{"id":194574,"date":"2025-02-24T09:25:37","date_gmt":"2025-02-24T09:25:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=194574"},"modified":"2025-02-24T09:25:51","modified_gmt":"2025-02-24T09:25:51","slug":"draw-the-addition-products-formed-when-one-equivalent-of-hbr-reacts-with-24-hexadiene-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/24\/draw-the-addition-products-formed-when-one-equivalent-of-hbr-reacts-with-24-hexadiene-2\/","title":{"rendered":"Draw the addition products formed when one equivalent of HBr reacts with 2,4-hexadiene"},"content":{"rendered":"\n

Draw the addition products formed when one equivalent of HBr reacts with 2,4-hexadiene. (Draw a single product for each.)<\/p>\n\n\n\n

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

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

when one equivalent of HBr<\/strong> reacts with 2,4-hexadiene<\/strong>. I’ll generate the image of the correct products and provide an explanation.<\/p>\n\n\n\n

Reaction of 2,4-Hexadiene with HBr<\/strong><\/h3>\n\n\n\n

2,4-Hexadiene is a conjugated diene<\/strong>, meaning it has alternating double bonds<\/strong> (C=C) at positions 2 and 4<\/strong> in a six-carbon chain. The reaction of one equivalent of HBr<\/strong> with a conjugated diene can proceed through two major pathways:<\/p>\n\n\n\n

    \n
  1. 1,2-Addition (Direct Addition)<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • HBr adds across the C2=C3 double bond<\/strong>, leading to a bromoalkene<\/strong> at position 3<\/strong>.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. 1,4-Addition (Conjugate Addition)<\/strong><\/li>\n<\/ol>\n\n\n\n
          \n
        • HBr adds across the conjugated system, leading to a bromoalkene<\/strong> at position 4<\/strong>.<\/li>\n<\/ul>\n\n\n\n

          Markovnikov\u2019s Rule<\/strong><\/h3>\n\n\n\n
            \n
          • The H+<\/strong> (proton) prefers to add to the carbon already bonded to more hydrogens<\/strong>, while the Br\u2212<\/strong> (bromide) attaches to the more substituted carbon<\/strong> in both cases.<\/li>\n<\/ul>\n\n\n\n

            Products<\/strong><\/h3>\n\n\n\n
              \n
            • 1,2-Addition Product<\/strong>: 3-Bromo-2-hexene<\/strong><\/li>\n\n\n\n
            • 1,4-Addition Product<\/strong>: 4-Bromo-2-hexene<\/strong><\/li>\n<\/ul>\n\n\n\n

              These two regioisomers form based on kinetic (1,2) vs. thermodynamic (1,4) control.<\/p>\n\n\n\n

              Now, I’ll generate the image of the correct products.<\/p>\n\n\n\n

              Here is the generated reaction diagram showing the addition of one equivalent of HBr<\/strong> to 2,4-hexadiene<\/strong>, forming the 1,2-addition product (3-Bromo-2-hexene)<\/strong> and the 1,4-addition product (4-Bromo-2-hexene)<\/strong>.<\/p>\n\n\n\n

              Explanation of the Reaction<\/strong><\/h3>\n\n\n\n
                \n
              1. 1,2-Addition (Kinetic Product)<\/strong><\/li>\n<\/ol>\n\n\n\n
                  \n
                • The proton (H\u207a)<\/strong> adds to C2<\/strong> (which has more hydrogens).<\/li>\n\n\n\n
                • The bromine (Br\u207b)<\/strong> attaches to C3<\/strong>, forming 3-Bromo-2-hexene<\/strong>.<\/li>\n\n\n\n
                • This product forms faster under low temperature (kinetic control).<\/strong><\/li>\n<\/ul>\n\n\n\n
                    \n
                  1. 1,4-Addition (Thermodynamic Product)<\/strong><\/li>\n<\/ol>\n\n\n\n
                      \n
                    • The proton adds to C2<\/strong>, leading to resonance stabilization.<\/li>\n\n\n\n
                    • The Br\u207b<\/strong> ultimately attaches at C4<\/strong>, forming 4-Bromo-2-hexene<\/strong>.<\/li>\n\n\n\n
                    • This product is more stable<\/strong> and dominates at higher temperatures (thermodynamic control).<\/strong><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

                      Draw the addition products formed when one equivalent of HBr reacts with 2,4-hexadiene. (Draw a single product for each.) The Correct Answer and Explanation is : when one equivalent of HBr reacts with 2,4-hexadiene. I’ll generate the image of the correct products and provide an explanation. Reaction of 2,4-Hexadiene with HBr 2,4-Hexadiene is a conjugated […]<\/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-194574","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194574","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=194574"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194574\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194574"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194574"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194574"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}