{"id":255813,"date":"2025-07-13T09:53:33","date_gmt":"2025-07-13T09:53:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=255813"},"modified":"2025-07-13T09:53:35","modified_gmt":"2025-07-13T09:53:35","slug":"nabro-hbr-oh-h2o-meoh-hbr-oh-sh-nash-naho2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/13\/nabro-hbr-oh-h2o-meoh-hbr-oh-sh-nash-naho2\/","title":{"rendered":"NaBrO HBr OH H2O MeOH HBr OH SH NaSH NaHO2"},"content":{"rendered":"\n<p>NaBrO HBr OH H2O MeOH HBr OH SH NaSH NaHO2<\/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\/image-457.png\" alt=\"\" class=\"wp-image-255814\"\/><\/figure>\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>The correct answer is that reactions&nbsp;<strong>I, II, and III<\/strong>&nbsp;are SN1 reactions.<\/p>\n\n\n\n<p>Here is a detailed explanation for each reaction:<\/p>\n\n\n\n<p><strong>Reaction I:<\/strong>&nbsp;This reaction involves a secondary alkyl halide,&nbsp;trans-1-bromo-4-isopropylcyclohexane, reacting with sodium acetate, a weak nucleophile. The SN1 mechanism is favored over SN2 for several reasons. The substrate&#8217;s most stable chair conformation places both the bulky isopropyl group and the bromine atom in equatorial positions. For an SN2 reaction to occur, the acetate nucleophile would need to perform a backside attack from a sterically hindered axial position. This makes the SN2 pathway slow. Instead, the reaction is more likely to proceed via the SN1 mechanism, where the C-Br bond breaks first to form a secondary carbocation intermediate, which is then attacked by the acetate nucleophile.<\/p>\n\n\n\n<p><strong>Reaction II:<\/strong>&nbsp;This is a classic example of an SN1 reaction. The substrate is a tertiary alcohol,&nbsp;1-methylcyclohexan-1-ol. SN2 reactions are impossible at sterically crowded tertiary centers. The reaction proceeds in two main steps: first, the hydroxyl (-OH) group is protonated by the strong acid (HBr) to form a good leaving group, water (-OH2+). Second, the water molecule departs, generating a stable tertiary carbocation. Finally, the bromide ion (Br-) acts as a nucleophile and attacks the carbocation to form the final product. The formation of a stable carbocation intermediate is the defining characteristic of an SN1 reaction.<\/p>\n\n\n\n<p><strong>Reaction III:<\/strong>&nbsp;This reaction shows a benzylic halide (benzyl bromide) undergoing solvolysis with methanol, which is a weak nucleophile and a polar protic solvent. Although the substrate is technically a primary halide, the SN1 pathway is highly favored. This is because the departure of the bromide leaving group forms a benzylic carbocation. This carbocation is significantly stabilized by resonance, as the positive charge can be delocalized over the adjacent benzene ring. The combination of a substrate that forms a highly stable carbocation and a weak nucleophile makes this a textbook SN1 reaction.<\/p>\n\n\n\n<p><strong>Reaction IV:<\/strong>&nbsp;This reaction is not an SN1 reaction; it displays the characteristics of an SN2 reaction. The substrate reacts with sodium hydrosulfide (NaSH), where the hydrosulfide ion (SH-) is a strong nucleophile. Most importantly, the reaction shows a clear inversion of stereochemistry: the&nbsp;<em>trans<\/em>&nbsp;starting material yields a&nbsp;<em>cis<\/em>&nbsp;product. This stereospecific inversion is the hallmark of the SN2 mechanism, where the nucleophile attacks from the side opposite the leaving group in a single, concerted step. An SN1 reaction would have proceeded through a planar carbocation, leading to a mixture of products (both retention and inversion), not a single inverted product.<\/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-banner6-498.jpeg\" alt=\"\" class=\"wp-image-255815\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>NaBrO HBr OH H2O MeOH HBr OH SH NaSH NaHO2 The Correct Answer and Explanation is: The correct answer is that reactions&nbsp;I, II, and III&nbsp;are SN1 reactions. Here is a detailed explanation for each reaction: Reaction I:&nbsp;This reaction involves a secondary alkyl halide,&nbsp;trans-1-bromo-4-isopropylcyclohexane, reacting with sodium acetate, a weak nucleophile. The SN1 mechanism is favored [&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-255813","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/255813","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=255813"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/255813\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=255813"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=255813"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=255813"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}