{"id":257515,"date":"2025-07-17T13:59:09","date_gmt":"2025-07-17T13:59:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257515"},"modified":"2025-07-17T13:59:11","modified_gmt":"2025-07-17T13:59:11","slug":"when-2-bromo-2-methylbutane-is-reacted-with-hot-alcoholic-potassium-hydroxide-one-major-product-k-and-a-minor-product-j-are-formed","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/when-2-bromo-2-methylbutane-is-reacted-with-hot-alcoholic-potassium-hydroxide-one-major-product-k-and-a-minor-product-j-are-formed\/","title":{"rendered":"When 2-bromo-2-methylbutane is reacted with hot alcoholic potassium hydroxide, one major product K and a minor product J are formed"},"content":{"rendered":"\n<p>When 2-bromo-2-methylbutane is reacted with hot alcoholic potassium hydroxide, one major product K and a minor product J are formed. Both K and J decolorize bromine dissolved in carbon tetrachloride. When product J is reacted with HBr in the presence of benzoyl peroxide or hydrogen peroxide, product H is formed. When H is treated with aqueous sodium hydroxide, compound I is formed. When compound I is treated with acidified potassium dichromate (VI), a compound L is formed. When compound I is treated with Lucas reagent (ZnCl\/HCl), the solution remains colorless unless it is subjected to heat. The solution forms an oily layer when heated. A solution of L turns litmus red. When a solution of L is added to a saturated solution of sodium bicarbonate, there is effervescence and carbon dioxide is evolved. 2.1 Write down a chemical equation for the reaction that results in the formation of compounds K and J. [2] 2.2 Give the IUPAC names of compound J. [1] 2.3 Give a reason why K is the major product when 2-bromo-2-methylbutane is reacted with hot alcoholic potassium hydroxide. [1] 2.4 Name and illustrate the reaction mechanism through which K is formed. Your illustration should have arrows showing electron movements as well as a clearly identified rate-determining step. [5] 2.5 Write an equation showing the reaction of J with HBr in the presence of benzoyl peroxide or hydrogen peroxide to form H. [2] 2.6 Draw the full structural formula of H and give its IUPAC name. [2] 2.7 Give the name of the reaction by which compound I is formed from the treatment of H with aqueous sodium hydroxide. [1] 2.8 Name and illustrate the mechanism of the reaction of H with aqueous sodium hydroxide. Your illustration should have arrows showing electron movements as well as a clearly identified rate-determining step. [5] 2.9 Give the identity of compound L and draw its line structure. [2] 2.10 What product would have been formed if the reaction between J and HBr was performed in the absence of benzoyl peroxide or hydrogen peroxide? Draw the structure of this product. [2] 2.11 What would you observe if the product you identified in 2.10 above was treated with dilute sulfuric acid followed by reaction with the Lucas reagent? Give a reason for your answer. [2] 3. Title_with_topic: Organic Chemistry Reactions and Mechanisms<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s break down the questions and explain each part:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2.1 Reaction that results in the formation of K and J:<\/h3>\n\n\n\n<p>When 2-bromo-2-methylbutane is reacted with hot alcoholic potassium hydroxide, the reaction undergoes an elimination reaction. In this case, the reaction is a <strong>\u03b2-elimination (E2)<\/strong> reaction, where a proton (H\u207a) is removed from the carbon adjacent to the carbon bonded to the bromine atom, and the bromine is eliminated. This results in the formation of alkenes.<\/p>\n\n\n\n<p>The products K and J are <strong>alkenes<\/strong> that form as a result of the elimination of HBr. The two possible products come from removing a hydrogen from different adjacent carbon atoms.<\/p>\n\n\n\n<p>For the reaction:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>K<\/strong> (major product): 2-methylbut-2-ene<\/li>\n\n\n\n<li><strong>J<\/strong> (minor product): 2-methylbut-1-ene<\/li>\n<\/ul>\n\n\n\n<p>The equation is:C5H11Br+KOH&nbsp;(alc)\u2192C5H10+HBr\\text{C}_5\\text{H}_{11}\\text{Br} + \\text{KOH (alc)} \\rightarrow \\text{C}_5\\text{H}_{10} + \\text{HBr}C5\u200bH11\u200bBr+KOH&nbsp;(alc)\u2192C5\u200bH10\u200b+HBr<\/p>\n\n\n\n<p>Where C\u2085H\u2081\u2081Br is 2-bromo-2-methylbutane and C\u2085H\u2081\u2080 represents the alkenes K and J.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2.2 IUPAC Name of Compound J:<\/h3>\n\n\n\n<p>Compound J is <strong>2-methylbut-1-ene<\/strong>. This is because the double bond is located between the first and second carbon atoms in the chain.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2.3 Reason for K being the Major Product:<\/h3>\n\n\n\n<p>The major product K, <strong>2-methylbut-2-ene<\/strong>, is formed because it is more stable. The stability of the alkene product is higher when the double bond is in the <strong>more substituted position<\/strong> (i.e., the second carbon is more substituted than the first). This is known as <strong>Zaitsev&#8217;s rule<\/strong>, which states that in an elimination reaction, the more substituted alkene is typically the major product.