{"id":194675,"date":"2025-02-24T11:19:45","date_gmt":"2025-02-24T11:19:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=194675"},"modified":"2025-02-24T11:19:46","modified_gmt":"2025-02-24T11:19:46","slug":"synthesis-of-2-chloro-2-methylbutane","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/24\/synthesis-of-2-chloro-2-methylbutane\/","title":{"rendered":"Synthesis of 2-chloro-2-methylbutane"},"content":{"rendered":"\n

TITLE: Synthesis of 2-chloro-2-methylbutane<\/p>\n\n\n\n

PURPOSE: In this experiment a tertiary alcohol ( 2-methyl -2-butanol) was converted to a tertiary alkyl chloride ( 2-chloro-2-methylbutane) using concentrated hidtochloric acid. The rxn. preceeds via a SN1 mech. The crude product is insolunle in water. Excess acid and any remaining 2-methyl-2-butanol are removed from the crude product by washing with water. Trace amounts of HCl can be removed by washing the product with a weak base, sodium bicarbonate. Trace amounts of water are removed by washing the product with saturated aqueous sodium chloride and drying the crude product over anhydrous sodium sulfate.<\/p>\n\n\n\n

The crude product was purifed using simple distillation. The purified & the starting material were analyzed by IR spectroscopy.<\/p>\n\n\n\n

1 a) PLEASE, calculate the theoretical yield of 2-chloro-2-methylbutane. Full calculation including Limiting Reagent must be clearly shown.REFER TO THE ATTACHMENT ( fill out the tb. if needed)<\/p>\n\n\n\n

b) Calculate the % yield of purified 2-chloro-2-methylbutane ( WHERE mass of purified product is 7.1<\/p>\n\n\n\n

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

To calculate the theoretical yield<\/strong> and percent yield<\/strong> of 2-chloro-2-methylbutane, follow these steps:<\/p>\n\n\n\n

\n\n\n\n

Step 1: Determine the balanced chemical equation<\/strong><\/h3>\n\n\n\n

C5H12O+HCl\u2192C5H11Cl+H2O\\text{C}_5\\text{H}_{12}\\text{O} + \\text{HCl} \\rightarrow \\text{C}_5\\text{H}_{11}\\text{Cl} + \\text{H}_2\\text{O}<\/p>\n\n\n\n

The reaction follows an SN1 mechanism<\/strong>, where 2-methyl-2-butanol (C\u2085H\u2081\u2082O)<\/strong> reacts with hydrochloric acid (HCl)<\/strong> to form 2-chloro-2-methylbutane (C\u2085H\u2081\u2081Cl)<\/strong> and water.<\/p>\n\n\n\n

\n\n\n\n

Step 2: Identify the Limiting Reagent<\/strong><\/h3>\n\n\n\n

Given data:<\/h4>\n\n\n\n
    \n
  • Mass of 2-methyl-2-butanol (C\u2085H\u2081\u2082O)<\/strong> = 12.0 g<\/strong><\/li>\n\n\n\n
  • Concentration of HCl = 6.0 M<\/strong><\/li>\n\n\n\n
  • Volume of HCl = 25.0 mL<\/strong><\/li>\n\n\n\n
  • Molar masses:<\/strong>\n
      \n
    • 2-methyl-2-butanol (C\u2085H\u2081\u2082O):<\/strong> 88.15 g\/mol<\/li>\n\n\n\n
    • 2-chloro-2-methylbutane (C\u2085H\u2081\u2081Cl):<\/strong> 106.60 g\/mol<\/li>\n\n\n\n
    • HCl:<\/strong> 36.46 g\/mol<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

      Step 2A: Moles of 2-methyl-2-butanol<\/strong><\/h4>\n\n\n\n

      Moles of C\u2085H\u2081\u2082O=12.0 g88.15 g\/mol=0.1362 moles\\text{Moles of C\u2085H\u2081\u2082O} = \\frac{12.0 \\text{ g}}{88.15 \\text{ g\/mol}} = 0.1362 \\text{ moles}<\/p>\n\n\n\n

      Step 2B: Moles of HCl<\/strong><\/h4>\n\n\n\n

      Moles of HCl=Molarity\u00d7Volume(L)\\text{Moles of HCl} = Molarity \\times Volume (L) =(6.0 M)\u00d7(0.0250 L)=0.150 moles= (6.0 \\text{ M}) \\times (0.0250 \\text{ L}) = 0.150 \\text{ moles}<\/p>\n\n\n\n

      Since the reaction requires a 1:1 molar ratio<\/strong>, and 2-methyl-2-butanol (0.1362 moles)<\/strong> is less than HCl (0.150 moles)<\/strong>, 2-methyl-2-butanol is the limiting reagent<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      Step 3: Calculate Theoretical Yield<\/strong><\/h3>\n\n\n\n

