{"id":192021,"date":"2025-02-17T06:01:02","date_gmt":"2025-02-17T06:01:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=192021"},"modified":"2025-02-17T06:01:05","modified_gmt":"2025-02-17T06:01:05","slug":"trans-hex-3-ene-ch2i2-zncu","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/17\/trans-hex-3-ene-ch2i2-zncu\/","title":{"rendered":"trans-hex-3-ene + CH2I2, Zn(Cu)"},"content":{"rendered":"\n

(a) trans-hex-3-ene + CH2I2, Zn(Cu)<\/p>\n\n\n\n

(b) cis-hept-2-ene + CH2I2, Zn(Cu)<\/p>\n\n\n\n

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

The reactions in question involve a key organic chemistry transformation known as a Wolff\u2013L\u00f6ffler\u2013Freytag reaction<\/strong> or halogenation with a zinc-copper catalyst<\/strong>, specifically using CH\u2082I\u2082<\/strong> (diiodomethane) and Zn(Cu)<\/strong> (zinc-copper couple). This reaction is commonly used to generate cyclopropane derivatives through a reaction between alkenes and diiodomethane. The role of Zn(Cu)<\/strong> is to induce the formation of a carbenoid species from CH\u2082I\u2082<\/strong>, which then reacts with the alkene to form a three-membered ring (cyclopropane).<\/p>\n\n\n\n

Reaction (a): trans-hex-3-ene + CH\u2082I\u2082, Zn(Cu)<\/strong><\/h3>\n\n\n\n
    \n
  • trans-hex-3-ene<\/strong> has the structure: CH\u2083-CH\u2082-CH=CH-CH\u2082-CH\u2083<\/strong>, with the double bond between carbon 3 and 4.<\/li>\n\n\n\n
  • Upon reaction with CH\u2082I\u2082<\/strong> in the presence of Zn(Cu)<\/strong>, a carbenoid (CH\u2082) is generated, which reacts with the alkene to form a cyclopropane ring.<\/li>\n\n\n\n
  • Due to the trans<\/strong> geometry of the alkene, the product will be a cis-cyclopropane<\/strong> because of the formation of a three-membered ring. The cis<\/strong> configuration arises because the carbenoid adds to the same side of the double bond, creating a cis-cyclopropane<\/strong> as the intermediate undergoes ring closure.<\/li>\n<\/ul>\n\n\n\n

    Product<\/strong>: cis-1,2-dimethylcyclopropane<\/strong><\/p>\n\n\n\n

    Reaction (b): cis-hept-2-ene + CH\u2082I\u2082, Zn(Cu)<\/strong><\/h3>\n\n\n\n
      \n
    • cis-hept-2-ene<\/strong> has the structure: CH\u2083-CH=CH-CH\u2082-CH\u2082-CH\u2083<\/strong>, with the double bond between carbons 2 and 3.<\/li>\n\n\n\n
    • Upon reaction with CH\u2082I\u2082<\/strong> in the presence of Zn(Cu)<\/strong>, a similar process occurs, generating the carbenoid (CH\u2082) that adds across the double bond.<\/li>\n\n\n\n
    • Due to the cis<\/strong> geometry of the starting alkene, the carbenoid will add to the same side, forming a cis-cyclopropane<\/strong> product, as with reaction (a).<\/li>\n<\/ul>\n\n\n\n

      Product<\/strong>: cis-1,2-dimethylcyclopropane<\/strong> (similar to (a) but with a different initial substrate).<\/p>\n\n\n\n

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

      The main aspect of this reaction lies in the formation of a cyclopropane<\/strong> ring via CH\u2082I\u2082<\/strong> and Zn(Cu)<\/strong>. The reaction proceeds by a radical-like<\/strong> mechanism, where zinc-copper is crucial in producing a carbenoid intermediate<\/strong> (CH\u2082), which is highly reactive and adds across the double bond of the alkene. The cis<\/strong> addition occurs because the carbenoid intermediate is formed on the same side of the double bond due to its sp\u00b2 hybridization, creating the cis-cyclopropane<\/strong>.<\/p>\n\n\n\n

      Let me generate an image of the mechanism for you.<\/p>\n\n\n\n

      Here is an image of the mechanism for the Wolff\u2013L\u00f6ffler\u2013Freytag reaction, showing how cis-hept-2-ene<\/strong> reacts with CH\u2082I\u2082<\/strong> and Zn(Cu)<\/strong> to form a cis-cyclopropane<\/strong> ring. This should help illustrate the process of the carbenoid intermediate<\/strong> forming and adding to the alkene. Let me know if you’d like further details!<\/p>\n\n\n\n

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

      (a) trans-hex-3-ene + CH2I2, Zn(Cu) (b) cis-hept-2-ene + CH2I2, Zn(Cu) The Correct Answer and Explanation is : The reactions in question involve a key organic chemistry transformation known as a Wolff\u2013L\u00f6ffler\u2013Freytag reaction or halogenation with a zinc-copper catalyst, specifically using CH\u2082I\u2082 (diiodomethane) and Zn(Cu) (zinc-copper couple). This reaction is commonly used to generate cyclopropane derivatives […]<\/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-192021","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192021","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=192021"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/192021\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=192021"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=192021"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=192021"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}