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2.4 Reaction Mechanism for Formation of K (E2 Mechanism):<\/h3>\n\n\n\n<p>The mechanism is <strong>E2 (bimolecular elimination)<\/strong>, where the base (OH\u207b) abstracts a proton from the carbon adjacent to the carbon bonded to the leaving group (Br), and the leaving group (Br\u207b) simultaneously leaves. The electron pairs from the C-H bond form the double bond.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Rate-determining step<\/strong>: The simultaneous breaking of the C-H and C-Br bonds.<\/li>\n<\/ul>\n\n\n\n<p>Mechanism diagram:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Base (OH\u207b) abstracts a proton (H\u207a) from carbon adjacent to the carbon bearing the bromine.<\/li>\n\n\n\n<li>The electron pair from the C-H bond forms the double bond.<\/li>\n\n\n\n<li>The bromine leaves as Br\u207b.<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">2.5 Reaction of J with HBr in Presence of Benzoyl Peroxide:<\/h3>\n\n\n\n<p>In the presence of benzoyl peroxide, the reaction undergoes <strong>radical substitution<\/strong>, where the <strong>Br<\/strong> radical adds to the <strong>more substituted carbon<\/strong> in J, forming an alkyl bromide.<\/p>\n\n\n\n<p>The equation:C5H10+HBr\u2192benzoyl&nbsp;peroxideC5H11Br\\text{C}_5\\text{H}_{10} + \\text{HBr} \\xrightarrow{\\text{benzoyl peroxide}} \\text{C}_5\\text{H}_{11}\\text{Br}C5\u200bH10\u200b+HBrbenzoyl&nbsp;peroxide\u200bC5\u200bH11\u200bBr<\/p>\n\n\n\n<p>This results in <strong>2-bromo-2-methylbutane<\/strong> (compound H).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2.6 Structural Formula of H and IUPAC Name:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The structure of <strong>H<\/strong> is 2-bromo-2-methylbutane. Its IUPAC name is 2-bromo-2-methylbutane.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">2.7 Name of the Reaction for Formation of I:<\/h3>\n\n\n\n<p>The reaction of H with aqueous sodium hydroxide is a <strong>hydrolysis reaction<\/strong>, where the bromine is replaced by a hydroxyl group (OH). This is a <strong>nucleophilic substitution<\/strong> (SN1) reaction.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2.8 Mechanism of the Reaction of H with Aqueous NaOH:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>SN1 Mechanism<\/strong>: The bromine leaves first, forming a carbocation. The hydroxide ion (OH\u207b) then attacks the carbocation.<\/li>\n\n\n\n<li><strong>Rate-determining step<\/strong>: The formation of the carbocation.<\/li>\n<\/ul>\n\n\n\n<p>Mechanism diagram:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>C-Br<\/strong> bond breaks, forming a carbocation.<\/li>\n\n\n\n<li><strong>OH\u207b<\/strong> attacks the carbocation, resulting in the formation of <strong>2-methylbutan-2-ol<\/strong> (compound I).<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">2.9 Identity and Line Structure of Compound L:<\/h3>\n\n\n\n<p>Compound L is <strong>2-methylbutanoic acid<\/strong> (carboxylic acid). The line structure is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">scssCopyEdit<code>     CH3-CH(CH3)-COOH\n<\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\">2.10 Product Formed if the Reaction of J with HBr is Without Benzoyl Peroxide:<\/h3>\n\n\n\n<p>Without benzoyl peroxide, the reaction between J and HBr would not undergo radical substitution. Instead, <strong>Markovnikov\u2019s rule<\/strong> would apply, leading to the addition of the hydrogen to the carbon with the fewest hydrogens. The product would be <strong>2-bromo-1-methylbutane<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2.11 Observation with Lucas Reagent:<\/h3>\n\n\n\n<p>When the product from 2.10 (2-bromo-1-methylbutane) is treated with dilute sulfuric acid and then with Lucas reagent, there would be no immediate reaction unless heat is applied. Upon heating, <strong>SN1 substitution<\/strong> occurs, forming a tertiary carbocation. The product would be <strong>tert-butyl alcohol<\/strong>.<\/p>\n\n\n\n<p>The solution would form an oily layer upon heating due to the formation of a more hydrophobic product.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>In summary, the key reactions involve <strong>E2 elimination<\/strong>, <strong>radical substitution<\/strong>, <strong>nucleophilic substitution<\/strong>, and <strong>carboxylic acid formation<\/strong> based on the reactivity of the compounds involved. Each step can be explained through mechanisms such as <strong>E2<\/strong> for elimination, <strong>SN1<\/strong> for substitution, and <strong>radical<\/strong> mechanisms for addition.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-782.jpeg\" alt=\"\" class=\"wp-image-257524\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>When 2-bromo-2-methylbutane is reacted with hot alcoholic potassium hydroxide, one major product K and a minor product J are formed. Both K and J decolorize bromine dissolved in carbon tetrachloride. When product J is reacted with HBr in the presence of benzoyl peroxide or hydrogen peroxide, product H is formed. When H is treated with [&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-257515","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257515","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=257515"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257515\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257515"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257515"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257515"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}