The major product of the reaction between 1-butene (CH3CH2CH=CH2) and hydrobromic acid (HBr) is CH3CH2CHBrCH3<\/strong>.<\/p>\n\n\n\n Correct Answer: d. CH3CH2CHBrCH3<\/strong><\/p>\n\n\n\n The reaction between an alkene and HBr is a classic example of a hydrohalogenation, where the alkene double bond reacts with a hydrogen halide (in this case, HBr) to form a halogenated alkane. The reaction mechanism involves the addition of HBr across the double bond.<\/p>\n\n\n\n This reaction is a straightforward example of how Markovnikov’s rule helps in determining the major product in the addition reactions of alkenes with hydrogen halides. The product formed is 2-bromobutane, which corresponds to choice d. in your question. None of the other options match the expected product based on the molecular structure and reaction mechanism.<\/p>\n\n\n\n I will now generate the image for the chemical reaction you’re studying.<\/p>\n\n\n\n Here is the illustrated chemical reaction diagram showing the addition of hydrobromic acid (HBr) to 1-butene, resulting in the formation of 2-bromobutane. The diagram includes each step of the reaction, illustrating the application of Markovnikov’s rule and the formation of the carbocation intermediate, followed by the attack by the bromide ion. This should help clarify the reaction process for you.<\/p>\n\n\n\n Select the major product that would result from the reaction below. CH3CH2CH=CH2 + HBr Hint: Make use of your Reaction Cards a. CH (CHBr),CH OCH b. CH2CH2CH, OCH, c. CHBICH,CH OCH d. CH2CH2CH Br The Correct Answer and Explanation is : The major product of the reaction between 1-butene (CH3CH2CH=CH2) and hydrobromic acid (HBr) is […]<\/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-193740","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193740","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=193740"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193740\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193740"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193740"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193740"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation<\/h3>\n\n\n\n
\n
The hydrogen atom (H+) from HBr will add to the carbon of the double bond that has the greater number of hydrogen atoms already attached. This is in accordance with Markovnikov’s rule, which states that in the addition of HX to an alkene, the H atom bonds to the less substituted carbon of the double bond, and the X (halide) bonds to the more substituted carbon. In the case of 1-butene, CH3CH2CH=CH2, the H+ ion will preferentially add to the carbon that is part of the CH2 group, as it has more hydrogens initially compared to the terminal CH2 carbon.<\/li>\n\n\n\n
When H+ adds to the less substituted carbon (second carbon from the left in 1-butene), it results in the formation of a secondary carbocation on the adjacent carbon (third carbon). This carbocation is more stable compared to a primary carbocation that would form if the hydrogen added to the terminal carbon.<\/li>\n\n\n\n
The bromide ion (Br-) then attacks the positively charged carbocation, resulting in the formation of CH3CH2CHBrCH3.<\/li>\n<\/ol>\n\n\n\n![\"\"\/](\"https:\/\/files.oaiusercontent.com\/file-Rw9gmeEFCPP6EBoBAcEtrZ?se=2025-02-20T20%3A46%3A52Z&sp=r&sv=2024-08-04&sr=b&rscc=max-age%3D604800%2C%20immutable%2C%20private&rscd=attachment%3B%20filename%3D3a7d03d6-c439-4d7c-b2e7-a883707b408c.webp&sig=I7qPw\/H6rHzxAH%2BapYNkHQOm0jYn1yTnLxY8e6vBYlU%3D\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"