{"id":42853,"date":"2025-06-29T11:29:05","date_gmt":"2025-06-29T11:29:05","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42853"},"modified":"2025-06-29T11:29:06","modified_gmt":"2025-06-29T11:29:06","slug":"identify-the-conformation-of-2-bromo-3-methylpentane-that-will-result-in-formation-of-trans-3-methylpent-2-ene-by-an-e2-reaction","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/identify-the-conformation-of-2-bromo-3-methylpentane-that-will-result-in-formation-of-trans-3-methylpent-2-ene-by-an-e2-reaction\/","title":{"rendered":"Identify the conformation of 2-bromo-3-methylpentane that will result in formation of trans-3-methylpent-2-ene by an E2 reaction"},"content":{"rendered":"\n<p>Identify the conformation of 2-bromo-3-methylpentane that will result in formation of trans-3-methylpent-2-ene by an E2 reaction. <\/p>\n\n\n\n<p>A I <\/p>\n\n\n\n<p>B II <\/p>\n\n\n\n<p>C III <\/p>\n\n\n\n<p>D IV<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct answer is <strong>C (III)<\/strong>.<\/p>\n\n\n\n<p>In an <strong>E2 (bimolecular elimination)<\/strong> reaction, the elimination of a leaving group (such as a bromine atom) and a hydrogen atom from adjacent carbon atoms occurs in a single, concerted step. The product of the E2 reaction depends on the <strong>antiperiplanar<\/strong> (trans-diaxial) geometry of the leaving group and the proton that is abstracted.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Conformation of 2-bromo-3-methylpentane<\/strong>: For 2-bromo-3-methylpentane to undergo an E2 elimination, the bromine (leaving group) on <strong>carbon 2<\/strong> and the hydrogen on <strong>carbon 3<\/strong> must be antiperiplanar. This means that both the leaving group and the hydrogen should be in opposite directions, ideally positioned in a <strong>trans-diaxial<\/strong> arrangement in a <strong>cyclic<\/strong> or <strong>planar<\/strong> conformation.<\/li>\n\n\n\n<li><strong>Trans-3-methylpent-2-ene<\/strong> is the product, which indicates the need for the elimination of a hydrogen atom from the \u03b2-carbon (carbon 3), along with the bromine from the \u03b1-carbon (carbon 2). The elimination leads to the double bond between carbon 2 and carbon 3, with a methyl group on carbon 3.<\/li>\n\n\n\n<li><strong>Conformation III<\/strong>: In this conformation, the <strong>bromine<\/strong> on carbon 2 and the <strong>hydrogen<\/strong> on carbon 3 are aligned in an <strong>antiperiplanar<\/strong> configuration. This allows the E2 reaction to proceed in a single step, resulting in the formation of the product <strong>trans-3-methylpent-2-ene<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>In summary, the correct conformation is one where the bromine on carbon 2 and the hydrogen on carbon 3 are antiperiplanar, which is conformation III (C). This arrangement enables the E2 elimination to occur efficiently and gives the desired trans product.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Identify the conformation of 2-bromo-3-methylpentane that will result in formation of trans-3-methylpent-2-ene by an E2 reaction. A I B II C III D IV The correct answer and explanation is: The correct answer is C (III). In an E2 (bimolecular elimination) reaction, the elimination of a leaving group (such as a bromine atom) and a [&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-42853","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42853","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=42853"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42853\/revisions"}],"predecessor-version":[{"id":42862,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42853\/revisions\/42862"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42853"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42853"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42853"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}