{"id":225763,"date":"2025-06-04T15:06:59","date_gmt":"2025-06-04T15:06:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=225763"},"modified":"2025-06-04T15:07:02","modified_gmt":"2025-06-04T15:07:02","slug":"the-compound-below-can-be-prepared-with-an-alkyl-iodide-and-a-suitable-nucleophile","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/04\/the-compound-below-can-be-prepared-with-an-alkyl-iodide-and-a-suitable-nucleophile\/","title":{"rendered":"The compound below can be prepared with an alkyl iodide and a suitable nucleophile"},"content":{"rendered":"\n<p>The compound below can be prepared with an alkyl iodide and a suitable nucleophile: Identify the alkyl iodide and the nucleophile that you would use: For an anionic nucleophile, you do not need to draw the counterion Alkyl iodide: Edit HSC Nucleophile: H H Edit OH H<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To answer this question accurately, we must first identify the target compound you&#8217;re referring to. However, since the compound was not included in your message, I will explain how to approach this kind of synthesis problem generally and provide a likely example based on the context: using an <strong>alkyl iodide<\/strong> and an <strong>anionic nucleophile<\/strong> such as <strong>OH\u207b<\/strong> (hydroxide).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Example Compound:<\/h3>\n\n\n\n<p>Let&#8217;s consider the preparation of <strong>ethanol (CH\u2083CH\u2082OH)<\/strong> using an alkyl iodide and a nucleophile:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Target compound<\/strong>: <strong>CH\u2083CH\u2082OH (ethanol)<\/strong><\/li>\n\n\n\n<li><strong>Alkyl iodide<\/strong>: <strong>CH\u2083CH\u2082I (ethyl iodide)<\/strong><\/li>\n\n\n\n<li><strong>Nucleophile<\/strong>: <strong>OH\u207b (hydroxide ion)<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Reaction:<\/h3>\n\n\n\n<p>This is a typical <strong>SN2 nucleophilic substitution<\/strong> reaction: CH\u2083CH\u2082I+OH\u2212\u2192CH\u2083CH\u2082OH+I\u2212\\text{CH\u2083CH\u2082I} + \\text{OH}\u207b \\rightarrow \\text{CH\u2083CH\u2082OH} + \\text{I}\u207b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>The synthesis of alcohols using alkyl halides and nucleophiles is a classic example of nucleophilic substitution reactions in organic chemistry, particularly via the <strong>SN2 mechanism<\/strong>. In this pathway, a strong nucleophile directly attacks the electrophilic carbon atom of the alkyl halide, displacing the halide (a good leaving group).<\/p>\n\n\n\n<p>In the example above, <strong>ethyl iodide (CH\u2083CH\u2082I)<\/strong> serves as the <strong>alkyl iodide<\/strong>. It contains a primary carbon attached to the iodine atom, making it highly reactive toward SN2 displacement because primary carbons are less hindered and more accessible to nucleophilic attack.<\/p>\n\n\n\n<p>The <strong>nucleophile<\/strong> is the <strong>hydroxide ion (OH\u207b)<\/strong>. As an anionic nucleophile, OH\u207b is strong and readily attacks the electrophilic carbon in the ethyl iodide. Iodide (I\u207b), being a large and polarizable ion, is an excellent leaving group and is easily displaced in the reaction.<\/p>\n\n\n\n<p>The result is the formation of <strong>ethanol (CH\u2083CH\u2082OH)<\/strong> and iodide ion as the byproduct. This method is useful in laboratory synthesis of simple alcohols.<\/p>\n\n\n\n<p>This reaction is also stereospecific; if the carbon center were chiral (not in this example), the inversion of configuration would be observed\u2014a hallmark of the SN2 process.<\/p>\n\n\n\n<p>This concept can be extended to more complex molecules by choosing appropriate alkyl halides and nucleophiles to construct desired functional groups in organic synthesis.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Alkyl iodide<\/strong>: <strong>CH\u2083CH\u2082I<\/strong> (ethyl iodide)<\/li>\n\n\n\n<li><strong>Nucleophile<\/strong>: <strong>OH\u207b<\/strong> (hydroxide ion)<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-351.jpeg\" alt=\"\" class=\"wp-image-225764\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The compound below can be prepared with an alkyl iodide and a suitable nucleophile: Identify the alkyl iodide and the nucleophile that you would use: For an anionic nucleophile, you do not need to draw the counterion Alkyl iodide: Edit HSC Nucleophile: H H Edit OH H The Correct Answer and Explanation is: To answer [&hellip;]<\/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-225763","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225763","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=225763"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225763\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=225763"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=225763"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=225763"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}