{"id":1201,"date":"2017-10-12T15:35:55","date_gmt":"2017-10-12T15:35:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&#038;p=1201"},"modified":"2018-10-05T18:19:23","modified_gmt":"2018-10-05T18:19:23","slug":"alkyne-acidity-formation-of-acetylide-anions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/alkyne-acidity-formation-of-acetylide-anions\/","title":{"raw":"Alkyne Acidity: Formation of Acetylide Anions","rendered":"Alkyne Acidity: Formation of Acetylide Anions"},"content":{"raw":"<div id=\"elm-main-content\" class=\"elm-content-container\">\r\n<div id=\"skills\">\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Objectives<\/h3>\r\nAfter completing this section, you should be able to\r\n<ol>\r\n \t<li>write an equation for the reaction that occurs between a terminal alkyne and a strong base, such as sodamide, NaNH<sub>2<\/sub>.<\/li>\r\n \t<li>rank a given list of compounds, including water, acetylene and ammonia, in order of increasing or decreasing acidity.<\/li>\r\n \t<li>rank a given list of hydrocarbons, such as acetylene, ethylene and ethane, in order of increasing or decreasing acidity.<\/li>\r\n \t<li>describe a general method for determining which of two given compounds is the stronger acid.<\/li>\r\n \t<li>provide an acceptable explanation of why terminal alkynes are more acidic than alkanes or alkenes.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Terms<\/h3>\r\n<div>\r\n\r\nMake certain that you can define, and use in context, the key terms below.\r\n<ul>\r\n \t<li>acetylide anion<\/li>\r\n \t<li>acidity order<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<div>\r\n<h3 class=\"boxtitle\">Study Notes<\/h3>\r\n<\/div>\r\n<div id=\"note\">\r\n\r\nAn <em>acetylide anion<\/em> is an anion formed by removing the proton from the end carbon of a terminal alkyne:\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/86904\/9-7a.png?origin=mt-web\" alt=\"reaction of propyne with sodium to form sodium propylide\" \/>\r\n\r\nAn <em>acidity <\/em><em>order<\/em>is a list of compounds arranged in order of increasing or decreasing acidity.\r\n\r\nThe general ideas discussed in this section should already be familiar to you from your previous exposure to chemistry and from the review in Section 2.8. A slightly different account of why terminal alkynes are stronger acids than are alkenes or alkanes is given below. However, the argument is still based on the differences between <em>sp<\/em>-, <em>sp<\/em><sup>2<\/sup>- and <em>sp<\/em><sup>3<\/sup>-hybrid orbitals.\r\n\r\nThe carbons of a triple bond are <em>sp<\/em>-hybridized. An <em>sp<\/em>\u2011hybrid orbital has a 50% <em>s<\/em> character and a 50% <em>p<\/em> character, whereas an <em>sp<\/em><sup>2<\/sup>\u2011hybrid orbital is 33% <em>s<\/em> and 67% <em>p<\/em>, and an <em>sp<\/em><sup>3<\/sup>\u2011hybrid orbital is 25% <em>s<\/em> and 75% <em>p<\/em>. The greater the <em>s<\/em> character of the orbital, the closer the electrons are to the nucleus. Thus in a C(<em>sp<\/em>)$\\ce{-}$H bond, the bonding electrons are closer to the carbon nucleus than they are in a C(<em>sp<\/em><sup>2<\/sup>)$\\ce{-}$H bond. In other words, compared to a C(<em>sp<\/em><sup>2<\/sup>)$\\ce{-}$H bond (or a C(<em>sp<\/em><sup>3<\/sup>)$\\ce{-}$H bond), a C(<em>sp<\/em>)$\\ce{-}$H bond is very slightly polar: C<sup><em>\u03b4<\/em>\u2212<\/sup>$\\ce{-}$H<sup><em>\u03b4<\/em>+<\/sup>. This slight polarity makes it easier for a base to remove a proton from a terminal alkyne than from a less polar or non-polar alkene or alkane.