{"id":247376,"date":"2025-07-07T16:32:07","date_gmt":"2025-07-07T16:32:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=247376"},"modified":"2025-07-07T16:32:09","modified_gmt":"2025-07-07T16:32:09","slug":"reagents-and-properties","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/reagents-and-properties\/","title":{"rendered":"Reagents and Properties"},"content":{"rendered":"\n<p>Reagents and Properties substance quantity molar mass mp (g\/mol) (&#8220;C) (1)-benzoin 7 mg 212.25 135-137 ethyl acetate* 2.2 mL Aniel 88.11 ~84 meso-hydrobenzoin 7 mg 214.27 137-139 n-hexane* 4 mL 86.18 -95 U&#8221; &#8216; &#8216;\/&#8221; 4 potassium bromide 100 mg 3JT #for TLC 3:7, 9.61 ! tfor KBr &#8216;pellet &#8220;4Kli n bp C) 76-77<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Corrected Information<\/h3>\n\n\n\n<p>The provided data describes a chemical experiment, specifically the stereoselective reduction of benzoin to produce meso-hydrobenzoin. The information, once corrected and organized, is presented in the table below.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>Substance<\/td><td>Role<\/td><td>Quantity<\/td><td>Molar Mass (&nbsp;g\/mol&nbsp;)<\/td><td>Melting Point (\u00b0C)<\/td><td>Boiling Point (\u00b0C)<\/td><\/tr><tr><td>Benzoin<\/td><td>Reactant<\/td><td>7 mg<\/td><td>212.25<\/td><td>135-137<\/td><td><\/td><\/tr><tr><td>meso-hydrobenzoin<\/td><td>Product\/Standard<\/td><td>7 mg<\/td><td>214.27<\/td><td>137-139<\/td><td><\/td><\/tr><tr><td>Ethyl Acetate<\/td><td>Solvent \/ TLC Eluent<\/td><td>2.2 mL<\/td><td>88.11<\/td><td>-84<\/td><td>76-77<\/td><\/tr><tr><td>n-Hexane<\/td><td>Solvent \/ TLC Eluent<\/td><td>4 mL<\/td><td>86.18<\/td><td>-95<\/td><td><\/td><\/tr><tr><td>Potassium Bromide<\/td><td>Analytical Reagent<\/td><td>100 mg<\/td><td>119.00<\/td><td><\/td><td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><em>Note: The garbled text &#8220;TLC 3:7&#8221; indicates that thin-layer chromatography will be performed using a 3:7 ratio of ethyl acetate to n-hexane as the eluent.<\/em><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>The information details the reagents and analytical procedures for a common organic chemistry laboratory experiment: the reduction of benzoin. This reaction converts the ketone functional group in the benzoin molecule into a secondary alcohol, resulting in the formation of hydrobenzoin.<\/p>\n\n\n\n<p>The reaction is stereoselective, meaning it preferentially creates one stereoisomer over others. Benzoin is a chiral molecule, and its reduction creates a second chiral center. The specific mention of meso-hydrobenzoin as a product indicates that the experiment is designed to yield this particular achiral diastereomer. A common reagent used to achieve this outcome, though not listed, is sodium borohydride (NaBH\u2084).<\/p>\n\n\n\n<p>The roles of the listed substances are distinct. Benzoin is the starting material, or reactant. Meso-hydrobenzoin is the target product; its inclusion in the table with a specified mass suggests it may also serve as an authentic reference standard for comparison. This comparison is crucial for confirming the identity of the synthesized product.<\/p>\n\n\n\n<p>Ethyl acetate and n-hexane are organic solvents. They likely serve multiple purposes. First, one of them (typically a polar solvent like ethyl acetate or an alcohol) would be the solvent in which the reduction reaction is carried out. Second, they are used in a 3:7 ratio as the mobile phase, or eluent, for thin-layer chromatography (TLC). TLC is an analytical technique used to monitor the reaction&#8217;s progress by separating the reactant from the product and to assess the purity of the final product.<\/p>\n\n\n\n<p>Finally, potassium bromide (KBr) is an analytical reagent used for infrared (IR) spectroscopy. The sample is ground with dry KBr and pressed into a thin pellet. Since KBr is transparent to infrared light, this method allows for the analysis of the product&#8217;s functional groups, confirming the disappearance of the benzoin\u2019s ketone group and the appearance of the hydrobenzoin\u2019s alcohol group.<\/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-banner5-1006.jpeg\" alt=\"\" class=\"wp-image-247377\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Reagents and Properties substance quantity molar mass mp (g\/mol) (&#8220;C) (1)-benzoin 7 mg 212.25 135-137 ethyl acetate* 2.2 mL Aniel 88.11 ~84 meso-hydrobenzoin 7 mg 214.27 137-139 n-hexane* 4 mL 86.18 -95 U&#8221; &#8216; &#8216;\/&#8221; 4 potassium bromide 100 mg 3JT #for TLC 3:7, 9.61 ! tfor KBr &#8216;pellet &#8220;4Kli n bp C) 76-77 The [&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-247376","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247376","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=247376"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247376\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=247376"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=247376"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=247376"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}