Draw the major organic product of the reaction between CH
3
\u200b
\u2212C\u2261CH and NaNH
2
\u200b
followed by 1\u2212chloro\u22123\u2212methylbutane.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the major organic product of the reaction between 1-butyne (( \\text{CH}_3\\text{C} \\equiv \\text{CH} )) and sodium amide (( \\text{NaNH}_2 )), followed by the reaction with 1-chloro-3-methylbutane, we need to analyze the reaction mechanism and the structure of the products formed.<\/p>\n\n\n\n The first step involves the deprotonation of 1-butyne by sodium amide. 1-butyne contains an acidic hydrogen atom attached to the terminal alkyne carbon (the carbon connected to the methyl group). Sodium amide is a strong base and will effectively remove this acidic hydrogen, generating a terminal alkyne anion:<\/p>\n\n\n\n [ The next step involves the nucleophilic substitution reaction between the resulting alkyne anion (( \\text{CH}_3\\text{C} \\equiv \\text{C}^- )) and 1-chloro-3-methylbutane (( \\text{CH}_3\\text{C}(\\text{CH}_3)\\text{CH}_2\\text{Cl} )). The anion acts as a nucleophile and attacks the electrophilic carbon attached to the chlorine in the alkyl halide, leading to the substitution of the chlorine atom.<\/p>\n\n\n\n The nucleophilic attack results in the formation of a new carbon-carbon bond between the terminal carbon of the alkyne and the carbon attached to the chlorine in 1-chloro-3-methylbutane. The chlorine is displaced as chloride ion (( \\text{Cl}^- )).<\/p>\n\n\n\n The overall reaction yields the product 2-heptyne (( \\text{CH}_3\\text{C} \\equiv \\text{C}\\text{CH}_2\\text{C}(\\text{CH}_3)\\text{CH}_2\\text{CH}_3 )), as the reaction extends the carbon chain by two carbons:<\/p>\n\n\n\n [ The major organic product of this reaction sequence is 3-methyl-1-heptyne. This product results from the nucleophilic substitution of the alkyne anion with 1-chloro-3-methylbutane, illustrating how a terminal alkyne can be utilized to form larger carbon frameworks through nucleophilic substitution reactions.<\/p>\n","protected":false},"excerpt":{"rendered":" Draw the major organic product of the reaction between CH3\u200b\u2212C\u2261CH and NaNH2\u200bfollowed by 1\u2212chloro\u22123\u2212methylbutane. The Correct Answer and Explanation is : To determine the major organic product of the reaction between 1-butyne (( \\text{CH}_3\\text{C} \\equiv \\text{CH} )) and sodium amide (( \\text{NaNH}_2 )), followed by the reaction with 1-chloro-3-methylbutane, we need to analyze the reaction […]<\/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-160732","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/160732","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=160732"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/160732\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=160732"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=160732"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=160732"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Deprotonation<\/h3>\n\n\n\n
\\text{CH}_3\\text{C} \\equiv \\text{CH} + \\text{NaNH}_2 \\rightarrow \\text{CH}_3\\text{C} \\equiv \\text{C}^- \\text{Na}^+ + \\text{NH}_3
]<\/p>\n\n\n\nStep 2: Nucleophilic Substitution<\/h3>\n\n\n\n
Product Formation<\/h3>\n\n\n\n
\\text{CH}_3\\text{C} \\equiv \\text{C}^- + \\text{CH}_3\\text{C}(\\text{CH}_3)\\text{CH}_2\\text{Cl} \\rightarrow \\text{CH}_3\\text{C}(\\text{C}\\text{C}(\\text{CH}_3)\\text{CH}_2\\text{CH}_3) + \\text{Cl}^-
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n