\r\n\r\nAs you will appreciate, the reaction between sodium amide and a terminal alkyne is an acid-base reaction. The sodium acetylide product is, of course, a salt. Terminal alkynes can also form salts with certain heavy-metal cations, notably silver(I) and copper(I). In the laboratory component of this course, you will use the formation of an insoluble silver acetylide as a method for distinguishing terminal alkynes from alkenes and non-terminal alkynes:\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/86905\/9-7b.png?origin=mt-web\" alt=\"three reactions showing silver cation as test for terminal alkyne\" \/>\r\n\r\nMetal acetylides are explosive when dry. They should be destroyed while still wet by warming with dilute nitric acid:\r\n\r\n<img class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/86906\/9-7c.png?origin=mt-web\" alt=\"silver propylide conversion to propyne with acid\" \/>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_1\">\r\n<h3><\/h3>\r\n<h3 class=\"editable\">Acidity of Terminal Alkynes: Formation of Acetylide\u00a0Anions<\/h3>\r\nTerminal alkynes are much more acidic than most other hydrocarbons. Removal of the proton leads to the formation of an acetylide anion, RC<u>=<\/u>C:<sup>-<\/sup>. The origin of the enhanced acidity can be attributed to the stability of the acetylide\u00a0anion, which has the unpaired electrons in an sp\u00a0hybridized orbital. The stability results from occupying an orbital with a high degree of s-orbital character. There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The enhanced acidity with greater s-character occurs despite the fact that the<a title=\"Homolytic C-H Bond Dissociation Energies of Organic Molecules\" href=\"https:\/\/chem.libretexts.org\/Core\/Organic_Chemistry\/Fundamentals\/Homolytic_C-H_Bond_Dissociation_Energies_of_Organic_Molecules\" rel=\"internal\"> homolytic\u00a0C-H BDE<\/a>\u00a0is larger.\r\n<table style=\"margin: auto;border-spacing: 1px\" border=\"1\" cellpadding=\"1\"><caption><em><strong>Table 9.7.1<\/strong>: Akynes<\/em><\/caption>\r\n<tbody>\r\n<tr>\r\n<td><strong>Compound<\/strong><\/td>\r\n<td><strong>Conjugate Base<\/strong><\/td>\r\n<td><strong>Hybridization<\/strong><\/td>\r\n<td><strong>\"s Character\"<\/strong><\/td>\r\n<td><strong>pKa<\/strong><\/td>\r\n<td><strong>C-H BDE (kJ\/mol)<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>CH<sub>3<\/sub>CH<sub>3<\/sub><\/td>\r\n<td>CH<sub>3<\/sub>CH<sub>2<\/sub><sup>-<\/sup><\/td>\r\n<td>sp<sup>3<\/sup><\/td>\r\n<td>25%<\/td>\r\n<td>50<\/td>\r\n<td>410<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>CH<sub>2<\/sub>CH<sub>2<\/sub><\/td>\r\n<td>CH<sub>2<\/sub>CH<sup>-<\/sup><\/td>\r\n<td>sp<sup>2<\/sup><\/td>\r\n<td>33%<\/td>\r\n<td>44<\/td>\r\n<td>473<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HCCH<\/td>\r\n<td>HCC<sup>-<\/sup><\/td>\r\n<td>sp<\/td>\r\n<td>50%<\/td>\r\n<td>25<\/td>\r\n<td>523<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nConsequently, acetylide\u00a0anions can be readily formed by deprotonation\u00a0using a sufficiently strong <a title=\"Acids and Bases\" href=\"https:\/\/chem.libretexts.org\/LibreTexts\/ChemTutor\/ACIDS_AND_BASES\" rel=\"internal\">base<\/a>. Amide anion (NH<sub>2<\/sub><sup>-<\/sup>), in the form of NaNH<sub>2<\/sub>\u200b is commonly used for the formation of acetylide anions.\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05144131\/clipboard_1395413234245.png\" alt=\"\" width=\"337px\" height=\"33px\" \/>\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05144133\/clipboard_1395411745257.png\" alt=\"\" width=\"353px\" height=\"40px\" \/>\r\n\r\n<\/div>\r\n<div id=\"section_2\">\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>Given that the p<em>K<\/em>a\u00a0of water is 15.74, would you expect hydroxide ion to be capable of removing a proton from each of the substances listed below? Justify your answers, briefly.