{"id":233576,"date":"2025-06-13T08:27:04","date_gmt":"2025-06-13T08:27:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233576"},"modified":"2025-06-13T08:29:23","modified_gmt":"2025-06-13T08:29:23","slug":"which-reagents-best-accomplish-this-transformation","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/which-reagents-best-accomplish-this-transformation\/","title":{"rendered":"Which reagents best accomplish this transformation"},"content":{"rendered":"\n

Which reagents best accomplish this transformation? OH enantiomer FOH a OsO4\/NMO b H2O2, H+ c. BHB\/THF followed by H2O2, HO- d. CHBCOBH followed by NaOH\/H2O2<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

The correct answer is d. CH3CO3H followed by NaOH\/H2O<\/strong>.<\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n

This reaction transforms an alkene, 1-methylcyclopentene, into a trans<\/em>-diol, (1R,2S)-1-methylcyclopentane-1,2-diol, and its enantiomer. This overall transformation is known as anti-dihydroxylation<\/strong>, meaning two hydroxyl (-OH) groups are added to opposite faces of the original double bond. Let’s analyze the reaction and the provided reagent options to understand why option (d) is the correct choice.<\/p>\n\n\n\n

Analysis of the Transformation:<\/strong><\/p>\n\n\n\n

    \n
  • Starting Material:<\/strong>\u00a01-methylcyclopentene, an alkene.<\/li>\n\n\n\n
  • Product:<\/strong>\u00a0A 1,2-diol where the two -OH groups have a\u00a0trans<\/em>\u00a0stereochemical relationship (one is on a wedge, the other on a dash). The reaction yields a racemic mixture (the product shown and its enantiomer).<\/li>\n\n\n\n
  • Required Reaction:<\/strong>\u00a0The conversion of a C=C double bond into two C-OH single bonds with\u00a0anti<\/em>\u00a0stereochemistry.<\/li>\n<\/ul>\n\n\n\n

    Evaluation of Reagent Options:<\/strong><\/p>\n\n\n\n

      \n
    • a. OsO4\/NMO:<\/strong>\u00a0Osmium tetroxide (OsO4) with a co-oxidant like N-methylmorpholine N-oxide (NMO) is a classic reagent for\u00a0syn-dihydroxylation<\/strong>. This reaction adds both hydroxyl groups to the\u00a0same face<\/em>\u00a0of the double bond, resulting in a\u00a0cis<\/em>-diol. This would produce a product different from the one shown.<\/li>\n\n\n\n
    • b. H2O, H+:<\/strong>\u00a0This is the reagent for\u00a0acid-catalyzed hydration<\/strong>. It adds one molecule of water across the double bond to form an alcohol, not a diol. The reaction follows Markovnikov’s rule, yielding 1-methylcyclopentanol.<\/li>\n\n\n\n
    • c. BH3\/THF followed by H2O2, HO\u2013:<\/strong>\u00a0This is the two-step procedure for\u00a0hydroboration-oxidation<\/strong>. This reaction also adds one molecule of water across the double bond to form an alcohol, not a diol. It proceeds with\u00a0anti<\/em>-Markovnikov regioselectivity and\u00a0syn<\/em>-addition stereochemistry, yielding\u00a0trans<\/em>-2-methylcyclopentanol.<\/li>\n\n\n\n
    • d. CH3CO3H followed by NaOH\/H2O:<\/strong>\u00a0This is a two-step process that achieves the desired\u00a0anti<\/em>-dihydroxylation.\n
        \n
      1. Step 1: Epoxidation.<\/strong>\u00a0The first reagent, peracetic acid (CH3CO3H, a type of peroxyacid), reacts with the alkene to form an epoxide. An epoxide is a three-membered ring containing an oxygen atom. This addition of the oxygen atom occurs in a\u00a0syn<\/em>\u00a0fashion, meaning it adds to one face of the double bond.<\/li>\n\n\n\n
      2. Step 2: Epoxide Ring-Opening.<\/strong>\u00a0The second step involves the ring-opening of the epoxide by a nucleophile (in this case, hydroxide ion, HO\u207b, from NaOH in water). The hydroxide ion attacks one of the carbons of the epoxide ring from the side opposite to the oxygen atom (a backside SN2 attack). This attack inverts the stereochemistry at the point of attack and opens the ring. After protonation by water, a\u00a0trans<\/em>-diol is formed.<\/li>\n<\/ol>\n<\/li>\n<\/ul>\n\n\n\n

        Because the initial alkene is planar, the peroxyacid can attack from the top or bottom face with equal probability. This leads to a racemic mixture of enantiomeric epoxides, which, upon ring-opening, yields a racemic mixture of the final trans<\/em>-diol products, as indicated by “+ enantiomer” in the question.<\/p>\n\n\n\n

        Therefore, the sequence of epoxidation followed by base-catalyzed hydrolysis is the correct method to accomplish the anti<\/em>-dihydroxylation shown in the transformation.thumb_upthumb_down<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Which reagents best accomplish this transformation? OH enantiomer FOH a OsO4\/NMO b H2O2, H+ c. BHB\/THF followed by H2O2, HO- d. CHBCOBH followed by NaOH\/H2O2 The Correct Answer and Explanation is: The correct answer is d. CH3CO3H followed by NaOH\/H2O. Explanation: This reaction transforms an alkene, 1-methylcyclopentene, into a trans-diol, (1R,2S)-1-methylcyclopentane-1,2-diol, and its enantiomer. This overall transformation […]<\/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-233576","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233576","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=233576"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233576\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233576"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233576"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233576"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}