{"id":1537,"date":"2017-10-12T13:44:37","date_gmt":"2017-10-12T13:44:37","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&#038;p=1537"},"modified":"2018-10-05T19:11:06","modified_gmt":"2018-10-05T19:11:06","slug":"the-halogenation-of-benzene","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/the-halogenation-of-benzene\/","title":{"raw":"The Halogenation of Benzene","rendered":"The Halogenation of Benzene"},"content":{"raw":"<div class=\"elm-header\">\r\n<div class=\"elm-header-custom\">Halogenation is an example of electrophillic aromatic substitution. In electrophilic aromatic substitutions, a benzene is attacked by an electrophile which results in substition of hydrogens. However, halogens are not electrophillic enough to break the aromaticity of benzenes, which require a catalyst to activate.<\/div>\r\n<\/div>\r\n<div id=\"elm-main-content\" class=\"elm-content-container\">\r\n<div id=\"section_1\">\r\n<h3 class=\"editable\">Activation of Halogen<\/h3>\r\n(where X= Br or Cl, we will discuss further in detail later why other members of the halogen family Flourine and Iodine are not used in halogenation of benzenes)\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152033\/no_reaction_2.jpg\" alt=\"no reaction (2).jpg\" width=\"272px\" height=\"170px\" \/>\r\n\r\nHence, Halogen needs the help and aid of Lewis Acidic Catalysts to activate it to become a very strong electrophile. Examples of these activated halogens are Ferric Hallides (FeX<sub>3<\/sub>) Aluminum Halides (AlX<sub>3<\/sub>) where X= Br or Cl. In the following examples, the halogen we will look at is Bromine.\r\n\r\nIn the example of bromine, in order to make bromine electrophillic enough to react with benzene, we use the aid of an aluminum halide such as aluminum bromide.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152034\/AlBr3.jpg\" alt=\"AlBr3.jpg\" width=\"162px\" height=\"127px\" \/>\r\n\r\nWith aluminum bromide as a Lewis acid, we can mix Br<sub>2<\/sub><sub>\u00a0<\/sub>with AlBr<sub>3<\/sub> to give us Br<sup>+<\/sup>. The presence of Br<sup>+<\/sup> is a much better electrophile than Br<sub>2<\/sub> alone. Bromination is acheived with the help of AlBr<sub>3<\/sub> (Lewis acid catalysts) as it polarizes the Br-Br bond. The polarization causes polarization causes the bromine atoms within the Br-Br bond to become more electrophillic. The presence of Br<sup>+<\/sup> compared to Br<sub>2<\/sub> alone is a much better electrophille that can then react with benzene.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152036\/electrophile_2.jpg\" alt=\"electrophile (2).jpg\" width=\"639px\" height=\"214px\" \/>\r\n\r\nAs the bromine has now become more electrophillic after activation of a catalyst, an electrophillic attack by the benzene occurs\u00a0 at the terminal bromine of Br-Br-AlBr<sub>3.\u00a0<\/sub>This allows the other bromine atom to leave with the AlBr<sub>3<\/sub><sub>\u00a0<\/sub>as a good leaving group, AlBr<sub>4<\/sub>-.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152038\/nuc_2.jpg\" alt=\"nuc (2).jpg\" width=\"406px\" height=\"203px\" \/>\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152040\/Benzene_Brominationfinal_1.jpg\" alt=\"Benzene Brominationfinal (1).jpg\" width=\"482px\" height=\"283px\" \/>\r\n\r\nAfter the electrophilic attack of bromide to the benzene, the hydrogen on the same carbon as bromine substitutes the carbocation in which resulted from the attack. Hence it being an electrophilic aromatic SUBSTITUTION. Since the by-product aluminum tetrabromide is a strong nucleophile,\u00a0 it pulls of a proton from the Hydrogen on the same carbon as bromine.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152042\/end_copy_1.jpg\" alt=\"end copy (1).jpg\" width=\"481px\" height=\"224px\" \/>\r\n\r\nIn the end, AlBr<sub>3<\/sub>was not consumed by the reaction and is regenerated. It serves as our catalyst in the halogenation of benzenes.