\r\n<ol>\r\n \t<li>ethanol (p<em>K<\/em>a = 16)<\/li>\r\n \t<li>acetic acid (p<em>K<\/em>a = 4.72)<\/li>\r\n \t<li>acetylene (p<em>K<\/em>a = 25)<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<h3>Answers<\/h3>\r\n1. [reveal-answer q=\"344133\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"344133\"]a. No, not very well. The pKa of ethanol is greater than that of water, thus the equilibrium lies to the left rather than to the right. $\\ce{\\sf{OH^- + {C}_2{H}_5{OH} &lt;=&gt; {C}_2{H}_5O^- + {H}_2{O}}}$[\/hidden-answer]\r\n\r\n[reveal-answer q=\"110629\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"110629\"]b. Yes, very well. There is a difference of 11 pKa units between the pKa of water and the pKa of acetic acid. The equilibrium lies well to the right. $\\ce{\\sf{CH_3CO_2H + OH^- &lt;=&gt; CH_2CO_2^- + H_2O}}$[\/hidden-answer]\r\n\r\n[reveal-answer q=\"126675\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"126675\"]c. No, hardly at all. The hydroxide ion is too weak a base to remove a proton from acetylene. The equilibrium lies well to the left. $\\ce{\\sf{H-C#C-H + OH^- &lt;=&gt; H-C#C^- + H_2O}}$[\/hidden-answer]\r\n<div id=\"s61714\">\r\n<div id=\"section_20\">\r\n<h3 id=\"Questions-61714\">Question<\/h3>\r\nIf KOH has a pKa of 15.7 in water, what pKa be required to deprotonate KOH?\r\n\r\n<\/div>\r\n<div id=\"section_21\">\r\n<h3 id=\"Solutions-61714\">Solutions<\/h3>\r\n<strong>1.<\/strong>\r\n\r\n[reveal-answer q=\"302772\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"302772\"]Need a stronger base, or a compound with a pKa &gt; 15.7 to deprotonate.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_3\">\r\n<h3 class=\"editable\">Contributors<\/h3>\r\n<ul>\r\n \t<li><a class=\"external\" title=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" href=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" target=\"_blank\" rel=\"external nofollow noopener\">Dr. Dietmar Kennepohl<\/a> FCIC (Professor of Chemistry, <a class=\"external\" title=\"http:\/\/www.athabascau.ca\/\" href=\"http:\/\/www.athabascau.ca\/\" target=\"_blank\" rel=\"external nofollow noopener\">Athabasca University<\/a>)<\/li>\r\n \t<li>Prof. Steven Farmer (<a class=\"external\" title=\"http:\/\/www.sonoma.edu\" href=\"http:\/\/www.sonoma.edu\" target=\"_blank\" rel=\"external nofollow noopener\">Sonoma State University<\/a>)<\/li>\r\n \t<li>William Reusch, Professor Emeritus (<a class=\"external\" title=\"http:\/\/www.msu.edu\/\" href=\"http:\/\/www.msu.edu\/\" target=\"_blank\" rel=\"external nofollow noopener\">Michigan State U.<\/a>), <a class=\"external\" title=\"http:\/\/www.cem.msu.edu\/~reusch\/VirtualText\/intro1.htm\" href=\"http:\/\/www.cem.msu.edu\/%7Ereusch\/VirtualText\/intro1.htm\" target=\"_blank\" rel=\"external nofollow noopener\">Virtual Textbook of\u00a0Organic\u00a0Chemistry<\/a><\/li>\r\n<\/ul>\r\n<ul>\r\n \t<li><a class=\"external\" title=\"http:\/\/www.chem.purdue.edu\/wenthold\/\" href=\"http:\/\/www.chem.purdue.edu\/wenthold\/\" target=\"_blank\" rel=\"external nofollow noopener\">Prof. Paul G. Wenthold <\/a>(<a class=\"external\" title=\"http:\/\/www.chem.purdue.edu\/\" href=\"http:\/\/www.chem.purdue.edu\/\" target=\"_blank\" rel=\"external nofollow noopener\">Purdue University<\/a>)<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>","rendered":"<div id=\"elm-main-content\" class=\"elm-content-container\">\n<div id=\"skills\">\n<div class=\"textbox learning-objectives\">\n<h3>Objectives<\/h3>\n<p>After completing this section, you should be able to<\/p>\n<ol>\n<li>write an equation for the reaction that occurs between a terminal alkyne and a strong base, such as sodamide, NaNH<sub>2<\/sub>.