{"id":196487,"date":"2025-03-05T04:55:38","date_gmt":"2025-03-05T04:55:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196487"},"modified":"2025-03-05T04:55:41","modified_gmt":"2025-03-05T04:55:41","slug":"how-would-you-use-the-reaction-of-an-amide-with-lialh4-as-the-key-step-is-going-from-bromo-cyclohexane-to-n-n-dimethylaminomethyl-cyclohexane","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/05\/how-would-you-use-the-reaction-of-an-amide-with-lialh4-as-the-key-step-is-going-from-bromo-cyclohexane-to-n-n-dimethylaminomethyl-cyclohexane\/","title":{"rendered":"How would you use the reaction of an amide with LiAlH4 as the key step is going from bromo cyclohexane to (N, N-dimethylaminomethyl) cyclohexane"},"content":{"rendered":"\n

How would you use the reaction of an amide with LiAlH4 as the key step is going from bromo cyclohexane to (N, N-dimethylaminomethyl) cyclohexane? Write all the steps in the reactionsequence.<\/p>\n\n\n\n

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

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

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

To synthesize (N,N-dimethylaminomethyl)cyclohexane from bromocyclohexane using the reduction of an amide with lithium aluminum hydride (LiAlH\u2084) as the key step, follow this sequence:<\/p>\n\n\n\n

    \n
  1. Formation of the Grignard Reagent:<\/strong>
    React bromocyclohexane with magnesium metal in dry ether to form cyclohexylmagnesium bromide, a Grignard reagent. [ \\text{C}6\\text{H}<\/em>{11}\\text{Br} + \\text{Mg} \\xrightarrow{\\text{ether}} \\text{C}6\\text{H}<\/em>{11}\\text{MgBr} ]<\/li>\n\n\n\n
  2. Carboxylation to Benzoic Acid:<\/strong>
    Add carbon dioxide (CO\u2082) to the Grignard reagent, followed by acidic work-up, to yield benzoic acid (C\u2086H\u2085COOH). [ \\text{C}6\\text{H}<\/em>{11}\\text{MgBr} + \\text{CO}_2 \\xrightarrow{\\text{acid work-up}} \\text{C}_6\\text{H}_5\\text{COOH} ]<\/li>\n\n\n\n
  3. Conversion to Benzoyl Chloride:<\/strong>
    React benzoic acid with thionyl chloride (SOCl\u2082) to form benzoyl chloride (C\u2086H\u2085COCl). [ \\text{C}_6\\text{H}_5\\text{COOH} + \\text{SOCl}_2 \\rightarrow \\text{C}_6\\text{H}_5\\text{COCl} + \\text{SO}_2 + \\text{HCl} ]<\/li>\n\n\n\n
  4. Synthesis of N,N-Dimethylbenzamide:<\/strong>
    React benzoyl chloride with dimethylamine (CH\u2083NH\u2082) to produce N,N-dimethylbenzamide. [ \\text{C}_6\\text{H}_5\\text{COCl} + \\text{CH}_3\\text{NH}_2 \\rightarrow \\text{C}_6\\text{H}_5\\text{CON(CH}_3\\text{)}_2 + \\text{HCl} ]<\/li>\n\n\n\n
  5. Reduction to (N,N-Dimethylaminomethyl)cyclohexane:<\/strong>
    Reduce N,N-dimethylbenzamide with LiAlH\u2084 in dry ether to obtain (N,N-dimethylaminomethyl)cyclohexane. [ \\text{C}6\\text{H}_5\\text{CON(CH}_3\\text{)}_2 + \\text{LiAlH}_4 \\xrightarrow{\\text{ether}} \\text{C}_6\\text{H}<\/em>{11}\\text{N(CH}_3\\text{)}_2 + \\text{LiAlH}_3\\text{O} ]<\/li>\n<\/ol>\n\n\n\n

    Explanation:<\/strong><\/p>\n\n\n\n

      \n
    • Grignard Reagent Formation:<\/strong> Bromocyclohexane reacts with magnesium in dry ether to form a Grignard reagent, which is a strong nucleophile.\ue206<\/li>\n\n\n\n
    • Carboxylation:<\/strong> The Grignard reagent adds CO\u2082 to form a magnesium salt of benzoic acid, which, upon acidic work-up, yields benzoic acid.\ue206<\/li>\n\n\n\n
    • Acylation:<\/strong> Benzoyl chloride reacts with dimethylamine to form N,N-dimethylbenzamide through nucleophilic acyl substitution.\ue206<\/li>\n\n\n\n
    • Reduction:<\/strong> LiAlH\u2084 reduces the amide group in N,N-dimethylbenzamide to a methylene group, yielding (N,N-dimethylaminomethyl)cyclohexane.\ue206<\/li>\n<\/ul>\n\n\n\n

      This sequence effectively utilizes the reduction of an amide with LiAlH\u2084 as the pivotal step to transform bromocyclohexane into (N,N-dimethylaminomethyl)cyclohexane.\ue206<\/p>\n","protected":false},"excerpt":{"rendered":"

      How would you use the reaction of an amide with LiAlH4 as the key step is going from bromo cyclohexane to (N, N-dimethylaminomethyl) cyclohexane? Write all the steps in the reactionsequence. The Correct Answer and Explanation is : To synthesize (N,N-dimethylaminomethyl)cyclohexane from bromocyclohexane using the reduction of an amide with lithium aluminum hydride (LiAlH\u2084) as […]<\/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-196487","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196487","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=196487"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196487\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196487"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196487"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196487"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}