\r\n<div id=\"section_2\">\r\n<h4 class=\"editable\">Dissociation Energies of Halogens and its Effect on Halogenation of Benzenes<\/h4>\r\nThe electrophillic bromination of benzenes is an exothermic reaction. Considering the exothermic rates of aromatic halogenation decreasing down the periodic table in the Halogen family, Flourination is the most exothermic and Iodination would be the least. Being so exothermic, a reaction of flourine with benzene is explosive! For iodine, electrophillic iodination is generally endothermic, hence a reaction is often not possible. Similar to bromide, chlorination would require the aid of an activating presence such as Alumnium Chloride or Ferric Chloride. The mechanism of this reaction is the same as with Bromination of benzene.\r\n<div>\r\n<div id=\"section_3\">\r\n<h3 class=\"editable\">Outside links<\/h3>\r\n<ul>\r\n \t<li><a class=\"external\" title=\"http:\/\/www.chemguide.co.uk\/mechanisms\/elsub\/halogenation.html\" href=\"http:\/\/www.chemguide.co.uk\/mechanisms\/elsub\/halogenation.html\" rel=\"freeklink\">http:\/\/www.chemguide.co.uk\/mechanism...ogenation.html<\/a><\/li>\r\n \t<li><a class=\"external\" title=\"http:\/\/en.wikipedia.org\/wiki\/Electrophilic_halogenation\" href=\"http:\/\/en.wikipedia.org\/wiki\/Electrophilic_halogenation\" rel=\"freeklink\">http:\/\/en.wikipedia.org\/wiki\/Electro...c_halogenation<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div id=\"section_4\">\r\n<h3 class=\"editable\">References<\/h3>\r\n<ol>\r\n \t<li>Vollhardt, Peter, and Neil Shore. <u>Organic Chemistry: Structure and Function<\/u>. 5th Edition. New York: W.H. Freeman and Company, 2007.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div id=\"section_5\">\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<div>\r\n<div id=\"section_5\">\r\n\r\n<strong>1. <\/strong>What reagents would you need to get the given product?\r\n\r\n<img class=\"internal default\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152045\/practice_1.jpg\" alt=\"practice 1.jpg\" width=\"528px\" height=\"173px\" \/>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n\r\n<strong>\u00a02. <\/strong>What product would result from the given reagents?\r\n\r\n<img class=\"internal default\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152047\/practice_2.jpg\" alt=\"practice 2.jpg\" width=\"663px\" height=\"221px\" \/>\r\n\r\n<strong>3. <\/strong>What is the major product given the reagents below?\r\n\r\n<img class=\"internal default\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152049\/practice_3.jpg\" alt=\"practice 3.jpg\" width=\"458px\" height=\"229px\" \/>\r\n\r\n<strong>4. <\/strong>Draw the formatin of Cl<sup>+ <\/sup>from AlCl<sub>3<\/sub>and Cl<sub>2<\/sub>\r\n\r\n<strong>5<\/strong>. Draw the mechanism of the reaction between Cl<sup>+<\/sup> and a benzene.\r\n<div id=\"section_6\">\r\n<h3 class=\"editable\">Solutions<\/h3>\r\n1.\r\n[reveal-answer q=\"610168\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"610168\"]Cl2\u00a0and AlCl3 or Cl2 and FeCl3[\/hidden-answer]\r\n\r\n2.\r\n[reveal-answer q=\"355301\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"355301\"]No Reaction[\/hidden-answer]\r\n\r\n3.\r\n[reveal-answer q=\"156759\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"156759\"]\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152051\/answer_3.jpg\" alt=\"answer 3.jpg\" width=\"154px\" height=\"216px\" \/>[\/hidden-answer]\r\n\r\n4.\r\n[reveal-answer q=\"328547\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"328547\"]<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152053\/alcl3.jpg\" alt=\"alcl3.jpg\" width=\"500px\" height=\"181px\" \/>[\/hidden-answer]\r\n\r\n5.\r\n[reveal-answer q=\"234130\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"234130\"]\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152055\/answer_4.jpg\" alt=\"answer 4.jpg\" width=\"567px\" height=\"331px\" \/>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div id=\"section_7\">\r\n<h3 class=\"editable\">Contributors<\/h3>\r\n<ul>\r\n \t<li>Catherine Nguyen<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"elm-header\">\n<div class=\"elm-header-custom\">Halogenation is an example of electrophillic aromatic substitution. In electrophilic aromatic substitutions, a benzene is attacked by an electrophile which results in substition of hydrogens. However, halogens are not electrophillic enough to break the aromaticity of benzenes, which require a catalyst to activate.