<\/li>\n<li>rank a given list of compounds, including water, acetylene and ammonia, in order of increasing or decreasing acidity.<\/li>\n<li>rank a given list of hydrocarbons, such as acetylene, ethylene and ethane, in order of increasing or decreasing acidity.<\/li>\n<li>describe a general method for determining which of two given compounds is the stronger acid.<\/li>\n<li>provide an acceptable explanation of why terminal alkynes are more acidic than alkanes or alkenes.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Terms<\/h3>\n<div>\n<p>Make certain that you can define, and use in context, the key terms below.<\/p>\n<ul>\n<li>acetylide anion<\/li>\n<li>acidity order<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<div>\n<h3 class=\"boxtitle\">Study Notes<\/h3>\n<\/div>\n<div id=\"note\">\n<p>An <em>acetylide anion<\/em> is an anion formed by removing the proton from the end carbon of a terminal alkyne:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/86904\/9-7a.png?origin=mt-web\" alt=\"reaction of propyne with sodium to form sodium propylide\" \/><\/p>\n<p>An <em>acidity <\/em><em>order<\/em>is a list of compounds arranged in order of increasing or decreasing acidity.<\/p>\n<p>The general ideas discussed in this section should already be familiar to you from your previous exposure to chemistry and from the review in Section 2.8. A slightly different account of why terminal alkynes are stronger acids than are alkenes or alkanes is given below. However, the argument is still based on the differences between <em>sp<\/em>-, <em>sp<\/em><sup>2<\/sup>&#8211; and <em>sp<\/em><sup>3<\/sup>-hybrid orbitals.<\/p>\n<p>The carbons of a triple bond are <em>sp<\/em>-hybridized. An <em>sp<\/em>\u2011hybrid orbital has a 50% <em>s<\/em> character and a 50% <em>p<\/em> character, whereas an <em>sp<\/em><sup>2<\/sup>\u2011hybrid orbital is 33% <em>s<\/em> and 67% <em>p<\/em>, and an <em>sp<\/em><sup>3<\/sup>\u2011hybrid orbital is 25% <em>s<\/em> and 75% <em>p<\/em>. The greater the <em>s<\/em> character of the orbital, the closer the electrons are to the nucleus. Thus in a C(<em>sp<\/em>)$\\ce{-}$H bond, the bonding electrons are closer to the carbon nucleus than they are in a C(<em>sp<\/em><sup>2<\/sup>)$\\ce{-}$H bond. In other words, compared to a C(<em>sp<\/em><sup>2<\/sup>)$\\ce{-}$H bond (or a C(<em>sp<\/em><sup>3<\/sup>)$\\ce{-}$H bond), a C(<em>sp<\/em>)$\\ce{-}$H bond is very slightly polar: C<sup><em>\u03b4<\/em>\u2212<\/sup>$\\ce{-}$H<sup><em>\u03b4<\/em>+<\/sup>. This slight polarity makes it easier for a base to remove a proton from a terminal alkyne than from a less polar or non-polar alkene or alkane.<\/p>\n<p>As you will appreciate, the reaction between sodium amide and a terminal alkyne is an acid-base reaction. The sodium acetylide product is, of course, a salt. Terminal alkynes can also form salts with certain heavy-metal cations, notably silver(I) and copper(I). In the laboratory component of this course, you will use the formation of an insoluble silver acetylide as a method for distinguishing terminal alkynes from alkenes and non-terminal alkynes:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/86905\/9-7b.png?origin=mt-web\" alt=\"three reactions showing silver cation as test for terminal alkyne\" \/><\/p>\n<p>Metal acetylides are explosive when dry. They should be destroyed while still wet by warming with dilute nitric acid:<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"http:\/\/chem.libretexts.org\/@api\/deki\/files\/86906\/9-7c.png?origin=mt-web\" alt=\"silver propylide