      Since the reaction has a 1:1 molar ratio<\/strong>, the moles of 2-chloro-2-methylbutane formed<\/strong> = moles of limiting reagent<\/strong> = 0.1362 moles<\/strong>. Mass of 2-chloro-2-methylbutane=Moles\u00d7Molar Mass\\text{Mass of 2-chloro-2-methylbutane} = \\text{Moles} \\times \\text{Molar Mass} =(0.1362 moles)\u00d7(106.60 g\/mol)= (0.1362 \\text{ moles}) \\times (106.60 \\text{ g\/mol}) =14.52 g(Theoretical Yield)= 14.52 \\text{ g} (\\textbf{Theoretical Yield})<\/p>\n\n\n\n

      \n\n\n\n

      Step 4: Calculate Percent Yield<\/strong><\/h3>\n\n\n\n

      Percent Yield=(Actual YieldTheoretical Yield)\u00d7100\\text{Percent Yield} = \\left(\\frac{\\text{Actual Yield}}{\\text{Theoretical Yield}}\\right) \\times 100<\/p>\n\n\n\n

      Given that the mass of purified product = 7.1 g<\/strong>: Percent Yield=(7.1 g14.52 g)\u00d7100\\text{Percent Yield} = \\left(\\frac{7.1 \\text{ g}}{14.52 \\text{ g}}\\right) \\times 100 =48.9%= 48.9\\%<\/p>\n\n\n\n

      \n\n\n\n

      Explanation (300 Words)<\/strong><\/h3>\n\n\n\n

      The reaction between 2-methyl-2-butanol<\/strong> and hydrochloric acid<\/strong> follows an SN1 mechanism<\/strong>, meaning it proceeds via the formation of a carbocation intermediate<\/strong>. The alcohol undergoes protonation<\/strong> by HCl, leading to the loss of water and the formation of a stable tertiary carbocation. The chloride ion then attacks the carbocation<\/strong>, producing 2-chloro-2-methylbutane<\/strong>.<\/p>\n\n\n\n

      To determine the theoretical yield<\/strong>, we first identified the limiting reagent<\/strong>. 2-methyl-2-butanol (12.0 g)<\/strong> was found to be the limiting reagent because it had fewer moles (0.1362 moles) than the available HCl (0.150 moles)<\/strong>. Since the reaction follows a 1:1 stoichiometry<\/strong>, the theoretical yield of the product was found by multiplying the moles of limiting reagent<\/strong> by the molar mass of 2-chloro-2-methylbutane (106.60 g\/mol)<\/strong>, giving a theoretical yield of 14.52 g<\/strong>.<\/p>\n\n\n\n

      The percent yield<\/strong> was calculated using the formula: Percent Yield=(Actual YieldTheoretical Yield)\u00d7100\\text{Percent Yield} = \\left(\\frac{\\text{Actual Yield}}{\\text{Theoretical Yield}}\\right) \\times 100<\/p>\n\n\n\n

      With an actual yield of 7.1 g<\/strong>, the percent yield was determined to be 48.9%<\/strong>.<\/p>\n\n\n\n

      The yield could be lower than expected<\/strong> due to multiple reasons, including incomplete reaction, product loss during purification (washing and drying), side reactions, or evaporation during distillation<\/strong>. Additionally, IR spectroscopy<\/strong> can confirm the formation of the product by detecting the absence of O-H stretching<\/strong> and the presence of C-Cl stretching<\/strong>.<\/p>\n\n\n\n

      This experiment successfully demonstrates the conversion of a tertiary alcohol<\/strong> to a tertiary alkyl halide<\/strong> and reinforces key principles of SN1 reactions, purification techniques, and yield calculation<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      Now, I will generate an image illustrating the reaction mechanism<\/strong> and the purification process.<\/p>\n\n\n\n

      I have generated an image illustrating the synthesis of 2-chloro-2-methylbutane<\/strong>, including the reaction mechanism, purification steps, and simple distillation setup. Let me know if you need any modifications or further explanations!<\/p>\n\n\n\n

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

      TITLE: Synthesis of 2-chloro-2-methylbutane PURPOSE: In this experiment a tertiary alcohol ( 2-methyl -2-butanol) was converted to a tertiary alkyl chloride ( 2-chloro-2-methylbutane) using concentrated hidtochloric acid. The rxn. preceeds via a SN1 mech. The crude product is insolunle in water. Excess acid and any remaining 2-methyl-2-butanol are removed from the crude product by washing […]<\/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-194675","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194675","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=194675"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194675\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194675"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194675"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194675"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}