When cyclohexene is dissolved in an appropriate solvent and allowed to react with chlorine, the product of the reaction, C6H10Cl2, has a melting point of \u2212 7 \u00b0 C and a boiling point (at 16 torr) of 74 \u00b0 C .
(a) Which stereoisomer is this?
(b) What does this experiment suggest about the mode of addition of chlorine to the double bond?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The product of the reaction, C6H10Cl2<\/strong>, formed when cyclohexene reacts with chlorine, is trans-1,2-dichlorocyclohexane<\/strong>.<\/p>\n\n\n\n The reaction of cyclohexene with chlorine is a classical example of an electrophilic addition<\/strong> reaction. Chlorine adds to the double bond of cyclohexene to form a dihalide<\/strong>. The fact that the product is trans-1,2-dichlorocyclohexane<\/strong> suggests that chlorine adds to the double bond through an anti addition<\/strong> mechanism.<\/p>\n\n\n\n In the first step of the reaction, the double bond in cyclohexene acts as a nucleophile and attacks a chlorine molecule (Cl\u2082). This leads to the formation of a chloronium ion<\/strong>, a three-membered cyclic structure where one of the chlorine atoms is attached to both carbon atoms from the original double bond.<\/p>\n\n\n\n In the second step, the negatively charged chloride ion (Cl\u207b) formed when chlorine was initially split apart acts as a nucleophile. However, due to steric hindrance and the positive charge on the chloronium ion, the chloride ion can only attack from the opposite side<\/strong> of where the initial chlorine is bound. This results in the anti addition<\/strong> of chlorine atoms across the double bond, leading to the formation of the trans-1,2-dichlorocyclohexane<\/strong> stereoisomer.<\/p>\n\n\n\n The experimental melting and boiling points provided (melting point of \u22127\u00b0C and boiling point of 74\u00b0C at 16 torr) are consistent with the trans<\/strong> isomer, which is more stable and typically has different physical properties compared to its cis counterpart.<\/p>\n\n\n\n In summary, the formation of the trans isomer indicates that chlorine adds to the cyclohexene via an anti addition mechanism, which is common in halogenation reactions of alkenes. The trans configuration minimizes steric strain and leads to a more stable product.<\/p>\n","protected":false},"excerpt":{"rendered":" When cyclohexene is dissolved in an appropriate solvent and allowed to react with chlorine, the product of the reaction, C6H10Cl2, has a melting point of \u2212 7 \u00b0 C and a boiling point (at 16 torr) of 74 \u00b0 C .(a) Which stereoisomer is this?(b) What does this experiment suggest about the mode of addition […]<\/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-147751","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/147751","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=147751"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/147751\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=147751"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=147751"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=147751"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(a) Which stereoisomer is this?<\/h3>\n\n\n\n
(b) What does this experiment suggest about the mode of addition of chlorine to the double bond?<\/h3>\n\n\n\n
Explanation:<\/h4>\n\n\n\n