Actual Yield Calculation and Explanation \u2013 Synthesis of p-Bromoaniline (Step 1: 4-bromoacetanilide)<\/strong><\/p>\n\n\n\n Moles=MassMolecular Weight=2.5\u2009g214.06\u2009g\/mol\u22480.01169\u2009mol\\text{Moles} = \\frac{\\text{Mass}}{\\text{Molecular Weight}} = \\frac{2.5\\,g}{214.06\\,g\/mol} \\approx 0.01169\\,mol<\/p>\n\n\n\n So, actual yield = 0.01169 mol<\/strong> of 4-bromoacetanilide.<\/p>\n\n\n\n If the initial amount of acetanilide used were given, we could calculate the theoretical yield. Since it\u2019s not provided, we calculate actual yield only<\/strong>, based on the final mass obtained.<\/p>\n\n\n\n In the synthesis of p-bromoaniline, a crucial intermediate is 4-bromoacetanilide<\/strong>, formed through the bromination of acetanilide<\/strong>. In Step 1 of the synthesis, bromine reacts with acetanilide in a 1:1 molar ratio<\/strong> to produce 4-bromoacetanilide via electrophilic aromatic substitution. This ratio implies that one mole of acetanilide yields one mole of 4-bromoacetanilide under ideal conditions.<\/p>\n\n\n\n From the data, the final mass of 4-bromoacetanilide obtained is 2.5 grams<\/strong>. Given its molecular weight is 214.06 g\/mol<\/strong>, we use the formula: Moles=MassMolecular Weight\\text{Moles} = \\frac{\\text{Mass}}{\\text{Molecular Weight}}<\/p>\n\n\n\n Substituting the values gives approximately 0.01169 mol<\/strong>. This value represents the actual yield<\/strong> in moles. If we had the initial amount of acetanilide, we could compare this with the theoretical yield to determine the percent yield<\/strong>, which is essential for evaluating the efficiency of the reaction.<\/p>\n\n\n\n This yield can be affected by several experimental factors such as incomplete reaction, side products (e.g., ortho isomers), or losses during purification (e.g., recrystallization). Nevertheless, knowing the actual amount of product obtained is important for planning subsequent steps \u2014 like hydrolysis of the acetamide group to finally yield p-bromoaniline<\/strong>.<\/p>\n\n\n\n This analysis not only confirms product formation but also allows researchers to evaluate how well the reaction proceeded and how to optimize it. Accurately determining the yield is thus fundamental in synthetic organic chemistry.<\/p>\n\n\n\n Synthesis of p-bromoaniline Result section From Step 1: Weight of 4-bromoacetanilide is 2.5g. Properties: C HBrNO, Molecular weight: 214.06g\/mol, Melting Point: 168\u00c2\u00b0C Please note that the ratio of acetanilide to 4-bromoacetanilide is 1:1. Calculate the actual yield? The Correct Answer and Explanation is: Actual Yield Calculation and Explanation \u2013 Synthesis of p-Bromoaniline (Step 1: 4-bromoacetanilide) […]<\/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-224599","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224599","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=224599"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224599\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=224599"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=224599"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=224599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given Data:<\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Calculate Moles of 4-Bromoacetanilide<\/h3>\n\n\n\n
\n\n\n\nStep 2: Determine Theoretical Yield (if needed)<\/h3>\n\n\n\n
\n\n\n\nExplanation<\/h3>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner7-114.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"