conversion to propyne with acid\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_1\">\n<h3><\/h3>\n<h3 class=\"editable\">Acidity of Terminal Alkynes: Formation of Acetylide\u00a0Anions<\/h3>\n<p>Terminal alkynes are much more acidic than most other hydrocarbons. Removal of the proton leads to the formation of an acetylide anion, RC<u>=<\/u>C:<sup>&#8211;<\/sup>. The origin of the enhanced acidity can be attributed to the stability of the acetylide\u00a0anion, which has the unpaired electrons in an sp\u00a0hybridized orbital. The stability results from occupying an orbital with a high degree of s-orbital character. There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The enhanced acidity with greater s-character occurs despite the fact that the<a title=\"Homolytic C-H Bond Dissociation Energies of Organic Molecules\" href=\"https:\/\/chem.libretexts.org\/Core\/Organic_Chemistry\/Fundamentals\/Homolytic_C-H_Bond_Dissociation_Energies_of_Organic_Molecules\" rel=\"internal\"> homolytic\u00a0C-H BDE<\/a>\u00a0is larger.<\/p>\n<table style=\"margin: auto;border-spacing: 1px\" cellpadding=\"1\">\n<caption><em><strong>Table 9.7.1<\/strong>: Akynes<\/em><\/caption>\n<tbody>\n<tr>\n<td><strong>Compound<\/strong><\/td>\n<td><strong>Conjugate Base<\/strong><\/td>\n<td><strong>Hybridization<\/strong><\/td>\n<td><strong>&#8220;s Character&#8221;<\/strong><\/td>\n<td><strong>pKa<\/strong><\/td>\n<td><strong>C-H BDE (kJ\/mol)<\/strong><\/td>\n<\/tr>\n<tr>\n<td>CH<sub>3<\/sub>CH<sub>3<\/sub><\/td>\n<td>CH<sub>3<\/sub>CH<sub>2<\/sub><sup>&#8211;<\/sup><\/td>\n<td>sp<sup>3<\/sup><\/td>\n<td>25%<\/td>\n<td>50<\/td>\n<td>410<\/td>\n<\/tr>\n<tr>\n<td>CH<sub>2<\/sub>CH<sub>2<\/sub><\/td>\n<td>CH<sub>2<\/sub>CH<sup>&#8211;<\/sup><\/td>\n<td>sp<sup>2<\/sup><\/td>\n<td>33%<\/td>\n<td>44<\/td>\n<td>473<\/td>\n<\/tr>\n<tr>\n<td>HCCH<\/td>\n<td>HCC<sup>&#8211;<\/sup><\/td>\n<td>sp<\/td>\n<td>50%<\/td>\n<td>25<\/td>\n<td>523<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Consequently, acetylide\u00a0anions can be readily formed by deprotonation\u00a0using a sufficiently strong <a title=\"Acids and Bases\" href=\"https:\/\/chem.libretexts.org\/LibreTexts\/ChemTutor\/ACIDS_AND_BASES\" rel=\"internal\">base<\/a>. Amide anion (NH<sub>2<\/sub><sup>&#8211;<\/sup>), in the form of NaNH<sub>2<\/sub>\u200b is commonly used for the formation of acetylide anions.<\/p>\n<p><img decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05144131\/clipboard_1395413234245.png\" alt=\"\" width=\"337px\" height=\"33px\" \/><\/p>\n<p><img decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05144133\/clipboard_1395411745257.png\" alt=\"\" width=\"353px\" height=\"40px\" \/><\/p>\n<\/div>\n<div id=\"section_2\">\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>Given that the p<em>K<\/em>a\u00a0of water is 15.74, would you expect hydroxide ion to be capable of removing a proton from each of the substances listed below? Justify your answers, briefly.\n<ol>\n<li>ethanol (p<em>K<\/em>a = 16)<\/li>\n<li>acetic acid (p<em>K<\/em>a = 4.72)<\/li>\n<li>acetylene (p<em>K<\/em>a = 25)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h3>Answers<\/h3>\n<p>1. <\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q344133\">Show Answer<\/span><\/p>\n<div id=\"q344133\" class=\"hidden-answer\" style=\"display: none\">a. No, not very well. The pKa of ethanol is greater than that of water, thus the equilibrium lies to the left rather than to the right. $\\ce{\\sf{OH^- + {C}_2{H}_5{OH} &lt;=&gt; {C}_2{H}_5O^- + {H}_2{O}}}$<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q110629\">Show Answer<\/span><\/p>\n<div id=\"q110629\" class=\"hidden-answer\" style=\"display: