{"id":222520,"date":"2025-05-31T12:22:11","date_gmt":"2025-05-31T12:22:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222520"},"modified":"2025-05-31T12:22:13","modified_gmt":"2025-05-31T12:22:13","slug":"jay-z-draw-the-mop-for-the-following-dci-30c-mop-a-dci-30c-same-chemical-reactivity-since-mop-b-both-are-hydrogens-h-h-protium-d","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/jay-z-draw-the-mop-for-the-following-dci-30c-mop-a-dci-30c-same-chemical-reactivity-since-mop-b-both-are-hydrogens-h-h-protium-d\/","title":{"rendered":"Jay-Z Draw the MOP for the following: DCI -30\u00b0C MOP A DCI -30\u00b0C same chemical reactivity since MOP B both are hydrogens H =H (protium) D ="},"content":{"rendered":"\n

Jay-Z Draw the MOP for the following: DCI -30\u00b0C MOP A DCI -30\u00b0C same chemical reactivity since MOP B both are hydrogens H =
H (protium) D =
H (deuterium) (only difference is in rate) Unlike the protium isotope of hydrogen (symbol H), deuterium is NOT understood and must be shown. Use the X-tool on JME toolbar to input D.
Jay-Z Draw the MOP for the following: DCI -30\u00b0C MOP A DCI -30\u00b0C same chemical reactivity since MOP B both are hydrogens H =
H (protium) D =
H (deuterium) (only difference is in rate) Unlike the protium isotope of hydrogen (symbol H), deuterium is NOT understood and must be shown. Use the X-tool on JME toolbar to input D.<\/p>\n\n\n\n

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

To properly address your question, let’s break it into its chemical context and what is expected.<\/p>\n\n\n\n

\n\n\n\n

Correct Answer: MOP A and MOP B for DCI at -30\u00b0C<\/strong><\/h3>\n\n\n\n

You are being asked to draw the Mechanism of Product formation (MOP)<\/strong> for a molecule called DCI<\/strong>, where:<\/p>\n\n\n\n

    \n
  • D = Deuterium (\u00b2H),<\/strong> an isotope of hydrogen<\/li>\n\n\n\n
  • C = Carbon<\/strong><\/li>\n\n\n\n
  • I = Iodine<\/strong><\/li>\n<\/ul>\n\n\n\n

    This likely refers to Deuterochloro-iodomethane<\/strong> or a similar compound undergoing a reaction at -30\u00b0C<\/strong>. You are asked to draw MOP A<\/strong> and MOP B<\/strong>, where:<\/p>\n\n\n\n

      \n
    • MOP A: DCI reacts with a protium (H) source<\/strong><\/li>\n\n\n\n
    • MOP B: DCI reacts with a deuterium (D) source<\/strong><\/li>\n<\/ul>\n\n\n\n

      Both MOPs have the same chemical reactivity<\/strong>, but the only difference is in rate of reaction<\/strong>, due to the isotopic substitution<\/strong> (H vs. D).<\/p>\n\n\n\n

      This means the reaction likely involves C-H or C-D bond cleavage<\/strong>, and you’re observing the Kinetic Isotope Effect (KIE)<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

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

      The Mechanism of Product (MOP) formation for DCI at \u201330\u00b0C involves a substitution<\/strong> or elimination<\/strong> reaction pathway where the C\u2013H or C\u2013D bond breaks<\/strong>. Deuterium (D) is an isotope of hydrogen with one neutron in addition to the proton, making it heavier than protium (H). Despite having nearly identical chemical properties, reactions involving D\u2013C bonds typically proceed more slowly<\/strong> than those involving H\u2013C bonds due to the Primary Kinetic Isotope Effect<\/strong>.<\/p>\n\n\n\n

      In MOP A<\/strong>, where DCI reacts in the presence of H (protium), the C\u2013D bond remains intact, and the lighter H reacts or is involved in the transition state. In MOP B<\/strong>, where DCI reacts in the presence of D (deuterium), the breaking or forming of a C\u2013D bond occurs, which requires more energy and thus slows the reaction rate.<\/p>\n\n\n\n

      This means both mechanisms are chemically identical<\/strong>, involving the same intermediates and transition states, but MOP A proceeds faster<\/strong> due to the lower bond dissociation energy of the C\u2013H bond compared to the C\u2013D bond.<\/p>\n\n\n\n

      From a mechanistic perspective, this difference doesn\u2019t result in different products, but does affect the reaction kinetics<\/strong>. It’s important to use the \u201cX\u201d tool in the JME Molecular Editor<\/strong> to explicitly indicate deuterium (D), as it\u2019s not assumed like H.<\/p>\n\n\n\n

      This concept is widely used in mechanistic studies<\/strong> to determine which bonds are broken in the rate-determining step of a reaction. Therefore, understanding the role of isotopes like D in MOPs provides crucial insight into reaction pathways<\/strong> and transition state dynamics<\/strong>.<\/p>\n\n\n\n

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

      Jay-Z Draw the MOP for the following: DCI -30\u00b0C MOP A DCI -30\u00b0C same chemical reactivity since MOP B both are hydrogens H =H (protium) D =H (deuterium) (only difference is in rate) Unlike the protium isotope of hydrogen (symbol H), deuterium is NOT understood and must be shown. Use the X-tool on JME toolbar […]<\/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-222520","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222520","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=222520"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222520\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222520"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222520"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222520"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}