<\/div>\n<\/div>\n<div id=\"elm-main-content\" class=\"elm-content-container\">\n<div id=\"section_1\">\n<h3 class=\"editable\">Activation of Halogen<\/h3>\n<p>(where X= Br or Cl, we will discuss further in detail later why other members of the halogen family Flourine and Iodine are not used in halogenation of benzenes)<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152033\/no_reaction_2.jpg\" alt=\"no reaction (2).jpg\" width=\"272px\" height=\"170px\" \/><\/p>\n<p>Hence, Halogen needs the help and aid of Lewis Acidic Catalysts to activate it to become a very strong electrophile. Examples of these activated halogens are Ferric Hallides (FeX<sub>3<\/sub>) Aluminum Halides (AlX<sub>3<\/sub>) where X= Br or Cl. In the following examples, the halogen we will look at is Bromine.<\/p>\n<p>In the example of bromine, in order to make bromine electrophillic enough to react with benzene, we use the aid of an aluminum halide such as aluminum bromide.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152034\/AlBr3.jpg\" alt=\"AlBr3.jpg\" width=\"162px\" height=\"127px\" \/><\/p>\n<p>With aluminum bromide as a Lewis acid, we can mix Br<sub>2<\/sub><sub>\u00a0<\/sub>with AlBr<sub>3<\/sub> to give us Br<sup>+<\/sup>. The presence of Br<sup>+<\/sup> is a much better electrophile than Br<sub>2<\/sub> alone. Bromination is acheived with the help of AlBr<sub>3<\/sub> (Lewis acid catalysts) as it polarizes the Br-Br bond. The polarization causes polarization causes the bromine atoms within the Br-Br bond to become more electrophillic. The presence of Br<sup>+<\/sup> compared to Br<sub>2<\/sub> alone is a much better electrophille that can then react with benzene.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152036\/electrophile_2.jpg\" alt=\"electrophile (2).jpg\" width=\"639px\" height=\"214px\" \/><\/p>\n<p>As the bromine has now become more electrophillic after activation of a catalyst, an electrophillic attack by the benzene occurs\u00a0 at the terminal bromine of Br-Br-AlBr<sub>3.\u00a0<\/sub>This allows the other bromine atom to leave with the AlBr<sub>3<\/sub><sub>\u00a0<\/sub>as a good leaving group, AlBr<sub>4<\/sub>-.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152038\/nuc_2.jpg\" alt=\"nuc (2).jpg\" width=\"406px\" height=\"203px\" \/><\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152040\/Benzene_Brominationfinal_1.jpg\" alt=\"Benzene Brominationfinal (1).jpg\" width=\"482px\" height=\"283px\" \/><\/p>\n<p>After the electrophilic attack of bromide to the benzene, the hydrogen on the same carbon as bromine substitutes the carbocation in which resulted from the attack. Hence it being an electrophilic aromatic SUBSTITUTION. Since the by-product aluminum tetrabromide is a strong nucleophile,\u00a0 it pulls of a proton from the Hydrogen on the same carbon as bromine.<\/p>\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152042\/end_copy_1.jpg\" alt=\"end copy (1).jpg\" width=\"481px\" height=\"224px\" \/><\/p>\n<p>In the end, AlBr<sub>3<\/sub>was not consumed by the reaction and is regenerated. It serves as our catalyst in the halogenation of benzenes.<\/p>\n<div id=\"section_2\">\n<h4 class=\"editable\">Dissociation Energies of Halogens and its Effect on Halogenation of Benzenes<\/h4>\n<p>The electrophillic bromination of benzenes is an exothermic reaction. Considering the exothermic rates of aromatic halogenation decreasing down the periodic table in the Halogen family, Flourination is the most exothermic and Iodination would be the least. Being so exothermic, a reaction of flourine with benzene is explosive! For iodine, electrophillic iodination is generally endothermic, hence a reaction is often not possible. Similar to bromide, chlorination would require the aid of an activating presence such as Alumnium Chloride or Ferric Chloride. The mechanism of this reaction is the same as with Bromination of benzene.