{"id":26171,"date":"2025-06-19T10:49:22","date_gmt":"2025-06-19T10:49:22","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=26171"},"modified":"2025-06-19T10:49:24","modified_gmt":"2025-06-19T10:49:24","slug":"draw-the-product-formed-by-the-reaction-of-potassium-tbutoxide-with-s-2s-bromo-methyl-i-phenylbutane-shown","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-the-product-formed-by-the-reaction-of-potassium-tbutoxide-with-s-2s-bromo-methyl-i-phenylbutane-shown\/","title":{"rendered":"Draw the product formed by the reaction of potassium t~butoxide with ([S.2S)-[~bromo ~methyl-I-phenylbutane (shown"},"content":{"rendered":"\n<p>Draw the product formed by the reaction of potassium t~butoxide with ([S.2S)-[~bromo ~methyl-I-phenylbutane (shown). Clearly show the stereochemistry 0f the producl (CHshCo K&#8217; Hjcun H,CHzc (CH;)COH<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-pale-pink-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The reaction of <strong>potassium tert-butoxide<\/strong> with <strong>(2S,3S)-3-bromo-2-methyl-1-phenylbutane<\/strong> undergoes an <strong>E2 elimination<\/strong> mechanism. Let\u2019s go step-by-step to determine the <strong>structure and stereochemistry<\/strong> of the <strong>alkene product<\/strong> formed.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Analyze the Substrate<\/strong><\/h3>\n\n\n\n<p>The substrate is <strong>(2S,3S)-3-bromo-2-methyl-1-phenylbutane<\/strong>, which has the following structure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A four-carbon chain with a <strong>bromine<\/strong> on carbon 3 and a <strong>methyl group<\/strong> on carbon 2<\/li>\n\n\n\n<li>A <strong>phenyl group<\/strong> attached to carbon 1<\/li>\n\n\n\n<li>Chiral centers at carbon 2 and carbon 3, both in the <strong>S configuration<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Understand the Base<\/strong><\/h3>\n\n\n\n<p><strong>Potassium tert-butoxide ((CH\u2083)\u2083CO\u207b)<\/strong> is a <strong>strong, bulky base<\/strong>, which promotes <strong>E2 elimination<\/strong> by abstracting a \u03b2-hydrogen <strong>anti-periplanar<\/strong> to the leaving group (Br).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Predict the Mechanism<\/strong><\/h3>\n\n\n\n<p>In an E2 reaction:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The base removes a \u03b2-hydrogen anti to the leaving group (Br)<\/li>\n\n\n\n<li>The electrons from the \u03b2-hydrogen form a \u03c0-bond between carbon 2 and 3<\/li>\n\n\n\n<li>The leaving group (Br\u207b) departs from carbon 3<\/li>\n<\/ul>\n\n\n\n<p>Since the base is bulky, it prefers to abstract the <strong>least hindered<\/strong> \u03b2-hydrogen \u2014 usually leading to the <strong>less substituted (Hofmann) alkene<\/strong>. However, if stereoelectronic factors favor the more substituted (Zaitsev) alkene, that will form.<\/p>\n\n\n\n<p>In this molecule:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03b2-hydrogens are on carbon 2<\/li>\n\n\n\n<li>The hydrogen anti to the Br must be positioned <strong>opposite<\/strong> in a <strong>staggered conformation<\/strong><\/li>\n\n\n\n<li>The resulting double bond forms between <strong>C-2 and C-3<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 4: Determine the Product<\/strong><\/h3>\n\n\n\n<p>The double bond forms between <strong>C-2 and C-3<\/strong>, and since carbon 2 has a methyl group and carbon 3 is now substituted with the chain, the product is:<\/p>\n\n\n\n<p><strong>(E)-2-methyl-1-phenyl-1-butene<\/strong><\/p>\n\n\n\n<p>The stereochemistry of the product is <strong>E (trans)<\/strong> because:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer: Structure<\/strong><\/h3>\n\n\n\n<p><strong>(E)-2-methyl-1-phenyl-1-butene<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>      Ph\n       \\\n        C=C\n       \/   \\\n    CH3     CH2CH3\n<\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Conclusion<\/strong><\/h3>\n\n\n\n<p>The strong bulky base potassium tert-butoxide promotes an E2 reaction by removing a \u03b2-hydrogen anti to the Br group. The result is formation of a double bond with <strong>E stereochemistry<\/strong>, giving <strong>(E)-2-methyl-1-phenyl-1-butene<\/strong> as the major product.The reaction of <strong>potassium tert-butoxide<\/strong> with <strong>(2S,3S)-3-bromo-2-methyl-1-phenylbutane<\/strong> undergoes an <strong>E2 elimination<\/strong> mechanism. Let\u2019s go step-by-step to determine the <strong>structure and stereochemistry<\/strong> of the <strong>alkene product<\/strong> formed.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"722\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-65.jpeg\" alt=\"\" class=\"wp-image-26179\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-65.jpeg 722w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner10-65-212x300.jpeg 212w\" sizes=\"auto, (max-width: 722px) 100vw, 722px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the product formed by the reaction of potassium t~butoxide with ([S.2S)-[~bromo ~methyl-I-phenylbutane (shown). Clearly show the stereochemistry 0f the producl (CHshCo K&#8217; Hjcun H,CHzc (CH;)COH The Correct Answer and Explanation is: The reaction of potassium tert-butoxide with (2S,3S)-3-bromo-2-methyl-1-phenylbutane undergoes an E2 elimination mechanism. Let\u2019s go step-by-step to determine the structure and stereochemistry of the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-26171","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26171","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=26171"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26171\/revisions"}],"predecessor-version":[{"id":26180,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26171\/revisions\/26180"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26171"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26171"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26171"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}