none\">b. Yes, very well. There is a difference of 11 pKa units between the pKa of water and the pKa of acetic acid. The equilibrium lies well to the right. $\\ce{\\sf{CH_3CO_2H + OH^- &lt;=&gt; CH_2CO_2^- + H_2O}}$<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q126675\">Show Answer<\/span><\/p>\n<div id=\"q126675\" class=\"hidden-answer\" style=\"display: none\">c. No, hardly at all. The hydroxide ion is too weak a base to remove a proton from acetylene. The equilibrium lies well to the left. $\\ce{\\sf{H-C#C-H + OH^- &lt;=&gt; H-C#C^- + H_2O}}$<\/div>\n<\/div>\n<div id=\"s61714\">\n<div id=\"section_20\">\n<h3 id=\"Questions-61714\">Question<\/h3>\n<p>If KOH has a pKa of 15.7 in water, what pKa be required to deprotonate KOH?<\/p>\n<\/div>\n<div id=\"section_21\">\n<h3 id=\"Solutions-61714\">Solutions<\/h3>\n<p><strong>1.<\/strong><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q302772\">Show Answer<\/span><\/p>\n<div id=\"q302772\" class=\"hidden-answer\" style=\"display: none\">Need a stronger base, or a compound with a pKa &gt; 15.7 to deprotonate.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"section_3\">\n<h3 class=\"editable\">Contributors<\/h3>\n<ul>\n<li><a class=\"external\" title=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" href=\"http:\/\/science.athabascau.ca\/staff-pages\/dietmark\" target=\"_blank\" rel=\"external nofollow noopener\">Dr. Dietmar Kennepohl<\/a> FCIC (Professor of Chemistry, <a class=\"external\" title=\"http:\/\/www.athabascau.ca\/\" href=\"http:\/\/www.athabascau.ca\/\" target=\"_blank\" rel=\"external nofollow noopener\">Athabasca University<\/a>)<\/li>\n<li>Prof. Steven Farmer (<a class=\"external\" title=\"http:\/\/www.sonoma.edu\" href=\"http:\/\/www.sonoma.edu\" target=\"_blank\" rel=\"external nofollow noopener\">Sonoma State University<\/a>)<\/li>\n<li>William Reusch, Professor Emeritus (<a class=\"external\" title=\"http:\/\/www.msu.edu\/\" href=\"http:\/\/www.msu.edu\/\" target=\"_blank\" rel=\"external nofollow noopener\">Michigan State U.<\/a>), <a class=\"external\" title=\"http:\/\/www.cem.msu.edu\/~reusch\/VirtualText\/intro1.htm\" href=\"http:\/\/www.cem.msu.edu\/%7Ereusch\/VirtualText\/intro1.htm\" target=\"_blank\" rel=\"external nofollow noopener\">Virtual Textbook of\u00a0Organic\u00a0Chemistry<\/a><\/li>\n<\/ul>\n<ul>\n<li><a class=\"external\" title=\"http:\/\/www.chem.purdue.edu\/wenthold\/\" href=\"http:\/\/www.chem.purdue.edu\/wenthold\/\" target=\"_blank\" rel=\"external nofollow noopener\">Prof. Paul G. Wenthold <\/a>(<a class=\"external\" title=\"http:\/\/www.chem.purdue.edu\/\" href=\"http:\/\/www.chem.purdue.edu\/\" target=\"_blank\" rel=\"external nofollow noopener\">Purdue University<\/a>)<\/li>\n<\/ul>\n<\/div>\n<\/div>\n","protected":false},"author":44985,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1201","chapter","type-chapter","status-publish","hentry"],"part":25,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1201","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/users\/44985"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1201\/revisions"}],"predecessor-version":[{"id":2322,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1201\/revisions\/2322"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/25"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1201\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/media?parent=1201"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1201"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/contributor?post=1201"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/license?post=1201"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}