<\/p>\n<div>\n<div id=\"section_3\">\n<h3 class=\"editable\">Outside links<\/h3>\n<ul>\n<li><a class=\"external\" title=\"http:\/\/www.chemguide.co.uk\/mechanisms\/elsub\/halogenation.html\" href=\"http:\/\/www.chemguide.co.uk\/mechanisms\/elsub\/halogenation.html\" rel=\"freeklink\">http:\/\/www.chemguide.co.uk\/mechanism&#8230;ogenation.html<\/a><\/li>\n<li><a class=\"external\" title=\"http:\/\/en.wikipedia.org\/wiki\/Electrophilic_halogenation\" href=\"http:\/\/en.wikipedia.org\/wiki\/Electrophilic_halogenation\" rel=\"freeklink\">http:\/\/en.wikipedia.org\/wiki\/Electro&#8230;c_halogenation<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div>\n<div id=\"section_4\">\n<h3 class=\"editable\">References<\/h3>\n<ol>\n<li>Vollhardt, Peter, and Neil Shore. <u>Organic Chemistry: Structure and Function<\/u>. 5th Edition. New York: W.H. Freeman and Company, 2007.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div>\n<div id=\"section_5\">\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<div>\n<div id=\"section_5\">\n<p><strong>1. <\/strong>What reagents would you need to get the given product?<\/p>\n<p><img decoding=\"async\" class=\"internal default\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152045\/practice_1.jpg\" alt=\"practice 1.jpg\" width=\"528px\" height=\"173px\" \/><\/p>\n<\/div>\n<\/div>\n<div>\n<p><strong>\u00a02. <\/strong>What product would result from the given reagents?<\/p>\n<p><img decoding=\"async\" class=\"internal default\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152047\/practice_2.jpg\" alt=\"practice 2.jpg\" width=\"663px\" height=\"221px\" \/><\/p>\n<p><strong>3. <\/strong>What is the major product given the reagents below?<\/p>\n<p><img decoding=\"async\" class=\"internal default\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152049\/practice_3.jpg\" alt=\"practice 3.jpg\" width=\"458px\" height=\"229px\" \/><\/p>\n<p><strong>4. <\/strong>Draw the formatin of Cl<sup>+ <\/sup>from AlCl<sub>3<\/sub>and Cl<sub>2<\/sub><\/p>\n<p><strong>5<\/strong>. Draw the mechanism of the reaction between Cl<sup>+<\/sup> and a benzene.<\/p>\n<div id=\"section_6\">\n<h3 class=\"editable\">Solutions<\/h3>\n<p>1.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q610168\">Show Answer<\/span><\/p>\n<div id=\"q610168\" class=\"hidden-answer\" style=\"display: none\">Cl2\u00a0and AlCl3 or Cl2 and FeCl3<\/div>\n<\/div>\n<p>2.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q355301\">Show Answer<\/span><\/p>\n<div id=\"q355301\" class=\"hidden-answer\" style=\"display: none\">No Reaction<\/div>\n<\/div>\n<p>3.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q156759\">Show Answer<\/span><\/p>\n<div id=\"q156759\" class=\"hidden-answer\" style=\"display: none\">\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152051\/answer_3.jpg\" alt=\"answer 3.jpg\" width=\"154px\" height=\"216px\" \/><\/div>\n<\/div>\n<p>4.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q328547\">Show Answer<\/span><\/p>\n<div id=\"q328547\" class=\"hidden-answer\" style=\"display: none\"><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152053\/alcl3.jpg\" alt=\"alcl3.jpg\" width=\"500px\" height=\"181px\" \/><\/div>\n<\/div>\n<p>5.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q234130\">Show Answer<\/span><\/p>\n<div id=\"q234130\" class=\"hidden-answer\" style=\"display: none\">\n<p><img decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/05152055\/answer_4.jpg\" alt=\"answer 4.jpg\" width=\"567px\" height=\"331px\" \/><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<div>\n<div id=\"section_7\">\n<h3 class=\"editable\">Contributors<\/h3>\n<ul>\n<li>Catherine Nguyen<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":44985,"menu_order":3,"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-1537","chapter","type-chapter","status-publish","hentry"],"part":28,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1537","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\/1537\/revisions"}],"predecessor-version":[{"id":2340,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1537\/revisions\/2340"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/28"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/1537\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/media?parent=1537"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=1537"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/contributor?post=1537"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/wp-json\/wp\/v2\/license?post=1537"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}