n

    [
    \\text{Moles of bromobenzene} = \\frac{6.75}{157.01} = 0.0430 \\text{ moles}
    ]<\/p>\n\n\n\n

    Step 2: Determine the Theoretical Yield<\/strong><\/h3>\n\n\n\n
      \n
    • The nitration reaction follows electrophilic aromatic substitution, forming ortho-<\/strong> and para-<\/strong> nitrobromobenzene.<\/li>\n\n\n\n
    • The molecular weight of both 2-Nitrobromobenzene<\/strong> and 4-Nitrobromobenzene<\/strong> = 202.96 g\/mol<\/strong><\/li>\n\n\n\n
    • Assuming complete reaction and 100% conversion:<\/li>\n<\/ul>\n\n\n\n

      [
      \\text{Theoretical mass of products} = 0.0430 \\times 202.96 = 8.73 \\text{ g}
      ]<\/p>\n\n\n\n

      Step 3: Determine the Actual Yield<\/strong><\/h3>\n\n\n\n
        \n
      • Mass of filter paper = 0.098 g<\/strong><\/li>\n\n\n\n
      • Mass of filter paper + 4-Nitrobromobenzene<\/strong> = 3.933 g<\/strong><\/li>\n\n\n\n
      • Actual mass of 4-Nitrobromobenzene<\/strong> = (3.933 – 0.098 = 3.835) g<\/li>\n\n\n\n
      • Given 2-Nitrobromobenzene<\/strong> = 1.4 mL<\/strong><\/li>\n\n\n\n
      • Given density of 2-Nitrobromobenzene<\/strong> = 1.719 g\/mL<\/strong><\/li>\n<\/ul>\n\n\n\n

        [
        \\text{Mass of 2-Nitrobromobenzene} = 1.4 \\times 1.719 = 2.407 \\text{ g}
        ]<\/p>\n\n\n\n

        [
        \\text{Total actual yield} = 3.835 + 2.407 = 6.242 \\text{ g}
        ]<\/p>\n\n\n\n

        Step 4: Determine the Percent Yield<\/strong><\/h3>\n\n\n\n

        [
        \\text{Percent yield} = \\left( \\frac{\\text{actual yield}}{\\text{theoretical yield}} \\right) \\times 100
        ]<\/p>\n\n\n\n

        [
        \\text{Percent yield} = \\left( \\frac{6.242}{8.73} \\right) \\times 100 = 71.5\\%
        ]<\/p>\n\n\n\n

        Step 5: Determine the Product Ratio<\/strong><\/h3>\n\n\n\n
          \n
        • Ortho (2-Nitrobromobenzene) product<\/strong> = 2.407 g<\/strong><\/li>\n\n\n\n
        • Para (4-Nitrobromobenzene) product<\/strong> = 3.835 g<\/strong><\/li>\n\n\n\n
        • Total product mass = 6.242 g<\/strong><\/li>\n<\/ul>\n\n\n\n

          [
          \\text{Ortho ratio} = \\left( \\frac{2.407}{6.242} \\right) \\times 100 = 38.56\\%
          ]<\/p>\n\n\n\n

          [
          \\text{Para ratio} = \\left( \\frac{3.835}{6.242} \\right) \\times 100 = 61.44\\%
          ]<\/p>\n\n\n\n

          Final Answer Summary:<\/strong><\/h3>\n\n\n\n
            \n
          • Theoretical yield<\/strong> = 8.73 g<\/strong><\/li>\n\n\n\n
          • Actual yield<\/strong> = 6.242 g<\/strong><\/li>\n\n\n\n
          • Percent yield<\/strong> = 71.5%<\/strong><\/li>\n\n\n\n
          • Ortho product ratio<\/strong> = 38.56%<\/strong><\/li>\n\n\n\n
          • Para product ratio<\/strong> = 61.44%<\/strong><\/li>\n<\/ul>\n\n\n\n

            Explanation (300 Words)<\/strong><\/h3>\n\n\n\n

            Nitration of bromobenzene follows an electrophilic aromatic substitution mechanism, in which the nitronium ion ((\\text{NO}_2^+)) attacks the benzene ring. Since bromine is an electron-donating group via resonance, it directs substitution to the ortho (2-position)<\/strong> and para (4-position)<\/strong> sites.<\/p>\n\n\n\n

            The experiment starts with 4.5 mL of bromobenzene, which corresponds to 0.0430 moles<\/strong>. Upon reaction with concentrated HNO\u2083 and H\u2082SO\u2084<\/strong>, bromobenzene forms two major products: 2-Nitrobromobenzene<\/strong> and 4-Nitrobromobenzene<\/strong>, with a combined theoretical yield of 8.73 g<\/strong>.<\/p>\n\n\n\n

            The experimental actual yield was 6.242 g<\/strong>, leading to a percent yield of 71.5%<\/strong>. This means some product loss occurred, possibly due to side reactions, incomplete nitration, or loss during purification.<\/p>\n\n\n\n

            The product distribution is not equal. Due to steric hindrance near the bromine atom, the para isomer<\/strong> is favored over the ortho isomer<\/strong>, leading to a 38.56% ortho-to-61.44% para ratio<\/strong>. This ratio aligns with expected regioselectivity trends for bromine-substituted benzene derivatives.<\/p>\n\n\n\n

            The results confirm that electrophilic nitration follows predictable regioselectivity principles, favoring para substitution due to steric effects. However, the significant formation of the ortho product suggests that reaction conditions allow both positions to be accessed efficiently.<\/p>\n","protected":false},"excerpt":{"rendered":"

            Bromobenzene (ml) density of bromobenzene (g\/mL) Conc HNO3 (mL) Conc H2SO4 (mL) filter paper (g) filter paper + 4-NO2-Br-Benzene actual yield of 4-NO2-Br-Benzene (g) actual yield of 2-NO2-Br-Benzene (ml) 2-NO2BrBenz-density (g\/mL) Theoretical yield (both isomers combined) Percent yield (both isomers combined) Ratio of ortho product Ratio of para product 4.5 1.5 4 4 0.098 3.933 […]<\/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-187706","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187706","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=187706"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187706\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187706"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187706"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187706"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}